Question 2.20: Tensile stresses of 100 MPa and 50 MPa are acting on two per...
Tensile stresses of 100 MPa and 50 MPa are acting on two perpendicular planes in a body. Determine the resultant stress inclined at 35° to the major principal plane using stress ellipsoid.
Draw two circles corresponding to the principal stress 100 MPa and 50 MPa as shown in Fig. 2.29. Draw a line ABO at 35° to the major principal stress. Intersection point C of a horizontal line from B and a vertical line from A, is the location of the resultant stress \sigma_r = OC = 86.8 MPa. Similarly, the resultant stress at any plane can be obtained and the locus of point C will be an ellipse.
Mathematically,
OD = \sigma_1 cos 35° = 100 cos 35°
CD = \sigma_2 sin 35° = 50 sin 35°
From Eqs (2.1a) and (2.2),
\sigma_{x^{\prime}}=\sigma _x \cos^2 \theta +\sigma _y \sin^2 \theta-\tau _{xy}\sin 2\theta (2.1a)
\tau_{x^{\prime}y^{\prime}}=\frac{\sigma _x-\sigma _y}{2}\sin 2\theta +\tau _{xy}\cos 2\theta (2.2)
\sigma_{x^{\prime}} = \sigma_1 \cos^2 \theta + \sigma_2 \sin^2 \theta (i)
and \tau _{x^{\prime} y^{\prime}} = (\sigma_1- \sigma_2) \times \sin \theta \times \cos \theta (ii)
Resultant stress of (i) and (ii),
\sigma_r=\sqrt{\left(\sigma_{x^{\prime}}\right)^2+\left(\tau_{x^{\prime} y^{\prime}}\right)^2}=\sqrt{\left(\sigma_1 \cos ^2 \theta+\sigma_2 \sin ^2 \theta\right)^2+\left(\left(\sigma_1-\sigma_2\right) \sin \theta \cos \theta\right)^2}
=\sqrt{\left(100\cos 35 ^\circ \right)^2 +\left(50\cos 35 ^\circ \right)^2}= 86.8 Mpa
