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A shaft is subjected to bending moment M and a torque T simultaneously. The ratio of the maximum bending stress to maximum shear stress developed in the shaft, is

A. $$\frac{{\text{M}}}{{\text{T}}}$$

B. $$\frac{{\text{T}}}{{\text{M}}}$$

C. $$\frac{{2{\text{M}}}}{{\text{T}}}$$

D. $$\frac{{2{\text{T}}}}{{\text{M}}}$$

Answer: Option C


This Question Belongs to Civil Engineering >> Theory Of Structures

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Comments ( 4 )

  1. Sudarshan Tuwar
    Sudarshan Tuwar :
    3 years ago

    32M/16T=2M/T

  2. علي چوني
    علي چوني :
    4 years ago

    Q.5. For the state of stress shown in Fig. (5), if the minimum normal stress is 10MPa; determine the following using Mohr's circle:
    1-The horizontal and vertical normal stresses and the associated shear stresses.
    2-The maximum and minimum shear stresses and the associated normal stresses.
    Show all results on sketches of properly oriented elements.

  3. Madhusmita Muduli
    Madhusmita Muduli :
    5 years ago

    Bending stress =32M/πD3
    Shear stress= 16T/πD3
    So
    Bending stress/shear stress =2M/T

  4. Matiur Rahaman
    Matiur Rahaman :
    5 years ago

    Max bending stress 32M/ πd^3
    Max shear stress 16T/ πd^3
    Ratio of bending to shear 2M/T

Related Questions on Theory of Structures

Y are the bending moment, moment of inertia, radius of curvature, modulus of If M, I, R, E, F and elasticity stress and the depth of the neutral axis at section, then

A. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$

B. $$\frac{{\text{I}}}{{\text{M}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$

C. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{F}}}{{\text{Y}}}$$

D. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{Y}}}{{\text{F}}}$$