Question

A structural steel rod has a radius of 10mm and a length of 1m. A 100k N force streaches it along its length. Calculate:(a) the stress(b) elongation(c) percentage.Given that the young's modulus of elasticity of the structural steel is 2.0*10(power)11 Nm(power)-2.chapter- SOLIDS

Answer 1

stress = force / area
Area on which force acted = (pi * d^2) / 4
where , d = diameter of rod

(a) stress = (100 x 1000) / (( pi * 10^2 )/4)
= 1273.23 N/mm^2

E = stress / strain
strain. = change in length / original length

therefore, (b) change in length (ie., Elongation)
= stress x original length / E
= (1273.23 x 1000) / ( 2 x 10^5)
= 6.366 mm

(c) percentage of elongation
= (6.366 / 1000) x 100
= 0.63 %

Answer 2

given a steel rod:    Radius = r = 10 mm = 0.010 m.    Length L = 1 m.
                               Tensile Force = F = 100 kN.
                                Young's Modulus = Y = 2.0 &* 10¹¹ Nm⁻²

a) Stress = Force/cross sectional area 
             σ = F / A
                = 100,000 /(π 0.010²)  Pa 
                = 318.309 MPa

b)  Y = (F/A) / (ΔL/L)
     Elongation ΔL = L F / (A Y)
                             = L σ / Y
                             = 1 * 318.309 * 10⁶ / 2.0 * 10¹¹   m
                             = 1.591 mm
     
c)  Percentage elongation :  ΔL/L * 100 = 0.1591 %