Question 1.6: A stepped bar, as shown in Figure 1.21, is acted upon by dif...
A stepped bar, as shown in Figure 1.21, is acted upon by different forces. If P_1 = 80 kN, P_2 = 60 kN, P_3 = 40 kN and Young’s modulus of the material is 200 GPa, find out the net change in length.
Let us consider different segments of the stepped bar one by one and try to draw the free-body diagrams separately (Figure 1.22).
As the whole body is in equilibrium, all the segments are also in equilibrium. Considering the equilibrium of each segment, we have obtained the free-body diagrams and it automatically shows that P_4 = 60 kN.
For each segment, we can write ΔL = PL/AE . All the segments being in tensile loading (as evident from free-body diagrams), the total change in length is the summation of individual elongations. So, overall change of the bar is given by
\begin{aligned} \Delta L_{ o } &=\Delta L_{ MN }+\Delta L_{ NO }+\Delta L_{ OR } \\ &=\frac{80 \times 600}{40 \times 40 \times 200 \times 10^3}+\frac{20 \times 800}{30 \times 30 \times 200 \times 10^3}+\frac{60 \times 1000}{20 \times 20 \times 200 \times 10^3} \\ &=0.15+0.089+0.75 \\ &=0.989 mm \end{aligned}
So, net change in length (elongation) is 0.989 mm.
