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:
311.41 degrees
Step-by-step explanation:
If 4 sin Ф = -3 and Ф is between 0 and 360 degrees, then we conclude that Ф must be either in Quadrant III or Quadrant IV (because the sine is negative).
Let's assume we're in Quadrant IV. Then sin Ф = opp / hyp = -3/4; that is, the opp side is negative and has length 3, and the hypo is positive 4.
According to the Pythagorean Theorem, (-3)^2 + x^2 = 4^2, or,
x^2 = 16 - 9 = 7.
Then x is either √7 or -√7.
To find the angle Ф, use the inverse sine function:
Ф = arcsin (-3/4). Using a calculator we get the angle -40.59 degrees, which corresponds to (360 degrees - 40.59 degrees), or 311.41 degrees. We can check this by finding the sine of 311.41 degrees; the result is -0.75, which matches "If 4sintheta = -3."
When you don't use or write the percent right.<span />
The answer is B . your welcome
1 × 1010+3 × 109+4 × 108+0 × 107+0 × 106+0 × 105+0 × 104+0 × 103+0 × 102+0 × 101+0 × 10<span>0
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