Two-dimensional stresses at a point on a plane P are 67.2 MPa (tension) and 84.64 MPa (shear, clockwise) and on another plane Q the stresses are 61.35 MPa (compression) and 68.57 MPa (shear, clockwise). Find the stresses on a plane R inclined at 20° clockwise to Q.

Question

Accepted Answer

Express the stresses on plane P in terms of Eqs (2.1) and (2.2).

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

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

[latex]67.2= \frac{-61.35+\sigma_y}{2}+\frac{-61.35-\sigma_y}{2} \cos 2 heta [/latex]

[latex]-68.57 \sin 2 heta [/latex] (i)

[latex]84.64= \frac{-61.35-\sigma_y}{2} \sin 2 heta+68.57 \cos 2 heta [/latex] (ii)

where [latex]\sigma_y[/latex] = normal stress on the plane perpendicular to Q,

[latex] heta [/latex] = angle between the planes P and Q.

From Eq. (i) and (ii),

[latex] an heta=-\frac{128.55}{153.21} [/latex]

or, [latex] heta \approx-40^{\circ} [/latex], i.e. plane P is at [latex]40^{\circ}[/latex] clockwise to plane Q.

Substituting in Eq. (iii), [latex]\sigma_y=86.36 MPa.[/latex]

Angle of inclination of plane R w.r.t. Q,

[latex]2 heta=2 imes\left(-20^{\circ} ight)=-40^{\circ} [/latex]

Normal stress on plane R,

[latex]\sigma_{\acute{x} }= \frac{-61.35+86.36}{2}-\frac{61.35+86.36}{2} imes \cos \left(-40^{\circ} ight)-68.57 \sin \left(-40^{\circ} ight) \simeq 0 [/latex]

[latex]\cos \left(-40^{\circ} ight)-68.57 \sin \left(-40^{\circ} ight) \simeq 0[/latex]

[latex] au_{\acute{x} \acute{y}}= \frac{-61.35-86.36}{2} \sin \left(-40^{\circ} ight)+68.57 \cos \left(-40^{\circ} ight)=100 MPa [/latex]

Question 2.7: Two-dimensional stresses at a point on a plane P are 67.2 MP...

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