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mixas84 [53]
3 years ago
10

The driveshaft of an automobile is being designed to transmit 134 hp at 2900 rpm. Determine the minimum diameter d required for

a solid steel shaft if the allowable shear stress in the shaft is not to exceed 6100 psi.
Engineering
1 answer:
Misha Larkins [42]3 years ago
4 0

Answer:

The minimum diameter is 1.344 in

Explanation:

The angular speed of the driveshaft is equal to:

w=\frac{2\pi N}{60}

Where

N = rotational speed of the driveshaft = 2900 rpm

w=\frac{2\pi *2900}{60} =303.69rad/s

The torque in the driveshaft is equal to:

\tau=\frac{P}{w}

Where

P = power transmitted by the driveshaft = 134 hp = 73700 lb*ft/s

\tau=\frac{73700}{303.69}  =242.68lb*ft

The minimum diameter is equal to:

d_{min} =(\frac{16T}{\pi *\tau } )^{1/3}

Where

T = shear stress = 6100 psi

τ = 242.68 lb*ft = 2912.16 lb*in

d_{min} =(\frac{16*2912.16}{\pi *6100} )^{1/3} =1.344in

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Answer:

See explanation

Explanation:

Solution:-

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3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete
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Answer:

Yes, fracture will occur

Explanation:

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To find the dimensionless parameter, we use critical stress crack propagation equation

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Y=\frac {K}{\sigma_c \sqrt {a\pi}}

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, \sigma_c  is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

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3 years ago
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When the material that exits is lesser in amount than that of the entering material in a black box experiment, the parts of the system need to be changed.

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