Case - 4 :Member subjected to normal stresses on one planes (σx or σy) accompanied by shear stress(q)

The element ABCD is subjected normal stress in X and shear stress on all faces.
Given values for normal stresses is σ_x in X direction and shear stress is q on all faces. Normal stress in Y direction is zero.

Normal stress on inclined plane

We know that,
$\sigma _{n}=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \left( \frac{\sigma _{x} - \sigma _{y} }{2} \right) \cdot cos2\theta + q\cdot sin2\theta$
here Normal stress in Y direction is zero,
$\therefore \sigma _{n}=\left( \frac{\sigma _{x} }{2} \right) + \left( \frac{\sigma _{x} }{2} \right) \cdot cos2\theta + q\cdot sin2\theta$

Tangential stress on inclined plane

We know that,
$\sigma _{t}=\frac{\left (\sigma _{x} - \sigma _{y}\right )}{2} \cdot sin2\theta -q cos2\theta$
$\therefore \sigma _{t}=\frac{\left (\sigma _{x} \right )}{2} \cdot sin2\theta -q cos2\theta$

Location of principal plane

We know that,
$tan2\theta=\frac{2q}{\left (\sigma _{x} - \sigma _{y}\right )}$
Here σ_y =0; $\therefore tan2\theta=\frac{2q}{\left (\sigma _{x} \right )}$
$\therefore 2\theta_{1}= tan^{-1} \frac{2q}{\left (\sigma _{x} \right )}$
$\therefore \theta_{1}= \frac{tan^{-1} \frac{2q}{\left (\sigma _{x} \right )}}{2}$
$\therefore \theta_{2}= \theta_{1}+90^{0}$

Vinayak Hulwane

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Case - 4 :Member subjected to normal stresses on one planes (σx or σy) accompanied by shear stress(q)

Vinayak Hulwane

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MSBTE

University

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Search in this blog

Case - 4 :Member subjected to normal stresses on one planes (σx or σy) accompanied by shear stress(q)

Vinayak Hulwane

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MSBTE

University

Flickr Images