This set of Pavement Design Multiple Choice Questions & Answers (MCQs) focuses on “Design of Joints”.
1. What type of reinforcement is generally provided in the expansion joints?
a) Tie bar
b) Dowel bar
c) Tri bar
d) Dowal bar
View Answer
Explanation: In order to hold the two adjacent slabs together, the reinforcement in the form of dowel bars is provided in the expansion joints. It also helps in relieving stress and keeping slab levels. Tie bars are provided in the longitudinal joints.
2. Tie bars are ______ that are used to ______
a) Round bars, transfer loads
b) Round bars, maintain aggregate interlock
c) Deformed rebars, transfer loads
d) Deformed rebars, maintain aggregate interlock
View Answer
Explanation: Tie bars are placed in between the longitudinal joints and their main purpose is to maintain the aggregate interlock. They are deformed rebars. Dowel bars, on the other hand, are round bars used primarily for the purpose of load transfer.
3. The dowel bar is embedded between the two slabs.
a) True
b) False
View Answer
Explanation: The dowel bar is a short steel bar. Half-length of it is bonded in one concrete slab while the other half is embedded in the other slab. This is done so to allow easy movement in between the slabs for the expansion and contraction.
4. As per IRC recommendations according to whose analysis must the dowel bars be designed?
a) Westergaard
b) Burmister
c) Bradbury
d) Teller
View Answer
Explanation: As per the recommendations for designing a dowel bar system, Bradbury’s analysis is to be used. The analysis includes the determination of load transfer capacity of a single dowel bar in shear, bending and bearing in concrete.
5. What is the equation used to compute the load transfer capacity of a dowel bar in shear?
a) \(P^{‘}=\frac{2d^3 F_f}{L_d+8.8δ^{‘}}\)
b) \(P^{‘}=\frac{F_b L_d^2 d}{12.5(L_d+1.5δ^{‘})}\)
c) \(P^{‘}=0.785d^2 F_s\)
d) \(P^{‘}=0.578d^2 F_s\)
View Answer
Explanation: This equation is as per the old IRC method i.e. the one prescribed in IRC 58:1988. The equation \(P^{‘}=0.785d^2 F_s\) is used to compute the load transfer in shear, \(P^{‘}=\frac{2d^3 F_f}{L_d+8.8δ^{‘}}\) in bending and \(P^{‘}=\frac{F_b L_d^2 d}{12.5(L_d+1.5δ^{‘})}\) in bearing.
6. The thickness of a PCC pavement is 20 cm and the width of the lane is 3.5 m. Find the area of a tie bar system along the longitudinal joints if the unit weight of concrete is 2400 kg/3, coefficient of friction is 1.2 and allowable tensile stress in deformed bars is 2000 kg/cm2.
a) 0.11 cm2
b) 1.01 cm2
c) 0.01 cm2
d) 1.10 cm2
View Answer
Explanation: The equation to find the area of cross-section is as follows:
\(A_s=\frac{fWbh×10^{-2}}{s_s}\)
Where f is the coefficient of friction = 1.2
W is the unit weight of concrete = 2400 kg/3
b is the width of the slab = 3.5 m
h is the thickness of the slab = 20 cm
SS is the allowable tensile stress in deformed bars = 2000 kg/cm2
Substituting these in the equation,
\(A_s=\frac{1.2×2400×3.5×20×10^{-2}}{2000}\)=1.01 cm2
7. Design the size of a dowel bar at the expansion joint for the following data:
Thickness of slab = 25 cm
Width of joint = 2.5 cm
Allowable flexural stress in bars = 1400 kg/cm2
Allowable shear stress in bars = 1000 kg/cm2
Allowable bearing stress in concrete = 100 kg/cm2
a) 50 cm
b) 51.7 cm
c) 41.7 cm
d) 40 cm
View Answer
Explanation: To find the length of the dowel bar the sum Ld + δ is used, where Ld is the embedded length and δ is the width of the joint. The equation to find the embedded length is given by:
\(L_d=5d\left(\frac{F_f}{F_b} ×\frac{L_d+1.5δ}{L_d+8.8δ}\right)^{1/2}\)
Where, Ff is the allowable flexural stress = 1400 kg/cm2
Fb is the allowable bearing stress in concrete = 100 kg/cm2
d is the diameter of the dowel bar and it can be assumed as 2.5 cm
δ is the width of the joint = 2.5 cm
After substituting all the values in the equation,
\(L_d=5×2.5\left(\frac{1400}{100}×\frac{L_d+1.5×2.5}{L_d+8.8×2.5}\right)^{1/2}\)
\(L_d=12.5\left(14×\frac{L_d+3.75}{L_d+22}\right)^{1/2}\)
The next step is to find the value of Lsub>d using trial and error method. This is in accordance with the method prescribed by IRC 58:1988.
Assuming Ld = 45 cm for the first trial,
\(L_d=12.5\left(14×\frac{45+3.75}{45+22}\right)^{1/2}\)=39.89 cm
39.89 is less than 45 cm, now assume 39.5 cm
\(L_d=12.5\left(14×\frac{39.5+3.75}{39.5+22}\right)^\frac{1}{2}\)=39.22 cm
39.22 cm is less than 39.5 cm, now assume 39.2 cm,
\(L_d=12.5\left(14×\frac{39.2+3.75}{39.2+22}\right)^{1/2}=39.18 ≈39.2\) cm
Therefore, the Ld is obtained as 39.2 cm.
The length of dowel bar would be = Ld + δ = 39.2 + 2.5 = 41.7 cm
8. The load capacity of the dowel system is assumed to be ______ of the design ______ load.
a) 40 %, bar
b) 40 %, wheel
c) 60 %, bar
d) 60 %, wheel
View Answer
Explanation: The load capacity of the dowel bar system is taken as 40% of the design wheel load. this assumption has been laid down in the IRC 58. The required load capacity factor for dowel bar system is obtained by dividing the capacity of the group by the capacity of one bar.
9. The thickness of a PCC pavement is 24 cm and the width of the lane is 3.5 m. Find the length of a tie bar system along the longitudinal joints if the allowable working stress in concrete is 1250 kg/cm2, unit weight of concrete is 2400 kg/m3, coefficient of friction is 1.5, allowable bond stress in deformed bars is 24.6 kg/cm2 and allowable tensile stress in deformed bars is 2000 kg/cm2.
a) 50 cm
b) 48 cm
c) 55 cm
d) 45 cm
View Answer
Explanation: There is a set procedure given by IRC to find the length of a tie bar system and it is carried out below.
Step 1: Cross-sectional area of the tie bar
\(A_s=\frac{fWbh×10^{-2}}{s_s}\)
Where As is the cross-sectional area
f is the coefficient of friction = 1.5
W is the unit weight of concrete = 2400 kg/m3
b is the width of the slab = 3.5 m
h is the thickness of the slab = 24 cm
Ss is the allowable tensile stress in deformed bars = 2000 kg/cm2
Substituting these in the equation,
\(A_s=\frac{1.5×2400×3.5×24×10^{-2}}{2000}\)=1.51 cm2
Step 2: Length of the tie bar
\(L_t=\frac{S_s d}{2S_b}\)
Where Lt is the length of the tie bar
Ss is the allowable tensile stress in deformed bars = 2000 kg/cm2
d is the diameter of the tie bar and it is assumed as 1.2 cm
Sb is the allowable bond stress in deformed bars = 24.6 kg/cm2
After substituting these in the equation,
\(L_t=\frac{2000×1.2}{2×24.6}=48.78≈50\) cm
10. Dowel bars need not be provided for slabs having a thickness less than ______
a) 15 cm
b) 20 cm
c) 10 cm
d) 25 cm
View Answer
Explanation: According to the specifications given in IRC 58:2002, the dowel bars are not satisfactory for slabs having a small thickness. For slabs having a thickness less than 15 cm, there is no need to provide dowel bars.
11. What is the maximum spacing for a plain 12 mm diameter tie bar to be used in a 25 cm thick slab?
a) 45 cm
b) 50 cm
c) 40 cm
d) 55 cm
View Answer
Explanation: There is a table giving the recommended values for the spacing and length of tie bars. It has been mentioned in Table 9 in IRC 58:2002. According to the table, for a plain tie bar of diameter 12 mm, the maximum spacing allowed for a 25 cm thick slab is 45 cm.
12. The thickness of a PCC pavement is 26 cm and the width of the lane is 3.5 m. Find the length of a tie bar system along the longitudinal joints if the allowable working stress in concrete is 1250 kg/cm2, unit weight of concrete is 2400 kg/m3, coefficient of friction is 1.4, allowable bond stress in deformed bars is 24.6 kg/cm2 and allowable tensile stress in deformed bars is 2000 kg/cm2.
a) 75 cm
b) 70 cm
c) 85 cm
d) 80 cm
View Answer
Explanation: The procedure given by IRC to find the spacing of a tie bar system is carried out as below.
Step 1: Cross-sectional area of the tie bar
\(A_s=\frac{fWbh×10^{-2}}{s_s}\)
Where As is the cross-sectional area
f is the coefficient of friction = 1.4
W is the unit weight of concrete = 2400 kg/m3
b is the width of the slab = 3.5 m
h is the thickness of the slab = 26 cm
Ssis the allowable tensile stress in deformed bars = 2000 kg/cm2
Substituting these in the equation,
\(A_s=\frac{1.4×2400×3.5×26×10^{-2}}{2000}\)=1.53 cm2
Step 2: Length of the tie bar
\(L_t=\frac{S_s d}{2S_b}\)
Where Lt is the length of the tie bar
Ss is the allowable tensile stress in deformed bars = 2000 kg/cm2
d is the diameter of the tie bar and it is assumed as 1.2 cm
Sb is the allowable bond stress in deformed bars = 24.6 kg/cm2
After substituting these in the equation,
\(L_t=\frac{2000×1.2}{2×24.6}=48.78≈50\) cm
Step 3: Spacing of the tie bar
No. of bars = \(\frac{Total \,area}{Area \,of \,one \,bar}=\frac{1.53}{\frac{π}{4}×1.2^2}\)=1.35
Spacing = \(\frac{1 \,m \,slab \,length}{No. \, of \,bars}=\frac{100 cm}{1.35}=74.07≈70\) cm
13. The joint width is taken as 2.5 cm when dowel bar is used.
a) True
b) False
View Answer
Explanation: The spacing of the contraction or expansion joint is taken as 2 cm for the stress computation in dowel bars. This is done so as to account for the grinding of concrete that is likely under the dowel which would lead to the consequent loss of support.
14. Design the spacing for a dowel bar system for the following data:
Design wheel load = 8000 kg
Percentage of load transfer = 40 %
Slab thickness = 30 cm
Joint width = 2 cm
Radius of relative stiffness = 105.45 cm
Characteristic compressive strength = 400 kg/cm2
a) 25 cm
b) 20 cm
c) 35 cm
d) 30 cm
View Answer
Explanation: The detailed procedure to find the spacing of the dowel bar system is explained in Appendix-3 of IRC 58:2002. The first step is to find the bearing stress in concrete as below,
\(F_b=\frac{(10.16-b) f_{ck}}{9.525}\)
The diameter of dowel bar, b is assumed to be 3.2 cm, so,
\(F_b=\frac{(10.16-3.2)×400}{9.525}\)=292 kg/cm2
The spacing between dowel bars is assumed a 30 cm and that the first dowel bar is placed at 15 cm from the edge of the pavement. The length of the dowel bar is assumed to be 50 mm. These are taken as per IRC 58:2002.
Only the number of dowel bars taking part in the load transfer is to be considered. Dowel bars up to a distance equal to the radius of subgrade reaction from the point of application of load are considered to be effective in the load transfer process.
Therefore, the number of dowel bars = 1+\(\frac{l}{spacing}=1+\frac{105.45}{30}=4.52≈5\) bars
Next, we have to assume that the load transfer by the first dowel bar is Pt and the load on dowel bar system is at a distance of 1 from the first dowel bar’s load to be zero. This is expressed as below:
\(\left(1+\frac{l-s}{l}+\frac{l-2s}{l}+⋯\right)P_t\)
i.e.
\(\left(1+\frac{105.45-30}{105.45}+\frac{105.45-60}{105.45}\frac{105.45-60}{105.45}\right)P_t = 2.293 P_t\)
The load carried by the outer dowel bar Pt, must be equated to 40% of design wheel load, i.e.
2.293 Pt=0.40×8000
Pt=1395.5 ≈1396 kg
Next step is to check the above design for bearing stress.
Bearing stress=\((P_t k)\frac{2+βz}{4β^3 EI}\)
Where k is the modulus of dowel/concrete interaction = 41500 kg/cm2/ cm
E is the modulus of elasticity and it is assumed as 2×106 kg/cm2
I is the moment of inertia of dowel bar and can be found out using the equation,
\(I=\frac{πb^4}{64}=\frac{π×3.2^4}{64}=5.147 cm^4\)
β is the relative stiffness of bar embedded in concrete which can be found as below:
\(β=\left[\frac{kb}{4EI}\right]^{1/4}=\left[\frac{41500×3.2}{4×2×10^6×5.147}\right]^{1/4}\)=0.238
Z is the joint width = 2 cm
Therefore, the bearing stress is obtained as
Bearing stress=1396×41500×\(\frac{2+0.238×2}{4×0.238^3×2×10^6×5.147}=258.41≈258\) kg/cm2
The bearing stress in dowel (258 kg/cm2) is found to be less than the bearing stress in concrete (292 kg/cm2), therefore the design is correct. The assumed spacing and diameter of the dowel bar are safe.
15. The maximum diameter of tie bars may be limited to 20 mm to ______
a) Permit warping
b) Resist warping
c) Permit bending
d) Resist bending
View Answer
Explanation: There are some specifications that are to be considered while designing the tie bars. The first is that the maximum diameter of the tie bar is limited to 20 mm in order to permit warping at the joints and reduce the stresses. The spacing must not be more than 75 cm to avoid the concentration of tensile stresses.
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