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## Q. 9.18

A point on a machine component has the following state of plane strain:

$∈_{x x}=-720 \mu, \quad ∈_{y y}=-400 \mu, \quad \gamma_{x y}=+660 \mu$

If E = 700 GPa and G = 28 GPa, calculate the principal planes and the principal stresses (a) by determining the corresponding state of plane stress and then using Mohr’s circle for stress, (b) by using Mohr’s circle for strain to determine the orientation and magnitude of the principal strains and then determine the corresponding stresses.

## Verified Solution

(a) We know from generalised Hooke’s law [refer to Eq. (9.140)]:

\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)

\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{v}{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}

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

$\sigma_{x x}=2 G ∈_{x x}+2 G\left(\frac{ν}{1+2 ν}\right)\left(∈_{x x}+\epsilon_{y y}\right)$

$=2 G\left[∈_{x x}+\frac{ν}{1-2 ν}\left(∈_{x x}+\epsilon_{y y}\right)\right]$

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

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

Similarly,

$\sigma_{y y}=2 G\left[∈_{y y}+\frac{ν}{1-2 ν}\left(∈_{x x}+\epsilon_{y y}\right)\right]$

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

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

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

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:

$\sigma_{\max }=(-62.72+20.54) MPa =-42.18 MPa$ ⦨ 57.93°

$\sigma_{\min }=(-62.72-20.54) MPa =-83.26 MPa$

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 $∈_1,∈_2, ∈_3\left(∈_1>∈_2>∈_3\right)$ 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 Gν}{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,

$\sigma_2=2 G\left[∈_2+\frac{ν}{1-2 ν}\left(∈_1+∈_2+∈_3\right)\right]$

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

or          $\sigma_2=-42.18 MPa$

and          $\sigma_3=2 G\left[∈_3+\left\lgroup \frac{ν}{1-2 ν} \right\rgroup\left(∈_1+∈_2+∈_3\right)\right]$

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

or      $\sigma_3=-83.26 MPa$

Thus, the ordered principal stresses are:

$\sigma_1=0 MPa$

$\sigma_2=-42.18 MPa$

and          $\sigma_3=-83.26 MPa$

As we note, both methods (a) and (b) give us the same results.