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A member ABCD is subjected to point loads P1, P2, P3 and P4 as shown in figure.

Calculate the force P3 ,necessary for equilibrium if P1 = 120 kN, P2 = 220 kN and P4 = 160 kN. Determine also the net change in length of the member. Take E = 200 GN/m2 .

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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)

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