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Wize University Statics Textbook (Master) > Force System Resultants (Moment, Couple, Dist. Load)

Moment (of a) Couple / Force Couple Moment

Moment (of a Force) about an axis or specified linePrevious Section
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Couple moments are a special type of moments. Whereas moments thus far have been caused by forces, leading to both translation and rotation; couple moments only cause rotation. A couple is defined as 2 parallel forces that have the SAME magnitude and OPPOSITE directions, separated by a perpendicular distance. Called either couple moment or moment couple.
There are several conditions for forces to qualify as a couple moment:
  • Two forces (the word couple implies TWO)
  • Equal magnitude
  • Opposite direction
  • Parallel lines of action
When two forces qualify as a couple moment, the computed moment vector is a free vector. This basically means that the moment vector caused by a couple moment is the same at every point.
To compute a couple moment:
  1. If you can find the perpendicular distance between the two forces, simply multiply the distance by one of the forces
  2. If perpendicular distance is difficult to compute, then.....
  • Compute a distance vector from one force line of action to the other force’s line of action
  • Take the cross product between the distance and one of the force vectors

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Couple moments are a special type of moments. Whereas moments thus far have been caused by forces, leading to both translation and rotation; couple moments only cause rotation.
There are several conditions for forces to qualify as a couple moment:
  • Two forces
  • Equal magnitude
  • Opposite direction
  • Parallel lines of action
When two forces qualify as a couple moment, the computed moment vector is a free vector. This basically means that the moment vector caused by a couple moment is the same at every point.
To compute a couple moment:
  1. If you can find the perpendicular distance between the two forces, simply multiply the distance by one of the forces
  2. If perpendicular distance is difficult to compute.
  • Compute a distance vector from one force line of action to the other force’s line of action
  • Take the cross product between the distance and one of the force vectors
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Couple moments are a special type of moments. Whereas moments thus far have been caused by forces, leading to both translation and rotation; couple moments only cause rotation.
There are several conditions for forces to qualify as a couple moment:
  • Two forces
  • Equal magnitude
  • Opposite direction
  • Parallel lines of action


When two forces qualify as a couple moment, the computed moment vector is a free vector. This basically means that the moment vector caused by a couple moment is the same at every point.
To compute a couple moment:
  1. If you can find the perpendicular distance between the two forces, simply multiply the distance by one of the forces
  2. If the perpendicular distance is difficult to compute which is typically the case for 3D problems. We can,



  • Compute a distance-vector from one force line of action to the other force’s line of action
  • Take the cross product between the distance and one of the force vectors

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Example

Determine the total couple moment on the triangle



3 Couples:3\ Couples:
F1=200lbF_1=200lb r1=4ftr_1=4ft M1=800 lb−ftM_1=800\ lb-ft
F2=150lbF_2=150lb r2=4ftr_2=4ft M2=600 lb−ftM_2=600\ lb-ft
F3=300lbF_3=300lb r3=4ftr_3=4ft M3=1200 lb−ftM_3=1200\ lb-ft

M=2600lb.ftM=2600lb.ft
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Example

Two couples act on the beam as shown. If F = 15 N, determine the resultant couple moment. Where on the beam does the resultant couple act?


Solution:


CCW+(MR)c=−15(45)(1.5m)−15(35)(4m)−20N(1.5m)=−84N−mCCW+\left(M_{R}\right)_{c}=-15\left(\frac{4}{5}\right)(1.5 \mathrm{m})-15\left(\frac{3}{5}\right)(4 \mathrm{m})-20 \mathrm{N}(1.5 \mathrm{m})=-84 N-m


The resultant moment can act anywhere on the beam.

Compute the resultant couple moment of the following:
1) Compute the resultant couple moment:



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  2. When you're done, check the "I have answered this question" box below.
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Calculate the distance 'd' required to cause no moment on the body.

Quiz: Couple Moments - Quiz Practice Questions (2D)
If α = 90o , what is the distance d that creates a zero resultant moment on the body.


What is the resultant couple moment?


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Example

Determine the resultant couple moment of the two couples that act on the assembly. Specify its magnitude and coordinate direction angles.

The resultant couple moment depends on the position vectors r1 and r2r_1\ and\ r_2or each applied couple F1 and F2F_1\ and\ F_2, as shown in the figure below,




(MC)R=r1×F1+r2×F2\left(M_{C}\right)_{R}=r_{1} \times F_{1}+r_{2} \times F_{2}

(MC)R=∣ijkr1xr1yr1zF1xF1yF1z∣+∣ijkr2xr2yr2zF2xF2yF2z∣\left(M_{C}\right)_{R}=\left|\begin{array}{ccc}{i} & {j} & {k} \\ {r_{1 x}} & {r_{1 y}} & {r_{1 z}} \\ {F_{1 x}} & {F_{1 y}} & {F_{1 z}}\end{array}\right|+\left|\begin{array}{ccc}{i} & {j} & {k} \\ {r_{2 x}} & {r_{2 y}} & {r_{2 z}} \\ {F_{2 x}} & {F_{2 y}} & {F_{2 z}}\end{array}\right|

(MC)R=∣ijk4cos⁡(30)5−4sin⁡(30)0060∣+∣ijk4cos⁡(30)0−4sin⁡(30)0800∣\left(M_{C}\right)_{R}=\left|\begin{array}{ccc}{i} & {j} & {k} \\ {4 \cos (30)} & {5} & {-4 \sin (30)} \\ {0} & {0} & {60}\end{array}\right|+\left|\begin{array}{ccc}{i} & {j} & {k} \\ {4 \cos (30)} & {0} & {-4 \sin (30)} \\ {0} & {80} & {0}\end{array}\right|

=(MC)R=+(5(60)−0)i−(4cos⁡(30)(60)−0)j+0k+(0−(−4sin⁡(30)(80))i−0j+(4cos⁡(30)(80)−0)k\begin{array}{l}{=\left(M_{C}\right)_{R}=+(5(60)-0) i-(4 \cos (30)(60)-0) j+0 k} \\ {+(0-(-4 \sin (30)(80)) i-0 j+(4 \cos (30)(80)-0) k}\end{array}

(MC)R=300i−207.85j+160i+277.13k(MC)R={460i−207.85j+277.13k}lb⋅in\begin{array}{l}{\left(M_{C}\right)_{R}=300 \mathrm{i}-207.85 \mathrm{j}+160 \mathrm{i}+277.13 \mathrm{k}} \\ {\left(M_{C}\right)_{R}=\{460 \mathrm{i}-207.85 \mathrm{j}+277.13 \mathrm{k}\} \mathrm{lb} \cdot \mathrm{in}}\end{array}

Magnitude,


MR=(460)2+(−207.85)2+(277.13)2=575.85=576lb⋅in(ans)M_{R}=\sqrt{(460)^{2}+(-207.85)^{2}+(277.13)^{2}}=575.85=576 \mathrm{lb} \cdot \mathrm{in}(\mathrm{ans})

Directional angles,


α=cos⁡−1([(Mc)R]x(Mc)R)α=cos⁡−1(460575.85)=37.0∘\begin{array}{c}{\alpha=\cos ^{-1}\left(\frac{\left[\left(M_{c}\right)_{R}\right]_{x}}{\left(M_{c}\right)_{R}}\right)} \\ {\alpha=\cos ^{-1}\left(\frac{460}{575.85}\right)=37.0^{\circ}}\end{array}


β=cos⁡−1([(Mc)R]y(Mc)R)β=cos⁡−1(−207.85575.85)=111∘\begin{array}{c}{\beta=\cos ^{-1}\left(\frac{\left[\left(M_{c}\right)_{R}\right]_{y}}{\left(M_{c}\right)_{R}}\right)} \\ {\beta=\cos ^{-1}\left(\frac{-207.85}{575.85}\right)=111^{\circ}}\end{array}


γ=cos⁡−1([(Mc)R]z(Mc)R)γ=cos⁡−1(277.13575.85)=61.2∘\begin{array}{c}{\gamma=\cos ^{-1}\left(\frac{\left[\left(M_{c}\right)_{R}\right]_{z}}{\left(M_{c}\right)_{R}}\right)} \\ {\gamma=\cos ^{-1}\left(\frac{277.13}{575.85}\right)=61.2^{\circ}}\end{array}


Compute the total moment on the wedge.





checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
Determine the total couple moment on the body.


Quiz: Couple Moments - 3D Quiz
Determine the resultant couple moment. Include magnitude and direction.