Case - 3 :Member subjected to normal stresses on two mutually perpendicular planes (σx and σy)

The element ABCD is subjected normal stress in X and Y directions. Given values for normal stresses is σ_x and σ_y in X and Y direction respectively. Shear stress 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 q = 0,
$\therefore \sigma _{n}$ $=\left( \frac{\sigma _{x} + \sigma _{y}}{2} \right) + \left( \frac{\sigma _{x} - \sigma _{y} }{2} \right) \cdot cos2\theta$

Tangential stress on inclined plane

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

Location of principal plane

We know that,
$tan2\theta=\frac{2q}{\left (\sigma _{x} - \sigma _{y}\right )}$
Here q = 0 therefore surface of element itself is principal plane and forces acting on it principal stresses.
Here, Plane AB and plane BC are principal planes and σ_x and σ_y are principal stresses

Vinayak Hulwane

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Case - 3 :Member subjected to normal stresses on two mutually perpendicular planes (σx and σy)

Vinayak Hulwane

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MSBTE

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Search in this blog

Case - 3 :Member subjected to normal stresses on two mutually perpendicular planes (σx and σy)

Vinayak Hulwane

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MSBTE

University

Flickr Images