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Andrews [41]
3 years ago
10

A) A cross-section of a solid circular rod is subject to a torque of T = 3.5 kNâ‹…m. If the diameter of the rod is D = 5 cm, wha

t is the maximum shear stress? include units.B) The maximum stress in a section of a circular tube subject to a torque is Tmax = 37 MPa . If the inner diameter is Di = 4.5 cm and the outer diameter is Do = 6.5 cm , what is the torque on the section? include units.
Engineering
1 answer:
Alisiya [41]3 years ago
7 0

Answer:

\tau_{max}  = 142.6 MPa

T = 1536.8 N m

Explanation:

Given data:

Torque = 3.5 k N m = 3.5*10^3 N.m

Diameter D = 5 cm = 0.05 m

a) from torsional equation we have

\frac[T}{J_{solid}} = \frac{\tau_{max}}{D/2}

\frac{T}{\pi/32 D^4} = \frac{\tau_{max}}{D/2}

solving for \tau_{max}

\tau_{max} = \frac{16 T}{\pi D^3} =\frac{16 \times 3.5*10^3}{\pi 0.05^3}

\tau_{max}  = 142.6 MPa

B)

\tau = 37 MPa = 37 \times  10^6 Pa

D_i = 4.5 cm = 0.045 m

D_o = 6.5 cm = 0.065 m

\frac{T}{J_{hollow}} = \frac{\tau_{max}}{D_o /2}

\frac{T}{(\pi/32) (0.065^4 - 0.045^4)} =\frac{37*10^6}{0.065/2}

T = 1536.8 N m

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Air expands through a turbine operating at steady state. At the inlet p1 = 150 lbf/in^2, T1 = 1400R and at the exit p2 = 14.8 lb
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The power developed in HP is 2702.7hp

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