What is the velocity ratio for creep in the belt drive system for \sigma_1 being the stress in tight side, \sigma_2 being the stress on slack side and E is the young’s modulus of elasticity of the belt material?

Correct Answer is : N1N2=d1d2×E+σ2E+σ1

N_1/N_2=d_1/d_2\times\frac{E-\sqrt\sigma_2}{E-\sqrt\sigma_1}

N_1/N_2=d_1/d_2\times\frac{E+\sqrt\sigma_2}{E-\sqrt\sigma_1}

N_1/N_2=d_1/d_2\times\frac{E+\sqrt\sigma_2}{E+\sqrt\sigma_1}

N_1/N_2=d_1/d_2\times\frac{E-\sqrt\sigma_2}{E+\sqrt\sigma_1}

SSC JE Mechanical Engineering 2018 Set 1 Paper

Question : 106 of 200

Marks: +1, -0

What is the velocity ratio for creep in the belt drive system for

σ_1

being the stress in tight side,

σ_2

being the stress on slack side and E is the young’s modulus of elasticity of the belt material?

Solution:

The velocity ratio for creep belt

N_1/N_2=d_1/d_2×{E+√σ_2}/{E+√σ_1}

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