Chapter 2.13, Problem 107P

(a)

To determine

Find the maximum deflection at point C of the cylindrical portions.

Answer to Problem 107P

The maximum deflection at point C of the cylindrical portions is $\underset{\_}{0.292\text{mm}}$.

Explanation of Solution

Given information:

The cross sectional area AC and BC of each portions is $1,750{\text{mm}}^{\text{2}}$.

The modulus of elasticity (E) for portion AC is $200\text{GPa}$

The yield stress $\left({\sigma }_y\right)$ for portion AC is $250\text{MPa}$

The modulus of elasticity (E) for portion CB is $200\text{GPa}$

The yield stress $\left({\sigma }_y\right)$ for portion CB is $345\text{MPa}$

Calculation:

Calculate the displacement at point C to cause yielding of AC using the relation:

$\begin{array}{c}{\delta }_{C,Y}=L_{AC}{\epsilon }_{Y,AC}\\ =\frac{L_{AC}{\sigma }_{Y,AC}}{E}\end{array}$ (1)

Here, $L_{AC}$ is length of portion AC and ${\epsilon }_{Y,AC}$ is strain yielding of AC.

Substitute $190\text{mm}$ for $L_{AC}$ and $250\text{MPa}$ for ${\sigma }_y$, and $200\text{GPa}$ for E in Equation (1).

$\begin{array}{c}{\delta }_{C,Y}=\frac{190\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\times 250\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)}{200\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)}\\ =\frac{0.190\times 250\times {10}^6}{200\times {10}^9}\\ =0.2375\times {10}^{-3}\text{m}\end{array}$

Find the corresponding force along AC using the relation as follows:

$F_{AC}=A{\sigma }_Y{,}_{AC}$ (2)

Substitute $1,750{\text{mm}}^{\text{2}}$ for A and $250\text{MPa}$ for ${\sigma }_Y{,}_{AC}$ in Equation (2).

$\begin{array}{c}{F}_{AC}=1,750{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\times 250\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\\ =1750\times {10}^{-6}\times 250\times {10}^6\\ =437.5\times {10}^3\text{N}\end{array}$

Find the corresponding force along CB using the relation as follows:

$F_{CB}=-\frac{EA{\delta }_c}{L_{CB}}$ (3)

Substitute $1,750{\text{mm}}^{\text{2}}$ for A, $190\text{mm}$ for $L_{CB}$ , $0.2375\times {10}^{-3}\text{m}$ for ${\delta }_c$, $200\text{GPa}$ for E in Equation (3).

$\begin{array}{c}{F}_{CB}=-\frac{200\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\times 1,750{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\times 0.2375\times {10}^{-3}\text{m}}{190\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =-\frac{200\times {10}^9\times 1750\times {10}^{-6}\times 0.2375\times {10}^{-3}}{0.190}\\ =-437.5\times {10}^3\text{N}\end{array}$

Sketch the element C as shown in Figure 1.

Refer to Figure 1.

Find the value of P using equilibrium element.

$\begin{array}{l}{F}_{AC}-\left(F_{CB}+P_Y\right)=0\\ P_Y=F_{AC}-F_{CB}\end{array}$ (4)

Substitute $-437.5\times {10}^3\text{N}$ for $F_{CB}$ and $437.5\times {10}^3\text{N}$ for $F_{AC}$ in Equation (4).

$\begin{array}{c}{P}_Y=437.5\times {10}^3-\left(-437.5\times {10}^3\right)\\ =875\times {10}^3\text{N}\end{array}$

Since the applied load,

$P=975\times {10}^3\text{N>875}\times {\text{10}}^3\text{N}$ Portion AC yields.

Refer to Figure 1.

Find the force along CB as follows:

$F_{CB}=F_{AC}-P$

Substitute $437.5\times {10}^3\text{N}$ for $F_{AC}$ and $975\times {10}^3\text{N}$ for P.

$\begin{array}{c}{F}_{CB}=437.5\times {10}^3-975\times {10}^3\text{N}\\ =-537.5\times {10}^3\text{N}\end{array}$

Determine the deflection at point C using the relation:

${\delta }_C=-\frac{F_{CB}{L}_{CD}}{EA}$ (5)

Substitute $-537.5\times {10}^3\text{N}$ for $F_{CB}$, $190\text{mm}$ for $L_{CD}$, $200\text{GPa}$ for E, and $1,750{\text{mm}}^{\text{2}}$ for A in Equation (5).

$\begin{array}{c}{\delta }_C=-\frac{-537.5\times {10}^3\times 190\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{200\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\times 1750{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2}\\ =\frac{537.5\times {10}^3\times 0.190}{200\times {10}^9\times 1750\times {10}^{-6}}\\ =0.29179\times {10}^{-3}\text{m}\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ =0.292\text{mm}\end{array}$

Thus, the maximum deflection at point C of the cylindrical portions is $\underset{\_}{0.292\text{mm}}$.

(b)

To determine

Find the maximum stress for each portion of rod.

Answer to Problem 107P

The maximum stress of rod AC is $\underset{\_}{250\text{MPa}}$.

The maximum stress of rod BC is $\underset{\_}{-307\text{MPa}}$.

Explanation of Solution

Calculation:

Refer part a.

The maximum stress of rod AC is $250\text{MPa}$

Therefore, the maximum stress of rod AC is $\underset{\_}{250\text{MPa}}$.

Determine the maximum stress at point BC using the relation:

${\sigma }_{BC}=\frac{F_{BC}}{A}$ (6)

Substitute $-537.5\times {10}^3\text{N}$ for $F_{CB}$ and $1,750{\text{mm}}^{\text{2}}$ for A in Equation (6).

$\begin{array}{c}{\sigma }_{BC}=\frac{-537.5\times {10}^3}{1750{\text{mm}}^2{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2}\\ =-307.14\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =-307\text{MPa}\end{array}$

Thus, the maximum stress of rod BC is $\underset{\_}{-307\text{MPa}}$.

(c)

To determine

Find the permanent deflection at point C.

Answer to Problem 107P

The permanent deflection at point C is $\underset{\_}{0.0272\text{mm}}$.

Explanation of Solution

Write the expression of deflection and force for unloading as follows:

$\begin{array}{c}{\delta }^{\prime }=\frac{{P^{\prime }}_{AC}{L}_{AC}}{EA}\\ =-\frac{{P^{\prime }}_{CB}{L}_{CB}}{EA}\end{array}$

$\begin{array}{c}{P^{\prime }}_{CB}=-{P^{\prime }}_{AC}\frac{L_{AC}}{L_{CB}}\\ =-{P^{\prime }}_{AC}\end{array}$

The value of $P^{\prime }$ is $975\times {10}^3\text{N}$.

$P^{\prime }={P^{\prime }}_{AC}-{P^{\prime }}_{CB}$

Substitute $-{P^{\prime }}_{AC}$ for ${P^{\prime }}_{CB}$. And $975\times {10}^3\text{N}$ for $P^{\prime }$.

$\begin{array}{l}975\times {10}^3={P^{\prime }}_{AC}-\left(-{P^{\prime }}_{AC}\right)\\ 975\times {10}^3=2{P^{\prime }}_{AC}\\ {P^{\prime }}_{AC}=487.5\times {10}^{-3}\text{N}\end{array}$

Determine the deflection using the relation.

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

Substitute $1,750{\text{mm}}^{\text{2}}$ for A, 190 mm for $L_{AC}$, $200\text{GPa}$ for E, and $487.5\times {10}^{-3}\text{N}$ for ${P^{\prime }}_{AC}$ in Equation (7).

$\begin{array}{c}{\delta }^{\prime }=\frac{487.5\times {10}^{-3}\times 190\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{200\text{GPa}\left(\frac{{10}^9\text{pa}}{1\text{GPa}}\right)\left(1750{\text{mm}}^{\text{2}}\times {\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\right)}\\ =\frac{487.5\times {10}^{-3}\times 0.190}{200\times {10}^9\left(1750\times {10}^{-6}\right)}\\ =0.26464\times {10}^3\text{m}\end{array}$

Find the permanent deflection using the relation:

${\delta }_P={\delta }_m-{\delta }^{\prime }$ (8)

Substitute $0.26464\times {10}^3\text{m}$ for ${\delta }^{\prime }$ and $0.29179\times {10}^{-3}$ for ${\delta }_m$ in Equation (8).

$\begin{array}{c}{\delta }_P=0.26464\times {10}^3-0.26464\times {10}^3\\ =0.02715\times {10}^{-3}\text{m}\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ =0.0272\text{mm}\end{array}$

Thus, the permanent deflection at point C is $\underset{\_}{0.0272\text{mm}}$.