# Equilibrium and Compatibility

## Contents

## Introduction

Equilibrium and compatibility are used when trying to find the unknown forces in a structure (beam, frame, or truss). Before equilibrium or compatibility can be used on a structure the determinacy of the structure must be found (see Determinacy, Indeterminacy and Stability). If the structure to be analysed is determinant then the unknown forces can be found using only the equilibrium equations for either a 2-D or 3-D body. If the structure is found to be indeterminate then compatibility conditions must be found before using equilibrium to find the unknown forces.^{[1]}

This section contains a description of equilibrium and compatibility as well as the needed equations and an example for both.

## Equilibrium

Equilibrium is when the sum of all forces and moments experienced by a structure is zero.^{[2]}^{[3]} A structure is in equilibrium if it is at rest before and after applying all forces and moments.^{[3]} The balance of forces restricts the structure from accelerating while the balance of moments restricts the structure from rotating. Equilibrium can be found by using the equations of equilibrium for either a 2-D or 3-D body.

**2-D Body**

A 2-D body is in equilibrium if it satisfies the three *equations of equilibrium of plane structures*.^{[3]}

$ \sum{F_{x}}=0\\ \sum{F_{y}}=0\\ \sum{M}=0\\ $

**3-D Body**

A 3-D body is in equilibrium if it satisfies the six *equations of equilibrium of space structures*.^{[3]}

$ \sum{F_{x}}=0\ \hspace{50mm}\sum{M_{x}}=0\\ \sum{F_{y}}=0\ \hspace{50mm} \sum{M_{y}}=0\\ \sum{F_{z}}=0\ \hspace{50mm} \sum{M_{z}}=0\\ $

As seen above, the object in question has equal and opposite forces and moments being applied to it. As a result, the sum of all forces will equal zero, and the sum of all moments taken from a specific point will also equal zero.

## Important Terms

**Force**

Force is described as an action causing any type of object to move (push or pull)^{[4]},change direction, deform (creating stress on the object), or essentially cause an object with a mass to accelerate. Force is a vector therefore having both direction and magnitude. Force is commonly represented by the symbol **F**^{[4]}^{[5]}. having SI units of a Newton [N] $ \Rightarrow kg \cdot m/s^{2} $. Newton’s second law states that a Force is equal to a mass multiplied by an acceleration: $ \vec{F}=m\vec{a} $. Where **m**, is mass in kilograms (kg)^{[4]}, and **a**, is acceleration in meters per second per second $ m/s^{2} $. ^{[4]}

**Couple Force**

A couple force also known as force couple, are two forces separated by a perpendicular distance of equal magnitude but in opposite directions.^{[4]}^{[5]} The resultant force or equivalent force of these couples will create rotation about a fixed point. This is referred to as a moment about a point,^{[4]} however since they are of equal magnitude, it produces a pure moment. This pure moment will not produce any translation of the object but rather rotation around the fixed point, which is to be found at the center of mass of the object.^{[4]} See figure 1.

*Figure 1: Couple Forces*

**Moment**

A moment also known as a torque,^{[4]} is based on the term of a couple force. Force couples produce pure moments, which lead to only rotation of the object around its center of mass.^{[5]} A moment creates rotation about a point of interest or axis, found at a perpendicular distance from where the force(s) is being applied.^{[4]} This being said, a force of any magnitude and direction being applied at a perpendicular distance from the point of interest will produce a moment where displacement, rotation or deformation can occur.

A distance **d**, in meters, times the cross-product of the applied force **F** is equal to the moment about a point.^{[5]}
$ M_{P}=d\times F $

Moments or torques are expressed by $ \Rightarrow kg \cdot m^{2}/s^{2} $ or simply newton meter [N∙m].^{[4]} See figure 2.

*Figure 2: Moment produced by force F*

**External Force**

An external force is a force that is applied outside the system, such as load on a structure and reaction forces. A load force is an object resting on the structure, such as snow on the roof. An applied force is another kind of force that acts outside the structure and can move a structure in the direction of the applied force. Since the structure needs to be in equilibrium state, supports are used to oppose the load and applied forces. These opposition forces are known as reaction force and keep a body or structure stable by having the ability to prevent motion.^{[2]} Some examples of external forces are force of gravity, pushing a box, wind force on a roller beam etc.

**Internal Force**

An internal force is a force that acts within a body, such as shear and normal forces, as well as torsion and bending moment. Normal forces act along the length of a member while shear forces act parallel to normal forces.^{[2]} Torsion and bending moments are created by normal and shear forces respectively. When examining the internal forces of a member split in two, these forces are shown in equal and opposite direction. This is because the member is initially in equilibrium therefore equal and opposite internal forces occur on either side to satisfy equilibrium conditions.^{[3]} This concept is demonstrated through the figure below.

*Figure 3:This diagram shows the external and internal forces on the beam. The forces outside the beam show the external forces and internal forces can be seen at the cut section*

**Coplanar Forces**

Forces found within a coplanar system (two axis) in an x-y plane.^{[2]}

**Concurrent Force System**

When all forces applied to a system pass through the same point. This satisfies the balance of moments by creating a point with zero moment.^{[2]}

*Figure 4: Concurrent Force System where all forces applied are passing through one origin.*

## Compatibility

Compatibility is used when solving indeterminate members (see Determinacy, Indeterminacy and Stability) because the equations of equilibrium do not allow us to solve for all of the unknowns within a system. Compatibility is a method used to provide extra equations when trying to find the unknown(s) in an indeterminate member. This is done by relating the geometry of the deformed member, with the unknown forces in the structure. ^{[3]} This method allows one or more of the unknown forces to be shown as a factor of another unknown force, resulting in a removal of one or more unknowns.

Compatibility conditions use equations to solve for displacement within a member caused by loading, or by change in temperature. ^{[2]} The displacement is very small in proportion to the length of the member because when loads or temperatures are applied to a member it is almost impossible to see a change in length.

**Compatibility Equations**^{[2]}

$ \ \delta=\frac{PL}{EA}\\ $

$ \ \hspace{25mm}P=Load\\ $

$ \ \hspace{25mm}L=Length \hspace{5mm} of \hspace{5mm} original \hspace{5mm} member\\ $

$ \ \hspace{25mm}E= Modulus \hspace{5mm} of \hspace{5mm} Elasticity\\ $

$ \ \hspace{25mm}A=Cross-sectional \hspace{5mm} area\\ $

$ \ \delta_{T}=\alpha_{T}\triangle{T}L\\ $

$ \ \hspace{25mm}\alpha_{T}=Linear \hspace{5mm} Coefficient \hspace{5mm} of \hspace{5mm} Expansion\\ $

$ \ \hspace{25mm}\triangle{T}=Change \hspace{5mm} in \hspace{5mm} temperature\\ $

$ \ \hspace{25mm}L=Length \hspace{5mm} of \hspace{5mm} original \hspace{5mm} member\\ $

$ \ \hspace{25mm}\delta_{T}=Change \hspace{5mm} in \hspace{5mm} length \hspace{5mm} of \hspace{5mm} the \hspace{5mm} member\\ $

## Example Problems

**Example 1: Equilibrium**

Solve the unknowns moment and reactions in the fixed-end beam below using equations of equilibrium. The length of the beam is in metres. Neglect the size of the beam.

**Step 1:**
Assume a coordinate system for the x and y directions. See figure below. Take counter-clockwise moments as positive.

**Step 2:**
Solve the forces in the x-direction.

$ \sum{F_x}=0\\ A_x-10kN=0\\ A_x=10kN\\ $

**Step 3:**
Solve the forces in the y-direction.

$ \sum{F_y}=0\\ A_y-100kN=0\\ A_y=100kN\\ $

**Step 4:**
Solve for the moment about point A.

$ \sum{M_A}=0\\ M_A-(100kN)(6m)=0\\ M_A=600kN{\cdot}m\\ $

**Example 2: Equilibrium & Compatibility**

(Example Modified from *Mechanics of Materials, R.c Hibbeler*, example 4.7, page 141) ^{[2]}

Three solid bars are pin connected to rigid beam, AC. If a uniform distributed load of 4kN/m and two point loads of 10 and 6kN are acting on the beam, determine the forces developed in all the members: AC, BE, and CF. Assume all dimensions in meters. Neglect the size of the beam. Counter-clockwise moment direction is positive.

**Given:**

AD: Stainless Steel $ E_{ss} = 193 GPa; \hspace{5mm} A = 30mm^2 $

BE: Titanium Alloy $ E_{ti} = 120 GPa; \hspace{5mm} A = 40mm^2 $

CF: A-36 Steel $ E_{st} = 200 GPa;\hspace{5mm} A = 50mm^2 $

**Step 1:** Draw Free Body Diagram

**Step 2:** Equilibrium Equations

Because there is 3 unknowns and only 2 applicable equations, this example is statically indeterminate to the first degree. Therefore will need to use a compatibility equations to derive a third equation, that relates the displacement of the three vertical supporting members.

$ + \rightarrow \sum{F_x} = 0 $

$ + \uparrow \sum{F_y} = 0; \Rightarrow F_{AD} + F_{BE} + F_{CF} - 8kN - 10kN - 6kN = 0 \hspace{10mm} $

$ F_{AD} + F_{BE} + F_{CF} = 24kN \hspace{10mm} (1) $

$ \sum{M_c} = 0; \Rightarrow -F_{AD}*(2m) + 10kN*(1.5m) + 8kN*(1m) - 6kN*(1m) + F_{CF}*(2m) = 0 $

Simplify $ \Rightarrow -2F_{AD} + 2F_{CF} = -17kN \hspace{10mm} (2) $

The displacement at A will be greater than at B or C due to the forces near A and a smaller cross-sectional area and elastic modulus in member AD. Member AC will remain straight due to the fact that it is a rigid beam.

**Step 3:** Relation of displacements using similar triangles

Using similar triangles we can find the displacement at A, therefore finding a relationship with the forces in all members.

$ \frac{\delta_{A}-\delta_{C}}{4m}\ = \frac{\delta_{B}-\delta_{C}}{2m}\ \Rightarrow \delta_{A} = 2\delta_{B}-\delta_{C} $

$ \therefore\frac{F_{AD}L_{AD}}{A_{AD}E_{AD}}\ = 2(\frac{F_{BE}L_{BE}}{A_{BE}E_{BE}}\) - \frac{F_{CF}L_{CF}}{A_{CF}E_{CF}}\ $

$ \therefore \frac{F_{AD}*(2m)}{(30mm^2)*(193GPa)}\ = 2*(\frac{F_{BE}*(2m)}{(40mm^2)*(120GPa)}\) - (\frac{F_{CF}*(2m)}{(50mm^2)(200GPa)}\) $

After simplifying: $ \Rightarrow F_{AD} = 2.41*F_{BE} - 0.58*F_{CF} (3) $

Units are in meters which makes sense because $ \delta $ is displacement measured by a distance.

**Step 4** Solving the set of equations

$ F_{AD} + F_{BE} + F_{CF} = 24kN \hspace{10mm} (1)\\ -2F_{AD} + 2F_{CF} = -17kN \hspace{10mm} (2)\\ F_{AD} = 2.41*F_{BE} - 0.58*F_{CF} (3) $

Solving equation (1) & (3)

$ 2.41*F_{BE} - 0.58*F_{CF} + F_{BE} + F_{CF} = 24kN\\ 3.41*F_{BE} + 0.42*F_{CF} = 24kN (4) $

Solving equation (2) & (3)

$ -2*(2.41*F_{BE} - 0.58*F_{CF}) + 2*F_{CF} = -17kN\\ -4.82*F_{BE} + 3.16*F_{CF} = -17kN\\ F_{CF} = -3.85*F_{BE} $

Solving equation (4) & (5)

$ 3.41*F_{BE} + 0.42*(-3.85*F_{BE}) = 24kN\\ 1.793*F_{BE} = 24kN \\ F_{BE} = 13.385kN \\ F_{CF} = -51.53kN $

Solving equation (1)

$ F_{AD} = 24 + 51.53 - 13.385 = 62.145kN\\ $

**Step 5** Overall Free Body Diagram & Forces

$ F_{AD} = 62.1kN\\ F_{BE} = 13.4kN\\ F_{CF} = 51.5kN (compression)\\ $

## References

- ↑ http://www.engineeringwiki.org/wiki/Determinacy,_Indeterminacy_and_Stability
- ↑
^{2.0}^{2.1}^{2.2}^{2.3}^{2.4}^{2.5}^{2.6}^{2.7}Hibbeler, R.C. (2010).*Mechanics of Materials: 8th Edition.*Upper Saddle River, NJ: Pearson Education. - ↑
^{3.0}^{3.1}^{3.2}^{3.3}^{3.4}^{3.5}Kassimali, A. (2011).*Structural Analysis: SI Edition (4th edition).*Stamford, CT: Cengage Learning. - ↑
^{4.0}^{4.1}^{4.2}^{4.3}^{4.4}^{4.5}^{4.6}^{4.7}^{4.8}^{4.9}Hibbeler, R.C. (2007).*Engineering Mechanics: Statics - Dynamics. 11th Edition*Prentice Hall/ Pearson Education - ↑
^{5.0}^{5.1}^{5.2}^{5.3}The Physics Classroom. Hosted by comPADRE*http://www.physicsclassroom.com/class/newtlaws/u2l2b.cfm*. Acessed Oct/12/2013