Question

 A rod ABCD rigidly fixed at the ends A and D is subjected to two equal and opposite forces P = 25 kN at B and C as shown in the fig. given below: Make calculations for the axial stresses in each section of the rod.

Answer 1

The following observations need to be made. 

(i) Due to symmetrical geometry and load, reaction at each of the fixed ends will be same both in magnitude and direction. That is P_a = P_d = P₁(say). 

(ii) Segments AB and CD are in tension and the segment BC is in compression. 

(iii) End supports are rigid and therefore total change in length of the rod is zero. The forces acting on each segment will be as shown in Fig. Using the relation δL = PL/AE

δ_Lab = (P₁ × 250)/( 250 × E) = P₁/E .......Extension

δ_Lbc = (P – P₁ ) × 400)/( 400 × E) = (P – P₁)/E .......Compression 

δ_Lcd = (P₁ × 250)/( 250 × E) = P₁/E .......Extension

Since net change in length = 0 i.e.,

δ_Lab – δL_bc + δL_cd = 0

P₁/E – (P – P₁)/E + P₁/E = 0

Or, P₁ – P + P₁ + P₁ = 0;

or, P₁ = P/3 = 25/3 KN

And; P – P₁ = 50/3KN

Now Stress in segment AB and CD

= 25 × 10³/3 × 250 = 33.33 N/mm² (tensile)

Stress in segment

BC = (50 × 10³)/(3 × 400) = 41.67 N/mm² (compressive)