Lecture 12

ENGR-1100 Introduction to Engineering Analysis

Lecture Outline

• Rigid body equilibrium.

• FBD- Free Body Diagram.

Rigid Bodies

-A rigid body is an ideal object that has dimensions and mass but does not deform under loading.

-The size and shape does significantly affect the response (reaction at supports) to applied force/s - A force applied to a rigid body may be translated along its line of action without altering its effects (principle of transmissibility)

Necessary and Sufficient Conditions of Equilibrium of a Rigid Body

The necessary AND sufficient condition for a rigid body to be in equilibrium is that the resultant force and couple acting on the rigid body must be zero.

0 FR1

i

rrr∑=

==n

i0 MM

1i

rrr∑=

==n

i

Particle equilibrium

Those are vector equations.How many scalar equations in the 2D?

in the 3D?

Rigid body equilibrium

0 FR xix ∑=

==n

i 1

0 FR yiy ∑=

==n

i 1

0 FR ziz ∑=

==n

i 1

0 MM xix ∑=

==n

i 1

0 MM yiy ∑=

==n

i 1

0 MM ziz ∑=

==n

i 1

0 FR xix ∑=

==n

i 1

0 FR yiy ∑=

==n

i 1

0 MM ziz ∑=

==n

i 1

3 independent equations in 2 D

6 independent equations in 3D

Isolation of bodies: Free Body Diagram

-In a system of interacting bodies, to be able to apply Newton’s laws properly, a systematic technique is to:

isolate each of the bodies or a collection of bodies identify the forces acting on each of the bodies, and then apply the equilibrium equations to each.

-The sketch of the isolated body or system of bodies considered as a single body, with all the external forces acting on it

by mechanical contact with other bodiesby gravitational attraction (weight)

when the rest of the bodies are imagined to be replaced by their actions, it is known as a free body diagram.

Isolation of bodies: Free Body Diagram

The free body diagram is the most important single step in the solution of problems in mechanics.

Remember:1. The forces on the isolated body (or system of

bodies) are to be considered.2. Apply Newton’s Third Law (every action has

an equal, opposite and collinear reaction) carefully

Steps of Drawing a FBD

1. Clearly identify the body (or system of bodies) to be isolated (the FREE body).

2. Draw a diagram of this “free body” completely isolated from the rest of the bodies.

3. Traverse the boundary of this “free body” and indicate ALL forces acting ON the free body (contact forces with other bodies forces)Known forces: Show vector arrows with proper magnitude (UNITS!!), direction and sense.Unknown magnitude but known direction of force: Show vector arrows with magnitude assumed as positive (if calculations show that the magnitude is negative, the minus sign indicates that the sense is opposite to the one assumed) Unknown magnitude and direction of force: Show x- and y- components of the vector with unknown magnitudes.

4. Show coordinate directions on the diagram

Two-dimensional reaction at supports and connections

1. Gravitational attraction

2. Flexible cord, rope, chain, or cable

3. Rigid link 4. Ball, roller, or rocker

5. Smooth surface 6. Smooth pin

7. Rough surface 8. Pin in a smooth pin guide

9a. Collar on a smooth shaftPin connection

9b. Collar on a smooth shaft:Fixed connection

Three-dimensional reaction at supports and connections

1. Ball and socket 2. Hinge

3. Ball bearing 4. Journal bearing

5. Thrust bearing 6. Smooth pin and bracket

7. Fixed support

Example P6-2

Draw complete free-body diagram of the beam shown in Fig. P6-2, which has a mass m.

Solution

Known forces

P

mg

Unknown forces Ay

Ax

By

x

Class Assignment: Exercise set 6-1, 6-3, 6-4, 6-10please submit to TA at the end of the lecture

p6-1

Ax

Ay

MA

p

W

x

y

W N2N1

x

y

6-3

p

6-4 x

y

Ax

Ay

pT

6-10

p

F1 N1

W

p

F1

N1

N2

x

y

Example P6-25Draw complete free-body diagram of the bent bar shown in Fig. P6-25. The support at A is a journal bearing and the supports at B and C are ball bearings.

Solution

Az

P1

P2

P3

x

y

z

Ay

Mz My

ByBx

Cz

Cx

Known forces

Unknown forces

Class Assignment: Exercise set 6-22 & 6-24please submit to TA at the end of the lecture

p2

p1

p2

p1

Ay

Ax

Az

MAZ

MAx

MAy

p1

p2

p1

p2

T

Ax Ay

Az