Have you ever had something break when you were trying to use it? Material will break when we apply more force than it can handle.
Everyday we use structures or physical objects to carry forces for us. This can be in the form of furniture, consumer products, transportation, buildings, toys, machines etc. As engineers we are expected to build new structures all the time. When designing the structure or object for a specific task we need the material to be strong enough to handle the forces involved with its intended use.
When these external forces are being applied with no movement/acceleration we can use an equilibrium analysis to understand how they will translate to internal forces that the material will have to support. Then we can select a material strong enough to withstand every day use with an added factor of safety.
In your original post, include the following:
- Choose a structure or object and explain how to construct a free-body diagram (FBD) of part or all of it.
- Show your structure/object and the free body diagram you created for it.
- Explain the difference between the external forces on the structure and the internal forces translating through the material itself.
- Make sure to choose a real-world example that is different from your peers!
MET211 Statics
equilibrium
Introduction
Free-body diagrams are diagrams of objects in static equilibrium
“Static” means the object is not moving but is “at rest.” “Equilibrium” means the forces acting on the object are in equilibrium or balanced against each other so that the object does not move.
A diagram that shows an object or body with all supports removed and replaced by forces in balance appears to be floating freely in space. It is called a free-body diagram (FBD).
Drawing it is a necessary first step in calculating the forces acting on the object.
Free Body Diagram
1. The rope tension is replaced by a force of 60 lb pulling upward on the crate.
2. The pull of gravity is 100 lb on the crate.
3. The floor had been partially supporting or pushing up on the crate. The support of the floor on the crate is shown as N.
The forces are of three categories:
Applied forces
Nonapplied forces such as weight
Forces replacing a support or sectioned member
Free Body Diagram Conventions
Forces drawn to replace various supports or connections on the main members.
Roller: The roller cannot exert a horizontal force; therefore, only a force perpendicular to the surface is present.
Roller: The only force present is that perpendicular to the roller surface
Free Body Diagram Conventions
Forces drawn to replace various supports or connections on the main members.
Smooth surface: Zero friction is assumed; therefore, only one force, that perpendicular to the surface, is present
Slot: The same principle applies as for a smooth surface: There is only one force present, that perpendicular to the slot
Free Body Diagram Conventions
Forces drawn to replace various supports or connections on the main members.
Pinned: Both horizontal and vertical components must be assumed at a pinned connection unless it is on a roller or smooth surface
The orientation of the support is immaterial; a horizontal and vertical force must still be assumed
Free Body Diagram Conventions
Forces drawn to replace various supports or connections on the main members.
Cable: There is always a single force pulling in the direction of the cable
Fixed support: The beam is embedded in the wall or support and therefore has three possible reactions: a moment, vertical force, and horizontal force.
Free Body Diagram Conventions
All objects or members will be assumed to be weightless unless the weight is specifically stated.
When considering weight later on, one can easily include it in the calculation by adding another force passing through the center of gravity of the object or member.
We will assume zero friction on smooth surfaces.
When friction is to be included later, the coefficient of friction or some other clear indication that it must be considered will be given.
If you assume an incorrect force direction, the calculated answer will be negative.
If this answer must be used in further calculations, substitute it into equations as a negative value and do not change any vector directions until the calculations are complete.
Free Body Diagram Conventions
Free Body Diagram Conventions
Free Body Diagram Conventions
Because cables can only be in tension (not compression), the free-body diagram of ring B shows three forces pulling on B.
Three Equations of Equilibrium
For complete static equilibrium, three requirements must be met:
1. Vertical forces balance.
2. Horizontal forces balance.
3. Moments balance; cw = ccw (about any point).
A concise form of stating the same points is:
∑Fy = 0
∑Fx = 0
∑M = 0
Three Equations of Equilibrium
Solve for the forces at A and B of the beam shown below
Three Equations of Equilibrium
Solve for the forces at A and B of the beam shown below
Two Force Members
A member that is acted upon by two forces is known as a two-force member.
A two-force member will always be in either tension or compression.
When a member is acted upon by at least three forces at several locations, there will likely be not only compression or tension but also bending.
In Figure shown, member BD is a two-force member, and members AC and CE are three-force members.
Two Force Members
Two Force Members
Pulleys
Cable tension due to the 20.4-kg mass is 20.4 (9.81) = 200 N
Pulleys
Cable tension due to the 20.4-kg mass is 20.4 (9.81) = 200 N
Coplanar Concurrent Forces
Find the load in each section of the cable system
1223 kg (9.81 m/s2) = 12 kN
Coplanar Concurrent Forces
∑Fx = 0
∑Fy = 0
Coplanar Concurrent Forces
∑Fx = 0
Coplanar Parallel Forces
A horizontal beam with only vertical loading on it is an example of a force system in which all the forces are parallel and in the same plane—a coplanar parallel system.
The beam is supported at two points. The free-body diagram replaces these supports with equivalent forces or reactions.
One of the reaction forces is determined by taking moments about the other point of support.
Coplanar Parallel Forces
A beam has concentrated loads applied as shown in the figure. Calculate the reactions at A and B. (Ignore the weight of the beam.)
Coplanar Parallel Forces
A beam has concentrated loads applied as shown in the figure. Calculate the reactions at A and B. (Ignore the weight of the beam.)
10RB = 600 + 2000
Coplanar Parallel Forces
Coplanar Nonconcurrent Force Systems
When dealing with coplanar nonconcurrent force systems, we are faced not only with vertical forces and moment equations but also with horizontal forces since the applied forces are no longer parallel.
Pinned connections will now have horizontal components.
Coplanar Nonconcurrent Force Systems
Solve for reactions RA and RB
ΣMA = 0
Coplanar Nonconcurrent Force Systems
Solve for reactions RA and RB
References
All figures and examples are taken from
APPLIED MECHANICS FOR ENGINEERING TECHNOLOGY, EIGHTH EDITION, Keith M. Walker, ISBN 978-0-13-172151-7