Question

A steel tie rod 50mm in diameter and 2.5m long is subjected to a pull of 100 KN. To what length the rod should be bored centrally so that the total extraction will increase by 15% under the same pull, the bore being 25mm diameter? For steel modulus of elasticity is 2 × 105 N/mm2.

Answer 1

Diameter of the steel tie rod = 50 mm = 0.05 m

Length of the steel rod, L = 2.5 m

Magnitude of the pull, P = 100 kN

Diameter of the bore = 25 mm = 0.025 m

Modulus of elasticity, E = 200 × 109 N/m2

Let length of the bore be ‘x’

Stress in the solid rod σ = P/A

= {(100 × 1000)/[(Π/4)(0.05)2]} = 50.92 × 106 N/m2

Elongation of the rod δL = σL/E

= (50.92 × 106 × 2.5) / (200 × 109)

= 0.000636m = 0.636mm

Elongation after the rod is bored = 1.15 × 0.636 = 0.731mm

Area of the reduction section = (Π/4) (0.052 – 0.0252) = 0.001472m2

Stress in the reduced section σb = (100 × 1000)/0.001472m2

= 67.93 × 106 N/m2

Elongation of the rod

= σ(2.5 – x)/E + σb.x/E = 0.731 × 10–3

= [50.92 × 106 (2.5 – x)]/(200 × 109) + (67.93 × 106.×)/(200 × 109 ) = 0.731 × 10–3

x = 1.12m

Hence length of the bore = 1.12m