Chapter 3.3, Problem 3.79P

To determine

Find the magnitude of the couple (M) and direction of its axis after replacing two couples in to single equivalent couple.

Answer to Problem 3.79P

The magnitude of the couple (M) and direction of its axis $\left({\theta }_x,{\theta }_y,{\theta }_z\right)$ are $\underset{\_}{2.72\text{N}\cdot \text{m}}$ and $\underset{\_}{134.9°,58°,61.9°}$ respectively.

Explanation of Solution

Given information:

The two vertical forces $\left(M_3\right)$ is 10 N.

The inclined force acting at point B $\left(F_B\right)$ is 50 N.

The inclined force acting at point C $\left(F_C\right)$ is 50 N.

The force acting at the point E $\left(F_E\right)$ is 12.5 N.

The force acting at the point D $\left(F_D\right)$ is 12.5 N.

The height between point C and F (CF) is 120 mm.

The height between point F and A (FA) is 120 mm.

The horizontal distance between point E and  D (ED) is 192 mm.

The height between point E and B (EB) is 144 mm.

The distance between point C and D (CD) is 160 mm.

Calculation:

Replace the couple in the ABCD plane with two couple’s P and Q.

Show the couple ABCD by replacing with two couple’s P and Q as Figure 1.

Calculate the distance CG using the formula:

Use Pythagoras theorem,

$CG=\sqrt{C{D}^2+C{F}^2}$

Substitute 160 mm for CD and 120 mm for CF.

$\begin{array}{c}CG=\sqrt{{\left(160\right)}^2+{\left(120\right)}^2}\\ =200\text{mm}\end{array}$

Calculate the force (P) acting at the point B using the formula:

$P=F_B\frac{CD}{CG}$

Substitute 50 N for $F_B$, 160 mm for CD and 200 mm for CG.

$\begin{array}{c}P=50\left(\frac{160}{200}\right)\\ =40\text{N}\end{array}$

Calculate the force (Q) acting at the point C using the formula:

$Q=F_C\frac{CF}{CG}$

Substitute 50 N for $F_C$, 120 mm for CF and 200 mm for CG.

$\begin{array}{c}Q=50\left(\frac{120}{200}\right)\\ =30\text{N}\end{array}$

Calculate the angle $\left(\theta \right)$ using the relation:

$\tan \theta =\frac{EB}{ED}$

Substitute 144 mm for EB and 192 mm for ED.

$\begin{array}{c}\tan \theta =\frac{144}{192}\\ \theta ={\tan }^{-1}\left(0.75\right)\\ =36.87°\end{array}$

Calculate the couple vector $\left(M_1\right)$ perpendicular to plane ABCD using the relation:

$M_1=\left[P\times \left(CF+FA\right)\right]-\left[Q\left(CD\right)\right]$

Substitute 40 N for P, 120 mm for CF, 120 mm for FA, 30 N for Q, 160 mm for CD.

$\begin{array}{c}{M}_1=\left[40\times \left(120+120\right)\right]-\left[30\left(160\right)\right]\\ =4800\text{N}\cdot \text{mm}\times \frac{1}{1000}\frac{\text{m}}{\text{mm}}\\ =4.80\text{N}\cdot \text{m}\end{array}$

Write the couple vector $\left(M_1\right)$ perpendicular to plane ABCD in vector form:

$M_1=\left(M_1\cos \theta \right)j+\left(M_1\sin \theta \right)k$

Substitute $4.80\text{N}\cdot \text{m}$ for $M_1$ and $36.87°$ for $\theta$.

$\begin{array}{c}{M}_1=\left(4.80\cos 36.87\right)j+\left(4.80\sin 36.87\right)k\\ =3.84j+2.88k\end{array}$

Calculate the couple vector $\left(M_2\right)$ in the xy plane using the relation:

$M_2=-F_D\times ED$

Substitute 12.5 N for $F_D$ and 192 mm for ED.

$\begin{array}{c}{M}_2=-12.5\times \left(\text{192mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)\\ =-2.4\text{N}\cdot \text{m}\end{array}$

Write the couple vector $\left(M_2\right)$ in the xy plane in vector form:

$M_2=\left(-2.4\text{N}\cdot \text{m}\right)j$

Calculate the position vector of BC $\left(r_{B/C}\right)$ using the relation:

$r_{B/C}=CDi+EBj-EDk$

Substitute 160 mm for CD, 144 mm for EB and 192 mm for ED.

$\begin{array}{c}{r}_{B/C}=\left(\text{160mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)i+\left(\text{144mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)j-\left(\text{192mm×}\frac{\text{1}}{\text{1000}}\frac{\text{m}}{\text{mm}}\right)k\\ =0.16i+0.144j-0.192k\end{array}$

Calculate the couple vector $\left(M_3\right)$ using the relation:

$\left(M_3\right)=r_{B/C}\times M_3$

Substitute $0.16i+0.144j-0.192k$ for $r_{B/C}$ and $\left(-10\text{N}\right)j$ for $M_3$.

$\begin{array}{c}\left(M_3\right)=0.16i+0.144j-0.192k\times \left(-10\right)j\\ =-1.6k-1.92i\end{array}$

Calculate magnitude of the couple (M) using the relation:

$M=M_1+M_2+M_3$

Substitute $3.84j+2.88k$ for $M_1$, $\left(-2.4\text{N}\cdot \text{m}\right)j$ for $M_2$ and $-1.6k-1.92i$ for $M_3$.

$\begin{array}{c}M=3.84j+2.88k-2.4j-1.6k-1.92i\\ =-1.92i+1.44j+1.28k\\ M=\sqrt{{\left(-1.92\right)}^2+{\left(1.44\right)}^2+{\left(1.28\right)}^2}\\ =2.72\text{N}\cdot \text{m}\end{array}$

Calculate the direction of axis along x direction $\left({\theta }_x\right)$ by considering the i component using the relation:

$\cos {\theta }_x=\frac{M_{\left(i\right)}}{M}$

Substitute $-1.92\text{N}\cdot \text{m}$ for $M_{\left(i\right)}$ and $2.72\text{N}\cdot \text{m}$ for $M$.

$\begin{array}{c}\cos {\theta }_x=\frac{-1.92}{2.72}\\ ={\cos }^{-1}\left(\frac{-1.92}{2.72}\right)\\ =134.9°\end{array}$

Calculate the direction of axis along y direction $\left({\theta }_y\right)$ by considering the j component using the relation:

$\cos {\theta }_y=\frac{M_{\left(j\right)}}{M}$

Substitute $1.44\text{N}\cdot \text{m}$ for $M_{\left(j\right)}$ and $2.72\text{N}\cdot \text{m}$ for $M$.

$\begin{array}{c}\cos {\theta }_y=\frac{1.44}{2.72}\\ ={\cos }^{-1}\left(\frac{1.44}{2.72}\right)\\ =58.0°\end{array}$

Calculate the direction of axis along z direction $\left({\theta }_z\right)$ by considering the k component using the relation:

$\cos {\theta }_z=\frac{M_{\left(k\right)}}{M}$

Substitute $1.28\text{N}\cdot \text{m}$ for $M_{\left(k\right)}$ and $2.72\text{N}\cdot \text{m}$ for $M$.

$\begin{array}{c}\cos {\theta }_z=\frac{1.28}{2.72}\\ ={\cos }^{-1}\left(\frac{1.28}{2.72}\right)\\ =61.9°\end{array}$

Therefore, the magnitude of the couple (M) and direction of its axis $\left({\theta }_x,{\theta }_y,{\theta }_z\right)$ are $\underset{\_}{2.72\text{N}\cdot \text{m}}$ and $\underset{\_}{134.9°,58°,61.9°}$ respectively.