Calculate force required for equilibrium:
A member ABCD is subjected to the point loads P_{1}, P_{2}, P_{3} and P_{4} as shown in the figure given below
Calculate force P_{3} required for equilibrium if P_{1} = 120 kN, P_{2} = 220 kN and P_{4} = 160 kN. Determine the net change in length of member. Take E = 200 GN/m^{2}.
Sol.: Modulus of elasticity E = 200 GN/ m^{2 }= 2 × 10^{5} N/mm^{2}.
By considering equilibrium of forces along axis of the member.
P_{1} + P_{3} = P_{2} + P_{4};
120 + P_{3} = 220 + 160
Force P_{3} = 220 + 160 - 120 = 260 kN
The forces which are acting on each segment of member are shown in free body diagrams shown below:
Let δL_{1}, δL_{2} and δL_{3}, be extensions in parts 1, 2 and 3 of steel bar respectively. Then,
As, Tension in AB and CD but compression in BC,
So, Total extension of the bar,
Extension of segment AB = [(120 × 10^{3}) × (0.75 × 10^{3})]/[1600 × (2 × 10^{5})] = 0.28125 mm
Compression of segment BC = [(100 ×10^{3}) × (1 × 10^{3})]/[625 × (2 × 10^{5})] = 0.8 mm
Extension of segment CD = [(160 ×103) × (1.2 × 10^{3})]/[900 × (2 × 10^{5})] = 1.0667 mm
Net change in length of the member = δl =0.28125 - 0.8 + 1.0667 = 0.54795 mm (increase) .......ANS