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8/3/2019 Structural Design of Mat Foundation
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Structural Design of MatFoundation
Conventional Rigid Method:
Step1: Calculate the total column load
Step2: Determine the pressure on the soil (q)below the mat at point A, B, Cby using the
equation
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Step 3: Compare the values of the soilpressures determine in step 2 with the net
allowable soil pressure to check if q
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Step 6: determine depth of the mat d. This canbe done by checking for diagonal tension
shear near various column. According to ACICode 318-95(section 11.122.1c). For criticalsection
Step 7: from the moment diagrams of all stripsin a given direction (that is X or Y), obtain the
maximum positive and negative moments perunit width M=M/B1
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G I
A B=500KN C
A=400KN C=450KN
7m
L=1500KN J=1200KN
K=1500KN
0.436m
7m0.095m
G=1500KN I=1200KN
H=1500KN
7m
F=400KN D=350KN
F H E=500KN J D
8m 8m
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Cu= 25 KPaqu= 128.5 KPa
qall= 42.83333 KPaB= 16.5 m
L= 21.5 mb= 0.5 m
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QA= 400 KN
QB= 500 KN
QC= 450 KN
QD= 350 KNQE= 500 KN
QF= 400 KN
QG= 1500 KNQH= 1500 KN
QI= 1200 KN
QJ= 1200 KNQK= 1500 KN
QL= 1500 KN
SQ= 11000 KN
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A=BL= 354.75 m2
13665.27 m4
8048.391 m4
x
x
y
y
I
YM
I
XM
A
Q
q
3
12
1BLIx
3
12
1LBIy
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7.813636 m
-0.43636 m
4800 KN.mxy QeM
Q
XQX ii
'
2'B
Xex
10.84545 m
0.095455 m
1050 KN.m
Q
YQY ii
'
2'
LYe y
yx QeM
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yKPaxyx
q 077.06.00.3113665
.1050
8048
.4800
75.354
000.11
qA= 36.7778 Kpa
qB= 31.8278 KPa
qC= 26.8778 KPaqD= 25.2223 KPa
qE= 30.1723 KPa
qF= 35.1223 KPa
The soil pressures at all point are less than the allowable bearing capacity
qall= 42.8333 Kpa OK
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Strip AGHF
qav= (qA+qF)/2 35.95 Kpa
Total soil reaction qav.B1.L 3284.93125 KNTotal column load on this strip 3800 KNAverage load 3542.4656 KNModified average soil pressure
qav.modified= 38.768434 KpaFactor modified column load F= 0.93223
QA= 372.8911 KNQL= 1398.342 KNQG= 1398.342 KNQF= 372.8911 KN
The load per unit beam is equal to 164.766 KN
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Strip GIJH
qav= (qB+qE)/2 31 Kpa
Total soil reaction qav.B1.L 5332 KN
Total column load on this strip 4000 KN
Average load 4666 KN
Modified average soil pressure
qav.modified= 27.128 Kpa
Factor modified column load F= 1.167
QB= 583.25 KNQK= 1749.75 KN
QH= 1749.75 KN
QE= 583.25 KN
The load per unit beam is equal to 217 KN
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Strip ICDJ
qav= (qC+qD)/2 26.05 Kpa
Total soil reaction qav.B1.L 2380.32 KN
Total column load on this strip 3200 KNAverage load 2790.159 KNModified average soil pressure
qav.modified= 30.53526 Kpa
Factor modified column load F= 0.871925
QC= 392.3662 KNQJ= 1046.31 KNQI= 1046.31 KNQD= 305.1737 KN
The load per unit beam is equal to 129.775m KN
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Shear on point
SA1= 41.19146 KN 0.25
SA2= -331.7 KN 0.25
SL1= 821.6612 KN 7.25m
SL2= -576.68 KN 7.25m
SG1= 576.6805 KN 14.25m
SG2= -821.661 KN 14.25mSF1= 331.6997 KN 21.25m
SF2= -41.1915 KN 21.25m
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Shear on point
SB1= 54.25581 KN 0.25
SB2= -528.994 KN 0.25
SK1= 990.17m KN 7.25m
SK2= -759.6m KN 7.25m
SH1= 759.58m KN 14.25m
SH2= -990.2m KN 14.25mSE1= 528.99m KN 21.25m
SE2= -54.26m KN 21.25m
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Shear on point
SC1= 32.4437 KN 0.25
SC2= -359.92 KN 0.25
SJ1= 548.502 KN 7.25m
SJ2= -497.81 KN 7.25m
SI1= 410.616 KN 14.25m
SI2= -635.69 KN 14.25m
SD1= 272.73 KN 21.25m
SD2= 32.4437 KN 21.25m
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Moment on point
X1= 2.013158 m
X2= 4.986842 m
MA= 5.148933 KN.m
Mmax(AL)= -328.733 KN.m
ML=
1720.014 KN.mX
5=
3.5mMmax(LG)= 710.824m KN.m X6= 3.5m
MG= 1720.014 KN.m
Mmax(GF)= -328.733 KN.mMF= -5.14893 KN.m
X3= 2.01316m m
X4= 4.986842 m
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Moment on point
X1= 2.4375 m
X2= 4.5625m m0.25 MB= 6.781977 KN.m
2.01316 Mmax(BK)= -637.93 KN.m
7.25m MK= 1620.892 KN.m X5= 3.5m10.75m Mmax(KH)= 291.62m KN.m X6= 3.5m
14.25m MH= 1620.9m KN.m
17.75m Mmax(HE)= -637.93 KN.m21.25m X3= 2.4375m m
X4= 4.5625m m
ME= -6.78198
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Moment on point
X1= 2.77344 m
X2= 4.23 mMC= 4.05546 KN.m
Mmax(CJ)= -495.06 KN.m
MJ= 664.08 KN.m X5= 3.84mMmax(JI)= -290.70 KN.m X6= 3.1641
MI= 358.91 KN.m
Mmax(ID)= -282.52 KN.mX3= 2.10156 m
X4= 4.89844 m
MD= 4.05546 KN.m
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Shear diagram AF
41.19146076
-331.6996577
821.6612435
-576.6804506
576.6804506
-821.6612435
331.6996577
-41.19146076
-1000
-800
-600
-400
-200
0
200
400
600
800
1000
0 5 10 15 20 25
Distance(m)
Strength(KN)
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Moment diagram AF
5.148932594
-328.7329597
1720.014483
710.8236945m
1720.014483
-5.148932594
-328.7329597-500
0
500
1000
1500
2000
0 5 10 15 20 25
Distance(m)
Moment(KN.m)
Top steel
Bottom steel
top steel
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Determination of the thickness of the Mat
Column load 1500KN bo= (0.5+d/2)+(0.5+d/2)+(0.5+d)=1.5+2d0.5+d/2 U=(bod)[f.0.34.f'c
0.5]
U= 1.7 x 1500= 2.55 MN
2.55=(1.5+2d)(d)[(0.85)(0.34)(20.7)0.5
]
(1.5+2d)d=1.94
0.5+d 2d2+1.5d-1.94=0
d= 0.68 m
minimum cover= 76 mm
steel bars= 25 mmh= 0.781 m
We take h= 0.8 m
Determination of Reinforcement
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Determination of Reinforcement
Maximum positive momement is located in strip AGHF
M'= 1720/B1= 404.706 KN.m/mMaximum negative momement is located in strip GIJH
M'= 637.93/B1= 150.101 KN.m/m
Mu=M'(load factor)=fAsfy(d-a/2)
For the positive moment
Mu=(404.71)(1.7)=fAs(413.7*1000)(0.68-a/2)688.007 = 351645 As(0.68-a/2)
0.00195654 = As(0.68-a/2)
= = 23.51As or As= 0.0425abf
fAa
c
yS
'85.0
)1)(7.20)(85.0(
7.413.s
A
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0.00195654 = 0.0425a(0.68-a/2)
= 0.0289a-0.02125a2
So a= 0.07147
0.0289a-0.02125a2
-0.001957=0 As= 0.00304 m2
/m-0.02125 As= 3037.58 mm
2/m
0.0289 For steel D=25mm A= 490.8739 mm2
-0.001957 So for spacing bars at 160 mm center to center
0.00066887 As provided= (490.874)(1000/160)= 3067.96 mm2/m
0.025862420.07147237
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For the negative momentMu=(150.1)(1.7)=fAs(413.7*1000)(0.68-a/2)
255.172 = 351645 As(0.68-a/2)
0.00073 = As(0.68-a/2)
Similarly
As= 0.0425a0.000726 = 0.0425a(0.68-a/2)
= 0.0289a-0.02125a2
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0.0289a-0.02125a2-0.00073=0 So a= 0.02575
-0.02125 As= 0.00109 m2/m
0.0289 As= 1094.25 mm2/m
-0.00073 For steel D=25mm A= 490.874 mm2
0.00077 So for spacing bars at 300mm center to center
0.02781 As provided= (490.874)(1000/300)= 1963.5 mm2/m
a= 0.02575