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SANDEEP JYANI
❏CIVIL ENGINEERING
EDUCATOR
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Fineness Modulus Test
Aggregate Type FM
Fine Aggregate (Sand)Fine Sand 2-2.5
Medium Sand 2.5-3.0
Coarse Sand 3.0-3.5
Coarse Aggregate 5.5-8.0
All in aggregate 3.5-7.5
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=
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⇒
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=
6.8−5.4
5.4−2.6
=0.5
•This test is not applicable of the
aggregates having size less than 6.3
mm.
•In order to perform this test,
sufficient quantity of aggregates are
taken such that 200 hundred pieces
of each fraction can be tested.
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Flakiness Index and Elongation Test
��=
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���
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���
Strength of Concrete
1.Water/Cement Ratio
•Water Cement Ratio means the ratio between the
weight of water to the weight of cementused in
concrete mix.
•Normally water cement ratio falls under 0.4 to 0.6
as per IS Code 10262 (2009)for nominal mix (M10,
M15 …. M25)
•In 1918 Abrams presented his classic law in the
form:
•�=
�
�
�
•where x =water/cement ratio by volume and for 28
days results the constants A and B are 14,000 lbs/sq.
in. and 7 respectively.
Strength of Concrete
2.Gel/Space Ratio
•Gel space ratio is defined as the ratio of the volume of the
hydrated cement paste to the sum of volumes of the hydrated
cement and of the capillary pores.
•Gel/Space ratio �=
�����������
��������������
•Many researchers argued that Abrams water/cement ratio law
can only be called a rule and not a law because Abrams’
statement does not include many qualifications necessary for its
validity to call it a law.
•Instead of relating the strength to water/cement ratio, the
strength can be more correctly related to the solid products of
hydration of cement to the space available for formation of this
product. Powers and Brownyardhave established the relationship
between the strength and gel/space ratio
•�������������������=���×
���
�����
�
•Where 240 = intrinsic strength of gel in MPa
Strength of Concrete
Que: If gel space ratio of a concrete is
given as 0.756, what is its theoretical
strength?
�������������������=���×
���
�����
�
Strength of Concrete
Que: If gel space ratio of a concrete is
given as 0.756, what is its theoretical
strength?
�������������������=���×
���
�����
�
=����.���
�
=���.���/��
�
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II. Strength Of Concrete
2. Tensile Strength of Concrete
•Tensile, strength of the concrete is tested indirectly, by noting its
modules of rupture that is determined by preparing a block of size 15
cm ×15 cm ×70 cm if the Maximum nominal size of aggregate is
greater than 20 mm; and of size 10 cm ×10 cm ×50 cm if the maximum
nominal size of aggregate is less than 20 mm.
•The beam is then placed over the roller supports and is subjected to
the load at which its failure takes place that is further used to find its
modulus of rupture (stress at which failure takes place)
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��
��
�
�=������������������
�=������������������������
�=���������������
�=���������������
�
����
����
����
�
Modulus of Rupture �
��=�.��
��
D��������������������=�.����.����
��
II. Strength Of Concrete
3. Split Tensile Strength of Concrete
•The length of cylinder varies from one to two
diameters. Normally the test cylinder is 150
mm diameter and 300 mm long
•The test consists of applying compressive line
loads along the diameter until it fails
•��������������������=
��
���
�=������������������
�=����������������
�=�������������
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Maturity Concept of Concrete
•The strength development of concrete depends on
both time and temperature it can be said that
strength is a function of summation of product of
time and temperature. This summation is called
maturity of concrete
Maturity=Σ(time×temperature)
•Hydration of concrete continues to take place upto
about –11C. Therefore, –11C is taken as a datum
line for computing maturity
Maturity Concept of Concrete
•Exp: A sample of concrete cured at 20C
for 28 days is taken as fully matured
concrete. Its maturity would be equal to…?
=(28×24)×[20–(–11)]
Maturity=Σ(time×temperature)
=20832°�ℎ
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SANDEEP JYANI
Tensile Strength of Concrete in Flexure
�
��=�.��
��
where �
��is the characteristic compressive
strength of concrete in N/mm
2
Effect of Creep on Young’s Modulus of Elasticity
•Long term Young’s Modulus of Elasticity of concrete
�
��=
�����
��
�+�
Where �is the creep coefficient
Creep Strain is strain which occurs due to continuous loading and temperature
effect for longer duration which may cause permanent deformation
Age of Loading �
7 days 2.2
28 days 1.6
1 year 1.1
Tensile Strength of Concrete in Flexure
�
��=�.��
��
Probabilistic Curve
Frequency
distribution
Or
Probability
distribution
(No. of times test
results obtained)
Strength of Concrete
�
���
�
�
�−�.��??????
5%
??????=�����������������
(������������������������)
�
�=���+�.��??????
=> �
�=�
��+�.��??????
�
�=������������
�
��=����������������������
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�=
���
�??????
���
GradeModular Ratio
M-15 19
M-20 13
M-25 11
M-30 9
M-35 8
M-40 7
Modular Ratio
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T -Beam
B
B
f
Flange
Web
D
f
Effective Width
B
w
Partial safety factors
2.Design Strength
•Design strength of material is the ratio of characteristic strength of material to
partial factor of safety
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�=
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���������������������(�
�)
•Clause 36.4.2 of IS 456 states that �
�for concreteand steelshould be taken as 1.5
and 1.15, respectively when assessing the strength of the structures or structural
members employing limit state of collapse.
•Partial safety factor for steel (1.15) is comparatively lower than that of concrete
(1.5) because the steel for reinforcement is produced in steel plants and
commercially available in specific diameters with expected better quality control
than that of concrete.
Partial safety factors
•���������
�=
�
�
�.��
=�.���
�
•������������
�=
�
��
�.�
=�.�����
•In case of concrete the characteristic strength is calculated on the basis of test results on
150 mm standard cubes. But the concrete in the structure has different sizes. To take the
size effect into account, it is assumed that the concrete in the structure develops a strength
of 0.67 or (1/1.50) times the characteristic strength of cubes.
•Accordingly, in the calculation of strength employing the limit state of collapse, the
characteristic strength (f
ck) is first multiplied with 0.67 (size effect) and then divided by 1.5
(�
�for concrete) to have 0.446 f
ckas the maximum strength of concrete in the stress block.
•������������
�=
�
��
�.�
�.��=�.����
��
Assumptions of Limit State Method
5.Tensile strength of concrete is neglected
6.Maximum strain in tension reinforcement in the section at failure
should not be less than
�
�≥
�.����
�
�
+�.������
�≥
�
�
�.���
�
+�.���
D
b
Neutral Axis
A
st d
�
�≥
�.����
�
�
+�.���
0.0035 0.446 f
ck
C
0.002
�.����
Actual Depth of Neutral Axis
The actual depth of Neutral axis is
calculated by equating the compression
force to tensile force
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⇒�.���
���
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��
��
⇒�
�=
�.���
��
��
�.���
���
Limiting Depth of Neutral Axis
Since stress diagram is not linear, hence we
use corresponding strain diagram
b
A
st d
�
�≥
�.����
�
�
+�.���
0.0035
0.002
�
����
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�.����
�
����
=
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�
�
+�.���
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⇒
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=
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+�.���
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⇒
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=
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+�.���
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+�
⇒�
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���
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�
⇒�
����=��
d
Maximum Depth of Neutral Axis
f
yN/mm
2
x
ulim
Fe 250 0.53 d
Fe415 0.48 d
Fe 500 0.46 d
Moment of Resistance
•�������������������������������������
•���=�.���
����
�(�−�.���
�)Valid for Balanced and over reinforced section
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•���=�.���
��
��(�−�.���
�)Valid for Balanced and under reinforced section
Design of Beam
Assumingoveralldepthas�=2����/10
Step1:LoadandBending moment
���=�����������
������������⇒�
�=�.��
Note:
(i)Total factored load =�
�
(ii)Iftotalworkingload=�, then convert into factored load �
�=1.5�
(iii)Whenonly live load is given, then dead load/self weight of the beam is considered to find
working load and convert into factored load i.e.
•�=��+��
•������������⇒�
�=�.��
��������������������=
�
��
�
�
Design of Beam
Step 2: Check for depth assumed:
�
����=���
�
⇒�
����=
�
����
��
�
����=�
����+��������������
If�
�������>�
����⇒������������
If �
�������<�
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Design Formula
We always try to design the limiting section
PROOF:
�
��=�.���
����
����(�−�.���
����)
=����(�−�.����)
=����
�
(�−�.���)
=����
�
�
=�����
�
=���
�
�=
�
�
��
Design Formula
Step 3: Area of Steel A
st
�
��=�.���
��
��(�−�.���
����)
⇒�
��=
�
��
�.����(�−�.����)
⇒�
��=
�
��
�.���
��(�−�.���)
⇒���=
�
��
�.���
���
Limiting values of Tension steel
•Minimum area of tension reinforcement
should not be less than
�
�
��
=
�.��
�
�
�
0=minimum area of tension r/f
f
y= characteristic strength of steel in N/mm
2
•Maximum area of tension reinforcement
should not be greater than 4 %of the gross
cross sectional area to avoid difficulty in
placing and compacting concrete properly
in framework
Effective Span of Beam
1)Simply Supported beam or Slab: The effective span of a simply
supported member is taken lesser of the following:
I.�=��+�
II.�(centre to centre distance between supports)
Where �=centre to centre distance between supports
�
�=clear span
�=��������������������������
�
�
�
Effective Span of Beam
2)Continuous Beam or Slab: The effective span of a continuous member
is taken as:
i.If width of support �
�≤
�
�
��
, then effective span is taken as lesser of …
a)�=��+�
b)�(centre to centre distance between supports)
Where�=centre to centre distance between supports
�
�=clear span
�=��������������������������
ii.If width of support �
�>
�
�
��
or 600 mm , then effective span is taken as lesser of
…
a)�=��+�.��
b)�=��+�.��
�
�
�
� �
�
�
�
�
T -Beam
B
B
f
Flange
Web
D
f
d
B
w
�
�
Flange
Web
�
�
B
w
ANALYSIS OF T-BEAM
CASE 1: When actual depth of Neutral
axis lies in the FLANGE PORTION
Step 1: Calculate actual depth of
Neutral Axis
�
�
��
�=�
⇒0.36�
���
��
�=0.87�
��
��
⇒�
�=
0.87�
��
��
0.36�
���
�
(�ℎ��ℎ�ℎ�����������ℎ���
�)
Step 2: Calculate Limiting depth of Neutral Axis
�
����
=ቐ
0.53������250
0.48������415
0.46������500
�
�
�.���
��
0.87�
�
�
�
Flange
Web
�
�
B
w
ANALYSIS OF T-BEAM
CASE 1: When actual depth of Neutral
axis lies in the FLANGE PORTION
Step 3: Calculate Moment of Resistance
�
�
��
�
��=�
�×0.45�
��×
3
7
�
�+
2
3
×0.45�
��×
4
7
�
��−0.42�
�
�
�
�.���
��
0.87�
�
⇒�
��=0.36�
���
��
��−0.42�
�
�
��=0.87�
��
���−0.42�
�
�
�
�
�
�
�
�
�
�
�
Web
�
�
B
w
ANALYSIS OF T-BEAM
CASE 2: When actual depth of Neutral axis lies in the
WEB PORTION
Sub Case A: When �
�>�
�����
�<
�
�
�
�
Step 1: Calculate actual depth of Neutral Axis
�
�
��
�=�
⇒0.36�
���
��
�+0.45�
��(�
�−�
�)�
�=0.87�
��
��
Step 2: Calculate Limiting depth of Neutral Axis
�
����
=ቐ
0.53������250
0.48������415
0.46������500
�
�
�.���
��
0.87�
�
Step 3: Calculate Moment of Resistance
�
��=0.36�
���
��
��−0.42�
�+0.45�
��(�
�−�
�)�
��−
�
�
2
�
�
�
�
�
�
�
�
�
�
Web
�
�
B
w
ANALYSIS OF T-BEAM
CASE 2: When actual depth of Neutral axis lies in the WEB
PORTION
Sub Case B: When �
�>�
�����
�>
�
�
�
�
According to IS code recommendation, when actual
depth of Neutral Axis lies in web portion and depth of
Flange is greater than
�
�
�
�then in that case equivalent
depth of Flange in stressed diagram
�
�=�.���
�+�.���
�
�
�
��
�=�
⇒0.36�
���
��
�+0.45�
���
�−�
��
�=0.87�
��
��
Step 2: Calculate Limiting depth of Neutral Axis
�
����
=ቐ
0.53������250
0.48������415
0.46������500
�
�
�.���
��
0.87�
�
Step 3: Calculate Moment of Resistance
�
��=0.36�
���
��
��−0.42�
�+0.45�
���
�−�
��
��−
�
�
2
�
�
�
�
Step 1: Calculate actual depth of Neutral Axis
�
�
�
�
�
�
�
�
Web
�
�
B
w
Design Formula OF T-BEAM
CASE 2: When actual depth of Neutral axis lies in the
WEB PORTION
Sub Case A: When �
�>�
�����
�<
�
�
�
�
�
�
��
�
�=�
1
⇒0.36�
���
��
�=0.87�
��
���
�
�
�.���
��
0.87�
�
�
�
�
�
�
�
�
�
⇒�
���=
0.36�
���
��
�
0.87�
�
�
�=�
2
⇒�
���=
�
�−�
�0.45�
���
�
0.87�
�
�
�
Web
�
�
B
w
ANALYSIS OF T-BEAM
CASE 2: When actual depth of Neutral axis lies in the WEB
PORTION
Sub Case A: When �
�>�
�����
�>
�
�
�
�
According to IS code recommendation, when actual
depth of Neutral Axis lies in web portion and depth of
Flange is greater than
�
�
�
�then in that case equivalent
depth of Flange in stressed diagram
�
�=�.���
�+�.���
�
�
�
��
�
�
�.���
��
0.87�
�
�
�
�
�
�
�
�
�
�
�
�
���=
0.36�
���
��
�
0.87�
�
�
���=
0.45�
���
�−�
��
�
0.87�
�
Steps for Design of Shear Reinforcement
1.Find maximum shear force
•SSB =
��
�
��
�
�
�
�
2.Nominal Shear Stress
•�
�=
�
��
��
�
�
��
and compare with �
����
3.Shear Strength of Concrete (�
�)
4.Net shear force resisted by Shear
reinforcement
•��=�
�−�
���
5.Design of Shear reinforcement
Steps for Design
6.Minimum shear reinforcement
7.Maximum spacing in shear
reinforcement
1.S
v< 0.75 d Vertical shear
Reinforcement
2.S
v< d Inclined shear Reinforcement
3.However, the spacing shall not exceed
300 mm in any case.
(whichever is minimum out of these)
BEAMS:
1.Minimum tension reinforcement:
�
�
��
=
�.��
�
�
2.The maximum reinforcement in tension or compression: should not exceed 0.04bD,
Where, D = overall depth of section (4 %of the gross cross sectional area)
3.Side face reinforcement: If depth of the web in a beam exceeds 750 mm, side-face
reinforcement should be provided along the two faces. The total area of such
reinforcement should not be less than 0.1 % of the web -area. It should be equally
distributed on each of the two faces; The spacing of such reinforcement should not
exceed 300 mm or web thickness whichever is less.
4.Shear Reinforcement: Clause 26.5.1.5 of IS 456 stipulates that the maximum spacing
of shear reinforcement measured along the axis of the member shall not be more than
0.75 d for vertical stirrups and d for inclined stirrups at 45°, where d is the effective
depth of the section.
•S
v< 0.75 d Vertical shear Reinforcement
•S
v< d Inclined shear Reinforcement
•However, the spacing shall not exceed 300 mm in any case.
REINFORCEMENT REQUIREMENTS
SLABS:
1.Minimum tension reinforcement: minimum reinforcement in either direction in slabs should not
be less than 0.15%of the total cross-sectional area when using mild steel reinforcement, and
0.12%of the total cross-sectional area when using high strength deformed reinforcement or
welded Wire fabric
•��������=�.��%�����������=�.��%×�×�
•��������=�.��%�����������=�.��%×�×�
2.No maximum reinforcementrecommendation is given in IS code so we consider the maximum
reinforcement criteria same as in case of beam i.e. 4% of bD
3.Maximum diameter of steel bar in slab
•∅
���=
���������������
�
4.Maximum spacing of the reinforcement
a)Main bar =��������ቊ
��
�����
b)Distribution bar =��������ቊ
��
�����
‘d’ is the effective depth of slab
REINFORCEMENT REQUIREMENTS
Anchorage Bond
•Let us consider a uniformly loaded
cantilever beam which has to resist
given BM and SF
•Let us assume that tension
reinforcement consists of single bar of
diameter ∅
•The tensile force at B is equal to
AB C
�
�
W
T
↽↽
↼
↽
↼↼
�
��
�=
�
�
∅
�
�
�
�.��
•Which must be transmitted to concrete
by bond stress in the embedded length
�
�=��
⇒�
��×�∅�
�=
�
�
∅
�
�
�
�.��
����
�=
�.���
�
∅
��
��
����������
�����������
�����������������
����
�=
??????
��
∅
��
��
•If �
��is the average bond stress acting
over the surface area �∅�
�then
��������������=���������������
Design Bond stress for concrete for PLAIN bars in TENSION
Grade of Concrete WSM LSM
M15 0.6
M20 0.8 1.2
M25 0.9 1.4
M30 1 1.5
M35 1.1 1.7
M40 and above 1.2 1.9
•For deformed bars conforming to IS 1786 these
values shall be increased by 60 percent
•For bars in compression, the values of bond stress
for bars in tension shall be increased by 25
percent.
•For fusion bonded epoxy coated deformed bars,
design bond stress values shall be taken as 80
percentof the values given in the above table
Fusion Bonded Epoxy is very fastcuring, thermosetting Protective Powder Coating. It is based on specially selected Epoxy resinsand hardeners. The epoxy
formulated in order to meet the specifications related to protection of steel bars as an anti-corrosion coating
87
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For Fe 415, �
���������=1.5×1.6=2.4
(����������60%�����������������415���������������)
�
�������������=1.5×1.6×1.25=3(����ℎ������������25%�������������������)
89
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For Fe 215, �
���������=1.5
�
�������������=1.5×1.25=1.87
91
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When Beam is subjected to Bending Moment and Torsional Moment
•Critical Section:
•Sections located less than a distance d from the face of the support may be designed for the same
torsion as computed at a distance d, where d is the effective depth
•Step 2: Equivalent Nominal Shear
Stress �
��
•�
��=
�
�
��
•This value of �
��should not exceed
maximum shear stress �
����
•If �
��> �
����, then revise the section
•Step 1: Equivalent shear force (41.3.1)
•The equivalent shear force �
�=�
�+�.�
�
�
�
•�
�=������������������ℎ��������
•�
�=���������ℎ��������
•�
�=���������������
•�=������ℎ���ℎ�����(�
�for flanged beam)
Design for Equivalent Shear
•Step 2: Equivalent Nominal Shear Stress �
��
•�
��=
�
�
��
•This value of �
��should not exceed maximum shear
stress �
����for safety against diagonal compression
failure
•If �
��> �
����, then revise the section
•If the equivalent nominal shear stress, �
��does not
exceed
�
�
(
�����������������������������,�������)
given, minimum shear reinforcement shall he provided
as per 26.5.1.6
Design for Equivalent Bending
•Step 3: Equivalent Moment
•�
�1=�
�+�
�
•�
�=
�
�
1.7
(1+
�
�
)
•�
�= equivalent bending moment
•�
�= equivalent moment due to torsion
•�
�= bending moment at cross section
•�
�= Torsional moment
•D = overall depth
•b = breadth of beam
•Step 4: Equivalent effective depth
•�=
�
�
??????�
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250��×250��
�
�=�.��−
�
���
�����
=�.��−
����
��(���)
=�.��
Design of Column
3.Circular Column
a)With separate ring used as a stirrup:
�=��{??????
��
�+??????
��
�}
Using LSM
•For short axially loaded column with �
���≤�.�����,
column can be designed as
•�
�=�.��
���
�+�.���
��
��
•For circular column,
•�
�=�.��(�.��
���
�+�.���
��
��)
Columns (LSM)
•Assumptions: Clause 39.1, Minimum Eccentricity 39.2,
Short Axially loaded Members in Compression 39.3
1.Minimum Eccentricity (whichever is maximum) (Cl 39.2)
�
���≥
�
500
+
�
30
>20��
2. If Minimum eccentricity value as per CL 39.2 is less than
or equal to 0.05×LeastLateralDimension, then the
column can be designed as per following equation of Short
Column
�
���≤�.�����(������������������������)
Columns (LSM)
3.
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350��×650��
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350��×650��
CONTROL OF DEFLECTION
•The deflection of a structure or its part should
not adversely affect the appearance or
efficiency of the structure or finishes or
partitions
•The deflection including the effects of
temperature, shrinkage and creep occurring
after the construction of partitions, and finishes
should not exceed span/350 or 20 mm,
whichever is lesser
•The total deflection due to all loads including
the effects of temperature, shrinkage and creep
should not exceed span/250 when measured
from the as cast level of the supports of floors,
roofs and all other horizontal-members,
CONTROL OF DEFLECTION
Deflection
Depth of Slab
•The depth of slab depends on bending moment and deflection criterion. the trail
depth can be obtained using:
•The effective depth d of two way slabs can also be
assumed using cl. 24.1,IS 456 provided short span
is ≤ 3.5m and loading class is < 3.5KN/m
2
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3�×3.5�����
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3�×3.5�����
For slabs spanning in two directions, the
shorter of the two spans should be taken
for calculating the span to effective depth
For 3�×3.5����������slab (using Fe
250),
����
�����������ℎ
=40
⇒
3�
�����������ℎ
=40
⇒�����������ℎ=
3
40
⇒�����������ℎ=0.075�
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Design of One Way Slab
Step 1: Find load over 1m of span
a)��������
b)����������=�������������
c)��������������=����������������������
Step 2: Bending Moment
a)��
���=
��
�
�
( for simply supported beam)
Step 3: Area of Steel
�
��=
�.����
�
�
�−�−
�.���
�
����
�
��
Step 4: Check for Shear
�
�=
�
��
≯�
����
�=
��
�
Criteria for Design
1.Depth of footing:
All foundations should be located at a minimum
depth of 0.5m below the ground surface
The depth is primarily governed by availability
of bearing capacity, minimum seasonal variation
like swelling and shrinkage of soil
Using rankine’sformula, minimum depth of
foundation is given by
�
�=
�
�
1−���∅
1+���∅
2
�=gross safe bearing capacity
�=unit weight of soil
∅=angle of friction
Design Steps for Footing
1.Size of Foundation:
a)Load from column = �
b)Weight of foundation =�
�=��%���
c)Total load �
�=�+�
�=�.��
d)Area of Foundation
����������������=
�
�
�
�
While calculating the area of footing required, self weight of footing is considered
e)Decide size of foundation
Square footing⇒����=�×�
Rectangular Footing⇒����=�×�
(���������������)
f)Net upward design soil pressure on foundation
�=
�(������)
�(�������)
���
�
�=�.��(���)
While calculating the upward soil pressure, the self weight of the footing is not considered
�
�
�
�
(���������������
������)
�
��
�
������������=��
����������������=��
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Area of Foundation
����������������=
�
�
�
�
While calculating the area of footing
required, self weight of footing is
considered
Net design soil pressure on foundation
�=
�(������)
�(�������)
���
�
�=�.��(���)
While calculating the upward soil pressure,
the self weight of the footing is not
considered
Design Steps for Footing
2.Check for Bending Moment
Critical section for bending moment is at
the face of the column.
➢Bending moment about x
1-x
1 axis
➢OX
1= overhang = (
�−�
�
)
➢M
x= BM
x1-x1
=w
�×��×
��
�
�
�
=w
�×��×
{
�−�
�
}
�
�
=�
�×��×
�−�
�
�
�
�
�
��������
��
��
�
�
�
�
�
�
�
�
�
Design Steps for Footing
2.Check for Bending Moment
Critical section for bending moment is at the face of
the column.
➢Bending moment about y
1-y
1 axis
➢M
y= BM
y1-y1
=w
�×��×
��
�
�
�
=w
�×��×
{
�−�
�
}
�
�
=�
�×��×
�−�
�
�
➢Maximum BM is maximum value among M
yand M
x
➢Depth required
d =
��
���
��
where B = 1000mm
�
�
�
��������
��
��
�
�
�
�
�
�
�
�
�
Design Steps for Footing
3.Check for One way Shear
Critical section for one way shear is at
a distance ‘d’ from the face of column.
➢OX
2= (
�−�
�
−�)
➢OY
2= (
�−�
�
−�)
Max shear about x
2-x
2
➢OX
2= (
�−�
�
−�)
➢V = w
�×��×OX
2
➢V = w
�×��×(
�−�
�
−�)
�
�
�
��
��
�
�
�
�
�
�
� �
��
�
��
�
��
�
Design Steps for Footing
3.Check for One way Shear
Critical section for one way shear is at a distance ‘d’
from the face of column.
•Max shear about y
2-y
2
➢OY
2= (
�−�
�
−�)
➢V = w
�×��×OY
2
➢V = w
�×��×(
�−�
�
−�)
•Nominal shear stress
➢�
�=
�
�.�
➢�
�=
�
�����
➢�
��houldnot be greater than k.�
�min
Where �
�min = 0.18N/mm
2
(WSM)
�
�min = 0.28N/mm
2
(LSM)
�
�
�
��
��
�
�
�
�
�
�
� �
��
�
��
�
��
�
THICKNESS>300 275 250 225 200 175 150
k 1 1.05 1.10 1.15 1.20 1.25 1.30
Design Steps for Footing
4.Check for Two way Shear
Punching shear shall be checked around
the column on a perimeter half the
effective depth of footing away from the
face of column
Net punching force
��=�
��×�−�
�(�+�)(�+�)
Resisting Area
=�[�+�+(�+�]�
�
��(���������)=
����������������
�������������
⇒�
��(���������)=
�
��×�−�
�(�+�)(�+�)
�[�+�+(�+�]�
Design Steps for Footing
4.Check for Two way Shear
⇒�
��(���������)=
�
��×�−�
�(�+�)(�+�)
�[�+�+(�+�]�
����
��(���������)�����������
������������
��(�����������)
•�
��(�����������)=ቐ
�
��.���
�����
�
��.���
�����
Where �
�=�.�+
�
�
<�.�
Design Steps for Footing
5. Area of Steel
�
��=
�
�
??????
����
WSM
�
��=
���
�.���
���
LSM
�
��=
�.��
��
�
�
�−�−
�.��
�
�
����
�
��
The above �
��is for ��width,
������
��=��
��
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�
�
�
�<�
��
��
�
a)�.��
��
b)�.���
��
c)�.���
��
d)�.��
��
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�
�
�
�<�
��
��
�
a)�.��
��
b)�.���
��
c)�.���
��
d)�.��
��
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a)��=
��(�−�)
�
b)��=
��(�−�)
�
�
c)��=
��(�−�)
�
�
d)��=
��(�+�)
�
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a)��=
��(�−�)
�
b)��=
��(�−�)
�
�
c)��=
��(�−�)
�
�
d)��=
��(�+�)
�
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PERMISSIBLE STRESS IN STEEL STRUCTURES
1.As per WSM
i.Maximum permissible AXIALstress in compression is given
by
??????
��=�.����
•Used in the design of columns and struts.
•Column is a compression member where bending moment
exists while in case of struts, also being a compression
member, bending moment is zero. Because strut is a
component of roof trusses and roof trusses are pin jointed
connection having bending moment equal to zero.
ii.Maximum permissible AXIALstress in tension is given by
??????
��=�.����
It is used in design of tension members
PERMISSIBLE STRESS IN STEEL STRUCTURES
1.As per WSM
iii.Maximum permissible bending stress in compression is given
•Used in design of flexural (bending) member that is beam, built up beam,
plate girder etc.
??????
��=�.����
iv.Maximum permissible bending stress in tension is given
•Used in the design of beams
??????
��=�.����
v.Maximum permissible average shear stress is given by
�
����=�.����
vi.Maximum permissible Maximum shear stress is given by
�
����=�.����
FOS=2.5 for average shear stress
FOS=2.2 for maximum shear stress
PERMISSIBLE STRESS IN STEEL STRUCTURES
1.As per WSM
vi.Maximum permissible bending stress in column base
is given by
??????=�.����
Increase of permissible stress
•When wind and earthquake load are considered, the
permissible stresses in steel structureare increased by
33.33%.
•When wind and earthquake load are considered, the
permissible stresses in connections (rivet and weld) are
increased by 25%.
PERMISSIBLE DEFLECTION IN STEEL STRUCTURES
•As per WSM, Maximum permissible horizontal and vertical
deflection is given by
�=
����
���
•As per LSM, Maximum permissible horizontal and vertical
deflection is given by
a)If supported elements are not susceptible to cracking
�=
����
���
b)If supported elements are susceptible to cracking
�=
����
���
PERMISSIBLE DEFLECTION IN GANTRY GIRDER
1.For manually operator crane, the
maximum permissible deflection is
�=
����
���
2.For electrically operator crane, the
maximum permissible deflection for a
capacity upto50T or 500kN
�=
����
���
3.For electrically operator crane, the
maximum permissible deflection for a
capacity more than 50T or 500kN
�=
����
����
Gantry girdersare laterally unsupported beams to carry. heavy loads from
place to place at the construction sites
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CONNECTIONS
1.RIVETED CONNECTIONS:
•Strength of riveted joint
•It is taken as minimum of shear strength, bearing strength and tearing
strength.
•FOR LAP JOINT:
1.FOR ENTIRE PLATE
a)SHEAR STRENGTH OF RIVETS
�
�=�×
�
�
�
�
��
Wheren →total number of rivets at joint
F
s→permissible shear stress in rivets
F
s = 100MPa (WSM)
F
u = ultimate shear stress in rivet
so in LSM =
��
�1.25
d →gross diameter of rivet (hole diameter)
Gross dia= nominal dia+ 1.5 mm, for nominal dia≤ 25 mm
Gross dia= nominal dia+ 2 mm, for nominal dia> 25 mm
P B P
CONNECTIONS
1.RIVETED CONNECTIONS:
•FOR LAP JOINT:
1.FOR ENTIRE LENGTH
b)BEARING STRENGTH OF ALL RIVETS
�
�=����
�
Wheren →total number of rivets at joint
t →thickness of thinner main plate
F
b→permissible shear stress in rivets (300MPa in WSM)
d →gross diameter of rivet (hole diameter)
Gross dia= nominal dia+ 1.5 mm, for nominal dia≤ 25 mm
Gross dia= nominal dia+ 2 mm, for nominal dia> 25 mm
c)TEARING STRENGTH OF PLATE
�
�=�−�
����
�
Where n
1→total number of rivets at critical section 1-1
t →thickness of thinner main plate
B →width of plate
F
t→permissible tensile stress (Axial = 0.6fy = 0.6250 = 150MPa)
d →gross diameter of rivet (hole diameter)
Gross dia= nominal dia+ 1.5 mm, for nominal dia≤ 25 mm
Gross dia= nominal dia+ 2 mm, for nominal dia> 25 mm
B
1
1
PP
1
1
B
CONNECTIONS
1.RIVETED CONNECTIONS:
•LAP JOINT:
2.FOR GAUGE LENGTH/PITCH LENGTH
a)SHEAR STRENGTH OF RIVETS
�
��=�×
�
�
�
�
�
�
Wheren →total number of rivets at joint in crossed gauge
length
F
s→permissible shear stress in rivets
F
s = 100MPa (WSM)
F
u = ultimate shear stress in rivet so in
LSM =
��
�1.25
d →gross diameter of rivet (hole diameter)
Gross dia= nominal dia+ 1.5 mm, for nominal dia≤ 25 mm
Gross dia= nominal dia+ 2 mm, for nominal dia> 25 mm
1
1
PP
g
CONNECTIONS
1.RIVETED CONNECTIONS:
•LAP JOINT:
2.FOR GAUGE LENGTH/PITCH LENGTH
b)BEARING STRENGTH OF RIVETS
�
��=�
����
�
Wheren →total number of rivets at joint in crossed gauge length
t →thickness of thinner main plate
F
b→permissible bearing stress in rivets (300MPa in WSM)
d →gross diameter of rivet (hole diameter)
c)TEARING STRENGTH OF PLATE
�
��=�−���
�
Where g →gauge length
t →thickness of thinner main plate
F
t→permissible tensile stress in plate(Axial = 0.6fy = 0.6250 =
150MPa)
When pitch distance is given then
�
�=�−�����
1
1
PP g
FIRST PLATE
SECOND PLATE
1
B
1
g
CONNECTIONS
1.RIVETED CONNECTIONS:
•DOUBLE COVER BUTT JOINT:
1.FOR ENTIRE WIDTH OF PLATE
•SHEAR STRENGTH OF RIVETS
�
��=��
�×
�
�
�
�
�
�
Wheren →total number of rivets at joint
F
s→permissible shear stress in rivets
F
s = 100MPa (WSM)
d →gross diameter of rivet (hole diameter)
2 →Double shear
1
1
P
P B
MAIN PLATE
PP
CONNECTIONS
1.RIVETED CONNECTIONS:
•DOUBLE COVER BUTT JOINT:
1.FOR ENTIRE WIDTH OF PLATE
•BEARING STRENGTH OF RIVETS
�
�=����
�
Wheren →total number of rivets at joint
t →min of (thickness of thinner main plate,
sum of cover plate thickness)
F
b→permissible bearing stress in rivets
d →gross diameter of rivet (hole diameter)
1
1
P
P B
MAIN PLATE
PP
1.RIVETED CONNECTIONS:
•DOUBLE COVER BUTT JOINT:
1.FOR ENTIRE WIDTH OF PLATE
•TEARING STRENGTH OF PLATES
�
�=�−�
����
�
Where n
1→total number of rivets at critical section 1-1
t →min of (thickness of thinner main plate,
sum of cover plate thickness)
B →width of plate
F
t→permissible strength of plate in tearing
d →gross diameter of rivet (hole diameter)
1
1
P
P
B
MAIN PLATE
P
P
CONNECTIONS
CONNECTIONS
1.RIVETED CONNECTIONS:
•DOUBLE COVER BUTT JOINT:
2.FOR GAUGE LENGTH
a)SHEAR STRENGTH OF RIVETS
�
��=�×�×
�
�
�
�
�
�
Wheren →total number of rivets at joint in crossed gauge
length (here 2)
F
s→permissible shear stress in rivets
F
s = 100MPa (WSM)
F
u = ultimate shear stress in rivet so in
LSM =
��
�1.25
d →gross diameter of rivet (hole diameter)
1
1
PP
g B
CONNECTIONS
1.RIVETED CONNECTIONS:
•DOUBLE COVER BUTT JOINT:
2.FOR GAUGE LENGTH/PITCH LENGTH
b)BEARING STRENGTH OF RIVETS
�
��=�
����
�
Wheren →total number of rivets at joint in crossed gauge length
t →min (thickness of thinner main plate, sum of cover plate thickness)
F
b→permissible bearing stress in rivets (300MPa in WSM)
d →gross diameter of rivet (hole diameter)
c)TEARING STRENGTH OF PLATE
�
��=�−�
����
�
Where g →gauge length
t →thickness of thinner main plate
F
t→permissible strength of plate in tearing
n →total number of rivets at in critical section 1-1 in crossed gauge length (here 1)
1
1
PP g
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•Number of Rivets required at a joint=
������������������
����������
�=
�
�
�
•Efficiency of joint
�=
�������������
�,�
�,�
�
������������������������
���
�
�=�����������������������
�
�=����������������������
�
�=�����������������������
•Efficiency for entire plate
•We have to ensure that �
�is less because rivet failure is more dangerous
•For Entire PLATE:
For Gauge Length:
�=
�������������
�,�
�,�
�
�������ℎ����������������
×100
⇒�=
�−�
1���
�
���
�
×100
⇒�=
�−�
1�
�
×100
⇒�=
�−�×�×�
�
���
�
×100
⇒�=
�−�
�
×100
Unwin’s formula
•It is used when diameter of rivet is not known
∅=�.���
Where t is thickness of thinner plate
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ASSUMPTIONS IN DESIGN OF RIVETED JOINT
�−���
�n�
�(MOST IMPORTANT CONSIDERATION)
1
P
gP
Where n is the number of rivets in shaded region
Analysis of Eccentric Connection
Step 1: Shear Force (�
1) in Rivet due to
Direct load P
�
�=
�
σ�
�
�
�
If diaof rivets are same, then the cross
section area would also be the same,
there fore direct shear load is
�
1=
�
��
�
�
�
⇒�
1=
�
�
�
�
����������������
Analysis of Eccentric Connection
Step 2: Shear Force (�
2) in Rivet due to
Twisting Moment T
�
2=
��
�
�
�
2
;�=��
(������ℎ��������������������)
Where �
�is the radial distance of each rivet
from centre of Rivet Group
�
�
����������������
Analysis of Eccentric Connection
Step 3: Resultant Shear Force in the
Rivet(�
�)
�=�
1
2
+�
2
2
+2�
1�
2cos�
�
�
����������������
CONNECTIONS
3.WELDED CONNECTIONS:
2.FILLET WELD
•The effective length of fillet weld should not be
less than ����������������(�)i.e.
•�
���=����������������(�)
•The size of normal fillets shall be taken as the
minimum weld leg size.
•Fillet weld should not be used if the angle
between fusion faces is less than 60°and greater
than and greater than 120°or we can say
•In weld, angle should be between 60°to 120°
•����������������
�=��(����������)
•�
�=��
���������������(�
�)
(�)Size or leg of weld
Size or leg of weld
�
��
�
�
�
�
�
������=����������
��=���������������(�
�)
���
IS RECOMMENDATIONS
5.EFFECTIVE CROSS SECTION AREA OF WELD (Throat area)
•Effective cross section area of weld = effective length of weld
throat thickness
����
���=�
����
�
6.LOAD CARRYING CAPACITY OF WELD/SHEAR STRENGTH OF
WELD
•P = Permissible shear stress effective area of weld
•�=�
��
����
�
•F
s→permissible shear stress
•F
s = 110MPa (WSM)
F
u = ultimate tensile stress in weld metal
so in LSM =
��
�1.25
(1.25 for shop weld and 1.5 for field weld)
7.PITCH OF WELD
•For weld in compression zone, max pitch p = 12t or 200mm
•In tension zone, max pitch p = 16t or 200mm
Eccentric Welded Connection
1.IN PLANE ECCENTRIC CONNECTION
a)Direct Shear Stress due to P at 1
b)Torsional Shear Stress at Point 1
c)Calculate Resultant Force
�
�
����������������
�
�
�
�
�
�
�
�
�=��
�
�
�
�
�
�
�
�
1=
�
(�
1+�
2+�
3)�
�
�
�
�
=
??????
�
���
??????=�
�=
��
���
�
�
�
�
Eccentric Welded Connection
1.IN PLANE ECCENTRIC CONNECTION
•The effect of eccentric load at the CG of
weld group will be direct load P and
twisting moment T i.e. �=��where �
is measured from CG of the weld group.
•Due to direct load the direct shear stress
�
�developed at point �
•Due to twisting moment, the torsional
shear stress �
�is developed at �
•Since these are two stresses are shear
stresses, we can find their resultant �
�
�
�=�
1
2
+�
2
2
+2�
1�
2cos�
�
�
����������������
�
�
�
�
�
�
�
�
�=��
�
�
�
�
�
�
�
�
�should not be greater than ??????
�(WSM, ??????
�=110���)
�
�should not be greater than �
�(LSM �
�=
��
3×1.25
)
LACINGS
•Lacing member are idealised as truss
element, i.e., they re subjected either to
tension or compression.
•B.M. in lacing member is zero, to ensure
that bending moment is zero, provide only
one rivet at each end as far as possible.
•�
���=
����
�
=
��
�
����
=
�
��
•Maximum slenderness ratio �for lacing
member is limited to 145.
•The angle of lacing w.r.t.vertical is 40°to
70°(welding 60°to 90°)
�
�
Arrangement in A is better than B, because if one rivet fails, spacing of lacing
member does not change in A while in B, spacing will be doubled. Hence there
will be possibility of buckling of connection in B.
c
�=
�
�
������
c
�=�����
A B
�=
��
�
����
FORCES IN LACING MEMBER
•Lacing system is designed to resist a transverse shear force
of �=�.�%������������.
•The transverse shear force �is shared by lacing system
both side equally, so the transverse shear force on each
lacing bar is
�
�
•2 denotes number of parallel planes
•For single lacing system of two parallel force system, the force in
each lacing bar �=
�
�sin�
•For double lacing system �=
�
�sin�
Plate Girder
•Compression Flange:
•It consists of flange plate, flange
angle and web equivalent
•Web equivalent is the web area
embedded between two flange angle
•In compression zone flange, web
equivalent is taken as
���������
6
or
��
6
•Tension Flange:
•It consists of flange plate, angle and
web equivalent
•In tension zone, web equivalent is
taken as
��
8
��������
�����������
����������������
�����������
���
�����������
�
Case1:LOAD CARRYING
CAPACITY FOR PLATE
•Load Carrying Capacity of tension member:
•Safe load carrying capacity:
•Calculation of �
���
•Chain riveting
•�
���=�−���
•Diamond riveting
•�
���=�−��
Case1:LOAD CARRYING
CAPACITY FOR PLATE
•Load Carrying Capacity of tension
member:
•Safe load carrying capacity:
•Staggered riveting
•The critical section would be …..
4−1−3−5
4−1−7
4−1−2−3−5
4−1−2−6
�����
�����
4
2
3
7
65
1
Case1:LOAD CARRYING
CAPACITY FOR PLATE
•Load Carrying Capacity of
tension member:
•Safe load carrying capacity:
•Staggered riveting
•The critical section would be the
minimum area that would be 4-1-2-3-5
•Along the critical sections, for each
inclined leg, correspondingly
�
�
�
��
term
is added to net area where �is the
staggered horizontal distance of the
inclined leg along the critical path or
section and g is the gauge distance
corresponding to the inclined leg
�����
�����
4
2
3
7
65
1
4−1−3−5
4−1−7
4−1−2−3−5
4−1−2−6
Case 2: LOAD CARRYING CAPACITY FOR ANGLE
A.If single angle tension member is
connected to gusset plate, then …
•Safe load carrying capacity:
•Calculation of �
���
•For angle
•�
���=�
�+��
�
•k =
���
��
�+�
�
•K= shear lag effect
•Where �
�is net area of connected leg
•�
�= (gross area of connected leg -area of
rivet hole)
•�
�is gross area of unconnected leg/outstand
leg
•�
�=�−�−
�
�
�
•�
�=�−
�
�
�
gusseteplate
angle
�
�
�
�=(�−�−
�
�
)�
�
�=(�−
�
�
)�
Case 2: LOAD CARRYING CAPACITY FOR ANGLE
B. If two angles are placed back
to back and connected to ONE
side of gusset plate
•�
���=�
�+��
�where K= shear
lag effect
k =
���
���+��
•If TACK rivets are not provided along
their flange then each angle behaves
independently hence factor k =
���
���+��
���������������
���������
Case 2: LOAD CARRYING CAPACITY FOR ANGLE
C. If two angles are placed back to
back and connected to both sides of
gusset plate
•�
���=�
�+��
�where K= 1
•�
���=�
�+�
�
•It is the most efficient way of
connecting, then load carrying
capacity is maximum.
•If the two angles do not have rivet,
then each angle behaves
independently hence factor k =
���
���+��
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SANDEEP JYANI
1.Arithmetic Progression Method
•Increase in population from
decade to decade is assumed to be
constant
Population Forecasting Methods
�
�=�
�+�ഥ�
�
�=Projected population after n decades
�
�=initial population/ last census
�=number of decades between now and future
ഥ�=average increase in population per decade
Que. 1 The population of 5 decades are
given. Find out the population after 1 and 6
decades beyond last census by arithmetic
progression method.
Population Forecasting Methods
YearPopulation
193025000
194028000
195034000
196042000
197047000
Que. 1 The population of 5 decades are
given. Find out the population after 1 and 6
decades beyond last census by arithmetic
progression method.
���������������=
�����
�
Population Forecasting Methods
YearPopulation
193025000
194028000
195034000
196042000
197047000
�
�=�
�+�ഥ�
ഥ�=average increase in population per decade
Increase
3000
6000
8000
5000
ഥ�=����
1.�����������������������
�=?
�
�=�
�+�ഥ�
⇒�
����=�����+(�)����
⇒�
����=�����
Que. 1 The population of 5 decades are
given. Find out the population after 1 and 6
decades beyond last census by arithmetic
progression method.
���������������=
����
�
Population Forecasting Methods
YearPopulation
193025000
194028000
195034000
196042000
197047000
�
�=�
�+�ഥ�
ഥ�=average increase in population per decade
Increase
3000
6000
8000
5000
ഥ�=����
2.������������������������
�=?
�
�=�
�+�ഥ�
⇒�
����=�����+(�)����
⇒�
����=�����
2.Geometric Progression Method or Geometric Increase Method
•In this method Percentage Increase in
population from decade to decade is
assumed to be constant
Population Forecasting Methods
�
�=�
��+
�
���
�
�
�=Projected population after n decades
�
�=population of last known decade
�=number of decades between now and future
�=geometric mean rate of increase in population per decade
�=
�
�
1�
2�
3�
4…��
����:����������������������������
����������������������������������
�����������������������������
Que. 2Determine the future population of a
town by Geometric Increase method in the
year 2011.
YearPopulation (in 1000)
195193
1961111
1971132
1981161
Que. 2Determine the future population of a
town by Geometric Increase method in the
year 2011.
YearPopulation (in 1000)
195193
1961111
1971132
1981161
Increase
18
21
29
Percentage Increase
19.35 %
18.91 %
21.96 %
��
��
���=
��
���
���=
��
���
���=
�
�=�
��+
�
���
�
�=
�
�
��
��
��
�…��
⇒�=
�
��.����.����.��
⇒�=��.�%���������
�
����=�
�����+
��.�
���
�
=����+
��.�
���
�
=���.���
c)Incremental Increase Method
•Combination of Arithmetic and Geometric Increase method
•Actual increase in each decade is found
•Average increment of increases is found
Population Forecasting Methods
�
�=�
�+�ഥ�+
�(�+�)
�
ഥ�
�
�=Projected population after n decades
�
�=population of last known decade
�=number of decades between now and future
ഥ�=average increase of population of known decades
Population after n decades from present is
given by
ഥ�=average of incremental increase of known decades
Que. 3Determine the future population of a
town by Incremental Increase method in the
year 2000.
Year Population
1940 23798624
1950 46978325
1960 54786437
1970 63467823
1980 69077421
�
�=�
�+�ഥ�+
�(�+�)
�
ഥ�
Que. 3Determine the future population of a
town by Incremental Increase method in the
year 2000.
Year Population
1940 23798624
1950 46978325
1960 54786437
1970 63467823
1980 69077421
Increase
23179701
7808112
8681386
5609598
�
�=�
�+�ഥ�+
�(�+�)
�
ഥ�
Incremental Increase
-15371589
873274
-3071788
ഥ�=
��������
�
ഥ�=average increase of
population of known decades
ഥ�=average of incremental
increase of known decades
ഥ�=
−��������
�
ഥ�=��������.�ഥ�=−�������
GOIManualRecommends..
1.Arithmetic Increase Method is
used for old cities, where growth
rate is constant
2.For new and younger cities, we will
use geometric Progression method
3.Whenever there is negative rate of
increase, incremental increase
method is used
4.Incremental Increase Method
generally gives values in between
Arithmetic progression method
and Geometric Progression
Method
Water Demand
5.Fire Demand
•For a total amount of water consumption,
for a city of 50 Lacs population, it hardly
amounts to 1 LPCD, but this water should
be easily available and kept always stored
in service reservoirs
5.Fire Demand
When population exceeds 50,000 the water required for fire demand can be
computed using the empirical formula:
�=��������������������������������
�=���������������������
1.���������
′
����������������������=�����
2.�������
′
���������=����
�
��
+��
3.������������������������������
′
��������
a)When Population is ≤�,��,����=�����(�−�.���)
b)When Population is >�,��,����=������/���
4.������
′
���������=�����
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Que 4. Compute the fire demand for a city having population of 140 000 as per
all formulae.
PRACTICE QUESTIONSPRACTICE QUESTIONS
1.���������
′
����������������������=�����
2.�������
′
���������=������+��
3.������������������������������
′
��������
a)When Population is ≤�,��,����=�����(�−�.���)
b)When Population is >�,��,����=������/���
4.������
′
���������=�����
Que 4. Compute the fire demand for a city having population of 140 000 as per
all formulae.
PRACTICE QUESTIONSPRACTICE QUESTIONS
Que 4. Compute the fire demand for a city having population of 140 000 as per
all formulae.
PRACTICE QUESTIONSPRACTICE QUESTIONS
1.Maximum Daily Demand
Various types of Demand for Design
=�.������������������������
=���������������������×
�������������������
��
2.Maximum Hourly Demand of Maximum Day/Peak Demand
=�.�×
�.��
��
=�.�×
�
��
=�.��
= 1.5 x Maximum daily demand/24
= 1.5 x (1.8 x average daily demand)/24
= 2.7 x average daily demand/24
= 2.7 x annual average hourly demand
3.Coincident Draft/Demand/Supply
Various types of Demand for Design
=������������������+����������
=������������������������,������������������������
4.Total Draft
2.Chemical Parameters
iii.Alkalinity
•For Alkalinity Measurement, 0.02N �
���
�is used in titration. 1 ml of acid
i.e. 0.02N �
���
�gives 1 mg/L value of Alkalinity expressed as ����
�
0.02
1000
×
100
2
×1000=1
��
��
����������
3
���������=
�����������������������
����������������
Que . Water contains 210g of ��
3
2−
,122g ���
3−
and 68g of ��
−
.What is the
total alkalinity of water expressed as ����
3?
PRACTICE QUESTIONSPRACTICE QUESTIONS
Que . Water contains 210g of ��
3
2−
,122g ���
3
−
and 68g of ��
−
.What is the
total alkalinity of water expressed as ����
3?
PRACTICE QUESTIONSPRACTICE QUESTIONS
Que . Water contains 210g of ��
3
2−
,122g ���
3−
and 68g of ��
−
.What is the
total alkalinity of water expressed as ����
3?
PRACTICE QUESTIONSPRACTICE QUESTIONS
2.Chemical Parameters
iv.Hardness
•The Hardness is expressed as ����
�equivalent of Calcium and
Magnesium present in water
��������������
���/�=
[������
�+
]
[��������
�+
]
×[����������
�]
+
[������
�+
]
[��������
�+
]
×[����������
�]
⇒��������������ℎ�����
2+
=
40
2
⇒��������������ℎ�������
3=
100
2
⇒��������������ℎ�����
2+
=
24
2
Hardness as CaCO3 in mg/LDegree of Hardness
0-55 Soft Water
56-100 Slightly Hard
101-200 Moderately Hard
201-500 Very Hard
Alkalinity and Hardness
�����������������=���������{�������������,����������}
Alkalinity and Hardness
�����������������=���������{�������������,����������}
320 120
3.Sedimentation
Design of Sedimentation tank
V
H= horizontal velocity or flow velocity
V
s= settling velocity
Time of horizontal flow =
�
�
�
=
�
�
��
=
���
�
���������������=
������
�
L
H
V
H
V
s
Time of Falling through height H => �=
�
�
�
If first assumption is valid, t=D
T
�
�=
���
�
=�=
�
�
�
=> �
�=
�
��
Actual settling velocity �
�=
��
�
−��
�
���
��������������������
�=
�
��
Surface overflow rate can be thought of a settling
velocity of that particle which if introduced at
the top most point at inlet will reach the bottom
most point at outlet. This V
sis shown by another
symbol V
0
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Que 1. If initial DO = 5 mg/L, final DO = 2 mg/L.
5ml of sample is mixed to form 100 ml of diluted sample.
Find BOD.
Que If initial DO = 5 mg/L, final DO = 2 mg/L.
5ml of sample is mixed to form 100 ml of diluted sample.
Find BOD.
5ml of sample is mixed with 95ml of aerated water
���=(5−2)×
100
5
⇒���=60��/�
Que 2. If initial DO = 5 mg/L, final DO =0 mg/L at the end of 5
days.
5ml of sample is mixed with 95 ml of aerated water. Find
BOD after 5 days.
Que 2. If initial DO = 5 mg/L, final DO =0 mg/L at the end of 5
days.
5ml of sample is mixed with 95 ml of aerated water. Find
BOD after 5 days.
���������������������������������ℎ����������0
Let L
t = amount of organic matter present at any time t
t = time in days
K=rate constant (unit=per day) or deoxygenation constant
L
0= maximum amount of organic matter present
���
��
=−���
=>
���
��
=−���
=>���
���
��
=−��
�−�
�
=> �
�=�
��
−��
���
�=�
�−�
�
=�
�−�
��
−��
���
�=�
��−�
−��
7.Biochemical Oxygen Demand (BOD)
•Deoxygenation constant for given system depends on type of
impurities present in waste water.
•For sample impurities exp-sugar, deoxygenation constant will be
more and for complex impurities like phenol, toulene, aldehydes,
ketones, etcK will be less.
•In general, deoxygenation constant can be under base e or base 10
���
�=�
��−�
−��
•When it is not given in question whether base e or base 10, then we
will use base e.
�
�
�°�
=�
�
��°�
�.���
�−��°
This equation is called as Vanthoff’sAeheniersEquation
7.Biochemical Oxygen Demand (BOD)
Population Equivalent
•It indicates strength of industrial waste water for estimating
the treatment required at the municipal treatment plant
•Average BOD of domestic sewage is 80g/capita/day
•��������������������=
��������
5
���ℎ�������������/���
0.08��/���
ENVIRONMENTAL ENGG -2
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��×
��
�
�/���
�����/� ����/���
��
�
��
�
Partial Flow Characteristics of Circular Sewer
•For maximum velocity,
�
�
=�.��
•For maximum discharge
�
�
=�.��
�
Que. Design a Sewer to serve a population of 36000 with a water supply of
135 lpcdout of which 80% goes into sewer. Slope = 1/625 and sewer is
designed to carry 4 times the average discharge under design condition.
What would be the velocity generated if n=0.012 and it is assumed to be
constant .
Natural forces of purification which affect self-purification process
1.Physical forces.
a)Dilution and dispersion: When sewage of concentration �
�flows at
the rate �
�into a river stream with concentration �
�flowing at the
rate �
�, the concentration C of the resulting mixture is given by
b)Sedimentation
c)Sun-Light: the sun light has a bleaching and stablishing effect of
bacteria. It acts through the biochemical reactions.
�=
�
��
�+�
��
�
�
�+�
�
ENVIRONMENTAL ENGG -2
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Floor Area Ratio (FAR)
��������������=
����������������������
�������������
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��������������=
����������������������
�������������
��������������=
3∗400
1000
��������������=1.2
ENVIRONMENTAL ENGG -2
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��������������=
����������������������
�������������
⇒1.8=
�×
60
100
×200
200
⇒�=3
VALUATION
•Sinking fund:-
•A certain amount of the gross rent is set aside
annually as sinking fund to accumulate the cost
of construction when the life of building is over.
�=
��
(�+�)
�
−�
I = Annual instalment required
�= Number of years required to create sinking fund.
�= Rate of interest expressed in decimal
S = Amount of sinking fund.
VALUATION
•Capitalized value:-
•It is defined as the amount of money whose
annual interest at the highest prevailing rate
will be equal to the net income received from
the property.
•To calculate the capitalized value, it is
necessary to know highest rate of interest
prevailing on such properties and net income
form the property.
•Annuity:-
•The return of capital investment in the shape
of annual instalments (monthly, quarterly,
half yearly & yearly) for a fixed number of
years is known as annuity.
•Depreciation:
•A structure, after sometimes gradually losses some
of its value due to its constant use and some other
similar reasons, such as
(a) The property in neglected condition
(b) The property being away from schools & market
(c) Design being out of fashion
(d) Poor specifications followed which requiring
maintenance.
•The loss thus involve in the value of properties
known as Depreciation.
�=�
���−��
���
�
D= Depreciated Value
P= Present Value
��= Fixed percentage of depreciation
�= number of year the building has been constructed
Valuation
•Year’s Purchase: It may be defined as the
figure which when multiplied by the net
income from a property gives capitalized
value of the property
•It can also be defined as “a certain amount
of capital whose annuity of Rs 1/-at a certain
rate of interest can be received”
•??????���
′
�����ℎ���=
100
��������������
=
1
�
•Where i= rate of interest in decimal
Que. Calculate the capillary rise in a glass tube of 2.5mm diameter when immersed vertically in
(a)Water and
(b)Mercury
Take Surface Tension of water as ??????=�.�����/�and for mercury ??????=�.���/�when in contact
with air. The specific gravity for mercury is 13.6 and take angle of contact for mercury as 130°.
Que. Calculate the capillary rise in a glass tube of 2.5mm diameter when immersed vertically in
(a)Water and
(b)Mercury
Take Surface Tension of water as ??????=�.�����/�and for mercury ??????=�.���/�when in contact
with air. The specific gravity for mercury is 13.6 and take angle of contact for mercury as 130°.
��������
⇒ℎ=
4??????����
���
⇒ℎ=
4×(0.0725)���(0°)
1000×9.81×(2.5×10
−3
)
⇒ℎ=0.118�
����������
⇒ℎ=
4??????����
���
⇒�=
�×(�.��)���(��0°)
��.�×����×�.��×(�.�×��
−�
)
⇒�=�.����
PRESSURE
�
•Pressure is defined as normal force
exerted by fluid per unit area.
⇒�=��
⇒�=�×�×�×�
�=
�
�
=
����
�
�=���
Note: The standard atmosphere (atm) is a unit of
pressure defined as 1.01325 bar (101325 Pa), equivalent
to 760 mm Hg (torr)
Pascal’s Law
•It states that pressure or
intensity of pressure at a point
in a static fluid is equal in all
directions
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�
�=����
����������
TYPES OF PRESSURE
1.Atmospheric pressure
•Atmospheric air exerts a normal pressure upon all surface with
which it is in contact and it is called atmospheric pressure.
•Measured by barometer.
2.Absolute pressure
•Pressure which is measured about with reference to complete
vacuum pressure.
•It is also called actual pressure at a given point.
3.Gauge and Vacuum pressure
•Vacuum pressure is the pressure below atmospheric pressure.
•Gauge pressure is the pressure measured above atmospheric
pressure.
•Gauge pressure is measured with the help of pressure
measuring instrument in which atmospheric pressure is taken as
datum i.e. atmosphere pressure on the scale is marked as 0.
Atmospheric
pressure
Gauge pressure
Absolute
pressure
Vacuum pressure
Pressure
Absolute zero Pressure
A
B
�
���=�
���+�
�����
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FLUID STATIC
Net force on body = �
�−�
�����
= ���
�−���
�����
= {���
�−�
�}����
=��������
=�����
F= ���
��
�
�
��
ENVIRONMENTAL ENGG -2
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³
³
METACENTRE
•It is the point of intersection of normal axis with
the new line of action of force of buoyancy, when
a small displacement is given.
•It is a point about which a body starts to oscillate
when it is tilted by small angle.
�.�.=�.�−�.�
��=
�
�
−��
•Where �
��=
��
�
��
; �
��=
��
�
��
;
�
���
�
= metacentric radius
•For stability G.M. should be positive, i.e.,
B.M>B.G
•From designing point of view,
•the moment of inertia �is taken about the
top view at the surface of liquid.
•Oscillation about longitudinal axis is called
rolling.
•Oscillation about transverse axis is called
pitching.
•TIME PERIOD FOR OSCILLATION:
�=��
�
�
��.�
•Where K is radius of gyration
•GM is metacentric height
•T is time of one complete oscillation
HYDROSTATIC FORCES ON PLANE SURFACES
•HYDROSTATIC FORCE:
•The force exerted by the static fluid, when a body is
exposed to it.
•CENTRE OF PRESSURE:
•It is the point from which the total hydrostatic force
is supposed to be acting.
�=
�+���
�
����
�=
���
�
����
�
���=��ഥ�
ഥ�=�������������������������������
Centre of pressure = ഥ�+
�
�
�ഥ�
(for vertical body)
CGCentre of pressure
�
�
�
�
�
�
�
�
�
�>�
�>�
�>�
�
h
ഥ�
HYDROSTATIC FORCES ON PLANE SURFACES
•CENTRE OF PRESSURE FOR INCLINED SURFACE:
Centre of pressure = ഥ�+
�
�
�ഥ�
sin
2
�
�
���=��ഥ�
For horizontal body, �= 0
h=ഥ�
�
�is moment of inertia about C-G axis
h is distance of centreof pressure from free surface
ഥ�is distance of centreof gravity from free surface
���
�
���
�
�
�
�
�
�
��
��
ഥ�
ഥ�
�
312
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CENTRE OF PRESSURE
BODY CENTRE OF PRESSURE (h)
d
d
d
d
d
Centre of pressure h = ഥ�+
�
�
�ഥ�
⇒�=
�
�
+
��
�
��
(��)(
�
�
)
b
⇒�=
�
�
+
�
�
⇒�=
��
�
CENTRE OF PRESSURE
BODY CENTRE OF PRESSURE (h)
�
�
�d
d
d
d
d
Centre of pressure h = ഥ�+
�
�
�ഥ�
⇒�=
�
�
+
��
�
��
(
�
�
��)(
�
�
)
b
⇒�=
�
�
+
�
�
⇒�=
�
�
CENTRE OF PRESSURE
BODY CENTRE OF PRESSURE (h)
�
�
�
�
�
�
d
d
d
d
d
Centre of pressure h = ഥ�+
�
�
�ഥ�
⇒�=
��
�
+
��
�
��
(
�
�
��)(
��
�
)
b
⇒�=
��
�
+
�
��
⇒�=
��
�
CENTRE OF PRESSURE
BODY CENTRE OF PRESSURE (h)
�
�
�
�
�
�
�
�
�
d
d
d
d
d
Centre of pressure h = ഥ�+
�
�
�ഥ�
⇒�=
�
�
+
��
�
��
(
��
�
�
)(
�
�
)
⇒�=
�
�
+
�
�
⇒�=
��
�
CENTRE OF PRESSURE
BODY CENTRE OF PRESSURE (h)
�
�
�
�
�
�
�
�
�
�
�
�
d
d
d
d
Horizontal Plane Surface Submerged in Liquid
•Consider a plane horizontal
surface immersed in a static fluid
•Since every point at the surface
is at equal distance from the free
surface of liquid, pressure
intensity shall be equal on entire
surface and is equal to
�=���
•So total pressure force
�=����
⇒�=��ഥ�
�
⇒�������������������������������������
ഥ�=�
⇒���������������������������������������
���=�
Liquids subjected to Constant HorizontalAcceleration
����=
�
�
�
�
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Liquids subjected to Constant VerticalAcceleration
Gauge Pressure at any point in a liquid mass which
is subjected to constant UPWARD ACCELERATION
�
�=����+
�
�
�
Gauge Pressure at any point in a liquid mass which is
subjected to constant DOWNWARD ACCELERATION
�
�=����−
�
�
�
3. LAMINAR AND TURBULENT FLOW
•For a pipe flow, type of flow is determined
by a non dimensional number called
Reynold’s number:
�
�=
��
�
•�=������������������������
•�=���������
•�=�������������������������
•�
�<���������������
•�
�>�����������������
•����<�
�<����May be laminar or Turbulent
CONSERVATION OF MASS
CASE –2: GENERALISED CONTINUITY EQUATION
��
��
+
�
��
��+
�
��
��+
�
��
��=�
•This equation is valid for all types of flow.
•Steady as well as unsteady
•Uniform as well as non uniform
•Compressible as well as incompressible
•If flow is incompressible (�=constant)
��
��
+
��
��
+
��
��
=�
Continuity Equation
•This equation is based on the principle of
conservation of mass
•So For a fluid flowing in pipe at all cross
sections, the quantity of fluid per second is
constant.
•Let us take 2 cross sections
�
��
�
�
� �
�
�
� �
�
�
� �
�
��������������������−�=�
��
��
�
�������
���������������
������������������
��������������������−�=�
��
��
�
����������������������������������,
��������������������−�=��������������������−�
⇒�
��
��
�=�
��
��
�
����������������������������������������
����������������������
Case 1: Steady and 1-Dimensional Flow
329
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�=��Ƹ�+��Ƹ�
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Twodimensionalflowisgivenby
�=�+��+�Ƹ�+�−�Ƹ�.
Theflowis
a)IncompressibleandIrrotational
b)IncompressibleandRotational
c)CompressibleandIrrotational
d)CompressibleandRotational
VORTEX FLOW
������������������������
�=
�
�
��
=
�
�
�
�
��
•Free surface of liquid is paraboloid.
•Volume of paraboloid formed=
��
�
�
�
•In case of Forced Vortex, the rise of liquid
at the ends is equal to the fall of liquid
level at the axis of rotation
VORTEX FLOW
Que. A tank of dia20cm and height 100 cm contains water uptoa height of 60cm.
It is rotated about its vertical axis at 300 rpm. Find the depth of parabola formed
at free surface of water
VORTEX FLOW
Que. A tank of dia20cm and height 100 cm contains water uptoa height of 60cm.
It is rotated about its vertical axis at 300 rpm. Find the depth of parabola formed
at free surface of water
�=
�
�
��
=
�
�
�
�
��
�=
���
��
�=
�����
��
�=��.�����/���
�=
��
�
=����
�=
�
�
�
�
��
⇒�=
(��.��)
�
(��)
�
�(���)
⇒�=��.����
VORTEX FLOW in CLOSED CYLINDRICAL VESSELS
•CASE 1:
�=
�
�
��
=
�
�
�
�
��
•CASE 2:
•For such closed vessels, we use 2 equations,
•�=
�
�
��
=
�
�
�
�
��
•Volume of air after rotation=Volume of vessel-Volume of
liquid in vessel
•Volume of air after rotation=volume of paraboloid
formed=
��
�
�
�
���������
′
���������
⇒
�
��
+�+
�
�
��
=��������
•���������
′
��������:
•It states that in a STEADY, IDEAL flow of an incompressible fluid, the total
energy at any point of the fluid is constant (pressure, kinetic, and potential
energy)
•Assumptions of Bernoulli’s Equation
1.Fluid is ideal (viscosity is zero)
2.The flow is incompressible
3.Flow is steady
4.The flow is irrotational
�����������������������
�������������
WEIRS AND NOTCHES
•Rectangular notch or weir:
�
�����������=
�
�
����
�/�
�
������=
�
�
�
�����
�/�
Where -
�is the total discharge
�
�is coefficient of discharge
�is length of notch or weir
�is head of water over the crest
WEIRS AND NOTCHES
•Triangular notch or weir:
�
�����������=
�
��
��tan
�
2
�
�/�
�
������=
�
��
�
���tan
�
2
�
�/�
Where -
�is the total discharge
�
�is coefficient of discharge
�is the angle of notch
�is head of water above the V-notch
�
�
344
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�
�
345
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�
�
�=
�
�
�
�����
�/�
�=
�
�
�.�����.��(�.�)
�/�
=�.���
�
/���
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�
�
•It is the ratio of inertia force to viscous force.
�
�=
�
�
�
�
•For flat plate or open flow,
�
�=
���
�
•For duct/pipe or closed flow,
�
�=
���
�
REYNOLD’S NUMBER
�−�������������
�−��������������
�
�<����;�����������
�
�>����;�������������
����<�
�<����;��������������
�������
����
DARCY WEISBACH EQUATION/FORMULA
•It is used to find out head loss due to friction, this equation is
applicable for laminar as well as turbulent flow but the flow must be
steady.
�
�=
��
′
��
�
���
�
�=
���
�
���
�
′
is friction coefficient or Fanning’s coefficient
�is friction factor or Darcy Weisbachcoefficient
Experimental
Formula for
�
�
LAMINAR FLOW THROUGH CIRCULAR PIPE
•HAGEN POISEUILLE LAW:
1.Shear Stress
���
�
−�+
��
��
∆���
�
−�×���∆�=�
−
��
��
∆���
�
−�×���∆�=�
−(
��
��
)�−��=�
�=−
��
��
.
�
�
•The shear stress �across a section varies with ‘r’.
•Hence shear stress distribution across a section is
linear.
��������
������������
�����������
������������
LAMINAR FLOW THROUGH CIRCULAR PIPE
2.Velocity Distribution
To obtain the velocity distribution across a section
substituting the value of shear stress �=�
��
��
�=�−������=−��
�=−
��
��
.
�
�
−�
��
��
=−
��
��
.
�
�
��
��
=
�
�
��
��
.
�
�
Integrating w.r.t.r
�=
�
��
��
��
�
�
+�
��������
������������
�����������
�������������=−
��
��
.
�
�
��������
������������
�����������
������������
LAMINAR FLOW THROUGH CIRCULAR PIPE
•HAGEN POISEUILLE LAW:
�=
�
��
��
��
�
�
+�
Here C is the constant of integration and its value
is obtained from boundary condition r=R and u=0
�=
�
��
��
��
�
�
+�
�=−
�
��
��
��
�
�
Now substituting value of C in the equation for u
�=
�
��
��
��
�
�
−
�
��
��
��
�
�
�=−
�
��
��
��
(�
�
−�
�
)
Values of �,
��
��
and R are constant, which means
the velocity u, varies with the square of r.
Hence Velocity distribution across the section of
pipe is parabolic.
LAMINAR FLOW THROUGH CIRCULAR PIPE
3.Maximum velocity
•The velocity is maximum when r=0
�=−
�
��
��
��
(�
�
−�
�
)
�
���=−
�
��
��
��
�
�
4.Discharge
It is calculated by considering the flow through a circular ring element of
radius ‘r’ and thickness ‘dr’. The fluid flowing per second through this
elementary ring-
��=����������������������������������
��=������
��=−
�
��
��
��
(�
�
−�
�
)�����
On integrating both sides, we get-
�=
�
��
−
��
��
�
�
LAMINAR FLOW THROUGH CIRCULAR PIPE
5.Average velocity
ഥ�=
�
����
ഥ�=
�
��
−
��
��
�
�
��
�
ഥ�=
�
��
−
��
��
�
�
6.Hence ratio of maximum and average velocity is-
����
ഥ�
=
�
��
−
��
��
�
�
�
��
−
��
��
�
�
=�.�
2. Maximum to average Velocity
�
���=
�
�
�
���
3. Drop of Pressure head for a given length
�
�=
�
�−�
�
��
=
����
����
���
�
Que. For an oil of viscosity 0.02Ns/m2 flowing between two stationary
parallel plates 1m wide maintained 10mm apart. The velocity midway
between the plates is 2 m/sec ,Calculate
a)Pressure gradient along flow
b)Average Velocity and
c)Discharge
�)⇒�
���=−
�
��
��
��
(�
�
)
⇒�=−
�
�(�.��)
��
��
(�.��)
�
⇒
��
��
=−�����/�
�
per m
�)⇒�
���=��/���
�
���=
�
�
�
���
⇒�
���=
�
�
(�)
⇒�
���=�.���/���
Que. For an oil of viscosity 0.02Ns/m2 flowing between two stationary
parallel plates 1m wide maintained 10mm apart. The velocity midway
between the plates is 2 m/sec ,Calculate
a)Pressure gradient along flow
b)Average Velocity and
c)Discharge
⇒�=����������×�
���
⇒�=�×�×�
���
⇒�=�×�.��×�.��
⇒�=�.�����
�
/���
MAJOR ENERGY LOSSES
a)Darcy –WeisbachFormula
•It is used to find out head loss due to
friction, this equation is applicable for
laminar as well as turbulent flow but the
flow must be steady.
�
�=
��′��
�
���
�′=coefficient of friction
�
�= loss of head due to friction
�=length of pipe
�=mean velocity of flow
�=diameter of pipe
�′=
��
��
(for �
�<2000)
�′=
�.���
�
�
�/�(for �
����������������
�
)
�
�=
���
�
���
�=��⇒�=
�
�
=
�
�
�
�
�
⇒�
�=
�����
�
�
�
���
�
�= friction factor
⇒�
�=
���
�
���
�
MAJOR ENERGY LOSSES
b)Chezy’s Formula
�=���
�=������������������=
����������
���������������
=
�
�
�
�
��
=
�
�
�=�����
′
���������
�=�����������������������������=
�
�
�
MINOR ENERGY LOSSES
1.LOSS OF HEAD DUE TO SUDDEN EXPANSION:
�
� �
�
�
�(���������)=
(�
�−�
�)
�
��
�
�(���������)=
�
�
�
��
�−
�
�
�
�
�
�
�(���������)=
�
�
�
��
�−
�
�
�
�
�
�=�
��
�=�
��
�
�
�
�
�
=
�
�
�
�
Loss from exit of pipe �
�=
��
�
��
(here A=infinity)
Que. Find the loss of head when a pipe of diameter 200mm is
suddenly enlarged to a diameter of 400mm. The rate of flow of water
through the pipe is 250 litres/sec.
MINOR ENERGY LOSSES
3.LOSS AT ENTRANCE OF PIPE
4.LOSS AT EXIT OF PIPE
5.LOSS DUE TO BEND IN PIPE
�
�=
�.��
�
�
���
�= velocity in pipe
�
�=
�
�
��
�= velocity at outlet of pipe
�
�=
��
�
��
�=velocity of flow
�=coefficient of bend (constant)
Depends upon angle of bend or radius of curvature of bend
Equivalent Pipe
•A hydraulically Equivalent pipe
means a pipe which can replace
existing compound pipe while
carrying same discharge under
same losses
�
�=
���
�
���
�=��⇒�=
�
�
=
�
�
�
�
�
⇒�
�=
�����
�
�
�
���
�
�= friction factor
⇒�
�=
���
�
���
�
Velocity Distribution in Turbulent Flow in Pipes
•�=�
���+�.��
∗
���
�(�/�)
(Prandtl's universal velocity distribution equation for turbulent flow in pipes)
It shows that in turbulent flow, the velocity varies directly with the logarithm of
the distance from the boundary
•�������������=�
∗
=
��
�
•
����−�
�
∗
=�.�����
��(�/�)
•�
���−�=��������������
Velocity Distribution in terms of Average Velocity in Turbulent Flow in Pipes
•Velocity distribution for turbulent flow is
•
ഥ�
�
∗
=�.�����
��(
�
∗
�
�
)+�.����������������
•
ഥ�
�
∗
=�.�����
��(
�
�
)+�.���������������
Difference of Local and Average Velocity for smooth and rough pipes
�−ഥ�
�
∗
=�.�����
��(�/�)+�.��
Que: Determine the wall shearing stress in a pipe of diameter 100 mm
which carries water. The velocities at the pipe centreand 30 mm from
the pipe centreare 2 m/s and 1.5 m/s respectively. The flow in pipe is
given as turbulent.
�
���−�
�
∗
=5.75log
10(�/�)
⇒
2−1.5
�
∗
=5.75log(
0.05
0.02
)
⇒�
∗
=
0.5
2.288
⇒�
∗
=0.2185�/���
�������������=�
∗
=
�
�
�
⇒�
∗
=
�
�
�
⇒�.����=
�
�
����
⇒�
�=��.���/�
�
Que: Determine the distance from the pipe wall at which the local
velocity is equal to the average velocity for turbulent flow in pipes.
�−ഥ�
�
∗
=�.�����
��(�/�)+�.��
⇒
ഥ�−ഥ�
�
∗
=�.�����
��(�/�)+�.��
⇒�=�.�����
��(�/�)+�.��
⇒�.�����
��(�/�)=−�.��
⇒�=�.�����
FLUID MECHANICS CIVIL ENGINEERING FLUID MECHANICS CIVIL ENGINEERING
FRICTION FACTOR AND REYNOLD’S NUMBER RELATIONSHIP
1.�=
64
��
For laminar flow
2.�=
�.���
�
�
�/�(for�
����������������
�
)
3.�=0.0032+
0.221
�
�
�.���for Turbulent Flow in smooth pipe (for
�
����������
�
<��<���
�
)
4.
1
�
=2log
10(
�
�
)+1.74For Turbulent Flow in rough pipe
R= radius of pipe
K= Average height of roughness
DRAG AND LIFT
•Drag Force = Force due to pressure in the direction of fluid motion +
Force due to shear stress in the direction of fluid motion.
•Lift Force= Force due to pressure in the direction perpendicular to
the direction of motion+ Force due to shear stress in the direction
perpendicular to the direction of motion
�
�=�
��
��
�
�
�
�=�
��
��
�
�
DRAG AND LIFT
•Drag and Lift for a body moving in a fluid of
•density �,
•at uniform velocity �,
•with�
�= Coefficient of drag,
•�
�= Coefficient of lift,
•A= Area of the body which is the projected area seen by a person looking towards the
object from a direction parallel to the velocity
is calculated experimentally as
�
�=�
��
��
�
�
�
�=�
��
��
�
�
•And Resultant force on the body is
�
�=�
�
�
+�
�
�
Que 1. A flat plate 1.5 m x 1.5 m
moves at 50 km/hour in
stationary air of density 1.15
kg/m
3
. If the coefficients of drag
and lift are 0.15 and 0. 75
respectively, determine :
(i)The lift force
(ii)The drag force
(iii)The resultant force, and
(iv)The power required to keep
the plate in motion
�
�=�
��
��
�
�
�
�=�
��
��
�
�
�����������=�.��.��
�
=�.���
�
���������������=������
=
������
����
=��.���/���
�
�=�.���
�=�.��
Que 1. A flat plate 1.5 m x 1.5 m
moves at 50 km/hour in
stationary air of density 1.15
kg/m
3
. If the coefficients of drag
and lift are 0.15 and 0. 75
respectively, determine :
(i)The lift force
(ii)The drag force
(iii)The resultant force, and
(iv)The power required to keep
the plate in motion
Drag Force
�
�=�
��
��
�
�
⇒�
�=(�.��)(�.��)
(�.��)(��.��)
�
�
⇒�
�=��.���
Lift Force
�
�=�
��
��
�
�
⇒�
�=(�.��)(�.��)
(�.��)(��.��)
�
�
⇒�
�=���.��
•And Resultant force on the body is
�
�=�
�
�
+�
�
�
⇒�
�=(��.��)
�
+(���.�)
�
⇒�
�=���.���
•Power required to keep the plate in motion
�=�����������������������������������
⇒�=�
��
⇒=��.���
⇒�=��.��×��.��
⇒�=����
3.Laminar flow and turbulent flow –
•If Reynolds no. is less than 500, in an open channel
then flow is Laminar.
•If Reynolds no. is more than 2000, in an open channel
then flow is Turbulentin open channel flow.
•If �
�lies between 500and 2000, it is in Transition
state
�
�=
���
�
�=
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���������������
�=�������
�=�����������������������������������
CLASSIFICATION OF FLOW IN OPEN CHANNELS
�=�������������������������
4. SUB CRITICAL, CRITICAL AND SUPER CRITICAL FLOW
�
�=
�
��
�−��������������=
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�����������������
�
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�
�=�;������������
�
�>�;��������������������������������������������������
CLASSIFICATION OF FLOW IN OPEN CHANNELS
4. SUB CRITICAL, CRITICAL AND SUPER CRITICAL FLOW
�
�=
�
��
������������������
′
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��������������������������������������������,��������"��������"
���������=��
CLASSIFICATION OF FLOW IN OPEN CHANNELS
���������=
��
�
�
�
����/������=�
����/�����+�
�����/������
(��������) (��������) (��������)
�����������������,�
�<����������
����������������������������������������
����������������������������������������
�−���������������������������������
⇒�
����/������=�−�
(��������)
�����������������.
�
�<�
�>�
�−�>�
������������������.
�
�>�
�<�
�−�<�
So for high velocity, disturbance will
not travel upstream, it will travel
downstream with a velocity of V-C
MOST ECONOMICAL SECTIONS OF OPEN CHANNELS
1.Rectangular:
•The condition for most economical section is for a given area, the
perimeter should be minimum.
a)�=����������������������
b)�=
�
�
(��������������������������������������)
2.Trapezoidal:
•The condition for most economical section
a)
�+���
�
=��+�
�
b)�=
�
�
c)Best side slope = 60
d)A semi circle drawn with O as centreand radius
equal to depth of flow will touch the three sides of channel
�
�
�
�
�
��
n
1
�
MOST ECONOMICAL SECTIONS OF OPEN
CHANNELS
3. Triangular Section
•Most economical triangular section has slanting sides
perpendicular to each other making an angle of 45°with the
vertical
•Hydraulic mean radius�=
�
22
4. Circular:
1.Condition for Maximum Velocity –
•For maximum velocity, the depth of water in circular
channel should be equal to 0.81times the diameter of
circular channel.
•For maximum velocity, the hydraulic mean depth is
0.3times the diameter of circular channel.
2.Condition for Maximum Discharge –
•For maximum discharge, the depth of flow in circular
channel should be equal to 0.95times the diameter of
circular channel.
SPECIFIC ENERGY
The total energy of a flowing liquid per unit
weight is given by
�����������=�+�+
�
�
��
If the channel bottom is taken as datum, total
energy of a flowing liquid per unit weight is
��������������=�+
�
�
��
SPECIFIC ENERGY –energy per unit weight of
the liquid w.r.t bottom of the channel
�=�����������������������
����������
�=���������������
�=������������������
SPECIFIC ENERGY CURVE
SPECIFIC ENERGY CURVE –The curve which
shows the variation of specific energy with
depth of flow.
�=�+
�
�
���
�
=�
�+�
��=���������������������=
�
�
�=�����������
�
�<�;���������������
�
�>�;�����������������
CRITICAL DEPTH �
�:
•The depth of flow of water at which the specific energy is minimum.
�
�=
�
�
�
�/�
CRITICAL VELOCITY �
�:
•The velocity of flow at critical depth.
�
�=��
�
Sub critical, critical and super critical flow
��������.=
�
�
��
�
CONDITION FOR MAXIMUM DISCHARGE FOR A GIVEN SPECIFIC ENERGY:
�=
��
�
�
�<�;���������������
�
�=�;������������
�
�>�;�����������������
393
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
�
�
/���
Important Definitions
1.GROSS HEAD
The difference between the head race
level and tail race level when no water
is flowing is called Gross Head
2.Net Head
Net Head or Effective Head is head
available at the inlet of turbine.
When water flows from head race to
turbine, a loss of head due to friction
(major loss) between water and
penstock occurs and other minor
losses due to bend, pipe fittings, loss
at entrance of penstock, etc occurs,
etc
�=�
�−ℎ
�
�ℎ���
�
�=
��
′
��
�
���
�=velocityofflowinpenstock
�=Lengthofpenstock
�=Diameterofpenstock
3.EFFICIENCY OF TURBINE
Following types of Efficiency are there:
1.Hydraulic Efficiency
�
ℎ=
����������������������
��������������������(���)
=
��
��
•The rotating part of a turbine on which VANES are fixed is
called as Runner
•Hydraulic Efficiency is ratio of power given by water to
the runner of turbine to the power supplied by water at
inlet of turbine
•Power at inlet of turbine is more, and as vanes are not
smooth, power keeps on decreasing, so power delivered
to runner is less than that at inlet
3.EFFICIENCY OF TURBINE
Following types of Efficiency are there:
2.Mechanical Efficiency
�
���ℎ������=
��������ℎ��ℎ������������
�����������������������������
=
��
��
•Power delivered by water to runner of a turbine is
transmitted to shaft of the turbine
•Due to mechanical losses, power available to shaft
of turbine is less than power delivered to the runner
of turbine
3.EFFICIENCY OF TURBINE
Following types of Efficiency are there:
3.Volumetric efficiency
�
����������=
����������������������������������
��������������������������.
•The volume of water striking the runner of a turbine is
slightly less than the volume of water supplied to the
turbine.
•Some of water is discharged away to the tail race without
striking the runner of turbine
3.EFFICIENCY OF TURBINE
Following types of Efficiency are there:
4.Overall efficiency
�
�=
�ℎ��������
��������������
•It is ratio of Power available at the shaft of the turbine to
the power supplied by water at the inlet of turbine
�
���=�
ℎ��������×�
���ℎ������×�
����������
Relationship between Speed, pole and
Frequency
�=
120�
�
N= rotational speed
�=frequency
�=number of poles in generator
For very low rpm, no of poles is very large and it is not feasible from
design point of view, so at low head in should not install a turbine
Design Aspects of Pelton Wheel
•Velocity of jet �
1=2��, If losses are not considered
•�
1=�
�2��, if losses are considered
•�=??????2��velocity of vane
•�=
??????��
60
�ℎ������������������
•Speed ratio ??????=
�
�
1
=0.43��0.48
•Jet ratio �=
�(�����ℎ�����������������)
�(��������)
(between 11 to 16)
•Number of buckets= 15+0.5�
=15+0.5
�
�
•SPECIFIC SPEED:
•The speed of a geometrically similar turbine which would develop unit
power when working under a unit head.
�
�=
��
�
�/�
•It plays an important role in the selection of the type of turbine.
•The suitability of turbine for a particular project depends on:
•Head of water
•Rotational speed
•Power development
•Lesser the specific speed, less will be the size of runner.
•Higher specific speed, lower efficiency of turbine
N=rpmP=powerH=head
CIVIL ENGINEERING
SANDEEP JYANI
When Vehicle is moving on Levelled Ground
•For up gradient,
•For Down Gradient,
���=��+
�
�
���+�.���
���=��+
�
�
���−�.���
���=��+
�
�
���
When Vehicle is moving on a Gradient
STOPPING SIGHT DISTANCE
IRC recommendations for SSD:
1.For single lane road with two way traffic, the
minimum SSD should be equal to 2SSD (for same
speed).
2.For undivided highway with two way traffic effect of
gradient is not considered while calculating SSD.
However for divided highway, gradient is
considered.
3.SSD on vertical curves is calculated along the centre
line of the curve from which a driver with an eye
level of 1.2m above the ground surface can see an
obstacle 0.15m above the ground.
4.If SSD cannot be provided in a particular stretch of
road, proper sign boards with speed restrictions
should be provided.
2. OVERTAKING SIGHT DISTANCE
•The minimum distance available for the driver to safely overtake the
slow vehicle in front of him by considering the traffic in the opposite
direction is called as the overtaking sight distance.
•This distance make us see whether the road is clear to undergo an
overtaking movement.
•The overtaking sight distance is also called as the passing sight
distance that will be measured along the center line of the road.
•This is the line level over which the driver keeping an eye level of 1.2
m above the road level can easily see the top of the object 1.2 m
above the road surface.
OVERTAKING SIGHT DISTANCE
Factors Affecting Overtaking Sight Distance
1.Velocities of the overtaking vehicle, overtaken
vehicle and of the vehicle coming in the opposite
direction.
2.Spacing between vehicles, which in-turn depends
on the speed
3.Skill and reaction time of the driver
4.Rate of acceleration of overtaking vehicle
5.Gradient of the road
OVERTAKING SIGHT DISTANCE
It is assumed that the vehicle A is forced to reduce its speed to ‘v
b,’ the speed of
the slow moving vehicle B and travels behind it during the reaction time t of the
driver.
So d
1is given by –
�
�=�
�t
Then the vehicle A starts to accelerate, shifts the lane, overtake and shift back to
the original lane. The vehicle
A maintains the spacing ‘s’ before and after overtaking. The spacing ‘s’ in m is
given by
�=�.��
�+�
(��������������������=���������������������������
=�.����)
OVERTAKING SIGHT DISTANCE
•Let ‘T’ be the duration of actual overtaking.
•The distance traveled by B during the overtaking
operation is �
��
•Also, during this time, vehicle A accelerated from
initial velocity v
band overtaking is completed
while reaching final velocity ‘v’.
•Hence the distance traveled is given by:
�
�=�
��+
�
�
��
�
��+�
��=�
��+
�
�
��
�
�=
�
�
��
�
�=
��
�
OVERTAKING SIGHT DISTANCE
IRC recommendations for OSD:
1.On divided highways and on roads with one way traffic
regulation –
OSD = d
1+d
2
2.On divided highways with four or more lanes, IRC suggests that
there is no need to provide OSD.
NOTE:
1.Effect of gradient is not considered while calculating OSD.
2.If OSD cannot be provided throughout the length of the road,
we provide overtaking zone at certain intervals.
3.Desirable length of overtaking zone is 5times OSD, subjected to
a minimum of 3times OSD.
HORIZONTAL ALIGNMENT DETAIL
Centrifugal force
�=
��
�
�
•
�
��
=
�
�
��
Where �is design velocity
R is radius of curve
P is centrifugal force
R
P
SUPER ELEVATION
•If the coefficient of friction is neglected or assumed to
be 0, i.e. �=�, so the centrifugal force has to be
countered by the Equilibrium Super Elevationonly
⇒�=
�
�
��
(�= m/sec)
⇒�=
(�.����)
�
�.���
(V = kmph)
⇒�=
�
�
����
In such a case the pressure on inner and outer wheels
will be equal
�+�=
�
�
��
Note
•When the pavement is horizontal (no superelevation) the pressure on
outer wheels will be higher
•When friction f=0, in such a case the pressure on inner and outer
wheels will be equal
•When the limiting condition of overturning occurs, then pressure at
inner wheel is 0.
SUPER ELEVATION
•Similarly in cases where super elevation can not be
provided, i.e. e=0, so the centrifugal force has to be
countered by frictional force only
�=
�
�
��
���=
�
�
����
�+�=
�
�
��
SUPER ELEVATION for Mixed Traffic Conditions
Superelevation
�=����≈����=
�
�
Calculate equilibrium superelevationcorresponding to
75% design speed
⇒�=
(�.���)²
��
If the value is less than or equal to 7%, then it is
correct, but if it exceeds 7% then
�
�
��
=
�+����
�−�����
=
�+�
�−��
∴
�²
��
=�+�=�.��+�.��=�.��
⇒�
���= �.����
������=��������������/���
�
���=��������������/���
�=����������������
•⇒�=
(�.���)²
��
•⇒�=
(�.��×�.����)²
�.���
(V –km/hr)
•⇒�=
�²
����
SUPER ELEVATION for Mixed Traffic Conditions
Steps for Super Elevation Design
1.The superelevation for 75 percent of
design speed is calculated neglecting
the friction
�=
(�.���)²
��
or ⇒�=
(�.��×�.����)²
�.���
(V –km/hr)
⇒�=
(�.���)²
����
⇒�=
�²
����
2.If the calculated value of �is less than
�%���.��, the value obtained is
provided as superelevation; else provide
�=�.��and proceed for step 3 and 4
Steps for Super Elevation Design
3.Check the value of coefficient of friction
developed at FULL VALUE OF DESIGN SPEED i.e.
�+�=
�
�
��
⇒�=
�
�
��
−�.��
Now if the value of �is less than 0.15, then super
elevation provided (0.07) is safe for design speed. If
not, calculate the restricted design speed.
4.As an alternative for safety, the safe speed is
restricted as…
�+�=�.��+�.��
⇒
�
�
��
=�.��
⇒�
���=�.����
RADIUS OF HORIZONTAL CURVE
•The maximum comfortable speed on a horizontal curve
depends on the radius of the curve.
•Although it is possible to design the curve with maximum
superelevationand coefficient of friction, it is not desirable
because re-alignment would be required if the design speed is
increased in future.
•Therefore, a ruling minimum radius can be derived by assuming
maximum superelevationand coefficient of friction.
�
������=
�
2
�(�+�)
EXTRA WIDENING OF CURVES
�
�=�
�+ �
��
�
�=
��
�
��
+
�
�.��
where n –number of traffic lanes
�–length of wheel base (normally 6m or 6.1m)
R –radius of horizontal curve (m)
V –design speed (km/hr)
LENGTH OF TRANSITION CURVES
1.Rate of change of centrifugal
acceleration
•Length of transition curve denoted by �
��
�
��=
�
�
��
As per IRC, value of rate of change of
centrifugal acceleration is
�=
��
��+�.��
�/���
�
�
���= 0.5
�
���= 0.8
�=∞
�=�
�
2
�
=0
�
2
�
����∗∗∗
2. Rate of change of superelevation
•Length of transition curve denoted by �
��
It is not desirable to raise outer edge of a pavement at a
larger rate than 1 in 150 , so length of transition curve
should be at least 150 times the total amount of raise of
outer edge with respect to inner edge
Length of transition curve may be reduced in case of :
Built up –1 in 100
And Built up –1 in 60 in hill roads
LENGTH OF TRANSITION CURVES
LENGTH OF TRANSITION CURVES
2. Rate of change of superelevation
•Length of transition curve denoted by �
��
�
��=����(Plain and rolling)
�
��=����(Built up areas)
�
��=���(Mountainous and steep)
Case 1:When super elevation is provided by raising the
outer edge with respect to inner edge�=�(
)
�+
��
Case 2 : When super elevation is provided by rotating
about centre�=
��+��
�
�=∞
�=�
�
2
�
=0
�
2
�
�
�
�
�
LENGTH OF TRANSITION CURVES
2.Rate of change of superelevation
•Length of transition curve denoted by �
��and e be the rate of superelevation
along the pavement width W
•Raise(E) of the outer edge with respect to inner edge is given by
E = eB= e(�+�
�)
•Rate of change should not be steeper than 1 in 150 for plain and rolling terrain,
•And should not be steeper than 1 in 60 for hilly terrain
•Case 1:When super elevation is provided by raising the outer edge with respect to
inner edge
�
�
��
=
�
�
⇒�=
�
��
�
…�
���������������=����=
�
�+�
�
…�
Equating 1 and 2 equations,
�
��=��(�+�
�)
�≥�����������������
�≥������������������
�
�
�
�
��
�
�
�
�
��
LENGTH OF TRANSITION CURVES
2.Rate of change of superelevation
•Length of transition curve denoted by �
��
•Case 2:When super elevation is provided by rotating about centre
�����������=����=
�
�+�
�
2
⇒�=�
�+�
�
2
…�
Also
�
�
��
=
�
�
⇒�=
�
��
�
…�
Equating 1 and 2 equations,
�
��=��
�+�
�
2
�≥�����������������
�≥������������������
�
�
�
�
�
��
�
�
�
�
��
LENGTH OF TRANSITION CURVES
•By Empirical Formula of IRC
•Length of transition curve denoted by �
��
•IRC suggest the length of the transition curve is minimum for a
plain and rolling terrain
�
��=
�.��
�
�
•and for steep and hilly terrain is:
�
��=
�
�
�
•And the shift ‘s’
�=
��
�
���
LENGTH OF TRANSITION CURVES = MAXIMUM OF �
���
���
��
VERTICAL ALIGNMENT
•GRADE COMPENSATION
•When horizontal curve exists together
with the up gradient, pulling power of
vehicle is reduced, hence to increase
the pulling power, the gradient is
reduced.
•This reduction in gradient to
compensate the loss of tractive effect
on horizontal curves is called GRADE
COMPENSATION.
•It is not required for grades flatter than
4%
•Grade compensation required is given
as
��+�
�
%subjected to a maximum of
��
�
%
•Compensated grade should not be less
than 4%
���������������
��������������������
��
�
��
��↷
VERTICAL ALIGNMENT
•CURVE
•Length of summit curve is
given by:
•When length of summit
curve greater than sight
distance (L>SSD):
•�=
��
�
�(�
�+�
�)
�
•When length of summit
curve lesser than sight
distance (L<SSD):
•�=��−
(���+���)
�
�
SSD
(�
�= 1.2 m, �
�=0.15m )
OSD/ISD
(�
�= 1.2 m, �
�=1.2 m )
�=���������������������(�)
�=�������������������(�) �=���������������������������������������
VERTICAL ALIGNMENT
•SUMMIT CURVE
•When stopping sight distance is considered the height of driver's eye
above the road surface (�
�) is taken as 1.2 metres, and height of object
above the pavement surface(�
�) is taken as 0.15metres. If overtaking
sight distance is considered, then the value of driver's eye height (�
�) and
the height of the obstruction (�
�) are taken as 1.2metres.
SSD
(�
�= 1.2 m, �
�=0.15m )
OSD/ISD
(�
�= 1.2 m, �
�=1.2 m )
�>�⇒�=
��
2
4.4
�=
��
2
9.6
�<�⇒�=2�−
4.4
�
�=2�−
9.6
�
VALLEY CURVE
•Length of transition curve for head light sight distance
�=
��
�
(�.�+�.����)
where L is length of valley curve(L>S) in metres
N is the deviation angle in radians(n
1+n
2), with slopes -n
1 and +n
2
S is SSD (m)
�=��−
(�.�+�.����)
�
where L is length of valley curve(L<S) in metres
N is the deviation angle in radians(n
1+n
2), with slopes -n
1 and +n
2
S is SSD (m)
CIVIL ENGINEERING
SANDEEP JYANI
Quality of Irrigation water
3.Proportion of Sodium Ion
concentration
•When percentage of Na ions in the total exchangeable cations
exceed to 10%, aggregation of soil grains breakdown and hence
soil becomes less permeable and of poor tilth
•Proportion of Na ion concentration is generally measured by
SAR(Sodium absorption ratio):
���=
��
+
��
+
2
+��
+
2
2
�����������������������������������������������
���������������������=
������������������
��������������ℎ�
436436
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
����/��
����/��
����/��
����/��
437437
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
����/��
����/��
����/��
����/��
Depth of water held in root zone
For Ease in calculation, water present in voids of the soil needs to be
expressed as depth of water
•Let , root zone depth = ‘�’ m
•Specific wt. of dry soil = �
d
•Cross-sectional area of the soil considered = A
•Equivalent depth of water present in voids of the soil = �m
�
�=
������������������������������
��������������������
⇒�
�=
���
w
���
d
⇒�=
�
d
�
w
���
•If is the depth of water stored in the root zone for
full field capacity but, this entire depth of water
cannot be extracted by the plants, hence available
moisture content will be given as;
�=
�
d
�
w
×�×(��−Ø)
•Equivalent depth of water readily available,
�=
�
d
�
w
×�×(��−M
0)
Que:Thefollowingdatawererecorded
fromanirrigatedfield:
1.Fieldcapacity:20%
2.Permanentwiltingpoint:10%
3.Permissibledepletionofavailable
soilmoisture:50%
4.Dryunitweightofsoil:1500kgf/m
3
5.Effectiverainfall:25mm
Basedonthesedata,thenetirrigation
requirementpermetredepthofsoilwill
be
a)75mm b)125mm
c)50mm d)25mm
�=
�
d
�
w
×
�×(��−Ø)
��=��%Ø=��%
Availablemois�����������
=��−��=��%
ReadilyAvailable
mois�����������
=
��
���
��=�%
�������������
�=
�
d
�
w
×�×(���)
Relationship between Duty (D) and Delta ∆
•Let ‘D’ hectare area is irrigated by supplying 1 m
3
/sec of water for the base period of
‘B’ days such that total water stored in the root zone is ∆ m
•Therefore, volume of water supplied = B Days x 1 m
3
/sec
= B x 24x60x60sec x 1 m
3
/sec
= 86400 B m
3
… (1)
•Amount of water stored = D (ha) x ∆ (m)
= D x 10
4
m
2
x ∆ m
=D x ∆ 10
4
m
3
…..(2)
Equating (1) and (2),
D x ∆ 10
4
m
3
= 86400 B m
3
=> D x ∆ = 8.64 B
D =ha/cumec∆ = mB= days
443443
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
(Δ)
444444
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
(Δ)
=> ∆ = 8.64 B/D
=> ∆ = 8.64 (120)/1400
=> ∆ = 0.740m
Important Definition
10. Irrigation efficiency;
•In general,
irrigation efficiency =
Wateravailableforuse
waterapplied
a)Water conveyance efficiency (n
c)
N
C=
�
�
�
�
×100
where,
w
ris water released from river reservoir for the field.
w
fis water given to the field
This efficiency account for water which possess seepage loss & evaporating
loss during conveyance from river (or) reservoir to the field.
Important Definition
10. Irrigation efficiency;
b)Water application efficiency (n
a)
N
a =
ws
w
f
×100
where,
w
sis the water stored in the field i.e, in the root zone
w
fis the water given to the field
This efficiency accounts for water losses such as surface run off
(R
f) & Deep Percolation (D
f) which occurs during application of
water in the field.
W
f= W
S+ R
f+ D
F
Important Definition
10. Irrigation efficiency;
c)Water use efficiency (n
u)
N
u =
W
u
w
f
×100
where,
w
fis the water given to the field
w
uwater use consumptively
w
u= leaching requirement + presowingrequirement
+ water stored
Important Definition
10. Irrigation efficiency;
d)Water storage efficiency (n
s)
N
s=
ws
wn
×100
where,
w
sis the water stored in the field, i.e, in the root zone
w
nis the amount of water to be stored in the field such that
moisture content is raised to field capacity (F
C)
w
nis the total water required up to F
c i.e., available moisture
content before irrigation.
IRRIGATION REQUIREMENTS OF CROP
•NET IRRIGATION REQUIREMENT
➢It is the amount of water required to be delivered at the field to meet CIR, i.e.
evapotranspiration as well as water required for presowingand leaching.
➢It is denoted by NIR
NIR = CIR + PRESOWING REQUIREMENT + LEACHING REQUIREMENT
IRRIGATION REQUIREMENTS OF CROP
•FIELD IRRIGATION REQUIREMENT
➢It is the amount of water required by the crop in the field plus
amount of water lost in application.
➢It is denoted by FIR
���=
NetIrrigationRequirement
Waterapplicationefficiency
���=
���
�
�
IRRIGATION REQUIREMENTS OF CROP
•GROSS IRRIGATION REQUIREMENT
➢It is the amount of water to be released in canals.
➢It is denoted by GIR
GIR =
FieldIrrigationrequirement
Waterconveyanceefficiency
GIR =
FIR
n
c
KENNEDY’S THEORY
Kennedy introduced the term critical
velocity (V
o) which will keep channel
free from silting & scouring
�
�=�.����
�.��
V
o→ critical velocity in m/sec
y → depth of flow in m
m → critical velocity ratio, whose
value will depend on type of soil
m > I → for coarse soil (1-1.2)
m < I → for fine soil (0.7-1)
Design Steps of Kennedy Theory
Step 1: Assume a trial depth of flow for the required discharge
Guidelines to assume trial depth of flow for the required discharge
Step 2: Determine the critical velocity �
�=�.����
�.��
��
�
/����−����−����−����−���>���
�(�) �����.�����.�−��
V
o→ critical velocity in m/sec
y → depth of flow in m
m → critical velocity ratio, whose value will
depend on type of soil
m > I → for coarse soil (1-1.2)
m < I → for fine soil (0.7-1)
Design Steps of Kennedy Theory
Step 3: Calculate the area using �=
�
�0
…(1)
Assuming channel to be trapezoidal channel and side slopes of as
1
2
����������to 1��������
Using 1���2,calculate �
�
�/2
�
�/2
�=��+2×
1
2
×
�
2
�
�=��+
�
2
2
…(2)
����������������=�+2�
2
+(�/2)
2
�=�+��
Step 4:Calculate Hydraulic Radius R
Step 5:Calculate Actual Mean Flow Velocity by using either Manning’s Equation
or Chezy’sEquation
a)Manning’s Equation �=
1
�
�
2/3
�
1/2
b)Chezy’sformula with Kutter’sEquation
�=���
Chezy’scoefficient or ��������������������������=
23+
0.00155
�
+
1
�
1+23+
0.00155
�
�
�
�=������
′
�����������������
Both of these equations can be used but Kennedy Has recommended use of
Kutter’sEquation
Design Steps of Kennedy Theory
�=
�
�
DESIGN STEPS FOR LACEY’S THEORY
For a given discharge, Q mean particle size
Step 1:Determine silt factor, f = 1.76�
��
Step 2:Determine Mean Flow velocity�=
��
�
���
�/�
Step 3: Find Hydraulic Radius using �=
�
�
�
�
�
Step 3: Determine Area of Cross Section �=
�
�
•Assuming, side slope
�
�
�∶�����=
�
�
�
Step 4: Determine Wetted Perimeter
�=�.���(����
�
/���)
Step 5: Determine Bed slope;
�=
�
����
×
�
�/�
�
�/�
�
�/2
�
�/2
457457
CIVIL ENGINEERING
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ALL FORMULAE OF CIVIL ENGINEERING
458458
CIVIL ENGINEERING
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ALL FORMULAE OF CIVIL ENGINEERING
f = 1.76�
��
f = 1.76�.��
f = 0.7
CIVIL ENGINEERING
SANDEEP JYANI
Hooke’s Law
•When a material is loaded within
elastic limit, the stress is
proportional to the strain produced
by the stress uptoProportional
Limit.
Or
•Ratio of the Stress to the
corresponding strain is constant
uptoProportional Limit.
•E= Young’s Modulus of Elasticity or
Modulus of Elasticity
•E
steel= 200 GPaE
rubber=50 Gpa
•E
rigid= infinte
??????∝�
??????=��
�=
??????
??????
??????
Stress
Strain �
Proportional Limit.
Elastic Limit.
△�=
��
��
Application of Hooke’s Law
2)Elongation of Prismatic bar due to Self Weight
�
�
���
�������������
(��/������)=�
�
??????�=
�
�−�×��
��
…(�)
��
�
�−�
�
�−�=�
�−�
����������
⇒�
�−�=���…(�)
??????�=
�����
��
⇒??????�=
����
�
��������������������������,
∆�
������=
��
�
��
∆�
������=න
�
�
??????�=න
�
�
����
�
Application of Hooke’s Law
3.Elongation of Tapered Bar �
�
��
�
∆�=
���
��
��
��
Application of Hooke’s Law
Que. What is the value of percentage elongation error in Tapered bar?
%�����������������= (∆L)
Tapered–(∆L)
Prismatic
By using mean dia of Tapered bar, make a prismatic bar of dia equal to mean dia
�
�+�∆�=
4PL
�D
1D
2�
�−�
∆�=
PL
��
D
(∆�)
Tapered
=
4PL
�(D+a)(D−a)�
4PL
�(D+a)(D−a)�
−
4PL
??????�
�
�
={
���
D
����
}
2
% of error
X 100
X 100
�
����=
�+�+�−�
�
=�
Application of Hooke’s Law
4)Elongation of Conical bar due to Self Weight
P
A
L
Elongation due to self weight of a conical bar in
terms of weight
Weight Density (Wt/Volume)= ʎ
∆�=
�
�
×
��
�
��
dx
x
IMPORTANT RELATIONS
E=���+�
E=���−��
(E, G, �, K)
�=
���
��+�
For metals
E>�>�
�=
��−��
��+��
Moment of Inertia of Some Important Sections
1.Rectangle
I
xx=
��
�
��
I
yy=
��
�
��
2.Circle
I
xx= I
yy=
�
��
�
�
x x
y
y
Moment of Inertia of Some Important Sections
3.Concentric Circles
I
xx= I
yy=
�
��
(�
�
�
−�
�
�
)
4.Triangle
I
CG=
�
��
��
�
I
base=
�
��
��
�
Section Modulus (z)
•It is the ratio of Moment of Inertia about the neutral axis to y
max
(fiber which is at maximum distance from Neutral Axis) i.e.
�=
�
��
�
���
•Section Modulus represents Bending strength of the Section
•Greater the value of z, greater the bending strength.
•The value of z depends upon Moment of Inertia and Distribution of
Area
Power
▪Power is rate of doing work
▪Work = force x displacement (linear)
▪Work= torque x angular displacement
▪ =T x ��
▪�=
����
����
= T x
�??????
��
▪�=
����
����
= T x ??????
•�=
����
����
= T x
��
��
•�=
����
����
= T x �
•�=
���(���)
��
•�=
����
��
Watt
•N=rpm, T=N-m
Resilience
U = Work done by IRF
= Work done by force P
U=(
�
�
)x ∆L
1.Resilience:
U=
�
�
�x
��
��
U=
�
�
�
���
U=
�
�
×
�
�
×
∆�
�
��
U=
�
�
������������������
U=
�
�
×??????×�×�
U=
�
�
×??????×
??????
�
�
U=
??????
�
��
�
3.Modulus of Resilience
•It is defined as Proof Resilience of
a material per unit volume
��=
���������������
������
��=
�������������������������
������
��=
??????
�
������������
��
��=
??????
�
��
�
�
Stress
Strain
??????
1
??????
2
??????
ELor ??????
3
�
��
��
��
Strain Energy of a Prismatic Bar due to Self Weight
�=
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W
A
L dx
Weight Density (Wt/Volume)= �
x
Thermal Stresses
Case 1: Bar is free to Expand
•∆�= Elongation on temperature
variation
•??????=Thermal Coefficient of
expansion i.e.
E, A, L
∆� �=
∆�
�
∆�
�=
�
�
∆�
�
�=�∆�
Elongation on temperature variation
∆�= �∆��
Thermal Stresses in free expansion are
zero
Thermal Stresses
Case 2: Completely Prevented Case
E, A, L, �
−��
��
=�∆��
Elongation due to temp variation = Elongation due to reaction
Fixed End
Fixed End
E, A, L
∆�
R
=>
??????
�������
�
=−�∆�
=>??????
�������=−�∆��
Thermal Stresses
Case 3: Partially Prevented Case
E, A, L, �
�∆��−�
�
=�
��
Fixed End
Fixed End
E, A, L
∆�
R
=>
??????
�������
�
=
�∆��−�
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�
N A
E F
yഥ�
D
B
Shaded Area A
Shear Stress Distribution
Let us consider a section x-x on
which shear force is F
Shear Stress on a fiber EF which
is located at a distance y from
the NA is given by
�=
��ഥ�
��
Important Relations
D
N A
�
�
B
AN
�
���= 1.5 �
�������
�
�
D/2
�
���= 1.5 �
�������
Important Relations
D
N A
�
�
AN
�
���= (9/8) �
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�
���= (4/3) �
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h
a
a
�
�
1.
2.
B
W
y
max
�
A �
���=
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B
y
B
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A �
���=
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SlopeDeflection
w N/m
3.
4.
B
y
max
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A �
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B
y
B
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A �
���=
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SlopeDeflection
M
w N/m
5.
6.
y
max�
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SlopeDeflection
W
�/2 �/2
BA
C
y
max�
�/2
�/2
BA
Cw N/m
9.
10.
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M
BA
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11.
12.
A
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A �
���B
W
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4.Super Position Theorem
13.
A
�
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⇒�
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⇒�
���=
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w N/m
Effective Length as Per End Conditions
1.Both End Hinged 3.One End Fixed and
other is free
2.Both End Fixed
4.One End Fixed and other is
Hinged
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Slenderness Ratio
�����������������=
���������������
����������������
1.Short/Stocky Column: Those columns have
slenderness less than 32are called short or
stocky struts.
2.Medium Column: Columns having
slenderness ratio between 32 and 120 are
known as Medium column or intermediate
column
3.Long Column: Columns having slenderness
ratio more than 120 are called long columns
Principle Stresses and Strains
•Principle stresses are maximum or minimum
normal stress which may be developed on a body
subjected to load
•On any plane, there are two kinds of stresses
a)Normal Stresses: The forces that are normal to the
plane are normal stresses and is denoted by ??????
b)Shear Stresses: The forces which are along to the
plane are responsible for shear
•Principle Plane or Principle Stresses
•It is plane on which shear stress is zeroand having only
normal stress is called as Principle Stress and denoted by
??????
Methods of Determining Stresses on Oblique Section
Analytical Method
Graphical Method
•At the maximum shear stress plane, maximum shear
stress is
•And the normal stresses ??????
�
′
and ??????
�
′
will be same in
magnitude and is equal to average of principle stresses
�
���=
??????
�−??????
�
�
??????
�
′
=
??????
�+??????
�
�
+
??????
�−??????
�
�
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??????
�
′
=
??????
�+??????
�
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−
??????
�−??????
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′
=−
??????
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??????
�
??????
�′
??????
�′
�
��′
•Case 2: The normal Stress ??????
����??????
�and shear
stress ??????
��are given on two mutually
perpendicular planes and it is required to find
out normal and shear stress on any plane which
is angle �from the vertical
�
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�
??????
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�
�
�
??????
�
??????
�′
??????
�′
�
��′
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Normal Stress on the plane AC
Shear Stress on the plane AC
??????
�
′
=
??????
�+??????
�
�
+
??????
�−??????
�
�
�����+�
�������
??????
�
′
=
??????
�+??????
�
�
−
??????
�−??????
�
�
�����−�
�������
�
��
′
=−
??????
�−??????
�
�
�����+�
�������
•Case 2: The normal Stress ??????
����??????
�and shear
stress ??????
��are given on two mutually
perpendicular planes and it is required to find
out normal and shear stress on any plane which
is angle �from the vertical
�
�
�
??????
�
�
�
�
??????
�
??????
�′
??????
�′
�
��′
�
��
�
��
Normal Stress on the plane AC
Shear Stress on the plane AC
??????
�
′
=
??????
�+??????
�
�
+
??????
�−??????
�
�
�����+�
�������
??????
�
′
=
??????
�+??????
�
�
−
??????
�−??????
�
�
�����−�
�������
�
��
′
=−
??????
�−??????
�
�
�����+�
�������
For AC plane to become principle plane,
�
��
′
=�
�=−
??????�−??????�
�
�����+�
�������
⇒�����
�=
�
��
??????
�−??????
�
�
LOCATION OF PRINCIPLE PLANE
•For Maximum Shear Stress Plane
�
���=
??????
�−??????
�
�
where ??????
����??????
���������������������
Case 2: The normal Stress ??????
����??????
�and shear stress ??????
��are given on two mutually perpendicular
planes and it is required to find out normal and shear stress on any plane which is angle �from the
vertical
??????
�=??????
�
′
=
??????
�+??????
�
�
+
??????
�−??????
�
�
2
+�
��
2
??????
�=??????
�
′
=
??????
�+??????
�
�
−
??????
�−??????
�
�
2
+�
��
2
⇒�
���=
??????�+??????�
�
+
??????�−??????�
�
2
+�
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2
−
??????�+??????�
�
−
??????�−??????�
�
2
+�
��
2
�
⇒�
���=
??????
�−??????
�
�
2
+�
��
2 �
���=
??????�−??????�
�
MOHR CIRCLE
Normal Stress on the plane AC
??????
�
′
=
??????
�+??????
�
�
+
??????
�−??????
�
�
�����+�
�������…(�)
�
��
′
=−
??????
�−??????
�
�
�����+�
�������…(�)
Shear Stress on the plane AC
Using equation 1,
??????
�
′
−
??????
�+??????
�
�
=
??????
�−??????
�
�
�����+�
�������…(�)
Using equation 2,
�
��
′
−�=−
??????
�−??????
�
�
�����+�
�������…(�)
Squaring and then adding equation 3 and 4 i.e�
�
+�
�
??????
�−
??????
�+??????
�
�
�
′
+�
��
′
−�
�
=
??????
�−??????
�
�
�
+�
��
�
…(�)
**Centre of Mohr Circle ⇒
??????
�+??????
�
�
,�
**Radius of Mohr Circle ⇒�=�
���=
??????
�−??????
�
�
2
+�
��
2
??????
�−
??????
�+??????
�
�
�
′
+�
��
′
−�
�
=
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�−??????
�
�
�
+�
��
�
…(�)
MOHR CIRCLE
In static fluid,
�
��=������=��=−�(����������������������)
⇒�=�
���=
�−�
�
2
+�
��
2
=0
Therefore Mohr Circle for Static Fluid is a Point Circle
•Guidelines to draw Mohr Circle
1.Mark the point A and B on the
face AB and BC and denote
the coordinates
2.Draw the x and y axis
respectively representing
Normal stress and Shear
stress
3.Mark the coordinate A and B
on the x-axis and y axis
4.Join the point A and B by the
line, and bisect it and draw
the circle
MOHR CIRCLE
�
�
�
??????
�
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??????
�
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��
�
��
(??????
�,�
��)
(??????
�,−�
��)
�(??????
�,�
��)
�(??????
�,−�
��)
Normal Stress
Shear Stress
�
��
(??????
�,�)(??????
�,�)
��
�,�������������������
Pressure vessels
•Due to internal pressure, three
types of stress may be
developed:
1.Hoop Stress or Circumferential
Stress ??????
�
2.Longitudinal Stress ??????
�
3.Radial Stress ??????
�
Analysis of Hoop Strain and Longitudinal Stress for
Cylindrical Vessel
1.The longitudinal stress is given by
??????
�=
��
��
2.Hoop Stress is given by
??????
�=
��
��
�=����������������
�=����������������
�=�ℎ�������
Analysis of Hoop Strain and Longitudinal Strain for Cylindrical Vessel
1.The longitudinal Strain is given by
�
�=
??????
�
�
−�
??????
�
�
⇒�
�=
��
���
−�
��
���
⇒�
�=
��
���
(�−��)
2.Hoop Strain is given by
�
�=
??????
�
�
−�
??????
�
�
⇒�
�=
��
���
−�
��
���
⇒�
�=
��
���
(�−�)
�=����������������
�=����������������
�=�ℎ�������
??????
�
??????
�
�
�
�
�
=
�−��
�−�
Stress and Strain In Sphere
In Sphere, Longitudinal and
Circumference direction are
interchangeable therefore longitudinal
and Hoop strain are the same
??????
�=??????
�=
��
4�
�
�=�
�=
��
4��
−??????
��
4��
⇒�
�=
��
���
(�−�)
??????
�=
��
4�
??????
�=
��
4�
1.Direct Vernier
•�=�������������������������������������
•�=�����������������������������������
•�=�������������������������
•�����ℎ���−1������������������������������������������������
�.�.��=�−1�
⇒�=
�−1
�
�
LeastCount=�−�
=�−
�−1
�
�
LeastCount=
�
�
Least Count is ‘one main scale division’ divided by total number of division on the
Vernier
Vernier Scale
Vernier is of two types
2.Retrograde Vernier
•In Retrograde Vernier smallest
division on Vernier (�)is longer
than the smallest division (�)on
Main Scale
•It extends in the opposite
direction as that of main scale
•It is constructed such that
�+1divisions on main scale
are equal in length of �divisions
of the Vernier
Vernier Scale
2.Retrograde Vernier
•�=�������������������������������������
•�=�����������������������������������
•�=�������������������������
•�����ℎ��(
)
�+
1������������������������������������������������
�.�.��=�+1�
⇒�=
�+1
�
�
LeastCount=�−�
=
�+1
�
�−�
LeastCount=
�
�
Least Count is ‘one main scale division’ divided by total number of
division on the Vernier
Vernier Scale
519
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Find number of divisions on the
vernier scale to be designed for a
theodolite circle which is divided
into degrees and one third degrees
to read 20”(LC).
a)20
b)30
c)50
d)60
520
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Find number of divisions on the
vernier scale to be designed for a
theodolite circle which is divided
into degrees and one third degrees
to read 20”(LC).
a)20
b)30
c)50
d)60
��=
�
�
�=
�
�
°=��′
��=��"=
��
��
���
��, �=
�
��
⇒�=
��′
��/��′
=�����������
521
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
The value of smallest division
of circle of theodolite is 10’.
Find number of divisions on
the vernier scale.
a)20
b)30
c)50
d)60
522
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
The value of smallest division
of circle of theodolite is 10’.
Find number of divisions on
the vernier scale.
a)20
b)30
c)50
d)60
��=
�
�
�=��′
��=��"=��/�����
⇒
��
��
=
��
�
⇒�=�����������
523
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
•�������������=
��������������
����������������
��������������
•�����������=
��������������
����������������
�
��������������
524
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
A surveyor measures a distance between
two points on a map of representative
fraction of 1:100 is 60 m. But later he found
that the used wrong representative
fraction of 1:50. What Is the correct
distance between the two points
(a)30
(b)45
(c)90
(d)120
525
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
A surveyor measures a distance between
two points on a map of representative
fraction of 1:100 is 60 m. But later he found
that the used wrong representative
fraction of 1:50. What Is the correct
distance between the two points
(a)30
(b)45
(c)90
(d)120
�������������=
��������������
����������������
��������������
⇒�������������=
�/��
�/���
��
⇒�������������=����
526
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
A Distance was measured between two
points on the plan drawn to scale
1cm=40mand the result was 468m. But it
was found that the surveyor used wrong
scale 1cm=20m. The true distance
between the points is
a)486 m
b)936 m
c)234 m
d)None of these
527
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
A Distance was measured between two
points on the plan drawn to scale
1cm=40mand the result was 468m. But it
was found that the surveyor used wrong
scale 1cm=20m. The true distance
between the points is
a)486 m
b)936 m
c)234 m
d)None of these
�������������=
��������������
����������������
��������������
⇒�������������=
�/����
�/����
���
⇒�������������=����
Shrunk Scale
•If graphical scale is not drawn on the plan and
the sheet on which the plan is drawn shrinks due
to variations in the atmospheric conditions, it
becomes essential to find the shrunk scale of the
plan.
�ℎ���������=�ℎ�������������×�������������
�ℎ����������������ℎ������������=
�ℎ���������ℎ
�����������ℎ
Probable error of mean of a number of observations
•The probable error of mean of a number of
observations of the same quantity is given
by
�
�=
�
�
�
•�
�=����������������������
•�
�=��������������������������������
•�=�������������������������������
530
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
In carrying a survey across a highway, eight
readings were taken with a level under
identical conditions. The probable error of
single observation is 0.01295m. Find the
probable error of the mean.
531
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
In carrying a survey across a highway, eight
readings were taken with a level under
identical conditions. The probable error of
single observation is 0.01295m. Find the
probable error of the mean.
�
�=
�
�
�
⇒�
�=
0.01295
�
⇒�
�=�.������
532
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
In carrying a line of levels across a river, following readings were taken with a level under identical
conditions : 2.322 2.346, 2.352, 2.306, 2.312, 2.300, 2.306, 2.326 metres. Calculate the
a)probable error of single observation
b)Probable error of the mean
533
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
In carrying a line of levels across a river, following readings were taken with a level under identical
conditions : 2.322 2.346, 2.352, 2.306, 2.312, 2.300, 2.306, 2.326 metres. Calculate the
a)probable error of single observation
b)Probable error of the mean
Reading (m) ��
�
2.322
2.346
2.352
2.306
2.312
2.3
2.306
2.326
Mean = σ�
�
=
Probable error of single observation is calculated as
�
�=±�.����
σ�
�
(�−�)
The probable error of mean of a number of observations
of the same quantity is given by
�
�=±�.����
σ�
�
�(�−�)
=
�
�
�
534
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
In carrying a line of levels across a river, following readings were taken with a level under
identical conditions : 2.322 2.346, 2.352, 2.306, 2.312, 2.300, 2.306, 2.326 metres. Calculate
the
a)probable error of single observation
b)Probable error of the mean
Reading (m) ��
�
2.3220.001 0.000001
2.3460.025 0.000625
2.3520.031 0.000961
2.3060.015 0.000225
2.3120.009 0.000081
2.30.021 0.000441
2.3060.015 0.000225
2.3260.005 0.000025
Mean = σ�
�
=0.002584
Probable error of single observation is calculated
as
�
�=±�.����
σ�
�
(�−�)
⇒�
�=±�.����
�.������
(�−�)
⇒�
�=±�.������
The probable error of mean of a number of
observations of the same quantity is given by
�
�=±�.����
σ�
�
�(�−�)
=
�
�
�
=±
�.�����
�
=±�.������
535
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
1.Correction for standardization: (C
a)
•It is also called as correction for Absolute length.
•If Actual length of the tape or not equal to Nominal length of
the tape, then correction for standardization is required.
•If tape is actually shorterthan Nominal length then error will
be positive because measured distance will be greater than
correct distance
•Therefore correctionwill be negative
•Nominal Designated length → l’ (20 m or 30 m)
•Actual / Absolute length of tape → l (19 m)
Measured distance = L’
Correct distance = L
Note: if Tape is shorter then correction is subtractive
536
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
1.Correction for standardization: (C
a)
•Nominal Designated length → l’ (20 m or 30 m)
•Actual / Absolute length of tape → l (19 m)
•Measured distance = L’
•Correct distance = L
���������������������������������=��������������������×
���������������������������������
�������������������������������
������=��×
��
��
���������������������=(������������)×
���������������������������������
�������������������������������
�
�����������������������=(��������������)×
���������������������������������
�������������������������������
�
537
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
The length of a line measured with a
20m chain was found to be 300m.
Calculate the true length of the line if
the chain was 10 cm too long
⇒����������������=���×
��.�
��
=���.��
���������������������������������=��������������������×
���������������������������������
�������������������������������
538
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
A line of True length 500 m when
measured by a 20 m Tape was
reported to be 502 m, then Actual
length of the Tape is ------?
500=�.�×
502
20
⇒�.�=19.92�.
���������������������������������=��������������������×
���������������������������������
�������������������������������
539
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Length of a line measured with a chain was found to be 250 m. Determine True length of the
line if;
a)Length was measured with a 30 chain and chain was 10 cm too long.
b)Length of the chain was 30 m in the beginning and 30.10 m at the end of the work.
a) �.�=��.��×
���
��
���������������������������=������������×
����������������
�������������
=���.����
b)�.�=��.��×
���
��
=���.���m
540
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
A 20m chain was found to be 10cm too long after chaining a distance of 1500m. It was
found to be 18cm too long at the end of day’s work after chaining a total distance of
2900m. Find the true distance if the chain was correct before the commencement of
work
541
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
2.Correction for slope (C
g)
•We know that slope distance will always be greater
than corresponding horizontal distance, therefore
correction for slope will always be negative
•Case : 1 When slope of the ground �is measured
�
�=�.�–�.�
=L cos�–L
�
�=−�(1−����)
L
�
h
L����
542542
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
2.Correction for slope (C
g)
•We know that slope distance will always be greater
than corresponding horizontal distance, therefore
correction for slope will always be negative
•Case : 1 When slope of the ground �is measured
�
�=�.�–�.�
=L cos�–L
�
�=−�(1−����)
L
�
h
L����
Civil Engineering by Sandeep Jyani
543543
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
2.Correction for slope (C
g)
Case II:-When height difference ‘h’ is measured.
�
�=�
�
−�
�
−�
�
�=��−
�
�
�
�
�
�
−�
�
�=��−
�
�
��
�
−
�
�
��
�
+⋯−�
�
�=
−�²
��
−
�
�
��
�
�
�=
−�²
��
L
�
h
�
2
−ℎ
2
Civil Engineering by Sandeep Jyani
544544
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 35 If downhill end of slope of a 20m tape is held 80cm too
low, then determine the correction for slope.
Sol.
Civil Engineering by Sandeep Jyani
545545
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 35 If downhill end of slope of a 20m tape is held 80cm too
low, then determine the correction for slope.
Sol. �
�=
−�²
��
�
�=
−�.�²
�(��)
= -0.016m
NOTE : SLOPE CORRECTION FOR 3IS ALWAYS NEGLECTED
Civil Engineering by Sandeep Jyani
546546
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
3. Correction for pull
�
�=
�−���
��
where
�
�→ standard pull
�→ Pull applied in the field
�→ Cross-sectional Area of the Tape
�→ Young’s modulus of Elasticity
�→ Nominal length of the tape/measured length of the line
If P > �
�→ then correction for pull will be positive and vice-versa.
Civil Engineering by Sandeep Jyani
547547
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 36 A steel tape 30 m long was standardized under a pull of 65 N. If pull at the time of
measurement was 80 N. Determine Correct Tape length if wt. of the tape is 10 N, young’s
modulus,
�=���
�
�/��²;�=��.����/�³
Civil Engineering by Sandeep Jyani
548548
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 36 A steel tape 30 m long was standardized under a pull of 65 N. If pull at the time of
measurement was 80 N. Determine Correct Tape length if wt. of the tape is 10 N, young’s
modulus,
�=���
�
�/��²;�=��.����/�³
Sol: �=���
⇒��=��.��×
��³
��
�
×�×��×��³
⇒�=�.����²
⇒�
�=
�−���
��
⇒�
�=
(��−��)×��
�.�����
�
=+�.���
−�
�
Correct length of Tape=��+�.����
−�
�
Civil Engineering by Sandeep Jyani
549549
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
4. Correction for temperature C
t
�
�=��−�
��
�→ thermal coefficient of linear expansion
�
�→ standard temperature
�→ field temperature
�→ Nominal length of the Tape/measured
length of the line
If t > t
o→ then �
�will be positive
Civil Engineering by Sandeep Jyani
550550
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 37 Determine correction for Temperature if measured
length of the lines is 1000 m, �=�.���
−�
/°C, standard
Temp is 27°C and field temp is 32°C
���. �
�=�(t-t
o) L
C
t=�.���
−�
(��−��)×����
=+�.����
Civil Engineering by Sandeep Jyani
551551
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
5.Correction for sagC
s
•When the Tape is supported between two ends, it takes the shape of
catenary.
•If the Tape is standardized on Flat and used in catenary then correction for
sag will be ��������, because chord length will always be less than Arc
length.
•For determination of correction for sag, the shape is assumed to be
parabolic, instead of catenary.
�
�=
−(��)²�
���²
=
−�²�
���²
Where, w → wt. per unit length of the Tape
l → length of the Tape suspended between two supports
W → Total wt. of the Tape
P → Pull applied at the ends.
P(force)P
552552
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
5.Correction for sagC
s
•To reduce sag correction we can increase no of supports.
•If Total length of the Tape suspended is ‘l’ which is
supported between ‘n’ no of bays
•�
�=−
�
�
�
���
�
�²
=
−(��)²�
���
�
�²
Note:-
•If Tape is standardized in catenary and used on flat then
correction for Sag will be positive
•Normal Tension (�
�) is that theoretical pull at which sag
correction and pull correction cancel outs each other–
(�
�−�
�)�
��
=
�
�
�
���
�
�
553553
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 38 Determine sag correction for a 30 m steel Tape under a
pull of 80 N. in 3 bays of 10 m each, cross-sectional Area of the
Tape is 8 mm² and unit wtof the steel may be taken as 77
KN/m³
Sol.
554554
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 38 Determine sag correction for a 30 m steel Tape under a pull
of 80 N. in 3 bays of 10 m each, cross-sectional Area of the Tape is 8
mm² and unit wtof the steel may be taken as 77 KN/m³
Sol. �
�=
−�²�
���
�
�²
�=���
=77×
10³
10
9×8×30×10
3
=18.48�
�
�=
−18.48²×30
24×3²×80²
=−7.411×10
−3
�.
555555
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 39. Determine Normal tension for a steel tape supported
between two supports 10 m apart, if the standard Tension is 65
N and wt. of the Tape is 0.62 N/m, �=2×10
5
and corss-
sectional Area of the Tape is 8 mm²
Sol:
556556
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 39. Determine Normal tension for a steel tape supported
between two supports 10 m apart, if the standard Tension is 65
N and wt. of the Tape is 0.62 N/m, �=2×10
5
and corss-
sectional Area of the Tape is 8 mm²
Sol:
(��−��)�
��
=
�
2
�
24�
�
2
⇒
(�
�−65)
8×2×10
5
=
(0.62×10)²
24�
�
2
=> P
n=162.29�
Civil Engineering by Sandeep Jyani
557557
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
5.Correction for MSL C
MSL
�
���=�
���−�
�
���=��−�+��
�
���=−��
�
���=
−��
�+�
Since, h is very small in comparison to R, therefore we can
neglect it,
�
���=
−��
�
L
L
msl
R
h
Civil Engineering by Sandeep Jyani
558558
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 40 Determine correction for MSL if measured length of the
line is 1000 m and Average elevation of the line from MSL is 300 m.
if R = 6370 km.
Sol:
Civil Engineering by Sandeep Jyani
559559
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 40 Determine correction for MSL if measured length of the
line is 1000 m and Average elevation of the line from MSL is 300 m.
if R = 6370 km.
Sol: �
���=
−ℎ�
�
=
−300×1000
6370×10
3
=−0.047�
Civil Engineering by Sandeep Jyani
560560
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
6. Correction for Misalignment C
m
�
�=�
1
2
−ℎ
2
+�
2
2
−ℎ
2
−�
1+�
2
Error due to misalignment is always
positive.
�
� �
�
h
Civil Engineering by Sandeep Jyani
561561
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 41. Calculate the sag correction for a 30m steel chain under a pull
of 100N in three equal spans of 10m each. Weight of one cubic cm of
steel=0.078 N. Area of cross section of tape=0.08 sqcm.
562562
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
•Fore Bearing :
•Fore Bearing of a line is Horizontal angle in
the direction of progress of survey.
•Back Bearing:
•Back Bearing of a line is the horizontal angle
in the direction opposite of the progress of
survey.
��=��±���°
Positive → FB → less than 180°
Negative → FB →greater than 180°
Civil Engineering by Sandeep Jyani
563563
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
•E = M.V –T.V (M.V > T.V)
•Error due to curvature is always positive.
•So correction will be negative.
•CORRECTION DUE TO CURVATURE is given by
C
C= -0.0785d
2
d = distance between instrument and staff
station (in km)
C
c= in m
Assumed Horizontal
Line of collimation
Level Line
���������
����������
•Effect of curvature occurs because the difference between level line and
horizontal line increases as the distance of the staff station from the instrument
station is increased.
•Level line is a curved line but line of collimation is a horizontal line which is
tangential to the level line.
•Due to curvature of Earth points appear to be lower than actually they are.
C
c=
�
�
��
−��⇒
R=6370 km
564564
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
•E = M.V –T.V (M.V < T.V)
•Error due to refraction is always negative.
•So correction will be positive.
•CORRECTION DUE TO REFRACTION is given by
C
r=
�
�
��
C
r= 0.0112d
2
d = distance between instrument and staff
station (in km)
C
r= in m
Assumed Horizontal
Line of collimation
Refracted Ray
����������
����������
•Density of air in the atmosphere decreases with the increase in altitude as air is
denser near the earth surface.
•A ray of light travels from thinner medium to denser medium, therefore it bends
towards the normal.
•Hence line of sight does not remains horizontal but it bends towards the centreof
the earth, i.e, downwards.
C
r=
�
�
�
�
��
+��⇒
R=6370 km
565565
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
C = C
c + C
r
C = -0.0673d
2
where
C is in m
d is in km
For a distance less than 250m, combined
correction is neglected in ordinary levelling.
C
r=
�
�
�
�
��
−��⇒
R=6370 km
566566
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
ℎ=0.0673�
2
�=
ℎ
0.0673
�=3.855ℎ
where
h is in m
d is in km
d
h
d
2
h
2
d
1
h
1
d = d
1+d
2= 3.855ℎ
1+3.855ℎ
2
Observer Light
house
567567
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 92. An observer standing on the deck of a ship just sees the
light house which is 40m above the sea level if the height of
observer’s eye is 8m above the sea level, determine the distance
of the observer from the light house.
568568
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 92. An observer standing on the deck of a ship just sees the
light house which is 40m above the sea level if the height of
observer’s eye is 8m above the sea level, determine the distance
of the observer from the light house.
Sol. d
1+d
2= 3.855 ( 8+ 40) = 35.28km
d
2
h
2
d
1
h
1
Observer
569569
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
107.InaclosedtraverseABC,following
readingsweretaken
StationAisfreefromlocalattraction.
CorrectbearingofCBis
a)275°
b)276°
c)281°
d)280°
570570
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
107.InaclosedtraverseABC,following
readingsweretaken
StationAisfreefromlocalattraction.
CorrectbearingofCBis
a)275°
b)276°
c)281°
d)280°
571571
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
108.Whatis∠ABCifFBoflineAB
is40°andBBoflineBCis280°?
a)90°
b)120°
c)240°
d)320°
572572
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
108.Whatis∠ABCifFBoflineAB
is40°andBBoflineBCis280°?
a)90°
b)120°
c)240°
d)320°
573573
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
109.Themagneticbearingofa
lineABisS30°E.Ifthedeclination
is6°West,thenwhatisthetrue
bearing?
a)S36°E
b)N36°W
c)S24°E
d)N24°W
574574
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
109.Themagneticbearingofa
lineABisS30°E.Ifthedeclination
is6°West,thenwhatisthetrue
bearing?
a)S36°E
b)N36°W
c)S24°E
d)N24°W
575575
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
PRINCIPLE OF STADIA METHOD
•�=�
�+�
•⇒����������������,
�
�
=
�
�
�
+
�
�
�
•Multiply by ff
1
•�
�= �+
�
�
�
�
�
•Put values of
�
�
�
�
in
�
�
�
�
=
�
�
•�
�= �+
�
�
�
•=> D = d + f
1
•⇒�=
�
�
(�)+(�+�)
⇒�=��+�
This equation is TACHEOMETRIC DISTANCE EQUATION
�∝�
�=
�
�
�
�
�
Vertical axis of instrument
�
�
�
�
�
��������������������=
�
�
(k=100)
�����������������=�+�(c=0)
Using similar triangle property,
�
�
�
�
=
�
�
576576
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 124 If the intercept on a vertical staff is observed as 0.75m from
a tacheometer, the horizontal distance between tacheometer and staff
station is
a)7.5m
b)25m
c)50m
d)75m
577577
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 124 If the intercept on a vertical staff is observed as 0.75m from a
tacheometer, the horizontal distance between tacheometer and staff
station is
a)7.5m
b)25m
c)50m
d)75m
D=KS + C (C=0)
=> D = 0.75 x 100 = 75m
578578
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 125 For a tacheometer, additive and multiplying constants are
respectively
a)0 and 100
b)100 and 0
c)0 and 0
d)100 and 100
579579
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que. 125 For a tacheometer, additive and multiplying constants are
respectively
a)0 and 100
b)100 and 0
c)0 and 0
d)100 and 100
580580
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
1.Mid ordinate rule
•This method is used with the
assumption that boundaries
between the extremities of
the ordinates (or offsets) are
straight lines.
•The base line is divided into a
number of divisions and the
ordinates are measured at
the mid points of each
divisions, so Area….
�
�
�
�
�
�
�
�
��
� �
�=��
����=���������������������������
⇒����=
�
�+�
�+�
�+⋯+�
�
�
�
⇒����=
�
�+�
�+�
�+⋯+�
�
�
��
⇒����=(�
�+�
�+�
�+⋯+�
�)�
�� �
581581
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
�
�
�
�
�
�
�
�
��
� �
�=��
�
⇒����=
�
�+�
�+�
�+�
�+⋯+�
�
�+�
�
2.Average Ordinate Rule
•This rule also assumes
that the boundaries
between the extremities
of the ordinates are
straight lines
•The offsets are measured
to each of the points of
the divisions of the base
line
�
�
���
582582
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
3.Trapezoidal Rule
•This rule is based on the assumption that the figures
are trapezoids. The rule is more accurate than the
previous two rules which are approximate versions of
the trapezoidal rule
•Area of the first trapezoid
�
�=
�
�+�
�
�
�
����������
�=
�
�+�
�
�
��������..
⇒�
�=
�
�−�+�
�
�
�
����������=�
�+�
�+�
�+⋯+�
�
=
�
�+�
�
�
�+
�
�+�
�
�
�+⋯+
�
�−�
+
�
�
�
�
=
�
�+�
�
�
+�
�+�
�+�
�+⋯+�
�−��
��
�
��
��
�
�
�
�
�
���
�=��
583583
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
4.Simpson’s One Third Rule
•This rule assumes that short lengths of boundary between the
ordinates are parabolic arcs. This method is more useful when
the boundary line departs considerably from the straight line
•����������=
�
�
+�
�
�
×��…(�)
•�����������������������=
�
�
������������������������������
⇒�������=
�
�
��������
=
�
�
����
=
�
�
�
�−
�
�+�
�
�
��
��
�
�
�
�
�
�
��
�
�
�
�
�
�
So area of first two intervals=
�
�
+�
�
�
��+
�
�
�
�−
�
�
+�
�
�
�
=
�
�
�
�+��
�+�
�
Similarly for the next two intervals =
�
�
�
�+��
�+�
��������
���������=
�
�
�
�+��+��
�+�
�+⋯+�
�−�+�(�
�+�
�+⋯+�
�−�)
Even ordinates odd ordinates
584584
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Note:
•Simpson’s one third rule is used when the number of Divisions are
even i.e. offsets are odd
•If in question it is given to solve by Simpson’s rule and number of
offsets are even, then
Area=A1+A2
•A1= area by simpson’sone third rule uptosecond last offset
•A2=Area between last two offsets
585585
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Question: 172 The following perpendicular offsets were taken at 10m interval from a survey line to an
irregular boundary line.
3.25, 5.60,4.20,6.65,8.75,6.20,3.25,4.20,5.65
Calculate the area enclosed between the survey line, the irregular boundary line and the first and last
offsets.
586586
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Question: 172 The following perpendicular offsets were taken at
10m interval from a survey line to an irregular boundary line.
3.25, 5.60,4.20,6.65,8.75,6.20,3.25,4.20,5.65
Calculate the area enclosed between the survey line, the
irregular boundary line and the first and last offsets.
Solution:
1.By Average Ordinate Rule:
����=
�
�+�
�+�
�+�
�+⋯+��
�+�
�
⇒����=
3.25+5.60+4.20+6.65+8.75+6.20+3.25+4.20+5.65
�+�
×(��×�)
⇒����=���.���
�
587587
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Question: 172 The following perpendicular offsets were taken at
10m interval from a survey line to an irregular boundary line.
3.25, 5.60,4.20,6.65,8.75,6.20,3.25,4.20,5.65
Calculate the area enclosed between the survey line, the
irregular boundary line and the first and last offsets.
Solution:
2.By Trapezoidal Rule:
⇒����=����
�
����=
�
0+��
2
+�
1+�
2+�
3+⋯+�
�−1�
⇒����=
3.25+5.65
2
+5.60+4.20+6.65+8.75+6.20+3.25+4.20×10
588588
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Question: 172 The following perpendicular offsets were taken at
10m interval from a survey line to an irregular boundary line.
3.25, 5.60,4.20,6.65,8.75,6.20,3.25,4.20,5.65
Calculate the area enclosed between the survey line, the
irregular boundary line and the first and last offsets.
Solution:
3.By Simpson’s Rule:
⇒����=���.���
�
⇒����=
��
�
×
3.25+5.65
2
+45.60+6.65+6.20+4.20+2(4.20+8.75+3.25)
���������=
�
�
�
�+�
�+��
�+�
�+⋯+�
�−�+�(�
�+�
�+⋯+�
�−�)
589589
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 173 : A series of offsets were taken from a chain line to a curved boundary line at intervals of 15 metres in
the following order.
0, 2.65, 3.80, 3.75, 4.65, 3.60, 4.95, 5.85 m .
Compute the area between the chain line, the curved boundary and the end offsets by (a) average ordinate rule,
(b) trapezoidal rule, and (c) Simpson's rule.
590590
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
•Calculation of Volume of various quantities are done for various
purposes
•Planning
•Design
1.Trapezoidal Formula or Average End Area Rule
•�=�
�
�
+�
�
�
+�
�+�
�+⋯+�
�−�
2.PrismoidalRule
•�=
�
�
{�
�+�
�+��
�+�
�+�
�+⋯+�(�
�+�
�+�
�…)}
591591
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
•Calculation of Volume of various quantities are done for various
purposes
•Planning
•Design
1.Trapezoidal Formula or Average End Area Rule
•�=�
�
�
+�
�
�
+�
�+�
�+⋯+�
�−�
•This method is based on the assumption that the mid area is the mean
of the end areas
592592
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Question 174:The cross section area of 3 sections of an
embankment at an interval of 40m are 10 m
2
, 15 m
2,
and 35 m
2
.
Calculate the quantity of Earth work for the embankment using
prismoidalmethod.
Solution:
�=
�
3
{�
1+�
�+4�
2+�
4+�
6+⋯+2(�
3+�
5+�
7…)}
⇒�=
40
3
{10+35+415+2(0)}
⇒�=1400�
3
593593
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 175. Calculate the
volume of an embankment in
cubic metre using trapezoidal
method if the cross sectional
area of three sections of an
embankment at an interval of
30m are 20sqm, 40sqm and
50sqm.
a)1100
b)1150
c)2250
d)2350
594594
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 175. Calculate the
volume of an embankment in
cubic metre using trapezoidal
method if the cross sectional
area of three sections of an
embankment at an interval of
30m are 20sqm, 40sqm and
50sqm.
a)1100
b)1150
c)2250
d)2350
595595
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 176. The area included by
contour lines for a proposed
dam are given as
Calculate the capacity in cubic
metre of the dam by trapezoidal
method
a)42 000 000
b)53 000 000
c)70 000 000
d)80 000 000
Contour (m)410 420 430 440 450
Area (ha) 205 120 145 95 135
596596
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 176. The area included by
contour lines for a proposed
dam are given as
Calculate the capacity in cubic
metre of the dam by trapezoidal
method
a)42 000 000
b)53 000 000
c)70 000 000
d)80 000 000
Contour (m)410 420 430 440 450
Area (ha) 205 120 145 95 135
597597
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
•To find the area of a plan, anchor point may be
placed either outside the plan (if area is small) or
inside the plan (if area is large).
•Then on the boundary of the plan a point is marked
and tracer is set on it. The initial reading of the
wheel is taken.
•After this tracer is carefully moved over the outline
of the plan in clockwise direction till the first point
is reached.
•Then the final reading is noted. Now the area of the
plan may be found as
����=�(�–�±���+�)
598598
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
����=�(�–�±���+�)
where
•F=FinalreadingI=Initialreading.
•N= The number of completed revolutions of disc. Plus sign to be used if the
zero mark of the dial passes index mark in clockwise direction and minus sign
if it passes in anticlockwise direction.
•�= a multiplying constant or planimeter constant and is equal to area per
revolution of the roller
•C= Constant of the instrument, which when multiplied with M, gives the area
of zero circle.
•The constant Cis added only when the anchor point is inside the area.
•Multiplying constant M is equal to the area of the plan (map) per revolution of
the roller i.e., area corresponding to one division of disc.
•Multiplying constant Mand Care normally written on the planimeter. The user
can verify these values by
•(i) Measuring a known area (like that of a rectangle) keeping anchor point outside
the area
•(ii) Again measuring a known area by keeping anchor point inside a known area.
The zero circle or the circle of correction is defined as the circle round the circumference of which if the tracing point ismoved, the wheel will simply slide without rotation on paper
without change in reading
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
A= initial reading
B= final reading
N= no. of completed revolutions of wheel
during one complete tracing. N is positive if
dial passes index in clockwise, N is negative if
dial rotates in anti-clock wise direction.
MandC= constants which values are
provided on the planimeter. Constant C is
used only when the anchor point is placed
inside the plan
�=100(4.254−9.918−10×1+23.521)
⇒�=100(4.254−9.918−10×1+23.521)
⇒�=785.7��
2
601601
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que: 180 Calculate the area of a figure from the following readings by a planimeter with
anchor point outside the figure.
Initial reading =7.875, final reading = 3.086 ; M = 10 sq. in.
The zero mark on the that passed the fixed index mark twice in the clockwise direction.
602602
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que: 180 Calculate the area of a figure from the following readings by a planimeter with anchor point outside the
figure.
Initial reading =7.875, final reading = 3.086 ; M = 10 sq. in.
The zero mark on the that passed the fixed index mark twice in the clockwise direction.
����=�(�–�±���+�)
•F=3.086I=7.875.
•N= +2
•�= 10 sq. in
•C= Since anchor point is outside, C is not to be used in the formula
⇒����=�(�–�±���+�)
=��×(�.���–�.���+��×�)
=���.������
603603
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
1.Length of curve (�):
•�=
??????�Δ
180°
(in radians)
2.Tangent Length (T)
•�=�
��=�
��=�tan
∆
�
3.Length of long chord (L) :
•�=�
1�
2=2�.���
Δ
2
4.Mid ordinate �:
•�=�1−cos
Δ
2
5.External distance (E):
•�=����
Δ
2
−1
•���
Δ
2
=
�
�+�
6.Chainages of T
1and T
2
•�ℎ���
1=�ℎ���−�����ℎ�
•�ℎ���
2=�ℎ���
1+�����ℎ�
�
∆
�
�
�
�
�
�
�
�
�
�
� �
�
�
�
∆
�
∆
�
�� ��
�
�
604604
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
���������ℎ��������������������������
180°=��������1������=
180°
�
1°=
�
180°
������
������������������������,����������
�
180°
������������������������,����������
180°
�
605605
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Designation of Curve
•A curve can be designated by radius R (��)Degree of curve (D).
•Degree of curve is the angle subtended by an Arc (��)a chord of
specified length at the centre.
1.Arc Definition:
•Case 1: Let arc length is 30m and radius of curve is R, the n degree of
curve is D
��
�
���
���
���°
=���
=>�=
�����
��
=>�=
����.��
�
∴�=
����
�
Remember
Civil Engineering by Sandeep Jyani
606606
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Designation of Curve
•For 30m �=
����
�
•For 20m�=
����
�
1.Arc Definition:
•Case 2: Let arc length is 20 m and radius of curve is R,
the n degree of curve is D
��
�
���
���
���°
=���
=>�=
�����
��
=>�=
����.��
�
∴�=
����
�
Remember
Civil Engineering by Sandeep Jyani
607607
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Designation of Curve
•For 30m �=
����
�
•For 20m�=
����
�
2.Chord Definition:
•Case I: for 30m chord
��
�
�
���
���
�
�
=
��
�
Since,
�
�
will be a small angle, therefore ����→�
=>
�
�
×
�
���°
=
��
�
=>�=
��×�×���°
��
=
����
�
������
�
�
•Case II: for 20m chord
���
�
�
=
��
�
Since,
�
�
will be a small angle, therefore ����→�
=>
�
�
×
�
���°
=
��
�
=>�=
��×�×���°
��
=
����
�
��
�
�
���
������
�
�
Civil Engineering by Sandeep Jyani
608608
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Note:
•Since Degree of curve is inversely proportional to Radius, for sharp
circles Degree of curve will be large, whereas for flat curve, Degree of
curve will be small.
609609
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 181: if Radius of curve is 1000 m, Δ = 60°, chainage of P.I = 2000m
Determine
i) length of curve
ii) Tangent Length
iii) Long chord
iv) mid ordinate (M)
v) Apex distance
vi) Chainages of �
1,�
2
vii) Degree of curve for 30 m Arc
i) �=
��??????
���°
=
�(����)��
���
=����.���
610610
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 181: if Radius of curve is 1000 m, Δ = 60°, chainage of P.I = 2000m
Determine
i) length of curve
ii) Tangent Length
iii) Long chord
iv) mid ordinate (M)
v) Apex distance
vi) Chainages of �
1,�
2
vii) Degree of curve for 30 m Arc
ii) �=����
??????
�
=���������°
=���.���
611611
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 181: if Radius of curve is 1000 m, Δ = 60°, chainage of P.I = 2000m
Determine
i) length of curve
ii) Tangent Length
iii) Long chord
iv) mid ordinate (M)
v) Apex distance
vi) Chainages of �
1,�
2
vii) Degree of curve for 30 m Arc
iii) �=�����
??????
�
=�×���������°=�����
=�×���������°
=�����
612612
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 181: if Radius of curve is 1000 m, Δ = 60°, chainage of P.I = 2000m
Determine
i) length of curve
ii) Tangent Length
iii) Long chord
iv) mid ordinate (M)
v) Apex distance
vi) Chainages of �
1,�
2
vii) Degree of curve for 30 m Arc
iv) �=��−���
??????
�
=�����−�����°
=���.���
613613
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 181: if Radius of curve is 1000 m, Δ = 60°, chainage of P.I = 2000m
Determine
i) length of curve
ii) Tangent Length
iii) Long chord
iv) mid ordinate (M)
v) Apex distance
vi) Chainages of �
1,�
2
vii) Degree of curve for 30 m Arc
v) �=����
??????
�
−�
=���������°−�=������
=������
614614
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 181: if Radius of curve is 1000 m, Δ = 60°, chainage of P.I = 2000m
Determine
i) length of curve
ii) Tangent Length
iii) Long chord
iv) mid ordinate (M)
v) Apex distance
vi) Chainages of �
1,�
2
vii) Degree of curve for 30 m Arc
vi) chof �
�=����−���.��=����.���
chof �
�=����.��+����.��=����.���
vii) �=
����
����
=�.���
615615
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Que 181: if Radius of curve is 1000 m, Δ = 60°, chainage of P.I = 2000m
Determine
i) length of curve
ii) Tangent Length
iii) Long chord
iv) mid ordinate (M)
v) Apex distance
vi) Chainages of �
1,�
2
vii) Degree of curve for 30 m Arc
ii) �=����
??????
�
=���������°=���.���
iii) �=�����
??????
�
=�×���������°=�����
iv) �=��−���
??????
�
=�����−�����°=���.���
v) �=����
??????
�
−�=���������°−�=������
vi) chof �
�=����−���.��=����.���
chof �
�=����.��+����.��=����.���
vii) �=
����
����
=�.���
i) �=
��??????
���°
=
�(����)��
���
=����.���
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Theinsideclearanceofsoil
sampleris
a)
�
�−�
�
��
���
b)
�
�−�
�
�
�
���
c)
�
�−�
�
��
���
d)
�
�−�
�
�
�
���
�
�=����������������������
�
�=����������������������
�
�=���������������������
�
�=���������������������
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Theinsideclearanceofsoil
sampleris
a)
�
�−�
�
��
���
b)
�
�−�
�
�
�
���
c)
�
�−�
�
��
���
d)
�
�−�
�
�
�
���
�
�=����������������������
�
�=����������������������
�
�=���������������������
�
�=���������������������
Soil samples
•The extent of disturbance of
sample due to the sampler
depends upon
a)Cutting edge
b)Inside wall friction
Civil Engineering by Sandeep Jyani
���������������=
�
�−�
�
�
�
�����������������������%
����������������=
�
�−�
�
�
�
�����������������������%
����������������>���������������
�
�=����������������������
�
�=����������������������
�
�=���������������������
�
�=���������������������
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
ForIdentificationofEXPANSIVESOIL,theFREESWELLTest
expressesthefreeSWELL%as
a)���������%=
����������������������������−����������������������
����������������������
b)���������%=
����������������������−���������������
���������������
c)���������%=
����������������������������−���������������
����������������������
d)���������%=
����������������������������−���������������
���������������
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
ForIdentificationofEXPANSIVESOIL,theFREESWELLTest
expressesthefreeSWELL%as
a)���������%=
����������������������������−����������������������
����������������������
b)���������%=
����������������������−���������������
���������������
c)���������%=
����������������������������−���������������
����������������������
d)���������%=
����������������������������−���������������
���������������
Identification of Expansive Soil
•Identification of EXPANSIVE SOIL is done by two
methods:
1.Free Swell Test
���������%=
����������������������������−���������������
���������������
Civil Engineering by Sandeep Jyani
Mineral Free Swell %
Montmorillonite1200-2000
Kaolinite 80
Illite 50-80
It is observed that free swell below
50% does not cause any problem to
the structure even under light
surcharge
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
ForIdentificationofEXPANSIVESOIL,theDIFFRENTIALFREE
SWELL%isexpressedas
a)
���������������−����������������������
����������������������
b)
����������������������������−���������������
����������������������
c)
����������������������������−����������������������
����������������������
d)
����������������������������−���������������
���������������
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
ForIdentificationofEXPANSIVESOIL,theDIFFRENTIALFREE
SWELL%isexpressedas
a)
���������������−����������������������
����������������������
b)
����������������������������−���������������
����������������������
c)
����������������������������−����������������������
����������������������
d)
����������������������������−���������������
���������������
Identification of Expansive Soil
2. Differential Free Swell Test
���%=
����������������������������−����������������������
����������������������
Civil Engineering by Sandeep Jyani
Degree of
Expansion
Differential
Free Swell %
Low <20
moderate 20-35
high 35-50
Very high >50
Single Pile Capacity in Sandy Soil
1.Driven Pile in Sand
•Unit Point resistance (�=ത??????)
increases in direct proportion to
the embedded length of the pile,
but Several field observations
indicate that unit point
resistance values increase only
uptoa limited depth known as
Critical Depth of Pile
•Critical Depth depends on the
angle of shearing resistance and
diameter of pile
Civil Engineering by Sandeep Jyani
���������
����
���������
����
Single Pile Capacity in Sandy Soil
1.Driven Pile in Sand
��������������������
The unit skin friction acting at any depth
is equal to the soil pressure acting normal
to the pile surface
Civil Engineering by Sandeep Jyani
Pile �K (Loose
sand)
K(Dense
Sand)
Steel 20 0.5 1
Concrete0.75∅ 1 2
Timber0.67∅ 1.5 4
�=���������������������������������
�=����������������������������
We know that ??????
ℎ=�??????
�
⇒�
�=������������������������������
⇒�
�=�
��
�where
�
�=�ത??????
���tan�
Single Pile Capacity in Sandy Soil
2.Pre Cast Bored Pile in Sand
•Due to boring condition, value of ‘�’
is small and load carrying capacity is
small as compared to driven piles
•Frictional resistance capacity is
calculated by taking
�=0.5and �=∅
•Point Bearing capacity of bored pile
is taken as Half of the Point Bearing
Capacity of Driven Pile.
Civil Engineering by Sandeep Jyani
�
�=������������������������������
⇒�
�=�
��
�where
�
�=�ത??????
���tan�
�=0.5and �=∅
�
��
���������
=
�
��
����������
�
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Singlepileloadcapacityinclayeysoil
forboredanddrivenpilesisgivenby
a)�=(??????�)�
����+(9�)�
�
b)�=(9�)�
����+(??????�)�
�
c)�=(??????�)�
����+(�ത??????
���tan�)�
�
d)�=(9�)�
�+(�ത??????
���tan�)�
����
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Singlepileloadcapacityinclayeysoil
forboredanddrivenpilesisgivenby
a)�=(??????�)�
����+(9�)�
�
b)�=(��)�
����+(??????�)�
�
c)�=(??????�)�
����+(�ത??????
���tan�)�
�
d)�=(9�)�
�+(�ത??????
���tan�)�
����
Group Pile
Ultimate load capacity of Pile
group
�
�����=����
�����,��
���
Disturbance in the soil due to
installation of Pile overlap the
stress and varies Pile Group
capacity as compared to
��
������������������
Civil Engineering by Sandeep Jyani
Group Pile
Stress Overlap in Sand
In case of Sand, strength of Soil increases
due to disturbance occurring during
installation, hence
�
�����>��
������
In case of Clay
In case of clay, the strength of soil decrease
on remoulding during installation of Pile
Group
�
�����<��
������
Civil Engineering by Sandeep Jyani
����
����
Efficiency of Pile Group
•For practical purpose we never take
efficiency greater than 1
•But efficiency can be greater than 1
for sandy soil and less than 1 for clay
soils
Civil Engineering by Sandeep Jyani
�=
�
�����
��
����������
Pile Capacity using Dynamic Pile Formula
1.Engineering News Formula
Civil Engineering by Sandeep Jyani
�
�����=
��
���(�+�)
���=6
�=
��
6�+�
�ℎ���
•�=����ℎ���ℎ�����
•�=ℎ���ℎ���������������
•�=����������������������ℎ��ℎ�������
�������������������5�����������ℎ�����
•�=���������������allowing reduction in theoretical
sets due to energy losses
ቊ
2.5���������ℎ�����
0.25����������������������ℎ�����
Pile Capacity using Dynamic Pile Formula
2.Modified Hilly Formula
Civil Engineering by Sandeep Jyani
�
�����=
���
�+
�
2
�=�����������������
�ℎ���
�=����ℎ���ℎ�����
�=ℎ���ℎ�����������
�=�����������������
�=�����������������������������
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Verticalstressgivenby
Newmark Chart is
describedas
a)??????
�=
�
�
��
b)??????
�=
�
�
��
c)??????
�=����
d)??????
�=
�
��
��
•m=no. of concentric circles
•n= no of radial lines
•q= intensity of load
•N= equivalent no. of areas
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Verticalstressgivenby
Newmark Chart is
describedas
a)??????
�=
�
�
��
b)??????
�=
�
�
��
c)??????
�=����
d)??????
�=
�
��
��
•m=no. of concentric circles
•n= no of radial lines
•q= intensity of load
•N= equivalent no. of areas
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of Earth pressure at
rest is given by
a)�
�=
�−�
�
b)�
�=
�
�+�
c)�
�=
�
�−�
d)�
�=
�+�
�
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of Earth pressure at
rest is given by
a)�
�=
�−�
�
b)�
�=
�
�+�
c)�
�=
�
�−�
d)�
�=
�+�
�
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of Active Earth pressure is
given by
a)�
�=
�+���??????
�−���??????
b)�
�=
�−���??????
�+���??????
c)�
�=
�−���??????
�+���??????
d)�
�=
�+���??????
�+���??????
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of Active Earth pressure is
given by
a)�
�=
�+���??????
�−���??????
b)�
�=
�−���??????
�+���??????
c)�
�=
�−���??????
�+���??????
d)�
�=
�+���??????
�+���??????
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of Passive Earth pressure
is given by
a)�
�=
�+���??????
�−���??????
b)�
�=
�−���??????
�+���??????
c)�
�=
�−���??????
�+���??????
d)�
�=
�+���??????
�+���??????
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of Passive Earth pressure
is given by
a)�
�=
�+���??????
�−���??????
b)�
�=
�−���??????
�+���??????
c)�
�=
�−���??????
�+���??????
d)�
�=
�+���??????
�+���??????
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Active Earth pressure, by Rankine
Theory for Dry and Moist Soil Mass
with No surcharge is given by
a)�
�=
1
2
�
���
2
at H/2 from base
b)�
�=
1
2
�
���
2
at H/3 from base
c)�
�=�
�??????
�at H/3 from base
d)�
�=�
�??????
�at H/2 from base
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Active Earth pressure, by Rankine
Theory for Dry and Moist Soil Mass
with No surcharge is given by
a)�
�=
1
2
�
���
2
at H/2 from base
b)�
�=
�
�
�
���
�
at H/3 from base
c)�
�=�
�??????
�at H/3 from base
d)�
�=�
�??????
�at H/2 from base
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Critical depth of unsupported
excavation is given by
a)�
�=
��
���
b)�
�=
��
���
c)�
�=
���
��
d)�
�=
���
��
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Critical depth of unsupported
excavation is given by
a)�
�=
��
���
b)�
�=
��
���
c)�
�=
���
��
d)�
�=
���
��
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Alaboratoryvanesheartestapparatusisusedtodeterminethe
shearstrengthofaclaysampleandonlyoneendofthevanetakes
partinsharingthesoil.IfT=appliedtorque,H=heightofvaneand
D=diameterofthevane,thenshearstrengthoftheclayisgivenby
a)
T
πD
2
H+
D
6
b)
T
πD
2
H
2
+
D
12
c)
T
πD
2
H+
D
10
d)
T
πD
2
H+
D
12
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Alaboratoryvanesheartestapparatusisusedtodeterminethe
shearstrengthofaclaysampleandonlyoneendofthevanetakes
partinsharingthesoil.IfT=appliedtorque,H=heightofvaneand
D=diameterofthevane,thenshearstrengthoftheclayisgivenby
a)
T
πD
2
H+
D
6
b)
�
??????�
�
�
�
+
�
��
c)
T
πD
2
H+
D
10
d)
T
πD
2
H+
D
12
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
If‘s’istheshearstrength,‘c’and
ϕareshearstrengthparameters,
and‘σ
n’isthenormalstressat
failure,thenCoulomb’sequation
forshearstrengthofthesoilcan
berepresentedby
a)c=s+σ
ntanϕ
b)c=s−σ
ntanϕ
c)s=σ
n+ctanϕ
d)s=c−σ
ntanϕ
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
If‘s’istheshearstrength,‘c’and
ϕareshearstrengthparameters,
and‘σ
n’isthenormalstressat
failure,thenCoulomb’sequation
forshearstrengthofthesoilcan
berepresentedby
a)c=s+σ
ntanϕ
b)�=�−??????
����??????
c)s=σ
n+ctanϕ
d)s=c−σ
ntanϕ
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
The relationship between major principal stress and
minor principal stress is given by
a)??????
�=??????
����
�
��+
∅
�
+�������+
∅
�
b)??????
�=??????
����
�
��−
∅
�
−�������−
∅
�
c)??????
�=??????
�
�+���∅
�−���∅
+��
�+���∅
�−���∅
d)All of the above
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
The relationship between major principal stress and
minor principal stress is given by
a)??????
�=??????
����
�
��+
∅
�
+�������+
∅
�
b)??????
�=??????
����
�
��−
∅
�
−�������−
∅
�
c)??????
�=??????
�
�+���∅
�−���∅
+��
�+���∅
�−���∅
d)All of the above
MOHR COULOMB CRITERIA
Civil Engineering by Sandeep Jyani
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
WhichoneofthefollowingpairsofparametersandexpressionisNOT
correctlymatched?
a)Coefficientofconsolidation…
���²
�
b)Coefficientofvolumecompressibility…
�0−�
(1+�0)(�−�0)
c)Overconsolidationratio…
��������������������������������
������������������������
d)Modulusofvolumechange…
�
�
1+�0
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
WhichoneofthefollowingpairsofparametersandexpressionisNOT
correctlymatched?
a)Coefficientofconsolidation…
���²
�
b)Coefficientofvolumecompressibility…
�0−�
(1+�0)(�−�0)
c)Overconsolidationratio…
��������������������������������
������������������������
d)Modulusofvolumechange…
�
�
1+�0
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
One dimensional consolidation is
given by following expression:
a)
∆�
��
=
∆�
�−��
b)
∆�
�
�
=
∆�
�
�
c)
∆�
�
�
=
∆�
�+�
�
d)
∆�
��
=
�+�
�
∆�
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
One dimensional consolidation is
given by following expression:
a)
∆�
��
=
∆�
�−��
b)
∆�
�
�
=
∆�
�
�
c)
∆�
�
�
=
∆�
�+�
�
d)
∆�
��
=
�+�
�
∆�
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Compression Index C
cis given by
a)�
�=
�1+�2
logഥ??????1+logഥ??????2
b)�
�=
�
1−�
2
logഥ??????1−logഥ??????2
c)�
�=
∆�
∆ഥ??????
d)�
�=
∆�
∆ഥ??????(1+�
0)
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Compression Index C
cis given by
a)�
�=
�1+�2
logഥ??????1+logഥ??????2
b)�
�=
�
�−�
�
���ഥ??????�−���ഥ??????�
c)�
�=
∆�
∆ഥ??????
d)�
�=
∆�
∆ഥ??????(1+�
0)
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of Compressibility a
v is
denoted by
a)�
�=
∆�
∆ഥ??????
b)�
�=
∆ഥ??????
∆�
c)�
�=
∆�
∆ഥ??????(1+�
0)
d)�
�=
∆ഥ??????
∆�(1+�
0)
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of Compressibility a
v is
denoted by
a)a
v=
∆e
∆ഥσ
b)�
�=
∆ഥ??????
∆�
c)�
�=
∆�
∆ഥ??????(1+�
0)
d)�
�=
∆ഥ??????
∆�(1+�
0)
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of volume compressibility
a)�
�=
∆�
∆ഥ??????(1+�
0)
b)�
�=
�
�
(1+�
0)
c)�
�=
1
������������������
d)All of the above
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of volume compressibility
a)�
�=
∆�
∆ഥ??????(1+�
0)
b)�
�=
�
�
(1+�
0)
c)�
�=
1
������������������
d)All of the above
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of consolidation is given by
a)�
�=
�??????(1+�)
��??????�
b)�
�=
�
??????�
�
??????
�
c)�
�=
��
2
�
�
d)All of the above
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Coefficient of consolidation is given by
a)C
v=
kz(1+e)
avγw
b)�
�=
�
??????�
�
??????
�
c)�
�=
��
2
�
�
d)All of the above
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Immediate Settlement in Sand is given by
a)�
�=
�0
��
���
10
ഥ??????0+??????ഥ??????
ഥ??????
�
�=�.�
��
ഥ??????�
�
�=��������������������
b)�
�=
��1−??????
2
�
�
��
c)�
�=
��1+??????
2
�
�
�
�
d)Δ�=�
0
Δe
1+�
0
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Immediate Settlement in Sand is given by
a)s
i=
H0
cs
log
10
ഥσ0+Δഥσ
ഥσ
�
�=�.�
��
ഥ??????�
�
�=��������������������
b)�
�=
��1−??????
2
�
�
��
c)�
�=
��1+??????
2
�
�
�
�
d)Δ�=�
0
Δe
1+�
0
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Computation of primary settlement
can be done by using
a)Δ�=�
0
Δe
1+�
0
b)Δ�=�
0�
�Δത??????
c)Δ�=
�
��
0
1+�
0
log
10
ഥ??????
0+Δഥ??????
??????
0
d)All of the above
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Computation of primary settlement
can be done by using
a)Δ�=�
0
Δe
1+�
0
b)Δ�=�
0�
�Δത??????
c)Δ�=
�
��
0
1+�
0
log
10
ഥ??????
0+Δഥ??????
??????
0
d)All of the above
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Seepage discharge through a flownetfor
one channel is given by
a)
���
�
��
b)
��
�
�
c)
���
�
�
�
d)
��
�
�
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Seepage discharge through
a flownetfor one channel is
given by
a)
���
�
��
b)
��
��
c)
���
�
�
�
d)
��
��
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Lab method variable head permeability
method gives the value of permeability as
a)�.���
��
��
���
��(
�
�
�
�
)
b)�=
��
��
c)�.���
��
��
���
��(
��
��
)
d)�=
��
��
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING
Lab method variable head permeability
method gives the value of permeability as
a)�.���
��
��
���
��(
�
�
�
�
)
b)�=
��
��
c)�.���
��
��
���
��(
��
��
)
d)�=
��
��
Lab Methods to determine Permeability
1.Constant Head Permeability Test
�
�
�
�=���������������,���������������
�=����������������������
���������,�����������������
⇒�=
���������������
����������������������
⇒�=
�
�
�������������
′
����,
�=���
⇒�=�
�
�
�
⇒�=
��
��
Civil Engineering by Sandeep Jyani
Lab Methods to determine Permeability
1.Constant Head Permeability Test
a)Constant head permeability test is done
for coarse grain soils only (Sand and
Gravel)
b)This test is not used for fine grain soil
because it is difficult to measure
discharge volume
�
�
�
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The actual velocity in a soil mass is given by
a)
�(�����������ℎ����)
�(���������)
b)
�(����ℎ������������)
�(��������)
c)�(�����������ℎ����)×����
d)None
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The actual velocity in a soil mass is given by
a)
�(����������������)
�(���������)
b)
�(�����������������)
�(��������)
c)�(����������������)����
d)None
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In Permeability, one dimensional flow, through a fully
saturated soil, Discharge, as mentioned by Darcy, is
a)�∝�������������������
b)�∝
�
�
���������������
���������������������
c)�∝�(�����������������)
d)All of the above
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In Permeability, one dimensional flow, through a fully
saturated soil, Discharge, as mentioned by Darcy, is
a)�∝�������������������
b)�∝
�
�
���������������
���������������������
c)�∝�(�����������������)
d)All of the above
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Critical Hydraulic Gradient is �
��given by
a)
??????
���
??????�
b)
??????
���
??????
�
−1
c)
�−1
1+�
d)All of the above
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Critical Hydraulic Gradient is �
��given by
a)
??????
���
??????�
b)
??????
���
??????
�
−1
c)
�−1
1+�
d)All of the above
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Effect of grain Size on Permeability is
given by Allen Hazen formula, which is
a)ഥ�=
�
����
b)�=��
��
�
c)�=��
��
d)All of the above
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Effect of grain Size on Permeability is
given by Allen Hazen formula, which is
a)ഥ�=
�
����
b)�=��
��
�
c)�=��
��
d)All of the above
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Empirical formula to calculate
Height of Capillary Rise is given
by
a)ഥ�=
�
����
b)�=��
��
�
c)�=��
��
d)All of the above
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Empirical formula to calculate
Height of Capillary Rise is given
by
a)ҧ�=
�
����
b)�=��
��
�
c)�=��
��
d)All of the above
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To find the quality of subgrade for Highway
Material, Group Index is given by
a)G.I = 0.2b + 0.005bd + 0.01a
b)G.I = 0.2a + 0.005ac + 0.05bd
c)G.I = 0.2a + 0.005ac + 0.01bd
d)G.I = 0.2a + 0.005bd + 0.01ac
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To find the quality of subgrade for Highway
Material, Group Index is given by
a)G.I = 0.2b + 0.005bd + 0.01a
b)G.I = 0.2a + 0.005ac + 0.05bd
c)G.I = 0.2a + 0.005ac + 0.01bd
d)G.I = 0.2a + 0.005bd + 0.01ac
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Uniformity coefficient is defined as
a)�
�=
�
��
���
b)�
�=
���
���
c)�
�=
�
��
�
��
d)�
�=
���
�
�
���
��
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Uniformity coefficient is defined as
a)�
�=
�
��
���
b)�
�=
���
���
c)�
�=
�
��
�
��
d)�
�=
���
�
�
���
��
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A plasticity chart, based on liquid limit and
plasticity index, expresses the “A-Line” as
a)�
�=0.73(�
�−20)
b)�
�=0.73�
�−20
c)�
�=0.20(�
�−73)
d)�
�=0.20(�
�−73)
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A plasticity chart, based on liquid limit and
plasticity index, expresses the “A-Line” as
a)�
�=0.73(�
�−20)
b)�
�=�.���
�−��
c)�
�=0.20(�
�−73)
d)�
�=0.20(�
�−73)
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Aclayeysoilhasliquidlimit=w
L;plasticlimit=
w
pandnaturalmoisturecontent=w.The
consistencyindexofthesoilisgivenby
a)
WL−W
WL−WP
b)W
L−W
P
c)W
L−W
d)
W
P−W
WL−WP
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Aclayeysoilhasliquidlimit=w
L;plasticlimit=
w
pandnaturalmoisturecontent=w.The
consistencyindexofthesoilisgivenby
a)
��−�
��−��
b)W
L−W
P
c)W
L−W
d)
W
P−W
WL−WP
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Thegivenfigureindicatethe
weights of different
pycnometers:
Thespecificgravityofthesolids
isgivenby
a)
W2
W4−W2
b)
W1−W2
W3−W4−(W2−W1)
c)
W2
W3−W4
d)
W2−W1
W2−W1−(W3−W4)
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Thegivenfigureindicatethe
weights of different
pycnometers:
Thespecificgravityofthesolids
isgivenby
a)
W2
W4−W2
b)
W1−W2
W3−W4−(W2−W1)
c)
W2
W3−W4
d)
��−��
��−��−(��−��)
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RelativeDensityisdefinedas
a)
����+�
����−����
b)
����−�
����+����
c)
����−�
����−����
d)
����+�
����+����
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RelativeDensityisdefinedas
a)
����+�
����−����
b)
����−�
����+����
c)
����−�
����−����
d)
����+�
����+����
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Activityofclayeysoilisexpressedas
a)�
�=
��������������
%������������������
b)�
�=
���������������
%������������������
c)�
�=
���������������
%������������������
d)�
�=
�����������
%������������������
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Activityofclayeysoilisexpressedas
a)�
�=
��������������
%������������������
b)�
�=
���������������
%������������������
c)�
�=
���������������
%������������������
d)�
�=
�����������
%������������������
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Aclayeysoilhasliquidlimit=w
L;plasticlimit=
w
pandnaturalmoisturecontent=w.The
consistencyindexofthesoilisgivenby
a)
WL−W
WL−WP
b)W
L−W
P
c)W
L−W
d)
WP−W
WL−WP
CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
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Aclayeysoilhasliquidlimit=w
L;plasticlimit=
w
pandnaturalmoisturecontent=w.The
consistencyindexofthesoilisgivenby
a)
�??????−�
�??????−�??????
b)W
L−W
P
c)W
L−W
d)
WP−W
WL−WP
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Aircontentistheratioof
a)�
�=
��
��
b)�
�=
�
�
�
c)�
�=
��
�
d)�
�=
��
�??????
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Aircontentistheratioof
a)�
�=
��
��
b)�
�=
�
�
�
c)�
�=
��
�
d)�
�=
��
��
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Degreeofsaturationmaybeobtainedfrom
a)�=
�
�
�??????
b)��=��
c)�=1−��
d)Anyoftheabove
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Degreeofsaturationmaybeobtainedfrom
a)�=
�
�
�??????
b)��=��
c)�=1−��
d)Anyoftheabove
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Porosityhasthefollowingrelationship:
a)�=
�
??????
�
b)�=
�
1+�
c)�=
��
�
�
d)Alloftheabove
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Porosityhasthefollowingrelationship:
a)�=
�
??????
�
b)�=
�
1+�
c)�=
��
�
�
d)Alloftheabove
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Voidratio(e)istheratioof
a)
�??????
�
b)
�??????
��
c)
��
��
d)
��
��
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Voidratio(e)istheratioof
a)
�??????
�
b)
��
��
c)
��
��
d)
��
��
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Volumeofsoildsisrelatedtototalvolumeas
a)�
�=
�
�+�
b)�
�=
�
�+�
c)�
�=
�
�+�
d)�
�=�(�+�)
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Volumeofsoildsisrelatedtototalvolumeas
a)�
�=
�
�+�
b)�
�=
�
�+�
c)�
�=
�
�+�
d)�
�=�(�+�)
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Dryunitweightmaybeexpressedas
a)�
�=
�−�
���
�
�+��
b)�
�=
�
�+�
c)�
�=
���
�+�
d)Anyoftheabove
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Dryunitweightmaybeexpressedas
a)�
�=
�−�
���
�
�+��
b)�
�=
�
�+�
c)�
�=
���
�+�
d)Anyoftheabove
Relation between different terms
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
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