Lecture 1 two way slab design (Corners not held down).pdf

Lecture 1 two way slab design (Corners not held down).pdf

1. The great aim of education is not knowledge but
action…
--Herbert Spencer

2. Structure Design VI (AR –306B)
Design of Concrete structures

3. Reference texts
1.Design of Reinforced concrete (WSM) by Harbhajan singh
2.Design of reinforced concrete (WSM) by B.C. Punmia

4. Teaching methodology
•Themethodologyincludesdemonstrationoflecturewiththe
helpofvideosandpowerpointpresentationsinthelogical
manner.
•Eachlecturewillbefollowedbytheexercisewhichhastobe
completedandsubmittedonorbeforethenextlecture.
•Therewillbetwominortestswhichwillbeconductedin
betweenthesession.
•Therewillalsobeaprovisionofsurprisetestswhichmaybe
conductedanyweekwithoutanypriornotice.

5. Bring to class your –
•Calculator
•IS Code
•Notebook
•Pen, Pencil, eraser etc.
No borrowing/sharing please!

6. Marking scheme (internal)
As per University syllabus –
Total weightage - 50 Marks
1.Minor test I - 20% (10)
2.Minor test II - 20% (10)
3.Assignment - 30% (15)
4.Surprise test/Quizzes- 20% (10)
5.Attendance - 10% (5)
Minimum pass marks- 40%*
*Subject to having minimum attendance.

7. Assignments
Objective –
•Help deepen understanding of the subject.
•Better performance in examinations.
Strongly advised to –
•Attempt all assignments.
•Submit all assignments.
•Not to copy. (But allowed to work in groups.)

8. Surprise tests/Quizzes
Objective –
•Promote continuous learning (and evaluation).
•To reduce the load of having to prepare for examination the night
before.
Methodology –
•UNANNOUNCED !!!
•20 –40 mins test on the topics covered in near past.

9. Attendance
Expected to attend all classes.
Absenteeism will lead to –
•Missing surprise tests.
•Poor understanding of the subject.
•Poor performance in examinations.
•Poor grades at the end.

10. Indian Standard
IS 456:2000 (Plain and Reinforced
concrete code of Practice)
Must have your own copy

11. Structure VI
Two way slab design
(Simply supported)
Lesson Plan 1
Gateway College of Architecture and Design
Jan 2018 Neeraj Kumar (Asst. Professor)

12. Content
•Introduction
•Some facts about slabs
•Classification
•Simply supported slab
•BM coefficients
•Design example

13. Instructional Objectives
At the end of this lecture, students should be able to:
•Understand the concept of slabs
•Classify two way slabs
•Understand the concept of simply supported slab
•Solve design problems for two way simply supported slabs

14. Introduction

15. Introduction
•In case a slab is supported on all four sides and the aspect
ratio of its length to width is less than or equal to two, it is
classified as two way slab.
•Such a slab when loaded will bend in the shape of a dish.
•Such slabs has to be reinforced at bottom for tension in two
directions perpendicular to each other.
•Because of two way bending action, such slabs carry more
load and exhibit less deflection as compared to one way
slabs.

16. Some facts about slabs
•Slabs are analyzed and designed as having a unit width, i.e. 1m strips.
•Compression reinforcement is used only in exceptional cases in a slab.
•Shear stress is usually very low and shear reinforcement is never
provided in slabs.
•Temperature reinforcement or distribution reinforcement is invariably
provided at right angles to the main reinforcement in a slab.
•Slabs are usually much thinner than beams.

17. Classification
•Two way slabs can be classified in two categories-
✓Simply supported Two way slab (corners held down)
✓Restrained two way slab (will discuss in next lecture)

18. Simply supported slab
•Such slabs are commonly used
as isolated roof slabs for
individual rooms in single
storey buildings.
•Corners of such slabs are
twisted and lift upward from
the support when the slabs are
loaded.
•Max BM per unit width are
obtained by the use of
equations and coefficients
listed in table 27 of IS:456-
2000

19. BM coefficients for slabs spanning in two directions
at right angle, simply supported on Four sides

20. Design example
•Design a simply supported roof slab for room 3mX5m
(internal dimension), wall thickness 230mm. Roof finish
weighs 2 kN/m
2
and roof is approachable for repair only. Use
M20 grade concrete and HYSD reinforcement.

21. Solution
Step 1-Material and design constant-
σ
cbc= 7N/mm
2
; σ
st= 230N/mm
2
m = 13.33
k = 0.29
j = 0.90
R = 0.91
Step 2-Estimate of slab thickness
Let d = Short span/30 (general thumb rule)
= 3000/30 = 100mm
Assuming moderate exposure (Clear cover = 30mm) and use of 8mm dia
bars
D = 100+30+8/2 = 134mm Say 140mm
hence,d
provided= 140 –30 –8/2 = 106mm

22. Solution
Step 3 –Effective short span (l)-
least of:
a) Short span (l
x)
i.c/c of supports
= 230/2 + 3000 + 230/2
= 3230mm = 3.23m
ii.clear span + d
= 3000 + 106
= 3106mm = 3.106m Say 3.11m
hence, l
x= 3.11m
b) Long span (l
y)
on same lines, long effective span = l
y= 5.11m

23. Solution
Step 4-Load per unit area (w kN/m
2
)-
Dead load
Roof finishing = 1x1x2 = 2kN
Self weight of slab = 1x1x0.14x25 = 3.50kN
Live load(roof is approachable for repair only)
Live load @ 0.75 kN/m
2
= 1x1x0.75 = 0.75kN
Total load = 6.25kN
Step 5-Aspect ratio (λ) = long span/short span
= l
y/l
x= 5.11/3.11 = 1.64 < 2
Slab will be designed as two way slab.
Note: As the slab is simply supported, moment coefficients will be
applicable.

24. Solution
Step 6-Max BM (M)
Designing for unit width along X-axis
and unit width along Y-axis.
a) Moment on strip of unit width
spanning l
x= M
x= α
xwl
x
2
α
x from table for l
y/l
x= 1.64
by interpolation,
α
x= [0.104+{(0.113-0.104)x(1.64-
1.5)}/(1.75-1.5)]
= 0.109
M
x= 0.109x6.25x(3.11)
2
= 6.59kN.m

25. Solution
b) Moment on strip of unit width
spanning l
y= M
y
M
y= α
ywl
x
2
α
yby interpolation is,
α
y= [0.046-{(0.046-0.037)x(1.64-
1.5)}/(1.75-1.5)]
= 0.041
M
y= 0.041x6.25x(3.11)
2
=
2.48kN.m
hence, Max BM (M) = 6.59kN.m
(larger of M
xand M
y)

26. Solution
Step 7-Effective depth required-
d
req= √(M/R.b)
= √(6.59x10
6
)/(0.91x1000)
= 85.10mm < d
provided hence, OK
Step 8-Reinforcement:
a) Along short span (Bottom layer):
A
st= M
x/σ
st.j.d
= 6.59x10
6
/230x0.90x106
= 300.34mm
2
spacing of 8 diabars = Aφx1000/A
st
= [{(πx8
2
/4)x1000}/300.34]
= 167.36mm

27. Solution
spacing of main bars < 3d and 300mm as per IS: 456
hence, provide 8φ@160mm c/c
A
stprovided= Aφx 1000/160
= 314.19mm
2
%age of main reinforcement at mid span of short span, = A
stprovided x100/bd =
0.3%
b)Along long span (top layer):
d’ for top layer = 106 –8 = 98mm
A
st= My/σ
st.j.d’
= 2.48x10
6
/230x0.90x98
= 122.25mm
2
spacing of 8 diabars = Aφx1000/Ast
= [{(πx8
2
/4)x1000}/122.25]
= 411mm

28. Solution
but spacing of main bars < 3d = 3x98 = 294mm and 300mm as per IS:
456
hence, provide 8φ@290mm c/c top layer.
c)Distribution bars:
A
st(dist)= 0.12% of total cross-sectional area
= (0.12xbxD)/100 = 168mm
2
spacing of 8 diadistribution bar = Aφx 1000/A
st(dist)
= 299.25mm < 5d < 450mm
hence, provide distribution bars as8φ@ 290mm c/c

29. Solution
Step 9-Check for deflection
Deflection check is carried out at mid span of short span and as per
IS:456-2000,
l
x/d < 20 x MF1
f
s= 0.58fy x Ast
req/Ast
provided
= 0.58x415x300.34/314.19
= 229.83
with percentage of steel at mid-span of short span = 0.3% as worked out
in last step,
MF1 from fig. = 1.5
hence, l
x/d < 20x1.5
or, 3110/106 < 30
29.34 < 30
hence, slab is safe in deflection.

31. Sketch of the reinforcement

32. Exercise
•Designasimplysupportedroofslabforroom4mX5m
(internaldimension),wallthickness230mm.Rooffinish
weighs2.5kN/m
2
androofisapproachableforrepaironly.
UseM25gradeconcreteandHYSDreinforcement.