Design of Mat/Raft Foundation
• Mat or raft foundation is a large concrete slab supporting
several columns in two or more rows.
• It is used where the supporting soil has low bearing capacity.
• The bearing capacity increased by combining all individual
footings in to one mat –since bearing capacity is proportional
to width and depth of foundations.
• In addition to increasing the bearing capacity, mat foundations
tend to bridge over irregularities of the soil and the average
settlement does not approach the extreme values of isolated
footings.
• Thus mat foundations are often used for supporting structures
that are sensitive to differential settlement.
• Design of uniform mat
• Design Assumptions
– mat is infinitely rigid
– planner soil pressure distribution under mat
• Design Procedure
I. Determine the line of action of the resultant of all the loads
acting on the mat
II. Determine the contact pressure distribution as under
– If the resultant passes through the center of gravity of the
mat, the contact pressure is given by
Q
A
– If the resultant has an eccentricity of ex and ey in the x and y
direction
Q Qe x Qe y
max/ min x y
A Iyy Ixx
The maximum contact pressure should be less than the
allowable soil pressure
– Divide the slab mat into strips in x and y directions. Each
strip is assumed to act as independent beam subjected to
the contact pressure and the columns loads.
– Determine the modified column loads
– Draw the shear force and bending moment diagrams for
each strip.
– Select depth of mat for shear requirement
– Select steel reinforcement for moment requirement
Y
ex
ey
X X
Y
• Example
• A mat foundation is to be design by the conventional method
(rigid method) for the loadings shown in Fig. below.
• All columns are 40X40cm
• Ultimate soil bearing pressure , qult = 100kPa
• fyk = 300MPa fyd = 300/1.15 = 260.87 Mpa
• C25 fck= 20MPafctk = 1.5 MPa,
X
600kN 750kN 600kN
6m
1800kN 1800kN 1320kN
6m
1800kN 1800kN 1320kN
6m Y
600kN 750kN 600kN
5m 5m
• Location of c.g. of loads
• P = (600 +750+ 600)*2 +(1800+1800+1320)*2 =13740kN
• 13740 X = (750 +1800+1800+750)*5 + (600 +1320+1320+600)* 10
X = 4.65m
ex = 5-4.65 = 0.35
X’ = 5 +0.35 = 5.35m
• B min = 2*( 5.35 +0.20+0.15 ) =11.40m
• 13740 Y = (600 +750+600)*18 + (1800 +1800+1320)* 12 + (1800
+1800+1320)* 6
Y = 9m
ey = 6+ 6/2 -9 = 0
• Lmin = 2* (9+0.20+0.15) = 18.70m
• Dimension of Mat 11.40 X 18.70m
•
• Actual contact pressure
= P/(BL) = 13740/(11.40*18.70) =64.45kPa < ult = 100kPa
• Thickness of the mat
• Punching shear
• Punching shear under 1800kN load
Take d= 0.70m and = min = 0.50/fyk = 0.50 /300 = 0.0017
k1 = ( 1+50) = (1 +50*0.0017) =1.085
0.4+3d
40
k2 = 1.6 – d =1.6 -0.70 = 0.90 , Take K2 =1
40
Pr =(0.85+0.4+1.105)2+(0.4+3(0.70)
=7.21m
• Net shear force developed 1.5d=1.105>0.85 1.5d
• Vd= 1800 -*(2.355* 2.50) , =64.45kP
• Vd= 1800 -64.45*(2.355* 2.50)=1420.55kN
• Punching shear resistance
Vup = 0.25fctd k1k2ud (MN)
• Vup = 0.25 *1000* 1.085*1.00*7.21*0.70
=1369.00kN < Vd .. NOT OK! Increase the depth
Take d= 0.75m and = min = 0.50/fyk = 0.50 /300 = 0.0017
k1 = ( 1+50) = (1 +50*0.0017) =1.085
k2 = 1.6 – d =1.6 -0.75 = 0.85 , Take K2 =1
Pr =(0.85+0.4+1.125)2+(0.4+3(0.75)
=7.40m
• Net shear force developed
• Vd= 1800 -*(2.375* 2.65) , =64.45kP
• Vd= 1800 -64.45*(2.375* 2.65)=1394.37kN
• Punching shear resistance
Vup = 0.25fctd k1k2ud (MN)
• Vup = 0.25 *1000* 1.085*1.00*7.40*0.75
=1505.44kN >Vd .. OK!
• Check punching shear under 1320kN
Pr =(1.125 +0.15+0.4)2+(0.4+3(0.75)) =6.00m
• Net shear force developed
0.4+3d
• Vd= 1320 -64.45*(1.675*2.65)=1033.92kN 40
40
• Punching shear resistance
Vup = 0.25fctd k1k2ud (MN)
• Vup = 0.25 *1000* 1.085*1.00*6.00*0.75 1.5d=1.125 0.15
=1220.63kN > Vd .. OK!
• Check punching shear under 600kN
• Pr =(1.125+0.15+0.4) +(1.125+0.15+0.4)
1.5d=1.125
=3.35m
• Net shear force developed 40
40
0.15
• Vd= 600 -64.45*(1.675*1.675)=419.18kN
• Punching shear resistance 1.5d=1.125 0.15
Vup = 0.25fctd k1k2ud (MN)
• Vup = 0.25 *1000* 1.085*1.00*3.35*0.75
=681.52kN > Vd .. OK!
• Soil reaction analysis:- Divide the slab mat into strips in x and y
directions 3.55m 5.00m 2.85m
3.35m
Strip 4
Strip 3
6.00m
Strip 2
6.00m
3.35m
Strip 1
Strip A Strip C
Strip B
• Strip A, (64.45)*3.55 = 228.80kN/m
• Strip B , (64.45)*5.00 = 322.25kN/m
• Strip C, (64.45)*2.85 = 183.68kN/m
• Strip 1 &Strip 4, (64.45)*3.35 = 215.91kN/m
• Strip 2 & Strip 3 (64.45)*6.00 = 386.70kN/m
• Shear force and Bending moment diagrams for each strip
• Strip A
600kN 1800kN 600kN
1800kN
228.80kN/m
0.35 6.00 6.00 6.00 0.35
4
P i
600 1800 1800 600 4800 kN
i 1
• R = 228.80* 18.70 =4278.56kN
• V = P- R = 4800-4278.56 =521.44 0
• Hence take average of P and R
• I.e., (4800 +4278.56 )/2 =4539.28kN
• avg = (4539.28)/18.70 =242.74kN/m
• P1avg = P4avg = (4539.28/4800) *600 =567.41kN
• P2avg = P3avg = (4539.28/4800) *1800 =1702.23kN
567.41kN 1702.23kN 567.41kN
1702.23kN
242.74kN/m
0.35 6.00 6.00 6.00 0.35
973.99kN
1.99m 728.24N
3.00m 482.45kN
84.96kN 4.01m
SFD
482.45kN 84.96kN
464.57kN-m 464.57kN-m
728.24kN 397.09kN-m
973.99kN
BMD
14.87kN-m 14.87kN-m
1489.48kN-m 1489.48kN-m
• Strip 1 &Strip 4, (64.45)*3.35 = 215.91kN/m
600kN 600kN
750kN
215.910kN/m
1.05 5.00 5.00 0.35
4
P i
600 750 600 1950 kN
i 1
• R = 215.91* 11.40=2461.37kN
• V = P- R = 1950-2461.37 = -511.37 0
• Hence take average of P and R
• I.e., (1950+2461.37 )/2 =2205.69kN
• avg = (2205.69)/11.40 =193.48kN/m
• P1avg = P3avg = (2205.69/1950) *600 =678.67kN
• P2avg = (2205.69/1950) *750 =848.34kN
678.67kN 678.67kN
848.34kN
193.48kN/m
1.05 5.00 5.00 0.35
491.88kN 610.94kN
1.84m
2.46m
203.15kN
SFD
67.73kN
475.52kN 477.68kN-m 356.46kN
180.78kN-m
BMD
11.85kN-m
147.56kN-m
106.66kN-m
