Chapter 2.3, Problem 41P

Fig. P2.41

2.41 Two cylindrical rods, one of steel and the other of brass, are joined at C and restrained by rigid supports at A and E. For the loading shown and knowing that E_s = 200 GPa and E_b = 105 GPa, determine (a) the reactions at A and E, (b) the deflection of point C.

To determine

Find the reaction at point A and E.

Answer to Problem 41P

The reaction at point A is $\underset{\_}{62.8\text{kN}\leftarrow }$.

The reaction at point E is $\underset{\_}{37.2\text{kN}\leftarrow }$.

Explanation of Solution

Given information:

The length $\left(l_{AB}\right)$ of rod AB is $180\text{mm}$.

The length $\left(l_{BC}\right)$ of rod BC is $120\text{mm}$.

The length $\left(l_{CD}\right)$ of rod CD is $100\text{mm}$.

The length $\left(l_{DE}\right)$ of rod DE  is $100\text{mm}$.

The young’s modulus for steel $\left(E_s\right)$ is $200\text{GPa}$.

The young’s modulus for steel $\left(E_b\right)$ is $105\text{GPa}$.

Calculation:

Find the area $\left(A_{Ac}\right)$ of rod AC using the relation:

$A_{AC}=\frac{\pi }{4}\left(d_{AC}\right)$ (1)

Here, $d_{AC}$ is diameter of rod AC.

Substitute $40\text{mm}$ $d_{AC}$ in Equation (1).

$\begin{array}{c}{A}_{AC}=\frac{\pi }{4}{\left(40\right)}^2\\ =1.25664\times {10}^3{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =1.25664\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Find the value of EA for AC section as follows:

Substitute $200\text{GPa}$ for E and $706.86\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{CE}$.

$\begin{array}{c}EA=1.25664\times {10}^{-3}\left(200\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)\\ =1.25664\times {10}^{-3}\times 200\times {10}^9\\ =251.327\times {10}^6\text{N}\end{array}$

Find the area $\left(A_{CE}\right)$ of rod CE using the relation:

$A_{CE}=\frac{\pi }{4}\left(d_{CE}\right)$ (2)

Here, $d_{CE}$ is diameter of rod CE.

Substitute $30\text{mm}$ $d_{CE}$ in Equation (2).

$\begin{array}{c}{A}_{CE}=\frac{\pi }{4}{\left(30\right)}^2\\ =706.85{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =706.86\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the value of EA for CE section as follows:

Substitute $105\text{GPa}$ for E and $706.86\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{CE}$.

$\begin{array}{c}EA=706.86\times {10}^{-6}\left(\left(105\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)\right)\\ =706.86\times {10}^{-6}\times 105\times {10}^9\\ =74.220\times {10}^6\text{N}\end{array}$

Sketch the free body diagram of cylindrical rod as shown in Figure 1.

Refer to Figure 1.

Take section A to B.

Find the change in length AB using the relation:

${\delta }_{AB}=\frac{PL}{EA}$ (3)

Here, P is the load, L is the length of rod, E is the young’s modulus, and A is the area of the section AB.

Substitute $R_A$ for P, $180\text{mm}$ for L, and $251.327\times {10}^6\text{N}$ for EA in Equation (3).

$\begin{array}{c}{\delta }_{AB}=\frac{R_A\times 180\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{251.327\times {10}^6}\\ =\frac{R_A\times 0.180}{251.327\times {10}^6}\\ =716.20\times {10}^{-12}{R}_A\end{array}$

Take section B to C.

Find the change in length BC using the relation:

${\delta }_{BC}=\frac{PL}{EA}$ (4)

Here, A is the area of the section BC.

Substitute $R_A-\left(60\times {10}^3\right)$ for P, $120\text{mm}$ for L, and $251.327\times {10}^6\text{N}$ for EA in Equation (4).

$\begin{array}{c}{\delta }_{BC}=\frac{R_A-\left(60\times {10}^3\right)\times 120\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{251.327\times {10}^6}\\ =\frac{{\left(R_A-60\times 10\right)}^3\times 0.120}{251.327\times {10}^6}\\ =447.47\times {10}^{-12}{R}_A-26.848\times {10}^{-6}\end{array}$

Take section C to D.

Find the change in length CD using the relation:

${\delta }_{CD}=\frac{PL}{EA}$ (5)

Here, A is the area of the section CD.

Substitute $R_A-\left(60\times {10}^3\right)$ for P, $100\text{mm}$ for L, and $74.220\times {10}^6$ for EA in Equation (5).

$\begin{array}{c}{\delta }_{CD}=\frac{R_A-\left(60\times {10}^3\right)\times 100\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{74.220\times {10}^6}\\ =\frac{\left(R_A-60\times {10}^3\right)\times 0.100}{74.220\times {10}^6}\\ =1.347\times {10}^{-9}{R}_A-80.841\times {10}^{-6}\end{array}$

Take section D to E.

Find the change in length DE using the relation:

${\delta }_{DE}=\frac{PL}{EA}$ (6)

Here, A is the area of the section CD.

Substitute $R_A-\left(100\times {10}^3\right)$ for P, $100\text{mm}$ for L, and $74.220\times {10}^6$ for EA in Equation (6).

$\begin{array}{c}{\delta }_{DE}=\frac{R_A-\left(100\times {10}^3\right)\times 100\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{74.220\times {10}^6}\\ =\frac{\left(R_A-100\times {10}^3\right)\times 0.100}{74.220\times {10}^6}\\ =1.347\times {10}^{-9}{R}_A-134.735\times {10}^{-6}\end{array}$

Take section A to E.

${\delta }_{AE}={\delta }_{AB}+{\delta }_{BC}+{\delta }_{CD}+{\delta }_{DE}$ (7)

Here, ${\delta }_{AB}$ is rotation of angle AB, ${\delta }_{CB}$ is rotation of angle BC, ${\delta }_{CD}$ is rotation of angle CD, ${\delta }_{DE}$ is rotation of angle DE.

Substitute $\left(1.347\times {10}^{-9}{R}_A-134.735\times {10}^{-6}\right)$ for ${\delta }_{DE}$, $1.347\times {10}^{-9}{R}_A-80.841\times {10}^{-6}$ for ${\delta }_{CD}$, $447.47\times {10}^{-12}{R}_A-26.848\times {10}^{-6}$ for ${\delta }_{CB}$, and $716.20\times {10}^{-12}{R}_A$ for ${\delta }_{AB}$ in Equation (7).

$\begin{array}{c}{\delta }_{AE}=\left[\begin{array}{l}716.20\times {10}^{-12}{R}_A+447.47\times {10}^{-12}{R}_A-26.848\times {10}^{-6}+\\ 1.347\times {10}^{-9}{R}_A-80.841\times {10}^{-6}+1.347\times {10}^{-9}{R}_A-134.735\times {10}^{-6}\end{array}\right]\\ =3.85837\times {10}^{-9}{R}_A-242.424\times {10}^{-6}\end{array}$

Since the point E cannot move relative to A.

${\delta }_{AE}=0$

Find the reaction at point A as follows:

$\begin{array}{l}3.85837\times {10}^{-9}{R}_A-242.424\times {10}^{-6}=0\\ 3.85837\times {10}^{-9}{R}_A=242.424\times {10}^{-6}\\ R_A=62.831\times {10}^3\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ R_A=62.8\text{kN}\leftarrow \end{array}$

Thus, the reaction at point A is $\underset{\_}{62.8\text{kN}\leftarrow }$.

Find the reaction at point E as follows:

$R_E=R_A-\left(100\times {10}^3\right)$ (8)

Substitute $62.831\times {10}^3\text{N}$ for $R_A$ in Equation (8).

$\begin{array}{c}{R}_E=62.831\times {10}^3-\left(100\times {10}^3\right)\\ =-37.2\times {10}^3\text{N}\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =37.2\text{kN}\leftarrow \end{array}$

Thus, the reaction at point E is $\underset{\_}{37.2\text{kN}\leftarrow }$.

To determine

Find the deflection of point C.

Answer to Problem 41P

The deflection of point C is $\underset{\_}{46.3\text{μm}}$

Explanation of Solution

Calculation:

Determine the deflection of point C using the relation:

${\delta }_C={\delta }_{AB}+{\delta }_{BC}$ (9)

Substitute $447.47\times {10}^{-12}{R}_A-26.848\times {10}^{-6}$ for ${\delta }_{CB}$ and $716.20\times {10}^{-12}{R}_A$ for ${\delta }_{AB}$ in Equation (9).

${\delta }_C=447.47\times {10}^{-12}{R}_A-26.848\times {10}^{-6}+716.20\times {10}^{-12}{R}_A$

Substitute $62.831\times {10}^3\text{N}$ for $R_A$ in Equation (9).

$\begin{array}{c}{\delta }_C=\left(\begin{array}{l}447.47\times {10}^{-12}\left(62.831\times {10}^3\right)-26.848\times {10}^{-6}\\ +716.20\times {10}^{-12}\left(62.831\times {10}^3\right)\end{array}\right)\\ =\left(2.8111\times {10}^{-05}-26.848\times {10}^{-6}+4.499\times {10}^{-5}\right)\\ =46.3\times {10}^{-6}\text{m}\left(\frac{1\text{μm}}{{10}^{-6}\text{m}}\right)\\ =46.3\text{μm}\end{array}$

Thus, the deflection of point C is $\underset{\_}{46.3\text{μm}\to }$.