lecture-04€¦ · prof. dr. qaisar ali reinforced concrete design –ii department of civil...
TRANSCRIPT
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Lecture-04
Analysis and Design of Two-way Slab System
(Part-I: Two Way Slabs Supported on Stiff Beams Or Walls)
By: Prof Dr. Qaisar Ali
Civil Engineering Department
UET Peshawar
www.drqaisarali.com
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Topics
Behavior
Moment Coefficient Method
Steps in Moment Coefficient Method
Design Example 1 (Typical House with 2 Rooms and Verandah)
Design Example 2 (100′ × 60′, 3-Storey Commercial Building)
Practice Examples
References
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab System (Long span/short span < 2)
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25′ 25′ 25′ 25′
20′
20′
20′
Behavior
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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One-Way Behavior Two-Way Behavior
Behavior
Two-Way Bending of Two-Way Slabs
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Behavior
Two-Way Bending of Two-Way Slabs
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Short Direction
Behavior
Short Direction Moments in Two-Way Slab
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Long Direction
Behavior
Long Direction Moments in Two-Way Slab
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
More Demand (Moment) in short direction due to size
of slab
Δcentral Strip = (5/384)wl4/EI
As these imaginary strips are part of monolithic slab, the deflection at
any point, of the two orthogonal slab strips must be same:
Δa = Δb
(5/384)wala4/EI = (5/384)wblb
4/EI
wa/wb = lb4/la
4 wa = wb (lb4/la
4)
Thus, larger share of load (Demand) is taken by the shorter direction.
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Behavior
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
The Moment Coefficient Method included for the first time in
1963 ACI Code is applicable to two-way slabs supported on
four sides of each slab panel by walls, steel beams relatively
deep, stiff, edge beams (h = 3hf).
Although, not included in 1977 and later versions of ACI code,
its continued use is permissible under the ACI 318-11 code
provision (13.5.1). Visit ACI 13.5.1.
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Moment Coefficient Method
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moments:
Ma, neg = Ca, negwula2
Mb, neg = Cb, negwulb2
Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la2 + Ca, pos, ll × wu, ll × la
2
Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb2 + Cb, pos, ll × wu, ll × lb
2
Where Ca, Cb = Tabulated moment coefficients
wu = Ultimate uniform load, psf
la, lb = length of clear spans in short and long directions respectively.
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Ma,neglb
la
Ma,neg
Ma,posMb,neg Mb,negMb,pos
Moment Coefficient Method
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Cases
Depending on the support conditions, several cases are possible:
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Moment Coefficient Method
4 spans @ 25′-0″
3 sp
ans @ 20
′-0″
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Moment Coefficient Method
Cases
Depending on the support conditions, several cases are possible:
4 spans @ 25′-0″
3 spans @ 20
′-0″
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Moment Coefficient Method
Cases
Depending on the support conditions, several cases are possible:
4 spans @ 25′-0″
3 sp
ans @ 20
′-0″
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Moment Coefficient Method
Cases
Depending on the support conditions, several cases are possible:
4 spans @ 25′-0″
3 spans @ 20
′-0″
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Moment Coefficient Method
Note: Horizontal sides of the figure represents longer side while vertical side
represents shorter side.
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Moment Coefficient Method
Note: Horizontal sides of the figure represents longer side while vertical side
represents shorter side.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Moment Coefficient Method
Note: Horizontal sides of the figure represents longer side while vertical side
represents shorter side.
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Moment Coefficient Method
Note: Horizontal sides of the figure represents longer side while vertical side
represents shorter side.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Moment Coefficient Method
Note: Horizontal sides of the figure represents longer side while vertical side
represents shorter side.
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
20
Moment Coefficient Method
Note: Horizontal sides of the figure represents longer side while vertical side
represents shorter side.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
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Moment Coefficient Method
Note: Horizontal sides of the figure represents longer side while vertical side
represents shorter side.
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
hmin = perimeter/ 180 = 2(la + lb)/180
Calculate loads on slab
Calculate m = la/ lb
Decide about case of slab
Use table to pick moment coefficients
Calculate Moments and then design
Apply reinforcement requirements (smax = 2hf, ACI 13.3.2)
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Steps in Moment Coefficient Method
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
3D Model of the House
In this building we will design a two-way slab (for rooms), a one-way
slab, beam and column (for verandah)
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Design Example 1(Typical House with 2 Rooms and Verandah)
16' X 12' 16' X 12'
9' Wide Verandah
B1B1
RCC Column
9" brick masonry wall
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Given Data:
Service Dead Load
4″ thick mud
2″ thick brick tile
Live Load = 40 psf
f′c = 3 ksi
fy = 40 ksi
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Design Example 1(Typical House with 2 Rooms and Verandah)
16' X 12' 16' X 12'
9' Wide Verandah
B1B1
RCC Column
9" brick masonry wall
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Sizes
For two way slab system
hmin = perimeter / 180 = 2(la + lb)/180
hmin = 2 (12 +16) /180 = 0.311 ft = 3.73 inch
Assume 5 inch slab
For one way slab system
For 5″ slab, span length l is min. of:
l = ln+ hf = 8 + (5/12) = 8.42′
c/c distance between supports = 8.875′
Slab thickness (hf) = (8.42/20) × (0.4+��/100000)
= 4.04″ (min. by ACI)
Taking 5 in. slab
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Design Example 1(Typical House with 2 Rooms and Verandah)
9′ Wide Verandah
ln = 8′
lc/c = (8 + (9/12) / 2 + 0.5) = 8.875′
9″ Brick Wall
Slab
h
16' X 12' 16' X 12'
9' Wide Verandah
B1B1
RCC Column
9" brick masonry wall
A
A
Section AA
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Loads
5 inch Slab = 5/12 x 0.15 = 0.0625 ksf
4 inch mud = 4/12 x 0.12 = 0.04 ksf
2 inch tile = 2/12 x 0.12 = 0.02 ksf
Total DL = 0.1225 ksf
Factored DL = 1.2 x 0.1225 = 0.147 ksf
Factored LL = 1.6 x 0.04 = 0.064 ksf
Total Factored Load = 0.211 ksf
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Design Example 1(Typical House with 2 Rooms and Verandah)
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Analysis
This system consist of both one way and two way slabs. Rooms
(two way slabs) are continuous with verandah (one way slab).
A system where a two way slab is continuous with a one way slab
or vice versa is called a mixed slab system.
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Design Example 1(Typical House with 2 Rooms and Verandah)
16' X 12' 16' X 12'
9' Wide Verandah
B1B1
RCC Column
9" brick masonry wall
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Analysis
The ACI approximate methods of analysis are not applicable to
such systems because:
In case of two way slabs, the moment coefficient tables are applicable
to two way slab system where a two way slab is continuous with a two
way slab.
In case of one ways slabs, the ACI approximate analysis is applicable
to one way slab system where a one way slab is continuous with a one
way slab.
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Design Example 1(Typical House with 2 Rooms and Verandah)
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Analysis
The best approach to analyze a mixed system is to use FE
software.
However, such a system can also be analyzed manually by making
certain approximations.
In the next slides, We will analyze this system using both of the
above mentioned methods.
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Design Example 1(Typical House with 2 Rooms and Verandah)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Analysis using FE Software (SAFE)
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Design Example 1(Typical House with 2 Rooms and Verandah)
Moments in Longer Direction in Two Way Slabs
Moments in Shorter Direction in Two Way Slabs & Moment in Verandah One Way Slab
Two Way Slab Moments (ft-kip/ft)(Rooms)
One Way Slab Moment (ft-kip/ft)(Verandah)
Ma,pos Mb,pos Ma,neg Mb,neg Mver (+ve)
1.58 1.17 2.10 1.67 1.10
Mb,pos Mb,posMb,neg Ma
,po
s
Ma
,po
s
Ma
,ne
g
Ma
,ne
g
Mv
er
(+v
e)
Mb,pos
Ma,p
os
Ma,n
eg
Mb,negMb,pos
Ma,n
eg
Ma,p
os
Mver
(+v
e)
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Analysis using Manual Approach
For approximate manual analysis of two way slab of rooms, we
know that two way slabs of rooms are not only continuous along the
long direction but also continuous along the short direction with the
verandah slab. We assume that the verandah slab is a two way slab
instead of one way slab in order to calculate Mb,neg
Now, using Moment Coefficient Method for analysis.
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Design Example 1(Typical House with 2 Rooms and Verandah)
Mb,pos
Ma,p
os
Ma,n
eg
Mb,negMb,pos
Ma,n
eg
Ma,p
os
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis
Case = 4
m = la/lb = 12/16 = 0.75
Ca,neg = 0.076
Cb,neg = 0.024
Ca,posDL = 0.043
Cb,posDL = 0.013
Ca,posLL = 0.052
Cb,posLL = 0.016
Design Example 1(Typical House with 2 Rooms and Verandah)
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16′
12′
Mb,negMb,pos
Ma
,po
s
Case 4
Ma,n
eg
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis
Case = 4
m = la/lb = 12/16 = 0.75
Ca,neg = 0.076
Cb,neg = 0.024
Ca,posDL = 0.043
Cb,posDL = 0.013
Ca,posLL = 0.052
Cb,posLL = 0.016
Design Example 1(Typical House with 2 Rooms and Verandah)
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16′
12′
Mb,negMb,pos
Ma
,po
s
Case 4
Ma,n
eg
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Case = 4
m = la/lb = 12/16 = 0.75
Ca,neg = 0.076
Cb,neg = 0.024
Ca,posDL = 0.043
Cb,posDL = 0.013
Ca,posLL = 0.052
Cb,posLL = 0.016
Two-Way Slab Analysis
Design Example 1(Typical House with 2 Rooms and Verandah)
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16′
12′
Mb,negMb,pos
Ma
,po
s
Case 4
Ma,n
eg
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Case = 4
m = la/lb = 12/16 = 0.75
Ca,neg = 0.076
Cb,neg = 0.024
Ca,posDL = 0.043
Cb,posDL = 0.013
Ca,posLL = 0.052
Cb,posLL = 0.016
Two-Way Slab Analysis
Design Example 1(Typical House with 2 Rooms and Verandah)
35
16′
12′
Mb,negMb,pos
Ma
,po
s
Case 4
Ma,n
eg
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Case = 4
m = la/lb = 12/16 = 0.75
Ca,neg = 0.076
Cb,neg = 0.024
Ca,posDL = 0.043
Cb,posDL = 0.013
Ca,posLL = 0.052
Cb,posLL = 0.016
Two-Way Slab Analysis
Design Example 1(Typical House with 2 Rooms and Verandah)
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16′
12′
Mb,negMb,pos
Ma
,po
s
Case 4
Ma,n
eg
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Case = 4
m = la/lb = 12/16 = 0.75
Ca,neg = 0.076
Cb,neg = 0.024
Ca,posDL = 0.043
Cb,posDL = 0.013
Ca,posLL = 0.055
Cb,posLL = 0.016
Two-Way Slab Analysis
Design Example 1(Typical House with 2 Rooms and Verandah)
37
16′
12′
Mb,negMb,pos
Ma
,po
s
Case 4
Ma,n
eg
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
38
Design Example 1(Typical House with 2 Rooms and Verandah)
Two-Way Slab Analysis
Calculating moments using ACI Coefficients:
Ma, neg = Ca, negwula2
Mb, neg = Cb, negwulb2
Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la2 + Ca, pos, ll × wu, ll × la
2
Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb2 + Cb, pos, ll × wu, ll × lb
2
Using above relations the moments calculated are:
Ca,neg = 0.076 Cb,neg = 0.024
Ca,posLL = 0.052 Cb,posLL = 0.016
Ca,posDL = 0.043 Cb,posDL = 0.013
wu, dl = 0.147 ksf, wu, ll = 0.064 ksf, wu = 0.211 ksf
Ma,neg = 2.31 ft-kip
Mb,neg = 1.29 ft-kip
Ma,pos = 1.39 ft-kip
Mb,pos = 0.76 ft-kip
16′
12′
Mb,negMb,pos
Ma
,po
s
Case 4
Ma,n
eg
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
One Way Slab Analysis
For calculation of Mver (+ve) along the short direction in the verandah
slab, we will pick the coefficients of continuous one way slabs having
two spans.
Now, using ACI Approximate analysis procedure (ACI 8.33) for analysis.
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Design Example 1(Typical House with 2 Rooms and Verandah)
Mver(
+v
e)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
40
Design Example 1(Typical House with 2 Rooms and Verandah)
One-Way Slab Analysis
For two span one way slab system, positive moment at midspan is given as
follows:
Mver (+ve) = wuln2/11
Mver (+ve) = 0.211 × (8)2/11
= 1.23 ft-k/ft
9′ Wide Verandah
ln = 8′
9″ Brick Wall
Wu = 0.211 ksf
h
1/11 1/11
1/91/9Simply Supported
Simply Supported
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
41
Design Example 1(Typical House with 2 Rooms and Verandah)
Design of Two-Way Slab
Comparison of Analysis Results from FE Analysis and Manual Analysis
Analysis results from both approaches are almost similar.
Hence the intelligent use of manual analysis yields fairly reasonable results in most cases.
Analysis Type
Ma,neg Mb,neg Ma,pos Mb,pos Mver (+ve)
SAFE 2.10 1.67 1.58 1.17 1.10
Manual 2.31 1.3 1.39 0.76 1.23
NOTE: All values are in ft-kip units
Mb,pos
Ma,p
os
Ma,n
eg
Mb,negMb,pos
Ma,n
eg
Ma,p
os
Mver
(+v
e)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
42
Design Example 1(Typical House with 2 Rooms and Verandah)
Design of Two-Way Slab
First determining capacity of min. reinforcement:
As,min = 0.002bhf = 0.12 in2
Using #3 bars: Spacing for As,min = 0.12 in2 = (0.11/0.12) × 12 = 11″ c/c
However ACI max spacing for two way slab = 2h = 2(5) = 10″ or 18″ = 10″ c/c
Hence using #3 bars @ 10″ c/c
For #3 bars @ 10″ c/c: As,min = (0.11/10) × 12 = 0.132 in2
Capacity for As,min: a = (0.132 × 40)/(0.85 × 3 × 12) = 0.17″
ΦMn = ΦAsminfy(d – a/2) = 0.9 × 0.132 × 40(4 – (0.17/2)) = 18.60 in-kip
Therefore, for Mu values ≤ 18.60 in-k/ft, use As,min (#3 @ 10″ c/c) & for Mu
values > 18.6 in-kip/ft, calculate steel area using trial & error procedure.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
43
Design Example 1(Typical House with 2 Rooms and Verandah)
Design of Two-Way Slab
For Ma,neg = 2.31 ft-kip = 27.71 in-kip > 18.60 in-kip: As = 0.20 in2 (#3 @ 6.6″ c/c)
Using #3 @ 6″ c/c
For Mb,neg = 1.29 ft-kip = 15.56 in-kip < 18.60 in-kip: Using #3 @ 10 c/c
For Ma,pos = 1.39 ft-kip = 16.67 in-kip < 18.60 in-kip: Using #3 @ 10 c/c
For Mb,pos = 0.76 ft-kip = 9.02 in-kip < 18.60 in-kip: Using #3 @ 10″ c/c
16′
12′
Mb,negMb,pos
Ma,p
os
Case 4
Ma,n
eg
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
44
Design Example 1(Typical House with 2 Rooms and Verandah)
Design of Two-Way Slab
Reinforcement at Discontinuous Ends
Reinforcement at discontinuous ends in a two way slab is 1/3 of the positive
reinforcement.
Positive reinforcement at midspan in this case is #3 @ 10 c/c. Therefore
reinforcement at discontinuous end may be provided @ 30 c/c.
However, in field practice, the spacing of reinforcement at discontinuous ends
seldom exceeds 18 c/c. The same is provided here as well.
Supporting Bars
Supporting bars are provided to support negative reinforcement.
They are provided perpendicular to negative reinforcement, generally at spacing
of 18 c/c.
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
45
Design Example 1(Typical House with 2 Rooms and Verandah)
Design of One-Way Slab
Main Reinforcement:
Mver (+ve) = 14.73 in-kip
As,min = 0.002bhf = 0.002(12)(5) = 0.12 in2
Using #3 bars, spacing = (0.11/0.12) × 12 = 11″ c/c
For one-way slabs, max spacing by ACI = 3h = 3(5) = 15″ or 18″ = 15″ c/c
For #3 bars @ 15″ c/c, As = (0.11/15) × 12 = 0.09 in2. Hence using As,min = 0.12 in
2
a = (0.12 × 40)/(0.85 × 3 × 12) = 0.16″
ΦMn = ΦAsminfy(d – a/2) = 0.9 × 0.12 × 40(4 – (0.16/2)) = 16.94 in-kip > Mver (+ve)
Therefore, using #3 @ 11″ c/c
However, for facilitating field work, we will use #3 @ 10″ c/c
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
46
Design Example 1(Typical House with 2 Rooms and Verandah)
Design of One-Way Slab
Shrinkage Reinforcement:
Ast = 0.002bhf = 0.12 in2 (#3 @ 11″ c/c)
However, for facilitating field work, we will use #3 @ 10″ c/c
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
47
Verandah Beam Design
Step 01: Sizes
Let depth of beam = 18″
ln + depth of beam = 15.875′ + (18/12) = 17.375′
c/c distance between beam supports
= 16.375 + (4.5/12) = 16.75′
Therefore l = 16.75′
Depth (h) = (16.75/18.5) × (0.4 + 40000/100000) × 12
= 8.69″ (Minimum requirement of ACI 9.5.2.2).
Take h = 1.5′ = 18″
d = h – 3 = 15″
b = 12″
Design Example 1(Typical House with 2 Rooms and Verandah)
Verandah Beam
16.375 16.375
ln = 16.375 – 0.5(12/12) = 15.875 ln = 16.375 – 0.5(12/12) = 15.875
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
48
Verandah Beam Design
Step 02: Loads
Load on beam will be equal to
Factored load on beam from slab + factored
self weight of beam web
Factored load on slab = 0. 211 ksf
Load on beam from slab = 0. 211 ksf x 5 =
1.055 k/ft
Factored Self load of beam web =
= 1.2 x (13 × 12/144) × 0.15 = 0.195 k/ft
Total load on beam = 1.055 + 0.195
= 1.25 k/ft
5′
Design Example 1(Typical House with 2 Rooms and Verandah)
ln = 8′
9″ Brick Wall
8/2 = 4′
12″ Column
4′ + 1′ = 5′
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Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
49
Verandah Beam
Design
Step 03: Analysis
Using ACI Moment
Coefficients for analysis
of verandah beam
Design Example 1(Typical House with 2 Rooms and Verandah)
16.375
ln = 16.375 – 0.5(12/12) = 15.875 ln = 16.375 – 0.5(12/12) = 15.87516.375
16.375
ln = 15.875 ln = 15.875
9.92 k
11.41 k
Vu(ext) = 8.34 k
Vu(int) = 9.61 k
343.66 in-kip 343.66 in-kip
420.03 in-kip
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Beam Design
Flexure Design:
Shear Design:
Smax is min. of: (1) Avfy/(50bw) = 14.67″ (2) d/2 =7.5″ (3) 24″ c/c (4) Avfy/ 0.75√(fc′)bw = 17.85″
50
Design Example 1(Typical House with 2 Rooms and Verandah)
Mu(in-kip)
d (in.)
b (in.)
As(in2)
Asmin(in2)
Asmax(in2)
As(governing)
Bar used
# of bars
343.66 (+) 15 28.75 (beff) 0.64 0.90 3.654 0.90 #4 5
#5 3
420.03 (-) 15 12 0.81 0.90 3.654 0.90 #4 5
#4 + #5 2 + 2
Location Vu (@ d)(kip)
ΦVc = Φ2 �′�bwd (kips) ΦVc > Vu, Hence
providing minimum reinforcement.
smax, ACI
S taken(#3 2-legged)
Exterior 8.34 14.78 7.5″ 7.5″
Interior 9.61 14.78 7.5″ 7.5″
-
26
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
51
Design Example 1(Typical House with 2 Rooms and Verandah)
Column Design
Sizes:
Column size = 12″ × 12″
Loads:
Pu = 11.41 × 2 = 22.82 kip
Verandah Column
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
52
Design Example 1(Typical House with 2 Rooms and Verandah)
Column Design
Main Reinforcement Design:
Nominal strength (ΦPn) of axially loaded column is:
ΦPn = 0.80Φ {0.85fc′ (Ag – Ast) + Astfy} {for tied column, ACI 10.3.6}
Let Ast = 1% of Ag (Ast is the main steel reinforcement area)
ΦPn = 0.80 × 0.65 × {0.85 × 3 × (144 – 0.01 × 144) + 0.01 × 144 × 40}
= 218.98 kip > Pu = 22.82 kip, O.K.
Ast =0.01 × 144 =1.44 in2
Using 3/4″ Φ (#6) with bar area Ab = 0.44 in2
No. of bars = 1.44/0.44 = 3.27 ≈ 4 bars
Use 4 #6 bars (or 8 #4 bars) and #3 ties @ 9″ c/c
-
27
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
53
Design Example 1(Typical House with 2 Rooms and Verandah)
Drafting Details for
Slabs
Panel Depth (in) Mark Bottom Reinforcement Mark Top reinforcement
S1 5"M1 #3 @ 10" c/c
MT1 #3 @ 10" c/c Continuous EndMT1 #3 @ 18" c/c Non Continuous End
M2 #3 @ 10" c/cMT2 #3 @ 6" c/c Continuous EndMT2 #3 @ 18" c/c Non Continuous End
S2 5"M1 #3 @ 10" c/c MT2 #3 @ 6" c/c Continuous EndM2 #3 @ 10" c/c MT2 #3 @ 18" c/c Non Continuous End
S2 9' wide
M1 M1
M2M2
MT1 MT1 MT1
MT2 MT2
MT2MT2
M1
M2
C1
AA
B
S1 16' x 12' S1 16' x 12'
B MT2
B1 B1
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
54
Design Example 1(Typical House with 2 Rooms and Verandah)
Drafting Details for Slabs
SECTION A-A
L /4 = 4'-0" L /3 = 5'-3"
L = 16'-0"
#3 @ 10" c/c
#3 @ 10" c/c
#3 @ 18" c/c #3 @ 18" c/c
1 1
h = 5"
#3 @ 10" c/c
#3 @ 10" c/c
L/3=2'-8" L/4=2'-0"
#3 @ 10" c/c
#3 @ 10" c/c
#3 @ 18" c/c#3 @ 10" c/c
9" Brick Masonry Wall
12" x 18" Beam
#3 @ 10" c/c
#3 @ 10" c/c#3 @ 10" c/c
#3 @ 18" c/c #3 @ 6" c/c
2
2
SECTION B-B
L = 12'-0" L = 8'-0"
L /4 = 4'-0"2 L /3 = 4'-0"
-
28
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
55
Design Example 1(Typical House with 2 Rooms and Verandah)
Drafting Details for Verandah Beam
2 #5 Bars
2 #4 Bars
SECTION B-B
12"
18"
5"
3 #5 Bars
#3,2 legged stirrups @ 7.5" c/c
SECTION A-A
12"
18"
5"
3 #5 Bars
#3,2 legged stirrups @ 7.5" c/c
2 #4 Bars
A
A
A
A
2 #4 + 2 #5 Bars
BEAM (B1)
B
B
L = 11'-10.5" L = 11'-10.5"
0.33L = 4'-0" 0.33L = 4'-0"
2 #4 Bars
3 #5 Bars #3, 2 legged stirrups @ 7.5" c/c
s /2=3.75"
3 #5 Bars 3 #5 Bars
B
B
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
56
Design Example 1(Typical House with 2 Rooms and Verandah)
Drafting Details for Verandah Column
OR
12"
4 #6 Bars
#3 ties @ 9" c/c
12"
12"
#3 Ties @ 9" c/c
4 #6 Bars
8#4 bars
#3 ties @ 9" c/c
12"
12"
12"
8 #4 Bars
#3 Ties @ 9" c/c
-
29
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
A 100′ 60′, 3-storey commercial building is to be designed. The grids of
column plan are fixed by the architect.
In this example, the slab of one of the floors of this 3-storey building will
be designed.
57
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Given Data:
Material Properties:
f′c = 3 ksi, fy = 40 ksi
Sizes:
Slab thickness = 7″
Columns = 14″ 14″
Beams = 14″ 20″
Loads:
S.D.L = Nil
Self Weight = 0.15 x (7/12) = 0.0875 ksf
L.L = 144 psf = 0.144 ksf ; wu = 0.336 ksf
58
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
-
30
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Complete analysis of the slab is done by analyzing four panels
59
Panel I Panel I
Panel I Panel I
Panel II Panel II
Panel III Panel III
Panel III Panel III
Panel IV Panel IV
4 spans @ 25′-0″
3 sp
ans @
20′-0″
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-I)
Case = 4
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posDL = 0.039
Cb,posDL = 0.016
Ca,posLL = 0.048
Cb,posLL = 0.020
Case 4la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,neg
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
60
-
31
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-I)
Case = 4
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posDL = 0.039
Cb,posDL = 0.016
Ca,posLL = 0.048
Cb,posLL = 0.020
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
61
Case 4la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,neg
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-I)
Case = 4
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posDL = 0.039
Cb,posDL = 0.016
Ca,posLL = 0.048
Cb,posLL = 0.020
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
62
Case 4la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,neg
-
32
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-I)
Case = 4
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posDL = 0.039
Cb,posDL = 0.016
Ca,posLL = 0.048
Cb,posLL = 0.020
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
63
Case 4la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,neg
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-I)
Case = 4
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posDL = 0.039
Cb,posDL = 0.016
Ca,posLL = 0.048
Cb,posLL = 0.020
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
64
Case 4la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,neg
-
33
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-I)
Case = 4
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.071
Cb,neg = 0.029
Ca,posDL = 0.039
Cb,posDL = 0.016
Ca,posLL = 0.048
Cb,posLL = 0.020
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
65
Case 4la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,neg
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
66
Two-Way Slab Analysis (Panel-I)
Calculating moments using ACI Coefficients:
Ma, neg = Ca, negwula2
Mb, neg = Cb, negwulb2
Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la2 + Ca, pos, ll × wu, ll × la
2
Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb2 + Cb, pos, ll × wu, ll × lb
2
Using above relations the moments calculated are:
Ca,neg = 0.071 Cb,neg = 0.029
Ca,posLL = 0.048 Cb,posLL = 0.020
Ca,posDL = 0.039 Cb,posDL = 0.016
wu, dl = 0.105 ksf, wu, ll = 0.2304 ksf, wu = 0.336 ksf
Ma,neg = 8.44 ft-k (101.2 in-k)
Mb,neg = 5.52 ft-k (66.2 in-k)
Ma,pos = 5.37 ft-k (64.4 in-k)
Mb,pos = 3.57 ft-k (42.8 in-k)
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Case 4la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,neg
-
34
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-II)
Case = 9
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.075
Cb,neg = 0.017
Ca,posDL = 0.029
Cb,posDL = 0.010
Ca,posLL = 0.042
Cb,posLL = 0.017
Case 9la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
67
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-II)
Case = 9
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.075
Cb,neg = 0.017
Ca,posDL = 0.029
Cb,posDL = 0.010
Ca,posLL = 0.042
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
68
Case 9la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
-
35
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-II)
Case = 9
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.075
Cb,neg = 0.017
Ca,posDL = 0.029
Cb,posDL = 0.010
Ca,posLL = 0.042
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
69
Case 9la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-II)
Case = 9
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.075
Cb,neg = 0.017
Ca,posDL = 0.029
Cb,posDL = 0.010
Ca,posLL = 0.042
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
70
Case 9la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
-
36
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-II)
Case = 9
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.075
Cb,neg = 0.017
Ca,posDL = 0.029
Cb,posDL = 0.010
Ca,posLL = 0.042
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
71
Case 9la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-II)
Case = 9
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.075
Cb,neg = 0.017
Ca,posDL = 0.029
Cb,posDL = 0.010
Ca,posLL = 0.042
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
72
Case 9la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
-
37
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
73
Two-Way Slab Analysis (Panel-II)
Calculating moments using ACI Coefficients:
Ma, neg = Ca, negwula2
Mb, neg = Cb, negwulb2
Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la2 + Ca, pos, ll × wu, ll × la
2
Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb2 + Cb, pos, ll × wu, ll × lb
2
Using above relations the moments calculated are:
Ca,neg = 0.075 Cb,neg = 0.017
Ca,posLL = 0.042 Cb,posLL = 0.017
Ca,posDL = 0.029 Cb,posDL = 0.010
wu, dl = 0.105 ksf, wu, ll = 0.2304 ksf, wu = 0.336 ksf
Ma,neg = 8.91 ft-k (106.9 in-k)
Mb,neg = 3.24 ft-k (38.8 in-k)
Ma,pos = 4.51 ft-k (54.1 in-k)
Mb,pos = 2.82 ft-k (33.8 in-k)
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Case 9la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-III)
Case = 8
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.055
Cb,neg = 0.041
Ca,posDL = 0.032
Cb,posDL = 0.015
Ca,posLL = 0.044
Cb,posLL = 0.019
Case 8la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
74
-
38
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-III)
Case = 8
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.055
Cb,neg = 0.041
Ca,posDL = 0.032
Cb,posDL = 0.015
Ca,posLL = 0.044
Cb,posLL = 0.019
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
75
Case 8la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-III)
Case = 8
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.055
Cb,neg = 0.041
Ca,posDL = 0.032
Cb,posDL = 0.015
Ca,posLL = 0.044
Cb,posLL = 0.019
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
76
Case 8la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
-
39
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-III)
Case = 8
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.055
Cb,neg = 0.041
Ca,posDL = 0.032
Cb,posDL = 0.015
Ca,posLL = 0.044
Cb,posLL = 0.019
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
77
Case 8la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-III)
Case = 8
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.055
Cb,neg = 0.041
Ca,posDL = 0.032
Cb,posDL = 0.015
Ca,posLL = 0.044
Cb,posLL = 0.019
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
78
Case 8la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
-
40
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-III)
Case = 8
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.055
Cb,neg = 0.041
Ca,posDL = 0.032
Cb,posDL = 0.015
Ca,posLL = 0.044
Cb,posLL = 0.019
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
79
Case 8la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
80
Two-Way Slab Analysis (Panel-III)
Calculating moments using ACI Coefficients:
Ma, neg = Ca, negwula2
Mb, neg = Cb, negwulb2
Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la2 + Ca, pos, ll × wu, ll × la
2
Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb2 + Cb, pos, ll × wu, ll × lb
2
Using above relations the moments calculated are:
Ca,neg = 0.055 Cb,neg = 0.041
Ca,posLL = 0.044 Cb,posLL = 0.019
Ca,posDL = 0.032 Cb,posDL = 0.015
wu, dl = 0.105 ksf, wu, ll = 0.2304 ksf, wu = 0.336 ksf
Ma,neg = 6.54 ft-k (78.4 in-k)
Mb,neg = 7.80 ft-k (93.6 in-k)
Ma,pos = 4.78 ft-k (57.4 in-k)
Mb,pos = 3.38 ft-k (40.5 in-k)
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Case 8la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
-
41
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-IV)
Case = 2
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.065
Cb,neg = 0.027
Ca,posDL = 0.026
Cb,posDL = 0.011
Ca,posLL = 0.041
Cb,posLL = 0.017
Case 2la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
81
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-IV)
Case = 2
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.065
Cb,neg = 0.027
Ca,posDL = 0.026
Cb,posDL = 0.011
Ca,posLL = 0.041
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
82
Case 2la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
-
42
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-IV)
Case = 2
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.065
Cb,neg = 0.027
Ca,posDL = 0.026
Cb,posDL = 0.011
Ca,posLL = 0.041
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
83
Case 2la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-IV)
Case = 2
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.065
Cb,neg = 0.027
Ca,posDL = 0.026
Cb,posDL = 0.011
Ca,posLL = 0.041
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
84
Case 2la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
-
43
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-IV)
Case = 2
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.065
Cb,neg = 0.027
Ca,posDL = 0.026
Cb,posDL = 0.011
Ca,posLL = 0.041
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
85
Case 2la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Two-Way Slab Analysis (Panel-IV)
Case = 2
m = la/lb = 18.83/23.83 = 0.78 ≈ 0.80
Ca,neg = 0.065
Cb,neg = 0.027
Ca,posDL = 0.026
Cb,posDL = 0.011
Ca,posLL = 0.041
Cb,posLL = 0.017
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
86
Case 2la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
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44
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
87
Two-Way Slab Analysis (Panel-IV)
Calculating moments using ACI Coefficients:
Ma, neg = Ca, negwula2
Mb, neg = Cb, negwulb2
Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la2 + Ca, pos, ll × wu, ll × la
2
Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb2 + Cb, pos, ll × wu, ll × lb
2
Using above relations the moments calculated are:
Ca,neg = 0.065 Cb,neg = 0.027
Ca,posLL = 0.041 Cb,posLL = 0.017
Ca,posDL = 0.026 Cb,posDL = 0.011
wu, dl = 0.105 ksf, wu, ll = 0.2304 ksf, wu = 0.336 ksf
Ma,neg = 7.72 ft-k (92.7 in-k)
Mb,neg = 5.14 ft-k (61.6 in-k)
Ma,pos = 4.31 ft-k (51.8 in-k)
Mb,pos = 2.88 ft-k (34.5 in-k)
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Case 2la = 18.83′
lb = 23.83′
Ma
,ne
gM
a,p
os
Mb,posMb,negMb,neg
Ma
,ne
g
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
88
Analysis Results (All values are in ft-kip)
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
8.42
4 spans @ 25′-0″
3 spans @ 2
0′-0″
5.375.523.57
8.91
8.91
3.244.51
2.82
6.54
7.80 7.804.783.38
7.72
7.72
5.145.144.31
2.881.50
1.79
1.19 1.19 Mneg at Non-
Continuous End =
1/3 of Mpos
NOTE: White: Longer Direction Moments, Yellow: Shorter Direction Moments
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45
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
89
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Slab Design
First determining the capacity of min. reinforcement:
As,min = 0.002bhf = 0.002 × 12 × 7 = 0.17 in2
For #4 bars, spacing = (0.20/0.17) × 12 = 14.2 c/c.
ACI max spacing for two-ways slabs = 2h = 2(7) = 14 or 18
Using #4 @ 12 c/c
For the 12 spacing: As = (0.20/12) × 12 = 0.20 in2. Hence As,min = 0.20 in
2
Capacity for As,min: a = (0.20 × 40)/(0.85 × 3 × 12) = 0.26
ΦMn = ΦAsminfy(d – a/2) = {0.9 × 0.20 × 40(6 – (0.26/2))}/12 = 3.52 ft-kip
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
90
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Slab Design
Positive Moments in Long Direction:
3.57 ft-kip/ft
3.38 ft-kip/ft
2.82 ft-kip/ft
2.88 ft-kip/ft
Since all the above moments values are almost equal to or less than
3.52 ft-kip/ft. Therefore using #4 @ 12 c/c for all positive moments in
short direction.
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46
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
91
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Slab Design
Positive Moments in Short Direction:
5.37 ft-kip/ft
4.51 ft-kip/ft
4.78 ft-kip/ft
4.31 ft-kip/ft
Using trial and success method for determining As for 5.37 ft-kip/ft:
Assume a = 0.2d= 0.2(6) = 1.2, As = (5.37 × 12)/{0.9 × 40(6 – (1.2/2))} = 0.33 in2
Now, a = (0.33 × 40)/(0.85 × 3 × 12) = 0.43, As = 0.31 in2
For #4 bars, spacing = (0.20/0.31) × 12 = 7.7
As all the above moments are almost same, we will use #4 @ 7 c/c for all above
moments.
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
92
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Slab Design
Negative Moments at Non-Continuous Ends in Short & Long Directions:
1.19 ft-kip/ft
1.79 ft-kip/ft
1.50 ft-kip/ft
Since, all the above moments are
less than 3.52 ft-kip/ft, therefore using
#4 @ 12 c/c
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47
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
93
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Slab Design
Negative Moments at Continuous Ends in Short Direction:
8.42 ft-kip/ft
8.91 ft-kip/ft
7.72 ft-kip/ft
6.54 ft-kip/ft
Using trial and success method:
For 8.91 ft-kip/ft: As = 0.52 in2
Using #4 bars, spacing = 4.5 c/c
As all above moments are almost same, we will use #4 @ 4.5 c/c
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
94
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Slab Design
Negative Moments at Continuous Ends in Long Direction:
5.52 ft-kip/ft
3.24 ft-kip/ft
7.80 ft-kip/ft
5.14 ft-kip/ft
Reinforcement:
7. 80 ft-kip/ft is almost equal to 8.91 ft-kip/ft:
Therefore using #4 bars @ 4.5 c/c
7. 80 ft-kip/ft is almost equal to 8.91 ft-kip/ft:
Therefore using #4 bars @ 7 c/c
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48
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Slab Reinforcement Details
95
A
BBB
C
C
A
C CB
A
A
BA
C
C
A
A B
A
A= #4 @ 12″B = #4 @ 7″C = #4 @ 4.5″
4 spans @ 25′-0″
3 sp
ans @
20
′-0″
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
NOTE: White: Longer Direction Moments, Yellow: Shorter Direction Moments
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
96
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Load Transfer from Slab to the Beam
Review of Load transfer to Beam from One-Way Slabs
• In case of one-way slab system the entire slab load is transferred in short direction.
• Load transfer in short direction = (Wu × l / 2 × 1) + (Wu × l / 2 × 1)
• Load transfer in long direction = Wu × l / 2 × 0
l/2
l/2
l/2 l/2
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49
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
97
Load transfer from Slab to Beam
Load Transfer to Beam from Two-Way Slab
• In case of two way slab system, entire slab load is NOT transferred in shorter direction.
• Load transfer in shorter direction = (Wu × l / 2 × Wa ) + (Wu × l / 2 × Wa )
Longer Direction
4 spans @ 25′-0″
3 sp
ans @
20′-0
″
Sh
orte
r Dire
ctio
n
Panel I Panel I
Panel I Panel I
Panel II Panel II
Panel III Panel III
Panel III Panel III
Panel IV Panel IV
This value will NOT be 1 in this case, It is specified by ACI Table• Load transfer in longer direction = (Wu × l / 2 × Wb ) + (Wu × l / 2 × Wb )
Wb = 1 - Wa
ACI Table for Wa values
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
98
Load transfer from Slab to Beam
Load Transfer to Beam from Two-Way Slab
4 spans @ 25′-0″
3 sp
an
s @ 2
0′-0
″
Panel I Panel I
Panel I Panel I
Panel II Panel II
Panel III Panel III
Panel III Panel III
Panel IV Panel IV
Load Transfer from Slab to Beam B1
• Load is transferred to B1 from Panel-I and Panel-II in
short direction
= Wu × l / 2 × Wa,Panel-I + Wu × l / 2 × Wa,Panel-II
= 0.336 × 20/2 × 0.71 + 0.336 × 20/2 × 0.83
= 5.17 k/ft
(For Wa Values, refer to the tables on next slides)• Wu = 0.336 ksf
• Wa,Panel-I = 0.71
• Wa,Panel-II = 0.83
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
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50
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
99
Load transfer from Slab to Beam
Load Transfer to Beam from Two-Way Slab (For Beam B1)
Sh
orte
r Dire
ction
20′
Wu = 0.336 ksf
Sh
orte
r Dire
ctio
n
20′
Wu = 0.336 ksf
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
100
Load transfer from Slab to Beam
Load Transfer to Beam from Two-Way Slab
4 spans @ 25′-0″
3 sp
an
s @ 2
0′-0
″
Panel I Panel I
Panel I Panel I
Panel II Panel II
Panel III Panel III
Panel III Panel III
Panel IV Panel IV
Load Transfer from Slab to Beam B2
• Load is transferred to B2 from Panel-I and Panel-III in
long direction
= Wu × l / 2 × Wb,Panel-I + Wu × l / 2 × Wb,Panel-III
= 0.336 × 25/2 × (1 - 0.71) + 0.336 × 25/2 × (1 - 0.55)
= 3.11 k/ft
(For Wa values, refer to the tables on next slide)• Wu = 0.336 ksf
• Wb,Panel-I = (1 – Wa,Panel-I) = 1 - 0.71 = 0.29
• Wb,Panel-III = (1 – Wa,Panel-III) = 1 - 0.55 = 0.45
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
-
51
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
101
Load transfer from Slab to Beam (For Beam B2)
Load Transfer to Beam from Two-Way Slab
= 0.336 × 25/2 × (1 - 0.71) + 0.336 × 25/2 ×
(1 - 0.55)
= 3.11 k/ft
Sh
orte
r Dire
ction
20′
Wu = 0.336 ksf
Sh
orte
r Dire
ctio
n
20′
Wu = 0.336 ksf
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Example 2
Load On Beams from coefficient tables
102
Table: Load on beam in Panel I, usingCoefficients
(wu = 0.336 ksf)
BeamLength
(ft)
Width (bs) of slab panel
supported by beam
Wa Wb
Load due to slab, Wwubs(k/ft)
B1 25 10 0.71 - 2.39
B2 25 10 0.71 - 2.39
B3 20 12.5 - 0.29 1.22
B4 20 12.5 - 0.29 1.22
Panel I
4 spans @ 25′-0″
3 spans @ 20
′-0″
B1
B1
B2
B2
B3 B3 B3 B4B4
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
-
52
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Table: Load on beam in Panel I, usingCoefficients
(wu = 0.336 ksf)
BeamLength
(ft)
Width (bs) of slab panel
supported by beam
Wa Wb
Load due to slab, Wwubs(k/ft)
B1 25 10 0.71 - 2.39
B2 25 10 0.71 - 2.39
B3 20 12.5 - 0.29 1.22
B4 20 12.5 - 0.29 1.22
Moment Coefficient Method: Example 2
Load On Beams from coefficient tables
B1
B1
B2
B2
B3 B3 B3 B4B4
4 spans @ 25′-0″
3 sp
ans @
20
′-0″
Panel I
103
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Panel I
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Example 2
Load On Beams from coefficient tables
104
4 spans @ 25′-0″
3 spans @ 20
′-0″
B1
B1
B2
B2
B3 B3 B3 B4B4
Table: Load on beam in Panel II, usingCoefficients
(wu = 0.336 ksf)
BeamLength
(ft)
Width (bs) of slab panel
supported by beam
Wa Wb
Load due to slab, Wwubs(k/ft)
B1 25 10 0.83 - 2.78
B3 20 12.5 - 0.17 0.714
B4 20 12.5 - 0.17 0.714
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Panel II
-
53
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Table: Load on beam in Panel II, usingCoefficients
(wu = 0.336 ksf)
BeamLength
(ft)
Width (bs) of slab panel
supported by beam
Wa Wb
Load due to slab, Wwubs(k/ft)
B1 25 10 0.83 - 2.78
B3 20 12.5 - 0.17 0.714
B4 20 12.5 - 0.17 0.714
Moment Coefficient Method: Example 2
Load On Beams from coefficient tables
B1
B1
B2
B2
B3 B3 B3 B4B4
Panel II
4 spans @ 25′-0″
3 sp
ans @
20
′-0″
105
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Panel II
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Example 2
Load On Beams from coefficient tables
106
4 spans @ 25′-0″
3 spans @ 20
′-0″
B1
B1
B2
B2
B3 B3 B3 B4B4
Table: Load on beam in Panel III, usingCoefficients
(wu = 0.336 ksf)
BeamLength
(ft)
Width (bs) of slab panel
supported by beam
Wa Wb
Load due to slab, Wwubs(k/ft)
B1 25 10 0.55 - 1.84
B2 25 10 0.55 - 1.84
B3 20 12.5 - 0.45 1.89
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Panel III
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54
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Table: Load on beam in Panel III, usingCoefficients
(wu = 0.336 ksf)
BeamLength
(ft)
Width (bs) of slab panel
supported by beam
Wa Wb
Load due to slab, Wwubs(k/ft)
B1 25 10 0.55 - 1.84
B2 25 10 0.55 - 1.84
B3 20 12.5 - 0.45 1.89
Moment Coefficient Method: Example 2
Load On Beams from coefficient tables
B1
B1
B2
B2
B3 B3 B3 B4B4
Panel III
4 spans @ 25′-0″
3 sp
ans @
20
′-0″
107
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Panel III
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Example 2
Load On Beams from coefficient tables
108
4 spans @ 25′-0″
3 spans @ 20
′-0″
B1
B1
B2
B2
B3 B3 B3 B4B4
Table: Load on beam in Panel IV, usingCoefficients
(wu = 0.336 ksf)
BeamLength
(ft)
Width (bs) of slab panel
supported by beam
Wa Wb
Load due to slab, Wwubs(k/ft)
B1 25 10 0.71 - 2.39
B3 20 12.5 - 0.29 1.22
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Panel IV
-
55
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Table: Load on beam in Panel IV, usingCoefficients
(wu = 0.336 ksf)
BeamLength
(ft)
Width (bs) of slab panel
supported by beam
Wa Wb
Load due to slab, Wwubs(k/ft)
B1 25 10 0.71 - 2.39
B3 20 12.5 - 0.29 1.22
Moment Coefficient Method: Example 2
Load On Beams from coefficient tables
B1
B1
B2
B2
B3 B3 B3 B4B4
Panel IV
4 spans @ 25′-0″
3 sp
ans @
20
′-0″
109
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Panel IV
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Example 2
Load On Beams
110
2.39 k/ft
1.22 k/ft
2.39 k/ft
1.22 k/ft
2.77 k/ft
2.77 k/ft
0.71 k/ft 0.71 k/ft
1.84 k/ft
1.89 k/ft
1.84 k/ft
1.89 k/ft
2.39 k/ft
1.22 k/ft
2.39 k/ft
1.22 k/ft
4 spans @ 25′-0″
3 spans @ 20
′-0″
B1
B1
B2
B2
B3 B3 B3 B4B4
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
-
56
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Example 2
Load On Beams
111
5.16 k/ft
1.22 k/ft
2.39 k/ft
3.11 k/ft
5.16 k/ft
0.71 k/ft 1.93 k/ft
4.23 k/ft
1.84 k/ft
3.11 k/ft
4.23 k/ft
1.93 k/ft
4 spans @ 25′-0″
3 sp
ans @
20
′-0″
B1
B1
B2
B2
B3 B3 B3 B4B4
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Example 2
Load On Beams
112
Self weight of beam
= 1.2 × (14 ×
13/144) × 0.15
= 0.23 kip/ft
Adding this with the
calculated loads on
beams:
14
20
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
5.16 k/ft
1.22 k/ft
2.39 k/ft
3.11 k/ft
5.16 k/ft
0.71 k/ft 1.93 k/ft
4.23 k/ft
1.84 k/ft
3.11 k/ft
4.23 k/ft
1.93 k/ft
4 spans @ 25′-0″
3 spans @ 20
′-0″
B1
B1
B2
B2
B3 B3 B3 B4B4
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57
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Example 2
Load On Beams (including self-weight)
113
1.45 k/ft
2.62 k/ft
5.39 k/ft
0.94 k/ft
3.34 k/ft
2.07 k/ft
2.16 k/ft
4.46 k/ft
4 spans @ 25′-0″
3 sp
ans @
20
′-0″
B1
B1
B2
B2
B3 B3 B3 B4B4
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
114
lnln ln
Simplesupport
Integral withsupport
wu
Spandrelsupport
NegativeMoment
x wuln2
1/24 1/10* 1/11 1/11 1/10* 0
*1/9 (2 spans)
* 1/11(on both faces of other interior supports)
Columnsupport
1/16
PositiveMomentx
1/14 1/16 1/11
wuln2
Note: For simply supported slab, M = wul2/8, where l = span length (ACI 8.9).
* 1/12 (for all spans with ln < 10 ft)
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
-
58
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
115
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Analysis of Beams
Interior Beam B1
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
116
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Analysis of Beams
Exterior Beam B2
-
59
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
117
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
Analysis of Beams
Interior Beam B3
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Analysis of Beams
Exterior Beam B4
118
Design Example 2(100′ × 60′, 3-Storey Commercial Building)
-
60
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
119
Pictures of a Multi-StoreyCommercial Building
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
120
Pictures of a Multi-StoreyCommercial Building
-
61
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Different Stages of Building Construction
121
Phases of Construction
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Home Work
Design the given slab system
122
Panel I Panel I
Panel I Panel I
Panel II Panel II
Panel III Panel III
Panel III Panel III
Panel IV Panel IV
4 spans @ 25′-0″
3 spans @ 20
′-0″
Slab thickness = 6″
SDL = 40 psf
LL = 60 psf
fc′ =3 ksi
fy = 40 ksi
Practice Examples
-
62
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
Moment Coefficient Method: Home Work
Design the given slab system
123
Panel I Panel I
Panel I Panel I
Panel II Panel II
Panel III Panel III
Panel III Panel III
Panel IV Panel IV
4 spans @ 20′-0″
3 sp
ans @
15′-0″
Slab thickness = 6″
SDL = 40 psf
LL = 60 psf
fc′ =3 ksi
fy = 40 ksi
Practice Examples
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
CRSI Design Handbook
ACI 318
Design of Concrete Structures 13th Ed. by Nilson, Darwin and
Dolan.
124
References
-
63
Prof. Dr. Qaisar Ali Reinforced Concrete Design – II
Department of Civil Engineering, University of Engineering and Technology Peshawar
The End
125