The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm and a wall thickness of 5 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear-stress distribution over the cross section.

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gkosworldreign gkosworldreign · Answer 1 · 2020-03-13T08:04:04+01:00

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

[tex]T_max =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}[/tex]

=14.5MPa

[tex]T_{min} =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}[/tex]

=10.3MPa

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AFOKE88 AFOKE88 · Answer 2 · 2022-03-29T17:49:53+02:00

The maximum shear stress in the tube when the cable force of 600 N is applied to the cable is; 14.5 MPa

From the diagram of the link which is part of the elevator attached, we can say that;

Twisting torque; T = 600 * 0.15 = 900 N.m

Formula for the maximum shear stress is;

τ_max = Tr/J

where;

T is twisting torque

r is distance from center to stressed surface

J is polar moment of inertia

Thus;

τ_max = (900 * 0.0175)/[(π/2)(0.0175⁴ - 0.0125⁴)]

τ_max = 14.5 MPa

Now, to get the shear distribution diagram, we need to calculate the initial shear stress as;

τ_i = (900 * 0.0125)/[(π/2)(0.0175⁴ - 0.0125⁴)]

τ_max = 10.3 MPa

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