3-71A 20mm diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15 degrees, what must be the length of the bar

Answers

Answer 1

Answer:

The right answer is "1.903 m".

Explanation:

Given that,

[tex]\tau =110 \ MPa[/tex]

[tex]G=80 \ GPa[/tex]

[tex]\Theta=15\times \frac{\pi}{180}[/tex]

   [tex]=\frac{\pi}{12}[/tex]

[tex]d=20 \ mm[/tex]

As we know,

⇒ [tex]\frac{\tau}{r}=\frac{G \Theta}{L}[/tex]

Or,

⇒ [tex]L=\frac{G \theta r}{\tau}[/tex]

       [tex]=\frac{80\times 10^3}{110}\times \frac{\pi}{12}\times 10[/tex]

       [tex]=1903.9 \ mm[/tex]

or,

       [tex]=1.903 \ m[/tex]


Related Questions

Air is compressed in a well insulated compressor from 95 kPa and 27 C to 600 kPa and 277 C. Use the air tables; assume negligible changes in kinetic and potential energy. Find the isentropic efficiency of the compressor. Find the exit temperature of the air if the compressor was reversible.

Answers

Answer:

a) 1.9%

b) T2s = 505.5 k = 232.5°C

Explanation:

P1 = 95 kPa

T1 = 27°C  = 300 k

P2 = 600 kPa

T1 = 277°c  = 550 k

Table used : Table ( A - 17 ) Ideal gas properties of air

a) determining the isentropic efficiency of the compressor

Л = ( h2s - h1 ) / ( h2a -  h1 ) ---- ( 1 )

where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k

To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)

 Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306  hence; h2s = 500.72 kJ/kg

back to equation1

Л = 0.019 = 1.9%

b) Calculate the exit temperature of the air if compressor is reversible

if compressor is reversible the corresponding exit temperature

T2s = 505.5 k = 232.5°C

given that h2s = 500.72 kJ/kg

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