A member ABCD is subjected to point loads P1 = 45 kN, P2 = 100 kN, and P4 = 130 kN. Determine the net elongation of the member. Assume E = 210 GPa. Refer Figure 2.35(a).

Question

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

Accepted Answer

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

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

∴ [latex] P_3=185 kN [/latex]

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

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

Chapter 2

Q. 2.22

Q. 2.22

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