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lozanna [386]
3 years ago
8

The 500 lb load is being hoisted at a constant speed using the motor M, which has a weight of 50 lb. The beam has a weight of 40

lb/ft and is fixed to the wall at A. Find axial, shear force and bending moment at point B. (Radius of the pulley wheel is 1 ft)

Engineering
1 answer:
pashok25 [27]3 years ago
6 0

Answer:

The answer and explanation is attached.

Explanation:

The 500 lb load is being hoisted at a constant speed using the motor M, which has a weight of 50 lb. The beam has a weight of 40 lb/ft and is fixed to the wall at A. Find axial, shear force and bending moment at point B. (Radius of the pulley wheel is 1 ft)

The answer and explanation is attached.

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A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the [100] di
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Answer:

For [1 1 0] and  [1 0 1] plane, σₓ = 6.05 MPa

For [0 1 1] plane, σ = 0; slip will not occur

Explanation:

compute the resolved shear stress in [111] direction on each of the [110], [011] and on the [101] plane.

Given;

Stress direction: [1 0 0] ⇒ A

Slip direction: [1 1 1]

Normal to slip direction: [1 1 1] ⇒ B

∅ is the angle between A & B

Step 1: cos∅ = A·B/|A| |B| = \frac{[100][111]}{\sqrt{1}.\sqrt{3}  } ⇒ cos∅ = 1/\sqrt{3}

σₓ = τ/cos ∅·cosλ

where τ is the critical resolved shear stress given as 2.47MPa

Step 2: Solve for the slip along each plane

(a) [1 1 0]

cosλ = [1 1 0]·[1 0 0]/(\sqrt{2}·\sqrt{1})        

note: cosλ = slip D·stress D/|slip D||stress D|

cosλ = 1/\sqrt{2}

∵ σₓ = τ/\frac{1}{\sqrt{2} } ·\frac{1}{\sqrt{3} } = \sqrt{6} * 2.47MPa = 6.05MPa

Hence, stress necessary to cause slip on [1 1 0] is 6.05MPa

(b) [0 1 1]

cosλ = [0 1 1]·[1 0 0]/(\sqrt{2}·\sqrt{1}) = 0

∵ σₓ = 2.47MPa/0, which is not defined

Hence, for stress along [1 0 0], slip will not occur along [0 1 1]

(c) [1 0 1]

cosλ = [0 1 1]·[1 0 0]/(\sqrt{2}·\sqrt{1})

cosλ = 1/\sqrt{2}

∵ σₓ = τ/\frac{1}{\sqrt{2} } ·\frac{1}{\sqrt{3} } = \sqrt{6} * 2.47MPa = 6.05MPa

See attachment for the space diagram

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A closed vessel of volume 80 litres contain gas at a gauge pressure of 150 kPa. If the gas is compressed isothermally to half it
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Answer:

The resulting pressure is 300 kilopascals.

Explanation:

Let consider that gas within the closed vessel behaves ideally. By the equation of state for ideal gases, we construct the following relationship for the isothermal relationship:

P_{1}\cdot V_{1} = P_{2} \cdot V_{2} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kilopascals.

V_{1}, V_{2} - Initial and final volume, measured in litres.

If we know that \frac{V_{1}}{V_{2}} = 2 and P_{1} = 150\,kPa, then the resulting pressure is:

P_{2} = P_{1}\times \frac{V_{1}}{V_{2}}

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The resulting pressure is 300 kilopascals.

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