Civil Engineering - Strength of Materials - Discussion

W=1000 kg convert this load into UDL of half span of the arch then UDL load,w=200 kg/m
The horizontal thrust formula for half UDL of the arch is =wl/16h.

So, H=wl/16h=200*10/16*1=1250 kg.

@Subhash.

By using this equation we get 125 kg.

Option B.

I got 2250kg of horizontal thrust by solving equilibrium method as we did for beams.

1st find the vertical forces at both ends. For that take moment about A ( left end). 1000 x 2.5 - By x 10 = 0.
By = 250 kg.

Now sum of all vertical forces = 1000 kg, we get Ay = 750 kg.
Now take section of the arch at crown where moment = 0 and consider left section.

750 x 5 - Ax * 1 - 1000 x 2.5 = 0. ( Ax is the horizontal force at hinge support A).

Ax = 1250 kg.

@All.

Here, it's not mentioned which side is the load is; Like they only told 2.5m from the crown.

Then, how to calculate further? Anyone explain me.

W = 1000 kg convert this load into UDL of half span of the arch then UDL load,
w = 200 kg/m.

The horizontal thrust formula for half UDL of the arch is = wl * l/16h.

So, H = w(l*l)/16h
= 200 * 10/16 * 1
= 1250 kg.