ECGD 4122 – Foundation EngineeringECGD 4122 – Foundation Engineering

Lecture 7Lecture 7

Faculty of Applied Engineering and Urban PlanningFaculty of Applied Engineering and Urban Planning

Civil Engineering DepartmentCivil Engineering Department

22ndnd Semester 2008/2009 Semester 2008/2009

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ContentContent

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Design of Strap Footings Design of Strap Footings

A strap footing is usually used to connect an A strap footing is usually used to connect an eccentrically loaded column footing to an eccentrically loaded column footing to an interior column footing.interior column footing.

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Design of Strap Footings Design of Strap Footings

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Design of Strap Footings Design of Strap Footings

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Design of Strap Footings Design of Strap Footings

Basic considerations for strap footing design:Basic considerations for strap footing design:

•Strap must be rigid to control rotation of the Strap must be rigid to control rotation of the exterior footingexterior footing

•Footing dimensions should be proportioned for Footing dimensions should be proportioned for approximately equal soil pressures and avoid approximately equal soil pressures and avoid large differences in large differences in BB to reduce differential to reduce differential settlementssettlements

•Strap should be out of contact with soil so that Strap should be out of contact with soil so that there are no soil reactions to modify the design there are no soil reactions to modify the design assumptionsassumptions

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Design of Strap Footings Design of Strap Footings

Design Procedure:Design Procedure:

1.1. Assume (e), i.e. assume LAssume (e), i.e. assume L11

2.2. Compute SCompute S11 , R , R11, R, R22

4. Calculate L4. Calculate L11 , B , B11 , L , L22, B, B22

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Design of Strap Footings Design of Strap Footings

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Example 1Example 1

Assume: e = 1.0 mAssume: e = 1.0 m

SS11 = S – e = 6.2 – 1.0 = 5.2 m = S – e = 6.2 – 1.0 = 5.2 m

ΣMΣMR2R2 = 0 = 0

5.2 R5.2 R1 1 - 6.2(580) = 0- 6.2(580) = 0 RR11 = 691.5 kN = 691.5 kN

ΣMΣMR1R1 = 0 = 0

-1.0 (580)-1.0 (580) + 5.2(900) – R+ 5.2(900) – R22(5.2)= 0(5.2)= 0 RR22 = 788.5 kN = 788.5 kN

LL11 = 2(e + x) = 2(1.0 + 0.2) = 2.4 m = 2(e + x) = 2(1.0 + 0.2) = 2.4 m

BB11 = R = R11/(L/(L11qqultult) = 691.5/[(2.4)(183)] = 1.57 (Use 1.60 m)) = 691.5/[(2.4)(183)] = 1.57 (Use 1.60 m)

BB22 = L = L22 = (R = (R22/q/qultult))1/21/2 = (788.5/183) = (788.5/183)1/21/2 = 2.08 (Use 2.10 m) = 2.08 (Use 2.10 m)9

Example 1 - SolutionExample 1 - Solution

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Types of Mat Foundations Types of Mat Foundations

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Types of Mat Foundations Types of Mat Foundations

For Saturated Clays with For Saturated Clays with ϕϕ = 0 = 0

qB

D0.41

L

0.195B15.14cq f

uu

Applied StressApplied Stress

fγDA

Qq

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Stress Analyses of Mat Foundations Stress Analyses of Mat Foundations

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Determine the net ultimate bearing capacity of a Determine the net ultimate bearing capacity of a

mat foundation sized 13.0 × 9.0 mmat foundation sized 13.0 × 9.0 m22 on a saturated on a saturated

clay deposit with clay deposit with ccuu = 94 kPa, = 94 kPa, = 0, and = 0, and DDff = 2.0 m. = 2.0 m.

Example 2Example 2

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Example 2 - SolutionExample 2 - Solution

kPa 597.13

(9)

(2)(0.4)1

(13)

(0.195)(9)1(5.14)(94)

B

D0.41

L

0.195B15.14c)(q

qq)(q

qB

D0.41

L

0.195B15.14cq

funetu

unetu

fuu

Compensated Mat FoundationsCompensated Mat Foundations

For no increase in the net soil pressure

γA

QD0γD

A

Qq ff

f

fu

f

uS

γDAQ

BD

0.41L

0.195B15.14c

γDAQ

qqF

15

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A mat foundation sized 30.0 × 40.0 mA mat foundation sized 30.0 × 40.0 m22 is to is to

withstand a total dead and live load of 200 MN and withstand a total dead and live load of 200 MN and

located on a soft clay deposit with located on a soft clay deposit with ccuu = 12.5 kPa = 12.5 kPa

and and = 18.8 kN/m = 18.8 kN/m33. Determine the depth of . Determine the depth of

foundation for:foundation for:

1)1) A fully compensated caseA fully compensated case

2)2) A factor of safety = 3.0A factor of safety = 3.0

Example 3Example 3

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Example 3 - SolutionExample 3 - Solution

m 8.87

(40)(18.8)(30)

)10(200

A

Q D 0 q

(1)

3

f

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Example 3 - SolutionExample 3 - Solution

m 7.43D Dfor Solving

(18.8)D(30)(40)

)10(200

(30)D

(0.4)1(40)

)(0.195)(3015)(5.14)(12.

3.0

γDAQ

BD

0.41L

0.195B15.14c

F

(2)

ff

f

3

f

f

fu

S

Pressure Under Mat FoundationsPressure Under Mat Foundations

yy

fapp. IIA

Q(q)

yMxMxy

Q'x

'33

'22

'11

xQxQxQ

xf

x

eQ

Bxe

yM

2'

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Pressure Under Mat FoundationsPressure Under Mat Foundations

app.netall. (q) )(q

Q'y

'33

'22

'11

yQyQyQ

yf

y

eQ

Lye

xM

2'

20

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Pressure Under Mat FoundationsPressure Under Mat Foundations