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Chapter 6 : Equilibrium of single rigid bodies

6.1 Introduction
6.2 Equilibrium equations and degrees of freedom
6.3 Types of supports
6.4 Procedure to analyze problems
6.5 Self-test
6.6 Problems, 2-D
6.7 Problems, 3-D

6.1 Introduction
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In the preceeding chapters we have considered various aspects as to how we can use a basic law of physics ( sum of the forces equal to zero) to determine unknown forces in connection with individual particles where rotational aspects were not of interest.

For single rigid bodies (like a bridge) we found in the preceeding chapter that these considerations are not sufficient but that the ability of forces to rotate a body has to be taken into account. This lead to the definition of the moment of a force and to the statement that the sum of the moment of ALL forces has to be zero in order for the body not to rotate.

In this chapter I summarize these results and the procedures of analysis as applied to determining forces acting on rigid bodies.

After briefly talking about several types of supports for rigid bodies I will bring several worked-out examples first for single-body and then for multi-body problem.

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6.2 Equilibrium Equations and degrees of freedom
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We have two types of equations resulting from :
  1. Vectorial sum of the forces equals zero

    (6.2a)      

    or in component form :

  2. Vectorial sum of the moments equals zero

    The sum of the moments around ANY point of ALL the forces acting on a rigid body has to be ZERO.

    (5.7a)      

    or in component form :

    where the moment vector is defined by :

    The vector is defined as a vector going from the point of rotation to anywhere on the line of action of the force vector .
    The point of rotation (the point around which you consider the sum of the moments) can be chosen arbitrarily. Some choices will require less computations than others. Look here if you want to know why your choice for the location of the point of rotation is arbitrary.

In the case of 2-dimensional problems (where we normally choose as coordinates the x- and y-axes ) forces have zero z-component. Hence, the sum of z-components of all forces is automatically satisfied and we have to consider only the sum of the forces in the x- and in the y-direction.

Furthermore, in 2-dimensional problems, rotation takes place only around an axis perpendicular to the display area (which is usually the z-axis). Or in other words, none of the forces has a moment around the x- and the y-axis. Therefore only the sum of the moments around the z-axis has to be considered. Often we speak in this case of a point of rotation lying in the plane of the display area.

Hence, as far as the equilibrium of a single rigid body is concerned the number of algebraic equations available is :

3 equations for 2-dimensional problems
6 equations for 3-dimensional problems
This puts restrictions on the types and number of forces which support a single rigid body (system). If the number of unknowns ( = the number of degrees of freedom ) resulting from the support forces is equal to the number of available equilibrium equations, we speak of a statically determinate system, if the number of unknowns exceeds the number of available equilibrium equations, the system is called statically indeterminate. In such a case the body's deformation for example (even if it is only minute) is needed to solve for all the unknown support forces. If the number of unknowns is less than the number of equations the body is said to be underconstrained, the body will simply move and/or rotate under influence of the given external loads.

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6.3 Types of supports
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Discussing the different ways as to how a body can be supported is important because the type of support tells you something about the support force itself and the number of unknowns resulting from it.

Two-dimensional supports
Symbols/text used Active Force # of unknowns
1
1
2


Three-dimensional supports
Symbols/text used Active Force # of unknowns
1
2
Ball/socket
3


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6.4 Procedure to analyze problems
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  1. Sketch the Free-Body Diagram of the rigid body carefully. Interior aspects of the body usually are not of interest. The F.B.D. should show ALL forces acting on the body from the outside.
    This means : all given loads and/or field forces ( gravity etc. ) and the unknown forces (often called the support forces). Quite often it is important to recognize that the problem statement includes information about for example the direction of a support force either in the text or in the figures.
    In many problems the weight of some or all items can be neglected. Reading the problem statement carefully will tell you whether or not that is the case.

    In real life you will have to err always on the safe side.

  2. Choose a coordinate system if not given in the problem statement. The location and orientation of your system do not effect the principal outcome of your calculations but influence the amount of work you will have to do in order to arrive at a solution.
    Look for dominant directions and align these with one of the coordinate axis. By in large only experience will teach you to find solutions with the least amount of work. Label uniquely all forces ad relevant distances as far as not given already in the problem statement. The same label must not be used for two different items.

  3. Set up the relevant equilibrium equations using the symbols you defined previously yourself and which were given in the problem statement.
    If the problem is 2-dimensional make use of the different ways you can determine the moment of a force.

  4. Solve the derived equations for the unknown forces and summarize your results.

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6.5 Self-Test
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Sorry, none available. Humor me though and try those available in previous chapters.

Chapter 2
Chapter 3
Chapter 4

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6.6 Problems, 2-D
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Problems in 2-dimensions are distinct in that for a single rigid body you have to consider :

Two equations for the sum of the forces, one for each direction.

The sum of the moment around an axis perpendicular to the plane in which the force lie.

for a total of three algebraic equations. It is often of advantage to substitute additional moment equation(s) for one or both sum-of-the-forces equation(s).

Problem 6.6a : Crate in equilibrium
Problem 6.6b : Ladder against wall, I
Problem 6.6c : Ladder against wall, II
Problem 6.6d : Beam on incline
Problem 6.6e : Supports for a truss
Problem 6.6f : Underconstrained plate
Problem 6.6g : Balancing Act I
Problem 6.6h : Balancing Act II

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6.7 Problems, 3-D
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Problem 6.7a : Hooked beam

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Zig Herzog; hgnherzog@yahoo.com

Last revised: 01/12/07