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STRESSES BENEATH LOADED AREAS

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STRESS TRANSFORMATION

n

y x y x

xy

2 2

2 2cos sin

n

y x

xy

2

2 2sin cos

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PRINCIPLE STRESSES

• Principle planes – planes for which the normal stresses take extreme valuesnormal stresses take extreme values, and shear stresses are zero. Stresses in principal planes are principal stresses.

• Major and minor principal stresses 2

2

y x y x

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1

2

2 2

y y

xy

2

2

2

2 2

y x y x

xy

MOHR’S CIRCLE (from Das, 2002)

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MOHR’S CIRCLE FOR KNOWN PRINCIPAL STRESSES

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MAXIMUM SHEAR STRESS

• Corresponds to the radius of the Mohr’s• Corresponds to the radius of the Mohr s circle

max

y x

xy2

2

2

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4

STRESSES BY A POINT LOADBOUSSINESQ (1885) SOLUTION

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STRESSES BY A POINT LOADBOUSSINESQ SOLUTION

X

P x z

R

x y

Rr R z

y z

R r

2

31 2

2

5

2 2

2

2

3 2( )( )

P y z y x x z

2

31 2

2

5

2 2

2

2

3 2( )( )

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y R Rr R z R r

2 5 2 3 2( )

( )

z

P z

R

P z

r z

3

2

3

2

3

5

3

2 2 5 2( ) /

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STRESSES BY A POINT LOADBOUSSINESQ SOLUTION

• Alternative formulation• Alternative formulation

z B

P

z r

z

P

zI

2 2 5 2 2

3

2

1

1

/

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z

STRESSES BY A POINT LOADWEESTERGARD’S SOLUTION

• Assumes presence of layers of thin• Assumes presence of layers of thin, infinitely rigid horizontal reinforcement between layers.

• Increase in pressure

Q P1

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z W

Q

z r

z

P

zI

2 2 3 2 2

1

2 1

/

6

Boussinesq more conservative!

COMPARISON OF BOUSSINESQ AND WESTERGAARD’S

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INFLUENCE COEFFICIENTS

(from Murthy, 2003)

VERTICAL STRESS BY A LINE LOAD (from Murthy, 2002)

q q

I

2 1

2 2z zz xz

z

1

2 2

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VERTICALVERTICAL STRESS BY A FLEXIBLE STRIP LOADING

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(from Das, 2002)

VERTICAL STRESS BY A FLEXIBLE STRIP LOADING

• From the solution for a line loading the• From the solution for a line loading, the vertical stress at P for a line load qdx

• The total stress increase at point P is

dq z

x x zz

2 3

2 2 2[( ) ]

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• The total stress increase at point P is

z

b

bq z

x x zdx

2 3

2 2 2[( ) ]

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VERTICAL STRESS BY A FLEXIBLE STRIP LOADING

• Solving the integral• Solving the integral

or

z

q z

x b

z

x b

bz x b z

x b z b z

2 2

41 1

2 2 2

2 2 2 2 2tan tan

( )

( )

z

q sin cos( )2

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• Principal stresses at P are

1 q

sin

3 q

sin

VERTICAL STRESS BY A FLEXIBLE STRIP LOADING (from

Murthy, 2002)

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PRESSURE RATIO FOR

STRIPSTRIP LOADING

(from Murthy, 2003)

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EXAMPLE – STRIP LOADING (Murthy 6.4)

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EXAMPLE – STRIP LOADING (Murthy 6.4)

Line load equation

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EXAMPLE – STRIP LOADING (Murthy 6.4)

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STRESSES BENEATH A CORNER OF A RECTANGULARRECTANGULAR

LOADING(from Murthy, 2003)

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dQ q db dl

STRESSES BENEATH A RECTANGULAR AREA

• Stress increase at P due to dQ• Stress increase at P due to dQ

• Integrating in x and y directions from 0 to b and 0 to l respectively or

d

dQ z

z rz

2

3 3

2 2 5 2/

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to b, and 0 to l, respectively, or

z d z

00

11

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STRESSES BENEATH A RECTANGUAR AREA

• The stress increase at P is• The stress increase at P is

The obtained result is in radians !!!!

where m=b/z and n=l/z or

z q

mn m n

m n m n

m n

m n

mn m n

m n m n

1

4

2 1

1

2

1

2 1

1

2 2 1/2

2 2 2 2

2 2

2 21

2 2 1/2

2 2 2 2

( )tan

( )

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where m=b/z and n=l/z, or

z qI

INFLUENCECOEFFICIENT

FOR RECTANGULAR

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RECTANGULAR LOADING

(after Fadum, 1934)

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COMPUTATION OF STRESSES BELOW RECTANGULAR

LOADING (from Murthy, 2003)

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Point 0 within the area

z=q(I1+I2+I3+I4)

Point 0 outside the area

z=q(I1-I2-I3+I4)

EXAMPLE – RECTANGULAR LOADING (Murthy 6.5)

Point 0 outside the area

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z=q(I1-I2-I3+I4)

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EXAMPLE – RECTANGULAR LOADING (Murthy 6.5)

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EXAMPLE –RECTANGULAR

LOADING

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LOADING(Murthy 6.5)

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STRESSES UNDER

CIRCULAR LOADINGS

(from Murthy, 2003)

dQ qdA qrd dr

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dq z r d dr

r zz

3

2

3

2 2 5 2( ) /

STRESSES BENEATH A CIRCULAR LOADING

• Integrating over the circular area• Integrating over the circular area

the stress at P under the center is

z zr

r R

r

r R

dqz rd dr

r z

0

3

0

2

2 2 5 200

20 03

2 ( ) /

3 2/

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z qz

R z

1

3

02 2 3 2( ) / or

zzq R

z

I

11

1 02

2

16

STRESSES UNDER THE CENTER OF

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THE CENTER OF CIRCULAR LOADING

(from Das, 2002)

INFLUENCE (NEWMARK) CHART FOR GENERAL LOADING

The sol tion for a circ lar loading• The solution for a circular loading

can be rearranged to

z qz

R z

1

3

02 2 3 2( ) /

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R

z qz0

2 3

1 1

/

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INFLUENCE (NEWMARK) CHART FOR GENERAL LOADING

R / and /q are dimensionless ratios• R0/z and z /q are dimensionless ratios

=> If geometry is presented in a dimensionless form, for a constant R0/z, z /q is a constant too.

• Newmark (1942) presented an influence

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Newmark (1942) presented an influence chart for general uniform loading.

STRESS INCREASE FOR UNITFOR UNIT

LOADING q AS A FUNCTION

OF z/R0 RATIO (from Das, 2002)

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( o as, 00 )

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INFLUENCE (NEWMARK) CHART FOR GENERAL LOADING

• Construction of charts:• Construction of charts:1. A set of concentric circles are constructed

for z /q =0, 0.1, 0.2, …., 1 (for z /q =0, R0/z=0; for z /q =1, R0/z=infinity).

2. The circles are divided by several equally spaced radial lines.

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p3. The number of created elements in the

chart is N, or the influence value of each element becomes 1/N.

R0/z RATIO AS A FUNCTION OF STRESS INCREASE RATIO

(from Murthy, 2003)

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INFLUENCE (NEWMARK) CHART FOR GENERAL LOADING

• Explanation:

0.1 0.2 0.3 0.4 0.5

Radius for

z/q=0.4

Loading between rings 0.3 and 0.4

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z

a d 0causes z/q=0.1.

INFLUENCE (NEWMARK) CHART FOR GENERAL LOADING

L di thLoading on the shaded area between rings 0.3 and 0.4 (20 radial lines) causes /q=0 1/20

0.1 0.2 0.3 0.4 0.5

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z/q=0.1/20 =0.005.

z

20

EVALUATION OF VERTICAL PRESSURE BY NEWMARK

CHART1. Determine the depth (z) of the point of

interest relative to the foundation.2. Plot the plan of the loaded area in a

scale where z equals to the unit length of the chart (line AB).

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( )3. Place the plan of the foundation so

that the point is located at the center of the chart.

EVALUATION OF VERTICAL PRESSURE BY NEWMARK

CHART4. Count the number of the elements (N)

of the chart covered by the loaded area. The increase in pressure is

z=CiNq

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where Ci is the influence value.

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EXAMPLE NEWMARK

CHARTCHART(from Murthy, 2003)

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EXAMPLE – NEWMARK CHARTA new building is going to be constructed next to an existing one. The bottom of the slab of the new building is 3 m above the bottom of the closest spread footing of the existing building. Find the maximum increase of the vertical pressure beneath the spread footing for the assumption that the new building slab will have a uniformly distributed loading of 150 kPa.

4 m

NEW BUILDING

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5 m 4 m 2 m

3 m

A

SPREAD FOOT.OLD BUILDING

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EXAMPLE –NEWMARK

CHARTC

z=20*0.005*150=15 kPa

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3m

PRESSURE ISOBARS (from Murthy, 2003)

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SIGNIFICANT DEPTH OF STRESSED ZONE (from Murthy, 2003)

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INFLUENCE CHARTS FOR SQUARE ANDSQUARE AND

STRIP LOADINGS

(from Murthy, 2003)

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INFLUENCE CHART

FORFOR CIRCULAR

LOADING(from Murthy, 2003)

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INFLUENCEINFLUENCE CHARTS FOR SQUARE AND

STRIP LOADINGS

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LOADINGS –WESTERGAARD

(from Murthy, 2003)

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EXAMPLE – PRESSURE ISOBARS (Murthy 6.12)

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EXAMPLE – PRESSURE ISOBARS (Murthy 6.12)

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EXAMPLE –

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PRESSURE ISOBARS

(Murthy 6.12)