Question

A uniform edge load of 500 lb/in. and 350 lb/in. is applied to the polystyrene specimen. If it is originally square and has dimensions of a = 2 in., b = 2 in., and a thickness of t = 0.25 in., determine its new dimensions a, b, and t after the load is applied. Ep = 597(10)3 psi, vp = 0.25.

Answer 1

The image of the load applied to the polystyrene is missing, so i have attached it.

Answer:

a_new = 2.00302 in

b_new = 2.00552

Explanation:

From the image attached, we can see that the load of 500 lb/in is applied in the x-direction while the load of 350 lb/in acts in the y-direction.

Now, formula for stress is;

Stress(σ) = Force/Area

We are not given force and area but the load and plate thickness.

Thus, stress = load/thickness

We are given;

Load in x - direction = 500 lb/in.

Load in y - direction = 350 lb/in.

Thickness; t = 0.25 in

Thus;

σ_x = 500/0.25

σ_x = 2000 ksi

σ_y = 350/0.25

σ_y = 1400 ksi

From Hooke's law for 2 dimensions, strain is given by the formula;

ε_x = (1/E)(σ_x - vσ_y)

ε_y = (1/E)(σ_y - vσ_x)

We are given v_p = 0.25 and Ep = 597 × 10³ psi

Thus;

ε_x = (1/(597 × 10^(3)))(2000 - (0.25 × 1400)

ε_x = 0.00276

ε_y = (1/(597 × 10^(3)))(1400 - (0.25 × 2000)

ε_y = 0.00151

From elongation formula, we know that;

Startin is: ε = ΔL/L

Thus; ΔL = Lε

We are given a = 2 and b = 2

Thus;

ΔL_x = 2 × 0.00276

ΔL_x = 0.00552

ΔL_y = 2 × 0.00151

ΔL_y = 0.00302

New dimensions are;

a_new = 2 + 0.00302

a_new = 2.00302 in

b_new = 2 + 0.00552

b_new = 2.00552