Respuesta :
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
The new dimensions are "2.003016, 2.005528, and 0.249644".
Dimensions:
Calculate the normal stress along the x-direction.
[tex]\to \sigma_x =\frac{500}{t}=\frac{500}{0.25}= 2000 \ \frac{lb}{in^2}\\\\[/tex]
Calculate the normal stress along the y-direction.
[tex]\to \sigma_y =\frac{350}{t}=\frac{350}{0.25}= 1400 \ \frac{lb}{in^2}[/tex]
Calculate the strain along the x-direction.
[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_y-\sigma_z)]\\\\[/tex]
[tex]=\frac{1}{597\times 10^3} [2000-0.25(1400+0)] \\\\= 2.764\times 10^{-3}\\\\[/tex]
Calculate the strain along the y-direction.
[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_x+\sigma_z)][/tex]
[tex]=\frac{1}{597\times 10^3}[1400 -0.25 (2000+0)]\\\\=1508\times 10^{-3}[/tex]
Calculate the strain along the z-direction.
[tex]\to \varepsilon_x= \frac{1}{E}[\sigma_z-v(\sigma_x +\sigma_y)][/tex]
[tex]=\frac{1}{597\times 10^3} [0-0.25(2000+1400)] \\\\=-1.424\times 10^{-3}[/tex]
Calculate the new dimensions.
[tex]\to b' =b+ \varepsilon_xb\\\\[/tex]
[tex]= 2+2.764\times 10^{-3} \times 2\\\\= 2.005528 \ in\\\\[/tex]
[tex]\to a' = a + \varepsilon_y a\\\\[/tex]
[tex]= 2+1.508\times 10^{-3} \times 2\\\\= 2.003016\ in\\\\[/tex]
[tex]\to c'= c + \varepsilon_x c\\\\[/tex]
[tex]= 0.25 +(-1.424\times 10^{-3}) \times 0.25\\\\= 0.249644\ in\\\\[/tex]
Find out more about the dimensions here:
brainly.com/question/15406884
