A beam weighing 12 tons lies on the brickwork. The width of the supporting area of the beam is 20 cm.

A beam weighing 12 tons lies on the brickwork. The width of the supporting area of the beam is 20 cm. What can be the smallest length of the supporting part of the beam if the permissible pressure is 1.2 MPa?

Given:

m = 12 tons = 12000 kilograms – beam weight;

b = 20 centimeters = 0.2 meters – the length of the supporting part of the beam;

g = 10 N / kg – acceleration of gravity;

P = 1.2 MPa = 1200000 Pascals – allowable pressure.

It is required to determine the smallest beam length a (meter).

Let’s find the force with which the beam is to press on the surface:

F = m * g = 12000 * 10 = 120000 Newton.

Then the permissible surface area of the beam will be equal to:

S = P / F = 1200000/120000 = 10m ^ 2.

Knowing the area, we calculate the length:

a = S / b = 10 / 0.2 = 50 meters.

Answer: the length of the beam is 50 meters.