1 / 25

# ENGI 1313 Mechanics I - PowerPoint PPT Presentation

ENGI 1313 Mechanics I. Lecture 25:Equilibrium of a Rigid Body. Lecture Objective. to illustrate application of 2D equations of equilibrium for a rigid body to examine concepts for analyzing equilibrium of a rigid body in 3D. Example 25-01.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

ENGI 1313 Mechanics I

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

#### Presentation Transcript

Lecture 25:Equilibrium of a Rigid Body

• to illustrate application of 2D equations of equilibrium for a rigid body

• to examine concepts for analyzing equilibrium of a rigid body in 3D

• Determine the force P needed to pull the 50-kg roller over the smooth step. Take θ = 60°.

 =

• What XY-coordinate System be Established?

y

x

 =

• Establish FBD

y

x

NB

 =

NA

w = mg = (50 kg)(9.807 m/s2) = 490 N

x

Example 25-01 (cont.)

• Determine Force Angles

• Roller self-weight

y

x

70

 = 20

NB

 =

 = 20

NA

w = 490 N

x

Example 25-01 (cont.)

• Determine Force Angles

• Normal reaction force at A

y

x

90

NB

 =

NA

w = 490 N

NA

x

Example 25-01 (cont.)

• Determine Force Angles

• Normal reaction force at B

y

x

r = 0.6 m

yB = (0.6 m – 0.1 m) = 0.5 m

NB

NB

 =

NA

w = 490 N

x

Example 25-01 (cont.)

• Draw FBD

w = 490 N

P

 = 20

y

 = 60

x

NB

NB

 =

NA

w = 490 N

NA= 0 N

x

Example 25-01 (cont.)

• What Equilibrium Equation should be Used to Find P?

• MB = 0

w = 490 N

P

 = 20

 = 60

xB = 0.3317 m

yB = 0.5 m

NB

NA

• If a support prevents rotation of a body about an axis, then the support exerts a ________ on the body about that axis.

• A) Couple moment

• B) Force

• C) Both A and B

• D) None of the above.

• Basic Equations

Moment equations can also be determined about any point on the rigid body. Typically the point selected is where the most unknown forces are applied. This procedure helps to simplify the solution.

• Engineering Design

• Basic analysis

• Check more rigorous methods

Axial Forces

• Design of Experimental Test Frame

Couple Forces

For Bending

• Ball and Socket

• Three orthogonal forces

• Single Journal Bearing

• Two forces and two couple moments

• Frictionless

• Circular shaft

• Orthogonal to longitudinal bearing axis

• Journal Bearing (cont.)

• Two or more (properly aligned) journal bearings will generate only support reaction forces

• Single Hinge

• Three orthogonal forces

• Two couple moments orthogonal to hinge axis

• Hinge Design

• Two or more (properly aligned) hinges will generate only support reaction forces

• What is the Common Characteristic?

• Statically determinate system

• Statically Indeterminate System

• Support reactions > equilibrium equations

• Rigid Body Instability

• 2-D problem

• Concurrent reaction forces

• Intersects an out-of-plane axis

• Rigid Body Instability

• 3-D problem

• Support reactions intersect a common axis

• Rigid Body Instability

• Parallel reaction forces

• Hibbeler (2007)

• http://wps.prenhall.com/esm_hibbeler_engmech_1