Answer:
The answer is below
Explanation:
The lengths of the rods are not given.
Let us assume the length of rod 1 = 1500 mm and the length of rod 2 = 800 mm
Solution:
The normal strain is defined as the change in member length δ divided by the initial member length L. The normal strain (ε) is:
ε = δ / L
δ = εL
For rod 1:
[tex]\delta_2=\epsilon_2 L_2\\\\\delta_2=0.0010\ mm/mm*1500\ mm=1.5\ mm[/tex]
The axial elongation of rod 2 is 1.5 mm. Since rigid bar ABC is attached to rod 2, the rigid bar move down by same amount.
The rigid bar moves down 1.8 mm but rods 1 will not be stretched by this amount. Because there is a gap between rod (1) and the rigid bar at B, the first deflection of 1 mm would not cause an elongation in rod 1. Therefore, the elongation in rods (1) is:
[tex]\delta_1=1.5\ mm-1\ mm=0.5\ mm[/tex]
The normal strain in rod 1 is:
[tex]\epsilon_1=\frac{\delta_1}{L_1} =\frac{0.5\ mm}{800\ mm} \\\\\epsilon_1=0.000625\ mm/mm[/tex]
