Structural Design of Mat Foundation

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

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Structural Design of Mat Foundation