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.

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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 [latex]\sigma_r [/latex] = 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 = [latex]\sigma_1 [/latex] cos 35° = 100 cos 35°

CD = [latex]\sigma_2 [/latex] sin 35° = 50 sin 35°

From Eqs (2.1a) and (2.2),

[latex]\sigma_{x^{\prime}}=\sigma _x \cos^2 heta +\sigma _y \sin^2 heta- au _{xy}\sin 2 heta [/latex] (2.1a)

[latex] au_{x^{\prime}y^{\prime}}=\frac{\sigma _x-\sigma _y}{2}\sin 2 heta + au _{xy}\cos 2 heta [/latex] (2.2)

[latex]\sigma_{x^{\prime}}[/latex] = [latex]\sigma_1 \cos^2 heta + \sigma_2 \sin^2 heta [/latex] (i)

and [latex] au _{x^{\prime} y^{\prime}} = (\sigma_1- \sigma_2) imes \sin heta imes \cos heta [/latex] (ii)

Resultant stress of (i) and (ii),

[latex]\sigma_r=\sqrt{\left(\sigma_{x^{\prime}} ight)^2+\left( au_{x^{\prime} y^{\prime}} ight)^2} [/latex]

[latex]=\sqrt{\left(\sigma_1 \cos ^2 heta+\sigma_2 \sin ^2 heta ight)^2+\left(\left(\sigma_1-\sigma_2 ight) \sin heta \cos heta ight)^2} [/latex]

[latex]=\sqrt{\left(100\cos 35 ^\circ ight)^2 +\left(50\cos 35 ^\circ ight)^2}=[/latex] 86.8 Mpa

Question 2.20: Tensile stresses of 100 MPa and 50 MPa are acting on two per...

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