Question

For the bar shown in Fig, calculate the reaction produced by the lower support on the bar. Take E = 200 GN/m². Find also the stresses in the bars.

Answer 1

Let R₁ = reaction at the upper support;

R₂ = reaction at the lower support when the bar touches it

If the bar MN finally rests on the lower support, we have

R₁ + R₂ = 55kN = 55000

N For bar LM, the total force = R₁ = 55000 – R₂ (tensile)

For bar MN, the total force =R₂ (compressive)

δL₁ =extension of LM = [(55000 – R₂) × 1.2]/[(110 × 10^–6) × 200 × 10⁹]

δL₂ = contraction of MN = [R₂ × 2.4]/[(220 × 10^–6) × 200 × 10⁹]

In order that N rests on the lower support, we have from compatibility equation

δL₁ – δL₂ = 1.2/1000 = 0.0012 m

Or, 

 [(55000 – R₂) × 1.2]/[(110 × 10^–6) × 200 × 10⁹]

– [R₂ × 2.4]/[(220 × 10^–6) × 200 × 10⁹] = 0.0012

on solving;

R₂ = 16500N or, 16.5 KN

R₁ = 55-16.5 = 38.5 KN

Stress in LM = R₁/A₁ = 38.5/110 × 10^–6 = 0.350 × 10⁶ kN/m² = 350 MN/m²

Stress in MN = R₂/A₂ = 16.5/220 × 10^–6 = 0.075 × 10⁶ kN/m² = 75 MN/m²