The surface of an elastic body is loaded by a uniform stress of 2000 psf over an area 30 ft by 60 ft, which has coordinates (0, 0) at the center.
Find:
(a) the vertical stress at a depth of 10 ft below a corner of the loaded area.
(b) the vertical stress at a depth of 20 ft below the center of the loaded area.
(c) the vertical stress at a depth of 30 ft below a point which has coordinates 10 ft greater than those of the corner of the loaded area in the first quadrant (first quadrant is the one with positive x and y coordinates).

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DeniceSandidge DeniceSandidge · Answer 1 · 2020-04-11T09:50:00+02:00

Answer:

q = 1.11 lb

[tex]\triangle \sigma _v = 1.66 \times 10^{-5}[/tex] psf

Explanation:

given data

uniform stress = 2000 psf

area = 30 ft by 60 ft

solution

we get here surface load uniform homogenous soil is express as

q = [tex]\frac{Q}{A}[/tex] ................1

q = [tex]\frac{2000}{30\times 60}[/tex]

q = 1.11 lb

and

now we apply here boussin vertical stress that is at depth 10 ft

[tex]\triangle \sigma _v[/tex] = [tex]\frac{q}{z^2} \times \frac{3}{2\pi (1+(\frac{d}{z})^2)^{5/2}}[/tex] ................2

[tex]\triangle \sigma _v = \frac{1.1}{10^2} \times \frac{3}{2\pi (1+(\frac{30}{10})^2)^{5/2}}[/tex]

[tex]\triangle \sigma _v = 1.66 \times 10^{-5}[/tex] psf

and

depth at 20 ft will be put value in equation 2

[tex]\triangle \sigma _v = \frac{1.1}{20^2} \times \frac{3}{2\pi (1+(\frac{0}{20})^2)^{5/2}}[/tex]

[tex]\triangle \sigma _v = 1.31 \times 10^{-3}[/tex] psf

and

at depth at 30 ft

depth at 20 ft will be put value in equation 2

[tex]\triangle \sigma _v = \frac{1.1}{30^2} \times \frac{3}{2\pi (1+(\frac{30+10}{30})^2)^{5/2}}[/tex]

[tex]\triangle \sigma _v = 4.53 \times 10^{-5}[/tex] psf