Modulus of elasticity E = 200 GN/ m2 = 2 × 105 N/mm2 .
Considering equilibrium of forces along the axis of the member.
P1 + P3 = P2 + P4;
120 + P3 = 220 + 160 Force
P3 = 220 + 160 – 120 = 260 kN
The forces acting on each segment of the member are shown in the free body diagrams shown below:
Let δL1, δL2 and δL3, be the extensions in the parts 1, 2 and 3 of the steel bar respectively. Then,
Extension of segment AB
= [(120 × 103 ) × (0.75 × 103 )][1600 × (2 × 105 )] = 0.28125 mm
Compression of segment BC
= [(100 ×103 ) × (1 × 103 )][625 × (2 × 105 )] = 0.8 mm
Extension of segment CD
= [(160 ×103 ) × (1.2 × 103 )][900 × (2 × 105 )] = 1.0667 mm
Net change in length of the member
= δl =0.28125 – 0.8 + 1.0667 = 0.54795 mm (increase)