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worty [1.4K]
4 years ago
7

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

rizontal plane. An unknown negative shear stress exists on the vertical plane. The absolute maximum shear stress at the point has a magnitude of 32 ksi. Determine the principal stresses and the shear stress on the vertical plane at the point.
Engineering
1 answer:
victus00 [196]4 years ago
4 0

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

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3 years ago
Steam enters an adiabatic condenser (heat exchanger) at a mass flow rate of 5.55 kg/s where it condensed to saturated liquid wat
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Answer:

The minimum mass flow rate will be "330 kg/s".

Explanation:

Given:

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m_{s}=5.55 \ kg/s

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For water,

\Delta T=10^{\circ}C

(Cp)_{w}=4.184 \ kJ/kg^{\circ}C

They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,

⇒  m_{s}\times (\Delta h)=M_{w}\times(Cp)_{w}\times \Delta T

On putting the estimated values, we get

⇒  5.55\times 2491=M_{w}\times 4.184\times 10\\

⇒  13825.05=M_{w}\times 41.84

⇒  M_{w}=330 \ kg/s

7 0
3 years ago
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