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erastovalidia [21]
2 years ago
4

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.
Engineering
1 answer:
brilliants [131]2 years ago
4 0

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 × 10^{-3} m

torque Mb = 6 N-m

spring index = 10

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

number of effective turns = 5.5

solution

first we get here bending stress that is express as

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

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

put here value and we get

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

\sigma = 305.577 MPa

and

angular deflection will be here

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

put here value and we get

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

angular deflection ∅  = 0.4889 radian  = 28.011°

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=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "

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