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PolarNik [594]
3 years ago
9

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

of 0.25 inches and an inner diameter of 8 inches. Use Mohr’s Circle to determine the absolute maximum shear stress in the pressure vessel when it is subjected to this pressure.

Physics
1 answer:
Mekhanik [1.2K]3 years ago
6 0

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

\sigma_h = \frac{Pd}{2t}

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

\sigma_l = \frac{Pd}{4t}

The letters have the same meaning as before.

Then he hoop stress would be,

\sigma_h = \frac{Pd}{2t}

\sigma_h = \frac{75 \times 8}{2\times 0.25}

\sigma_h = 1200psi

And the longitudinal stress would be

\sigma_l = \frac{Pd}{4t}

\sigma_l = \frac{75\times 8}{4\times 0.25}

\sigma_l = 600Psi

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

\tau_{max} = \frac{\sigma_h}{2}

\tau_{max} = \frac{1200}{2}

\tau_{max} = 600Psi

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

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Determine the speed of the 50-kg cylinder after it has descended a distance of 2 m, starting from rest. Gear A has a mass of 10
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This question is incomplete, the missing image is uploaded along this answer below.

Answer:

the speed of the 50-kg cylinder after it has descended is 3.67 m/s

Explanation:

 Given the data in the question and the image below;

relation between velocity of cylinder and velocity of the drum is;

V_D = ω_c × r_c  ----- let this be equ 1

where V_D is velocity of cylinder,  ω_c is the angular velocity of drum C and r_c is the radius of drum C

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ω_B = ω_C

ω_B = V_D / r_c  -------- let this equ 2

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V_D / 0.1 m = 10V_D

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from equation 2; ω_B = V_D / r_c

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substitutive in value of radius r_c (0.1 m)

ω_A( 0.15 m ) = (V_D / 0.1 m ) 0.2 m

ω_A( 0.15 ) = 0.2V_D / 0.1

ω_A =  2V_D  / 0.15

ω_A = 13.333V_D   ----- let this be equation 3

To get the speed of the cylinder, we use energy conversation;

assuming that the final position is;

T₁ + ∑U_{1-2 = T₂

0 + m_Dgh = \frac{1}{2}m_DV²_D + \frac{1}{2}I_Aω²_A + \frac{1}{2}I_Bω²_B

so

m_Dgh = \frac{1}{2}m_DV²_D + \frac{1}{2}(m_Ak_A²)(13.333V_D)² + \frac{1}{2}(m_Bk_B²)(10V_D)²

we given that; m_D = 50 kg, h = 2 m, m_A = 10 kg, k_A 125 mm = 0.125 m, m_B = 30 kg, k_B = 150 mm = 0.15 m.

we know that; g = 9.81 m/s²

so we substitute

50 × 9.81 × 2 = ( \frac{1}{2} × 50 × V_D²) + \frac{1}{2}( 10 × (0.125)² )(13.333V_D)² + \frac{1}{2}( 30 × (0.15)²)(10V_D)²

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981 = 72.638V_D²

V_D² = 981 / 72.638

V_D² = 13.5053

V_D = √13.5053

V_D = 3.674955 ≈ 3.67 m/s

Therefore,  the speed of the 50-kg cylinder after it has descended is 3.67 m/s

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