Answer:
The value we got is 1.423 ksi which is less than 30 and so the material hasn't failed and yielding hasn't occured.
Explanation:
This is a cylindrical thin walled vessel. Thus;
Hoop stress; σ1 = pr/t
Where ;
p is internal pressure
r_o is radius = 4/2 = 2 in
t is thickness of wall = 0.1 in
Thus;
σ1 = 70 x 2/0.1 = 1400 ksi
Longitudinal stress; σ2 = pr/2t
σ2 = 70 x 2/(0.1 x 2) = 700 ksi
Now, we want to find the normal stress. The inner radius of the circle will be; r_i = r_o - t = 2 - 0.1 = 1.9 in
So, normal stress by axial force is given by;
σ_fx = F/A = F/(π(r_o² - r_i²))
We are given that F = 500 lb
σ_fx = 500/(π(2² - 1.9²))
σ_fx = 408.1 ksi
We can now find the torsion from the formula;
τ = (T•r_o)/J
We are given that T = 70 lb.ft = 70 x 12 lb.in = 840 lb.in
J is the polar moment of inertia and has a formula; ((π/2)(r_o⁴ - r_i⁴))
So,J = ((π/2)(2⁴ - 1.9⁴)) = 4.662 in⁴
Thus,τ_xy = (840 x 2)/4.662 = 360.4 ksi
σ1 is in the y direction and σ2 is in the x direction. Thus;
σ_x = σ1 + σ_fx = 700 + 408.1 = 1108.1 ksi
Also, σ_y = σ1 = 1400 ksi
Now distortion energy can be expressed as;
σ_Y² = σ_x - σ_x•σ_y + (σ_y)² + 3(τ_xy)²
Plugging in the relevant values, we obtain ;
σ_Y² = 1108.1² - (1108.1*1400) + 1400² + 3(360.4)²
So, σ_Y² = 2.02 x 10^(6) psi²
σ_Y = √2.02 x 10^(6)
σ_Y = 1423 psi = 1.423 ksi
The question says the yield strength of the material is 30 ksi.
The value we got is less than 30 and so the material hasn't failed and yielding hasn't occured.
