design of residential building

DESIGN OF SIMPLE RESIDENTIAL BUILDING SLAB, BEAM, COLUMN ♀♀♀♀

2016

HAMMAD BASHIR BSC CIVIL ENGINEER

HAMMAD BASHIR GONDAL (JUTT) (B.SC CIVIL ENGINEER) +923430817733

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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





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





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





BENDING MOMENT CALCULATION

ASPECT RATIO:

m= la/lb

la= shorter length…. lb= longer length

BENDING MOMENT Co-efficients





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





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





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





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″





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





BENDING MOMENT COEFFIENTS





COLUMN DESIGN

Sizes:

Column size = 12″ × 12″

Loads:

Pu = 11.41 × 2 = 22.82 kip





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





DRAFTING DETAILS

FRONT VIEW





3D VIEW

SLAB DRAFTING DETAILS





SLAB DRAFTING DETAILS





BEAM DETAILS





COLUMN DESIGN DETAIL

Reference

I. Design of concrete structure by NIlson

II. Notes of PROF. ZIAAUDDIN MIAN UET LAHORE PAKISTAN



design of residential building

Documents

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ftkip b

ll lb2

ksf total factored load

ll la2 mb

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