Raft foundation design .. incomplete .
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Analysis and Design of the mat foundation: There are 3 methods for the design of mat foundations: 1. The Conventional method of design. 2. The Approximate Flexible method of design. 3. Finite-Difference and Finite Element methods of design. We’ve chosen the Conventional method of design to use in both of our-handmade and computerized calculations since it’s simple and accurate too. Step1 - Calculate the total Loads on raft ( from: columns , walls, water, soils, beams). Service ‘Pn’ Service ‘Pn’ Service ‘Pn’ Service ‘Pn’ Service ‘Pn’ Service ‘Pn’ Service ‘Pn’ Service ‘Pn’ Service ‘Pn’ Service ‘Pn’ Service ‘Pn’ Pn 0 277.6 529.2 529.2 529.2 529.2 529.2 529.2 529.2 277.6 0 4259.6 0 62.5 186.5 2 186.5 2 186.5 2 125 186.5 2 186.5 2 186.5 2 62.5 0 1369.12 359.5 0 0 0 0 0 0 0 0 0 359.5 719 529.2 186.5 2 365.7 365.7 365.7 163.7 5 365.7 365.7 365.7 186.5 2 529.2 3789.39 529.2 3561. 5 7115. 7 7115. 7 7115. 7 163.7 5 7115. 7 7115. 7 7115. 7 3561. 5 529.2 51039.6 3 529.2 186.5 2 365.7 365.7 365.7 163.7 5 365.7 365.7 365.7 186.5 2 529.2 3789.39 359.5 0 0 0 0 0 0 0 0 0 359.5 719 0 62.5 186.5 2 186.5 2 186.5 2 125 186.5 2 186.5 2 186.5 2 62.5 0 1369.12 0 277.6 529.2 529.2 529.2 529.2 529.2 529.2 529.2 277.6 0 4259.6 2306. 6 4614. 8 9278. 6 9278. 6 9278. 6 1799. 7 9278. 6 9278. 6 9278. 6 4614. 8 2306. 6 71313.8 5 Pn= ∑ P=P 1+P 2 +P 3 + …+ Pn. ¿ 71313.85 KN Pu Pu Pu Pu Pu Pu Pu Pu Pu Pu Pu Pu 0 333.2 635 635 635 635 635 635 635 333.2 0 5111.4 0 87.5 258.47 258.47 258.47 266.7 258.47 258.47 258.47 87.5 0 1992.5 2 431. 0 0 0 0 0 0 0 0 0 431. 862.8
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Raft foundation design .. incomplete .
Transcript of Raft foundation design .. incomplete .
Analysis and Design of the mat foundation:
There are 3 methods for the design of mat foundations:
1. The Conventional method of design.2. The Approximate Flexible method of design.3. Finite-Difference and Finite Element methods of design.
We’ve chosen the Conventional method of design to use in both of our-handmade and computerized calculations since it’s simple and accurate too.
Step1 - Calculate the total Loads on raft ( from: columns , walls, water, soils, beams).
Service ‘Pn’
Service ‘Pn’
Service ‘Pn’
Service ‘Pn’
Service ‘Pn’
Service ‘Pn’
Service ‘Pn’
Service ‘Pn’
Service ‘Pn’
Service ‘Pn’
Service ‘Pn’ Pn
0 277.6 529.2 529.2 529.2 529.2 529.2 529.2 529.2 277.6 0 4259.60 62.5 186.52 186.52 186.52 125 186.52 186.52 186.52 62.5 0 1369.12
359.5 0 0 0 0 0 0 0 0 0 359.5 719529.2 186.52 365.7 365.7 365.7 163.75 365.7 365.7 365.7 186.52 529.2 3789.39529.2 3561.5 7115.7 7115.7 7115.7 163.75 7115.7 7115.7 7115.7 3561.5 529.2 51039.63529.2 186.52 365.7 365.7 365.7 163.75 365.7 365.7 365.7 186.52 529.2 3789.39359.5 0 0 0 0 0 0 0 0 0 359.5 719
0 62.5 186.52 186.52 186.52 125 186.52 186.52 186.52 62.5 0 1369.120 277.6 529.2 529.2 529.2 529.2 529.2 529.2 529.2 277.6 0 4259.6
2306.6 4614.8 9278.6 9278.6 9278.6 1799.7 9278.6 9278.6 9278.6 4614.8 2306.6 71313.85Pn=∑ P=P 1+P 2+P 3+…+P n.
¿71313.85 KNPu Pu Pu Pu Pu Pu Pu Pu Pu Pu Pu
SPu0 333.2 635 635 635 635 635 635 635 333.2 0 5111.40 87.5 258.47 258.47 258.47 266.7 258.47 258.47 258.47 87.5 0 1992.52
431.4 0 0 0 0 0 0 0 0 0 431.4 862.8635 258.47 502.5 502.5 502.5 266.7 502.5 502.5 502.5 258.47 635 5068.64635 258.47 8602.5 8602.5 8602.5 8326.7 8602.5 8602.5 8602.5 258.47 635 61728.64635 258.47 502.5 502.5 502.5 266.7 502.5 502.5 502.5 258.47 635 5068.64
431.4 0 0 0 0 0 0 0 0 0 431.4 862.80 87.5 258.47 258.47 258.47 266.7 258.47 258.47 258.47 87.5 0 1992.520 333.2 635 635 635 635 635 635 635 333.2 0 5111.4
2767.8 1616.81 11394.44 11394.44 11394.44 10663.5 11394.44 11394.44 11394.44 1616.81 2767.8 87799.36 u=87799.36 KN
Step2 - Determine the soil’s pressure - distributed under the foundation – and specify the maximum value of it:
q= PnA
+ MxIx
y+ MyIy
x≤ q allow ,net
Where: 1) A: Area of the raft = (41.95 )∗(21.95 )=920.8 m2
2) Ix: moment of inertia of the raft about the x-axis .
Ix= B L3
12=41 .95∗21.953
12=36970.4 m4
3) Iy: moment of inertia of the raft about the y-axis .
Iy= L B3
12=21.95∗41.953
12=135035.9 m4
The eccentricities (ex & ey) are found as shown below:
1. ex=x'− B2
=20− 402
=0 m
Where:
x '=P 1. x' 1+P2. x ' 2+P 3. x ' 3+…P
x '=5∗(2∗186.52+3∗365.74 )+10∗(2∗186.52+3∗365.74 )+15∗(2∗186.52+3∗365.74 )+20∗(2∗125+3∗163.75 )+¿+25∗(2∗186.52+3∗365.74 )+30∗(2∗186.52+3∗365.74 )+35∗(2∗186.52+3∗365.74 )
+40∗(2∗62.5+3∗186.52 )+¿10931.93
¿
x '=218638.610931.93
=20 m
2. ey= y '−L2=10−20
2=0 m
Where:
y '= P1. y ' 1+P 2. y ' 2+P 3. y' 3+…P
y '=
5∗(2∗186.52+6∗365.74+163.75 )+10∗(2∗186.52+6∗365.74+163.75 )+15∗(2∗186.52+6∗365.74+163.75 )+20∗(2∗62.5+6∗186.52+125)
10931.93
y '=109319.310931.93
=10.0 m
Raft moment values in both x & y directions:
Mx=P . ey=64943.93∗0=0 KN .mMy=P . ex=64943.93∗0=0 KN .m
q= PA
+ MxIx
y+ MyIy
x ≤ q allow , net
q=71313.85920.8
+ 036970.4
y+ 0135035.9
x≤ q allow ,net
q=70.52+0 y+0x.
Step3 : Compare the values of the soil pressure to determine weather q ≤ q a≪.net .
Point P/A x 0x y 0y q
A 70.53 -20.98 0 10.98 0 70.53B 70.53 -20 0 10.98 0 70.53C 70.53 -17.5 0 10.98 0 70.53D 70.53 -15 0 10.98 0 70.53E 70.53 -12.5 0 10.98 0 70.53F 70.53 -10 0 10.98 0 70.53G 70.53 -7.5 0 10.98 0 70.53H 70.53 -5 0 10.98 0 70.53I 70.53 -2.5 0 10.98 0 70.53J 70.53 0 0 10.98 0 70.53K 70.53 2.5 0 10.98 0 70.53L 70.53 5 0 10.98 0 70.53M 70.53 7.5 0 10.98 0 70.53N 70.53 10 0 10.98 0 70.53O 70.53 12.5 0 10.98 0 70.53P 70.53 15 0 10.98 0 70.53Q 70.53 17.5 0 10.98 0 70.53R 70.53 20 0 10.98 0 70.53S 70.53 20.98 0 10.98 0 70.53T 70.53 20.98 0 -10.98 0 70.53U 70.53 20 0 -10.98 0 70.53V 70.53 17.5 0 -10.98 0 70.53W 70.53 15 0 -10.98 0 70.53X 70.53 12.5 0 -10.98 0 70.53Y 70.53 10 0 -10.98 0 70.53Z 70.53 7.5 0 -10.98 0 70.53
A1 70.53 5 0 -10.98 0 70.53A2 70.53 2.5 0 -10.98 0 70.53A3 70.53 0 0 -10.98 0 70.53A4 70.53 -2.5 0 -10.98 0 70.53
A5 70.53 -5 0 -10.98 0 70.53A6 70.53 -7.5 0 -10.98 0 70.53A7 70.53 -10 0 -10.98 0 70.53A8 70.53 -12.5 0 -10.98 0 70.53A9 70.53 -15 0 -10.98 0 70.53A10 70.53 -17.5 0 -10.98 0 70.53A11 70.53 -20 0 -10.98 0 70.53A12 70.53 -20.98 0 -10.98 0 70.53
It’s given that( q a≪.net=5 kg/cm2=¿ 500KN/m2), and it’s found that (qmax.= 70.53KN/m2)
qmax . ≤ qa≪. net . ( is correct)
Step4 : Divide the mat into strips in both x and y directions with a width=B1 for each of them.
Here, strips are taken as center to center between columns and walls. Each internal strip either in x or y direction is 5m- wide and as fore external edge strips, the strip width is 3.42m in both x and y directions.
Step5 : Draw the shear V and the moment M Diagrams for each strip in both x and y directions.
Note : Strip (M N O Q Y Z) – width= 5m- is chosen to be analyzed and designed as a sample.
I. Soil pressure beneath the strip:
qN=q(atM)+q(atN )+q(atO)
3=70.53+70.53+70.53
3=70.53 KN /m2
qY=q(atX)+q(atY )+q(atZ)
3=70.53+70.53+70.53
3=70.53 KN /m2
II. Shear and Moment Diagrams:
Step6 : Determine the effective depth‘d’ of the foundation by “checking for diagonal tension shear” near various columns:
A critical internal column of (30cm*30cm) with ( Pd=206.74 KN , Pl=159 KN ) loads is sufficient to give a controlled value for the thicKness of the raft as it’s loaded with the ultimate loading value .
According to (ACI 318-95 section 9.2.1),
U=1.4(Pd) +1.6(Pl)
U=1.4*(206.74) +1.6*(159) =543.836 KN.
b0=4∗(d+300)
. Vc ≥ Vu
.Vc=¿ (4 )∗√ f c '∗b0∗d=0.85∗4∗√24∗( 4∗(d+300 ) )∗d ≥543.836
d2+300 d=8.16
d2+300 d−8.16=0
d=300mm
h=300+75+(142 )=382mm
accordingly, use h=400 mm
According to the equations of shear design (ACI 318-08) :
β=300300
=1.0 , d=400−50−142
=343 mm , b0=4∗(300+ 3432 )=1886 mm
α s=40−interior column .
Vc=16 (1+ 2
β )√ f c ' b0 d where, 16 (1+ 2
β )=0.333
Vc= 112
(α s d
b0
+2)√ f c ' b0 d where, 1
12 (α s d
b0
+2)=0.773
Vc=13√ f c ' b0 d where,
13=0.333 -control
Ø Vc=13
√ f c' b0 d=0.75∗0.333∗√24∗1886∗343∗10−3=791.5 KN
VU=502−(70.53+70.53)
2∗(0.3+0.343 )∗(0.3∗0.343 )=472.84 NN
VU=472.8 KN< Ø Vc=791.5 KN … (Safe against punching shear).
So, the thickness (40 cm) is adequate .
Step7 : Obtain the maximum value of both positive and negative moments in each
strip in both directions(x and y) per unit width (i.e. ...M'= M
B 1).
Step8 : (Design) finally determine the steel area per unit width for positive and negative reinforcement in x and y directions.
Check for one way shear ( beam shear) :
At the central column:
Pu=1.2 ( Pd )+1.6 ( Pl )=502 Kn .
d=400−50−142
=343 mm
Ø Vc=16
√ f c ' bw d=0.75∗16
∗√24∗5000∗343∗10−3=1050.2 KN
- Vu at distance d¿¿ of column
x=21.95−0.98−5−5+0.175+0.343=¿11.49m
W ( x )=W (11.49 )=70.53∗5=352.65 Kn /m
Vu=502−(352.65+352.65)
2∗(21.95−11.49)=472.84 Kn
