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## Q. 2.7

For the couple shown in Fig. (a), determine (1) the corresponding couple-vector; and (2) the moment of the couple about the axis $GH$.

## Verified Solution

Part 1
One method for determining the couple-vector is to multiply the magnitude of the couple by the unit vector in its direction. The magnitude of the couple is

$Fd=100(0.6)=60\:kN\cdot m$

The sense of the couple is shown in Fig. (b)—counterclockwise looking down on the plane of the couple. Letting $λ$ be the unit vector perpendicular to the plane of the couple, as shown in Fig. (c), the couple-vector $C$ may be written as $C=60λ\:kN\cdot m$. Because $λ$ is perpendicular to the line $AB$, it can be seen that $λ=(3j+4k)/5$ (recalling that perpendicular lines have negative reciprocal slopes). Therefore, the couple-vector is

$C=60λ=60\left(\frac{3j+4k}{5}\right)=36j+48k\:kN\cdot m$

Alternative Solution Because the couple-vector is equal to the moment of the couple about any point, it can also be determined by adding the moments of the two forces forming the couple about any convenient point, such as point $B$. Letting $F$ be the $100-kN$ force that acts along the line $DE$, we have

$F=100λ_{DE}=100\frac{\overrightarrow{DE}}{|\overrightarrow{DE}|}=100\left(\frac{-0.4j+0.3k}{0.5}\right)=−80j+60k\:kN$

Equating $C$ to the moment of $F$ about point $B$ (the other force of the couple passes through $B$), we obtain

$C=r_{BD}×F=\left|\begin{matrix}i&j&k\\−0.6&0&0\\0&−80&60\end{matrix}\right|=36j+48k\:kN\cdot m$

which agrees with the answer determined previously.

In this solution, the choice of point $B$ as the moment center was arbitrary.
Because the moment of a couple is the same about every point, the same result would have been obtained no matter which point had been chosen as the moment center.

Part 2
The most direct method for determining the moment of the couple about the axis $GH$ is $M_{GH}=C\cdot λ_{GH}$. Because $C$ has already been computed, all we need to do is compute the unit vector $λ_{GH}$ and evaluate the dot product. Referring to Fig. (a), we have

$λ_{GH}=\frac{\overrightarrow{GH}}{|\overrightarrow{GH}|}=\frac{−0.3i+0.3k}{0.3\sqrt2}=−0.7071i+0.7071k$

Hence the moment of the couple about axis $GH$ is

$M_{GH}=C\cdot λ_{GH}=(36j+48k)\cdot (−0.7071i+0.7071k)=+33.9\:kN\cdot m$

The result is illustrated in Fig. (d). If you need help in interpreting the positive sign in the answer, you should refer back to Fig. 2.9.