Question 2.18: Figure 2.28 shows the state of stress at a point. Determine ...
Figure 2.28 shows the state of stress at a point. Determine the principal normal and shear stresses and the orientation at which they act. Also determine the octahedral normal and shear stresses.
Stress tensor, \left[\tau _{ i j}\right]=\begin{vmatrix} {3} & {2} & {0} \\ {2} & {4} & {1} \\ {0} & {1} & {2} \end{vmatrix} MPa
Stress invariants,
I_1 =\tau_{x x}+\tau_{y y}+\tau_{z z}=3+4+2=9I_2 =\tau_{x x} \tau_{y y}+\tau_{y y} \tau_{z z}+\tau_{z z} \tau_{x x}-\tau_{x y}^2-\tau_{y z}^2-\tau_{z x}^2
=3 \times 4+4 \times 2+2 \times 3-2^2-1^2-0 =21
I_3=\begin{vmatrix} {\tau_{x x}} & {\tau_{x y}} & {\tau_{z x}} \\ {\tau_{x y}} & {\tau_{x y}} & {\tau_{z y}} \\ {\tau_{x z}} & {\tau_{y z}} & {\tau_{z z}} \end{vmatrix}= \begin{vmatrix} {3} & {2} & {0} \\ {2} & {4} & {1} \\ {0} & {1} & {2} \end{vmatrix}=13
Substituting in Eq. (2.13b),
\tau_{n n}^3-I_1 \tau_{n n}^2+I_2 \tau_{n n}-I_3=0or, \tau_{n n}^3-9 \tau_{n n}^2+21 \tau_{n n}-13=0
Solving by hit and trial, \tau_{n n}=1,2.27,5.73
Using Eqs (2.11) and (2.12), for direction cosines of the plane, when, principal normal stress \sigma_I=1 MPa,
\tau _{xy}\times c_{nx}+(\tau _{yy}-\tau _{nn})\times c_{ny}+\tau _{zy}\times c_{nz}=0 (2.11b)
\left(c_{n x}\right)^2+\left(c_{n y}\right)^2+\left(c_{n z}\right)^2=1 (2.12)
(3-1) \times c_{n x}+2 \times c_{n y}=0
2 \times c_{n x}+(4-1) \times c_{n y}+1 \times c_{n z}=0
1 \times c_{n y}+(2-1) \times c_{n z}=0
Solution of these equations gives,
c_{n x}=c_{n y}=-c_{n z}=\pm 1 / \sqrt{3}Again using Eqs (2.11) and (2.12), for direction cosines of the plane, when, principal normal stress \sigma_{II} = 2.27 MPa
\left(c_{n x}\right)^2+\left(c_{n y}\right)^2+\left(c_{n z}\right)^2=1
(3-2.27) \times c_{n x}+2 \times c_{n y}=0
2 \times c_{n x}+(4-2.27) \times c_{n y}+1 \times c_{n z}=0
1 \times c_{n y}+(2-2.27) \times c_{n z}=0
Solution of these equations gives,
-c_{n x}=\pm 0.5811,
c_{n y}=\pm 0.2121,
and c_{n z}=\pm 0.7855
Similarly, when principal normal stress \sigma_{III}= 5.73 MPa, the direction cosines are,
c_{n x}=\pm 0.5776,
c_{n y}=\pm 0.7884,
and, c_{n z}=\pm 0.2113
Normal octahedral stress, Eq. (2.14),
\tau_{n n, oct } =\frac{\tau _{xx}+\tau _{yy}+\tau _{zz}}{}=\frac{\sigma_I+\sigma_{I I}+\sigma_{I I I}}{3} (2.14)
\tau_{n n, oct } =\frac{\sigma_I+\sigma_{I I}+\sigma_{I I I}}{3}=\frac{1+2.27+5.73}{3} = 3 MPa
Octahedral shear stress is given by Eq. (2.18),
\tau _{n s,oct}=\frac{1}{3} \sqrt{(\sigma _{I}-\sigma _{II})^2 +(\sigma _{II}-\sigma _{III})^2+(\sigma _{III}-\sigma _{I})^2 }=\frac{1}{3} \sqrt{(1-2.27)^2+(2.27-5.73)^2 +(5.73-1)^2}=2 MPa
Principal shear stress and normal stress on the planes of principal shear stress are given by Eq. (2.20)
\tau _{I}=\frac{\sigma _{II}-\sigma _{III}}{2} \ \text{and} \ \sigma _n=\frac{\sigma _{II}+\sigma _{III}}{2} (2.20a)
\tau _{II}=\frac{\sigma _{III}-\sigma _{I}}{2} \ \text{and} \ \sigma _n=\frac{\sigma _{III}+\sigma _{I}}{2} (2.20b)
\tau _{III}=\frac{\sigma _{I}-\sigma _{II}}{2} \ \text{and} \ \sigma _n=\frac{\sigma _{I}+\sigma _{II}}{2} (2.20c)
\tau_I =\frac{2.27-1}{2}=\pm 0.635 MPa ,
\sigma_n=\frac{1+2.27}{2}=1.635 MPa
\tau_{I I}=\frac{5.73-2.27}{2}=\pm 1.73 MPa ,
\sigma_n=\frac{2.27+5.73}{2}=4 MPa
\tau_{III}=\frac{5.73-1}{2}=\pm 2.365 MPa ,
\sigma_n=\frac{5.73+1}{2}=3.365 MPa
The planes of principal shear stress bisect the planes of principal normal stress.