Session 9 – 10 MAT FOUNDATION

Course : S0484/Foundation Engineering

Year : 2007

Version : 1/0

MAT FOUNDATION

Topic:

• Introduction

• Type of Mat Foundation

• Bearing Capacity

• Settlement

INTRODUCTION

A mat foundation (or raft foundation) is continuous in two directions capable of supporting multiple columns, wall or floor loads. It has dimensions from 20 to 80 ft or more for houses and hundreds of feet for large structures such as multi-story hospitals and some warehouses, or consisting of stiffening beams placed below a flat slab are useful in unstable soils such as expansive, collapsible or soft materials where differential movements can be significant (exceeding 0.5 inch).

TYPE OF MAT FOUNDATION

FLAT PLATE

TYPE OF MAT FOUNDATION

FLAT PLATE THICKENED UNDER COLUMN

TYPE OF MAT FOUNDATION

BEAMS AND SLAB

TYPE OF MAT FOUNDATION

SLAB WITH BASEMENT WALL

BEARING CAPACITY

GROSS ULTIMATE BEARING CAPACITY

idsqiqdqscicdcsu FFFNBFFFNqqFFFNccq .....).5.0(........

FOR SATURATED CLAYS WITH = 0 AND VERTICAL LOADING CONDITION

qFFFNccq cicdcsu ....

WITH:

B

DF

L

B

L

B

N

N

L

BF

fcd

c

qcs

4.01

195.01

14.5

111

(=0 Nc = 5.14 ; Nq = 1.00 ; N =0.00)

SO:

qB

D

L

Bcq fuu

4.01

195.0114.5

BEARING CAPACITY

NET ULTIMATE BEARING CAPACITY

B

D

L

Bcq

qqq

fuunet

uunet

4.01195.0

114.5)(

)(

NET ALLOWABLE BEARING CAPACITY

B

D

L

Bc

FS

qq f

uunet

netall 4.01195.0

1713.1)()( (FS = 3)

BEARING CAPACITY

PLOT OF qall(net)/cu against Df/B with factor of safety = 3

BEARING CAPACITY

QDfUnit weight =

)(. netallf qDA

Qq

Where:

Q = dead weight of the structure and the live load

A = area of the raft

EXAMPLE 1

Problem:

Determine the net ultimate bearing capacity of a mat foundation measuring 45 ft x 30 ft on a saturated clay with cu = 1950 lb/ft2, = 0 and Df = 6.5 ft

Solution:

B

D

L

Bcq fuunet 4.01

195.0114.5)(

2/307,12

30

5.64.01

45

30195.01195014.5

ftlb

xx

FACTOR OF SAFETY

f

unetunet

DAQ

q

q

qFS

.

)(

f

fu

DAQ

B

D

LB

c

FS.

4.01195.0

114.5

For Saturated Clay

EXAMPLE 2 – PROBLEM

QDfUnit weight =

The mat dimension is 60 ft x 100 ft. The total dead and live load on the mat is 25x103 kip. The mat is placed over a saturated clay having a unit weight of 120 lb/ft3 and cu = 2800 lb/ft2. Given Df = 5 ft, determine the factor of safety against bearing capacity failure

EXAMPLE 2 - SOLUTION

f

fu

DAQ

B

D

LB

c

FS.

4.01195.0

114.5

66.4

5120100601025

605

4.01100

60195.01280014.5

6

xx

FS

EXAMPLE 3 – PROBLEM

Consider a mat foundation 90 ft x 120 ft in plan, as shown in the following figure. The total dead load and live load on the raft is 45x103 kip. Estimate the consolidation settlement at the center of the foundation

Q

90 ft x 120 ft6 ft

5 ft

40 ft

18 ft

sand

sand

sand

Normally consolidated clay sat = 118 lb/ft3 Cc = 0.28 ; eo = 0.9

Water table

= 100 lb/ft3

sat = 121.5 lb/ft3

EXAMPLE 3 – SOLUTION

o

avo

o

ccc p

pp

e

HCS log

1

.

2/39644.621182

184.625.1214010011 ftlbpo

For Q = 45x106 lb, the net load per unit area is

26

/3567610012090

1045. ftlb

x

xD

A

Qq f

EXAMPLE 3 – SOLUTION

In order to calculate pav, the loaded area can be divided into four areas, each measuring 45 ft x 60 ft. The average stress increase in the clay layer below the corner of each rectangular area can be calculated by using the following formula:

12

12/

12

12 HH

IHIHqp HaHa

HHav

18

405184053567 12

12 /HaHa

HHav

IIp

Where:

H1 = the depth of top elevation of clay layer

H2 = the depth of bottom elevation of clay layer

Ia(H1) ; Ia(H2) = influence factor

90 ft

120 ft

45 ft x 60 ft

EXAMPLE 3 – SOLUTION

21.0

95.063

60

71.018405

45

2

2

2

HaI

H

Ln

H

Bm

225.0

33.145

60

1405

45

1

1

1

HaI

H

Ln

H

Bm

EXAMPLE 3 – SOLUTION

EXAMPLE 3 – SOLUTION

2

/ /3.61518

225.04521.0633567

12ftlbp HHav

in

xSc 68.6

3964

2.24613964log

9.01

121828.0