A compressive load P is transmitted through a rigid plate to

A compressive load P is transmitted through a rigid plate to three alloy bars that are identical expect that initially the middle bar is slightly shorter than the other two bars (see figure below). The length L of the outer bars is 2.5 ft and the cross-sectional area of the bars A is 4.91 in2. The modulus of elasticity of the alloy bars is E-6.5x108 psi and the initial gap (s) is 0.05 in Determine: a. the load P required to close the gap. b. the downward displacement of the rigid plate (8) when the load P 150kips

Solution

(a) The load P is divided equally between the two outer bars.

Load on each bar = P/2

Initial length of each bar (L) = 2.5 ft = 30 in

Cross-sectional of each bar (A) = 4.91 in²

Stress acting in each bar = P/2A

Reduction in length of bar (S) = 0.05 in

Strain acting in bar = S/L = 0.05/30 = 0.001667

Modulus of elasticity (E) = 6.5 x 10⁶ psi

Now, Stress = Modulus of elasticity x Strain

=> P/2A = E x 0.001667

=> P = 2EA x 0.001667 = 106383.3 lbs = 106.383 kips

(b) When P = 150 kips is applied, the rigid plate will first move distance S=0.05 in. Then the plate will be in contact with three bars.

Load carried by each bar = 150/3 = 50 kips = 50000 lbs

Stress acting = 50000 / 4.91 = 10183.3 psi

Let the decrease in length of each bar be L₁ in.

Strain acting in each bar = L₁/(L-S) = L₁/(30-0.05) = L₁/29.95

Now, Stress = Modulus of elasticity x Strain

=> 10183.3 = 6.5 x 10⁶ x (L₁/29.95)

=> L₁ = 0.0469 in

Total downward displacement of the rigid plate = 0.05 + 0.0469 = 0.0969 in