Civil Engineering - Highway Engineering - Discussion

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.

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.

Thanks @Disaku.

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.

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).

Someone please explain it.

As per my knowledge, I think the right answer is 40cm.

Explanation please.

Thanks @Disaku.

Can anyone explain the answer? please.