Answer and Explanation:
For computing the displacement at point B and D we need to determine the following calculations
[tex]P_Net = P_C + P_E + P_B[/tex]
= 250 + 350 - 50
= 550 N
Now the deflection for bar AB is
[tex]\delta_{AB} = \frac{PL_{AB}}{AE} \\\\ = \frac{550 \times 500}{6,000 \times 200 \times 10^{3}}[/tex]
Now for bar BC it is
[tex]\delta_{BC} = \frac{PL_{BC}}{AE} \\\\ = \frac{(550 + 50) \times 250}{5,000 \times 200 \times 10^{3}} \\\\ = 1.5 \times 10^{-04} mm[/tex]
And for bar CD it is
[tex]\delta_{CD} = \frac{PL_{CD}}{AE} \\\\ = \frac{(550 -250 + 50) \times 250}{5,000 \times 200 \times 10^{3}} \\\\ = 0.875 \times 10^{-4} mm[/tex]
Now the displacement is as follows
For B
2.292 × 10^{-4} mm
For D, it is
[tex]= 2.292 \times 10^{-4} + 1.5 \times 10^{-4} + 0.875 \times 10^{-4} mm \\\\ = 4.667 \times 10^{-4} mm[/tex]
We simply applied the above formulas for determining the displacements at points B, D and the same is to be considered
