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Question 2.22: A member ABCD is subjected to point loads P1 = 45 kN, P2 = 1......

A member ABCD is subjected to point loads P_1 = 45 kN, P_2 = 100 kN, and P_4  = 130 kN. Determine the net elongation of the member. Assume E = 210 GPa. Refer Figure 2.35(a).

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For the external equilibrium, algebraic sum of horizontal forces on the bar must be

\begin{array}{r} -P_1+P_2-P_3+P_4=0 \\ -45+100-P_3+130=0 \end{array}

∴        P_3=185  kN

Now, we draw FBD (Figures 2.35b, c and d) of three portions of the member and superpose their deformations

Therefore, Net deformation = Algebraic sum of deformations in three portions

\begin{aligned} & =\frac{10^3}{210 \times 10^3}\left[\frac{130 \times 1000}{1250}-\frac{55 \times 600}{2500}+\frac{45 \times 1200}{625}\right] \\ & =\frac{1}{210}(177.2)=0.844  mm \end{aligned}

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