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Vertical Stress Increases

Chapter 8

Point Load

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Point Load

Point Load

32 ⁄

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Point Load

32

1⁄

⁄ 1 ⁄

‐4 ‐3 ‐2 ‐1 0 1 2 3 4

Point Load

I1

r/z

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ExampleA vertical point load of 25 kN is applied at the ground surface. What is the stressincrease in the ground 2 m below the ground surface directly beneath the load?

∆252

0.4775 2.98 ⁄

r = 0, z = 2

ExampleA vertical point load of 25 kN is applied at the ground surface. What is the stressincrease in the ground 2 m below the ground surface and 2 m to the side?

∆252

0.0844 0.5275 ⁄

r = 2, z = 2

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Line Load

NOTE:qhasunitsofkN/m

Line Load

2 1⁄ 1

2⁄⁄ 1

∆⁄

2⁄⁄ 1

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‐4 ‐3 ‐2 ‐1 0 1 2 3 4

I

x/z

2⁄⁄ 1

Line Load

ExampleA vertical line load of 8 kN/m is applied at the ground surface. What is the stressincrease in the ground 4 m below the ground surface directly beneath the load?

∆8 /4

0.637 1.274 ⁄

x = 0, z = 4

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ExampleA vertical line load of 8 kN/m is applied at the ground surface. What is the stressincrease in the ground 4 m below the ground surface and 2 m off to the side?

∆8 /4

0.407 0.814 ⁄

x = 2, z = 4

Circular Loaded Area

NOTE:qhasunitsofkN/m2 orkPa

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Circular Loaded Area

∆1

1⁄ 1 /

Circular Loaded Area

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.2 0.4 0.6 0.8 1 1.2

/q

z/R

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ExampleA uniform load of 80 kPa is applied at the ground surface over a circular areawith a diameter of 8 m. What is the stress increase directly beneath the centerof the loaded area at a depth of 8 m?

∆ 80 0.2845 22.76

z = 8, R = 4

Circular Loaded Area

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ExampleA uniform load of 80 kPa is applied at the ground surface over a circular areawith a diameter of 8 m. What is the stress increase 2 m below the ground surfaceat a radial distance of 4 m from the load centerline?

∆ 80 0.417 33.36

z = 8, R = 4

r = 4 = R

Rectangular Loaded Area

∆

2 1, 2 1

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ExampleA uniform load of 80 kPa is applied at the ground surface over a rectangular areawith a length of 5 m and a width of 2 m. What is the stress increase beneath thecorner of the loaded area at a depth of 10 m?

210

0.2

510

0.5

∆ 80 0.04 3.2

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ExampleA uniform load of 80 kPa is applied at the ground surface over a rectangular areawith a length of 5 m and a width of 2 m. What is the stress increase beneath thecorner of the loaded area at a depth of 10 m?

510

0.5

210

0.2

∆ 80 0.04 3.2

L

B

Rectangular Loaded Area

A'

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L

B

Rectangular Loaded Area

A'

L/2 L/2

B/2

B/2

= L/B

z/(B/2) =

∆

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Example

Example

44

1.0

24

0.5

∆ 150 0.12 18

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Example

14

2 1 0.5 1 0.25 11 0.25 0.25 1

1 0.25 21 0.25 1

tan2 1 0.5 1 0.25 11 0.25 0.25 1

0.120175

′ 1 1 ′ 0.5 0.25

1 0.5 0.5

′ ′ 1 0.25 0.25

Example

14

0.25

24

0.5

∆ 150 0.048 7.2

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Example

∆ 18.0 7.2 10.8