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tiny-mole [99]
3 years ago
8

A manufacturing process that unintentionally introduces cracks to the surface of a part was used to produce load-bearing compone

nts. The design requires that the component be able to withstand a stress of 450MPa. A component failed catastrophically in service. You are asked to do a failure analysis to determine whether the component failed due to an overload in service or flaws from the manufacturing process. The manufacturer claims that the components were polished to remove the cracks and inspected to ensure that no surface cracks were larger than 0.5mm, which means the component could stand a stress greater than 450MPa. The manufacturer believes the component failed due to operator error. It has been independently verified that the 5cm diameter cylindrical part was subjected to a axial tensile load of 1x106N.The component is made from a material, which has a fracture toughness of and an ultimate 75????Pamtensile strength of 600MPa. Assume Y=1.12 for the external cracks.Who is at fault for the component failure, the manufacturer or the operator, or both? Show your work to support your answer.
Engineering
1 answer:
ElenaW [278]3 years ago
5 0
You connect the motherboard to the astronomical blow up device
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When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str
Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 mm^4

So the shear stress at point  A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

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(Specific weight) A 1-ft-diameter cylindrical tank that is 5 ft long weighs 125 lb and is filled with a liquid having a specific
lina2011 [118]

Answer:

The answer to the question is 514.17 lbf

Explanation:

Volume of cylindrical tank = πr²h = 3.92699 ft³

Weight of tank = 125 lb

Specific weight of content = 66.4 lb/ft³

Mass of content =  66.4×3.92699 = 260.752 lb

Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg

=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N

To have an acceleration of 10.7 ft/s² = 3.261 m/s²

we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf

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4 years ago
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