Question

a hollow steel tube 3.5m long has external diameter 120mm and is subjected to a tensile load of 400kN and extension measured is 2mm. Determine the internal diameter. Take E=200GPa
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Answer 1

Answer:

= 2 m = 2 000 mm

D = 50 mm

d = 30 mm

W = 25 kN = 25 000 N = F

E = 100 GPa = 100 000 MPa

Req:

σ = ?

D L = ?

Solution:

A = (p/4) D2 – (p/4) d2 = (p/4) (D2 –d2)

= (p/4) (502 –302) = 1257 mm2

σ = F/A = 25 000 / 1257 = 19.9 MPa (Ans.)

E = σ / e = σ / (D L / L) = σ L /

20 000 = 19.9 x 2 000 / D L

D L = 19.9 x 2 000 / 100 000 = 0.398 mm

= 0.4 mm (Ans.)

σ = 19.9 MPa

D L = 0.4 mm