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riadik2000 [5.3K]
3 years ago
5

Two parallel surfaces move relative to each other are separated by a gap of 0.375 in. The gap is filled by a fluid of with .0043

1 lb-sec/in2 viscosity. The relative motion is resisted by a shear stress of 0.25 lb/in2 due to the viscosity of the fluid. What is the velocity of the parallel surfaces above assuming the velocity gradient in the space between the surfaces is constant?

Engineering
1 answer:
jasenka [17]3 years ago
8 0

Answer:

Velocity will be 21.75in/sec

Diagram well explains the calculation.

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A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
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Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

                             τ = 16*T / pi*d^3

                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

                             T_s = τ_s*pi*d_s^3 / 16

                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

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