A plane element is subjected to stress in x — y plane as follows,

$\sigma _x$ = 40 MPa (Compression),

$\sigma _y$ = 80 MPa (Tension),

and, $\tau _{x y}$ = 60 MPa (Clockwise).

Determine the stresses at a plane having its normal at 45° anticlockwise to the direction of the tensile stress 80 MPa. Also determine principal stresses and maximum shear stress.

Question

A plane element is subjected to stress in x — y plane as follows,

[latex]\sigma _x[/latex]= 40 MPa (Compression),

[latex]\sigma _y[/latex] = 80 MPa (Tension),

and, [latex] au _{x y}[/latex]= 60 MPa (Clockwise).

Determine the stresses at a plane having its normal at 45° anticlockwise to the direction of the tensile stress 80 MPa. Also determine principal stresses and maximum shear stress.

Accepted Answer

Rotation angle of x- y plane (Fig. 2.13),

[latex]2 heta = 2 imes 135[/latex] = 270° (anticlockwise)

Normal stress and shear stress on x'- y 'plane are given by Eqs. (2.1b) and (2.2),

[latex]\sigma_{\acute{x} } =\frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2} \cos 2 heta- au_{x y} \sin 2 heta [/latex]

[latex]=\frac{-40+80}{2}+\frac{-40-80}{2} \cos 270^{\circ}-60 [/latex]

[latex] \sin 270^{\circ}=80 MPa [/latex]

[latex] au {\acute{_x} \acute{_y} } =\frac{\sigma_x-\sigma_y}{2} \sin 2 heta+ au_{x y} \cos 2 heta[/latex]

[latex]= \frac{-40-80}{2} \sin 270^{\circ}+60 \cos 270^{\circ}= 60 MPa[/latex]

Principal plane angle (Eq. (2.3),
[latex] an 2 heta_p=-\frac{ au_{x y}}{\left(\sigma_x-\sigma_y ight) / 2}=\frac{60}{60}=1[/latex]

or, [latex]2 heta_p=45^{\circ}, 225^{\circ}[/latex] (anticlockwise)

Principal stresses are given by Eq. (2.4),

[latex]\sigma_{\acute{x} \max / \min } =\frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2} ight)^2+ au_{x y}^2} [/latex]

[latex]=\frac{-40+80}{2} \pm \sqrt{\left(\frac{-40-80}{2} ight)^2+60^2}[/latex]

= 104.85 MPa (Tension),

- 64.85 MPa (Compression)

Maximum shear stress by Eq. (2.6),

[latex] au_{\acute{_x} \acute{_y}, \max } =\pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2} ight)^2+ au_{x y}^2} [/latex]
[latex]=\pm \sqrt{\left(\frac{-40-80}{2} ight)^2+60^2} =+84.85 MPa[/latex]

Normal stress at the plane of maximum shear stress (Eq. (2.7)),

[latex]\sigma_{\acute{x}}=\frac{\sigma_x+\sigma_y}{2}=\frac{-40+80}{2}=20 MPa[/latex]

Maximum shear plane angle,
[latex]2 heta_s=45^{\circ}+90^{\circ}=135^{\circ}[/latex]

Question 2.3: A plane element is subjected to stress in x — y plane as fol...

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