Question

a marble column of cross-sectional area 1.2 m^2 supports a mass of 25,000 kg. (a) what is the stress within the column? (b) what is the strain. (c) by how much is the column shortened if it is 9.6 m high

Answer 1

Stress = F/A
Stress = 2.5x 10^4 x 9.8/1.2
Stress = 20.4 x 10^4 N/m^2

Strain = Stress/Young's modulus
Strain = 20.4 x 10^4/5 x 10^10
Strain = 4.08 x 10^-6

Strain = Change in length/Original length
Change in length = Strain x original length
Change in length = 4.08 x 10^-6 x 9.6
Change in length = 3.91 x 10^-5m

Answer 2

a) Since: stress is force/area

force is 25,000 kg * 9.8 N/kg = 245000 N.

So stress is 204166.66 N/m^2.

Answer 3

Taking g=10,

a) Stress = (25000*10)/1.2 = 208333.33 N/(m^2)
b) Young's Modulus of Marble = 50 * 10^9

Strain = 4.166 * 10^(-6)
c) change in length = 9.6 * 4.166 * 10^(-6) = 4 * 10^ (-5) m

Answer 4

(a) stress = force/area = 25000*g/1.2 = 208333.33 Kg/ms^2

(b) & (c) => for strain calculations , value of young moudulus of elasticity of marble is required..

Answer 5