Chapter 2.13, Problem 122P

Bar AB has a cross-sectional area of 1200 mm² and is made of a steel that is assumed to be elastoplastic with E = 200 GPa and σ_Y = 250 MPa. Knowing that the force F increases from 0 to 520 kN and then decreases to zero, determine (a) the permanent deflection of point C, (b) the residual stress in the bar.

Fig. P2.122

To determine

The permanent deflection of point C.

Answer to Problem 122P

The permanent deflection of point C is $\underset{\_}{0.1042\text{mm}}$.

Explanation of Solution

Given information:

The cross sectional area A of section AB is $1,200{\text{mm}}^{\text{2}}$.

The modulus of elasticity E is $200\text{GPa}$.

The yield stress $\left({\sigma }_Y\right)$ is $250\text{MPa}$.

The force F is $520\text{kN}$.

Calculation:

Determine the force at yield portion AC using the relation:

$P_{AC}=A{\sigma }_Y$ (1)

Substitute $1,200{\text{mm}}^{\text{2}}$ for A and $250\text{MPa}$ for $\left({\sigma }_y\right)$ in Equation (1).

$\begin{array}{c}{P}_{AC}=1200{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\times 250\\ =1200\times {10}^{-6}\times 250\\ =300\times {10}^3\text{N}\end{array}$

Sketch the bar ACB as shown in Figure 1.

Find the load $P_{CB}$ using equilibrium as follows:

$\begin{array}{l}F+P_{CB}-P_{AC}=0\\ P_{CB}=P_{AC}-F\end{array}$ (2)

Substitute $300\times {10}^3\text{N}$ for $P_{AC}$ and $520\text{kN}$ for F in Equation (2).

$\begin{array}{c}{P}_{CB}=300\times {10}^3-520\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\\ =300,000-520\times {10}^3\\ =-220\times {10}^3\text{N}\end{array}$

Find the length $L_{CB}$ of the bar as follows:

Refer to Figure 1.

$\begin{array}{c}{L}_{CB}=440-120\\ =320\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =0.32\text{m}\end{array}$

Find the deflection at point C using the relation:

${\delta }_C=-\frac{P_{CB}{L}_{CB}}{EA}$ (3)

Here, $L_{CB}$ is length of the rod and $P_{CB}$ is force to yield portion CB.

Substitute $-220\times {10}^3\text{N}$ for $P_{CB}$ and $0.32\text{m}$ for $L_{CB}$, $200\text{GPa}$ for E, and $1,200{\text{mm}}^{\text{2}}$ for A in Equation (3).

$\begin{array}{c}{\delta }_C=-\frac{-220\times {10}^3\times 0.32}{200\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\times 1,200{\text{mm}}^{\text{2}}\times \frac{1\text{m}}{{10}^3\text{mm}}}\\ =\frac{70,400}{240,000,000}\\ =0.29333\times {10}^{-3}\text{m}\end{array}$

Find the stress in rod along CB using the relation:

${\sigma }_{CB}=\frac{P_{CB}}{A}$ (4)

Substitute $-220\times {10}^3\text{N}$ for $P_{CB}$ and $1,200{\text{mm}}^{\text{2}}$ for A in Equation (4).

$\begin{array}{c}{\sigma }_{CB}=\frac{-220\times {10}^3}{1,200}\\ =-183.33\times {10}^6\text{Pa}\end{array}$

Show the expression of deflection at point C for unloading to find the load $\left({P^{\prime }}_{AC}\right)$ using the relation:

$\begin{array}{l}{{\delta }^{\prime }}_C=\frac{{P^{\prime }}_{AC}{L}_{AC}}{EA}=\frac{{P^{\prime }}_{AC}{L}_{AC}}{EA}\\ {P^{\prime }}_{AC}\left(\frac{L_{AC}}{EA}+\frac{L_{BC}}{EA}\right)=\frac{F{L}_{CB}}{EA}\\ {P^{\prime }}_{AC}=\frac{F{L}_{CB}}{L_{AC}+L_{CB}}\end{array}$ (5)

Here, $L_{AC}$ is length of bar AC.

Substitute $440\text{mm}$ for $L_{AC}+L_{CB}$, $520\text{kN}$ for F, $0.32\text{m}$ for $L_{CB}$ in Equation (5).

$\begin{array}{c}{P^{\prime }}_{AC}=\frac{520\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\left(0.32\right)}{440\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =\frac{520\times {10}^3\times 0.32}{0.440}\\ =378.18\times {10}^3\text{N}\end{array}$

Find the load $\left({P^{\prime }}_{CB}\right)$ along CB using the relation:

${P^{\prime }}_{CB}={P^{\prime }}_{AC}-F$ (6)

Substitute $378.18\times {10}^3\text{N}$ for ${P^{\prime }}_{AC}$ and $520\text{kN}$ for F in Equation (6).

$\begin{array}{c}{P^{\prime }}_{CB}=378.18\times {10}^3-520\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\\ =-141.820\times {10}^3\text{N}\end{array}$

Calculate the stress at point along AC using the relation:

${{\sigma }^{\prime }}_{AC}=\frac{{P^{\prime }}_{AC}}{A}$ (7)

Substitute $378.18\times {10}^3\text{N}$ for ${P^{\prime }}_{AC}$ and $1200{\text{mm}}^{\text{2}}$ for A in Equation (7).

$\begin{array}{c}{{\sigma }^{\prime }}_{AC}=\frac{378.18\times {10}^3}{1200{\text{mm}}^2\times {\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2}\\ =\frac{378.18\times {10}^3}{1200\times {10}^{-6}}\\ =315.150\times {10}^6\text{Pa}\end{array}$

Calculate the stress at point along BC using the relation:

${{\sigma }^{\prime }}_{BC}=\frac{{P^{\prime }}_{BC}}{A}$ (8)

Substitute $-141.820\times {10}^3\text{N}$ for ${P^{\prime }}_{BC}$ and $1200{\text{mm}}^{\text{2}}$ for A in Equation (8).

$\begin{array}{c}{{\sigma }^{\prime }}_{BC}=-\frac{141.820\times {10}^3}{1200{\text{mm}}^2\times {\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2}\\ =-\frac{141.820\times {10}^3}{1200\times {10}^{-6}}\\ =-118.183\times {10}^6\text{Pa}\end{array}$

Determine the deflection at point C using the relation:

${{\delta }^{\prime }}_C=\frac{{P^{\prime }}_{AC}\times a}{EA}$ (9)

Substitute $378.18\times {10}^3\text{N}$ for ${P^{\prime }}_{AC}$, $1200{\text{mm}}^{\text{2}}$ for A, $0.120\text{m}$ for a. $200\text{GPa}$ for E in Equation (9).

$\begin{array}{c}{{\delta }^{\prime }}_C=\frac{{P^{\prime }}_{AC}\times a}{EA}\\ =\frac{378.18\times {10}^3\times 0.120}{\left(200\times {10}^9\times 1200\times {10}^{-6}\right)}\\ =0.189090\times {10}^{-3}\text{m}\end{array}$

Determine the permanent deflection at point C using the relation:

${\delta }_{C,P}={\delta }_C-{{\delta }^{\prime }}_C$ (10)

Substitute $0.189090\times {10}^{-3}\text{m}$ for ${{\delta }^{\prime }}_C$ and $0.29333\times {10}^{-3}$ for ${\delta }_C$ in Equation (10).

$\begin{array}{c}{\delta }_{C,P}=0.29333\times {10}^{-3}-0.189090\times {10}^{-3}\\ =0.104240\times {10}^{-3}\text{m}\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ =0.1042\text{mm}\end{array}$

Thus, the permanent deflection of point C is $\underset{\_}{0.1042\text{mm}}$.

To determine

Find the residual stress in bar AC and CB.

Explanation of Solution

The residual stress in bar AC is $\underset{\_}{-65.2\text{MPa}}$

The residual stress in bar CB is $\underset{\_}{-65.2\text{MPa}}$

Calculation:

Find the residual stress in bar AC using the relation:

${\sigma }_{AC,\text{res}}={\sigma }_Y-{{\sigma }^{\prime }}_{AC}$ (11)

Substitute $250\text{MPa}$ for ${\sigma }_Y$ and $315.150\times {10}^6\text{Pa}$ for ${{\sigma }^{\prime }}_{AC}$ in Equation (11).

$\begin{array}{c}{\sigma }_{AC,\text{res}}=250\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)-315.150\times {10}^6\text{Pa}\\ \text{=250}\times {\text{10}}^6-315.150\times {10}^6\\ =-65.150\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =-65.2\text{MPa}\end{array}$

Thus, the residual stress in bar AC is $\underset{\_}{-65.2\text{MPa}}$

Find the residual stress in bar BC using the relation:

${\sigma }_{BC,\text{res}}={\sigma }_{CB}-{{\sigma }^{\prime }}_{BC}$ (12)

Substitute $-183.33\times {10}^6\text{Pa}$ for ${\sigma }_{CB}$ and $118.183\times {10}^6\text{Pa}$ for ${{\sigma }^{\prime }}_{CB}$ in Equation (12).

$\begin{array}{c}{\sigma }_{BC,\text{res}}=-183.33\times {10}^6+118.183\times {10}^6\\ =-65.150\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =-65.2\text{MPa}\end{array}$

Thus, the residual stress in bar CB is $\underset{\_}{-65.2\text{MPa}}$