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!!
9514 1404 393
Answer:
0.261 ohm
Step-by-step explanation:
If the radius increases by a factor of 0.7/0.4= 7/4, the square of this factor is (7/4)^2 = 49/16. The inverse of this square is 16/49, which is the factor by which the resistance changed.
The resistance of the larger wire is ...
(16/49)(0.80 ohm) ≈ 0.261 ohm
You write ten thousand in numbers like this 10,000
Answer:
Step-by-step explanation:
It is unclear from the phrasing what dimension 8.2 ft represents.
If 8.2 ft is the direct distance from the edge of the roof to the top of the pitch, then the <em>horizontal</em> distance from the edge to the top is √(8.2²-2.4²) ≅ 7.84 ft, and the slope is 2.4/7.84 ≅ 0.31
If 8.2 ft is the <em>horizontal </em>distance from the edge of the root to the top of the pitch, then the slope is 2.4/8.2 ≅ 0.29
