The answer to your quesition is b
Answer:
1790 μrad.
Explanation:
Young's modulus, E is given as 10000 ksi,
μ is given as 0.33,
Inside diameter, d = 54 in,
Thickness, t = 1 in,
Pressure, p = 794 psi = 0.794 ksi
To determine shear strain, longitudinal strain and circumferential strain will be evaluated,
Longitudinal strain, eL = (pd/4tE)(1 - 2μ)
eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)
eL = 3.64 x 10-⁴ radians
Circumferential strain , eH = (pd/4tE)(2-μ)
eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)
eH = 1.79 x 10-³ radians
The maximum shear strain is 1790 μrad.
Snap rings, and bearings can be used to keep a gear on a shaft, hope this helps!!
Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:

Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:

Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:

Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero