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iris [78.8K]
3 years ago
5

The two shafts of a Hooke’s coupling have their axes inclined at 20°.The shaft A revolves at a uniform speed of 1000 rpm. The sh

aft B carries a flywheel of mass 30 kg. If the radius of gyration of the flywheel is 100 mm, find the maximum torque in shaft B.
Engineering
1 answer:
lapo4ka [179]3 years ago
4 0

Answer:

33.429 N-m

Explanation:

Given :

Inclination angle of two shaft, α = 20°

Speed of shaft A, N_{A} = 1000 rpm

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

                                   = 0.1 m

Now we know that for maximum velocity,

\frac{N_{B}}{N_{A}} = \frac{cos\alpha }{1 - sin^{2}\alpha }

\frac{N_{B}}{1000} = \frac{cos20}{1 - sin^{2}20 }

N_{B} = 1064.1 rpm

Now we know

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

                                   = 0.1 m

Therefore moment of inertia of flywheel, I = m.k^{2}

                                                                      =30 X 0.1^{2}

                                                                     = 0.3 kg-m^{2}

Now torque on the output shaft

T₂ = I x ω

    = 0.3 X 1064.2 rpm

    = 0.3\times \frac{2\pi \times 1064.1}{60}

     = 33.429 N-m

Torque on the Shaft B is 33.429 N-m

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Explain why the failure of a garden hose occurred near its end and why the tear occurred along its length. Use numerical values
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Explanation:

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3 years ago
The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

K = \sigma Y \sqrt{\pi a}

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fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

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crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

\sigma = 1728.489 MPa

Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

P = A × σ

P = 1728.289 × 4.5

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3 years ago
If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
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Answer:

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Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

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y = 3.96 in

\sigma_b = \frac{M}{I} \times y

putting all value to get bending stress

\sigma_b = \frac{50.112}{144} \times 3.96  

\sigma_b =  1.37 psi

Bending stress at point 3.96 is \sigma_b = 1.37 psi

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