A beam of same cross section is used in two different orientations as shown in figure. Bending moment applied to the beam in both cases are same. The maximum bending stress induced in case A and B are related to
F2 Vinanti Engineering 12.07.23 D2

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  1. σa = 2σb
  2. σa = σ​b
  3. σa = σb/2
  4. σa = σb​/4

Answer (Detailed Solution Below)

Option 1 : σa = 2σb
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Concept:

From the simple bending theory equation:

\(\begin{array}{l} \frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\\ Z = \frac{I}{y} = \frac{{b{h^3}}}{{12}} \times \frac{2}{h} = \frac{{b{h^2}}}{6} \end{array}\)

If σb is the maximum bending stresses due to bending moment M on shaft:

\({\sigma _b} = \frac{M}{Z} = \frac{{6M}}{{b{h^2}}}\)

Explanation:

For same bending moment:

\({\sigma _b} \propto \frac{1}{{b{h^2}}} \Rightarrow \frac{{{\sigma _{b1}}}}{{{\sigma _{b2}}}} = \frac{{{b_2}h_2^2}}{{{b_1}h_1^2}} = \frac{{\left( {\frac{b}{2}} \right){b^2}}}{{b{{\left( {\frac{b}{2}} \right)}^2}}} = 2\)

So σA = 2 σB
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