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.
Answer:
square cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.
Required:
a. What is the nominal maximum tensile stress on the bar?
b. If there were an initial 1.2 mm deep surface crack on the right surface of the bar, what would the critical stress needed to cause instantaneous fast fracture of the bar be?
Answer:
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