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 1 of 2.
Jogendra said:
9 years ago
Explanation please.
Jehangir said:
9 years ago
Someone please explain it.
(1)
Disaku said:
9 years ago
2400*7.5*0.2*1=1400*Ast.
Ast = 2.57 c.m^2,
Area of each bar. ast=(3.14*1^2)/4 =0.78,
Diameter taken as 1c.m,
Spacing= (ast*100)/Ast = 30.3 nearly 30 C.M.
Ast = 2.57 c.m^2,
Area of each bar. ast=(3.14*1^2)/4 =0.78,
Diameter taken as 1c.m,
Spacing= (ast*100)/Ast = 30.3 nearly 30 C.M.
(3)
Chhaya said:
8 years ago
Thanks @Disaku.
(2)
Shahul said:
7 years ago
Thanks @Disaku.
Nasir said:
7 years ago
Can anyone explain the answer? please.
Harsh Shukla said:
7 years ago
Density = Weight/Volume.
The weight of slab = Density of concrete * Volume.
Density of concrete= 2400kg/m^3.
Volume = L*B*H = 1*7.5*0.2=1.5 m^3.
Weight of slab = 2400*1.5 = 3600kg.
This weight of slab should be equal to the force generated due to resisting stresses in the concrete slab.
Force generated = stresses * Area of steel (Ast).
Force generated = 1400*Ast = Weight of slab.
Ast = 3600/1400 =2.57 cm^2.
Considering 1cm diameter bars.
Area of 1 bar = (3.14/4)*1^2=0.785.
Spacing =( Area of 1 bar /Ast )*100 = 30.35cm ~ 30cm.
The weight of slab = Density of concrete * Volume.
Density of concrete= 2400kg/m^3.
Volume = L*B*H = 1*7.5*0.2=1.5 m^3.
Weight of slab = 2400*1.5 = 3600kg.
This weight of slab should be equal to the force generated due to resisting stresses in the concrete slab.
Force generated = stresses * Area of steel (Ast).
Force generated = 1400*Ast = Weight of slab.
Ast = 3600/1400 =2.57 cm^2.
Considering 1cm diameter bars.
Area of 1 bar = (3.14/4)*1^2=0.785.
Spacing =( Area of 1 bar /Ast )*100 = 30.35cm ~ 30cm.
(3)
Kcube said:
5 years ago
Here the coefficient of Friction between pavement and subgrade is not given or assumed to be 1 but usually, it is taken as 1.5, then answer becomes 20 cm.
Ast=b*h*W*f / 100*working stress (b is in meter, h is in cm, W is in Kg/m3, working stress is in Kg/cm2).
Number of Bars, N= Ast / A(of i bar),
Spacing (in cm) = 100/N.
Ast=b*h*W*f / 100*working stress (b is in meter, h is in cm, W is in Kg/m3, working stress is in Kg/cm2).
Number of Bars, N= Ast / A(of i bar),
Spacing (in cm) = 100/N.
(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).
(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).
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