# Pavement Design Questions and Answers – Highway Pavements Design – Joint Spacing

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This set of Pavement Design Multiple Choice Questions & Answers (MCQs) focuses on “Highway Pavements Design – Joint Spacing”.

1. What is the spacing allotted for the expansion joints in summer in the rigid pavement?
a) 50 – 60 m
b) 50 – 80 m
c) 90 – 120 m
d) 90 – 100 m

Explanation: As recommended by the IRC codes, the spacing between the adjacent expansion joints during the summertime is 90 to 120 m. During winter, the spacing is 50 to 60 m. these values are valid for a smooth interface.

2. What is the expansion joint spacing design based on?
a) Length of pavement
b) Length of joint
c) Width of pavement
d) Width of joint

Explanation: The design of the expansion joint is based on two factors. One is the maximum temperature variation expected and the other is the width of the joint. Width of the expansion joint depends on the length of the slab.

3. What is the equation used to find the spacing of the expansion joint?
a) $$L_e=\frac{δ}{100C(T_2-T_1)}$$
b) $$L_e=\frac{δ}{10C(T_2-T_1)}$$
c) $$L_e=\frac{2δ}{100C(T_2-T_1)}$$
d) $$L_e=\frac{2δ}{10C(T_2-T_1)}$$

Explanation: $$L_e=\frac{δ}{100C(T_2-T_1)}$$ is the equation used to find the spacing of the expansion joints. In the equation, the term δ represents half the width of the joint, C is the coefficient of thermal expansion of concrete and the T2-T1 indicates the rise in temperature during the expansion.
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4. The maximum spacing between the contraction joints in a reinforced cement concrete slab is ______
a) 15 cm
b) 15 m
c) 14 cm
d) 14 m

Explanation: As per IRC recommendations, the maximum spacing between the contraction joints in a reinforced cement concrete pavement of thickness 20 cm is 14 m. in case of plain cement concrete slabs, the spacing is less than that of in reinforced one. The maximum spacing is taken to be 4.5 m.

5. A thinner slab requires a ______ joint spacing.
a) Longer
b) Shorter
c) Deeper
d) Narrower

Explanation: The thickness of the slab is an important factor that affects the spacing of the joint. A thinner slab has high warping stress or curling stress. Hence to resist the same, the spacing must be shorter, otherwise, the slab would warp more.

6. What is the equation to compute the contraction joint spacing for a plain cement concrete pavement?
a) $$L_c=\frac{2×10^2 SA}{fWbh}$$
b) $$L_c=\frac{2×10^4 S}{fW}$$
c) $$L_c=\frac{2×10^2 S}{fW}$$
d) $$L_c=\frac{2×10^4 SA}{fWbh}$$

Explanation: For calculating the spacing between the contraction joints in plain cement concrete pavement, the equation $$L_c=\frac{2×10^4 S}{fW}$$ is used. In the case of a reinforced cement concrete pavement, the equation to be used is $$L_c=\frac{2×10^2 SA}{fWbh}$$.

7. At what length of the slab does the maximum stress due to friction occur in the case of contraction joints?
a) One fourth
b) Half
c) Three fourth
d) Full

Explanation: The slab contracts when there is a fall in temperature and it also undergoes shrinkage during curing periods. To resist the same, frictional stresses are developed. The maximum stress is developed at half the length of the slab.

8. Find the spacing between expansion joints for the following data given below.
Width of expansion joint gap = 2.5 cm
Temperature during laying the pavement = 14°C
Maximum temperature during summer = 50°C
Coefficient of thermal expansion = 10×10-6 /°C
a) 35.34 m
b) 35.72 m
c) 34.34 m
d) 34.72 m

Explanation: $$L_e=\frac{δ}{100C(T_2-T_1)}$$ is the equation used to find the spacing of the expansion joints. As per the question,
δ =2.5/2=1.25 cm C=10×10-6/°C T1=14°C T2=50°C
Therefore, $$L_e=\frac{δ}{100C(T_2-T_1)} =\frac{1.25}{100×10×10^{-6}×(50-14)}$$=34.72 m

9. Determine the spacing between contraction joints of a 4 m wide slab having 15 cm thickness, coefficient of friction is 1.3 and allowable stress in concrete is 0.8 kg/cm2.
a) 4.13 m
b) 5.5 m
c) 5.13 m
d) 4.5 m

Explanation: The equation to find the spacing between contraction joints for plain cement concrete slab is $$L_c=\frac{2×10^4 S}{fW}$$. The unit weight of concrete, W is taken as 2400 kg/m3. Therefore, spacing is found as:
$$L_c=\frac{2×10^4×0.8}{1.3×2400}$$=5.13 m
As per IRC 58:2002, the maximum spacing for an unreinforced slab of thickness 15 cm is 4.5 m, therefore, the spacing has to be restricted to 4.5 m.

10. Find the spacing between expansion joints for the following data given below.
Width of expansion joint gap = 2.5 cm
Temperature during laying the pavement = 16°C
Maximum temperature during summer = 55°C
Coefficient of thermal expansion = 10×10-6 /°C
a) 32.05 m
b) 33.05 m
c) 34.05 m
d) 35.05 m

Explanation: The equation used to find the spacing of the expansion joints is $$L_e=\frac{δ}{100C(T_2-T_1)}$$ . After substituting the values from the question,
δ=2.5/2=1.25 cm C=10×10-6/°C T1=16°C T2=55°C
Therefore, $$L_e=\frac{δ}{100C(T_2-T_1)} =\frac{1.25}{100×10×10^{-6}×(55-16)}$$=32.05 m

11. Find the spacing for contraction joint to be included in an RCC slab in the rigid pavement having the following details.
Permissible tensile stress in the reinforcement, SS = 1250 kg/cm2
Slab width, b = 3.5 m
Thickness of slab, h = 22 cm
Coefficient of friction, f = 0.15
Unit weight of concrete, W = 2400 kg/m3
The steel bars of 1 cm diameter are placed at 0.25 m spacing.
a) 9 m
b) 10.92 m
c) 10 m
d) 9.92 m

Explanation: In the case of an RCC slab, the equation to find spacing is $$L_c=\frac{2×10^2 SA}{fWbh}$$. The term A represents the area of reinforcement, i.e. the steel bars and it needs to be computed.
The width of the slab is 3.5 m and the spacing between bars is 0.25 m. Therefore, the no. of bars would be obtained as 3.5/0.25=14 nos. Now the area of reinforcement can be obtained as
A=no.of bars $$×\frac{π}{4}×d^2=14×\frac{π}{4}×1^2=10.99≈11$$ m2
The spacing can now be obtained as
$$L_c=\frac{2×10^2×1250×11}{1.5×2400×3.5×22}$$=9.92 m

12. For finding the spacing between contraction joints in an RCC slab, it is assumed that the reinforcement takes up all the compressive force in the slab.
a) True
b) False

Explanation: The reinforcement is provided in the form of steel bars. The steel bars are good in taking up the tensile stresses and not compressive stresses. Hence, the assumption taken is that the reinforcement takes up all the tensile force in the slab.

13. The maximum spacing of contraction joint in a 30 cm thick PCC slab is taken as 5 m.
a) True
b) False

Explanation: The maximum spacing of contraction joints in the case of PCC slab is provided based on the thickness of the slab. Thee Table 7 in IRC 58:2002 has given the details on the same. According to that, for a slab of thickness 30 cm, the maximum spacing is 5 m.

14. Find the spacing for contraction joint to be included in an RCC slab in the rigid pavement having the following details.
Permissible tensile stress in the reinforcement, SS = 1200 kg/cm2
Slab width, b = 4 m
Thickness of slab, h = 25 cm
Area of reinforcement, A = 9.35 m2
Coefficient of friction, f = 0.15
Unit weight of concrete, W = 2400 kg/m3
a) 2.27 m
b) 2.72 m
c) 7.72 m
d) 7.27 m

Explanation: The equation to find spacing of contraction joint in an RCC slab is $$L_c=\frac{2×10^2 SA}{fWbh}$$. The explanation and the values of the terms are given in the question.
The spacing can now be obtained as
$$L_c=\frac{2×10^2×1400×9.35}{1.5×2400×4×25}$$=7.27 m

15. The joint filler in the expansion joint is assumed to be compressed by how much percentage?
a) 80 %
b) 75 %
c) 50 %
d) 35 %

Explanation: The filler in the expansion joint is considered to be compressed by 50 % of its original thickness. It is due to this reason that the width or gap of the joint is taken as half in case of computing the spacing of an expansion joint.

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