DESIGN OF SIMPLE RESIDENTIAL BUILDING SLAB, BEAM, COLUMN ♀♀♀♀
2016
HAMMAD BASHIR BSC CIVIL ENGINEER
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DESIGN OF RESIDENTIAL BUILDING
EXAMPLE:
DESIGN TYPICAL HOUSE WITH TWO ROOMS AND VERANDAH….
Solution:
In this house, we will Design
1. two-way slab
2. one-way slab, Beam and column
DESIGN
1) SLAB
2) BEAM
3) COLUMN
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SLAB:
THICKNESS: According to ACI 318 the minimum thickness of slab should be 5 inch.
Load calculation:
Service dead loads:
Material THICKNESS (INCH)
THICKNESS (FEET)
DENSITY LOAD CALCULATION KSF
SLAB 5” 5”/12 0.15 0.0625 MUD 4” 4”/12 0.12 0.04 TILE 2” 2”/12 0.12 0.02
TOTAL DEAD LOADS= 0.1225 KSF
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FACTORED DEAD LOAD 1.2 0.1225 0.147 Ksf
Service dead loads: FOR RESIDENTIAL BUILDING LIVE LOAD WILL BE ACCORDING TO LOADING
CRITERIA……
Sr.no
Occupancy or use
Live load
Kgs/m2 Pascal N/m2
lb/ft2
1 Private rooms, school class rooms.
200 1900 40
2 Offices. 250 to 425 2400 to 4000
50 to 85
3 Fixed-seats, assembly halls, library reading rooms.
300 2900 60
4 Corridors in public building 400 3800 80 5 Movable seats assembly hall 500 4800 100 6 Wholesales stores, light storage
warehouses. 610 6000 125
7 Library stack rooms 730 7200 150 8 Heavy manufacturing, heavy storage
warehouses, side walks and driveways subject to truckling
1200 12000 250
9 Stairs, general 500 4800 100 10 Stairs, upto two-family residences,
50% more than specifications. 300 2900 60
WE USE 40 Psf = 0.040 Ksf
FACTORED LIVE LOAD 1.6 0.04 0.064 Ksf
TOTAL FACTORED LOAD= FACTORED LIVE LOAD + FACTORED DEAD LOAD
= 0.1225 + 0.064 = 0.211 Ksf
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BENDING MOMENT CALCULATION
ASPECT RATIO:
m= la/lb
la= shorter length…. lb= longer length
BENDING MOMENT Co-efficients
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TWO WAY SLAB DESIGN…..
m = la/lb = 12/16 = 0.75
Ca,neg = 0.076 Cb,neg = 0.024
Ca,pos DL = 0.043 Cb,pos DL = 0.013
Ca,pos LL = 0.052 Cb,pos LL = 0.016
Calculating moments using ACI Coefficients:
Ma, neg = Ca, neg wula2
Mb, neg = Cb, neg wulb2
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
a,neg = 2.31 ft-kip
b,neg = 1.29 ft-kip
a,pos = 1.39 ft-kip
b,pos = 0.76 ft-kip
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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.
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
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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 in2
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
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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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″
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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 kip/ft
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BENDING MOMENT COEFFIENTS
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COLUMN DESIGN
Sizes:
Column size = 12″ × 12″
Loads:
Pu = 11.41 × 2 = 22.82 kip
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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
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DRAFTING DETAILS
FRONT VIEW
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3D VIEW
SLAB DRAFTING DETAILS
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SLAB DRAFTING DETAILS
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BEAM DETAILS
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COLUMN DESIGN DETAIL
Reference
I. Design of concrete structure by NIlson
II. Notes of PROF. ZIAAUDDIN MIAN UET LAHORE PAKISTAN
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