civil is a very beautiful subject plz read.pdf
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About This Presentation
it is just civil
Slide Content
DESIG OF CONCRETE STRUCTURE
G.C.BEHERA
12-12-2022 1GCEKK, BHAWANIPATNA
G.C.BEHERA
12-12-2022 1GCEKK, BHAWANIPATNA
A: Programme B.Tech
Course Code PCI4I103
Course Name DEISGN OF CONCRETE
STRUCTURE
Semester 5th
Course Type Theory
Contact Hour 40
IA Marks 50
University Exam Marks 100
12-12-2022 2GCEKK, BHAWANIPATNA
Course Code PCI4I103
Course Name DEISGN OF CONCRETE
STRUCTURE
Semester 5th
Course Type Theory
Contact Hour 40
IA Marks 50
University Exam Marks 100
12-12-2022 2GCEKK, BHAWANIPATNA
B: Course Prerequisites
Code Course Name Description Semester
PCME4202 Mechanics of
Solid
Mod-II-Shear
Force and
Bending Moment
for Simple Beams
3rd
PCME4202 Mechanics of
Solid
Mod-II-
Deflection of
Beams
3rd
12-12-2022 3GCEKK, BHAWANIPATNA
Code Course Name Description Semester
PCME4202 Mechanics of
Solid
Mod-II-Shear
Force and
Bending Moment
for Simple Beams
3rd
PCME4202 Mechanics of
Solid
Mod-II-
Deflection of
Beams
3rd
12-12-2022 3GCEKK, BHAWANIPATNA
C: Course Objective
•Tostatethefour objectivesofthe designof
reinforcedconcretestructuresandnamethethree
methodsofdesignofconcrete structure and
identifythebestmethodofdesign,
•Tostatethebasisoftheanalysisofthestructure,
•Namethedifferentloads,forcesandeffectstobe
consideredinthedesign,
•Statethebasisofdeterminingthecombinationof
differentloadsactingonthestructure.
•Togain basic knowledgetoanalyzethebeams,
slabs,staircasesandColumns.
12-12-2022 4GCEKK, BHAWANIPATNA
•Tostatethefour objectivesofthe designof
reinforcedconcretestructuresandnamethethree
methodsofdesignofconcrete structure and
identifythebestmethodofdesign,
•Tostatethebasisoftheanalysisofthestructure,
•Namethedifferentloads,forcesandeffectstobe
consideredinthedesign,
•Statethebasisofdeterminingthecombinationof
differentloadsactingonthestructure.
•Togain basic knowledgetoanalyzethebeams,
slabs,staircasesandColumns.
12-12-2022 4GCEKK, BHAWANIPATNA
D: Course Syllabus:
•Design of Concrete Structures (3-0-0)
•Module I (10 Classes)
•Properties of concrete and reinforcing steel, philosophy, concept and methods of reinforced
concrete design, introduction to limit state method, limit state of collapse and limit state of
serviceability, application of limit state method to rectangular beams for flexure, shear,
bond and torsion
•Module II (8 Classes)
•Design of doubly reinforced beams, design of T and L beams, design of one way and two
way slabs, design of staircases.
•Module III (8 Classes)
•Design of short and long columns with axial and eccentric loadings, Design of isolated and
combined column footings
•Module IV (8 Classes)
•Retaining walls, various forces acting on retaining wall, stability requirement, design of
cantilever and counterfort retaining walls,
•Module V ( 6 Classes)
•Design of water tanks, design requirements, design of tanks on ground, under ground and
elevated water tanks.
12-12-2022 5GCEKK, BHAWANIPATNA
•Design of Concrete Structures (3-0-0)
•Module I (10 Classes)
•Properties of concrete and reinforcing steel, philosophy, concept and methods of reinforced
concrete design, introduction to limit state method, limit state of collapse and limit state of
serviceability, application of limit state method to rectangular beams for flexure, shear,
bond and torsion
•Module II (8 Classes)
•Design of doubly reinforced beams, design of T and L beams, design of one way and two
way slabs, design of staircases.
•Module III (8 Classes)
•Design of short and long columns with axial and eccentric loadings, Design of isolated and
combined column footings
•Module IV (8 Classes)
•Retaining walls, various forces acting on retaining wall, stability requirement, design of
cantilever and counterfort retaining walls,
•Module V ( 6 Classes)
•Design of water tanks, design requirements, design of tanks on ground, under ground and
elevated water tanks.
12-12-2022 5GCEKK, BHAWANIPATNA
BOOKS AND DIGITAL LEARNING
SOURCE
12-12-2022 GCEKK, BHAWANIPATNA 6
Books:
1. Design of Reinforced Concrete Structue by N. Subramanian, Oxford University
Press
2. Limit State Design by A.K.Jain, Neemchand& Bros
3. Reinforced Concrete Design by S U Pillai & D. Menon, McGraw Hill
4. Design of concrete structures by J.N.Bandyopadhyay, PHI
5. Limit State Design of Reinforced Concrete -P.C Verghese
6. Reinforced Concrete Design by S.N.Sinha, McGraw Hill
7. RCC Design-B.C.Punmia, A.K.Jainand A.K.Jain-LaxmiPublications
Digital Learning Resources:
Course Name Design of Reinforced Concrete Structures 12 weeks
Course Link ( https://nptel.ac.in/courses/105/105/105105105/)
Course Instructor PROF. NIRJHAR DHANG, ,IIT Kharagpur
SOURCE
12-12-2022 GCEKK, BHAWANIPATNA 6
Books:
1. Design of Reinforced Concrete Structue by N. Subramanian, Oxford University
Press
2. Limit State Design by A.K.Jain, Neemchand& Bros
3. Reinforced Concrete Design by S U Pillai & D. Menon, McGraw Hill
4. Design of concrete structures by J.N.Bandyopadhyay, PHI
5. Limit State Design of Reinforced Concrete -P.C Verghese
6. Reinforced Concrete Design by S.N.Sinha, McGraw Hill
7. RCC Design-B.C.Punmia, A.K.Jainand A.K.Jain-LaxmiPublications
Digital Learning Resources:
Course Name Design of Reinforced Concrete Structures 12 weeks
Course Link ( https://nptel.ac.in/courses/105/105/105105105/)
Course Instructor PROF. NIRJHAR DHANG, ,IIT Kharagpur
E: Course Outcome
•Students who successfully complete this course will be
able to
–Identify and compute the main mechanical properties of
concrete and steel.
–Identify and calculate the design loads and distribution.
–Apply the strength method to design R.C. structural
members.
–Analyze and design R.C. beams and slabs for flexure and
shear.
–Analyze and design short and slender R.C. columns and
their footings.
–Analyze and design retaining structures.
12-12-2022 7GCEKK, BHAWANIPATNA
•Students who successfully complete this course will be
able to
–Identify and compute the main mechanical properties of
concrete and steel.
–Identify and calculate the design loads and distribution.
–Apply the strength method to design R.C. structural
members.
–Analyze and design R.C. beams and slabs for flexure and
shear.
–Analyze and design short and slender R.C. columns and
their footings.
–Analyze and design retaining structures.
12-12-2022 7GCEKK, BHAWANIPATNA
F: Programme Outcomes Addressed to the students
PO-1 Engineering Knowledge CO-1
PO-2- Problem Analysis CO4,5&6,
PO-3- Design / developmentofSolution CO4,5 & 6,
PO-4- Conduct Analysisoncomplex analysis CO-2,CO-3
PO-5 Modern Tool Usages
PO-6 EngineerandSociety CO-3
PO-7- EnvironmentandSustainability CO-1
PO-8- Ethics
PO-9- IndividualandTeam work 2,3,4,5 & 6
PO-10-Communication
PO-11- Project Mang. And Finance
Po-12- LifeLongLearning CO-2
12-12-2022 8GCEKK, BHAWANIPATNA
PO-1 Engineering Knowledge CO-1
PO-2- Problem Analysis CO4,5&6,
PO-3- Design / developmentofSolution CO4,5 & 6,
PO-4- Conduct Analysisoncomplex analysis CO-2,CO-3
PO-5 Modern Tool Usages
PO-6 EngineerandSociety CO-3
PO-7- EnvironmentandSustainability CO-1
PO-8- Ethics
PO-9- IndividualandTeam work 2,3,4,5 & 6
PO-10-Communication
PO-11- Project Mang. And Finance
Po-12- LifeLongLearning CO-2
12-12-2022 8GCEKK, BHAWANIPATNA
G. Gaps in the Syllabus (if Any) to meet Industry
Requirement
Gap-1- Complete Designofa structure not includedinthe
syllabus.
Gap-2-Not more emphasisisgivenonfield calculations.
12-12-2022 GCEKK, BHAWANIPATNA 9
Requirement
Gap-1- Complete Designofa structure not includedinthe
syllabus.
Gap-2-Not more emphasisisgivenonfield calculations.
12-12-2022 GCEKK, BHAWANIPATNA 9
H: Topics beyond Syllabus/ Advanced
Topics
DesignofLintels, Chhaza
Acomplete design methodology.
12-12-2022 GCEKK, BHAWANIPATNA 10
Topics
DesignofLintels, Chhaza
Acomplete design methodology.
12-12-2022 GCEKK, BHAWANIPATNA 10
I: Assessment Methodologies:
S.N Description Type
1 Assignment Direct
2 Internal Exam Direct
3 University ExamDirect
4 Tutorial Direct
5 Presentation Direct
Student FeedbackIndirect
6 Employer feedbackIndirect
12-12-2022 GCEKK, BHAWANIPATNA 11
S.N Description Type
1 Assignment Direct
2 Internal Exam Direct
3 University ExamDirect
4 Tutorial Direct
5 Presentation Direct
Student FeedbackIndirect
6 Employer feedbackIndirect
12-12-2022 GCEKK, BHAWANIPATNA 11
J: Course Plan:
S.NDayChapter/
Module
Topics To be Covered Topics
Covered
01 I Propertiesofconcrete ,comp, tensile and other properties
02 I
Objective of design , concept and methods of reinforced concrete design
03 I
Introduction to limit state method, Limit state of collapse , Limit state of serviceability
04 I Analysis of Limit state method to rectangular beams for flexure, Determination of NA
axis, MOR, Percentage of steel
05
MOR of under reinforced, balance and over reinforced beam
06 I
Application of Limit state method to rectangular beams for shear
07 I
Shear design of beams
08 I
ApplicationofLimit state methodtorectangular beams for bond, bondlength
calculation,Hooks
09 I
Design of a complete rectangular Simply supported beam, effective length
calculations, design parameters
10 I
Application of Limit state method to rectangular beams for torsion.
12-12-2022 GCEKK, BHAWANIPATNA 12
S.NDayChapter/
Module
Topics To be Covered Topics
Covered
01 I Propertiesofconcrete ,comp, tensile and other properties
02 I
Objective of design , concept and methods of reinforced concrete design
03 I
Introduction to limit state method, Limit state of collapse , Limit state of serviceability
04 I Analysis of Limit state method to rectangular beams for flexure, Determination of NA
axis, MOR, Percentage of steel
05
MOR of under reinforced, balance and over reinforced beam
06 I
Application of Limit state method to rectangular beams for shear
07 I
Shear design of beams
08 I
ApplicationofLimit state methodtorectangular beams for bond, bondlength
calculation,Hooks
09 I
Design of a complete rectangular Simply supported beam, effective length
calculations, design parameters
10 I
Application of Limit state method to rectangular beams for torsion.
12-12-2022 GCEKK, BHAWANIPATNA 12
J: Course Plan:
S.NDa
y
Chapter/
Module
Topics To be Covered Topics
Covered
11 II
Analysis of doubly reinforced beams
11
12 II
Design of doubly reinforced beams
12
13 II
Analysis of Flanged section.
13
14 II
Design of T beam, L beam
14
15 II
Analysis of one way slabs
15
16 II
Design of two way slab
16
17 II
Analysis of two way slabs
17
18 II
Calculation of loads of staircases.
18
12-12-2022 GCEKK, BHAWANIPATNA 13
S.NDa
y
Chapter/
Module
Topics To be Covered Topics
Covered
11 II
Analysis of doubly reinforced beams
11
12 II
Design of doubly reinforced beams
12
13 II
Analysis of Flanged section.
13
14 II
Design of T beam, L beam
14
15 II
Analysis of one way slabs
15
16 II
Design of two way slab
16
17 II
Analysis of two way slabs
17
18 II
Calculation of loads of staircases.
18
12-12-2022 GCEKK, BHAWANIPATNA 13
J: Course Plan:
S.NDa
y
Chapter/
Module
Topics To be Covered Topics
Covered
19 III
Analysis of short with axial loading,
19
20 III
Design of short with axial loading,
20
21 III
Analysis of short columns with eccentric loading,
21
22 III
Analysis of long columns
22
23 III
Analysis of long columns
23
24 III
Design of isolated column footing.
24
25 III
Problems on combined footing
25
26 III
Problems on combined footing
26
12-12-2022 GCEKK, BHAWANIPATNA 14
S.NDa
y
Chapter/
Module
Topics To be Covered Topics
Covered
19 III
Analysis of short with axial loading,
19
20 III
Design of short with axial loading,
20
21 III
Analysis of short columns with eccentric loading,
21
22 III
Analysis of long columns
22
23 III
Analysis of long columns
23
24 III
Design of isolated column footing.
24
25 III
Problems on combined footing
25
26 III
Problems on combined footing
26
12-12-2022 GCEKK, BHAWANIPATNA 14
J: Course Plan:
S.
N
Da
y
Chapte
r/
Module
Topics To be Covered Topics
Covere
d
27-
28
IV
Retaining Walls and forces acting , Stability
27-28
29-
30
IV
Design of Cantilever type retaining wall
29-30
31-
32
IV
Design of Cantilever type retaining wall
31-32
33-
34
IV
Design of Counterfort retaining wall
33-34
35-
36
V
Design of ground water tank
35-36
37-
38
V
Design of Ground water tank
37-38
39-
40
V
Design of Elevated water tank
39-40
12-12-2022 GCEKK, BHAWANIPATNA 15
S.
N
Da
y
Chapte
r/
Module
Topics To be Covered Topics
Covere
d
27-
28
IV
Retaining Walls and forces acting , Stability
27-28
29-
30
IV
Design of Cantilever type retaining wall
29-30
31-
32
IV
Design of Cantilever type retaining wall
31-32
33-
34
IV
Design of Counterfort retaining wall
33-34
35-
36
V
Design of ground water tank
35-36
37-
38
V
Design of Ground water tank
37-38
39-
40
V
Design of Elevated water tank
39-40
12-12-2022 GCEKK, BHAWANIPATNA 15
MODULE-1 CLASS -1
•Concrete highest consumed material.
•Italian architect Pontionce remarked that concrete liberated
us from the rectangle
•Due to its flexibility in form and superiority in performance,
it has replaced, to a large extent, the earlier materials like
stone, timber and steel.
•Further, architect's scope and imaginations have widened to
a great extent due to its mouldabilityand monolithicity.
Thus, it has helped the architects and engineers to build
several attractive shell forms and other curved structures.
However, its role in several straight line structural forms like
multistoriedframes, bridges, foundations etc. is enormous.
12-12-2022 GCEKK, BHAWANIPATNA 16
•Concrete highest consumed material.
•Italian architect Pontionce remarked that concrete liberated
us from the rectangle
•Due to its flexibility in form and superiority in performance,
it has replaced, to a large extent, the earlier materials like
stone, timber and steel.
•Further, architect's scope and imaginations have widened to
a great extent due to its mouldabilityand monolithicity.
Thus, it has helped the architects and engineers to build
several attractive shell forms and other curved structures.
However, its role in several straight line structural forms like
multistoriedframes, bridges, foundations etc. is enormous.
12-12-2022 GCEKK, BHAWANIPATNA 16
Objectives of the Design of Reinforced
Concrete Structures
•The structures so designed should have an
acceptable probability of performing
satisfactorily during their intended life.
•The designed structure should sustain all
loads and deform within limits for
construction and use.
•The designed structures should be durable.
•The designed structures should adequately
resist to the effects of misuse and fire.
12-12-2022 GCEKK, BHAWANIPATNA 17
Concrete Structures
•The structures so designed should have an
acceptable probability of performing
satisfactorily during their intended life.
•The designed structure should sustain all
loads and deform within limits for
construction and use.
•The designed structures should be durable.
•The designed structures should adequately
resist to the effects of misuse and fire.
12-12-2022 GCEKK, BHAWANIPATNA 17
How to fulfil the objectives?
Tofulfilobjectiveitisnecessary
•Tounderstand the strength and deformation
characteristicsofthematerialsusedinthedesign
asalso their deterioration under hostile
exposure.
Whatarethematerials:Concrete,Reinforcement
Whatisconcrete
Whyreinforcement
Whatisreinforcement
12-12-2022 GCEKK, BHAWANIPATNA 18
Tofulfilobjectiveitisnecessary
•Tounderstand the strength and deformation
characteristicsofthematerialsusedinthedesign
asalso their deterioration under hostile
exposure.
Whatarethematerials:Concrete,Reinforcement
Whatisconcrete
Whyreinforcement
Whatisreinforcement
12-12-2022 GCEKK, BHAWANIPATNA 18
CLASS-1.1
Properties of concrete ,comp, tensile
and other properties
G.C.Behera
Monday, December 12, 2022 19G.C.BEHERA GCEK BHAWANIPATNA
Properties of concrete ,comp, tensile
and other properties
G.C.Behera
Monday, December 12, 2022 19G.C.BEHERA GCEK BHAWANIPATNA
Comp. Strength of Concrete
•The strength and deformation characteristicsofconcrete
thusdependonthegradeandtypeofcement,aggregates,
admixtures, environmental conditions and curing. The
increaseofstrengthwithitsageduringcuringisconsidered
tobemarginalafter28days.Blendedcements(likeflyash
cement) have slower rateofstrength gain than ordinary
Portland cementasrecognizedbycode, Dependingon
several factors during its preparation, placement and
curing, concrete has a wide rangeofcompressive strength
and the materialisgradedonthe basisofits compressive
strengthon28
th
dayalsoknownas"characteristicstrength"
asdefined below while discussing various strength and
deformationproperties.
Monday, December 12, 2022 20G.C.BEHERA GCEK BHAWANIPATNA
•The strength and deformation characteristicsofconcrete
thusdependonthegradeandtypeofcement,aggregates,
admixtures, environmental conditions and curing. The
increaseofstrengthwithitsageduringcuringisconsidered
tobemarginalafter28days.Blendedcements(likeflyash
cement) have slower rateofstrength gain than ordinary
Portland cementasrecognizedbycode, Dependingon
several factors during its preparation, placement and
curing, concrete has a wide rangeofcompressive strength
and the materialisgradedonthe basisofits compressive
strengthon28
th
dayalsoknownas"characteristicstrength"
asdefined below while discussing various strength and
deformationproperties.
Monday, December 12, 2022 20G.C.BEHERA GCEK BHAWANIPATNA
Characteristic strength
•Characteristicstrengthisdefinedasthestrengthbelowwhichnot
more than fivepercentofthe test results are expectedtofall.
Concreteisgradedonthe basisofits characteristic compressive
strengthof150mmsizecubeat28daysandexpressedinN/mm
2
.
•ThegradesaredesignatedbyoneletterM(formix)andanumber
from10to80indicating the characteristic compressive strength
(f
ck)inN/mm
2
.AsperIS456 (Table 2),concretehasthree groups
as
•(i)ordinaryconcrete(M10toM20),
•(ii)standardconcrete(M25toM55)and
•(iii)highstrengthconcrete(M60toM80).
Thesizeofspecimen for determining characteristic strength may
bedifferentindifferentcountries.
Monday, December 12, 2022 21G.C.BEHERA GCEK BHAWANIPATNA
•Characteristicstrengthisdefinedasthestrengthbelowwhichnot
more than fivepercentofthe test results are expectedtofall.
Concreteisgradedonthe basisofits characteristic compressive
strengthof150mmsizecubeat28daysandexpressedinN/mm
2
.
•ThegradesaredesignatedbyoneletterM(formix)andanumber
from10to80indicating the characteristic compressive strength
(f
ck)inN/mm
2
.AsperIS456 (Table 2),concretehasthree groups
as
•(i)ordinaryconcrete(M10toM20),
•(ii)standardconcrete(M25toM55)and
•(iii)highstrengthconcrete(M60toM80).
Thesizeofspecimen for determining characteristic strength may
bedifferentindifferentcountries.
Monday, December 12, 2022 21G.C.BEHERA GCEK BHAWANIPATNA
Procedure for Compressive strength
•IS 516-1959
•Temp-(27±3)
0
C
•Mixing by Machine power loaded
•Add all the mixing water into the drum before solid material.
•solid materials: the skip shall be loaded with about half of coarse
aggregate, then with the fine aggregate, then with the cement and
then with the remaining coarse aggregate on top.
Mixing by drum hand operated
•materials in a similar manner, and the water shall be added
immediately before the rotation of the drum is started.
•The period of mixing shall be not less than 2 minutes after all the
materials are in the drum, and shall continue till the resulting
concrete is uniform in appearance. When using
•pan mixers, the concrete shall be heaped together before sampling.
Monday, December 12, 2022 22G.C.BEHERA GCEK BHAWANIPATNA
•IS 516-1959
•Temp-(27±3)
0
C
•Mixing by Machine power loaded
•Add all the mixing water into the drum before solid material.
•solid materials: the skip shall be loaded with about half of coarse
aggregate, then with the fine aggregate, then with the cement and
then with the remaining coarse aggregate on top.
Mixing by drum hand operated
•materials in a similar manner, and the water shall be added
immediately before the rotation of the drum is started.
•The period of mixing shall be not less than 2 minutes after all the
materials are in the drum, and shall continue till the resulting
concrete is uniform in appearance. When using
•pan mixers, the concrete shall be heaped together before sampling.
Monday, December 12, 2022 22G.C.BEHERA GCEK BHAWANIPATNA
Hand Mixing --
Theconcrete batch shallbemixedona watertight,non-absorbent
platform with a shovel, trowelorsimilar suitable implement, using
thefollowingprocedure:
•a)The cement and fine aggregate shallbemixed dry until the
mixture
isthoroughlyblendedandisuniformincolour,
•b)Thecoarse aggregate shall thenbeadded and mixed with the
cement and fine aggregate until the coarse aggregateisuniformly
distributedthroughoutthebatch,and
•c)Thewater shall thenbeadded and the entire batch mixed until
the concrete appearstobehomogeneous and has the desired
consistency.Ifrepeatedmixingisnecessary,becauseoftheaddition
ofwaterinincrements while adjusting the consistency, the batch
shallbediscardedanda freshbatch made without interruptingthe
mixingtomaketrialconsistencytests.
Monday, December 12, 2022 23G.C.BEHERA GCEK BHAWANIPATNA
Theconcrete batch shallbemixedona watertight,non-absorbent
platform with a shovel, trowelorsimilar suitable implement, using
thefollowingprocedure:
•a)The cement and fine aggregate shallbemixed dry until the
mixture
isthoroughlyblendedandisuniformincolour,
•b)Thecoarse aggregate shall thenbeadded and mixed with the
cement and fine aggregate until the coarse aggregateisuniformly
distributedthroughoutthebatch,and
•c)Thewater shall thenbeadded and the entire batch mixed until
the concrete appearstobehomogeneous and has the desired
consistency.Ifrepeatedmixingisnecessary,becauseoftheaddition
ofwaterinincrements while adjusting the consistency, the batch
shallbediscardedanda freshbatch made without interruptingthe
mixingtomaketrialconsistencytests.
Monday, December 12, 2022 23G.C.BEHERA GCEK BHAWANIPATNA
•Testspecimenscubicalinshapeshallbe15X15X
15cm. If the largest nominal sizeofthe
aggregate does notexceed 2cn,10 cmcubes
maybeusedasanalternative.
Cylindrical test
•specimensshallhavealengthequaltotwicethe
diameter.Theyshallbe15cmindiameterand30
cmlong.
•Tamping Bar - The tamping bar shallbea steel
bar16mmdiameter, 0·6 m long and bullet
pointedatthelowerend.
Monday, December 12, 2022 24G.C.BEHERA GCEK BHAWANIPATNA
15cm. If the largest nominal sizeofthe
aggregate does notexceed 2cn,10 cmcubes
maybeusedasanalternative.
Cylindrical test
•specimensshallhavealengthequaltotwicethe
diameter.Theyshallbe15cmindiameterand30
cmlong.
•Tamping Bar - The tamping bar shallbea steel
bar16mmdiameter, 0·6 m long and bullet
pointedatthelowerend.
Monday, December 12, 2022 24G.C.BEHERA GCEK BHAWANIPATNA
MOULD
•The mould shallbe ofmetal, preferably steelorcast iron,andstout
enoughtoprevent distortion.Itshallbeconstructedinsuch a manneras
tofacilitate the removalofthe moulded specimen without damage.
•The heightofthe mould and the distance between opposite faces shall
bethe specified size + 0·2mm.
•The angle between adjacent internal faces and between internal faces
andtop and bottom planesofthe mould shallbe 90°+0-5°. The interior
facesofthe mould shallbeplane surfaces with a permissible variationof
0·03mm.
•Cylinders--The cylindrical mould shallbe ofmetal which shallbenot
less than 3mmthick. Each mould shallbecapableofbeing opened
longitudinallytofacilitate removalofthe specimen and shallbeprovided
with a meansofkeepingitclosed whileinuse. The ends shallnotdepart
from a plane surface, perpendiculartothe axisofthe mould,bymore
than 0·05mm. When assembled ready for use, the mean internal
diameterofthe mould shallbe 15·0cm±0·2mmandinnodirection
shall the internal diameterbeless than14·95cmormore than15·05
cm.The height shallbe 30-0cm±0-1cm.
Monday, December 12, 2022 25G.C.BEHERA GCEK BHAWANIPATNA
•The mould shallbe ofmetal, preferably steelorcast iron,andstout
enoughtoprevent distortion.Itshallbeconstructedinsuch a manneras
tofacilitate the removalofthe moulded specimen without damage.
•The heightofthe mould and the distance between opposite faces shall
bethe specified size + 0·2mm.
•The angle between adjacent internal faces and between internal faces
andtop and bottom planesofthe mould shallbe 90°+0-5°. The interior
facesofthe mould shallbeplane surfaces with a permissible variationof
0·03mm.
•Cylinders--The cylindrical mould shallbe ofmetal which shallbenot
less than 3mmthick. Each mould shallbecapableofbeing opened
longitudinallytofacilitate removalofthe specimen and shallbeprovided
with a meansofkeepingitclosed whileinuse. The ends shallnotdepart
from a plane surface, perpendiculartothe axisofthe mould,bymore
than 0·05mm. When assembled ready for use, the mean internal
diameterofthe mould shallbe 15·0cm±0·2mmandinnodirection
shall the internal diameterbeless than14·95cmormore than15·05
cm.The height shallbe 30-0cm±0-1cm.
Monday, December 12, 2022 25G.C.BEHERA GCEK BHAWANIPATNA
COMPACTION
•CompactingbyHand–InuniformmannerFor
cubical specimens,innocase shall the
concretebesubjectedtolessthan35strokes
per layer for15cmcubesor25strokes per
layer for10cmcubes. For cylindrical
specimens,thenumberofstrokesshallnotbe
less than thirty per layer. The strokes shall
penetrate into the underlying layer and the
bottom layer shallberodded throughout its
depth. Where voids arc leftbythe tamping
bar,thesidesofthemouldshallbetappedto
closethevoids.
Monday, December 12, 2022 26G.C.BEHERA GCEK BHAWANIPATNA
•CompactingbyHand–InuniformmannerFor
cubical specimens,innocase shall the
concretebesubjectedtolessthan35strokes
per layer for15cmcubesor25strokes per
layer for10cmcubes. For cylindrical
specimens,thenumberofstrokesshallnotbe
less than thirty per layer. The strokes shall
penetrate into the underlying layer and the
bottom layer shallberodded throughout its
depth. Where voids arc leftbythe tamping
bar,thesidesofthemouldshallbetappedto
closethevoids.
Monday, December 12, 2022 26G.C.BEHERA GCEK BHAWANIPATNA
Compaction by vibration
•When compactingbyvibration, eachlayer
shallbevibratedbymeansofanelectricor
pneumatichammerorvibratororbymeansof
a suitable vibrating table until the specified
conditionisattained.
•CappingSpecimens-Theendsofallcylindrical
testspecimensthatarenotplanewithin0·05
mmshallbecapped.
Monday, December 12, 2022 27G.C.BEHERA GCEK BHAWANIPATNA
•When compactingbyvibration, eachlayer
shallbevibratedbymeansofanelectricor
pneumatichammerorvibratororbymeansof
a suitable vibrating table until the specified
conditionisattained.
•CappingSpecimens-Theendsofallcylindrical
testspecimensthatarenotplanewithin0·05
mmshallbecapped.
Monday, December 12, 2022 27G.C.BEHERA GCEK BHAWANIPATNA
Curing -
•Thetestspecimen-shallbestoredinaplace,freefromvibration,in
moist airofatleast90percent relative humidity andata
temperatureof27° ± 2°e for24hours ± l hour from the timeof
additionofwatertothedryingredients.
•Afterthisperiod,thespecimensshallbemarkedandremovedfrom
the moulds and, unless required for test within24hours,
immediately submergedinclean, fresh waterorsaturated lime
solutionandkeptthereuntiltakenoutjustpriortotest.
•Thewaterorsolutioninwhich the specimens are submerged shall
berenewed every seven days and shallbemaintainedata
temperatureof27° ± 2°C.Thespecimens shall notbealk.weoto
becomedryatanytimeuntiltheyhavebeentested.
Monday, December 12, 2022 28G.C.BEHERA GCEK BHAWANIPATNA
•Thetestspecimen-shallbestoredinaplace,freefromvibration,in
moist airofatleast90percent relative humidity andata
temperatureof27° ± 2°e for24hours ± l hour from the timeof
additionofwatertothedryingredients.
•Afterthisperiod,thespecimensshallbemarkedandremovedfrom
the moulds and, unless required for test within24hours,
immediately submergedinclean, fresh waterorsaturated lime
solutionandkeptthereuntiltakenoutjustpriortotest.
•Thewaterorsolutioninwhich the specimens are submerged shall
berenewed every seven days and shallbemaintainedata
temperatureof27° ± 2°C.Thespecimens shall notbealk.weoto
becomedryatanytimeuntiltheyhavebeentested.
Monday, December 12, 2022 28G.C.BEHERA GCEK BHAWANIPATNA
TESTING
•ThetestingmachineCTMORUTM
•The permissible error shallbenot greater than ± 2
percentofthemaximumload.
•specimens,themostusualbeing7and28days.Agesof
13weeks and one year are recommendediftestsat
greateragesarerequired.Whereitmaybenecessary
toobtaintheearlystrengths.testsmaybemadeatthe
agesof24hours±1/2hourand72hours±2hours.
•The ages shallbecalculated from the timeofthe
additionofwatertothedryingredients.
•NumberofSpecimen -Atleast three specimens,
preferably from different batches, shallbemade for
testingateachselectedage.
Monday, December 12, 2022 29G.C.BEHERA GCEK BHAWANIPATNA
•ThetestingmachineCTMORUTM
•The permissible error shallbenot greater than ± 2
percentofthemaximumload.
•specimens,themostusualbeing7and28days.Agesof
13weeks and one year are recommendediftestsat
greateragesarerequired.Whereitmaybenecessary
toobtaintheearlystrengths.testsmaybemadeatthe
agesof24hours±1/2hourand72hours±2hours.
•The ages shallbecalculated from the timeofthe
additionofwatertothedryingredients.
•NumberofSpecimen -Atleast three specimens,
preferably from different batches, shallbemade for
testingateachselectedage.
Monday, December 12, 2022 29G.C.BEHERA GCEK BHAWANIPATNA
•Inthe caseofcubes, the specimen shallbeplacedInthe
machineInsuchamannerthattheloadshallbeappliedto
opposite sidesofthe cubesascast, nottothe top and
bottom.
•The load shallbeapplied without shock and Increased
continuouslyata rateofapproximately 140 kg/sqcm/min
until the resistanceofthe specimentothe Increasing load
breaksdownandnogreaterloadcanbesustained.
•The measured compressive strengthofthe specimen shall
becalculatedbydividingthemaximumloadappliedtothe
specimenduringthetestbythecross-sectionalarea.
•Averageofthreevaluesshallbetakenastherepresentative
ofthe batch provided the Individual variationisnot more
than±15percentoftheaverage.Otherwiserepeattests.
Monday, December 12,
2022
30G.C.BEHERA GCEK BHAWANIPATNA
machineInsuchamannerthattheloadshallbeappliedto
opposite sidesofthe cubesascast, nottothe top and
bottom.
•The load shallbeapplied without shock and Increased
continuouslyata rateofapproximately 140 kg/sqcm/min
until the resistanceofthe specimentothe Increasing load
breaksdownandnogreaterloadcanbesustained.
•The measured compressive strengthofthe specimen shall
becalculatedbydividingthemaximumloadappliedtothe
specimenduringthetestbythecross-sectionalarea.
•Averageofthreevaluesshallbetakenastherepresentative
ofthe batch provided the Individual variationisnot more
than±15percentoftheaverage.Otherwiserepeattests.
Monday, December 12,
2022
30G.C.BEHERA GCEK BHAWANIPATNA
(b) Other strengths of concrete
•Inadditiontoitsgoodcompressivestrength,
concrete has flexural and splitting tensile
strengthstoo.Theflexuralandsplittingtensile
strengthsareobtainedasdescribedinIS516
andIS5816, respectively. However, the
following expression givesanestimationof
flexural strength (f
cr)ofconcrete fromits
characteristiccompressivestrength(cl.6.2.2)
Monday, December 12, 2022 31G.C.BEHERA GCEK BHAWANIPATNA
•Inadditiontoitsgoodcompressivestrength,
concrete has flexural and splitting tensile
strengthstoo.Theflexuralandsplittingtensile
strengthsareobtainedasdescribedinIS516
andIS5816, respectively. However, the
following expression givesanestimationof
flexural strength (f
cr)ofconcrete fromits
characteristiccompressivestrength(cl.6.2.2)
Monday, December 12, 2022 31G.C.BEHERA GCEK BHAWANIPATNA
•Thestandardsizeshallbe15xISx70cmor
Alternatively,ifthelargestnominalsizeofthe
aggregatedocsnotexceed19mm,specimens
10x10x5Ocmmaybeused.
•Tamping Bar -The tamping bar shall be a steel
bar weighing 2 kg, 40 cm long, and shall have
a ramming face 25 mm square.
Monday, December 12, 2022 32G.C.BEHERA GCEK BHAWANIPATNA
Alternatively,ifthelargestnominalsizeofthe
aggregatedocsnotexceed19mm,specimens
10x10x5Ocmmaybeused.
•Tamping Bar -The tamping bar shall be a steel
bar weighing 2 kg, 40 cm long, and shall have
a ramming face 25 mm square.
Monday, December 12, 2022 32G.C.BEHERA GCEK BHAWANIPATNA
•Thebedofthetestingmachineshallbeprovidedwithtwo
steelrollers,38mmindiameter,onwhichthespecimenis
tobesupported,andthese
•rollersshallbesomountedthatthedistancefromcentreto
centreis60cmfor15·0cmspecimensor40cmfor10·0cm
specimens, The load shallbeapplied through two similar
roller",mountedatthethirdpointsofthesupportingspan,
thatis,spacedat20or13·3cmcentretocentre.
The load shall be applied
•without shock and increasing continuouslyata rate such
that the extreme fibre stress increasesatapproximately 7
kg/sqcn.rmm.thatis,atarateofloadingof400kg/minfor
the15·0cmspecimen-andatarateof180kg/minforthe
10·0cmspecimens.
Monday, December 12, 2022 33G.C.BEHERA GCEK BHAWANIPATNA
steelrollers,38mmindiameter,onwhichthespecimenis
tobesupported,andthese
•rollersshallbesomountedthatthedistancefromcentreto
centreis60cmfor15·0cmspecimensor40cmfor10·0cm
specimens, The load shallbeapplied through two similar
roller",mountedatthethirdpointsofthesupportingspan,
thatis,spacedat20or13·3cmcentretocentre.
The load shall be applied
•without shock and increasing continuouslyata rate such
that the extreme fibre stress increasesatapproximately 7
kg/sqcn.rmm.thatis,atarateofloadingof400kg/minfor
the15·0cmspecimen-andatarateof180kg/minforthe
10·0cmspecimens.
Monday, December 12, 2022 33G.C.BEHERA GCEK BHAWANIPATNA
•when aisgreater than20cmfor15·0cm
specimen,orgreaterthan13.3cmfora10·0
cmspecimen,
•whenaislessthan20cmbutgreaterthan17
cmfor15·0cmspecimen,orlessthan13.3cm
butgreaterthanfora10·0cmspecimen
Monday, December 12, 2022 34G.C.BEHERA GCEK BHAWANIPATNA
specimen,orgreaterthan13.3cmfora10·0
cmspecimen,
•whenaislessthan20cmbutgreaterthan17
cmfor15·0cmspecimen,orlessthan13.3cm
butgreaterthanfora10·0cmspecimen
Monday, December 12, 2022 34G.C.BEHERA GCEK BHAWANIPATNA
Elastic Modulus
•E
c= initial tangent
modulusatthe origin,
also knownasshort term
staticmodulus
•E
s= secant modulus at A
•E
t= tangent modulus atA
•ε
e= elastic strain at A
•ε
i= inelastic strain at A
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 35
•E
c= initial tangent
modulusatthe origin,
also knownasshort term
staticmodulus
•E
s= secant modulus at A
•E
t= tangent modulus atA
•ε
e= elastic strain at A
•ε
i= inelastic strain at A
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 35
(d) Shrinkage of concrete
•Shrinkageisthe time dependent deformation,
generally compressiveinnature. The constituentsof
concrete, sizeofthe member and environmental
conditionsarethefactorsonwhichthetotalshrinkage
ofconcrete depends. However, the total shrinkageof
concreteismost influencedbythe total amountof
waterpresentintheconcreteatthetimeofmixingfor
a given humidity and temperature. The cement
content, however, influences the total shrinkageof
concretetoa lesser extent. The approximate valueof
thetotalshrinkagestrainfordesignistakenas0.0003
intheabsenceoftestdata.
Monday, December 12, 2022 36G.C.BEHERA GCEK BHAWANIPATNA
•Shrinkageisthe time dependent deformation,
generally compressiveinnature. The constituentsof
concrete, sizeofthe member and environmental
conditionsarethefactorsonwhichthetotalshrinkage
ofconcrete depends. However, the total shrinkageof
concreteismost influencedbythe total amountof
waterpresentintheconcreteatthetimeofmixingfor
a given humidity and temperature. The cement
content, however, influences the total shrinkageof
concretetoa lesser extent. The approximate valueof
thetotalshrinkagestrainfordesignistakenas0.0003
intheabsenceoftestdata.
Monday, December 12, 2022 36G.C.BEHERA GCEK BHAWANIPATNA
CREEP
•Creepisanothertimedependentdeformationofconcreteby
whichitcontinuestodeform, usually under compressive
stress.Thecreepstrainsrecoverpartlywhenthestressesare
released. creep recovery in two parts. The elastic recoveryis
immediateandthecreeprecoveryisslowinnature.
The creep of concrete is influenced by
• Properties of concrete
• Water/cement ratio
• Humidity and temperature of curing
• Humidity during the period of use
• Age of concrete at first loading
• Magnitude of stress and its duration
• Surface-volume ratio of the member
Monday, December 12, 2022 37G.C.BEHERA GCEK BHAWANIPATNA
•Creepisanothertimedependentdeformationofconcreteby
whichitcontinuestodeform, usually under compressive
stress.Thecreepstrainsrecoverpartlywhenthestressesare
released. creep recovery in two parts. The elastic recoveryis
immediateandthecreeprecoveryisslowinnature.
The creep of concrete is influenced by
• Properties of concrete
• Water/cement ratio
• Humidity and temperature of curing
• Humidity during the period of use
• Age of concrete at first loading
• Magnitude of stress and its duration
• Surface-volume ratio of the member
Monday, December 12, 2022 37G.C.BEHERA GCEK BHAWANIPATNA
Workability and Durability of Concrete
•Workability
•Itisthe property which determines the ease and
homogeneity with which concrete canbemixed,
placed, compacted and finished. A workable concrete
willnothaveanysegregationorbleeding.Segregation
causes large voids and hence concrete becomes less
durable.Bleedingresultsinseveralsmallporesonthe
surfaceduetoexcess water comingup. Bleeding also
makesconcretelessdurable.Thedegreeofworkability
ofconcreteisclassifiedfromverylowtoveryhighwith
thecorrespondingvalueofslumpinmm
Monday, December 12, 2022 38G.C.BEHERA GCEK BHAWANIPATNA
•Workability
•Itisthe property which determines the ease and
homogeneity with which concrete canbemixed,
placed, compacted and finished. A workable concrete
willnothaveanysegregationorbleeding.Segregation
causes large voids and hence concrete becomes less
durable.Bleedingresultsinseveralsmallporesonthe
surfaceduetoexcess water comingup. Bleeding also
makesconcretelessdurable.Thedegreeofworkability
ofconcreteisclassifiedfromverylowtoveryhighwith
thecorrespondingvalueofslumpinmm
Monday, December 12, 2022 38G.C.BEHERA GCEK BHAWANIPATNA
Durability of concrete
•A durable concrete performs satisfactorilyin
the working environment during its
anticipated exposure conditions during
service.Thedurableconcreteshouldhavelow
permeability with adequate cement content,
sufficient low free water/cement ratio and
ensuredcompletecompactionofconcreteby
adequatecuring.Formoreinformation,please
refertocl.8ofIS456.
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 39
•A durable concrete performs satisfactorilyin
the working environment during its
anticipated exposure conditions during
service.Thedurableconcreteshouldhavelow
permeability with adequate cement content,
sufficient low free water/cement ratio and
ensuredcompletecompactionofconcreteby
adequatecuring.Formoreinformation,please
refertocl.8ofIS456.
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 39
Design mix and nominal mix concrete
•Indesign mix, the proportionsofcement,
aggregates (sand and gravel), water and
mineral admixtures,ifany, are actually
designed, whileinnominal mix, the
proportions are nominally adopted. The
designmixconcreteispreferredtothe
nominalmixastheformerresultsinthegrade
ofconcrete having the specified workability
and characteristic strength (videcl. 9ofIS
456).
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 40
•Indesign mix, the proportionsofcement,
aggregates (sand and gravel), water and
mineral admixtures,ifany, are actually
designed, whileinnominal mix, the
proportions are nominally adopted. The
designmixconcreteispreferredtothe
nominalmixastheformerresultsinthegrade
ofconcrete having the specified workability
and characteristic strength (videcl. 9ofIS
456).
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 40
Batching
•Mass and volume are the two typesof
batchingformeasuringcement,sand,coarse
aggregates, admixtures and water. Coarse
aggregatesmaybegravel,gradestonechipsor
othermanmadeaggregates.Thequantitiesof
cement, sand, coarse aggregates and solid
admixturesshallbemeasuredbymass.Liquid
admixturesandwateraremeasuredeitherby
volumeorbymass(cl.10ofIS456).
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 41
•Mass and volume are the two typesof
batchingformeasuringcement,sand,coarse
aggregates, admixtures and water. Coarse
aggregatesmaybegravel,gradestonechipsor
othermanmadeaggregates.Thequantitiesof
cement, sand, coarse aggregates and solid
admixturesshallbemeasuredbymass.Liquid
admixturesandwateraremeasuredeitherby
volumeorbymass(cl.10ofIS456).
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 41
Properties of Steel
•Steelisusedasthereinforcingmaterialinconcreteto
makeitgoodintension. Steelassuchisgoodin
tensionaswellasincompression. Unlike concrete,
steel reinforcement rods are producedinsteel plants.
Moreover, the reinforcing barsorrods are
commercially availableinsome specific diameters.
Normally, steel barsupto12mmindiameter are
designatedasbars which canbecoiled for
transportation.Barsmorethan12mmindiameterare
termedasrods and they are transportedinstandard
lengths.
•Like concrete, steel also has several typesorgrades.
The four typesofsteel usedinconcrete structuresas
specifiedincl.5.6ofIS456aregivenbelow:
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 42
•Steelisusedasthereinforcingmaterialinconcreteto
makeitgoodintension. Steelassuchisgoodin
tensionaswellasincompression. Unlike concrete,
steel reinforcement rods are producedinsteel plants.
Moreover, the reinforcing barsorrods are
commercially availableinsome specific diameters.
Normally, steel barsupto12mmindiameter are
designatedasbars which canbecoiled for
transportation.Barsmorethan12mmindiameterare
termedasrods and they are transportedinstandard
lengths.
•Like concrete, steel also has several typesorgrades.
The four typesofsteel usedinconcrete structuresas
specifiedincl.5.6ofIS456aregivenbelow:
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 42
(i) Mild steel and medium tensile steel bars conforming to
IS 432 (Part 1)
(ii) High yield strength deformed (HYSD) steel bars
conforming to IS 1786
(iii) Hard-drawn steel wire fabric conforming to IS 1566
(iv) Structural steel conforming to Grade A of IS 2062.
Mild steel bars had been progressively replacedby
HYSDbarsandsubsequentlyTMTbarsarepromotedin
our country. The implicationsofadopting different
kindsofblended cement and reinforcing steel should
beexaminedbeforeadopting.
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 43
IS 432 (Part 1)
(ii) High yield strength deformed (HYSD) steel bars
conforming to IS 1786
(iii) Hard-drawn steel wire fabric conforming to IS 1566
(iv) Structural steel conforming to Grade A of IS 2062.
Mild steel bars had been progressively replacedby
HYSDbarsandsubsequentlyTMTbarsarepromotedin
our country. The implicationsofadopting different
kindsofblended cement and reinforcing steel should
beexaminedbeforeadopting.
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 43
•Until the relevant Indian Standard specification
for reinforcing steel are modifiedtoinclude the
conceptofcharacteristic strength, the
characteristic value shallbeassumedasthe
minimum yield stressor0.2% proof stress
specifiedinthe relevant Indian Standard
specification.Thecharacteristicstrengthofsteel
designatedbysymbolfy(N/mm
2
)
•
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 44
for reinforcing steel are modifiedtoinclude the
conceptofcharacteristic strength, the
characteristic value shallbeassumedasthe
minimum yield stressor0.2% proof stress
specifiedinthe relevant Indian Standard
specification.Thecharacteristicstrengthofsteel
designatedbysymbolfy(N/mm
2
)
•
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 44
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 45
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 46
•Figuresshowtherepresentativestress-straincurvesforsteel
havingdefiniteyieldpointandnothavingdefiniteyieldpoint,
respectively. The characteristic yield strengthf
yofsteelis
assumedastheminimumyieldstressor0.2percentofproof
stressforsteelhavingnodefiniteyieldpoint.Themodulusof
elasticityofsteelistakentobe200000N/mm
2
.
•Formildsteel(Fig.Slide-26),thestressisproportionaltothe
strainuptothe yield point. Thereafter, post yield strain
increases faster while the stressisassumedtoremainat
constantvalueoff
y.
•For cold-worked bars(Slide27),the stressisproportionalto
the strainuptoa stressof0.8f
y. Thereafter, the inelastic
curveisdefinedasgivenbelow:
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 47
havingdefiniteyieldpointandnothavingdefiniteyieldpoint,
respectively. The characteristic yield strengthf
yofsteelis
assumedastheminimumyieldstressor0.2percentofproof
stressforsteelhavingnodefiniteyieldpoint.Themodulusof
elasticityofsteelistakentobe200000N/mm
2
.
•Formildsteel(Fig.Slide-26),thestressisproportionaltothe
strainuptothe yield point. Thereafter, post yield strain
increases faster while the stressisassumedtoremainat
constantvalueoff
y.
•For cold-worked bars(Slide27),the stressisproportionalto
the strainuptoa stressof0.8f
y. Thereafter, the inelastic
curveisdefinedasgivenbelow:
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 47
•For cold-worked bars (slide-27),
the stressisproportionaltothe
strainuptoa stressof0.8f
y.
Thereafter, the inelastic curveis
definedasgiveninfig:
•Linearinterpolationistobedone
for intermediate values. The two
gradesofcold-worked bars used
assteelreinforcementareFe415
andFe500with the valuesoff
y
as415 N/mm
2
and500 N/mm
2
,
respectively.
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 48
the stressisproportionaltothe
strainuptoa stressof0.8f
y.
Thereafter, the inelastic curveis
definedasgiveninfig:
•Linearinterpolationistobedone
for intermediate values. The two
gradesofcold-worked bars used
assteelreinforcementareFe415
andFe500with the valuesoff
y
as415 N/mm
2
and500 N/mm
2
,
respectively.
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 48
MILD STEEL
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 49
Monday, December 12, 2022 G.C.BEHERA GCEK BHAWANIPATNA 49
DESIGN PRINCIPLE
Monday, December 12, 2022 50G.C.BEHERA GCEK BHAWANIPATNA
Monday, December 12, 2022 50G.C.BEHERA GCEK BHAWANIPATNA
DESIGN PRINCIPLE
Monday, December 12, 2022 51G.C.BEHERA GCEK BHAWANIPATNA
Monday, December 12, 2022 51G.C.BEHERA GCEK BHAWANIPATNA
CLASS-2
Objective of design , concept and
methods of reinforced concrete
design
Objective of design , concept and
methods of reinforced concrete
design
Objectives of the Design of Reinforced
Concrete Structures
•The structures so designed should have an
acceptable probability of performing
satisfactorily during their intended life.
•The designed structure should sustain all
loads and deform within limits for
construction and use.
•The designed structures should be durable.
•The designed structures should adequately
resist to the effects of misuse and fire.
12-12-2022 GCEKK, BHAWANIPATNA 53
Concrete Structures
•The structures so designed should have an
acceptable probability of performing
satisfactorily during their intended life.
•The designed structure should sustain all
loads and deform within limits for
construction and use.
•The designed structures should be durable.
•The designed structures should adequately
resist to the effects of misuse and fire.
12-12-2022 GCEKK, BHAWANIPATNA 53
How to fulfil the objectives?
Tofulfilobjectiveitisnecessary
•Tounderstand the strength and deformation
characteristicsofthematerialsusedinthedesign
asalso their deterioration under hostile
exposure.
Whatarethematerials:Concrete,Reinforcement
Whatisconcrete
Whyreinforcement
Whatisreinforcement
12-12-2022 GCEKK, BHAWANIPATNA 54
Tofulfilobjectiveitisnecessary
•Tounderstand the strength and deformation
characteristicsofthematerialsusedinthedesign
asalso their deterioration under hostile
exposure.
Whatarethematerials:Concrete,Reinforcement
Whatisconcrete
Whyreinforcement
Whatisreinforcement
12-12-2022 GCEKK, BHAWANIPATNA 54
In any method of design, the following are the common steps to be
followed:
(i) To assess the dead loads and other external loads and forces likely
to be applied on the structure,
(ii) To determine the design loads from different combinations of
loads,
(iii) To estimate structural responses (bending moment, shear force,
axial thrust etc.) due to the design loads,
(iv) To determine the cross-sectional areas of concrete sections and
amounts of reinforcement needed.
followed:
(i) To assess the dead loads and other external loads and forces likely
to be applied on the structure,
(ii) To determine the design loads from different combinations of
loads,
(iii) To estimate structural responses (bending moment, shear force,
axial thrust etc.) due to the design loads,
(iv) To determine the cross-sectional areas of concrete sections and
amounts of reinforcement needed.
Method of Design
•Three methodsofdesign are acceptedincl.
18.2ofIS456:2000(IndianStandardPlainand
Reinforced Concrete - CodeofPractice,
publishedbytheBureauofIndianStandards,
NewDelhi).Theyareasfollows:
•LIMITSTATEMETHOD
•WORKKINGSTATEMETHOD
•METHODBASEDONEXPERIMENTALRESULT
•Three methodsofdesign are acceptedincl.
18.2ofIS456:2000(IndianStandardPlainand
Reinforced Concrete - CodeofPractice,
publishedbytheBureauofIndianStandards,
NewDelhi).Theyareasfollows:
•LIMITSTATEMETHOD
•WORKKINGSTATEMETHOD
•METHODBASEDONEXPERIMENTALRESULT
•Manyofthe above steps have lotofuncertainties.
Estimationofloads and evaluationofmaterial properties
aretoname a few. Hence, some suitable factorsofsafety
shouldbetaken into consideration dependingonthe
degreesofsuchuncertainties.
•Limitstatemethodisoneofthethreemethodsofdesignas
perIS456:2000. The code hasputmore emphasisonthis
methodbypresentingit ina full section (Section 5),while
accommodating the working stress methodinAnnex Bof
the code (IS456). Considering rapid developmentin
concrete technology and simultaneous developmentin
handlingproblemsofuncertainties,thelimitstatemethod
isasuperiormethodwherecertainaspectsofrealitycanbe
explainedinabettermanner.
Estimationofloads and evaluationofmaterial properties
aretoname a few. Hence, some suitable factorsofsafety
shouldbetaken into consideration dependingonthe
degreesofsuchuncertainties.
•Limitstatemethodisoneofthethreemethodsofdesignas
perIS456:2000. The code hasputmore emphasisonthis
methodbypresentingit ina full section (Section 5),while
accommodating the working stress methodinAnnex Bof
the code (IS456). Considering rapid developmentin
concrete technology and simultaneous developmentin
handlingproblemsofuncertainties,thelimitstatemethod
isasuperiormethodwherecertainaspectsofrealitycanbe
explainedinabettermanner.
LIMIT STATE
•Limit states are the acceptable limits for the safety and
serviceability requirementsofthe structure before failure
occurs. The designofstructuresbythis method will thus
ensure that theywillnot reach limit states andwillnot
becomeunfitfortheuseforwhichtheyareintended.Itis
worth mentioning that structures will not just failor
collapsebyviolating (exceeding) the limit states. Failure,
therefore, implies that clearly defined limit statesof
structuralusefulnesshasbeenexceeded.
•Limit stateofcollapse was found / detailedinseveral
countriesincontinent fifty years ago.In1960Soviet Code
recognized three limit states: (i) deformation, (ii) cracking
and(iii)collapse.
•Limit states are the acceptable limits for the safety and
serviceability requirementsofthe structure before failure
occurs. The designofstructuresbythis method will thus
ensure that theywillnot reach limit states andwillnot
becomeunfitfortheuseforwhichtheyareintended.Itis
worth mentioning that structures will not just failor
collapsebyviolating (exceeding) the limit states. Failure,
therefore, implies that clearly defined limit statesof
structuralusefulnesshasbeenexceeded.
•Limit stateofcollapse was found / detailedinseveral
countriesincontinent fifty years ago.In1960Soviet Code
recognized three limit states: (i) deformation, (ii) cracking
and(iii)collapse.
(i)Limitstateofcollapsedealswiththe
strength and stabilityofstructures
subjectedtothe maximum design
loads out ofthe possible
combinationsofseveral typesof
loads. Therefore, this limit state
ensures that neither any part nor
the whole structure should collapse
orbecome unstable under any
combinationofexpectedoverloads.
(ii) Limit stateofserviceability deals
with deflection and crackingof
structures under service loads,
durability under working
environment during their
anticipated exposure conditions
during service, stabilityofstructures
asawhole,fireresistanceetc.
The two main limit states
strength and stabilityofstructures
subjectedtothe maximum design
loads out ofthe possible
combinationsofseveral typesof
loads. Therefore, this limit state
ensures that neither any part nor
the whole structure should collapse
orbecome unstable under any
combinationofexpectedoverloads.
(ii) Limit stateofserviceability deals
with deflection and crackingof
structures under service loads,
durability under working
environment during their
anticipated exposure conditions
during service, stabilityofstructures
asawhole,fireresistanceetc.
The two main limit states
LIMIT STATE METHOD
•Theterm“Limitstates”isofcontinentaloriginwheretherearetwo
limit states - serviceability / collapse. For reasons not very clear,in
Englishliteraturelimitstateofcollapseistermedaslimitstate.
•limit state methodofdesign has been foundtobethe best for the
designofreinforced concrete members. More detailsofthis
methodareexplainedinModule3(Lesson4).However,becauseof
itssuperioritytoothertwomethods(seesections2.3.2and2.3.3of
Lesson 3),IS456:2000 hasbeenthoroughly updatedinits fourth
revisionin2000takinginto considerationtherapiddevelopmentin
the fieldofconcrete technology and incorporating important
aspects like durability etc.Thisstandard hasputgreater emphasis
tolimit state methodofdesignbypresentingitina full section
(section 5),while the working stress method has been givenin
Annex Bofthe same standard. Accordingly, structuresorstructural
elementsshallnormallybedesignedbylimitstatemethod.
•Theterm“Limitstates”isofcontinentaloriginwheretherearetwo
limit states - serviceability / collapse. For reasons not very clear,in
Englishliteraturelimitstateofcollapseistermedaslimitstate.
•limit state methodofdesign has been foundtobethe best for the
designofreinforced concrete members. More detailsofthis
methodareexplainedinModule3(Lesson4).However,becauseof
itssuperioritytoothertwomethods(seesections2.3.2and2.3.3of
Lesson 3),IS456:2000 hasbeenthoroughly updatedinits fourth
revisionin2000takinginto considerationtherapiddevelopmentin
the fieldofconcrete technology and incorporating important
aspects like durability etc.Thisstandard hasputgreater emphasis
tolimit state methodofdesignbypresentingitina full section
(section 5),while the working stress method has been givenin
Annex Bofthe same standard. Accordingly, structuresorstructural
elementsshallnormallybedesignedbylimitstatemethod.
Design Loads
•The design loads are determined separately for
thetwomethodsofdesignasmentionedbelow
after determining the combinationofdifferent
loads.
•Inthelimitstatemethod,thedesignloadisthe
characteristic load with appropriate partial
safetyfactor(videsec.2.3.2.3forpartialsafety
factors).
•Intheworkingstressmethod,thedesignloadis
thecharacteristicloadonly.
•The design loads are determined separately for
thetwomethodsofdesignasmentionedbelow
after determining the combinationofdifferent
loads.
•Inthelimitstatemethod,thedesignloadisthe
characteristic load with appropriate partial
safetyfactor(videsec.2.3.2.3forpartialsafety
factors).
•Intheworkingstressmethod,thedesignloadis
thecharacteristicloadonly.
LOAD COMBINATION
CHARACTERISTIC LOADS
•The characteristic valuesofloads are basedon
statistical data.Itisassumed thatinninety-fiveper
centcasesthecharacteristicloadswillnotbeexceeded
duringthelifeofthestructures.
•However,structuresaresubjectedtooverloadingalso.
Hence, structures shouldbedesigned with loads
obtainedbymultiplying the characteristic loads with
suitable factorsofsafety dependingonthe natureof
loadsortheir combinations, and the limit state being
considered. These factorsofsafety for loads are
termedaspartialsafetyfactors(γ
f)forloads.Thus,the
designloadsarecalculatedas
•Designload=CharacteristicLoadXPartialLoadfactor
•The characteristic valuesofloads are basedon
statistical data.Itisassumed thatinninety-fiveper
centcasesthecharacteristicloadswillnotbeexceeded
duringthelifeofthestructures.
•However,structuresaresubjectedtooverloadingalso.
Hence, structures shouldbedesigned with loads
obtainedbymultiplying the characteristic loads with
suitable factorsofsafety dependingonthe natureof
loadsortheir combinations, and the limit state being
considered. These factorsofsafety for loads are
termedaspartialsafetyfactors(γ
f)forloads.Thus,the
designloadsarecalculatedas
•Designload=CharacteristicLoadXPartialLoadfactor
CHARACTERISTIC LOADS
Characteristic load = Average/mean load + K ( s=standard deviation for load)
The valueofKisassumed such that the actual load does not exceed the
characteristic load during the lifeofthe structurein95per centofthe cases.
In absence of any data,loadsgiven in various standards shall be assumed as
thecharacteristic loads.
Characteristic load = Average/mean load + K ( s=standard deviation for load)
The valueofKisassumed such that the actual load does not exceed the
characteristic load during the lifeofthe structurein95per centofthe cases.
In absence of any data,loadsgiven in various standards shall be assumed as
thecharacteristic loads.
DESIGN LOAD
Values of Partial load factors
Design load= Characteristic Load X Partial Load factor
Load
Combinati
on
LIMIT State of CollapseLIMIT State of Serviceability
DL LL WL DL LL WL
DL+LL 1.5 1.5 1.0 1.0
DL+WL 1.5or
0.9
1
1.5 1.0 1.0
DL+LL+WL1.2 1.2 1.2 1.0 0.8 0.8
Values of Partial load factors
Design load= Characteristic Load X Partial Load factor
Load
Combinati
on
LIMIT State of CollapseLIMIT State of Serviceability
DL LL WL DL LL WL
DL+LL 1.5 1.5 1.0 1.0
DL+WL 1.5or
0.9
1
1.5 1.0 1.0
DL+LL+WL1.2 1.2 1.2 1.0 0.8 0.8
DIFFERENT LOADS
•Loads
•The following are the different types of loads and forces acting on the structure. As mentioned
earlier, their values have been assumed based on earlier data and experiences. It is worth
mentioning that their assumed values as stipulated in IS 875 have been used successfully.
•Dead loads
•These are the self weight of the structure to be designed. Needless to mention that the
dimensions of the cross section are to be assumed initially which enable to estimate the dead
loads from the known unit weights of the materials of the structure. The accuracy of the
estimation thus depends on the assumed values of the initial dimensions of the cross section.
The values of unit weights of the materials are specified in Part 1 of IS 875.
•Imposed loads
•They are also known as live loads and consist of all loads other than the dead loads of the
structure. The values of the imposed loads depend on the functional requirement of the
structure. Residential buildings will have comparatively lower values of the imposed loads
than those of school or office buildings. The standard values are stipulated in Part 2 of IS 875.
•Wind loads
•These loads depend on the velocity of the wind at the location of the structure, permeability of
the structure, height of the structure etc. They may be horizontal or inclined forces depending
on the angle of inclination of the roof for pitched roof structures. They can even be suction
type of forces depending on the angle of inclination of the roof or geometry of the buildings.
Wind loads are specified in Part 3 of IS 875.
•
•Loads
•The following are the different types of loads and forces acting on the structure. As mentioned
earlier, their values have been assumed based on earlier data and experiences. It is worth
mentioning that their assumed values as stipulated in IS 875 have been used successfully.
•Dead loads
•These are the self weight of the structure to be designed. Needless to mention that the
dimensions of the cross section are to be assumed initially which enable to estimate the dead
loads from the known unit weights of the materials of the structure. The accuracy of the
estimation thus depends on the assumed values of the initial dimensions of the cross section.
The values of unit weights of the materials are specified in Part 1 of IS 875.
•Imposed loads
•They are also known as live loads and consist of all loads other than the dead loads of the
structure. The values of the imposed loads depend on the functional requirement of the
structure. Residential buildings will have comparatively lower values of the imposed loads
than those of school or office buildings. The standard values are stipulated in Part 2 of IS 875.
•Wind loads
•These loads depend on the velocity of the wind at the location of the structure, permeability of
the structure, height of the structure etc. They may be horizontal or inclined forces depending
on the angle of inclination of the roof for pitched roof structures. They can even be suction
type of forces depending on the angle of inclination of the roof or geometry of the buildings.
Wind loads are specified in Part 3 of IS 875.
•
•Snow loads
•These are important loads for structures located in areas having snow fall, which gets
accumulated in different parts of the structure depending on projections, height, slope etc. of
the structure . The standard values of snow loads are specified in Part 4 of IS 875.
•Earthquake forces
•Earthquake generates waves which move from the origin of its location (epicenter) with
velocities depending on the intensity and magnitude of the earthquake. The impact of
earthquake on structures depends on the stiffness of the structure, stiffness of the soil media,
height and location of the structure etc.. Accordingly, the country has been divided into
several zones depending on the magnitude of the earthquake. The earthquake forces are
prescribed in IS 1893. Designers have adopted equivalent static load approach or spectral
method.
•Shrinkage, creep and temperature effects
•Shrinkage, creep and temperature (high or low) may produce stresses and cause deformations
like other loads and forces . Hence, these are also considered as loads which are time
dependent. The safety and serviceability of structures are to be checked following the
stipulations of cls. 6.2.4, 5 and 6 of IS 456:2000 and Part 5 of IS 875.
•Other forces and effects
•It is difficult to prepare an exhaustive list of loads, forces and effects coming onto the
structures and affecting the safety and serviceability of them. However, IS 456:2000
stipulates the following forces and effects to be taken into account in case they are liable to
affect materially the safety and serviceability of the structures. The relevant codes as
mentioned therein are also indicated below:
DIFFERENT LOADS
•These are important loads for structures located in areas having snow fall, which gets
accumulated in different parts of the structure depending on projections, height, slope etc. of
the structure . The standard values of snow loads are specified in Part 4 of IS 875.
•Earthquake forces
•Earthquake generates waves which move from the origin of its location (epicenter) with
velocities depending on the intensity and magnitude of the earthquake. The impact of
earthquake on structures depends on the stiffness of the structure, stiffness of the soil media,
height and location of the structure etc.. Accordingly, the country has been divided into
several zones depending on the magnitude of the earthquake. The earthquake forces are
prescribed in IS 1893. Designers have adopted equivalent static load approach or spectral
method.
•Shrinkage, creep and temperature effects
•Shrinkage, creep and temperature (high or low) may produce stresses and cause deformations
like other loads and forces . Hence, these are also considered as loads which are time
dependent. The safety and serviceability of structures are to be checked following the
stipulations of cls. 6.2.4, 5 and 6 of IS 456:2000 and Part 5 of IS 875.
•Other forces and effects
•It is difficult to prepare an exhaustive list of loads, forces and effects coming onto the
structures and affecting the safety and serviceability of them. However, IS 456:2000
stipulates the following forces and effects to be taken into account in case they are liable to
affect materially the safety and serviceability of the structures. The relevant codes as
mentioned therein are also indicated below:
DIFFERENT LOADS
CHARACTERISTIC STRENGTH
•Characteristic strengthofa materialasobtained from the statistical approach
isthestrengthofthatmaterialbelowwhichnotmorethanfivepercentofthe
test results are expectedtofall .However, such characteristic strengths may
differ from sampletosample also. Accordingly, the design strengthis
calculated dividing the characteristic strength furtherbythe partial safety
factor for the material
m, where
mdependsonthematerial and the limit
statebeingconsidered.Thus,Kis1.65for5%probability
•StandardDeviation=
Characteristic strength= Average/mean strength -K (standard deviation for strength)
Design strengthofmaterial= Characteristic strength /Partial safety
factor for material
Where, δ=deviation of the individual test strength from the average or
mean strength of n samples.
n= number of test results.
IS 456:2000 has recommended minimum value of n=30.
•Characteristic strengthofa materialasobtained from the statistical approach
isthestrengthofthatmaterialbelowwhichnotmorethanfivepercentofthe
test results are expectedtofall .However, such characteristic strengths may
differ from sampletosample also. Accordingly, the design strengthis
calculated dividing the characteristic strength furtherbythe partial safety
factor for the material
m, where
mdependsonthematerial and the limit
statebeingconsidered.Thus,Kis1.65for5%probability
•StandardDeviation=
Characteristic strength= Average/mean strength -K (standard deviation for strength)
Design strengthofmaterial= Characteristic strength /Partial safety
factor for material
Where, δ=deviation of the individual test strength from the average or
mean strength of n samples.
n= number of test results.
IS 456:2000 has recommended minimum value of n=30.
CHARACTERISTIC STRENGTH
Characteristic strengths may differ from sampletosample also. Accordingly, the
design strengthiscalculated dividing the characteristic strength furtherbythe partial
safetyfactorforthe material.
Clause36.4.2ofIS456states that
mfor concreteandsteel shouldbetakenas1.5
and1.15, respectively when assessing the strengthofthe structuresorstructural
membersemploying limit stateofcollapse.
Characteristic Strength= mean strength- k.S=f
m-kS
S=standard deviation, k=1.65 for 5% probability
Characteristic strengths may differ from sampletosample also. Accordingly, the
design strengthiscalculated dividing the characteristic strength furtherbythe partial
safetyfactorforthe material.
Clause36.4.2ofIS456states that
mfor concreteandsteel shouldbetakenas1.5
and1.15, respectively when assessing the strengthofthe structuresorstructural
membersemploying limit stateofcollapse.
Characteristic Strength= mean strength- k.S=f
m-kS
S=standard deviation, k=1.65 for 5% probability
WHY LIMIT STATE
•Concept of separate partial safety factors of loads of different combinations in
the two limit state methods.
•(ii) Concept of separate partial safety factors of materials depending on their
quality control during preparation. Thus, γ
m
for concrete is 1.5 and the same for
steel is 1.15. This is more logical than one arbitrary value in the name of safety
factor.
•(iii) A structure designed by employing limit state method of collapse and
checked for other limit states will ensure the strength and stability requirements
at the collapse under the design loads and also deflection and cracking at the
limit state of serviceability. This will help to achieve the structure with acceptable
probabilities that the structure will not become unfit for the use for which it is
intended.
•(iv) The stress block represents in a more realistic manner when the structure is
at the collapsing stage (limit state of collapse) subjected to design loads.
•Concept of separate partial safety factors of loads of different combinations in
the two limit state methods.
•(ii) Concept of separate partial safety factors of materials depending on their
quality control during preparation. Thus, γ
m
for concrete is 1.5 and the same for
steel is 1.15. This is more logical than one arbitrary value in the name of safety
factor.
•(iii) A structure designed by employing limit state method of collapse and
checked for other limit states will ensure the strength and stability requirements
at the collapse under the design loads and also deflection and cracking at the
limit state of serviceability. This will help to achieve the structure with acceptable
probabilities that the structure will not become unfit for the use for which it is
intended.
•(iv) The stress block represents in a more realistic manner when the structure is
at the collapsing stage (limit state of collapse) subjected to design loads.
Working stress method
•This methodofdesign, consideredasthe
methodofearlier times, has several
limitations.However,insituationswherelimit
statemethodcannotbeconvenientlyapplied,
workingstressmethodcanbeemployedasan
alternative.Itisexpected thatinthe near
future the working stress method willbe
completely replacedbythe limit state
method. Presently, this methodisputin
AnnexBofIS456:2000.
•This methodofdesign, consideredasthe
methodofearlier times, has several
limitations.However,insituationswherelimit
statemethodcannotbeconvenientlyapplied,
workingstressmethodcanbeemployedasan
alternative.Itisexpected thatinthe near
future the working stress method willbe
completely replacedbythe limit state
method. Presently, this methodisputin
AnnexBofIS456:2000.
METHOD BASED ON EXPERIMENTAL APPROACH
•The designer may perform experimental investigationson
modelsorfull size structuresorelements and accordingly
designthestructuresorelementsandalsoitshouldsatisfy
designobjectives.Moreover,theengineer-in-chargehasto
approve the experimental details and the analysis
connectedtherewith.
•Thoughthechoiceofthemethodofdesignisstilllefttothe
designeraspercl.18.2ofIS456:2000, the superiorityof
the limit state methodisevident from the emphasis given
tothismethodbypresentingitinafullsection(Section5),
whileaccommodatingtheworkingstressmethodinAnnex
BofIS456:2000, from its earlier placeofsection 6inIS
456:1978.Itisexpectedthat a gradual change overtothe
limit state methodofdesignwilltake placeinthe near
future after overcoming the inconveniencesofadopting
thismethodinsomesituations.
•The designer may perform experimental investigationson
modelsorfull size structuresorelements and accordingly
designthestructuresorelementsandalsoitshouldsatisfy
designobjectives.Moreover,theengineer-in-chargehasto
approve the experimental details and the analysis
connectedtherewith.
•Thoughthechoiceofthemethodofdesignisstilllefttothe
designeraspercl.18.2ofIS456:2000, the superiorityof
the limit state methodisevident from the emphasis given
tothismethodbypresentingitinafullsection(Section5),
whileaccommodatingtheworkingstressmethodinAnnex
BofIS456:2000, from its earlier placeofsection 6inIS
456:1978.Itisexpectedthat a gradual change overtothe
limit state methodofdesignwilltake placeinthe near
future after overcoming the inconveniencesofadopting
thismethodinsomesituations.
STRESS STRAIN DIGRAM FOR
CONCRETE
CONCRETE
MILD STEEL
HYSD BARS
FLEXURAL STRENGTH OF BEAMS
T
C
C
T
C
C
T
ASSUMPTIONS
The following are the assumptions of the design of flexural members employing limit
state of collapse:
(i) Plane sections normal to the axis remain plane after bending.
This assumption ensures that the cross-sectionofthe member does not warp due
tothe loads applied.Itfurther means that the strainatany pointonthe cross-
sectionisdirectly proportionaltoits distance from the neutral axis.
(ii) The maximum strain in concrete at the outer most compression fibre is taken as
0.0035 in bending .
Thisisa clearly defined limiting strainofconcreteinbending compression beyond
which the concrete willbetakenasreaching the stateofcollapse.Itisvery clear
that the specified limiting strainof0.0035does not dependonthe strengthof
concrete.
(iii) The acceptable stress-strain curve of concrete is assumed to be parabolic .
The maximum compressive stress-strain curveinthe structureisobtainedby
reducing the valuesofthe top parabolic curve (Figs.21 ofIS456:2000)intwo
stages. First, dividingby1.5duetosize effect and secondly, again dividingby1.5
considering the partial safety factorofthe material. The middle and bottom curves
(Fig.21 ofIS456:2000) represent these stages. Thus, the maximum compressive
stressinbendingislimitedtothe constant valueof0.446f
ckfor the strain ranging
from 0.002to0.0035( Figs.21and22ofIS456:2000).
The following are the assumptions of the design of flexural members employing limit
state of collapse:
(i) Plane sections normal to the axis remain plane after bending.
This assumption ensures that the cross-sectionofthe member does not warp due
tothe loads applied.Itfurther means that the strainatany pointonthe cross-
sectionisdirectly proportionaltoits distance from the neutral axis.
(ii) The maximum strain in concrete at the outer most compression fibre is taken as
0.0035 in bending .
Thisisa clearly defined limiting strainofconcreteinbending compression beyond
which the concrete willbetakenasreaching the stateofcollapse.Itisvery clear
that the specified limiting strainof0.0035does not dependonthe strengthof
concrete.
(iii) The acceptable stress-strain curve of concrete is assumed to be parabolic .
The maximum compressive stress-strain curveinthe structureisobtainedby
reducing the valuesofthe top parabolic curve (Figs.21 ofIS456:2000)intwo
stages. First, dividingby1.5duetosize effect and secondly, again dividingby1.5
considering the partial safety factorofthe material. The middle and bottom curves
(Fig.21 ofIS456:2000) represent these stages. Thus, the maximum compressive
stressinbendingislimitedtothe constant valueof0.446f
ckfor the strain ranging
from 0.002to0.0035( Figs.21and22ofIS456:2000).
ASSUMPTIONS
(iv) The tensile strengthofconcreteisignored.
Concrete has some tensile strength (very smallbutnot zero). Yet, this tensile
strengthisignored and the steel reinforcementisassumedtoresist the tensile
stress. However, the tensile strengthofconcreteistaken into accounttocheck the
deflection and crack widthsinthe limit stateofserviceability.
(v) The design stressesofthe reinforcement are derived from the representative stress-
strain curvesasshowninFigs.23A and BofIS456:2000, for the typeofsteel used
using the partial safetyfactor
mas1.15.
Inthe reinforced concrete structures, two typesofsteel are used: one with definite
yield point (Figs.23BofIS456:2000) and the other where the yield points are not
definite (cold work deformed bars). The representative stress-strain diagram (Fig.
23AofIS456:2000) defines the points between 0.8f
yand1.0 f
yincaseofcold work
deformedbars where the curveisinelastic.
(vi) The maximum straininthe tension reinforcementinthe sectionatfailure shall not
belessthanf
y/(1.15E
s) + 0.002,
This assumption ensures ductile failureinwhich the tensile reinforcement
undergoes a certain degreeofinelastic deformation before concrete failsin
compression.
(iv) The tensile strengthofconcreteisignored.
Concrete has some tensile strength (very smallbutnot zero). Yet, this tensile
strengthisignored and the steel reinforcementisassumedtoresist the tensile
stress. However, the tensile strengthofconcreteistaken into accounttocheck the
deflection and crack widthsinthe limit stateofserviceability.
(v) The design stressesofthe reinforcement are derived from the representative stress-
strain curvesasshowninFigs.23A and BofIS456:2000, for the typeofsteel used
using the partial safetyfactor
mas1.15.
Inthe reinforced concrete structures, two typesofsteel are used: one with definite
yield point (Figs.23BofIS456:2000) and the other where the yield points are not
definite (cold work deformed bars). The representative stress-strain diagram (Fig.
23AofIS456:2000) defines the points between 0.8f
yand1.0 f
yincaseofcold work
deformedbars where the curveisinelastic.
(vi) The maximum straininthe tension reinforcementinthe sectionatfailure shall not
belessthanf
y/(1.15E
s) + 0.002,
This assumption ensures ductile failureinwhich the tensile reinforcement
undergoes a certain degreeofinelastic deformation before concrete failsin
compression.
BEAM SECTIONS
A
st= area of tension steel
b = width of the beam
C = total compressive force of
concrete
D=Overall depth
d = effective depth of the beam
L = centre to centre distance
between supports
T = total tensile force of steel
x
u= depth of neutral axis from
the most compressed fibre
d-x
u
A
st= area of tension steel
b = width of the beam
C = total compressive force of
concrete
D=Overall depth
d = effective depth of the beam
L = centre to centre distance
between supports
T = total tensile force of steel
x
u= depth of neutral axis from
the most compressed fibre
d-x
u
SECTION ABOVE NEUTRAL AXIS
A
1=b*(3/7)x
u
x
u
d-x
u
A
st
A
1=b*(3/7)x
u
A
2=b*(4/7)x
u
A
1=b*(3/7)x
u
x
u
d-x
u
A
st
A
1=b*(3/7)x
u
A
2=b*(4/7)x
u
EQULIBRIUM EQUATIONS
The cross-sections of the beam under the applied loads as shown in Figure has three types of
combinations of shear forces and bending moments:
(i) only shear force is there at the support and bending moment is zero,
(ii) both bending moment (increasing gradually) and shear force (constant = P) are there between the
support and the loading point and
(iii) a constant moment (= PL/3) is there in the middle third zone i.e. between the two loads where
the shear force is zero.
Since the beam is in static equilibrium, any cross-section of the beam is also in static equilibrium.
Considering the cross-section in the middle zone ,
The three equations of equilibrium are the following
(i) Equilibrium of horizontal forces: Σ H = 0 gives T = C
(ii) Equilibrium of vertical shear forces: Σ V = 0
This equation gives an identity 0 = 0 as there is no shear in the middle third zone of the beam.
(iii) Equilibrium of moments: Σ M = 0,
This equation shows that the applied moment at the section is fully resisted by moment of the
resisting couple T a = C a , where a is the operating lever arm between T and C (Figs. 3.4.19 and 20).
The cross-sections of the beam under the applied loads as shown in Figure has three types of
combinations of shear forces and bending moments:
(i) only shear force is there at the support and bending moment is zero,
(ii) both bending moment (increasing gradually) and shear force (constant = P) are there between the
support and the loading point and
(iii) a constant moment (= PL/3) is there in the middle third zone i.e. between the two loads where
the shear force is zero.
Since the beam is in static equilibrium, any cross-section of the beam is also in static equilibrium.
Considering the cross-section in the middle zone ,
The three equations of equilibrium are the following
(i) Equilibrium of horizontal forces: Σ H = 0 gives T = C
(ii) Equilibrium of vertical shear forces: Σ V = 0
This equation gives an identity 0 = 0 as there is no shear in the middle third zone of the beam.
(iii) Equilibrium of moments: Σ M = 0,
This equation shows that the applied moment at the section is fully resisted by moment of the
resisting couple T a = C a , where a is the operating lever arm between T and C (Figs. 3.4.19 and 20).
CALCULATION OF FORCES
•C = Total compressive force of concrete = C
1+ C
2
•C
1= Compressive force of concrete due to the constant stress of
0.446 f
ckand up to a depth of x
3from the top fibre
•C
2 = Compressive force of concrete due to the convex parabolic
stress block of values ranging from zero at the neutral axis to 0.446
f
ckat a distance of x
4
•x
1= Distance of the line of action of C
1from the most compressed
fibre
•x
2= Distance of the line of action of C
2from the most compressed
fibre
•x
3= Distance of the fibre from the most compressed fibre to where
the strain = 0.002 and stress = 0.446 f
ck
•x
4= Distance from N.A to the point where strain is 0.002 in
compression zone
•x
u= Distance of the neutral axis from the most compressed fibre .
•C = Total compressive force of concrete = C
1+ C
2
•C
1= Compressive force of concrete due to the constant stress of
0.446 f
ckand up to a depth of x
3from the top fibre
•C
2 = Compressive force of concrete due to the convex parabolic
stress block of values ranging from zero at the neutral axis to 0.446
f
ckat a distance of x
4
•x
1= Distance of the line of action of C
1from the most compressed
fibre
•x
2= Distance of the line of action of C
2from the most compressed
fibre
•x
3= Distance of the fibre from the most compressed fibre to where
the strain = 0.002 and stress = 0.446 f
ck
•x
4= Distance from N.A to the point where strain is 0.002 in
compression zone
•x
u= Distance of the neutral axis from the most compressed fibre .
CALCULATION OF COMPRESSIVE FORCE
•x
3+x
4=x
u
•x
4=(4/7)x
u
x
3=(3/7)x
u
•x
3+x
4=x
u
•x
4=(4/7)x
u
x
3=(3/7)x
u
CALCULATION OF x
c
Equating Sum of All horizontal forces equal to zero
C=T
c
Equating Sum of All horizontal forces equal to zero
C=T
CALCULATION OF MOR
As ona particular section Compressive
force C and Tension T are created and
both are equalinmagnitude and
oppositeindiction and their lineof
action notinsame point ,it iscreating a
couple . This coupleormomentis
MomentofResistance(MOR)ofthe
section which resists the external
moment.
IfthisMORisgreaterorequaltothe
external moment,the sectionissafe
MOR=( CorT ) * Lever arm
Lever arm= Distance between C and T=z
z-=d-0.42x
u
c
T
0.42 x
u
d
d-0.42 x
u
d’
As ona particular section Compressive
force C and Tension T are created and
both are equalinmagnitude and
oppositeindiction and their lineof
action notinsame point ,it iscreating a
couple . This coupleormomentis
MomentofResistance(MOR)ofthe
section which resists the external
moment.
IfthisMORisgreaterorequaltothe
external moment,the sectionissafe
MOR=( CorT ) * Lever arm
Lever arm= Distance between C and T=z
z-=d-0.42x
u
c
T
0.42 x
u
d
d-0.42 x
u
d’
OVER ,BALANCE AND UNDER
REINFORCED SECTIONS
G.C. BEHERA
REINFORCED SECTIONS
G.C. BEHERA
CALCULATION OF x
c
Equating Sum of All horizontal forces equal to zero
C=T
c
Equating Sum of All horizontal forces equal to zero
C=T
CALCULATION OF MOR AND
PERCENTAGE OF STEEL
c
T
0.42 x
u
d
d-0.42 x
u
d’
PERCENTAGE OF STEEL
c
T
0.42 x
u
d
d-0.42 x
u
d’
PECENTAGE OF STEEL
ACTUAL STRESS STRAIN AND DESIGN
STRESS STRAIN CURVE
ACTUAL STRESS STRAIN
STRESS STRAIN CURVE
ACTUAL STRESS STRAIN
BALANCE SECTION
When maximum stressesinconcrete and steel
reach simultaneously, the sectionisknownas
balance section. The straininconcreteis0.0035
while straininsteelis(0.87f
y/E
s)+0.002
Taking E
s=2.0*10
5
N/mm
2
X
umaxalso known as x
ulimit
Independent of grade of concrete
When maximum stressesinconcrete and steel
reach simultaneously, the sectionisknownas
balance section. The straininconcreteis0.0035
while straininsteelis(0.87f
y/E
s)+0.002
Taking E
s=2.0*10
5
N/mm
2
X
umaxalso known as x
ulimit
Independent of grade of concrete
•For a balance section, x
uactual Neutral axis is
equal to critical neutral axis x
umaxor x
ulimit.
•
BALANCE SECTION (MOR)
uactual Neutral axis is
equal to critical neutral axis x
umaxor x
ulimit.
•
BALANCE SECTION (MOR)
BALANCE SECTION (MOR)
f
ck Values of p
tlim Values of M
ulim
Fe250 Fe415 Fe500Fe250 Fe415 Fe500
20 1.76 0.96 0.75 2.96bd
2
2.76 bd
2
2.66bd
2
25 2.20 1.20 0.94 3.70bd
2
3.45bd
2
3.33bd
2
30 2.64 1.43 1.13 4.44bd
2
4.14bd
2
3.99bd
2
35 3.08 1.67 1.32 5.21bd
2
4.83bd
2
4.65 bd
2
40 3.52 1.91 1.51 5.92bd
2
5.52bd
2
5.32bd
2
f
ck Values of p
tlim Values of M
ulim
Fe250 Fe415 Fe500Fe250 Fe415 Fe500
20 1.76 0.96 0.75 2.96bd
2
2.76 bd
2
2.66bd
2
25 2.20 1.20 0.94 3.70bd
2
3.45bd
2
3.33bd
2
30 2.64 1.43 1.13 4.44bd
2
4.14bd
2
3.99bd
2
35 3.08 1.67 1.32 5.21bd
2
4.83bd
2
4.65 bd
2
40 3.52 1.91 1.51 5.92bd
2
5.52bd
2
5.32bd
2
UNDER REINFORCED SECTION
Section fails in ductile manner
Section fails in ductile manner
OVER EINFORCED SECTION
OVER EINFORCED SECTION
•For over reinforced case, there are two possibilities.
Either straininconcreteismore than 0.0035 (The maximum straininconcreteatthe outer
most compression fibreistakenas0.0035inbending (whichisnot possibleasthereis
fractureinconcreteatstrain 0.0035)
or
straininsteel less than yield strain whichisagainst assumptionThemaximum straininthe
tension reinforcementinthe sectionatfailure shall notbeless thanf
y/(1.15E
s) + 0.002,This
assumption ensures ductile failureinwhich the tensile reinforcement undergoes a certain
degreeofinelastic deformation before concrete failsincompression.
So,overreinforcedsectioninlimitstatemethodisnotallowed.
Failure is sudden, failure due to concrete.
Brittle failure
Moment of Resistance=MOR is calculated by putting x
u=x
umax
•For over reinforced case, there are two possibilities.
Either straininconcreteismore than 0.0035 (The maximum straininconcreteatthe outer
most compression fibreistakenas0.0035inbending (whichisnot possibleasthereis
fractureinconcreteatstrain 0.0035)
or
straininsteel less than yield strain whichisagainst assumptionThemaximum straininthe
tension reinforcementinthe sectionatfailure shall notbeless thanf
y/(1.15E
s) + 0.002,This
assumption ensures ductile failureinwhich the tensile reinforcement undergoes a certain
degreeofinelastic deformation before concrete failsincompression.
So,overreinforcedsectioninlimitstatemethodisnotallowed.
Failure is sudden, failure due to concrete.
Brittle failure
Moment of Resistance=MOR is calculated by putting x
u=x
umax
OVER ,BALANCE AND UNDER
REINFORCED SECTIONS
PROBLEMS
G.C. BEHERA
REINFORCED SECTIONS
PROBLEMS
G.C. BEHERA
TYPES OF PROBLEMS
•TYPE-1-TO DETERMINE MOR OF A GIVEN
SECTION
•Itmayberequiredtoestimatethemomentof
resistance.
GIVEN
•Section,b,D,effectivecover,
•A
st,
•GradeofConcreteandGradeofSteel
•TYPE-1-TO DETERMINE MOR OF A GIVEN
SECTION
•Itmayberequiredtoestimatethemomentof
resistance.
GIVEN
•Section,b,D,effectivecover,
•A
st,
•GradeofConcreteandGradeofSteel
SOLUTION FOR TYPE-1
•STEPS
1. Find b, D, d, A
st, f
ck, f
y
2.Calculate actual NA depth x
u
3.Calculate X
umax= X
ulimit
•STEPS
1. Find b, D, d, A
st, f
ck, f
y
2.Calculate actual NA depth x
u
3.Calculate X
umax= X
ulimit
SOLUTION FOR TYPE-1
•STEPS
4.A) If x
ux
ulimit
Section is under-reinforced.
MOR is calculated taking tensile force into
consideration.
•STEPS
4.A) If x
ux
ulimit
Section is under-reinforced.
MOR is calculated taking tensile force into
consideration.
SOLUTION FOR TYPE-1
•STEPS
4.B) If x
u=x
ulimit
Section is Balance section.
MOR is calculated taking tensile force or
compression force into consideration.
M
u=M
ulimit
•STEPS
4.B) If x
u=x
ulimit
Section is Balance section.
MOR is calculated taking tensile force or
compression force into consideration.
M
u=M
ulimit
SOLUTION FOR TYPE-1
•STEPS
4.C) If x
u> x
ulimit
Section is over reinforced.
MOR is calculated taking compression force
into consideration.
MOR is equal to M
ulimit
Put x
u=x
ulimit
M
u=M
ulimit
•STEPS
4.C) If x
u> x
ulimit
Section is over reinforced.
MOR is calculated taking compression force
into consideration.
MOR is equal to M
ulimit
Put x
u=x
ulimit
M
u=M
ulimit
PROBLEM-TYPE-1
Determine the service imposed loadsoftwo simply supported beamofsame effective
spanof8 m and same cross-sectional dimensions,buthavingtwodifferent amountsof
reinforcement. Both the beams are madeofM20andFe415.
Determine the service imposed loadsoftwo simply supported beamofsame effective
spanof8 m and same cross-sectional dimensions,buthavingtwodifferent amountsof
reinforcement. Both the beams are madeofM20andFe415.
SOLUTION
Given data:
b = 300 mm,
d = 550 mm,
D = 600 mm,
A
st= 4*(
/4)*(20)
2
=1256 mm
2
(4-20 T),
L
eff= 8 m
f
ck=20 MPa
f
y=415 MPa
and boundary condition = simply supported
Given data:
b = 300 mm,
d = 550 mm,
D = 600 mm,
A
st= 4*(
/4)*(20)
2
=1256 mm
2
(4-20 T),
L
eff= 8 m
f
ck=20 MPa
f
y=415 MPa
and boundary condition = simply supported
SOLUTION
Step 1: Calculate actual NA depth x
u
X
u= 209.9439 mm
Step 2:Calculate x
ulim
For Fe 415 x
umax=x
ulim=0.48d= 264 mm
Step 3: x
u<x
ulim
Section is under reinforced
MOR=209427197 Nmm=209.43 kNm
If load is w,
factor load=wu,
then external factor BM=M
u=wu*(l
eff)
2
Mu= MOR= 209.43 kNm
Step 1: Calculate actual NA depth x
u
X
u= 209.9439 mm
Step 2:Calculate x
ulim
For Fe 415 x
umax=x
ulim=0.48d= 264 mm
Step 3: x
u<x
ulim
Section is under reinforced
MOR=209427197 Nmm=209.43 kNm
If load is w,
factor load=wu,
then external factor BM=M
u=wu*(l
eff)
2
Mu= MOR= 209.43 kNm
•Factor BM=M
u=209.43=[w
u*(l
eff)
2
]/8
•w
u=26.1784 kN/m
•w=26.1784/1.5 = 17.452266 kN/m
•w=DL+LL
•DL=0.3*0.6*1*25=4.5 kN/m
•LL=17.452266 -4.5=12.952266 kN/m
•Maximum Service LL or Imposed Load that can
be given=12.952266 kN/m
SOLUTION
u=209.43=[w
u*(l
eff)
2
]/8
•w
u=26.1784 kN/m
•w=26.1784/1.5 = 17.452266 kN/m
•w=DL+LL
•DL=0.3*0.6*1*25=4.5 kN/m
•LL=17.452266 -4.5=12.952266 kN/m
•Maximum Service LL or Imposed Load that can
be given=12.952266 kN/m
SOLUTION
PROBLEM-TYPE-1
Determine the service imposed loadsoftwo simply supported beamofsame effective
spanof8 m and same cross-sectional dimensions,buthavingtwodifferent amountsof
reinforcement. Both the beams are madeofM20andFe415.
Determine the service imposed loadsoftwo simply supported beamofsame effective
spanof8 m and same cross-sectional dimensions,buthavingtwodifferent amountsof
reinforcement. Both the beams are madeofM20andFe415.
SOLUTION
Given data:
b = 300 mm,
d = 550 mm,
D = 600 mm,
A
st= 4*(
/4)*(20)
2
+2*( /4)*(16)
2
=1658 mm
2
(4-20 T, 2-16 T),
L
eff= 8 m
f
ck=20 Mpa
f
y=415 MPa
and boundary condition = simply supported
Given data:
b = 300 mm,
d = 550 mm,
D = 600 mm,
A
st= 4*(
/4)*(20)
2
+2*( /4)*(16)
2
=1658 mm
2
(4-20 T, 2-16 T),
L
eff= 8 m
f
ck=20 Mpa
f
y=415 MPa
and boundary condition = simply supported
SOLUTION
Step 1: Calculate actual NA depth x
u
X
u= 277.1393 mm
Step 2:Calculate x
ulim
For Fe 415 x
umax=x
ulim=0.48d= 264 mm
Step 3: x
u> x
ulim
Section is over reinforced
MOR=250403789 Nmm=250.40 kNm
If load is w,
factor load=wu,
then external factor BM=M
u=(1/8)*wu*(l
eff)
2
Mu= MOR= 250.40 kNm
Step 1: Calculate actual NA depth x
u
X
u= 277.1393 mm
Step 2:Calculate x
ulim
For Fe 415 x
umax=x
ulim=0.48d= 264 mm
Step 3: x
u> x
ulim
Section is over reinforced
MOR=250403789 Nmm=250.40 kNm
If load is w,
factor load=wu,
then external factor BM=M
u=(1/8)*wu*(l
eff)
2
Mu= MOR= 250.40 kNm
•Factor BM=M
u=250.40=w
u*(l
eff)
2
•w
u=31.3 kN/m
•w=31.3/1.5 = 20.866667 kN/m
•W=DL+LL
•DL=0.3*0.6*1*25=4.5 kN/m
•LL=20.866667 -4.5=16.366667 kN/m
•Maximum Service LL or Imposed Load that can
be given=16.366667 kN/m
SOLUTION
u=250.40=w
u*(l
eff)
2
•w
u=31.3 kN/m
•w=31.3/1.5 = 20.866667 kN/m
•W=DL+LL
•DL=0.3*0.6*1*25=4.5 kN/m
•LL=20.866667 -4.5=16.366667 kN/m
•Maximum Service LL or Imposed Load that can
be given=16.366667 kN/m
SOLUTION
•Factor BM=M
u=250.40=w
u*(l
eff)
2
•w
u=31.3 kN/m
•w=31.3/1.5 = 20.866667 kN/m
•w=DL+LL
•DL=0.3*0.6*1*25=4.5 kN/m
•LL=20.866667 -4.5=16.366667 kN/m
•Maximum Service LL or Imposed Load that can
be given=16.366667 kN/m
SOLUTION
u=250.40=w
u*(l
eff)
2
•w
u=31.3 kN/m
•w=31.3/1.5 = 20.866667 kN/m
•w=DL+LL
•DL=0.3*0.6*1*25=4.5 kN/m
•LL=20.866667 -4.5=16.366667 kN/m
•Maximum Service LL or Imposed Load that can
be given=16.366667 kN/m
SOLUTION
•Factor BM=M
u= As per steel
SOLUTION
BM=M
u=262.87 kNm
This is not true. All the steel in tension
zone may not be yielded
u= As per steel
SOLUTION
BM=M
u=262.87 kNm
This is not true. All the steel in tension
zone may not be yielded
TYPES OF PROBLEMS-II
•To Compute Amount of Steel for a given load
•Given
Section (b,D, d)
Grade of Concrete and Steel
Service load,
Length of Beam and End Condition
•To Compute Amount of Steel for a given load
•Given
Section (b,D, d)
Grade of Concrete and Steel
Service load,
Length of Beam and End Condition
TYPES OF PROBLEMS-II
•Solution:
•Step-1
Compute Mu
Find Load w, Calculate w
u
Find M
ufrom beam length and end condition
•Step-2
Compute Mulimit
Step-3 A)
If M
u< M
ulimit
Design it as under reinforced
M
u=MOR
In the equation Mu known, only unknown is A
stInthe equation Mu
As it a quadratic equation, solution will provide you two values
Find out the feasible one.
•Solution:
•Step-1
Compute Mu
Find Load w, Calculate w
u
Find M
ufrom beam length and end condition
•Step-2
Compute Mulimit
Step-3 A)
If M
u< M
ulimit
Design it as under reinforced
M
u=MOR
In the equation Mu known, only unknown is A
stInthe equation Mu
As it a quadratic equation, solution will provide you two values
Find out the feasible one.
TYPES OF PROBLEMS-II
Step-3 B)
If M
u= M
ulimit
Design it as Balance section.
M
u=MOR
In the equation Mu known, only unknown is A
stInthe equation Mu
As it a quadratic equation, solution will provide you two values
Find out the feasible one.
OR find
xulimit
In the equation Mu known, only unknown is A
st
Step-3 B)
If M
u= M
ulimit
Design it as Balance section.
M
u=MOR
In the equation Mu known, only unknown is A
stInthe equation Mu
As it a quadratic equation, solution will provide you two values
Find out the feasible one.
OR find
xulimit
In the equation Mu known, only unknown is A
st
TYPES OF PROBLEMS-II
Step-3 C)
If M
u> M
ulimit
Design it as over reinforced
This section is not allowed in Limit state method.
Redesign the section.
Increase the section or make it doubly reinforced
section
Step-3 C)
If M
u> M
ulimit
Design it as over reinforced
This section is not allowed in Limit state method.
Redesign the section.
Increase the section or make it doubly reinforced
section
TYPES OF PROBLEMS-II
Problem:Arectangularbeam200mmwideand400mmdeep
upto centreofreinforcement. Design the beam if it hasto
resistamoment25kNm.UseM20andfe415
200 mm
400
mm
Solution:
Step-1 Computation of M
u
M=25 kNm
M
u= Factor BM=1.5*25= 37.5 kNm
Step-2 Computation of M
ulimit
M
ulimit=.36*20*.48*(1-0.42*.48)*200*400
2
=88.30 kNm
As M
u< M
ulimit
Design it as under reinforced
Problem:Arectangularbeam200mmwideand400mmdeep
upto centreofreinforcement. Design the beam if it hasto
resistamoment25kNm.UseM20andfe415
200 mm
400
mm
Solution:
Step-1 Computation of M
u
M=25 kNm
M
u= Factor BM=1.5*25= 37.5 kNm
Step-2 Computation of M
ulimit
M
ulimit=.36*20*.48*(1-0.42*.48)*200*400
2
=88.30 kNm
As M
u< M
ulimit
Design it as under reinforced
TYPES OF PROBLEMS-II
M
u= Factor BM=1.5*25= 37.5 kNm
A
st= 279.99= 280 mm
2
A
st= 3575.4048 mm
2
Discard this result
M
u= Factor BM=1.5*25= 37.5 kNm
A
st= 279.99= 280 mm
2
A
st= 3575.4048 mm
2
Discard this result
ASSIGNMENT –POBLEM TYPE-II
Problem:Arectangularbeam200mmwideand400mmdeep
upto centreofreinforcement. Design the beam if it hasto
resistamoment60kNm.UseM25andfe500
200 mm
400
mm
Problem:Arectangularbeam200mmwideand400mmdeep
upto centreofreinforcement. Design the beam if it hasto
resistamoment60kNm.UseM25andfe500
200 mm
400
mm
DESIGN OF BEAMS
G.C. BEHERA
G.C. BEHERA
EFFECTIVE LENGTH
•Clear distance between walls = 6m
•Thickness of one wall =300 mm
•Thickness of another wall =400 mm
•Effective depth of beam= 450 mm
•Simply supported Beam
•Find effective length of beam?
•L
eff=Clear span+deff=6m+.45m=6.45 m
•L
eff=C/C bet supports=6m+.3/2+0.4/2=6.35 m
•Thickness of one wall =300 mm
•Thickness of another wall =400 mm
•Effective depth of beam= 450 mm
•Simply supported Beam
•Find effective length of beam?
•L
eff=Clear span+deff=6m+.45m=6.45 m
•L
eff=C/C bet supports=6m+.3/2+0.4/2=6.35 m
EFFECTIVE LENGTH
•B)Continuous BeamorSlab -Inthecaseofcontinuous beamor
slab,ifthe widthofthe supportisless thanl/12ofthe clear span,
the effective span shallbeasin22.2 (a).Ifthe supports are wider
thanI/12ofthe clear spanor600mmwhicheverisless, the
effectivespanshallbetakenasunder:
•1)Forendspanwithoneendfixedandtheothercontinuousorfor
intermediate spans, the effective span shallbethe clear span
betweensupports;
•2) Forendspan with oneendfree and the other continuous, the
effectivespanshallbeequaltotheclearspanplushalftheeffective
depthofthe beamorslaborthe clear spanplushalf the widthof
thediscontinuoussupport,whicheverisless;
•3)Inthe caseofspans with rollerorrocket bearings, the effective
spanshallalwaysbethedistancebetweenthecentresofbearings
•B)Continuous BeamorSlab -Inthecaseofcontinuous beamor
slab,ifthe widthofthe supportisless thanl/12ofthe clear span,
the effective span shallbeasin22.2 (a).Ifthe supports are wider
thanI/12ofthe clear spanor600mmwhicheverisless, the
effectivespanshallbetakenasunder:
•1)Forendspanwithoneendfixedandtheothercontinuousorfor
intermediate spans, the effective span shallbethe clear span
betweensupports;
•2) Forendspan with oneendfree and the other continuous, the
effectivespanshallbeequaltotheclearspanplushalftheeffective
depthofthe beamorslaborthe clear spanplushalf the widthof
thediscontinuoussupport,whicheverisless;
•3)Inthe caseofspans with rollerorrocket bearings, the effective
spanshallalwaysbethedistancebetweenthecentresofbearings
c)Cantilever-The effective lengthofa cantilever shall betakenasits lengthtothe
faceofthe support plus half the effective depth except whereitforms theendofa
continuous beam where the lengthtothe centreofsupport shallbetaken.
•D) Frames-Inthe analysis of a continuous frame, centre to centre distance shall be
used.
faceofthe support plus half the effective depth except whereitforms theendofa
continuous beam where the lengthtothe centreofsupport shallbetaken.
•D) Frames-Inthe analysis of a continuous frame, centre to centre distance shall be
used.
SHEAR AND BM COEFFICIENTS
TENSILE REINFORCEMENT
Minimum reinforcement not less than
Minimum reinforcement not less than
DESIGN FOR SHEAR
G.C. BEHERA
G.C. BEHERA
FAILURE OF BEAMS
•This lesson explains the three failure modesduetoshear forceinbeams
anddefinesdifferentshearstressesneededtodesignthebeamsforshear.
Thecriticalsectionsforshearandtheminimumshearreinforcementtobe
providedinbeams are mentionedasperIS456. The designofshear
reinforcement willbeillustrated through several numerical problems
includingthecurtailmentoftensionreinforcementinflexuralmembers.
•Bendinginreinforced concrete beamsisusually accompaniedbyshear,
theexactanalysisofwhichisverycomplex.However,experimentalstudies
confirmed the following three different modesoffailureduetopossible
combinationsofshearforceandbendingmomentatagivensection
•This lesson explains the three failure modesduetoshear forceinbeams
anddefinesdifferentshearstressesneededtodesignthebeamsforshear.
Thecriticalsectionsforshearandtheminimumshearreinforcementtobe
providedinbeams are mentionedasperIS456. The designofshear
reinforcement willbeillustrated through several numerical problems
includingthecurtailmentoftensionreinforcementinflexuralmembers.
•Bendinginreinforced concrete beamsisusually accompaniedbyshear,
theexactanalysisofwhichisverycomplex.However,experimentalstudies
confirmed the following three different modesoffailureduetopossible
combinationsofshearforceandbendingmomentatagivensection
Flexural Tension steel yields
Flexural Compression Concrete crushes in compression
Failure modes
Flexural Compression Concrete crushes in compression
Failure modes
SHEAR STRESS VARIATION
Diagonal iksubjected
to tension, crack along
JL
Diagonal iksubjected
to tension, crack along
JL
SHEAR STRESS CALCULATION
•The distributionofshear
stressinreinforced
concreterectangular,Tand
L-beamsofuniformand
varyingdepthsdependson
the distributionofthe
normalstress.However,for
the sakeofsimplicity the
nominal shear stressτ
vis
considered which is
calculatedasfollows(IS
456,cls.40.1and40.1.1):
•The distributionofshear
stressinreinforced
concreterectangular,Tand
L-beamsofuniformand
varyingdepthsdependson
the distributionofthe
normalstress.However,for
the sakeofsimplicity the
nominal shear stressτ
vis
considered which is
calculatedasfollows(IS
456,cls.40.1and40.1.1):
SHEAR STRESS CALCULATION
In a beam of uniform depth
For varying Depth
•where Vu = Factored shear force due to design loads,
•b = breadth of rectangular beams and breadth of the web
b
wfor flanged beams, and
•d = effective depth.
•M
u= Factored bending moment at the section, and
•β = angle between the top and the bottom edges.
•
vuis the nominal shear stress
•The positive sign is applicable when the bending moment
M
udecreases numerically in the same directionasthe
effectivedepthincreases,andthenegativesignisapplicable
when the bending moment M
uincreases numerically in the
samedirectionastheeffectivedepthincreases.
In a beam of uniform depth
For varying Depth
•where Vu = Factored shear force due to design loads,
•b = breadth of rectangular beams and breadth of the web
b
wfor flanged beams, and
•d = effective depth.
•M
u= Factored bending moment at the section, and
•β = angle between the top and the bottom edges.
•
vuis the nominal shear stress
•The positive sign is applicable when the bending moment
M
udecreases numerically in the same directionasthe
effectivedepthincreases,andthenegativesignisapplicable
when the bending moment M
uincreases numerically in the
samedirectionastheeffectivedepthincreases.
Design Shear Strength of Reinforced Concrete Beams
•Recent laboratory experiments confirmed that
reinforced concreteinbeams has shear strength
even without any shear reinforcement. This shear
strength (τ
c) dependsonthe gradeofconcreteand
the percentageoftension steelinbeams.Onthe
otherhand,theshearstrengthofreinforcedconcrete
with the reinforcementisrestrictedtosome
maximum valueτ
cmaxdependingonthe gradeof
concrete. These minimumandmaximum shear
strengthsofreinforced concrete (IS 456, cls.40.2.1
and40.2.3,respectively)aregivenbelow:
•Recent laboratory experiments confirmed that
reinforced concreteinbeams has shear strength
even without any shear reinforcement. This shear
strength (τ
c) dependsonthe gradeofconcreteand
the percentageoftension steelinbeams.Onthe
otherhand,theshearstrengthofreinforcedconcrete
with the reinforcementisrestrictedtosome
maximum valueτ
cmaxdependingonthe gradeof
concrete. These minimumandmaximum shear
strengthsofreinforced concrete (IS 456, cls.40.2.1
and40.2.3,respectively)aregivenbelow:
SHEAR RESISTING MECHANISM
•Freebody diagramofa segment
ofreinforced concrete beam
separatedbya diagonal tension
crack.
•Thecomponentsofsheartransfer
mechanismare
•a)Shear resistancebyuncracked
concreteincompressionVcz
•b)Vertical componentof
aggregate interlock across crack
surfaceV
ay
•c)Dowel forceintension
reinforcementV
d
•d)ShearresistancebystirrupsV
s
Mechanism of Shear Transfer in
Cracked concrete Beam
•Freebody diagramofa segment
ofreinforced concrete beam
separatedbya diagonal tension
crack.
•Thecomponentsofsheartransfer
mechanismare
•a)Shear resistancebyuncracked
concreteincompressionVcz
•b)Vertical componentof
aggregate interlock across crack
surfaceV
ay
•c)Dowel forceintension
reinforcementV
d
•d)ShearresistancebystirrupsV
s
Mechanism of Shear Transfer in
Cracked concrete Beam
Shear strength of Reinforced Concrete
Table19ofIS456 stipulates the design shear strengthofconcreteτ
c
for different
gradesofconcrete with a wide rangeofpercentagesofpositive tensile steel
reinforcement.Itisworthmentioningthatthereinforcedconcretebeamsmustbe
provided withtheminimum shear reinforcementaspercl.40.3 even whenτ
vis
lessthanτ
cgiveninTable.
Design shear strength of concrete, τ
cin N/mm
2
A
sisthe areaoflongitudinal tension reinforcement which continuesatleast one effective
depth beyond the section considered exceptatsupport where the full areaoftension
reinforcement maybeused provided the detailingisasperIS456, cls.26.2.2and26.2.3.
Table19ofIS456 stipulates the design shear strengthofconcreteτ
c
for different
gradesofconcrete with a wide rangeofpercentagesofpositive tensile steel
reinforcement.Itisworthmentioningthatthereinforcedconcretebeamsmustbe
provided withtheminimum shear reinforcementaspercl.40.3 even whenτ
vis
lessthanτ
cgiveninTable.
Design shear strength of concrete, τ
cin N/mm
2
A
sisthe areaoflongitudinal tension reinforcement which continuesatleast one effective
depth beyond the section considered exceptatsupport where the full areaoftension
reinforcement maybeused provided the detailingisasperIS456, cls.26.2.2and26.2.3.
Maximum shear stress τ
cmaxwith shear reinforcement (cls.
40.2.3, 40.5.1 and 41.3.1)
Table20ofIS456stipulatesthemaximumshearstress
ofreinforcedconcreteinbeamsτ
cmaxasgivenbelowin
Table. Undernocircumstances, the nominal shear
stressinbeamsτ
vshall exceedτ
cmaxgiveninTable
belowfordifferentgradesofconcrete.
Maximum shear stress, τ
cmaxin N/mm
2
cmaxwith shear reinforcement (cls.
40.2.3, 40.5.1 and 41.3.1)
Table20ofIS456stipulatesthemaximumshearstress
ofreinforcedconcreteinbeamsτ
cmaxasgivenbelowin
Table. Undernocircumstances, the nominal shear
stressinbeamsτ
vshall exceedτ
cmaxgiveninTable
belowfordifferentgradesofconcrete.
Maximum shear stress, τ
cmaxin N/mm
2
Critical Section for Shear
•Clauses22.6.2 and22.6.2.1 stipulate
the critical section for shear and areas
follows:
•For beams generally subjectedto
uniformly distributed loadsorwhere
the principal loadislocated further
than 2d fromthefaceofthe support,
where disthe effective depthofthe
beam, the critical sections dependon
the conditionsofsupportsasshownin
Figs.arementionedbelow.
•(i)Whenthereactioninthedirectionof
theappliedshearintroducestension(a)
into the end regionofthe member, the
shear forceistobecomputedatthe
faceofthe supportofthe memberat
thatsection.
•Clauses22.6.2 and22.6.2.1 stipulate
the critical section for shear and areas
follows:
•For beams generally subjectedto
uniformly distributed loadsorwhere
the principal loadislocated further
than 2d fromthefaceofthe support,
where disthe effective depthofthe
beam, the critical sections dependon
the conditionsofsupportsasshownin
Figs.arementionedbelow.
•(i)Whenthereactioninthedirectionof
theappliedshearintroducestension(a)
into the end regionofthe member, the
shear forceistobecomputedatthe
faceofthe supportofthe memberat
thatsection.
Critical Section for Shear
•(ii) When the reaction in the
directionoftheapplied shear
introduces compression into the
end regionofthe member (Figs. b
andc),theshearforcecomputedat
a distanced from the faceofthe
support istobeused for the design
ofsectionslocatedatadistanceless
thandfromthefaceofthesupport.
The enhanced shear strengthof
sections closetosupports, however,
maybeconsideredasdiscussed in
thefollowingsection.
•(ii) When the reaction in the
directionoftheapplied shear
introduces compression into the
end regionofthe member (Figs. b
andc),theshearforcecomputedat
a distanced from the faceofthe
support istobeused for the design
ofsectionslocatedatadistanceless
thandfromthefaceofthesupport.
The enhanced shear strengthof
sections closetosupports, however,
maybeconsideredasdiscussed in
thefollowingsection.
Enhanced Shear Strength of Sections Close to supports
•40.5.1 General
Shear failureatsectionsofbeams
and cantilevers without shear
reinforcementwillnormally occuron
plane inclinedatanangle30”tothe
horizontal.Ifthe angleoffailure
planeisforcedtobeinclined more
steeply than this [because the
section considered(X-X)inFig.is
closetoa supportorfor other
reasons], the shear force equiredto
producefailureisincreased.
The enhancementofshear strength
maybetaken into accountinthe
designofsections near a supportby
increasing design shear strengthof
concreteto2d
c/a
v, provided that
design shear stressatthe faceofthe
support remains less than the values
giveninTable20.
Account maybetakenofthe enhancement
inany situation where the section
consideredisclosertothe faceofa support
orconcentrated load than twice the
effective depth,d.Tobeeffective, tension
reinforcementshould extendoneach sideof
the point whereit isintersectedbya
possible failure plane for a distanceatleast
equaltothe effective depth,orbeprovided
withanequivdent anchorage.
•40.5.1 General
Shear failureatsectionsofbeams
and cantilevers without shear
reinforcementwillnormally occuron
plane inclinedatanangle30”tothe
horizontal.Ifthe angleoffailure
planeisforcedtobeinclined more
steeply than this [because the
section considered(X-X)inFig.is
closetoa supportorfor other
reasons], the shear force equiredto
producefailureisincreased.
The enhancementofshear strength
maybetaken into accountinthe
designofsections near a supportby
increasing design shear strengthof
concreteto2d
c/a
v, provided that
design shear stressatthe faceofthe
support remains less than the values
giveninTable20.
Account maybetakenofthe enhancement
inany situation where the section
consideredisclosertothe faceofa support
orconcentrated load than twice the
effective depth,d.Tobeeffective, tension
reinforcementshould extendoneach sideof
the point whereit isintersectedbya
possible failure plane for a distanceatleast
equaltothe effective depth,orbeprovided
withanequivdent anchorage.
Minimum Shear Reinforcement (cls. 40.3, 26.5.1.5 and
26.5.1.6 of IS 456)
Minimum shear reinforcement hasto
beprovided even whenτ
v
islessthanτ
c
giveninTable 6.1asrecommendedin
cl.40.3ofIS456. The amountof
minimum shear reinforcement,as
givenincl.26.5.1.6,isgiven below.
•The minimum shear
reinforcement intheformof
stirrups shallbeprovided such
that:
•whereAsv= total cross-
sectionalareaofstirruplegs
effectiveinshear,
•s
v= stirrup spacing along
thelengthofthemember,
•b= breadthofthe beamor
breadthofthe webofthe
webofflanged beambw,
and
•f
y= characteristic strength
ofthestirrupreinforcement
inN/mm
2
whichshallnotbe
taken greater than415
N/mm
2
.
26.5.1.6 of IS 456)
Minimum shear reinforcement hasto
beprovided even whenτ
v
islessthanτ
c
giveninTable 6.1asrecommendedin
cl.40.3ofIS456. The amountof
minimum shear reinforcement,as
givenincl.26.5.1.6,isgiven below.
•The minimum shear
reinforcement intheformof
stirrups shallbeprovided such
that:
•whereAsv= total cross-
sectionalareaofstirruplegs
effectiveinshear,
•s
v= stirrup spacing along
thelengthofthemember,
•b= breadthofthe beamor
breadthofthe webofthe
webofflanged beambw,
and
•f
y= characteristic strength
ofthestirrupreinforcement
inN/mm
2
whichshallnotbe
taken greater than415
N/mm
2
.
•The minimum shear reinforcement is provided for the following:
•(i) Any sudden failure of beams is prevented if concrete cover bursts and the bond
to the tension steel is lost.
•(ii) Brittle shear failure is arrested which would have occurred without shear
reinforcement.
•(iii) Tension failure is prevented which would have occurred due to shrinkage,
thermal stresses and internal cracking in beams.
•(iv) To hold the reinforcement in place when concrete is poured.
•(v) Section becomes effective with the tie effect of the compression steel.
Further,cl.26.5.1.5ofIS456stipulates that the maximum spacingofshear
reinforcement measured along the axisofthe member shall notbemore than 0.75d
for vertical stirrupsandd for inclined stirrupsat45
o
, where disthe effective depthof
the section. However, the spacing shall notexceed300mminany case.
•(i) Any sudden failure of beams is prevented if concrete cover bursts and the bond
to the tension steel is lost.
•(ii) Brittle shear failure is arrested which would have occurred without shear
reinforcement.
•(iii) Tension failure is prevented which would have occurred due to shrinkage,
thermal stresses and internal cracking in beams.
•(iv) To hold the reinforcement in place when concrete is poured.
•(v) Section becomes effective with the tie effect of the compression steel.
Further,cl.26.5.1.5ofIS456stipulates that the maximum spacingofshear
reinforcement measured along the axisofthe member shall notbemore than 0.75d
for vertical stirrupsandd for inclined stirrupsat45
o
, where disthe effective depthof
the section. However, the spacing shall notexceed300mminany case.
Design of Shear Reinforcement (cl. 40.4 of IS 456)
•When τ
vis more than τ
c, shear reinforcement shall be
provided in any of the three following forms:
•(a) Vertical stirrups,
•(b) Bent-up bars along with stirrups, and
•(c) Inclined stirrups.
•Inthe caseofbent-upbars,it istobeseen that the
contributiontowardsshearresistanceofbent-upbarsshould
notbemore than fifty per centofthatofthe total shear
reinforcement.
•The amount of shear reinforcement to be provided is
determined to carry a shear force V
usequal to
•V
us=V
u-V
c=V
u-τ
cbd
•where b is the breadth of rectangular beams or b
win the
case of flanged beams.
The strengths of shear reinforcement Vusfor the three types
of shear reinforcement are as follows
•When τ
vis more than τ
c, shear reinforcement shall be
provided in any of the three following forms:
•(a) Vertical stirrups,
•(b) Bent-up bars along with stirrups, and
•(c) Inclined stirrups.
•Inthe caseofbent-upbars,it istobeseen that the
contributiontowardsshearresistanceofbent-upbarsshould
notbemore than fifty per centofthatofthe total shear
reinforcement.
•The amount of shear reinforcement to be provided is
determined to carry a shear force V
usequal to
•V
us=V
u-V
c=V
u-τ
cbd
•where b is the breadth of rectangular beams or b
win the
case of flanged beams.
The strengths of shear reinforcement Vusfor the three types
of shear reinforcement are as follows
SHEAR REINFORCEMENT
•(a) Vertical stirrups:
•(b) For inclined stirrups or a series of bars bent-up at different cross-sections:
•(c) For single bar or single group of parallel bars, all bent-up at the same cross-
section:
•where A
sv= total cross-sectional area of stirrup legs or bent-up bars within a distance
s
v,
•s
v= spacing of stirrups or bent-up bars along the length of the member,
•τ
v = nominal shear stress,
•τ
c= design shear strength of concrete,
•(a) Vertical stirrups:
•(b) For inclined stirrups or a series of bars bent-up at different cross-sections:
•(c) For single bar or single group of parallel bars, all bent-up at the same cross-
section:
•where A
sv= total cross-sectional area of stirrup legs or bent-up bars within a distance
s
v,
•s
v= spacing of stirrups or bent-up bars along the length of the member,
•τ
v = nominal shear stress,
•τ
c= design shear strength of concrete,
SHEAR REINFORCEMENT
•b = breadth of the member which for the flanged beams shall be taken
as the breadth of the web b
w,
•f
y= characteristic strength of the stirrup or bent-up reinforcement
which shall not be taken greater than 415 N/mm
2
,
•α = angle between the inclined stirrup or bent-up bar and the axis of
the member, not less than 45
o
, and d = effective depth.
The following two points are to be noted:
•(i) The total shear resistance shall be computed as the sum of the
resistance for the various types separately where more than one type
of shear reinforcement is used.
•(ii) The area of stirrups shall not be less than the minimum specified in
cl. 26.5.1.6.
•b = breadth of the member which for the flanged beams shall be taken
as the breadth of the web b
w,
•f
y= characteristic strength of the stirrup or bent-up reinforcement
which shall not be taken greater than 415 N/mm
2
,
•α = angle between the inclined stirrup or bent-up bar and the axis of
the member, not less than 45
o
, and d = effective depth.
The following two points are to be noted:
•(i) The total shear resistance shall be computed as the sum of the
resistance for the various types separately where more than one type
of shear reinforcement is used.
•(ii) The area of stirrups shall not be less than the minimum specified in
cl. 26.5.1.6.
Shear Reinforcement for Sections Close to Supports
As stipulated in cl. 40.5.2 of IS 456, the total area of the
required shear reinforcement A
sis obtained from:
This reinforcement should be provided within the middle three quarters of a
v
, where a
vis less
than d, horizontal shear reinforcement will be effective than vertical.
Alternatively, one simplified method has been recommended in cl. 40.5.3 of IS 456 and the
same is given below.
The following methodisfor beams carrying generally uniform loadorwhere the principal
loadislocated further than 2d from the faceofsupport.Theshear stressiscalculatedata
section a distance d from the faceofsupport. The valueofτ
ciscalculatedinaccordance with
Table 6.1andappropriate shear reinforcementisprovidedatsections closertothe support.
Nofurther check forshearatsuch sectionsisrequired.
As stipulated in cl. 40.5.2 of IS 456, the total area of the
required shear reinforcement A
sis obtained from:
This reinforcement should be provided within the middle three quarters of a
v
, where a
vis less
than d, horizontal shear reinforcement will be effective than vertical.
Alternatively, one simplified method has been recommended in cl. 40.5.3 of IS 456 and the
same is given below.
The following methodisfor beams carrying generally uniform loadorwhere the principal
loadislocated further than 2d from the faceofsupport.Theshear stressiscalculatedata
section a distance d from the faceofsupport. The valueofτ
ciscalculatedinaccordance with
Table 6.1andappropriate shear reinforcementisprovidedatsections closertothe support.
Nofurther check forshearatsuch sectionsisrequired.
CurtailmentofTension ReinforcementinFlexural
Members(cl.26.2.3.2ofIS456)
Curtailmentoftension reinforcementisdonetoprovide the required reduced
areaofsteel with the reductionofthe bending moment. However, shear force
increases with the reductionofbending moment. Therefore,it isnecessaryto
satisfy any oneoffollowing three conditions while terminating the flexural
reinforcementintensionzone:
•(i)Theshearstressτ
vatthecut-offpointshouldnotexceedtwo-
thirdsofthepermitted value which includes theshear strength
ofthewebreinforcement.Accordingly,
Members(cl.26.2.3.2ofIS456)
Curtailmentoftension reinforcementisdonetoprovide the required reduced
areaofsteel with the reductionofthe bending moment. However, shear force
increases with the reductionofbending moment. Therefore,it isnecessaryto
satisfy any oneoffollowing three conditions while terminating the flexural
reinforcementintensionzone:
•(i)Theshearstressτ
vatthecut-offpointshouldnotexceedtwo-
thirdsofthepermitted value which includes theshear strength
ofthewebreinforcement.Accordingly,
Curtailment of Tension Reinforcement in
Flexural Members (cl. 26.2.3.2 of IS 456)
•(ii) For eachofthe terminated bars, additional stirrup area
shouldbeprovided over a distanceofthree-fourthof
effectivedepthfromthecut-offpoint.Theadditionalstirrup
areashallnotbelessthan0.4bs/f
y,wherebisthebreadth
ofrectangular beams andisreplacedbyb
w, the breadthof
thewebforflangedbeams,
•s=spacingofadditionalstirrupsand
•f
yisthe characteristic strengthofstirrup reinforcementin
N/mm
2
.Thevalueofsshallnotexceedd/(8
β
b),
whereβ
bistheratioofareaofbarscut-offtothetotalarea
ofbarsatthatsection,anddistheeffectivedepth.
Flexural Members (cl. 26.2.3.2 of IS 456)
•(ii) For eachofthe terminated bars, additional stirrup area
shouldbeprovided over a distanceofthree-fourthof
effectivedepthfromthecut-offpoint.Theadditionalstirrup
areashallnotbelessthan0.4bs/f
y,wherebisthebreadth
ofrectangular beams andisreplacedbyb
w, the breadthof
thewebforflangedbeams,
•s=spacingofadditionalstirrupsand
•f
yisthe characteristic strengthofstirrup reinforcementin
N/mm
2
.Thevalueofsshallnotexceedd/(8
β
b),
whereβ
bistheratioofareaofbarscut-offtothetotalarea
ofbarsatthatsection,anddistheeffectivedepth.
Curtailment of Tension Reinforcement in
Flexural Members (cl. 26.2.3.2 of IS 456)
•(iii) For barsofdiameters36mmand smaller, the
continuingbarsprovidedoublethearearequiredfor
flexureatthe cut-off point. The shear stress should
notexceedthree-fourthsthatpermitted.
Flexural Members (cl. 26.2.3.2 of IS 456)
•(iii) For barsofdiameters36mmand smaller, the
continuingbarsprovidedoublethearearequiredfor
flexureatthe cut-off point. The shear stress should
notexceedthree-fourthsthatpermitted.
BOND, ANCHORAGE
G.C. BEHERA
G.C. BEHERA
BOND
•The bond between steel and concreteisvery
importantandessentialsothattheycanacttogether
withoutanyslipinaloadedstructure.
•With the perfect bond between them, the plane
sectionofabeamremainsplaneevenafterbending.
Thelengthofamemberrequiredtodevelopthefull
bondiscalledtheanchoragelength.
•Thebondismeasuredbybondstress.Thelocalbond
stress varies along a member with the variationof
bending moment. The average value throughout its
anchoragelengthisdesignatedastheaveragebond
stress.Inourcalculation,theaveragebondstresswill
beused.
•The bond between steel and concreteisvery
importantandessentialsothattheycanacttogether
withoutanyslipinaloadedstructure.
•With the perfect bond between them, the plane
sectionofabeamremainsplaneevenafterbending.
Thelengthofamemberrequiredtodevelopthefull
bondiscalledtheanchoragelength.
•Thebondismeasuredbybondstress.Thelocalbond
stress varies along a member with the variationof
bending moment. The average value throughout its
anchoragelengthisdesignatedastheaveragebond
stress.Inourcalculation,theaveragebondstresswill
beused.
BOND
•Thetermbondreferstotheadhesionbetweentheconcreteand
thesteelwhichresiststheslippingofsteelbarfromconcrete.It
isthisbond whichisresponsible for transfertostresses from
steeltoconcrete and thereby providing composite actionof
steelandconcreteinR.C.C.Thebonddevelopsduetosettingof
concreteondryingwhichresultsingrippingofthesteelbars.
•The bond resistanceinreinforced concreteisobtainedby
followingmechanism:
➢Chemical adhesion: it isduetogum like propertyofthe
substances,formedaftersettingofconcrete.
➢Frictional resistance: it isduetoadhesion between steel and
concrete.
➢Grippingaction:itisduetogrippingofsteelbytheconcreteon
drying.
➢Mechanical interlock:Itis providedbythe corrugationsorribs
presentonthesurfaceofthedeformedbars.
•Thetermbondreferstotheadhesionbetweentheconcreteand
thesteelwhichresiststheslippingofsteelbarfromconcrete.It
isthisbond whichisresponsible for transfertostresses from
steeltoconcrete and thereby providing composite actionof
steelandconcreteinR.C.C.Thebonddevelopsduetosettingof
concreteondryingwhichresultsingrippingofthesteelbars.
•The bond resistanceinreinforced concreteisobtainedby
followingmechanism:
➢Chemical adhesion: it isduetogum like propertyofthe
substances,formedaftersettingofconcrete.
➢Frictional resistance: it isduetoadhesion between steel and
concrete.
➢Grippingaction:itisduetogrippingofsteelbytheconcreteon
drying.
➢Mechanical interlock:Itis providedbythe corrugationsorribs
presentonthesurfaceofthedeformedbars.
BOND
•The bondisassumedtobeperfectinthe
designofreinforced concrete. The bond
betweensteelandconcretecanbeincreased
bythefollowingmethods:
➢Usingtwistedordeformedbar.
➢Usingrichmixofconcrete.
➢Adequatecompactionandcuringofconcrete
forpropersetting.
➢Providinghooksattheendofreinforcingbars.
•The bondisassumedtobeperfectinthe
designofreinforced concrete. The bond
betweensteelandconcretecanbeincreased
bythefollowingmethods:
➢Usingtwistedordeformedbar.
➢Usingrichmixofconcrete.
➢Adequatecompactionandcuringofconcrete
forpropersetting.
➢Providinghooksattheendofreinforcingbars.
BOND
•Theconceptofdevelopment length and anchorageofthe steel
reinforcement has replaced the earlier practiceofchecking and
satisfying the permissible flexural bond stress.Itisobserved that
theflexuralbondstressdoesnotprovideanappropriatemethodof
ensuring safety against bond failure. The development length
criteria gives a better estimateofthe strengthofthe bond.In
simply supported beams, the critical section existsatthe pointof
the maximum stressorata section where bars are curtailed.In
continuous beam,inadditiontopointsofmaximum stress and
curtailment,pointsofcontraflexureshouldbecheckedforbond.
•There are two types of bond failure:
➢Anchorage bond failure
➢Flexural bond failure
•Theconceptofdevelopment length and anchorageofthe steel
reinforcement has replaced the earlier practiceofchecking and
satisfying the permissible flexural bond stress.Itisobserved that
theflexuralbondstressdoesnotprovideanappropriatemethodof
ensuring safety against bond failure. The development length
criteria gives a better estimateofthe strengthofthe bond.In
simply supported beams, the critical section existsatthe pointof
the maximum stressorata section where bars are curtailed.In
continuous beam,inadditiontopointsofmaximum stress and
curtailment,pointsofcontraflexureshouldbecheckedforbond.
•There are two types of bond failure:
➢Anchorage bond failure
➢Flexural bond failure
BOND ANCHORAGE
•Thus,atensilememberhastobeanchoredproperlyby
providingadditionallengthoneithersideofthepoint
ofmaximumtension,whichisknownas‘Development
lengthintension’.Similarly,forcompressionmembers
also,wehave‘DevelopmentlengthL
dincompression’.
•Deformedbarsareknowntobesuperiortothesmooth
mildsteel bars duetothe presenceofribs.Insuch a
case,itisneededtocheck for the sufficient
developmentlengthL
donly rather than checking both
for the local bond stressanddevelopment lengthas
requiredforthesmoothmildsteelbars.Accordingly,IS
456,cl.26.2 stipulates the requirementsofproper
anchorageofreinforcementintermsofdevelopment
lengthL
d
onlyemployingdesignbondstressτ
bd.
•Thus,atensilememberhastobeanchoredproperlyby
providingadditionallengthoneithersideofthepoint
ofmaximumtension,whichisknownas‘Development
lengthintension’.Similarly,forcompressionmembers
also,wehave‘DevelopmentlengthL
dincompression’.
•Deformedbarsareknowntobesuperiortothesmooth
mildsteel bars duetothe presenceofribs.Insuch a
case,itisneededtocheck for the sufficient
developmentlengthL
donly rather than checking both
for the local bond stressanddevelopment lengthas
requiredforthesmoothmildsteelbars.Accordingly,IS
456,cl.26.2 stipulates the requirementsofproper
anchorageofreinforcementintermsofdevelopment
lengthL
d
onlyemployingdesignbondstressτ
bd.
Design Bond Stress τ
bd
•Thedesign bond stressτ
bdisdefinedasthe shear force perunit
nominal surface areaofreinforcingbar.Thestressisactingonthe
interface between bars and surrounding concreteandalong the
directionparalleltothebars.
•Thisconceptofdesign bond stress finally resultsinadditional
lengthofabarofspecifieddiametertobeprovidedbeyondagiven
criticalsection.Though, theoverall bond failuremaybeavoidedby
thisprovisionofadditionaldevelopmentlengthL
d,slippageofabar
may not always resultinoverall failureofa beam.Itis, thus,
desirabletoprovide end anchorages alsotomaintain the integrity
ofthe structureandthereby,toenableitcarrying the loads. Clause
26.2ofIS456 stipulates,“Thecalculated tensionorcompressionin
anybaratanysectionshallbedevelopedoneachsideofthesection
byanappropriate development lengthorend anchorageorbya
combinationthereof.”
bd
•Thedesign bond stressτ
bdisdefinedasthe shear force perunit
nominal surface areaofreinforcingbar.Thestressisactingonthe
interface between bars and surrounding concreteandalong the
directionparalleltothebars.
•Thisconceptofdesign bond stress finally resultsinadditional
lengthofabarofspecifieddiametertobeprovidedbeyondagiven
criticalsection.Though, theoverall bond failuremaybeavoidedby
thisprovisionofadditionaldevelopmentlengthL
d,slippageofabar
may not always resultinoverall failureofa beam.Itis, thus,
desirabletoprovide end anchorages alsotomaintain the integrity
ofthe structureandthereby,toenableitcarrying the loads. Clause
26.2ofIS456 stipulates,“Thecalculated tensionorcompressionin
anybaratanysectionshallbedevelopedoneachsideofthesection
byanappropriate development lengthorend anchorageorbya
combinationthereof.”
Design bond stress –values
•Thelocal bond stress varies along the lengthofthe reinforcement
while the average bond stress gives the average value throughout
itsdevelopmentlength.
•Thisaverage bond stressisstill usedinthe working stress method
andIS456has mentioned aboutit incl. B-2.1.2. However,inthe
limit state methodofdesign, the average bond stress has been
designatedasdesign bond stressτ
bdand the values are givenincl.
26.2.1.1.Thesameisgivenbelowasareadyreference.
•
For deformed bars conformingtoIS1786, these values shallbeincreasedby60
percent. For barsincompression, the valuesofbond stressintension shallbe
increasedby25percent.
•Thelocal bond stress varies along the lengthofthe reinforcement
while the average bond stress gives the average value throughout
itsdevelopmentlength.
•Thisaverage bond stressisstill usedinthe working stress method
andIS456has mentioned aboutit incl. B-2.1.2. However,inthe
limit state methodofdesign, the average bond stress has been
designatedasdesign bond stressτ
bdand the values are givenincl.
26.2.1.1.Thesameisgivenbelowasareadyreference.
•
For deformed bars conformingtoIS1786, these values shallbeincreasedby60
percent. For barsincompression, the valuesofbond stressintension shallbe
increasedby25percent.
DESIGN BOND STRESS
Grade of concrete τ
bdfor plain bars (N/mm
2
) τ
bdfor deformed bars (N/mm
2
)
M20 1.2 1.92
M25 1.4 2.24
M30 1.5 2.4
M35 1.7 2.72
M40 and above 1.9 3.04
Grade of concrete τ
bdfor plain bars (N/mm
2
) τ
bdfor deformed bars (N/mm
2
)
M20 1.2 1.92
M25 1.4 2.24
M30 1.5 2.4
M35 1.7 2.72
M40 and above 1.9 3.04
ACNHORAGE BOND (DEVELOPMENT LENGTH)
Figure shows a simply supported beam
subjectedtouniformly distributed load. Because
ofthe maximum moment, theA
strequiredisthe
maximumatx = L/2. For any section 1-1ata
distance x <L/2, someofthetensile bars canbe
curtailed. Letusthen assume that section 1-1is
the theoretical cut-off pointofone bar.
However,it isnecessarytoextend the bar for a
lengthL
dasexplained earlier. Letusderive the
expressiontodetermine L
dofthisbar.
The free body diagramofthe segmentABofthe bar.
AtB,the tensile forceT tryingtopull out the barisof
the value T =(π/4)*(
)
2
*s
S, whereisthe nominal
diameterofthebar andσ
sisthe tensile stressinbar at
the section consideredatdesign loads.Itisnecessary
tohave the resistance forcetobedevelopedbyτ
bdfor
the length L
dtoovercome the tensile force.
Figure shows a simply supported beam
subjectedtouniformly distributed load. Because
ofthe maximum moment, theA
strequiredisthe
maximumatx = L/2. For any section 1-1ata
distance x <L/2, someofthetensile bars canbe
curtailed. Letusthen assume that section 1-1is
the theoretical cut-off pointofone bar.
However,it isnecessarytoextend the bar for a
lengthL
dasexplained earlier. Letusderive the
expressiontodetermine L
dofthisbar.
The free body diagramofthe segmentABofthe bar.
AtB,the tensile forceT tryingtopull out the barisof
the value T =(π/4)*(
)
2
*s
S, whereisthe nominal
diameterofthebar andσ
sisthe tensile stressinbar at
the section consideredatdesign loads.Itisnecessary
tohave the resistance forcetobedevelopedbyτ
bdfor
the length L
dtoovercome the tensile force.
ACNHORAGE BOND (DEVELOPMENT LENGTH)
T = (π/4)*()
2
*s
S
s
Sis the stress in steel and maximum is 0.87f
y
T = (π/4)*(
)
2
*0.87f
y
It is necessary to have the resistance force to be
developed by τ
bdfor the length L
dto overcome the
tensile force.
The resistance force is developed on the periphry
of the rod .
Area of this bond stress acting over a length of L
d=
π *
*L
d
Bond force over this length Ld= π *
*L
d*
bdBond
force must be greater than tension T
T = (π/4)*()
2
*s
S
s
Sis the stress in steel and maximum is 0.87f
y
T = (π/4)*(
)
2
*0.87f
y
It is necessary to have the resistance force to be
developed by τ
bdfor the length L
dto overcome the
tensile force.
The resistance force is developed on the periphry
of the rod .
Area of this bond stress acting over a length of L
d=
π *
*L
d
Bond force over this length Ld= π *
*L
d*
bdBond
force must be greater than tension T
FLEXURAL BOND
•Flexuralbondisalsoknownaslocalbond.
•Localbondatapointistherateofchangeoftensionin
Steelatagivenlocation.
•Inasimplebeam,atthecriticalsectioni.e.attheface
ofsupport,atthepointsofinflectionandatthepoints
ofhighshearforce,highbondstressaydevelopdueto
the large variationsinbending moment. These bond
stressesarecalledflexuralbondstressesandshouldbe
checkedcarefullyatallcriticalsections.
•Consider a beam subjectedtoflexural loading.
Considertwosectionsatadistanceofχandχ+Δχalong
thelengthofthebeamsubjectedtoamomentMand
M+ΔMrespectively.
•LetTandT+ΔTisthetensiondevelopedther
•Flexuralbondisalsoknownaslocalbond.
•Localbondatapointistherateofchangeoftensionin
Steelatagivenlocation.
•Inasimplebeam,atthecriticalsectioni.e.attheface
ofsupport,atthepointsofinflectionandatthepoints
ofhighshearforce,highbondstressaydevelopdueto
the large variationsinbending moment. These bond
stressesarecalledflexuralbondstressesandshouldbe
checkedcarefullyatallcriticalsections.
•Consider a beam subjectedtoflexural loading.
Considertwosectionsatadistanceofχandχ+Δχalong
thelengthofthebeamsubjectedtoamomentMand
M+ΔMrespectively.
•LetTandT+ΔTisthetensiondevelopedther
FLEXURAL BOND
•Itzistheleverarm,T=M/z
•T+ΔT=(M+ΔM)/z
•ΔT=ΔM/z
•ΔTmustberegisteredbybondstress.
•ΔT=π*
*Δx*
bd=ΔM/z
•Itzistheleverarm,T=M/z
•T+ΔT=(M+ΔM)/z
•ΔT=ΔM/z
•ΔTmustberegisteredbybondstress.
•ΔT=π*
*Δx*
bd=ΔM/z
•From anchorage bond we have
•Equating both we will get
•Where M
1is the moment of Resistance considering all bars there
•As per IS 456
•L
0isthe sumofanchorage beyond the centreofsupport and equivalent anchorage
valueofany hookorany mechanical anchorageatthe support.
•This additional lengthisprovided forsafety.
•L
oislimitedtothe effectivedepthofthe memberor12which everisgreater.
•L
discalledasthe development length.Itisthe minimum lengthofthe bar which
mustbeembeddedinthe concrete beyond any sectiontodevelop its full strength.
Thisisalso calledasanchorage lengthincaseofaxial tensionoraxial compression
and development lengthincaseofflexural tensionorflexural compression.
•Equating both we will get
•Where M
1is the moment of Resistance considering all bars there
•As per IS 456
•L
0isthe sumofanchorage beyond the centreofsupport and equivalent anchorage
valueofany hookorany mechanical anchorageatthe support.
•This additional lengthisprovided forsafety.
•L
oislimitedtothe effectivedepthofthe memberor12which everisgreater.
•L
discalledasthe development length.Itisthe minimum lengthofthe bar which
mustbeembeddedinthe concrete beyond any sectiontodevelop its full strength.
Thisisalso calledasanchorage lengthincaseofaxial tensionoraxial compression
and development lengthincaseofflexural tensionorflexural compression.
•To get the L
dvalue satisfactory,
➢Decrease bar dia,
➢Increase L
0
➢By reducing bent up bars
Checking of L
dvalue
➢At simple supports
➢At Cantilever Support
➢At point of Contra flexure
➢At bar cut off points
dvalue satisfactory,
➢Decrease bar dia,
➢Increase L
0
➢By reducing bent up bars
Checking of L
dvalue
➢At simple supports
➢At Cantilever Support
➢At point of Contra flexure
➢At bar cut off points
ANCHORING OF BARS
•25.2.2 ANCHORING REINFORCING BARS25.2.2.1
Any deficiencyinthe required development
length canbemadeup byanchoring the
reinforcing bars suitably. Deformed bars have
superior bond properties owingtomechanical
bearing and, therefore, provisiabsolutely
essential. However, plain bars should preferably
endinhooks,astheremaybesomeuncertainty
regarding the full mobilizationofbond strength
throughadhesionandfriction.onofhooksisnot
•25.2.2 ANCHORING REINFORCING BARS25.2.2.1
Any deficiencyinthe required development
length canbemadeup byanchoring the
reinforcing bars suitably. Deformed bars have
superior bond properties owingtomechanical
bearing and, therefore, provisiabsolutely
essential. However, plain bars should preferably
endinhooks,astheremaybesomeuncertainty
regarding the full mobilizationofbond strength
throughadhesionandfriction.onofhooksisnot
Anchoring Reinforcing Bars
The salient points are:
•Deformed bars may notneedend anchoragesif
the development length requirementissatisfied.
•Hooks should normallybeprovided for plain
barsintension.
•Standard hooks andbendsshouldbeasperIS
2502 orasgiveninTable67ofSP-16, which are
showninFigs.
•The anchorage valueofstandardbendshallbe
consideredas4 times the diameterofthe bar for
each45
o
bendsubjecttoa maximum valueof16
times the diameterofthe bar.
•The anchorage valueofstandard U-type hook
shallbe16times the diameterofthe bar.
The salient points are:
•Deformed bars may notneedend anchoragesif
the development length requirementissatisfied.
•Hooks should normallybeprovided for plain
barsintension.
•Standard hooks andbendsshouldbeasperIS
2502 orasgiveninTable67ofSP-16, which are
showninFigs.
•The anchorage valueofstandardbendshallbe
consideredas4 times the diameterofthe bar for
each45
o
bendsubjecttoa maximum valueof16
times the diameterofthe bar.
•The anchorage valueofstandard U-type hook
shallbe16times the diameterofthe bar.
BARS IN COMPRESSION
•The anchorage length of straight compression
bars shall be equal to its development length
as mentioned.
• The development length shall include the
projected length of hooks, bends and straight
lengths beyond bends, if provided.
•The anchorage length of straight compression
bars shall be equal to its development length
as mentioned.
• The development length shall include the
projected length of hooks, bends and straight
lengths beyond bends, if provided.
Reinforcement Splicing (cl. 26.2.5 of IS 456)
•Reinforcementisneededtobejoinedtomakeitlongerbyoverlapping sufficient length
orbyweldingtodevelop its full design bond stress. They shouldbeaway from the
sectionsofmaximum stress andbestaggered.IS456(cl.26.2.5) recommends that
splicesinflexural members should notbeatsections where the bending momentis
more than50per centofthe momentofresistance and not more than half the bars
shallbesplicedata section.
•(a) Lap Splices (cl. 26.2.5.1 of IS 456)
•The following are the salient points:
•• They should be used for bar diameters up to 36 mm.
•• They should be considered as staggered if the centre to centre distance of the splices
is at least 1.3 times the lap length calculated as mentioned below.
•• The lap length including anchorage value of hooks for bars in flexural tension shall be
L
dor 30φ, whichever is greater. The same for direct tension shall be 2L
dor 30φ,
whichever is greater.
•• The lap length in compression shall be equal to L
din compression but not less than
24φ.
•• The lap length shall be calculated on the basis of diameter of the smaller bar when
bars of two different diameters are to be spliced.
•• Lap splices of bundled bars shall be made by splicing one bar at a time and all such
individual splices within a bundle shall be staggered.
•Reinforcementisneededtobejoinedtomakeitlongerbyoverlapping sufficient length
orbyweldingtodevelop its full design bond stress. They shouldbeaway from the
sectionsofmaximum stress andbestaggered.IS456(cl.26.2.5) recommends that
splicesinflexural members should notbeatsections where the bending momentis
more than50per centofthe momentofresistance and not more than half the bars
shallbesplicedata section.
•(a) Lap Splices (cl. 26.2.5.1 of IS 456)
•The following are the salient points:
•• They should be used for bar diameters up to 36 mm.
•• They should be considered as staggered if the centre to centre distance of the splices
is at least 1.3 times the lap length calculated as mentioned below.
•• The lap length including anchorage value of hooks for bars in flexural tension shall be
L
dor 30φ, whichever is greater. The same for direct tension shall be 2L
dor 30φ,
whichever is greater.
•• The lap length in compression shall be equal to L
din compression but not less than
24φ.
•• The lap length shall be calculated on the basis of diameter of the smaller bar when
bars of two different diameters are to be spliced.
•• Lap splices of bundled bars shall be made by splicing one bar at a time and all such
individual splices within a bundle shall be staggered.
•(b) Strength of Welds (cl. 26.2.5.2 of IS 456)
•The strengthofwelded splices and mechanical
connections shallbetakenas100 per centofthe
designstrengthofjoinedbarsforcompressionsplices.
•Fortensionsplices,suchstrengthofweldedbarsshall
betakenas80percentofthe design strengthof
welded bars. However,itcangoevenupto100per
centifweldingisstrictlysupervisedandifatanycross-
sectionofthe member not more than20percentof
the tensile reinforcementiswelded. For mechanical
connectionoftension splice, 100 per centofdesign
strengthofmechanicalconnectionshallbetaken.
•The strengthofwelded splices and mechanical
connections shallbetakenas100 per centofthe
designstrengthofjoinedbarsforcompressionsplices.
•Fortensionsplices,suchstrengthofweldedbarsshall
betakenas80percentofthe design strengthof
welded bars. However,itcangoevenupto100per
centifweldingisstrictlysupervisedandifatanycross-
sectionofthe member not more than20percentof
the tensile reinforcementiswelded. For mechanical
connectionoftension splice, 100 per centofdesign
strengthofmechanicalconnectionshallbetaken.
PROBLEM
Determine the anchorage length of 4-20T reinforcing bars going into the
support of the simply supported beam shown in Fig. The factored shear
force V
u= 280 kN, width of the column support = 300 mm. Use M 20 concrete
and Fe 415 steel.
Determine the anchorage length of 4-20T reinforcing bars going into the
support of the simply supported beam shown in Fig. The factored shear
force V
u= 280 kN, width of the column support = 300 mm. Use M 20 concrete
and Fe 415 steel.
SOLUTION
•τ
bdfor M 20 and Fe 415 (with 60% increased) = 1.6(1.2) = 1.92 N/mm
2
•L
d=47.01
•τ
bdfor M 20 and Fe 415 (with 60% increased) = 1.6(1.2) = 1.92 N/mm
2
•L
d=47.01
So the anchorage is 100 mm beyond centre of support as shown in Fig.
DESIGN OF BEAMS
G.C. BEHERA
G.C. BEHERA
EFFECTIVE LENGTH
•Clear distance between walls = 6m
•Thickness of one wall =300 mm
•Thickness of another wall =400 mm
•Effective depth of beam= 450 mm
•Simply supported Beam
•Find effective length of beam?
•L
eff=Clear span+deff=6m+.45m=6.45 m
•L
eff=C/C bet supports=6m+.3/2+0.4/2=6.35 m
•Thickness of one wall =300 mm
•Thickness of another wall =400 mm
•Effective depth of beam= 450 mm
•Simply supported Beam
•Find effective length of beam?
•L
eff=Clear span+deff=6m+.45m=6.45 m
•L
eff=C/C bet supports=6m+.3/2+0.4/2=6.35 m
EFFECTIVE LENGTH
•B)Continuous BeamorSlab -Inthecaseofcontinuous beamor
slab,ifthe widthofthe supportisless thanl/12ofthe clear span,
the effective span shallbeasin22.2 (a).Ifthe supports are wider
thanI/12ofthe clear spanor600mmwhicheverisless, the
effectivespanshallbetakenasunder:
•1)Forendspanwithoneendfixedandtheothercontinuousorfor
intermediate spans, the effective span shallbethe clear span
betweensupports;
•2) Forendspan with oneendfree and the other continuous, the
effectivespanshallbeequaltotheclearspanplushalftheeffective
depthofthe beamorslaborthe clear spanplushalf the widthof
thediscontinuoussupport,whicheverisless;
•3)Inthe caseofspans with rollerorrocket bearings, the effective
spanshallalwaysbethedistancebetweenthecentresofbearings
•B)Continuous BeamorSlab -Inthecaseofcontinuous beamor
slab,ifthe widthofthe supportisless thanl/12ofthe clear span,
the effective span shallbeasin22.2 (a).Ifthe supports are wider
thanI/12ofthe clear spanor600mmwhicheverisless, the
effectivespanshallbetakenasunder:
•1)Forendspanwithoneendfixedandtheothercontinuousorfor
intermediate spans, the effective span shallbethe clear span
betweensupports;
•2) Forendspan with oneendfree and the other continuous, the
effectivespanshallbeequaltotheclearspanplushalftheeffective
depthofthe beamorslaborthe clear spanplushalf the widthof
thediscontinuoussupport,whicheverisless;
•3)Inthe caseofspans with rollerorrocket bearings, the effective
spanshallalwaysbethedistancebetweenthecentresofbearings
c)Cantilever-The effective lengthofa cantilever shall betakenasits lengthtothe
faceofthe support plus half the effective depth except whereitforms theendofa
continuous beam where the lengthtothe centreofsupport shallbetaken.
•D) Frames-Inthe analysis of a continuous frame, centre to centre distance shall be
used.
faceofthe support plus half the effective depth except whereitforms theendofa
continuous beam where the lengthtothe centreofsupport shallbetaken.
•D) Frames-Inthe analysis of a continuous frame, centre to centre distance shall be
used.
SHEAR AND BM COEFFICIENTS
TENSILE REINFORCEMENT
Minimum reinforcement not less than
Minimum reinforcement not less than
PROBLEM
•Design aRCrectangular beam for a simply supported beam supportedontwo
masonry walls230mmthick having centretocentre distance 6mt. The beamis
carrying imposed laod15kN/m. Design the beam with M20andFe415.
•SOLUTION:
•Step-1: AssumptionofDimensions:
•Take a trial depth=L/10toL/15forsimply supported case.(L/d ratio20).
Here letustake L/d=12d=6m/12=500mm,TakeD=500mm,d
effassumed=450mm
Assumeb=D/2=250mm
Step-2: CalculationofEffective length:
Clear Span=L
c=6mt-0.23m=5.77m
a)Leff=c/c distanceofsupports=6.0 m
b)Leff=L
c+d
eff= 5.77m+0.45m=6.22m
which everissmaller= 6.0mt.
Step-3: Calculation loadandmoment:
DLofthe beam=1m*0.5m*0.25m*25=3.125kN/m
ImposedLaod=15kN/m
Total Load=DL+LL=3.125+15=18.125kN/m
Factored Load=wu=18.125*1.5=27.1875=27.2kN/m
Factored Moment=w
ul
eff
2
/8=122.36kNm
230 mm
6000 mm
L
c=5770 mm
•Design aRCrectangular beam for a simply supported beam supportedontwo
masonry walls230mmthick having centretocentre distance 6mt. The beamis
carrying imposed laod15kN/m. Design the beam with M20andFe415.
•SOLUTION:
•Step-1: AssumptionofDimensions:
•Take a trial depth=L/10toL/15forsimply supported case.(L/d ratio20).
Here letustake L/d=12d=6m/12=500mm,TakeD=500mm,d
effassumed=450mm
Assumeb=D/2=250mm
Step-2: CalculationofEffective length:
Clear Span=L
c=6mt-0.23m=5.77m
a)Leff=c/c distanceofsupports=6.0 m
b)Leff=L
c+d
eff= 5.77m+0.45m=6.22m
which everissmaller= 6.0mt.
Step-3: Calculation loadandmoment:
DLofthe beam=1m*0.5m*0.25m*25=3.125kN/m
ImposedLaod=15kN/m
Total Load=DL+LL=3.125+15=18.125kN/m
Factored Load=wu=18.125*1.5=27.1875=27.2kN/m
Factored Moment=w
ul
eff
2
/8=122.36kNm
230 mm
6000 mm
L
c=5770 mm
PROBLEM
Step-3: Calculation loadandmoment:
MaximumSFwillbecalculatedata distance d
efffromFaceofthe support.
Vu= w
ul
eff/2-w
u*565=66.20kN
Step-4: Depth Calculation:
Toresist a factored bending moment122.36kNm making a balance section with
breadth250mm,werequired a depth
d
required=421.11mm
d
assumed> d
required
Nowwewill taked=450mm,b=250mm
Step-5: AreaofMain steel:
Asdepth requiredis421.11mm,wehave taken450mm,itwillbedesignedasunder-
reinforced section.
230 mm
6000 mm
L
c=5770 mm
.23/2+.45=.565m
Step-3: Calculation loadandmoment:
MaximumSFwillbecalculatedata distance d
efffromFaceofthe support.
Vu= w
ul
eff/2-w
u*565=66.20kN
Step-4: Depth Calculation:
Toresist a factored bending moment122.36kNm making a balance section with
breadth250mm,werequired a depth
d
required=421.11mm
d
assumed> d
required
Nowwewill taked=450mm,b=250mm
Step-5: AreaofMain steel:
Asdepth requiredis421.11mm,wehave taken450mm,itwillbedesignedasunder-
reinforced section.
230 mm
6000 mm
L
c=5770 mm
.23/2+.45=.565m
A
st=903.80mm
2
or4517.885mm
2
So, take A
st=3 nos.of20mmdia. Bars, A
st(provided)=942mm
2
Step-5: Check forShear:
Calculationof
c
st=903.80mm
2
or4517.885mm
2
So, take A
st=3 nos.of20mmdia. Bars, A
st(provided)=942mm
2
Step-5: Check forShear:
Calculationof
c
SOLUTION
p
t
c
0.75 0.56
1.00 0.62
0.837
Calculation of shear resistance of concrete
c
By interpolation
So, provide shear reinforcement
As V
suis less , let us take 2 legged 8 mm dia. Bars as stirrup
p
t
c
0.75 0.56
1.00 0.62
0.837
Calculation of shear resistance of concrete
c
By interpolation
So, provide shear reinforcement
As V
suis less , let us take 2 legged 8 mm dia. Bars as stirrup
•Maximum spacing as per minimum reinforcement
•Spacing should be least of the following
•a) 0.75d=0.75*450=337.5 mm
•b) 300 mm
•So, provide 2 legged 8 mm dia. Stirrups 300 mm c/c throughout the length of the beam.
Step-6: Check fordevelopment length:
Vu= w
ul
eff/2-w
u*565=66.20kN
Taking no anchorage length L
0=0
Condition is satisfied
•Spacing should be least of the following
•a) 0.75d=0.75*450=337.5 mm
•b) 300 mm
•So, provide 2 legged 8 mm dia. Stirrups 300 mm c/c throughout the length of the beam.
Step-6: Check fordevelopment length:
Vu= w
ul
eff/2-w
u*565=66.20kN
Taking no anchorage length L
0=0
Condition is satisfied
•Step-7: Check for deflection:
•For simply supported beam less than 10 m, k=20
•m
l= modification factor for length if l> 10 m
•m
t= modification factor for tensile reinforcement
•m
c= modification factor for compressive reinforcement
•m
f= modification factor for flanged beam
Here as length is < 10 m, do not consider m
l, it is not a flanged
section and also there is no compressive reinforcement.
Calculation of m
t: p
t=0.837
f
s=0.58*f
y*(A
strequired/ A
stprovided)=0.58*415*(903.8/ 942)=230.93
P
t=0.837, f
s=230.93
•For simply supported beam less than 10 m, k=20
•m
l= modification factor for length if l> 10 m
•m
t= modification factor for tensile reinforcement
•m
c= modification factor for compressive reinforcement
•m
f= modification factor for flanged beam
Here as length is < 10 m, do not consider m
l, it is not a flanged
section and also there is no compressive reinforcement.
Calculation of m
t: p
t=0.837
f
s=0.58*f
y*(A
strequired/ A
stprovided)=0.58*415*(903.8/ 942)=230.93
P
t=0.837, f
s=230.93
m
t=1.05,
l
eff/d
eff=6000/450=13.33
t=1.05,
l
eff/d
eff=6000/450=13.33
Longitudinal and cross section of Beam
CLASS-10 TORSION
G.C. BEHERA
G.C. BEHERA
TORSION
•Beams and slabs are subjectedtotorsioninadditionto
bending moment and shear force. Loads acting normalto
theplaneofbendingwillcausebendingmomentandshear
force.However,loadsawayfrom theplaneofbendingwill
inducetorsional momentalongwithbending momentand
shear. Space frames, invertedL-beamsasinsupporting
sunshades and canopies, beams curvedinplan, edge
beamsofslabs are someofthe examples where torsional
momentsarealsopresent.
•Skewbendingtheory,space-trussanalogyandNadai’ssand
heap theory are someofthe theories developedto
understand the behaviourofreinforced concrete under
torsioncombinedwithbendingmomentandshear.
•Beams and slabs are subjectedtotorsioninadditionto
bending moment and shear force. Loads acting normalto
theplaneofbendingwillcausebendingmomentandshear
force.However,loadsawayfrom theplaneofbendingwill
inducetorsional momentalongwithbending momentand
shear. Space frames, invertedL-beamsasinsupporting
sunshades and canopies, beams curvedinplan, edge
beamsofslabs are someofthe examples where torsional
momentsarealsopresent.
•Skewbendingtheory,space-trussanalogyandNadai’ssand
heap theory are someofthe theories developedto
understand the behaviourofreinforced concrete under
torsioncombinedwithbendingmomentandshear.
TORSION
•These torsional moments are of two types:
•(i) Primary or equilibrium torsion, and
•(ii) Secondary or compatibility torsion.
•Theprimary torsionisrequired for the basic staticequilibriumof
mostofthe statically determinate structures. Accordingly, this
torsionalmoment mustbeconsideredinthe designasit isa major
component.
•Thesecondary torsionisrequiredtosatisfy the compatibility
condition between members. However, statically indeterminate
structuresmayhaveanyofthetwotypesoftorsions.
•Minor torsional effects maybeignoredinstatically indeterminate
structuresduetotheadvantageofhavingmorethanoneloadpath
for the distributionofloadstomaintain theequilibrium.Thismay
produce minor cracks without causing failure. However, torsional
moments shouldbetaken into accountinthe statically
indeterminate structuresifthey areofequilibrium type and where
the torsional stiffnessofthe members hasbeenconsideredinthe
structural analysis.Itisworth mentioning that torsion mustbe
consideredinstructuressubjectedtounsymmetricalloadingsabout
axes.
•These torsional moments are of two types:
•(i) Primary or equilibrium torsion, and
•(ii) Secondary or compatibility torsion.
•Theprimary torsionisrequired for the basic staticequilibriumof
mostofthe statically determinate structures. Accordingly, this
torsionalmoment mustbeconsideredinthe designasit isa major
component.
•Thesecondary torsionisrequiredtosatisfy the compatibility
condition between members. However, statically indeterminate
structuresmayhaveanyofthetwotypesoftorsions.
•Minor torsional effects maybeignoredinstatically indeterminate
structuresduetotheadvantageofhavingmorethanoneloadpath
for the distributionofloadstomaintain theequilibrium.Thismay
produce minor cracks without causing failure. However, torsional
moments shouldbetaken into accountinthe statically
indeterminate structuresifthey areofequilibrium type and where
the torsional stiffnessofthe members hasbeenconsideredinthe
structural analysis.Itisworth mentioning that torsion mustbe
consideredinstructuressubjectedtounsymmetricalloadingsabout
axes.
TORSION
•Clause41 ofIS 456 stipulates theabove stating
that,"Instructures,wheretorsionisrequiredto
maintainequilibrium,membersshallbedesigned
for torsioninaccordance with41.2,41.3 and
41.4.However,forsuchindeterminatestructures
where torsion canbeeliminatedbyreleasing
redundant restraints,nospecific design for
torsionisnecessary,providedtorsionalstiffnessis
neglectedinthe calculationofinternal forces.
Adequate controlofany torsional crackingis
providedbythe shear reinforcementaspercl.
40".
•Clause41 ofIS 456 stipulates theabove stating
that,"Instructures,wheretorsionisrequiredto
maintainequilibrium,membersshallbedesigned
for torsioninaccordance with41.2,41.3 and
41.4.However,forsuchindeterminatestructures
where torsion canbeeliminatedbyreleasing
redundant restraints,nospecific design for
torsionisnecessary,providedtorsionalstiffnessis
neglectedinthe calculationofinternal forces.
Adequate controlofany torsional crackingis
providedbythe shear reinforcementaspercl.
40".
ANALYSIS FOR TORSION
•Theexact analysisofreinforced concrete members subjectedto
torsional moments combined with bending moments and shear
forcesisbeyond the scope here. However, the codal provisionsof
designingsuchmembersarediscussedbelow.
•ApproachofDesign for Combined Bending, Shear and Torsionas
perIS456
•Asperthe stipulationsofIS456, the longitudinal and transverse
reinforcements are determined taking into account the combined
effectsofbendingmoment,shearforceandtorsionalmoment.
•Two empirical relationsofequivalentshearandequivalentbending
momentaregiven.
•These fictitious shear force and bending moment, designatedas
equivalent shear and equivalent bending moment, are separate
functionsofactual shear and torsion, and actual actualbending
momentandtorsion,respectively.
•Theexact analysisofreinforced concrete members subjectedto
torsional moments combined with bending moments and shear
forcesisbeyond the scope here. However, the codal provisionsof
designingsuchmembersarediscussedbelow.
•ApproachofDesign for Combined Bending, Shear and Torsionas
perIS456
•Asperthe stipulationsofIS456, the longitudinal and transverse
reinforcements are determined taking into account the combined
effectsofbendingmoment,shearforceandtorsionalmoment.
•Two empirical relationsofequivalentshearandequivalentbending
momentaregiven.
•These fictitious shear force and bending moment, designatedas
equivalent shear and equivalent bending moment, are separate
functionsofactual shear and torsion, and actual actualbending
momentandtorsion,respectively.
•a)The equivalent shear, a functionofthe
actual shear and torsional momentis
determined from the following empirical
relation:
•V
e= V
u + 1.6(T
u/b)
•where V
e= equivalent shear,
•V
u = actual shear,
•T
u= actual torsional moment,
•b = breadth of beam
ANALYSIS FOR TORSION
actual shear and torsional momentis
determined from the following empirical
relation:
•V
e= V
u + 1.6(T
u/b)
•where V
e= equivalent shear,
•V
u = actual shear,
•T
u= actual torsional moment,
•b = breadth of beam
ANALYSIS FOR TORSION
•If , then provide the minimum stirrup.
•If , then provide both longitudinal and
transverse steel.
•Cl. 41.4.2 Longitudinal Reinforcement
•The longitudinal reinforcement shallbedesignedtoresistanequivalent
bendingmoment,M
e1,givenby
•MuisthefactoredBM,
•M
tistheequivalentmomentduetotorsion.
•T
uisthefactoredTorsion,Distheoveralldepth,bisthewidthofthebeam.
ANALYSIS FOR TORSION
•If , then provide both longitudinal and
transverse steel.
•Cl. 41.4.2 Longitudinal Reinforcement
•The longitudinal reinforcement shallbedesignedtoresistanequivalent
bendingmoment,M
e1,givenby
•MuisthefactoredBM,
•M
tistheequivalentmomentduetotorsion.
•T
uisthefactoredTorsion,Distheoveralldepth,bisthewidthofthebeam.
ANALYSIS FOR TORSION
•Cl.41.4.2.1
•IfthenumericalvalueofM
t,asdefinedin41.4.2exceedsthe
numerical valueofthe momentMu,longitudinal
reinforcementshallbeprovidedontheflexuralcompression
face, such that the beam can also withstandanequivalent
M
e2givenby
•M
e2=M
t-M
u,themomentM
e2,beingtakenasactinginthe
oppositesensetothemomentM
u.
•IfthenumericalvalueofM
t,asdefinedin41.4.2exceedsthe
numerical valueofthe momentMu,longitudinal
reinforcementshallbeprovidedontheflexuralcompression
face, such that the beam can also withstandanequivalent
M
e2givenby
•M
e2=M
t-M
u,themomentM
e2,beingtakenasactinginthe
oppositesensetothemomentM
u.
DESIGN FOR TORSION
•Cl.41.4.3 Transverse Reinforcement
•Two legged closed hoops enclosing the corner longitudinal
barsshallhaveanareaofcross-sectionA
sv
•How ever total transverse reinforcement should notbeless than
•T
u=torsional moment
•V
u= FactoredSF
•S
v=spacingofthestirrup reinforcement,
•b
1=centre-to-centre distance between corner bars in the direction of the width,
•d
1=centre-to-centre distance between comer in the direction of the depth
•b=breadth of the member,
•f
y=characteristic strength of the stirrup reinforcement
•Cl.41.4.3 Transverse Reinforcement
•Two legged closed hoops enclosing the corner longitudinal
barsshallhaveanareaofcross-sectionA
sv
•How ever total transverse reinforcement should notbeless than
•T
u=torsional moment
•V
u= FactoredSF
•S
v=spacingofthestirrup reinforcement,
•b
1=centre-to-centre distance between corner bars in the direction of the width,
•d
1=centre-to-centre distance between comer in the direction of the depth
•b=breadth of the member,
•f
y=characteristic strength of the stirrup reinforcement
The transverse reinforcement shall consistofrectangular close stirrups placed
perpendiculartothe axisofthe member. The spacingofstirrups shall notbemore
than the leastofx
1,(x
1+ y
1)/4and300mm,where x
1andy
1are the shortandlong
dimensionsofthe stirrups
Longitudinal reinforcements shouldbeplacedascloseaspossibletothe cornersof
the cross-section.
perpendiculartothe axisofthe member. The spacingofstirrups shall notbemore
than the leastofx
1,(x
1+ y
1)/4and300mm,where x
1andy
1are the shortandlong
dimensionsofthe stirrups
Longitudinal reinforcements shouldbeplacedascloseaspossibletothe cornersof
the cross-section.
PROBLEM
Determine the reinforcement required of a ring beam of b = 400 mm, d = 650 mm, D = 700
mm and subjected to factored Mu = 200 kNm, factored Tu= 50 kNmand factored Vu = 100
kN.Use M 20 and Fe 415 for the design.
Step 1: Check for the depth of the beam
Calculation of equivalent shear force due to torsion moment
V
et=1.6(T
u/b)=1.6*(50/.04)=200 kN
V
e=V
u+V
et=100+200=300kN
Equivalent shear stress=
ve=V
e/bd=300*1000/(400*600)=1.154 N/mm
2
cmaxfor M20 concrete=Table 20 IS-456 2.8N/mm
2
cmax>
ve, No redesign is required.
Determine the reinforcement required of a ring beam of b = 400 mm, d = 650 mm, D = 700
mm and subjected to factored Mu = 200 kNm, factored Tu= 50 kNmand factored Vu = 100
kN.Use M 20 and Fe 415 for the design.
Step 1: Check for the depth of the beam
Calculation of equivalent shear force due to torsion moment
V
et=1.6(T
u/b)=1.6*(50/.04)=200 kN
V
e=V
u+V
et=100+200=300kN
Equivalent shear stress=
ve=V
e/bd=300*1000/(400*600)=1.154 N/mm
2
cmaxfor M20 concrete=Table 20 IS-456 2.8N/mm
2
cmax>
ve, No redesign is required.
•Step 2: Longitudinal tension reinforcement
•Calculation of Bending Moment
•M
t=Equivalent bending moment due to torsion=
(T
u/1.7) {1 + (D/b)} = (50/1.7) {1 + (700/400)} =80.88 kNm
•M
e1= M
u+ M
t = 200 + 80.88 = 280.88 kNm
As M
u>M
t, M
e2not required.
M
e1/bd
2
= (280.88)(10
6
)/(400)(650)(650) = 1.66 N/mm
2
From Table 2 of SP-16, corresponding to
Mu/bd
2
= 1.66 N/mm
2
, p
t= 0.5156.
A
st=p
t*b*d/100= 0.5156(400)(650)/100 = 1340.56 mm
2
. Provide
2-25T and 2-16T = 981 + 402 = 1383 mm
2
.
p
tprovided=A
st*100/bd=1383*100/(400*650)= 0.532
A
stmin=0.85*bd/fy=0.85*400*650/415=532.53 mm2
A
stmax=0.04bD=0.04*400*700=11200 mm2
(A
stmin=532.53)<(A
stprovided=1383)<(A
stmax=11200), So,ok
•Calculation of Bending Moment
•M
t=Equivalent bending moment due to torsion=
(T
u/1.7) {1 + (D/b)} = (50/1.7) {1 + (700/400)} =80.88 kNm
•M
e1= M
u+ M
t = 200 + 80.88 = 280.88 kNm
As M
u>M
t, M
e2not required.
M
e1/bd
2
= (280.88)(10
6
)/(400)(650)(650) = 1.66 N/mm
2
From Table 2 of SP-16, corresponding to
Mu/bd
2
= 1.66 N/mm
2
, p
t= 0.5156.
A
st=p
t*b*d/100= 0.5156(400)(650)/100 = 1340.56 mm
2
. Provide
2-25T and 2-16T = 981 + 402 = 1383 mm
2
.
p
tprovided=A
st*100/bd=1383*100/(400*650)= 0.532
A
stmin=0.85*bd/fy=0.85*400*650/415=532.53 mm2
A
stmax=0.04bD=0.04*400*700=11200 mm2
(A
stmin=532.53)<(A
stprovided=1383)<(A
stmax=11200), So,ok
Step 3: Longitudinal compressive reinforcement
As M
t<M
u, No compressive reinforcement is required.
Step 4: Longitudinal side face reinforcement
IS-456-CL-26.5.1.3 Side face reinforcement Where the depthofthe
webina beam exceeds750mm,side face reinforcement shallbe
provided along the two faces.Thetotal areaofsuch reinforcement
shallbenot less than 0.1 percentofthe web area and shallbe
distributedequallyontwofacesataspacingnotexceeding300mmor
webthicknesswhicheverisless.
Side face reinforcement shallbeprovidedasthe depthofthe beam
exceeds450mm.
Areaofside face reinforcement=0.1*bd/100=0.1*400*650/100=260
mm
2
,
Providing 2-10mmdiameter bars (area =157mm2 )oneach face the
totalareafortwofaceswillbe157*2=314mm
2
.Henceo.k.
As M
t<M
u, No compressive reinforcement is required.
Step 4: Longitudinal side face reinforcement
IS-456-CL-26.5.1.3 Side face reinforcement Where the depthofthe
webina beam exceeds750mm,side face reinforcement shallbe
provided along the two faces.Thetotal areaofsuch reinforcement
shallbenot less than 0.1 percentofthe web area and shallbe
distributedequallyontwofacesataspacingnotexceeding300mmor
webthicknesswhicheverisless.
Side face reinforcement shallbeprovidedasthe depthofthe beam
exceeds450mm.
Areaofside face reinforcement=0.1*bd/100=0.1*400*650/100=260
mm
2
,
Providing 2-10mmdiameter bars (area =157mm2 )oneach face the
totalareafortwofaceswillbe157*2=314mm
2
.Henceo.k.
Step 5: TRANSVERSE REINFORCEMENT
Calculation of
c
For pt=0.532,
c =0.48+(0.56-0.48)*(0.532-0.5)/(0.75-0.5)=0.488 N/mm2
Calculation of
c
For pt=0.532,
c =0.48+(0.56-0.48)*(0.532-0.5)/(0.75-0.5)=0.488 N/mm2
•Fig.
10 mm bar
12 mm bar
16 mm bar
b1
25 mm bar
25 mm
cover
concrete
10 mm bar stirrup
25 mm
b=400 mm
x
1mm
D=700 mm
25 mm
10 mm
25 mm
d1
y
1
b1=400-25-10-25/2-25-10-25/2=305mm
d1=700-25-10-12/2-25-10-25/2=611.5mm
x1=400-25-10/2-25-10/2=340 mm
y1=700-25-10/2-25-10/2=640 mm
10 mm bar
12 mm bar
16 mm bar
b1
25 mm bar
25 mm
cover
concrete
10 mm bar stirrup
25 mm
b=400 mm
x
1mm
D=700 mm
25 mm
10 mm
25 mm
d1
y
1
b1=400-25-10-25/2-25-10-25/2=305mm
d1=700-25-10-12/2-25-10-25/2=611.5mm
x1=400-25-10/2-25-10/2=340 mm
y1=700-25-10/2-25-10/2=640 mm
Using 2 legged 10 mm dia bars
s
v=169.97 mm
Provide 2 legged 10 mm dia bars 150 mmc/c
Again A
sv≥(
ve-
c)b*s
v/(0.87*f
y)
A
sv≥[(
ve-
c)b*s
v/(0.87*f
y)={(1.154-0.488)*400*150/(0.87*415)}=110.67 mm
2
]
[A
sv=157]≥110.67 mm
2
Design is ok.
Spacing can not be greater than x1=340 mm
Spacing can not be greater than( x1+y1)/4=(340+640)/4=245 mm
Spacing can not be greater than 300 mm
Stirrup spacing provided 150 mm satisfies all these three conditions.
157=
50∗10
6
∗�
�
305∗611.5 ∗ (0.87∗415)
+
100∗10
3
∗ �
�
2.5∗611.5(0.87∗415)
�
��=
�
��
�
�
1�
1(0.87�
�)
+
�
��
�
2.5�
1(0.87�
�)
s
v=169.97 mm
Provide 2 legged 10 mm dia bars 150 mmc/c
Again A
sv≥(
ve-
c)b*s
v/(0.87*f
y)
A
sv≥[(
ve-
c)b*s
v/(0.87*f
y)={(1.154-0.488)*400*150/(0.87*415)}=110.67 mm
2
]
[A
sv=157]≥110.67 mm
2
Design is ok.
Spacing can not be greater than x1=340 mm
Spacing can not be greater than( x1+y1)/4=(340+640)/4=245 mm
Spacing can not be greater than 300 mm
Stirrup spacing provided 150 mm satisfies all these three conditions.
157=
50∗10
6
∗�
�
305∗611.5 ∗ (0.87∗415)
+
100∗10
3
∗ �
�
2.5∗611.5(0.87∗415)
�
��=
�
��
�
�
1�
1(0.87�
�)
+
�
��
�
2.5�
1(0.87�
�)
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