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:
1) m=2/9; y-intercept= 17/3
2) m=7 ; y-intercept= 22
3) m= -2/11 ; y-intercept= 100/11
4) m= 13/19 ; y-intercept= -53/13
Step by Step Explanation:
Use the slope formula to find the slope m.
First off both triangles form a 90° angle there both congruent. you can tell they form a 90° angle because of square box
Answer:
Step-by-step explanation:
Apply the Midsegment Theorem:
![\displaystyle \frac{1}{2}[96 - 4x]° = [5x + 6]° → [48 - 2x]° = [5x + 6]° → -42 = -7x; 6 = x \\ \\ [48 - 2(6)]° = [48 - 12]° = 36° \\ \\ AND/OR \\ \\ [5(6) + 6]° = [30 + 6]° = 36°](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B2%7D%5B96%20-%204x%5D%C2%B0%20%3D%20%5B5x%20%2B%206%5D%C2%B0%20%E2%86%92%20%5B48%20-%202x%5D%C2%B0%20%3D%20%5B5x%20%2B%206%5D%C2%B0%20%E2%86%92%20-42%20%3D%20-7x%3B%206%20%3D%20x%20%5C%5C%20%5C%5C%20%5B48%20-%202%286%29%5D%C2%B0%20%3D%20%5B48%20-%2012%5D%C2%B0%20%3D%2036%C2%B0%20%5C%5C%20%5C%5C%20AND%2FOR%20%5C%5C%20%5C%5C%20%5B5%286%29%20%2B%206%5D%C2%B0%20%3D%20%5B30%20%2B%206%5D%C2%B0%20%3D%2036%C2%B0)
I am joyous to assist you at any time.
