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:
A) -15, 5, 2
Step-by-step explanation:
Answer:
F(d) = 30 + 0.50d
Step-by-step explanation:
Given
Charges = P8.00 ---- first 4 km
Additional = P0.50
Required
Write a function to address the scenario.
Represent the whole distance covered with d.
First,we need to determine the total charges for the first four hours.
Charges = 8.00 * 4
Charges = 32.00
Next, we determine the charges for additional distance.
Charges = 0.50 * (d - 4)
d - 4 is the remaining distance after the first 4.
Charges = 0.50d - 2
The function is then written as;
F(d) = 32 + 0.50d - 2
F(d) = 32 - 2 + 0.50d
F(d) = 30 + 0.50d
Please, Karla, explain what "C" and "n" represent. Are you talking about combinations (for example, n+2 objects taken n at a time? Or is C some kind of function? I don't quite see the relationship of this problem to 'data management.'
