ME 201Engineering Mechanics: Statics

Unit 6.1

Conditions for Rigid-Body Equilibrium

Free-Body Diagrams

Equilibrium

For equilibrium

Balance of forces

Prevents translation

Balance of moments

Prevents rotation

∑ F =0

∑ M =0

We will use both scalar (2D) and vector (3D) analysis methods

Free Body Diagrams

1. Sketch shape free from constraints &

connections

2. Include external forces and couples

Applied loadings

Reactions at supports

Weight of the body

3. Include dimensions

Support Reactions

General conditions:

If a support prevents translation in a given

direction, then a force is developed on the

member in that direction

If rotation is prevented, a couple moment is

exerted on the member

Common Support Reactions

Simple Support

Pin or Hinge Support

Fixed Support

Common Support Reactions

Simple support (roller, smooth)

1 unknown

Reaction normal to surface

No moment

Common Support Reactions

Pin or Hinge Support

2 unknowns

Components along X & Y

No moment

Common Support Reactions

Fixed Support

3 unknowns

Components along X & Y

Moment

Example Problem

The beam has as mass of 100 kg.

Draw FBD of beam

Example Problem Solution

Ax

Ay

2 m

3 m

1200 N

100 kg * 9.81

981 N

MA

A G

The beam has as mass of 100 kg.

In Class Exercise

Draw the FBD of the truss

Solution

Ax

Ay

Bx

4 kN

5 kN

In Class Exercise

Draw the FBD of

member AB

Solution

Bx

By

200 N

34

5

NA

30°

In Class Exercise

Draw the FBD of member AB

Solution

Ax’

By’

MA

400 lb ft

In Class Exercise

Draw the FBD of member AB

Solution

Ax

Ay 30°

T

500 Nm

Equilibrium

For equilibrium

Balance of forces

Prevents translation

Balance of moments

Prevents rotation

∑ F =0

∑ M =0

We will use both scalar (2D) and vector (3D) analysis methods

Free Body Diagrams

1. Sketch shape free from constraints &

connections

2. Include external forces and couples

Applied loadings

Reactions at supports

Weight of the body

3. Include dimensions

Support Reactions

General conditions:

If a support prevents translation in a given

direction, then a force is developed on the

member in that direction

If rotation is prevented, a couple moment is

exerted on the member

Common Support Reactions

Simple Support

Pin or Hinge Support

Fixed Support

Common Support Reactions

Simple support (roller, smooth)

1 unknown

Reaction normal to surface

No moment

Common Support Reactions

Pin or Hinge Support

2 unknowns

Components along X & Y

No moment

Common Support Reactions

Fixed Support

3 unknowns

Components along X & Y

Moment

More Support Reactions

See Text, pages 202-203

Solving Problems

Begin with FBD (Free Body Diagram)

Use Equations of Equilibrium to formulate

equations containing unknown forces,

reactions, & moments

∑ Fx =0 ∑ Fy =0 ∑ Mo =0

Solve simultaneous equations (or look for

equation with single unknown)

Example Problem

The beam has as mass of 100 kg.

Draw FBD of beam

Example Problem Solution

Ax

Ay

2 m

3 m

1200 N

100 kg * 9.81

981 N

MA

A G

The beam has as mass of 100 kg.

In Class Exercise

Vertical force applied to pedal

Force in link at B is 20 lb.

Spring is stretched 1.5 in.

Draw FBD of pedal

Solution

F

20 lb

20 lb/in * 1.5 in

30 lb

Ax

Ay

5”

1.5”

1”

Vertical force applied to pedal

Force in link at B is 20 lb.

Spring is stretched 1.5 in.

Draw FBD of pedal

In Class Exercise

Platform has a mass of 200 kg

Draw FBD of platform

Solution

Ax

Ay

1.4 m 0.8 m

1 m

70°

200 kg * 9.81

A

T

Platform has a mass of 200 kg

Draw FBD of platform

In Class Exercise

Draw the FBD of

member AB

Solution

Bx

By

200 N

34

5

NA

30°

In Class Exercise

Draw the FBD of the truss

Solution

Ax

Ay

Bx

4 kN

5 kN

In Class Exercise

Draw the FBD of member AB

Solution

Ax

By

MA

400 lb ft

In Class Exercise

Draw the FBD of member AB

Solution

Ax

Ay 30°

T

500 Nm

Example Problem

Pipes have a mass

of 300 kg each.

Draw FBD of each

pipe and both pipes

together

Solution

300 kg * 9.81

2943 N

30°

30°R

NB

300 kg * 9.81

2943 N

N

NA

BA R

30°

30°30°

Pipes have a mass of 300 kg each.

Draw FBD of each pipe and both

pipes together

Solution

300 kg * 9.81

2943 N

30°

NB

300 kg * 9.81

2943 N

N

NA

B

A

30°

30°

Pipes have a mass of 300 kg each.

Draw FBD of each pipe and both

pipes together