Home Design Slab,Beam,Stairs
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Transcript of Home Design Slab,Beam,Stairs
ENGR.HAMMAD BASHIR
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ENGR.HAMMAD BASHIR
:
Waqas Rahmat
B.Sc. Electrical Engineer
:
This residential building structural component is
designed according to ACI 14 Code.
Hammad BashirBSC Civil ENGINEERPakistan, IslamabadEmail: [email protected]
Website: www.hammadjatt.weebly.com
Cell: +92343-0817733
:
Engr. Sajjid rasheed khokar M.SC CIVIL ENGINEER PEC. NO CIVIL/7998Engr.Naveed RashidPec no civil /5984Structural engineer Naveed associates
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PLOT LOCATION:
GROUND FLOOR PLAN:
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DESIGN
1) SLAB
2) BEAM
3) STAIRS
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FIRST FLOOR..YOU CAN MAKE IT ON YOUR CHOICE..HOWEVER YOU CAN PUT SAMESTRUCTURE ON FIRST FLOOR.
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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
CALCULATIONKSF
SLAB 5” 5”/12 0.15 0.0625MUD 4” 4”/12 0.12 0.04TILE 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 BUILDING LIVE LOAD WILL BE ACCORDING TO LOADING CRITERIA……
Sr.no
Occupancy or use Live loadKgs/m2 Pascal
N/m2lb/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 805 Movable seats assembly hall 500 4800 1006 Wholesales stores, light storage
warehouses.610 6000 125
7 Library stack rooms 730 7200 1508 Heavy manufacturing, heavy
storage warehouses, side walks and driveways subject to truckling
1200 12000 250
9 Stairs, general 500 4800 10010 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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Calculating moments using ACI Coefficients:
Ma, neg = Ca, neg wula2
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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
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.
Shrinkage Reinforcement IF NEEDED:
Ast = 0.002bhf = 0.12 in2 (#3 @ 11″ c/c)
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However, for facilitating field work, we will use #3 @ 10″ c/c
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SLAB
STEEL
DRAWING
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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/f
Factored Self load of beam web =
= 1.2 x (13 × 12/144) × 0.15 = 0.195 k/f
Total load on beam = 1.055 + 0.195
= 1.25 kip/ft
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STAIR DRAWING:
RISER= 175 mm
Tread = 300mm
Landing: 1.2m
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ReferenceI. Design of concrete structure by NIlson
II. Notes of PROF. ZIAAUDDIN MIAN UET LAHORE PAKISTANIII. Design of concrete structures by z.a siddique UET lahore
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