Question 2.22: Figure 2.30(a) shows an element under axial stress 20 MPa an...
Figure 2.30(a) shows an element under axial stress 20 MPa and 50 MPa along with shear stress 40 MPa in x-y plane. Determine the state of stress in x^{\prime} – y^{\prime} plane at 30° counter-clockwise to x-y plane.
Equation (2.11b) can be used to cal-culate stresses in \acute{x}-y^{\prime} plane and Eq. (2.13a) will give principal stresses. The directive cosines in transfer matrix are drived from Fig. 2.30(b) as follows.
\tau_{x y} \times c_{n x}+\left(\tau_{y y}-\tau_{n n}\right) \times c_{n y}+\tau_{z y} \times c_{n z}=0 (2.11b)
\begin{vmatrix} {\left(\tau _{x x} – \tau _{n n}\right)} & {\tau _{y x}} & {\tau _{z x}} \\ {\tau _{x y}} & {\left(\tau _{y y} – \tau _{n n}\right)} & {\tau _{z y}} \\ {\tau _{x z }} & {\tau _{z y}} & {\left(\tau _{z z} – \tau _{n n}\right)} \end{vmatrix}=0 (2.13a)
c_{x^{\prime} x}=\cos \thetac_{x^{\prime} y}=\cos \left(90^{\circ}+\theta\right)=-\sin \theta
c_{y^{\prime} x}=\cos \left(90^{\circ}-\theta\right)=\sin \theta
c_{y^{\prime} y}=\cos \theta
Substitute these values in Eq. (2.10b) for stresses in x^{\prime} – y^{\prime} reference axes as follows.
\tau_{n n}= \tau_{x x} \times\left(c_{n x}\right)^2+\tau_{y y} \times\left(c_{n y}\right)^2+\tau_{z z} \times\left(c_{n z}\right)^2 +2 \tau_{x y}\left(c_{n x} c_{n y}\right)+2 \tau_{x z}\left(c_{n x} c_{n z}\right)+2 \tau_{y z}\left(c_{n y} c_{n z}\right) (2.10b)
\begin{vmatrix} {\sigma_{x^{\prime}}}& {\tau _{x^{\prime}y^{\prime}}} \\ {\tau _{y^{\prime} x^{\prime}}} & {\sigma_{y^{\prime}y^{\prime}}} \end{vmatrix} =\begin{vmatrix} {c_{x^{\prime}x}} & {c _{x^{\prime}y}} \\ {c_{y^{\prime} x}} & {c_{y^{\prime} y}} \end{vmatrix} \begin{vmatrix} {\sigma _{x x} } & {\tau _{x y}} \\ {\tau _{y x}} & {\sigma _{y y}} \end{vmatrix} \begin{vmatrix} {c_{x^{\prime}x}} & {c_{y^{\prime} x}} \\ {c_{x^{\prime}x}} & {c_{y^{\prime} y}} \end{vmatrix}
=\begin{vmatrix} {0.866 } & {-0.5} \\ {0.5} & {0.866} \end{vmatrix} \begin{vmatrix} {20 } & {40} \\ {40} & {-50} \end{vmatrix} \begin{vmatrix} {0.866 } & {0.5} \\ {-0.5} & {0.866} \end{vmatrix} =\begin{vmatrix} {-32.14 } & {50.3} \\ {50.3} & {2.14} \end{vmatrix}
Principal stresses are given by Eq.(2.13a),
=\begin{vmatrix} {-32.14-\sigma } & {50.3} \\ {50.3} & {2.14 – \sigma } \end{vmatrix}=0which gives principal stresses
\sigma= 38.14 MPa and -68.14 MPa