Question

A steel bar is subjected to loads as shown in fig. Determine the change in length of the bar ABCD of 18 cm diameter. E = 180 kN/mm².

Answer 1

Since d = 180mm 

E = 180 × 10³ N/mm²

L_AB = 300mm 

L_BC = 310mm 

L_CD = 310mm

From the fig 

Load on portion AB = P_AB = 50 × 10³ N

Load on portion BC = P_BC = 20 × 10³ N

Load on portion CD = P_CD = 60 × 10³ N

Area of portion AB = Area of portion BC = Area of portion CD = A = Πd²/4

= Π(180)²/4 = 25446.9 mm

Using the relation δl = Pl/AE

δL_ab = (50 × 10³ × 300)/( 25446.9 × 180 × 10³) = 0.0033mm ---------------- Compression

δL_bc = (20 × 10³ × 310)/( 25446.9 × 180 × 10³) = 0.0012mm ---------------- Compression 

δL_cd = (60 × 10³ × 310)/( 25446.9 × 180 × 10³) = 0.0041mm ---------------- Compression 

Since net change in length = – δL_ab – δL_bc – δL_cd

= – 0.0033 – 0.0012 – 0.0041 

= – 0.00856 mm.

Decrease in length = 0.00856mm