Question 9.18: A point on a machine component has the following state of pl...

\begin{aligned} \sigma_{x x} & =\left\lgroup \frac{E}{1+ν} \right\rgroup ∈_{x x}+\frac{ν}{1-2 ν} \cdot \frac{E}{1+ν}\left(∈_{x x}+∈_{y y}+∈_{z z}\right) \\ \sigma_{y y} & =\left\lgroup \frac{E}{1+ν} \right\rgroup ∈_{y y}+\frac{ν}{1-2 ν} \cdot \frac{E}{1+ν}\left(∈_{x x}+∈_{y y}+∈_{z z}\right) \\ \sigma_{z z} & =\left\lgroup \frac{E}{1+ν} \right\rgroup ∈_{z z}+\frac{v}{1-2 ν} \cdot \frac{E}{1+ν}\left(∈_{x x}+∈_{y y}+∈_{z z}\right) \\ \tau_{x y} & =G \gamma_{x y}, \quad \tau_{y z}=G \gamma_{y z}, \quad \tau_{z x}=G \gamma_{z x} \end{aligned}

\begin{aligned} \sigma_{x x} & =\frac{E \varepsilon_{x x}}{(1+ν)}+\frac{E ν}{(1+ν)(1-2 ν)}\left(\varepsilon_{x x}+\varepsilon_{y y}+\varepsilon_{z z}\right) \\ \sigma_{y y} & =\frac{E ∈_{x x}}{(1+ν)}+\frac{E ν}{(1+ν)(1-2 ν)}\left(∈_{x x}+∈_{y y}+∈_{z z}\right) \\ \sigma_{z z} & =\frac{E ∈_{z z}}{(1+ν)}+\frac{E ν}{(1+ν)(1-2 ν)}\left(∈_{x x}+∈_{y y}+∈_{z z}\right) \\ \tau_{x y} & =G \gamma_{x y}, \quad \tau_{y z}=G \gamma_{y z}, \quad \tau_{z x}=G \gamma_{z x} \end{aligned}              (9.140)

with G = E/2(1 + ν) and ν as Poisson’s ratio. For our case

ν=\left\lgroup \frac{E}{2 G} \right\rgroup -1=\frac{70}{(2)(28)}-1=0.25

Now considering ∈_{z z}=\gamma_{z x}=\gamma_{z y}=0 in the above equation, we get

\begin{aligned} \sigma_{x x} & =2 G ∈_{x x}+2 G\left\lgroup \frac{ν}{1+2 ν} \right\rgroup\left(∈_{x x}+∈_{y y}\right) \\ & =2 G\left[∈_{x x}+\frac{ν}{1-2 ν}\left(∈_{x x}+∈_{y y}\right)\right] \end{aligned}

=(2)(28)\left(10^3\right)\left[(-720)+\frac{0.25}{1-(2)(0.25)}(-720-400)\right]\left(10^{-6}\right)

or        \sigma_{x x}=-71.68  MPa          (1)

Similarly,

\begin{aligned} \sigma_{y y} & =2 G\left[∈_{y y}+\frac{ν}{1-2 ν}\left(∈_{x x}+∈_{y y}\right)\right] \\ & =(2)(28)\left(10^{-3}\right)\left[-400+\frac{0.25}{1-2(0.25)}(-720-400)\right]\left(10^{-6}\right) \end{aligned}

or          \sigma_{y y}=-53.76  MPa             (2)

and          \sigma_{z z}=26\left(\frac{ν}{1-2 ν}\right)\left(∈_{x x}+∈_{y y}\right)=(2)(28)\left(10^3\right) \frac{0.25}{1-2(0.25)}(-720-400)\left(10^{-6}\right)

or        \sigma_{z z}=-31.36  MPa             (3)

Also,    \tau_{x y}=(28)\left(10^3\right)(660)\left(10^{-6}\right)=18.48  MPa              (4)

\tau_{y z}=\tau_{z x}=0

Clearly, \sigma_{z z}=-31.36 MPa is one of the principal stresses (out-of-plane principal stress). To determine the two in-plane principal stresses, we need to draw Mohr’s circle for stresses where we will assume tensorial sign convention has been followed. Accordingly, we draw the following Mohr’s circle of stresses in Figure 9.50.

From the above figure

\tan 2 \theta=\frac{(-18.48)}{71.68-62.72} \Rightarrow \theta=57.93^{\circ} ⦨

which locates the principal planes of stresses. The principal stresses are given by:

\begin{aligned} & \sigma_{\max }=(-62.72+20.54)  MPa =-42.18  MPa  ⦨  57.93^{\circ} \\ & \sigma_{\min }=(-62.72-20.54)  MPa =-83.26  MPa \end{aligned}

The ordered principal stresses are:

\sigma_1=-31.36  MPa , \sigma_2=-42.18  MPa , \sigma_3=-83.26  MPa

(b) As we followed tensorial notation, we draw the following Mohr’s circle of strain as shown in Figure 9.51:

Here,

\tan 2 \phi=\frac{-330}{720-560} \Rightarrow \phi=57.93^{\circ} ⦨

which locates the principal planes of strains. The in-plane principal strains are:

\begin{aligned} & ∈_{\max }=(-560+366.74) \mu=-193.26 \mu \\ & ∈_{\min }=(-560-366.74) \mu=-926.74 \mu \end{aligned}

Thus, the ordered principal strains ∈_1, ∈_2, ∈_3\left(∈_1>∈_2>∈_3\right) \text { are: }

0, -193.26μ, -926.74μ

The corresponding principal stresses \sigma_1, \sigma_2, \sigma_3\left(\sigma_1>\sigma_2>\sigma_3\right) are given by:

\sigma_1=2 G\left[∈_1+\frac{ν}{1-2 ν}\left(∈_2+∈_3+∈_1\right)\right]           [refer to Eq. (9.139)]

\begin{aligned} & \sigma_{x x}=2 G ∈_{x x}+\frac{2 G ν}{1-2 ν}\left(∈_{x x}+∈_{y y}+∈_{z z}\right) \\ & \sigma_{y y}=2 G ∈_{y y}+\frac{2 G ν}{1-2 ν}\left(∈_{x x}+∈_{y y}+∈_{z z}\right) \\ & \sigma_{z z}=2 G ∈_{z z}+\frac{2 G ν}{1-2 ν}\left(∈_{x x}+∈_{y y}+∈_{z z}\right) \\ & \tau_{x y}=G \gamma_{x y}, \quad \tau_{y z}=G \gamma_{y z}, \quad \tau_{z x}=G \gamma_{z x} \end{aligned}            (9.139)

=(2)(28)\left(10^3\right)\left[0+\frac{0.25}{1-2(0.25)}(-193.26-926.74+0)\right]\left(10^{-6}\right)  MPa

or        \sigma_1=-31.36  MPa

Similarly,

\begin{aligned} \sigma_2 & =2 G\left[∈_2+\frac{ν}{1-2 ν}\left(∈_1+∈_2+∈_3\right)\right] \\ & =(2)(28)\left(10^3\right)\left[-193.26+\frac{0.25}{1-2(0.25)}(0-926.74-193.26)\right]\left(10^{-6}\right) \end{aligned}

or          \sigma_2=-42.18  MPa

and

  \begin{aligned} \sigma_3 & =2 G\left[∈_3+\left(\frac{ν}{1-2 ν}\right)\left(∈_1+∈_2+∈_3\right)\right] \\ & =(2)(28)\left(10^3\right)\left[-926.74+\frac{0.25}{1-2(0.25)}(0-193.26-926.74)\right]\left(10^{-6}\right) \end{aligned}