# Foundation Engineering Questions and Answers – Stresses in Flexible Pavements

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This set of Foundation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Stresses in Flexible Pavements”.

1. The maximum stress occurs on __________
a) vertical plane passing through radial distance of one from the point of load application
b) vertical plane passing through radial distance of two from the point of load application
c) vertical plane passing through the point of load application
d) vertical plane passing through radial distance of three from the point of load application

Explanation: The stress equation is given by,
$$σ_z=\frac{3Q}{2πz^2} \left[\frac{1}{1+(\frac{r}{z})^2}\right]^\frac{5}{2},$$
if r=0, we get the maximum stress value. Hence, the maximum stress occurs on vertical plane passing through the point of load application.

2. The stress σz at any point (r,z) below the pavement is __________
a) $$σ_z=\frac{3P}{2πz^2}\left[\frac{1}{1+(\frac{r}{z})^3}\right]^\frac{5}{2}$$
b) $$σ_z=\frac{3P}{2πz^2}\left[\frac{1}{(\frac{r}{z})^2}\right]^\frac{5}{2}$$
c) $$σ_z=\left[\frac{1}{1+(\frac{r}{z})^2}\right]^\frac{5}{2}$$
d) $$σ_z=\frac{3P}{2πz^2}\left[\frac{1}{1+(\frac{r}{z})^2}\right]^\frac{5}{2}$$

Explanation: The stress σz at any point (r,z) below the pavement is given by,
$$σ_z=\frac{3P}{2πz^2}\left[\frac{1}{1+(\frac{r}{z})^2}\right]^\frac{5}{2},$$
Where, σz is the vertical stress
P is the point load acting at the ground surface
r is the radial horizontal distance
z is the vertical distance.

3. For the loaded circular area of radius a, the vertical stress on a vertical line passing through the center of the loaded area is __________
a) $$σ_z=p\left[1-\left[\frac{1}{1+(\frac{a}{z})^2}\right]^\frac{3}{2}\right]$$
b) $$σ_z=p\left[1+\left[\frac{1}{1+(\frac{a}{z})^2}\right]^\frac{3}{2}\right]$$
c) $$σ_z=p\left[\frac{1}{1+(\frac{a}{z})^2}\right]^\frac{3}{2}$$
d) $$σ_z=p\left[1-\left[\frac{1}{1+(\frac{a}{z})^2}\right]^\frac{5}{2}\right]$$

Explanation: The vertical pressure σz under a uniformly loaded circular area is given by,
$$σ_z=p\left[1-\left[\frac{1}{1+(\frac{a}{z})^2}\right]^\frac{3}{2}\right]$$
where, p=load intensity per unit area
z= depth of point.
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4. For a loaded circular area of radius ‘a’, μ=Poisson’s ratio, the horizontal radial stress is ________
a) $$σ_r=\frac{p}{2} \left[1+2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}\right]$$
b) $$σ_r=\frac{p}{2} \left[\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+\frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right]$$
c) $$σ_r=\frac{p}{2} \left[1-2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+\frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right]$$
d) $$σ_r=\frac{p}{2} \left[1+2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+\frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right]$$

$$σ_r=\frac{p}{2} \left[1+2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+\frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right],$$
μ=Poisson’s ratio.

5. The vertical strain ∈z due to triaxial load (σzrtr) under the center of plate is _________
a) $$∈_z=\frac{1}{E}[σ_z-μ(σ_r)]$$
b) $$∈_z=\frac{1}{E}[σ_z-2μ(σ_r)]$$
c) $$∈_z=\frac{1}{E}[σ_z-23μ(σ_r)]$$
d) $$∈_z=\frac{1}{E}[σ_z-0.43μ(σ_r)]$$

Explanation: The vertical strain ∈z due to triaxial load (σzrtr) under the center of plate is given by,
$$∈_z=\frac{1}{E}[σ_z-2μ(σ_r)],$$
where,
σz=vertical stress
E=elastic modulus
μ= Poisson’s ratio.

6. The elastic strain ∆ of the subgrade is __________
a) $$∆=\frac{p}{E}\left[(2-2μ^2)(a^2+z^2)^\frac{1}{2}-\frac{(1+μ) z^2}{(a^2+z^2)^\frac{1}{2}}+(μ+2μ^2-1)z\right]$$
b) $$∆=\frac{p}{2}\left[\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+ \frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right]$$
c) $$∆=\frac{p}{2}\left[1-2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+ \frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right]$$
d) $$∆=\frac{p}{2}\left[1+2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+ \frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right]$$

Explanation: The vertical strain is given by,
$$∆=\frac{1}{E}[σ_z-2μ(σ_r)],$$ also the vertical and horizontal stresses are given by,
$$σ_z=p\left[1-\left[\frac{1}{1+(\frac{a}{z})^2}\right]^\frac{3}{2}\right]\, and\, σ_r=\frac{p}{2}\left[1+2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+ \frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right],$$
therefore from the three equations, we get,
$$∆=\frac{p}{E}\left[(2-2μ^2)(a^2+z^2)^\frac{1}{2}-\frac{(1+μ) z^2)}{(a^2+z^2)^\frac{1}{2}}+(μ+2μ^2-1)z\right]$$

7. For a Poisson’s ratio of 0.5, the elastic strain ∆ is ___________
a) $$∆=\frac{3pa^2}{2E}$$
b) $$∆=\frac{pa^2}{2E(a^2+z^2)^\frac{1}{2}}$$
c) $$∆=\frac{3pa^2}{2E(a^2+z^2)^\frac{1}{2}}$$
d) $$∆=\frac{a^2}{2E(a^2+z^2)^\frac{1}{2}}$$

Explanation: The elastic strain is given by,
$$∆=\frac{p}{E}\left[(2-2μ^2)(a^2+z^2)^\frac{1}{2}-\frac{(1+μ) z^2)}{(a^2+z^2)^\frac{1}{2}}+(μ+2μ^2-1)z\right]$$substituting μ=0.5,
∴ $$∆=\frac{3pa^2}{2E(a^2+z^2)^\frac{1}{2}}.$$

8. For a Poisson’s ratio of 0.5, the elastic strain ∆ in terms of Boussinesq settlement factor is __________
a) $$∆=\frac{pa}{E} F_B$$
b) $$∆=\frac{3pa^2}{2E} F_B$$
c) $$∆=\frac{a^2}{2E} F_B$$
d) $$∆=\frac{pa^2}{2E} F_B$$

Explanation: When the Poisson’s ratio is 0.5, the elastic strain ∆ is given by,
$$∆=\frac{3pa^2}{2E(a^2+z^2)^\frac{1}{2}},$$ the Boussinesq settlement factor is $$F_B=\frac{3}{2} \frac{1}{(1+(\frac{z}{a})^2)^\frac{1}{2}},$$
∴ substituting in the equation,
$$∆=\frac{pa}{E} F_B.$$

9. The Boussinesq settlement factor is given by _______
a) $$F_B=\frac{3}{2} \frac{1}{1+(\frac{z}{a})^2}$$
b) $$F_B=\frac{3}{2} \frac{1}{(1+(\frac{z}{a})^2)^\frac{5}{2}}$$
c) $$F_B=\frac{3}{2} \frac{1}{(1+(\frac{z}{a})^2)^\frac{1}{2}}$$
d) $$F_B=\frac{1}{(1+(\frac{z}{a})^2)^\frac{1}{2}}$$

Explanation: The Boussinesq settlement factor is given by equation,
$$F_B=\frac{3}{2} \frac{1}{(1+(\frac{z}{a})^2)^\frac{1}{2}},$$
where, FB= Boussinesq settlement factor
z=depth of point

10. If the load is at the surface of subgrade, the Boussinesq settlement factor is given by ________
a) 1.5
b) 2
c) 2.5
d) 3

∴ $$F_B=\frac{3}{2}\frac{1}{(1+(\frac{z}{a})^2)^\frac{1}{2}} =\frac{3}{2}\frac{1}{(1+(\frac{0}{a})^2)^\frac{1}{2}}$$
∴ FB=1.5.

11. If the load is at the surface of subgrade, the elastic strain is ________
a) $$∆=\frac{pa}{E} 1.5$$
b) $$∆=\frac{3pa^2}{2E} 2.5$$
c) $$∆=\frac{a^2}{2E} 3$$
d) $$∆=\frac{pa^2}{2E} 3.5$$

Explanation: When the Poisson’s ratio is 0.5, the elastic strain ∆ is given by,
$$∆=\frac{pa}{E} F_B,$$ and at the surface, z=0,
∴ FB=1.5
Hence, $$∆=\frac{pa}{E} 1.5.$$

12. The Burmister’s influence coefficient is ________
a) $$C_B=\frac{σ_z}{p}$$
b) CBz-p
c) CBz*p
d) CBz+p

Explanation: The vertical stress at any depth z is,
σz=CB p. Therefore, the Burmister’s influence coefficient is given by,
$$C_B=\frac{σ_z}{p}.$$

13. If F is a dimensionless factor, the defection in rigid plate of radius a is__________
a) $$∆=\frac{pa}{E} 1.5F$$
b) $$∆=\frac{pa}{E} 1.18F$$
c) $$∆=\frac{a^2}{2E} 1.18F$$
d) $$∆=\frac{pa}{E} 2F$$

Explanation: The defections for a two layer system is obtained from,
$$∆=\frac{pa}{E} 1.5F,$$ for flexible pavement
$$∆=\frac{pa}{E} 1.18F,$$ for rigid pavement.

14. In Burmister analysis, the subgrade layers are non-homogeneous and anisotropic.
a) True
b) False

Explanation: In Burmister analysis, the subgrade layers are homogeneous and isotropic. The surface layer is free from shearing and normal stress outside the loaded area. Also, the subgrade layers are elastic.

15. In Burmister analysis, the subgrade layer is infinite only in vertical direction.
a) True
b) False

Explanation: In Burmister analysis, the subgrade layer is infinite both in vertical as well as horizontal directions. Both the layers are in continuous contact with each other. The surface layer is infinite in extent in lateral direction, but of finite depth.

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