CHAPTER 4: Equilibrium of Rigid Bodies

CHAPTER 4: Equilibrium of Rigid Bodies

Equilibrium of 2-D BodiesEquilibrium of 2-D Bodies

2

IntroductionIntroduction

For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body.

The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces is equivalent to zero,

( )∑ ∑ =∑ ×== 00 FrMF Orrrr

∑ =∑ =∑ =∑ =∑ =∑ =

000000

zyx

zyxMMMFFF

In rectangular components,

3

Reactions at Supports & ConnectionsReactions at Supports & Connections

A smooth roller resists motion perpendicular to the surface only.

Normal force

Reactions equivalent to a force with known line of action.

4

Reactions at Supports & ConnectionsReactions at Supports & Connections

Reactions equivalent to a force with known line of action.

A cable can resist a pull only directed along its length.

Tensileforce

Guess the reaction of a short link and frictionless pin in slot:

axialforce

Normalforce

5

Reactions at Supports & ConnectionsReactions at Supports & Connections

Reactions equivalent to a force of unknown direction and magnitude.

A smooth pin resists horizontal and vertical translation but not rotation

6

Reactions at Supports & ConnectionsReactions at Supports & Connections

Reactions equivalent to a force of unknown direction and magnitude and a couple of unknown magnitude

Guess for a fixed or cantilever support

7

An illustration of the object showing ALL forces acting on it.

The Free Body Diagram (FBD)The Free Body Diagram (FBD)

Indicate all forcesacting on the object and dimensions of the object.

IMPORTANT:

Clearly identify the object.

FBD of the STATIONARY PICK-UP

x b

da

8

FBD ExampleFBD ExampleDetermine the FBD of the beam.The mass of the beam is 100 kg.

1200N2m

6m

A

1200N2m

A3m

G

981N

Ay

Ax

MA

Due to applied force

Due to beam weight

Due to fixed support

FBD of beam

9

FBD ExampleFBD Example

FBD of the pedal

10

FBD ExampleFBD ExampleTwo smooth pipes, each having a mass of 300kg, are

supported by the forks of the tractor. Draw the FBD for:

(a) each pipe and(b) both pipes together.

11

FBD ExampleFBD Example3rd law of motion:“Forces come in pair of action and reaction.”

FBD of pipe A

FBD of pipe B

12

FBD ExampleFBD ExampleFBD of pipes A and B together

13

Some simple cases of rigid body equilibriumSome simple cases of rigid body equilibrium

Two force body – a body held in equilibrium by forces acting at only two points in the body. Two force body

(LOAs coincide)

NOTE:LOAs of a two force body is known.

14

Some simple cases of rigid body equilibriumSome simple cases of rigid body equilibrium

Three force body – a body held in equilibrium by three concurrent forces.

Three force body(concurrent LOAs)

The forces in a three force body form a closed triangle.

15

Equilibrium Conditions on a PlaneEquilibrium Conditions on a PlaneF x 0

F y 0

Mo 0

3 available equations3 unknown quantities (max.)

When the equilibrium equations are enough to solve the reactions, the reactions are “determinate.”

Reactions become indeterminate when there are more unknown quantities than equations.

The 3 equations can not be augmented with additional equations, but they can be replaced

∑ ∑ ∑ === 000 BAx MMF

16

Determinate vs. IndeterminateDeterminate vs. Indeterminate

=

• More unknowns than equations

• 3 eqns & 3 unknowns• Fewer unknowns but

partially constrained

17

STEPS IN SOLVING PROBLEMS INVOLVING EQUILIBRIUMSTEPS IN SOLVING PROBLEMS INVOLVING EQUILIBRIUM

1. Identify the object, its surrounding and boundary.

2. Draw the necessary FBD.3. Identify the required quantities.4. Set-up the necessary equilibrium

equations based on the FBD.5. Solve for the unknown quantities

based on the equations.6. Verify when in doubt.

18

Example 33Example 33Draw the FBD of the platform and solve the unknown reactions on it.The weight of the platform is 1967N.

G70º

1.4m0.8m

1.0mA

B T = tensionW = weight of platfromAx = horizontal reaction at AAy = vertical reaction at A

G70º

1.4m0.8m

1.0m

A

B

Ax

Ay

T

W

FBD of platform

19

Solution (scalar approach)Solution (scalar approach)

o

o

oo

r

r

r

70sin19670

0)(

70cos0

0)(

)70cos1()70sin2.2()19674.1(0

0)(

TNA

F

TA

F

mTmTNm

Mccw

y

y

x

x

A

+−=

=↑+

+=

=→+

⋅+⋅+⋅−=

=+

∑

∑

∑

G70º

1.4m0.8m

1.0m

A

B

Ax

Ay

T

W

FBD of platform

20

Solution (scalar approach)Solution (scalar approach)

NTmNmT

TmNm

1.1143)(4.28.2753

4.28.27530

+=

=

+−=

NA

NNA

y

y

8.892)(

70sin)1.1143(19670

+=

++−= o

G70º

1.4m0.8m

1.0m

A

B

Ax

Ay

T

W

FBD of platform

NANA

x

x

391)(70cos)1.1143(0

+=++= o

21

Solution (scalar approach)Solution (scalar approach)

G70º

1.4m0.8m

1.0m

A

B

Ax

Ay

T

W

FBD of platform

→=

↑=

=

NA

NANT

Summary

x

y

391

8.8921.1143:

22

Solution (vector approach)Solution (vector approach)

)70sin70cos(

)1967()(00

0

jTiT

NjjAiATWA

F

yxrr

rrr

rrr

r

oo +−+

−++=

++=

=∑

G70º

1.4m0.8m

1.0m

A

B

Ax

Ay

T

W

FBD of platform

070sin70cos012.2)19674.1(0

)()(0

0

oo

rrr

rr

rrrr

r

TTmm

kjiNjmi

TrWr

M

Tw

A

−+−×=

×+×=

=∑

23

Solution (vector approach)Solution (vector approach)

G70º

1.4m0.8m

1.0m

A

B

Ax

Ay

T

W

FBD of platformN

mNmT

TmNmmkTmkTNmk

1.11434.28.2753

4.28.2753070cos70sin2.2)19674.1(0

==

+−=++−⋅=

rrroo

NA

NNA

y

y

8.892)(

70sin)1.1143(19670

+=

++−= o

NANA

x

x

391)(70cos)1.1143(0

+=++= o

24

Example 34Example 34The link shown is pin connected at A and rests against a smooth support at B. Compute the vertical and horizontal reaction at the pin A.

FBD of link

25

Solution (vector approach)Solution (vector approach)

NmkNmkkNm

NmkNjrNr

M

NjjAiAjNiN

NjAN

F

B

BB

A

yxBB

B

rrr

rrrrr

r

rrrrr

rrr

r

oo

9060)75.0(0

)90()60()(0

0

60)30cos30sin(0

)60(0

0

1

−−⋅=

−+−×+×=

=

−++−−=

−++=

=

∑

∑

FBD of link

26

Solution (vector approach)Solution (vector approach)

NA

NNA

NjjAjN

NANA

iAiN

NjjAiAjNiN

NNNmkNmkkNm

y

y

yB

x

x

xB

yxBB

NB

B

2.233)(

6030cos)200(

6030cos0

100)(30sin)200(

30sin0

60)30cos30sin(0

2009060)75.0(0

75.0150

+=

+=

−+−=

+==

+−=

−++−−=

==

−−⋅=

o

o

o

o

oo

rrr

rr

rrrrr

rrr

27

Solution (vector approach)Solution (vector approach)

NjA

NiA

NNSummary

y

x

B

r

r

2.233

100

200:

=

=

=FBD of link

28

Example 35Example 35The lever ABC is pin-supported at A and connected to a short link BD as shown. The weight of the members are negligible. Determine the reaction of the pin to the lever at A.

29

FBD of lever(3-force body)

FBD of link(two-force body)

SolutionSolution

30FBD of lever(3-force body)

( ) o26.60tan 4.02.05.01 == +−θ

FA

F

400

45º

o26.60o74.119

15.26º

45º

NF

F

A

A

FA

6.1074

40026.15sin

45sin

40026.15sin45sin

=

=

=

o

o

oo

SolutionSolution