Chapter 11.5, Problem 69P

The 20-mm-diameter steel rod CD is welded to the 20-mm- diameter steel shaft AB as shown. End C of rod CD is touching the rigid surface shown when a couple T_B is applied to a disk attached to shaft AB. Knowing that the bearings are self aligning and exert no couples on the shaft, determine the angle of rotation of the disk when T_B = 400 N ∙ m. Use E = 200 GPa and G = 77.2 GPa. (Consider the strain energy due to both bending and twisting in shaft AB and to bending in arm CD.)

Fig. P11.69

To determine

The angle of rotation of the disk when $T_B=400\text{N}\cdot \text{m}$.

Answer to Problem 69P

The angle of rotation of the disk at B is $\underset{\_}{{\phi }_B=5.28°}$.

Explanation of Solution

Given information:

The diameter of the shaft AB and the steel rod CD is $d=20\text{mm}$.

The modulus of rigidity $G=77.2\text{GPa}$.

The torque applied at B is $T_B=400\text{N}\cdot \text{m}$

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

The length of steel rod CD is $L_{CD}=300\text{mm}$.

The length of shaft AB is $L_{AB}=270\text{mm}$.

Calculation:

Calculate the moment of inertia $\left(I\right)$ of the lever as shown below.

$I=\frac{\pi d^4}{64}$

Substitute $20\text{mm}$ for d.

$\begin{array}{c}I=\frac{\pi \times {20}^4}{64}\\ =7,853.98{\text{mm}}^4\end{array}$

Consider the bending of rod CD.

Sketch the Free Body Diagram as shown in Figure 1.

Refer to Figure 1.

Take moment about rod D is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_{AB}}=0\\ F_C{L}_{CD}-T_B=0\\ F_C=\frac{T_B}{L_{CD}}\end{array}$

Substitute $400\text{N}\cdot \text{m}$ for $T_B$ and $300\text{mm}$ for $L_{CD}$.

$\begin{array}{c}{F}_C=\frac{400\text{N}\cdot \text{m}}{300\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}}\\ =1,333.33\text{N}\end{array}$

Summation of forces along y direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ 1,333.33-F_D=0\\ F_D=1,333.33\text{N}\end{array}$

Calculate the bending moment at a distance x from C as shown below.

$M=1,333.33x$

Calculate the strain energy as shown below.

$U={\displaystyle {\int }_0^a\frac{M^2}{2EI}dx}$ (1)

For the steel rod CD.

Substitute $1,333.33x$ for $M$, $200\text{GPa}$ for E, $7,853.98{\text{mm}}^4$ for I, and apply the limits in Equation (1).

$\begin{array}{c}{U}_{CD}=\frac{1}{2\times 200\text{GPa}\times \frac{{10}^3\text{N}/{\text{mm}}^2}{1\text{GPa}}\times 7,853.98{\text{mm}}^4}{\displaystyle {\int }_0^{300}{\left(1,333.33x\right)}^{2}dx}\\ =\frac{1,777.769\times {10}^3}{3,141.592\times {10}^6}{\displaystyle {\int }_0^{300}{x}^{2}dx}\\ =5.6588\times {10}^{-4}{\left(\frac{x^3}{3}\right)}_0^{300}\\ =5.6588\times {10}^{-4}\left(\frac{{300}^3}{3}\right)\end{array}$

$\begin{array}{l}=5,092.9\text{N}\cdot \text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ =5.0929\text{J}\end{array}$

Consider the bending of shaft ADB.

Sketch the Free Body Diagram of the shaft as shown in Figure 2.

Refer to Figure 2.

Take moment about A is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -F_B\times L_{AB}+F_{D}a=0\\ F_B{L}_{AB}=F_{D}a\\ F_B=\frac{F_{D}a}{L_{AB}}\end{array}$

Take moment about B is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ -F_A\times L_{AB}+F_{D}b=0\\ F_A{L}_{AB}=F_{D}b\\ F_A=\frac{F_{D}b}{L_{AB}}\end{array}$

Bending moment at a distance x from A $M=\left(\frac{F_{D}b}{L_{AB}}\right)x$.

Bending moment at a distance x from B $M=\left(\frac{F_{D}a}{L_{AB}}\right)x$.

Calculate the strain energy for shaft AB using Equation (1) as shown below.

$\begin{array}{c}{U}_{AB}=\frac{1}{2EI}\left({\displaystyle {\int }_0^a{\left(\frac{F_{D}b}{L_{AB}}x\right)}^{2}dx}+{\displaystyle {\int }_0^b{\left(\frac{F_{D}a}{L_{AB}}x\right)}^{2}dx}\right)\\ =\frac{F_D^2}{2EI{L}_{AB}^2}\left({\displaystyle {\int }_0^a{b}^2{x}^{2}dx}+{\displaystyle {\int }_0^b{a}^2{x}^{2}dx}\right)\\ =\frac{F_D^2}{2EI{L}_{AB}^2}\left\{{\left(b^2\frac{x^3}{3}\right)}_0^a+{\left(a^2\frac{x^3}{3}\right)}_0^b\right\}\\ =\frac{F_D^2}{6EI{L}_{AB}^2}\left\{b^2{a}^3+a^2{b}^3\right\}\end{array}$

$=\frac{F_D^2{a}^2{b}^2}{6EI{L}_{AB}^2}\left\{a+b\right\}$

Substitute $L_{AB}$ for $\left(a+b\right)$.

$\begin{array}{c}{U}_{AB}=\frac{F_D^2{a}^2{b}^2}{6EI{L}_{AB}^2}{L}_{AB}\\ =\frac{F_D^2{a}^2{b}^2}{6EI{L}_{AB}}\end{array}$

Substitute $1,333.33\text{N}$ for $F_D$, $200\text{GPa}$ for E, $7,853.98{\text{mm}}^4$ for I, $270\text{mm}$ for $L_{AB}$, $200\text{mm}$ for b, and $70\text{mm}$ for a.

$\begin{array}{c}{U}_{AB}=\frac{{\left(1,333.33\text{N}\right)}^2\times {\left(70\text{mm}\right)}^2\times {\left(200\text{mm}\right)}^2}{6\times 200\text{GPa}\times \frac{{10}^3\text{N}/{\text{mm}}^2}{1\text{GPa}}\times 7,853.98{\text{mm}}^4\times 270\text{mm}}\\ =\frac{348.4427\times {10}^{12}}{254.46895\times {10}^{10}}\\ =136.93\text{N}\cdot \text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ =0.137\text{J}\end{array}$

Consider the portion DB of shaft ADB carries the torque.

Calculate the polar moment of inertia $\left(J\right)$ as shown below.

$J=2I$

Substitute $7,853.98{\text{mm}}^4$ for I.

$\begin{array}{c}J=2\times 7,853.98\\ =15,707.96{\text{mm}}^4\end{array}$

Calculate the strain energy $\left(U_{DB}\right)$ as shown below.

$U_{DB}=\frac{T_B^2{L}_{DB}}{2GJ}$

Substitute $400\text{N}\cdot \text{m}$ for $T_B$, $200\text{mm}$ for $L_{DB}$, $77.2\text{GPa}$ for G and $15,707.96{\text{mm}}^4$ for J.

$\begin{array}{c}{U}_{DB}=\frac{{\left(400\text{N}\cdot \text{m}\right)}^2\times 200\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}}{2\times 77.2\text{GPa}\times \frac{{10}^9\text{N}/{\text{m}}^2}{1\text{GPa}}\times 15,707.96{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =\frac{32,000}{2,425.3}\\ =13.19\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ =13.19\text{J}\end{array}$

Calculate the total strain energy $\left(U\right)$ as shown below.

$U=U_{CD}+U_{AB}+U_{DB}$

Substitute $5.0929\text{J}$ for $U_{CD}$, $0.137\text{J}$ for $U_{AB}$, and $13.19\text{J}$ for $U_{DB}$.

$\begin{array}{c}U=5.0929+0.137+13.19\\ =18.4199\text{J}\times \frac{1\text{N}\cdot \text{m}}{1\text{J}}\\ =18.4199\text{N}\cdot \text{m}\end{array}$

Calculate the angle $\left({\phi }_B\right)$ as shown below.

Provide the work energy equation at disk B as shown below.

$\begin{array}{l}\frac{1}{2}{T}_B{\phi }_B=U\\ {\phi }_B=\frac{2U}{T_B}\end{array}$

Substitute $18.4199\text{N}\cdot \text{m}$ for U and $400\text{N}\cdot \text{m}$ for $T_B$.

$\begin{array}{c}{\phi }_B=\frac{2\times 18.4199}{400}\\ =0.0920995\text{rad}\times \frac{\frac{180°}{\pi }}{1\text{rad}}\\ =5.28°\end{array}$

Therefore, the angle of rotation of the disk at B is $\underset{\_}{{\phi }_B=5.28°}$.