Answer:
82.79MPa
Step-by-step explanation:
Minimum yield strength for AISI 1040 cold drawn steel as obtained from literature, Sᵧ = 490 MPa
Given, outer radius, r₀ = 25mm = 0.025m, thickness = 6mm = 0.006m, internal radius, rᵢ = 19mm = 0.019m,
Largest allowable stress = 0.8(-490) = -392 MPa (minus sign because of compressive nature of the stress)
The tangential stress, σₜ = - ((r₀²p₀)/(r₀² - rᵢ²))(1 + (rᵢ²/r²))
But the maximum tangential stress will occur on the internal diameter of the tube, where r = rᵢ
σₜₘₐₓ = -2(r₀²p₀)/(r₀² - rᵢ²)
p₀ = - σₜₘₐₓ(r₀² - rᵢ²)/2(r₀²) = -392(0.025² - 0.019²)/2(0.025²) = 82.79 MPa.
Hope this helps!!
Answer:
yes
Step-by-step explanation:
Substitute x = 2 and y = 5 into the left side of the equations and if equal to the right side then they are a solution.
x + 2y = 2 + 2(5) = 2 + 10 = 12 ← True
3x + 6y = 3(2) + 6(5) = 6 + 30 = 36 ← True
Thus x = 2, y = 5 is a solution to both equations
Answer:
Let's solve for x.
y=4x+2
Step 1: Flip the equation.
4x+2=y
Step 2: Add -2 to both sides.
4x+2+−2=y+−2
4x=y−2
Step 3: Divide both sides by 4.
4x/4=y−2/4
x=1/4y+ −1/2
Answer:
x=1/4y+ −1/2
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Step-by-step explanation:
