Chapter 2.13, Problem 123P

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

*2.123 Solve Prob. 2.122, assuming that a = 180 mm.

To determine

The permanent deflection of point C.

Answer to Problem 123P

The permanent deflection of point C is $\underset{\_}{0.00788\text{mm}}$.

Explanation of Solution

Given information:

The distance between member AC (a) is $180\text{mm}$.

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 to 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.

$\begin{array}{c}{L}_{CB}=440-180\\ =260\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =0.26\text{m}\end{array}$

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 $0.26\text{m}$ for $L_{CB}$, $200\text{GPa}$ for E, and $1,200{\text{mm}}^{\text{2}}$ for A in Equation (3).

$\begin{array}{c}{\delta }_C=-\frac{-220\times {10}^3\times 0.26}{200\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\times 1,200{\text{mm}}^{\text{2}}\times \frac{1\text{m}}{{10}^3\text{mm}}}\\ =\frac{57,200}{240,000,000}\\ =0.23833\times {10}^{-3}\text{m}\end{array}$

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).

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

Calculate the load $\left({P^{\prime }}_{AC}\right)$ for unloading 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, $0.26\text{m}$ for $L_{CB}$ in Equation (5).

$\begin{array}{c}{P^{\prime }}_{AC}=\frac{520\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\left(0.26\right)}{440\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =\frac{135,200}{0.440}\\ =307.27\times {10}^3\text{N}\end{array}$

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

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

Substitute $307.27\times {10}^3\text{N}$ for ${P^{\prime }}_{AC}$ and $520\text{kN}$ for F in Equation (6).

$\begin{array}{c}{P^{\prime }}_{CB}=307.27\times {10}^3-520\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\\ =-212.73\times {10}^3\text{N}\end{array}$

Determine the deflection at point C $\left({{\delta }^{\prime }}_C\right)$ using the relation:

${{\delta }^{\prime }}_C=\frac{{P^{\prime }}_{AC}\times a}{EA}$ (7)

Substitute $307.27\times {10}^3\text{N}$ for ${P^{\prime }}_{AC}$, $1200{\text{mm}}^{\text{2}}$ for A, $0.180\text{m}$ for a. $200\text{GPa}$ for E in Equation (7).

$\begin{array}{c}{{\delta }^{\prime }}_C=\frac{{P^{\prime }}_{AC}\times a}{EA}\\ =\frac{307.27\times {10}^3\times 0.180}{\left(200\times {10}^9\times 1200\times {10}^{-6}\right)}\\ =0.23045\times {10}^{-3}\text{m}\end{array}$

Calculate the stress at along AC using the relation:

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

Substitute $307.27\times {10}^3\text{N}$ for ${P^{\prime }}_{AC}$ and $1200{\text{mm}}^{\text{2}}$ for A in Equation (8).

$\begin{array}{c}{{\sigma }^{\prime }}_{AC}=\frac{307.27\times {10}^3}{1200{\text{mm}}^2\times {\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2}\\ =\frac{307.27\times {10}^3}{1200\times {10}^{-6}}\\ =256.058\times {10}^6\text{Pa}\end{array}$

Calculate the stress along BC using the relation:

${{\sigma }^{\prime }}_{BC}=\frac{{P^{\prime }}_{BC}}{A}$ (9)

Substitute $-212.73\times {10}^3\text{N}$ for ${P^{\prime }}_{BC}$ and $1200{\text{mm}}^{\text{2}}$ for A in Equation (9).

$\begin{array}{c}{{\sigma }^{\prime }}_{BC}=\frac{-212.73\times {10}^3}{1200{\text{mm}}^2\times {\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2}\\ =\frac{-212.73\times {10}^3}{1200\times {10}^{-6}}\\ =-177.275\times {10}^6\text{Pa}\end{array}$

Determine the permanent deflection at point C using the relation:

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

Substitute $0.23045\times {10}^{-3}\text{m}$ for ${{\delta }^{\prime }}_C$ and $0.2383\times {10}^{-3}$ for ${\delta }_C$ in Equation (10).

$\begin{array}{c}{\delta }_{C,P}=0.2383\times {10}^{-3}-0.23045\times {10}^{-3}\text{m}\\ =0.00788\times {10}^{-3}\text{m}\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ =0.00788\text{mm}\end{array}$

Thus, the permanent deflection of point C is $\underset{\_}{0.00788\text{mm}}$.

To determine

Find the residual stress in bar AC and CB.

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

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

Explanation of Solution

Calculation:

Find the residual stress in bar AC using the relation:

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

Substitute $250\text{MPa}$ for ${\sigma }_{AC}$ and $256.058\times {10}^6\text{Pa}$ for ${{\sigma }^{\prime }}_{AC}$ in Equation (11).

$\begin{array}{c}{\sigma }_{AC,\text{res}}=250\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)-256.058\times {10}^6\\ \text{=250}\times {\text{10}}^6-256.058\times {10}^6\text{Pa}\\ =-6.0580\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =-6.06\text{MPa}\end{array}$

Thus, the residual stress in bar AC is $\underset{\_}{-6.06\text{MPa}}$

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 $177.275\times {10}^6\text{Pa}$ for ${{\sigma }^{\prime }}_{CB}$ in Equation (12).

$\begin{array}{c}{\sigma }_{BC,\text{res}}=-183.33\times {10}^6+177.275\times {10}^6\\ =-6.0580\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =-6.06\text{MPa}\end{array}$

Thus, the residual stress in bar CB is $\underset{\_}{-6.06\text{MPa}}$