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Forces and Moments: Simplification of force and couple systems

Resultant moment

Resultant moment of the moments of forces

Like forces (vectors), moments can be added together to determine their collective effect known as the resultant moment. Let $\bold F_1, \bold F_2\dots,\bold F_n$ be a system of forces acting on a body (Fig. 3.23). The resultant moment of the forces about a point $O$ (anywhere in the space) is determined by a vector addition of the moment of each force about $O$ ,

(3.23) $(\bold M_R)_O=\sum_{i=1}^n \bold r_i\times \bold F_i$

or in a concise notation,

$(\bold M_R)_O=\sum \bold r\times \bold F$

where $\bold r_i$ is the quasi-moment arm vector of $\bold F_i$ , and $(\bold M_R)_O$ is the resultant moment about point $O$ .

Fig. 3.23 The resultant of the moments of forces about a point in space.

Remark: the resultant moment of forces is a moment acting about a point (or axis). This moment conveys the collective rotational tendencies of the forces about an specific point (or axis). Indeed, changing the location of the pivot point changes the resultant moment of the forces.

In the case of a coplanar (two-dimensional) force system, Eq. 3.23 reduces to an algebraic sum of the moments. In this case, the sign of the moments of the forces and their resultant follow the convention that associates a positive moment with a counterclockwise rotation.

The resultant moment of a system of coplanar forces $\bold F_1, \bold F_2 \dots\bold F_n$ about a point $O$ in the plane of the forces (Fig. 3.24a) is,

(3.24) $(M_R)_O=\circlearrowleft{+}\sum_{i=1}^n F_id_i=\sum_{i=1}^n (M_O)_i$

where $F_1, F_2, \dots F_n$ are the magnitude of the forces, $d_1,\dots d_n$ are the moment arms, and the notation $\circlearrowleft{+}$ indicates the direction of positive moments. The result of the equation is a scalar. If $(M_R)_O$ is positive, the sense of the rotation of the moment is counterclockwise, otherwise it is clockwise.

Fig. 3.24 The resultant moment of a system of coplanar forces.

In a Cartesian coordinate system, decomposing each force to its rectangular components and adding the moments of the components of the forces (according to the principle of moments) facilitate calculating the resultant moment.

EXAMPLE 3.4.1

Calculate the resultant moment of the force system about point $O$ .

SOLUTION

Setting the positive directions (shown in the figure), the rectangular components of the forces are determined.

The sum of moments of the components of the forces about point $O$ is,

$\begin{split}(M_R)_O&=\circlearrowleft{+} \sum Fd= (2)(4) -(5)(4)-(8.66)(1.5)=-25 \text{ kN.m}\\\therefore (M_R)_O&=-25 \text{ kN.m}\quad \text{or} \quad (M_R)_O=25\text{ kN.m}\circlearrowright\end{split}$

Resultant couple moment

If several couples (couples of forces) act on a body (Fig. 3.25), the couple moments can be added together to determine the resultant couple moment. Since the couple moments are free vectors (independent of location), their resultant is also a free vector. In a mathematical notation, the resultant of the couple moments $\bold M_1, \bold M_2,\dots,\bold M_n$ is,

(3.25) $\bold M_R=\sum_{i=1}^n \bold M_i$

where $\bold M_R$ is the resultant couple moment. The calculation of a resultant couple moment does not depend on a pivot point, therefore $\bold M_R$ has no subscription denoting a pivot point. Equation 3.25 is a vector sum in general, however, it becomes a scalar sum in two-dimensional problems. The sign of a couple moment in two-dimensions is positive if it is counterclockwise.

Fig. 3.25 The resultant of couple moments.

Resultant Moment of a system of forces and couple moments

A system containing forces and couple moments acting on a body is referred to as a system of forces and couple moments or a force-couple system.

If forces $\bold F_1, \bold F_2\dots,\bold F_n$ and couple moments $\bold M_1, \bold M_2,\dots,\bold M_n$ act on a body, the resultant moment of the forces and couple moments about a point $O$ is defined as the sum of the moments of the forces about $O$ in addition to the sum of the couple moments. In a mathematical notation,

(3.26) $(\bold M_R)_O=\sum_{i=1}^n \bold r_i\times \bold F_i + \sum_{i=1}^n \bold M_i$

or in a concise notation,

$(\bold M_R)_O=\sum \bold r\times \bold F + \sum \bold M$

This definition physically conveys the rotational tendencies produced by the forces about the pivot point $O$ plus the rotational tendencies of the couple moments produced on the body.

Equivalent force-couple systems and simplification of systems

A force-couple system can be reduced or simplified to an equivalent system that consists of only a resultant force and a resultant couple moment, or even a single force, applied at a specific point of the body (or space). The reduction of forces and moments simplifies the analysis of the system in question.

A simplified system makes it convenient to quickly observe and understand the overall disclosure of a force-couple system on the body. For example, each limb of a human body has its own weight, but it is convenient and more understandable (in many problems) to consider the total weight and shown it as a single force on the body rather than showing each limb’s weight. In this section, the concept of equivalent force-couple systems is first defined and then simplification of different force-couple systems are presented.

Equivalent systems

Two force-couple systems are called equivalent if their external effects on a rigid body are the same. The following two conditions needs to be satisfied.

1- The resultant forces of the two systems are equal, i.e.,

$\underset{\text{System 1}}{\bold F_R}=\underset{\text{System 2}}{\bold F_R}\implies \sum_\text{System 1}\bold F=\sum_\text{System 2}\bold F$

2- The resultant moments of the systems about any fixed arbitrary point, $O$ , in the space are equal, i.e.,

$\underset{\text{System 1}}{(\bold M_R)_O}=\underset{\text{System 2}}{(\bold M_R)_O}\implies\sum_\text{System 1}\bold r\times \bold F+ \sum_\text{System 1} \bold M=\sum_\text{System 2}\bold r\times \bold F+ \sum_\text{System 2} \bold M$

Note that the resultant moment of a system is the sum of moments produced by the system forces about a fixed point plus the couple moments.

Context: Simplification of force and couple systems

Engineers often simplify systems by combining forces and moments to make calculations easier.
When simplifying systems, engineers need to ensure that the simplified system has external effects equivalent to the original system.
This approach is effective for determining support reactions as part of rigid body equilibrium (Section 5.2), particularly in structures subject to many combinations of point forces and/or structures subject to distributed loads (Section 3.5).
This simplification approach is limited as it may not allow internal effects (e.g. tension, shear) in a member to be assessed. Internal effects are discussed more in Chapter 7 of this book.

Applications: Where do engineers use simplifications of force and couple systems?

In addition to designing new structures, engineers also evaluate existing structures in case they are being changed to resist different types of loads or if they are degrading because of problems like corrosion. Looking at the column in Fig. 3.26, we see that forces are transferred into the column from a few sources including the steel beams resting on the column ledges (formally called ‘corbels’) as well as forces and moments coming from the above floor.

Fig. 3.26 (a) Column with a corbel in a parking garage (b) illustration of forces and moments acting on the top of the column (c) simplified system that has the same external effect as (b). Photo by D. Tomlinson.

If you were to check this column’s adequacy at resisting the applied forces without failing the easiest approach is to reduce the system to a single force, $\bold F_{\rm simplified}$ , acting at an eccentricity, $d_{\rm simplified}$ . With these simplified forces, you can compare these forces to the material capacity of the column accounting for the combination of forces and moments. Engineers often use design resources (e.g. tables and charts) that allow them to quickly evaluate the adequacy of these simplified systems. Material capacities and these design resources are discussed in upper year engineering courses.

EXAMPLE 3.4.2

Consider two systems of non-equal but collinear forces (acting on the same bar) as shown in the figure. Show that both systems are equivalent.

Solution

1- The same resultant forces.

$\begin{split}\text {system 1: } F_R&=\overset{+}{\rightarrow}\sum F_x= 10+20+30=60\text{ N}\\\text {system 2: } F_R&=\overset{+}{\rightarrow}\sum F_x= 70 -10=60\text{ N}\\\therefore \underset{\text{System 1}}{\bold F_R}&=\underset{\text{System 2}}{\bold F_R}\end{split}$

2- The same resultant moments.

Selecting a point on the line of action of the forces leads to zero resultant moments for both systems.

Therefore the two force-couple systems are equivalent.

EXAMPLE 3.4.3

Prove that the coplanar force-couple systems shown in the figure are equivalent.

1- The same resultant forces.

$\begin{split}\text {system 1: } F_R&=+\uparrow\sum F_y= -1 -1=-2 \implies F_R=2 \text{ N}\downarrow\\\text {system 2: } F_R&=+\uparrow\sum F_y= -2 \implies F_R=2 \text{ N}\downarrow\\\therefore \underset{\text{System 1}}{\bold F_R}&=\underset{\text{System 2}}{\bold F_R}\end{split}$

2- The same resultant moments.

The resultant moments of both systems about an arbitrary point should be equal. We choose point $O$ on right end of the bar and calculate the resultant moments.

$\begin{split}\text {system 1: } (M_R)_O&= \circlearrowleft{+}\sum Fd= -(1)(4) -(1)(6) = -10\text { N.m}\\\implies (M_R)_O&=10\text { N.m}\circlearrowright\\\text {system 2: } (M_R)_O&= \circlearrowleft{+}\sum Fd + \sum M= (2)(0) -10 = -10\text { N.m}\\\implies (M_R)_O&=10\text { N.m}\circlearrowright\\\therefore \underset{\text{System 1}}{(\bold M_R)_O}&=\underset{\text{System 2}}{(\bold M_R)_O}\end{split}$

Therefore the systems are equivalent.

In the last example, we observe that a system of two forces is equivalently represented by a system of one force and a couple moment. Any system can be simplified to a system including only a force and a couple. This concept is presented in the next subsection.

General simplification of force-couple systems

The concept of equivalent systems can be used to simplify or reduce a force-couple system to an equivalent system that generally includes only a resultant force and a resultant couple moment at one specific point. This process is called the simplification of a force-couple system.

The following example shows how to move a force to a different application point without changing its external effect on the same rigid body. Consider a force $\bold F$ in the x-y plane as shown in Fig 3.27a. The force is acting at point $A$ . We intend to find an equivalent system in which $\bold F$ acts at (or relocated to) point $O$ .

We can add (place) both $\bold F$ and $-\bold F$ to the at point $O$ without changing the original system (Fig. 3.27b); because $\bold F+(-\bold F)=\bold 0$ . The system of forces shown in Fig. 3.27b indicates that $\bold F$ at $A$ and $-\bold F$ at $O$ form a couple and can be replaced by a couple moment with the magnitude $M=Fd$ . The couple moment is a free vector and can be placed anywhere; we choose to locate it at point $O$ (Fig. 3.27c). The resultant system as shown in Fig. 3.27c is an equivalent system consisting of a force and a couple moment.

Upon relocating $\bold F$ , the added couple moment with the magnitude $M=Fd$ is added (Fig. 3.27c). This states that if a force is relocated from point $A$ to point $O$ (not on the line of action of the force), a couple moment $\bold M$ should be added to the equivalent system to preserve the moment that the force in the original system creates about point $O$ .

Fig. 3.27 Equivalent systems regarding relocating a force from point $A$ to point $O$ .

The general procedure of the simplification of a force-couple system to an equivalent system has the following steps:

Choose a point, $O$ , on the body (or in the space) for all forces to be relocated there.
Calculate the resultant moment of the forces about point $O$ and the couple moments, i.e. $(\bold M_R)_O=\sum \bold r\times \bold F + \sum \bold M$ .
Move all the forces to the new location ( $O$ ) and calculate the resultant force $\bold F_R=\sum \bold F$ .
Place, at $O$ , a couple moment $\bold M$ equal to $(\bold M_R)_O$ .

Remark: to relocate a force $\bold F$ from point $A$ to a new point $O$ , whenever visualization of forces are convenient (like in coplanar systems), you can first show the forces $\bold F$ and $-\bold F$ at point $O$ . Then, create a couple moment using $\bold F$ at $A$ and $-\bold F$ at $O$ , and show the couple moment together with $\bold F$ at $O$ .

EXAMPLE 3.4.4

Reduce (simplify) the following system to an equivalent system consisting of a force and a couple moment acting at point $O$ .

The following steps are take.

1- Calculate the resultant moment of the forces about point $O$ and the couple moments, i.e. $(\bold M_R)_O=\sum Fd + \sum M$ (scalar formulation in two dimensions).

$\begin{split}(M_R)_O&= \circlearrowleft{+}\sum Fd + \sum M= -(2)(1.5) -(2)(2.5) -(2)(3.5) -(2)(4.5) - (2)(5.5) + 5\\&= -30\text { N.m}\end{split}$

The negative sign of $-30\text { N.m}$ means the resultant moment is clockwise. Therefore,

$(M_R)_O=30\text { N.m} \circlearrowright$

2- Calculate the resultant force $F_R=\sum F$ (scalar formulation in two dimensions).

$\begin{split}F_R=+\uparrow\sum F\implies F_R&=-2-2-2-2-2=-10 \text { N}\\\therefore \ F_R&=10 \text { N} \downarrow\end{split}$

3- Place, at $O$ , the resultant force $\bold F_R$ and a couple moment $\bold M$ equal to $(\bold M_R)_O$ .

Systems with further simplification

Although all systems can be reduced to an equivalent system consisting of a force and a couple moment, there are systems that can be further reduced to a single resultant force or a wrench as presented below.

1- Concurrent force systems

In a concurrent force system (Fig. 3.28a), the lines of action of the forces meet (intersect) at a common point $O$ . Therefore, the forces produce no moment about $O$ and the system can be reduced to an equivalent system consisting of one resultant force $\bold F_R=\sum \bold F$ as shown in Fig. 3.28b.

Fig. 3.28 Reduction of a concurrent force system to an equivalent system consisting of one resultant force.

2- Coplanar force systems

In a coplanar force system (Fig. 3.29a), the lines of action of the forces are in the same plane, say x-y plane. The moments of the forces about any point on the plane are perpendicular to the plane, so are their resultant moment. This system can be initially reduced to a resultant force $\bold F_R=\sum \bold F$ located at any point $O$ and a resultant couple moment, $\bold M_R=\sum\bold r\times \bold F$ of the forces about point $O$ (Fig. 3.29b).

Then, the system can be further simplified to a system that contains only $\bold F_R$ as shown in Fig. 3.29c. This is achieved by offsetting in the plane $\bold F_R$ by distance $d$ from $O$ (Fig 3.29c). Offsetting the force can replicate the same moment, $\bold M_R$ , about point $O$ . This leads to a further simplified system shown in Fig. 3.29c.

To find $d$ , the resultant moments of the systems shown in Fig. 3.29b and c must be equal. Choosing point $O$ and using the scalar formulation, we write,

$\begin{split}\underset{\text{System b}}{(M_R)_O}&=\underset{\text{System c}}{( M_R)_O}\implies M_R = F_Rd\\\therefore d&=\frac{M_R}{F_R}\end{split}$

It should be noted that $\bold F_R$ should be offset in the correct side of $O$ , in order to replicate the couple moment ( $\bold M_R$ ). This means that $\bold F_R$ at its final location, $O'$ , should create a moment that is equal to $\bold M_R$ about $O$ .

Fig. 3.29 Further simplification of a coplanar force system to an equivalent system consisting of a resultant force.

EXAMPLE 3.4.5

Reduce (simplify) the system of coplanar forces to a system consisting of a single resultant force. Measured from point $A$ , determine $d$ , the location of the resultant force.

SOLUTION

1- Calculate $F_R$ .

$\begin{split}F_R=+\uparrow\sum F\implies F_R&=-10-8-6-4-2=-30 \text { N}\\\therefore \ F_R&=30 \text { N} \downarrow\end{split}$

2- Make the resultant moments of the original and the simplified systems equal.

$\begin{split}\text {Original system } (M_R)_A&= \circlearrowleft{+}\sum Fd= -(10)(3) -(8)(4) -(6)(6) -(4)(8) -(2)(11)\\&= -152\text { N.m}\end{split}$

The negative sign of $-152\text { N.m}$ means the resultant moment is clockwise. Therefore, the downward resultant force should be right to the rotation point to replicate such a moment. Therefore,

$\begin{split}\text {Original system } (M_R)_A&= 152\text { N.m} \circlearrowright\\\text {Simplified system } (M_R)_A&= \circlearrowleft{+} F_Rd= -(30)(d)\text { N.m}\\&\implies F_Rd= (30)(d)\text { N.m}\circlearrowright\end{split}$

Letting $\text {Original system } (M_R)_A=\text {Simplified system } (M_R)_A$ leads to,

$d = \frac{152 \text{ N.m}}{30 \text{ N}}=5.1\text { m}$

3- Parallel force systems (non-coplanar)

A parallel force system contains forces that are parallel to each other, say along the $z$ axis as shown in Fig. 3.30a. Parallel forces can be slid to a common perpendicular plane like the x-y plane in this case (Fig 3.30a). A system of parallel forces can be initially simplified to a system consisting of a resultant force $\bold F_R=\sum \bold F$ applied at a point $O$ in the x-y plane, and a resultant couple moment $\bold M_R=\sum\bold r\times \bold F$ as shown in Fig. 3.30b.

Since all forces are parallel, $\bold F_R$ is parallel to the $z$ axis. Also, all moments of the forces are perpendicular to the axis $z$ and therefore $\bold M_R$ is perpendicular to the resultant force.

Similar to the further simplification of a coplanar system, where $\bold F_R$ and $\bold M_R$ are perpendicular, a system of parallel forces can be further reduced to a single resultant force (Fig. 3.30c). To achieve this, $\bold F_R$ should be offset (to point $O'$ ) in the plane containing $\bold F_R$ and perpendicular to the axis of $\bold M_R$ as shown in Fig 3.30c. The offset distance $d$ from $O$ is determined as,

$d=\frac{M_R}{F_R}$

where $F_R=|\bold F_R|$ and $M_R=|\bold M_R|$ . It should be noted that the $\bold F_R$ should be offset in the correct side of $O$ in order to replicate $\bold M_R$ about $O$ .

Fig. 3.30 Further simplification of a parallel force system to an equivalent system consisting of the resultant force.

4- Systems reducing to a wrench

A general force-couple system in three dimensions (Fig. 3.31a) can be reduced to an equivalent system (Fig. 3.31b) consisting of a resultant force $\bold F_R=\sum \bold F$ at a point $O$ and a resultant couple moment $\bold M_R= \sum \bold r\times \bold F+\sum \bold M$ .

Since $\bold F_R$ and $\bold M_R$ are not generally perpendicular to each other (as in a case of a parallel force system), the system cannot be further reduced to a single resultant force. However, this system can be further simplified to an equivalent system called a wrench. To achieve this, $\bold M_R$ is resolved into components parallel and perpendicular to the resultant force $\bold F_R$ . The components of $\bold M_R$ are shown in Fig. 3.31c.

The parallel component ( $\bold M_{R\parallel}$ ) and perpendicular component ( $\bold M_{R\perp}$ ) of $\bold M_R$ can be obtained using the dot product as (see Section 2.6.4),

$\bold M_{R\parallel}=(\bold M_R \cdot \bold u_{F_R})\bold u_{F_R},\quad \bold M_{R\perp}=\bold M_R - \bold M_{R\parallel}$

where $\bold u_{F_R}=\frac{\bold F_R}{|\bold F_R|}$ is the unit vector of $\bold F_R$ .

For the purpose of clear visualization, $\bold F_R$ and the components of $\bold M_R$ are demonstrated in Fig. 3.31d.

Since $\bold F_R$ and $\bold M_{R\perp}$ are perpendicular to each other, offsetting $\bold F_R$ in the plane containing $\bold F_R$ and perpendicular to $\bold M_{R\perp}$ can replicate $\bold M_{R\perp}$ . As demonstrate in Fig 3.31e, offsetting $\bold F_R$ to point $O'$ by distance $d=\frac{M_{R\perp}}{F_R}$ , where $F_R=|\bold F_R|$ and $M_R=|\bold M_{R\perp}|$ , reduces the system to $\bold F_R$ and $\bold M_{R\parallel}$ as shown in Fig. 3.31f.