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:
Step-by-step explanation:
We have to convert 2% in to a decimal then multiply it into the x-value which is the months with the addition of 500 because that is the start value.If you were to graph this it would be a linear fuction, y=0.2x+500. The rate of change is found by slope, so rise over run. taking 2 points on the graph and do this
(y 2-y 1)/(x 2-x 1) and graph the equation and get two of the points and find the slope for the rate of change.
Answer:
12, 4, 4/3
Step-by-step explanation:
The pattern is that its 1/3 the previous term...
3 is no cuz obtuse angle is >90 degrees, all the angles together must equal 180 degrees, and if you try it, you get no. 5 is 25% of 45, which is $11.25.
