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A helical torsion spring of mean diameter 60 mm is made of a round wire of 6 mm diameter. If a torque of 6 N-m is applied on the spring, find the bending stress induced and the angular deflection of the spring in degrees. The spring index is 10 and modulus of elasticity for the spring material is 200 kN/mm2 . The number of effective turns may be taken as 5.5.

Respuesta :

Answer:

bending stress = 305.577 MPa

angular deflection   = 28.011°

Explanation:

given data

mean diameter D = 60 mm = 0.6 m

wire diameter d = 6 mm = 6 × [tex]10^{-3}[/tex] m

torque Mb = 6 N-m

spring index = 10

elasticity for the spring material = 200 kN/mm² = 200 ×  [tex]10^{9}[/tex] N/m²

number of effective turns = 5.5

solution

first we get here bending stress that is express as

[tex]\sigma = \frac{k\times 32\times Mb}{\pi \times d^3}[/tex]     .....................1

here k is stress factor i.e 1.08 for round wire

put here value and we get

[tex]\sigma = \frac{1.08\times 32\times 6}{\pi \times (6\times 10^{-3})^3}[/tex]    

[tex]\sigma[/tex] = 305.577 MPa

and

angular deflection will be here

angular deflection ∅ = [tex]\frac{64\times Mb \times D\times y }{E\times d^4}[/tex]     ............2

put here value and we get

angular deflection ∅ =  [tex]\frac{64\times 6 \times 0.06\times 5.5}{200\times 10^9\times (6\times 10^{-3})^4}[/tex]  

angular deflection ∅  = 0.4889 radian  = 28.011°

annyksl

The bending stress is equal to 305.577 MPa, while the angular deflection is equal to 28.011°.

How is it possible to arrive at this result?

  • First, we will have to find the value of the bending stress. this value must be found using the following formula:

[tex]\delta=\frac{k*32*Mb}{\pi *d^3}[/tex]

K represents a constant and its value is equal to 1.08. Therefore, we can solve the equation as follows:

[tex]\delta= \frac{1.8*32*6}{\pi *(6*10^-^3)^3} = 305.577 MPa[/tex]

  • To calculate the angular deflection, we will use the formula:

[tex]\emptyset=\frac{64*Mb*D*y}{E*d^4} \\\emptyset= \frac{64*6*0.06*5.5}{200*10^9*(6*10^-^3)^4} = 0.4889 ------->28.011[/tex]

It is important that you recognize the symbols in the equations these symbols are:

  • D = 60 mm = [tex]0.6m[/tex]
  • d = 6 mm =[tex]6*10^-^3m[/tex]
  • Mb =[tex]6N-m[/tex]
  • spring index = [tex]10[/tex]
  • elasticity for the spring material = 200 kN/mm² = [tex]200*10^9 N/m^{2}[/tex]
  • number of effective turns =[tex]5.5[/tex]

More information on solving equations in the link:

https://brainly.com/question/17177510

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