## 10 Oct These assignments introduce the concept of equilibrium.? For a body to ?be at rest (in equilibrium) all forces must be balanced.? We use this ?fact to solve for unknown forces in the

These assignments introduce the concept of equilibrium. For a body to be at rest (in equilibrium) all forces must be balanced. We use this fact to solve for unknown forces in the x and y directions. Any moments calculated about a point on the body must also balance. We use the equilibrium equations to balance these forces.

complete and submit the problems

MET211 Statics

Unit 2 Quiz

Problem 1

Draw free-body diagrams of members AC and BG

Problem 2

Determine the reactions at A and B for the beams loaded as shown. Beam weight may be neglected.

Problem 3

Determine the reactions at A and B for the beams loaded as shown. Beam weight may be neglected.

Problem 4

Determine the reactions at A and B for the beams loaded as shown. Beam weight may be neglected.

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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

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