This set of Pavement Design Multiple Choice Questions & Answers (MCQs) focuses on “Wheel Load Stresses in Rigid Pavement”.
1. ______ indicated that the failure of most concrete slabs occurred at the corner of the slab.
a) H. M. Westergaard
b) A. T. Goldbeck
c) Chester McDowell
d) Norman McWell
View Answer
Explanation: A. T. Goldbeck was the engineer who figured out that most of the concrete slabs failed at the corners. He understood this and developed an equation to compute the stresses due to the wheel loads at the corner of the rigid pavement.
2. Which of the below is not a critical load position?
a) Interior
b) Corner
c) Edge
d) Center
View Answer
Explanation: When the rigid pavement is subjected to wheel loads and temperature changes, it experiences the stresses in different positions. There are three critical positions namely, interior, corner and edge.
3. The pressure deformation characteristics is a function of a parameter that is termed as ______
a) Radius of stiffness
b) Radius of relative stiffness
c) Relative stiffness radius
d) Relative radius of stiffness
View Answer
Explanation: The slab and subgrade are related to each other in terms of their stiffness. The pressure deformation characteristics of a rigid pavement is a function of the relative stiffness of the slab to that of the subgrade. This parameter is termed as the radius of relative stiffness.
4. Slab deflection is considered to be the direct measurement of the pressure of the subgrade.
a) True
b) False
View Answer
Explanation: The subgrade offers a certain degree of resistance to the deflection of the slab. The deformation of the subgrade is the same as the deflection of the slab. Hence, the slab deflection can be taken as the direct measurement of the magnitude of subgrade pressure.
5. Find the radius of relative stiffness for the following data:
Modulus of elasticity of cement concrete, E = 2×105 kg/cm2
Thickness of slab, h = 15 cm
Poisson’s ratio, μ = 0.15
Modulus of subgrade reaction, k = 5 kg/cm3
a) 55.82 cm
b) 58.25 cm
c) 58.52 cm
d) 52.85 cm
View Answer
Explanation: The equation to find the radius of relative stiffness is given by \(l=\left[\frac{Eh^3}{12k(1-μ^2)}\right]^{1∕4}\). All the terms and their values are given in the question. So, after substituting the same,
\(l=\left[\frac{2×10^5×15^3}{12×5×(1-0.15^2)}\right]^{1∕4}\)=58.25 cm
6. When the value of the radius of wheel load distribution is ______, it is equal to the equivalent radius of resisting section.
a) Greater than 1.724
b) Lesser than 1.724
c) Greater than 1.724h
d) Lesser than 1.724h
View Answer
Explanation: Whenever the load is acting on a small area of the pavement, it is necessary to determine the sectional area of the pavement that is helpful in resisting the same. The term equivalent radius of resisting section is used to represent the same. When the radius of wheel load distribution is greater than 1.724 times the thickness h, it is equal to the equivalent radius of resisting section.
7. Compute the equivalent radius of resisting section if the radius of wheel load distribution is 15 cm and the pavement thickness is 15 cm.
a) 16.04 cm
b) 14.06 cm
c) 14.60 cm
d) 16.40 cm
View Answer
Explanation: In order to find the equivalent radius of resisting section represented by b, the formula \(b=\sqrt{1.6a^2+h^2}-0.675h\) can be used. The term a represents the radius of wheel load distribution of 15 cm and h represents the pavement thickness of 15 cm. It can be used when a < 1.724h i.e. 15 < 1.724 X 15, which is true. Therefore, \(b=\sqrt{1.6×15^2+15^2}\)-0.675×15=14.06 cm
8. What is the critical stress at the interior of rigid pavement for the below data?
Wheel load, P = 5100 kg
Modulus of elasticity, E = 3×105 kg/cm2
Pavement thickness, h = 17 cm
Poisson’s ratio, μ = 0.15
Modulus of subgrade reaction, k = 7 kg/cm3
Radius of contact area, a = 15 cm
a) 22.80 kg/cm2
b) 20.85 kg/cm2
c) 23. 85 kg/cm2
d) 21.80 kg/cm2
View Answer
Explanation: The equation to find the critical stress at the interior is \(S_i=\frac{0.316P}{h}^2[4 logl/b+1.069]\). First, the radius of relative stiffness, l and equivalent radius of resisting section, b are to be found.
\(l=\left[\frac{Eh^3}{12k(1-μ^2)}\right]^{1∕4}=\left[\frac{3×10^5×17^3}{12×7×(1-0.15^2)}\right]^{1∕4}\)=65.09 cm
a < 1.724h i.e. 15 < 1.724 X 17, therefore,
\(b=\sqrt{1.6a^2+h^2}-0.675h =\sqrt{1.6×15^2+17^2}\)-0.675×17=14 cm
Now,
\(S_i=\frac{0.316P}{h}^2[4 logl/b+1.069]=\frac{0.316×5100}{17^2}[4 log65.09/14+1.069]\)=20.85 kg/cm2
9. What is the modified equation for maximum stress at the corner?
a) \(s_c=\frac{3P}{h^2} \left[1-(\frac{a\sqrt{2}}{l})\right]\)
b) \(s_c=\frac{3P}{h^2} \left[1-{(\frac{a\sqrt{2}}{l})}^{1.5}\right]\)
c) \(s_c=\frac{3P}{h^2} \left[1-{(\frac{a\sqrt{2}}{l})}^{1.2}\right]\)
d) \(s_c=\frac{3P}{h^2} \left[1-{(\frac{a\sqrt{2}}{l})}^{0.2}\right]\)
View Answer
Explanation: \(s_c=\frac{3P}{h^2} \left[1-{(\frac{a\sqrt{2}}{l})}^{1.2}\right]\) is the equation that has been modified by Kelly. The original equation given by Westergaard did not have any power to the term \((\frac{a\sqrt{2}}{l})\). IRC has given two modifications for the equations by Westergaard, one for corner stress and one for edge stress.
10. Find the corner stress for the data given below.
Wheel load, P = 5500 kg
Pavement thickness, h = 20 cm
Radius of relative stiffness, l = 65 kg/cm3
Radius of contact area, a = 15 cm
a) 29.97 kg/cm2
b) 29.79 kg/cm2
c) 27.97 kg/cm2
d) 27.79 kg/cm2
View Answer
Explanation: \(s_c=\frac{3P}{h^2} \left[1-(\frac{a\sqrt{2}}{l})\right]\) is the equation generally used for computing corner stress. All the values to be substituted in the equation are given in the question. Therefore,
\(s_c=\frac{3×5500}{20^2} \left[1-(\frac{15\sqrt{2}}{65})\right]\) =27.79 kg/cm2
11. What is the critical stress at the edge of rigid pavement for the below data?
Wheel load, P = 5100 kg
Modulus of elasticity, E = 2.1×105 kg/cm2
Pavement thickness, h = 16 cm
Poisson’s ratio, μ = 0.15
Modulus of subgrade reaction, k = 7 kg/cm3
Radius of contact area, a = 15 cm
a) 31.82 kg/cm2
b) 38.12 kg/cm2
c) 31.28 kg/cm2
d) 38.21 kg/cm2
View Answer
Explanation: The equation for edge stress is \(S_e=\frac{0.572P}{h^2}[4 logl/b+0.359]\). First, the radius of relative stiffness, l and equivalent radius of resisting section, b have to be found out.
\(l=\left[\frac{Eh^3}{12k(1-μ^2)}\right]^{1∕4}=\left[\frac{2.1×10^5×16^3}{12×7×(1-0.15^2)}\right]^{1∕4}\)=56.89 cm
a < 1.724h i.e. 15 < 1.724 X 16, therefore,
\(b=\sqrt{1.6a^2+h^2}-0.675h =\sqrt{1.6×15^2+16^2}\)-0.675×16=14.02 cm
Now,
\(S_e=\frac{0.572P}{h}^2[4 logl/b+0.359]=\frac{0.572×5100}{16^2}[4 log56.89/14.02+0.359]\)=31.82kg/cm2
12. In the design chart for finding corner stress by IRC, what parameter below is not constant?
a) Poisson’s ratio
b) Modulus of subgrade reaction
c) Modulus of elasticity
d) Wheel load
View Answer
Explanation: IRC has designed two design charts, one for finding the edge stress and one for the corner stress. While developing these, the wheel load was fixed as 5100 kg, the modulus of elasticity as 3×105 kg/cm2, Poisson’s ratio as 0.15 and the radius of contact area as 15 cm. If the slab thickness and modulus of subgrade reaction are known, the corresponding stress can be found out.
13. At what distance from the apex of the slab corner does the maximum stress occur in a rigid pavement?
a) \(X=2.58\sqrt{a}\)
b) \(X=\sqrt{al}\)
c) \(X=2.58\sqrt{al}\)
d) \(X=2.58\sqrt{l}\)
View Answer
Explanation: \(X=2.58\sqrt{al}\) is the formula used to compute the distance from the apex of the slab corner to the section of maximum stress along the corner bisector. In the equation, a is the radius of contact area and l is the radius of relative stiffness.
14. Find the distance at which the maximum corner stress would occur if the following data is available.
Modulus of elasticity of cement concrete, E = 2×105 kg/cm2
Thickness of slab, h = 18 cm
Poisson’s ratio, μ = 0.15
Modulus of subgrade reaction, k = 6 kg/cm3
Radius of contact area, a = 15 cm
a) 80.89 cm
b) 79.81 cm
c) 76.87 cm
d) 81.78 cm
View Answer
Explanation: First, the parameter \(l=\left[\frac{Eh^3}{12k(1-μ^2)}\right]^{1∕4}\) is found out.
\(l=\left[\frac{2×10^5×18^3}{12×6×(1-0.15^2)}\right]^{1∕4} =63.80 cm\)
Now, the distance X can be found as
\(X=2.58\sqrt{al}=2.58\sqrt{15×63.80}\)=79.81 cm
15. The edge stress can be found out for a slab thickness of range 10-30 cm using design charts.
a) True
b) False
View Answer
Explanation: The edge stress can be found out using design charts by IRC if the slab thickness is known. But the design charts have been developed for a specific range of slab thickness from 14 to 25 cm only. Therefore, the stresses for any thickness beyond this range has to be computed using equations.
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