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Rigid bar ABC is supported by three symmetrically-positioned vertical rods, which are initially unstrained. After load P is applied, the normal strain in rods (2) is 0.0010 mm/mm. Determine the normal strain in rod (1) if there is a gap of 1.0 mm in the connection between rod (1) and the rigid bar at B. Report the strain in mm/mm

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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]

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