Question

A steel bar is 3 m in length and is subjected to an axial pull of 1000 kN. The bar
is 40 mm in diameter for a length of 1.5 m, 30 mm in diameter for a length of 0.8 m
and 20 mm in diameter for the remaining length. Find the total extension of
the bar. Take modulus of Elasticity as 200 GPa.​

Answer 1

Answer:

Give:

b = 20 mm

h = 20 mm

L = 1 m = 1 000 mm

F = 20 kN = 40 000 N

E = 200 GPa = 200 000 MPa

Req.:

DL = ?

Solution:

σ = F / A

A = b x h = 20 x 20 = 400 mm2

σ = F/A = 40 000/400 = 100 MPa

E = σ / e

e = DL / L

E = σ /(DL / L) = σ L / DL

200 000 = 100 x 1 000 / DL

DL = 100 x 1 000 / 200 000 = 0.5 mm (Ans.)

DL = 0.5 mm