Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 5 (Q.No. 33)
33.
If the width of a pavement slab is 7.5 m, thickness 20 cm and working stress 1400 kg/cm2, spacing of 10 mm tie bars for the longitudinal joint, is
Discussion:
12 comments Page 2 of 2.
Saqib said:
4 years ago
The answer would be 40mm.
Explanation :
Consider the slab of width B with tie bar placed at Centre. The left half of the tie bar will resist the frictional force acting on the left half of the slab (B/2) and the same will happen to the right side.
Taking left half into the picture:
Frictional force = coeff of fric X weight of left half of slab.
= coeff of fric X (B/2)X h X 1m X Unit weight of CC.
= 1.5 x 3.75 x 0.2 x 1 x 2400 = 2700kg.
Resisting force = 1400xArea of steel.
Therefore Area of steel per 1m length = 2700/1400 = 1.92 cm^2.
Spacing = 100 * 3.14 * 1^2/(4x1.92) = 40.88 cm (taking 10mm bar as 1cm in the formula).
Explanation :
Consider the slab of width B with tie bar placed at Centre. The left half of the tie bar will resist the frictional force acting on the left half of the slab (B/2) and the same will happen to the right side.
Taking left half into the picture:
Frictional force = coeff of fric X weight of left half of slab.
= coeff of fric X (B/2)X h X 1m X Unit weight of CC.
= 1.5 x 3.75 x 0.2 x 1 x 2400 = 2700kg.
Resisting force = 1400xArea of steel.
Therefore Area of steel per 1m length = 2700/1400 = 1.92 cm^2.
Spacing = 100 * 3.14 * 1^2/(4x1.92) = 40.88 cm (taking 10mm bar as 1cm in the formula).
Shoaib khan said:
7 months ago
Can we solve it by 100WL/(At) t?
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