Normal stress on cantilever beam

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SUMMARY

The discussion focuses on calculating normal stress on a cantilever beam using the formula \(\sigma_x = -\frac{M(x)y}{I}\) and the moment of inertia \(I = \frac{bh^3}{3}\). The user calculated the moment of inertia for point A as \(I = \frac{8th^3}{3}\) and derived the normal stress as \(\sigma_x = -\frac{3\tau_0}{4th}\). The user questions the equivalence of principal stresses at points A and B, noting the influence of applied traction creating a moment around the beam's fixed end.

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  • Understanding of cantilever beam mechanics
  • Familiarity with stress and strain concepts
  • Knowledge of moment of inertia calculations
  • Proficiency in applying the principal stress theory
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  • Learn about the effects of applied moments on beam stress distribution
  • Research the differences in stress states at various points along a cantilever beam
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Homework Statement



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Homework Equations



\sigma_x = -\frac{M(x)y}{I}

I = \frac{bh^3}{3}

The Attempt at a Solution



For point A, I did the following:

I = \frac{t(2h)^3}{3}

I = \frac{8th^3}{3}

\sigma_x = -\frac{3\tau_0}{4th}

\tau_{yx} = -\tau_0

\sigma_y = 0

Are my principal stresses correct for point A? Also, would these principal stresses be the same for Point B?
 
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A principal stress state contains no shear. Also, it's hard to imagine why A and B would be equivalent when the applied traction is creating a moment around the fixed end of the beam.
 

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