Question 7.2: A plane-stress condition exists at a point on the surface of...

A plane-stress condition exists at a point on the surface of a loaded structure, where the stresses have the magnitudes and directions shown on the stress element of Fig. 7-8a.
Determine the stresses acting on an element that is oriented at a clockwise angle of 15° with respect to the original element.

7.8a
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The stresses acting on the original element (Fig. 7-8a) have the following values:

\sigma_{x} = -46 MPa              \sigma_{y} = 12 MPa        \tau_{xy} = -19 MPa

An element oriented at a clockwise angle of 15° is shown in Fig. 7-8b, where the x_{1} axis is at an angle θ = -15° with respect to the x axis. (As an alternative, the x_{1} axis could be placed at a positive angle θ = 75° .)
Transformation equations. We can readily calculate the stresses on the x_{1} face of the element oriented at θ = -15° by using the transformation equations (Eqs. 7-4a and 7-4b). The calculations proceed as follows:

\sigma_{x_{1}}=\frac{\sigma_{x}+\sigma_{y}}{2}+\frac{\sigma_{x}-\sigma_{y}}{2}\cos 2_\ \theta+\tau_{xy}\sin 2_\ \theta             (7-4a)

\tau_{x_{1}y_{1}} =-\frac{\sigma_{x}-\sigma_{y}}{2}\sin 2_\ \theta +\tau_{xy}\cos 2_\ \theta                      (7-4b)

\frac{\sigma_{x}+\sigma_{y}}{2} = -17 MPa                   \frac{\sigma_{x}-\sigma_{y}}{2} = -29 MPa

sin 2 θ = sin (-30°) = -0.5            cos 2 θ = cos (-30°) = 0.8660

Substituting into the transformation equations, we get

\sigma_{x_{1}}=\frac{\sigma_{x}+\sigma_{y}}{2}+\frac{\sigma_{x}-\sigma_{y}}{2}\cos 2_\ \theta +\tau_{xy}\sin 2_\ \theta

 

= -17 MPa + (-29 MPa)(0.8660) + (-19 MPa)(-0.5)

= -32.6 MPa

\tau_{x_{1}y_{1}} =-\frac{\sigma_{x}-\sigma_{y}}{2}\sin 2_\ \theta +\tau_{xy}\cos 2_\ \theta

 

= -(-29 MPa)(-0.5) + (-19 MPa)(0.8660)

= -31.0 MPa

Also, the normal stress acting on the y_{1} face (Eq. 7-5) is

\sigma_{y_{1}}=\frac{\sigma_{x}+\sigma_{y}}{2}-\frac{\sigma_{x}-\sigma_{y}}{2}\cos 2_\ \theta-\tau_{xy}\sin 2_\ \theta

 

= -17 MPa – (-29 MPa)(0.8660) – (-19 MPa)(-0.5)

= -1.4 MPa

This stress can be verified by substituting θ = 75° into Eq. (7-4a). As a further check on the results, we note that \sigma_{x_{1}} + \sigma_{y_{1}} = \sigma_{x} + \sigma_{y}.
The stresses acting on the inclined element are shown in Fig. 7-8b, where the arrows indicate the true directions of the stresses. Again we note that both stress elements shown in Fig. 7-8 represent the same state of stress.

7.8b

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