2. Objects in Equilibrium - Chulalongkorn University: …pioneer.netserv.chula.ac.th/~mkuntine/42-111/files/ch2... · · 2012-01-272 2 Objectives Students must be able to #1 Course

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2. Objects in Equilibrium

2142111 Statics, 2011/2

© Department of Mechanical Engineering, Chulalongkorn University

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Objectives Students must be able to #1

Course ObjectiveAnalyze rigid bodies in equilibrium

Chapter ObjectivesState the equations of equilibrium for various bodies (particles, 2D/3D rigid bodies)Draw free body diagrams (FBDs) of various bodies (select the body, draw the isolated body, apply loads and support reactions, add axes and dimensions, use 3 colors to differentiate the body, loads/reactions and other information)Substitute information from FBDs into equation of equilibrium in scalar & vector forms and solve for support reactions

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Objectives Students must be able to #2

Chapter ObjectivesState the characteristics and identify 2 and 3-force membersDetermine from FBDs whether a body in equilibrium is statically determinate, statically indeterminate with redundant supports or statically indeterminate with improper supports

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ContentsEquilibrium of Objects

Particles2D Rigid Bodies3D Rigid Bodies

SD and SI Problems

This is the of Statics

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Equilibrium Definition

An object is in equilibrium when it is stationary or in steady translation relative to an inertial reference frame.

Equilibrium

= =

= =

∑∑

v v v

v v v0

0O

R

R O

F F

M M

When a body is in equilibrium, the resultant force and

the resultant couple about any point are both zero.R

R

F

M O

v

v

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Equilibrium Procedure

Equilibrium

Formulate problems from physical situations. Simplify problems by making appropriate assumptionsDraw the free body diagram (FBD) of objects under considerationState the condition of equilibriumSubstitute variables from the FBD into the equilibrium equationsSubstitute the numbers and solve for solutions

Delay substitute numbersUse appropriate significant figures

Technical judgment and engineering senseTry to predict the answersIs the answer reasonable?

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Equilibrium Free Body Diagram (FBD)

Equilibrium

FBD is the sketch of the body under consideration that is isolated from all other bodies or surroundings.

The isolation of body clearly separate cause and effectsof loads on the body.

A thorough understanding of FBD is most vital for solving problems.

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Equilibrium FBD Construction

Equilibrium

No FBDs Cannot apply equilibrium conditions Stop...........

→→

Select the body to be isolatedDraw boundary of isolated body, excluding supportsIndicate a coordinate system by drawing axesAdd all applied loads (forces and couples) on the isolated body.Add all to support reactions (forces and couples) represent the supports that were removed.Beware of loads or support reactions with specific directions due to physical meanings Add dimensions and other information that are required in the equilibrium equation

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Equilibrium Helps on FBD

Use different colours in FBDsBody outline - blueLoad (force and couple) - redMiscellaneous (dimension, angle, etc.) - black

Equilibrium

Establish the x, y axes in any suitable orientation.Label all the known and unknown applied load and support reaction magnitudes Beware of loads or support reactions with specific directions due to physical meanings Otherwise, directions of unknown loads and support reactions can be assumed.

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Equilibrium On FBD Analyses

Objective: To find support reactionsApply the equations of equilibrium: state equations and substitute in the variables from the FBD under consideration

Load components are positive if they are directed along a positive direction, and vice versaIt is possible to assume positive directions for unknown forces and moments. If the solution yields a negative result, the actual load direction is opposite of that shown in the FBD.

, , ,

0 or 0, 0, 0

0 or 0, 0, 0x y z

O O x O y O z

F F F F

M M M M

= = = =

= = = =

∑ ∑ ∑ ∑∑ ∑ ∑ ∑

v v

v v

Equilibrium

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Particles Definition of Equilibrium

A particle is in equilibrium when it is stationary or in steady translation relative to an inertial reference frame.

=∑v v

0F

= = =∑ ∑ ∑0, 0, 0x y zF F F

Particle Equilibrium

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Example Hibbeler Ex 3-1 #1

The sphere has a mass of 6 kg and is supported as shown.Draw a free-body diagram of the sphere, the cord CE, the knot at C and the spring CD.


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State the physical meanings of these 9 forcesDo any of these forces have specific directions?

1

2

3 4

5

6

7

8

9

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Determine the tension in cables ABand AD for equilibrium of the 250-kg engine shown.


Physical meanings of forcesSpecific directions?

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Particle is in equilibrium.

0 cos30 0 (1)

0 sin30 0.25 kN 0 (2)

Solve (2) and subst. into (1) 4.9035 kN, 4.2466 kN4.90 kN, 4.25 kN Ans

x B D

y B

B D

B D

A

F T T

F T g

T TT T

⎡ ⎤= ° − =⎣ ⎦⎡ ⎤= ° − =⎣ ⎦

= =

= =

∑∑

29.807 m/sg =


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If the sack at A has a weight of 20 lb, determine the weight of the sack at Band the force in each cord needed to hold the system in the equilibrium position shown.


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⎡ ⎤= ° − ° =⎣ ⎦⎡ ⎤= ° − ° − =⎣ ⎦

= =

= =

∑∑

Consider ring in equilibrium

0 sin30 cos45 0 (1)

0 cos30 sin45 (20 lb) 0 (2)

Solve (1) & (2), 38.637 lb, 54.641 lb38.6 lb, 54.6 lb Ans

x EG EC

y EG EC

EC EG

EC EG

E

F T T

F T T

T TT T


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⎡ ⎤= ° − =⎣ ⎦⎡ ⎤= + ° − =⎣ ⎦

= =

= =

∑∑

FBD of in equilibrium

0 ( lb)cos 45 (4 5) 0 (3)

0 (3 5) ( lb)sin45 0 (4)

Solve (3) & (4), 34.151 lb, 47.811 lb34.2 lb, 47.8 lb Ans

x EC CD

y CD EC B

CD B

CD B

C

F T T

F T T W

T WT W


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Particles Equilibrium in 3D

=

+ + =

∑∑

v v

v0

ˆ ˆ ˆ( ) 0x y z

F

F i F j F k

=

=

=

∑∑∑

0

0

0

x

y

z

F

F

F

v1F

v2F

v3F

xy

z


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A 90-N load is suspended from the hook. The load is supported by two cables and a spring having a stiffness k = 500 N/m. Determine the force in the cables and the stretch of the spring for equilibrium. Cable ADlies in the x-y plane and cable AClies in the x-z plane.


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[ ]

Equilibrium of FBD of

0 sin30 (4 5) 0 (1)

0 cos30 0 (2)

0 (3 5) (90 N) 0 (3)

Solve (3), (1), (2) : 150 N,240 N, 207.85 N 208 N Ans

/ (207.85 N) /(50

x D C

y D B

z C

C

D B

s B AB

AB B

A

F F F

F F F

F F

FF FF ks F ks

s F k

⎡ ⎤= ° − =⎣ ⎦⎡ ⎤= − ° + =⎣ ⎦⎡ ⎤= − =⎣ ⎦

=

= = =

= =

= =

∑∑∑

0 N/ m)0.41569 m 0.416 m AnsABs = =


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Determine the force developed in each cable used to support the 40-kN crate.


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Example Hibbeler Ex 3-7 #2Position vectors

ˆ ˆ ˆ3 4 8 mˆ ˆ ˆ3 4 8 m

Force vectors

ˆ ˆ ˆ( 3 4 8 )89

ˆ ˆ ˆ( 3 4 8 )89

ˆ ˆ40 kN

AB

AC

AB BB B

AB

AC CC C

AC

D D

r i j k

r i j k

r FF F i j krr F

F F i j kr

F F i

W k

= − − +

= − + +

= = − − +

= = − + +

=

= −

v

v

vv

vv

v

v


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⎡ ⎤=⎣ ⎦+ + + =

− − + + − + + + + − =

− −− + + + + + − =

∑v v

v v v v v

v

v

Equilibrium of

0

0

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( 3 4 8 ) ( 3 4 8 ) ( ) ( 40 kN) 089 89

3 4 83 4 8ˆ ˆ ˆ( ) ( ) ( 40 kN) 089 89 89 89 89 89

B C D

CBD

C C CB B BD

A

F

F F F WFF i j k i j k F i k

F F FF F FF i j k


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33 dir.: 0 (1)89 89

44 dir.: 0 (2)89 89

88 dir.: (40 kN) 0 (3)89 89

From (2),16From (3), (40 kN) 0 or 23.585 kN

896From (1), 0 or 15.0 kN89

23.6 kN, 15.0 kN Ans

CBD

CB

CB

B C

CC

CD D

C B D

FFx F

FFy

FFz

F FF

F

FF F

F F F

−− + =

−+ =

+ − =

=

− = =

−+ = =

= = =


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Equilibrium of 2D Rigid Bodies

Use similar analyses as those for particlesAdditional consideration

Action forces in supports/constraintsFree-body diagram (FBD) of 2D rigid bodiesEquilibrium equations (scalar form) for rigid bodiesTwo-force and three-force members

2D Equilibrium

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=

=

=

∑∑∑

0

0

0

x

y

O

F

F

M

Equilibrium 2D EquationsScalar Form

The sum of the moment about any point O is zero.

Particle

Rigid Body

2D Equilibrium

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2D Supports Pin Joints

The pin supportprevents the translation by exerting a reaction force on the barallows a free rotation so it does not exert a couple on the bar.

2D Equilibrium

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2D Supports Roller Supports

The roller supportprevents the translation in y-direction by exerting a reaction force to the bar.allows a translation in x direction and a free rotation so it does not exert and a couple to the bar.

2D Equilibrium

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2D Supports Fixed Supports

The fixed or built-in supportprevents all translation and rotation by exerting a reaction force and a couple to the bar.

2D Equilibrium

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2D Supports Summary #1

2D Equilibrium

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2D Equilibrium

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2D Equilibrium

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Example Hibbeler Ex 5-1 #1Draw the free-body diagram of the uniform beam shown.The beam has the weight of 981 N.

2D Equilibrium

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2D Equilibrium

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2D Equilibrium

Find: , ,

Equilibrium of

0 0, 0 Ans

0 (1200 N) (981 N) 0

2181 N 2.18 kN Ans

0 (1200 N)(2 m) (981 N)(3 m) 0

5343 N m 5.34 kN m Ans

y y A

x x x

y y

y

A A

A

A A M

AG

F A A

F A

B

M M

M

⎡ ⎤= − = =⎣ ⎦⎡ ⎤= − − =⎣ ⎦

= =

⎡ ⎤= − − =⎣ ⎦= ⋅ = ⋅

∑∑

∑

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Two smooth pipes, each having the weight of 2943 N, are supported by the forks of the tractor. Draw the FBDs for each pipe and both pipes together.

2D Equilibrium

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2D Equilibrium

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2D Equilibrium

Pipes in equilibriumFind , , , T P F R

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2D Equilibrium

0

(2943 N)sin30° 0 1471.5 N 1.47 kN Ans

0

(2943 N)cos30° 0 2548.7 N 2.55 kN Ans

x

y

F

RR

F

PP

⎡ ⎤=⎣ ⎦− == =

⎡ ⎤=⎣ ⎦− == =

∑

∑

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2D Equilibrium

0 (2943 N)sin30° 0

2943 N 2.94 kN Ans

0 (2943 N)cos30° 0

2548.7 N 2.55 kN Ans

x

y

F R T

T

F F

F

⎡ ⎤= − − + =⎣ ⎦= =

⎡ ⎤= − =⎣ ⎦= =

∑

∑

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Determine the horizontal and vertical components of reaction for the beam loaded as shown. Neglect the weight of the beam in the calculation.

2D Equilibrium

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⎡ ⎤= ° − =⎣ ⎦= =

⎡ ⎤= + ° −⎣ ⎦° − =

= =

⎡ ⎤= − ° −⎣ ⎦

∑

∑

∑

Find: , ,

Equilibrium of

0 (600 N)cos 45 0

424.26 N 424 N Ans

0 (100 N)(2 m) (600 N)sin45 (5 m)

(600 N)cos 45 (0.2 m) (7 m) 0

319.50 319 N Ans

0 (600 N)sin45 (100

y x y

x x

x

B

y

y

y y

A B B

ADB

F B

B

M

A

A

F A − + =

= =

N) (200 N) 0

404.76 N 405 N Ansy

y

B

B

2D Equilibrium

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The link shown is pin-connected at A and rests against a smooth support at B. Compute the horizontal and vertical components of reactions at the pin A.

2D Equilibrium

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FBD of link BA

2D Equilibrium

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⎡ ⎤= + − ⋅ − + =⎣ ⎦=

⎡ ⎤= → + − ° = =⎣ ⎦⎡ ⎤= ↑ + − ° − =⎣ ⎦

= =

∑

∑∑

Equilibrium of FBD of link

0 (90 N m) (60 N)(1 m) (0.75 m) 0

200 N Ans

0 sin30 0, 100 N Ans

0 cos30 (60 N) 0

233.21 N 233 N Ans

A B

B

x x B x

y y B

y

BA

M N

N

F A N A

F A N

A

2D Equilibrium

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Special Member 2-Force Member #1

The object is a two-force memberwhen subjected to two equivalent forces acting at different points.

2D EquilibriumDefinition

For a two-force member in equilibrium, forcesare equal in magnitude.are opposite in direction.have the same line of action.The are no other loads, e.g. weight

The ability to recognize 2-force members is important in the analyses of structures.

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Special Member 2-Force Member #2Example

2D Equilibrium

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The object is a three-force member when subjected to three equivalent forces acting at different points.

If a three force member is in equilibrium, the threeforces are:

coplanar andeither parallel or concurrent.

Definition2D Equilibrium

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1Fv

2Fv

3FvO

3Fv

2Fv

1Fv

Concurrent Forces Parallel Forces

2D Equilibrium

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The lever ABC is pin-supported at A and connected to a short link BD as shown. If the weight of the members is negligible, determine the force of the pin on the lever at A.

2D Equilibrium

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2D Equilibrium


0

(400 N)(0.7 m) cos 45 (0.2 m)sin45 (0.1 m) 0

1319.9 N 1.32 kN Ans

0

(400 N) cos 45 0533.33 N 533 kN Ans

0

sin45 0

933.33 N 933

A

BD

BD

BD

x

BD x

x

y

BD y

y

ABC

M

FF

F

F

F AA

F

F A

A

⎡ ⎤=⎣ ⎦− + ° +

° =

= =

⎡ ⎤=⎣ ⎦− ° + =

= =

⎡ ⎤=⎣ ⎦− ° + =

= =

∑

∑

∑

kN Ans

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BD is a two force member

ABC is a three-force member.Must be concurrent at O

1 0.7tan 60.2550.4

θ − ⎛ ⎞= = °⎜ ⎟⎝ ⎠

2D Equilibrium

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⎡ ⎤= → +⎣ ⎦° − ° + =

⎡ ⎤= ↑ +⎣ ⎦° − ° =

= =

= =

∑

∑


0

cos60.255 cos 45 (400 N) 0 (1)

0

sin60.255 sin45 0 (2)

Solve (1) & (2)1074.9 N 1.07 kN

1319.9 N 1.32 kN Ans

x

A

y

A

A

ABC

F

F F

F

F F

FF

2D Equilibrium

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Draw the FBD of the unloaded platform that is suspended off the edge of the oil rig shown. The platform has the weight of 1962 N.

2D Equilibrium

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2D Equilibrium

Platform in equilibrium: Find , ,

0

cos70 (1 m) sin70 (2.2 m)(1962 N)(1.4 m) 01140.1 N 1.14 kN Ans

0

cos70 0 389.92 N 390 N Ans

0

sin70 (1962 N) 0

890.69 N 89

y y

A

x

x

x

y

y

y

A A T

M

T T

T

F

A TA

F

A T

A

⎡ ⎤=⎣ ⎦° + ° −

== =

⎡ ⎤=⎣ ⎦− ° =

= =

⎡ ⎤=⎣ ⎦+ ° − =

= =

∑

∑

∑

1 N Ans

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2D Equilibrium

Platform in equilibrium: Find , ,

tan70 , 2.1980 m0.8 m

1 mtan , 66.3571.4 m0

cos cos70 0 (1)

0

sin sin70 (1962 N) 0 (2)Solve (1) & (2)

1140.1 N, 972.30 N1.14 k

A

x

A

y

A

A

F Td d

d

F

F T

F

F T

T FT

θ

θ θ

θ

θ

° = =

+= = °

⎡ ⎤=⎣ ⎦− ° =

⎡ ⎤=⎣ ⎦+ ° − =

= =

=

∑

∑

N, 972 N, 66.4 AnsAF θ= = °

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Equilibrium of 3D Rigid Bodies3D Equilibrium

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3D Equilibrium

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3D Equilibrium

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3D Equilibrium

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Comparison with 2D supports

3D Equilibrium

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3D Equilibrium

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3D Equilibrium

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3D Equilibrium

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3D Equilibrium

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3D Equilibrium Rigid Bodies

Action forces in supportsEquilibrium conditions

=

=

∑∑

v v

v v0

0O

F

MVector form

= =

= =

= =

∑ ∑∑ ∑∑ ∑

0 0

0 and 0

0 0

x x

y y

z z

F M

F M

F MScalar form

3D Equilibrium

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The homogeneous plate shown has a mass of 100 kg and is subjected to a force and couple moment along its edges. If it is supported in the horizontal plane by means of a roller at A, a ball-and-socket joint at B, and a cord at C, determine the components of reaction at the supports.

3D Equilibrium

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FBD of plate

⎡ ⎤= =⎣ ⎦⎡ ⎤= =⎣ ⎦

∑∑

Equilibrium of plate

0 0

0 0x x

y y

ABC

F B

F B

3D Equilibrium

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⎡ ⎤= + + − − =⎣ ⎦⎡ ⎤= − + =⎣ ⎦⎡ ⎤= + −⎣ ⎦

− − ⋅ =

∑∑∑

Equilibrium of plate

0 (300 N) (980.7 N) 0 (1)

0 (2 m) (980.7 N)(1 m) (2 m) 0 (2)

0 (300 N)(1.5 m) (980.7 N)(1.5 m) (3 m)

(3 m) (200 N m) 0 (3)

z z z C

x C z

y z

z

ABC

F A B T

M T B

M B

A

= = −

=

= =

= = = −

Solve (1), (2) & (3)790.35 N, 216.67 N, 707.02 N

790 N, 707 N, 0, 217 N Ans

z z

C

z C

x y z

A BT

A TB B B

3D Equilibrium

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′

′

⎡ ⎤= + + − − =⎣ ⎦⎡ ⎤= + − =⎣ ⎦⎡ ⎤= − − − ⋅⎣ ⎦

− =

∑∑∑

Alternatively, for equilibrium of plate

0 (300 N) (980.7 N) 0 (1)

0 (980.7 N)(1 m) (300 N)(2 m) (2 m) 0 (4)

0 (300 N)(1.5 m) (980.7 N)(1.5 m) (200 N m)

(3 m) 0 (5)

z z z C

x z

y

C

ABC

F A B T

M A

M

T

=

= −

=

Solve (1), (4) & (5)790.35 N, 216.67 N,

707.02 N

z

z

C

ABT

3D Equilibrium

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The windlass shown is supported by a thrust bearing at A and a smooth journal bearing at B, which are properly aligned on the shaft. Determine the magnitude of the vertical force P that must be applied to the handle to maintain equilibrium of the 100 kg bucket. Also calculate the reactions at the bearings.

3D Equilibrium

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FBD of windlass

3D Equilibrium

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⎡ ⎤= − ° =⎣ ⎦=

⎡ ⎤= − + + =⎣ ⎦=

⎡ ⎤= − = → =⎣ ⎦

∑

∑

∑

Equilibrium of windlass

0 (981 N)(0.1 m) (0.3 m)cos30 0

377.59 N

0 (981 N)(0.5 m) (0.8 m) (0.4 m) 0

424.33 N

0 (0.8 m) 0 0

x

y z

z

z y y

M P

P

M A P

A

M A A

3D Equilibrium

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⎡ ⎤= =⎣ ⎦⎡ ⎤= + = → =⎣ ⎦⎡ ⎤= − + − =⎣ ⎦

− + − =

→ =

= = = = = =

∑∑∑

0 0

0 0 0

0 (981 N) 0

(424.33 N) (981 N) (377.59 N) 0934.26 N

378 N, 0, 424 N, 0, 934 N Ans

x x

y y y y

z z z

z

z

x y z y z

F A

F A B B

F A B P

BB

P A A A B B

3D Equilibrium

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Rod AB shown is subjected to the 200-N force. Determine the reactions at the ball-and-socket joint A and the tension in cables BDand BE.

3D Equilibrium

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FBD of rod AB

= + +

= =

= = + −

= − = + −

v

v v v

v v

v v

ˆ ˆ ˆ( ) Nˆ N 0.5ˆ ˆ ˆ ˆ N (0.5 1 ) m

ˆ ˆ ˆ ˆ200 N ( 2 2 ) m

x y z

E E C B

D D C

B

A A i A j A k

T T i r r

T T j r i j k

F k r i j k

3D Equilibrium

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⎡ ⎤=⎣ ⎦+ + + =

+ + + + + =

⎡ ⎤= + =⎣ ⎦⎡ ⎤= + =⎣ ⎦⎡ ⎤= − =⎣ ⎦

∑

∑∑∑

v v

v v v v v

v

Equilibrium of body

0

0ˆ ˆ ˆ( ) ( ) ( ) 0

0 0 (1)

0 0 (2)

0 (200 N) 0 (3)

E D

x E y D z

x x E

y y D

z z

ACB

F

A T T F

A T i A T j A F k

F A T

F A T

F A

3D Equilibrium

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⎡ ⎤=⎣ ⎦× + × + =

× + × + =

× + + =

+ − × + + =

⎡ ⎤= − ⋅ =⎣ ⎦⎡ ⎤= − + ⋅⎣ ⎦

∑

∑∑

v v

v v v vv v

v v v vv v

v v v vv

v v v v

Equilibrium of body

0

( ) ( ) 0

(0.5 ) ( ) 0

(0.5 ) 0ˆ ˆ ˆ( 2 2 ) m (0.5 ) 0

0 (2 m) (200 N m) 0 (1)

0 ( 2 m) (100 N m

A

C B E D

B B E D

B E D

E D

x D

y E

ACB

M

r F r T T

r F r T T

r F T T

i j k F T T

M T

M T =

⎡ ⎤= − =⎣ ⎦∑) 0 (2)

0 (1 m) (2 m) 0 (3)z D EM T T

3D Equilibrium

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=

=

= −

= −

=

Solve eqns (1) through (6)100 N50 N

50 N100 N

200 N Ans

The negative sign indicates that and have a direction which is opposite to those shown in the FBD.

D

E

x

y

z

x y

TTAA

A

A A

3D Equilibrium

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The bent rod is supported at Aby a journal bearing, at D by a ball-and-socket joint,at B by means of cable BC. Using only one equilibrium equation, obtain a directsolution for the tension in cable BC. The bearing at A is capable of exerting forcecomponents only in the z and ydirections since it is properly aligned on the shaft.

3D Equilibrium

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83


FBD of ABED

3D Equilibrium

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84


= ⋅ × =∑ ∑vv

Summing moments about an axis passing through points and .

ˆ ( ) 0DA DA

D AM u r F

= = − −

v

v

any pointany pointReminder: is a position vector from on the axis to

in the line of action of the force.1 1ˆ ˆˆ2 2

DADA

DA

r DA

ru i jr

3D Equilibrium

85

85


× = × + ×

= − × − + +

− × −

= − + + ⋅

= ⋅ × =

= − − ⋅

=

−

∑

∑ ∑

v v vv v v

v

v

v

vv

( )0.2 0.3 0.6ˆ ˆ ˆ( 1 ) m ( ) N0.7 0.7 0.7

ˆ( 0.5 ) m ( 981 ) Nˆ[( 0.85714 490.5) 0.28571 ] N m

Equilibrium of body ˆ ( ) 01 1ˆ ˆ[ ] [( 0.857142 2

B B E

B B B

B B

DA DA

r F r T r W

j T i T j T k

j k

T i T kABED

M u r F

i j T + + ⋅

= − − + ⋅ =

= =

vˆ490.5) 0.28571 ] N m

1( )( 0.85714 490.5) N m 02

572.25 N 572 N Ans

B B

B

B

i T k

T

T

3D Equilibrium

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86

Statically determinate (SD) problems can be analyzed using the equilibrium condition alone.

Statically Determinate ObjectsSD/SI

= =

= =

∑∑

v v v

v v v0

0O

R

R O

F F

M M

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87

An object with redundant supports

An object with improper supports

Statically Indeterminate ObjectsSD/SI

Statically indeterminate (SI) problems can not be analyze with equilibrium equations.

88

88

SI Objects Redundant Supports #1

An object has redundant supports, when it has more supports than the minimum number necessary to maintain it in equilibrium.

ResultMore unknown forces or couples than the number of independent equilibrium equations.

SD/SI

89

89


5 Unknowns3 Equilibrium Equations

SD/SI

90

90


8 Unknowns6 Equilibrium Equations

SD/SI

91

91

Example Redundant Supports #1

⎡ ⎤= =⎣ ⎦⎡ ⎤= − + =⎣ ⎦⎡ ⎤= − − + =⎣ ⎦

∑∑∑

Equilibrium of beam

0 0

0 0

0 (2 ) 0

x x

y y

A

AB

F A

F A F B

M M FL B L

3 equations vs 4 unknown reactions

SD/SI

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92

Example Redundant Supports #2

⎡ ⎤= + =⎣ ⎦⎡ ⎤= + − =⎣ ⎦⎡ ⎤= − + =⎣ ⎦

∑∑∑

Equilibrium of beam

0 0

0 0

0 (2 ) 0

x x x

y y y

A y

AB

F A B

F A B F

M FL B L

Three equations and four unknown reactions:Ay and By can be found.Ax = −Bx

SD/SI

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93


Degree of redundancy =# of unknown reactions − # of independent equations

The unknown reactions can be determined by supplementing the equilibrium equations with equations relating forces/couples with deformation.

SD/SI

See you in Mechanics of Materials I

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94

SI Objects Improper Supports #1

An object has improper supports when it has inadequate(not enough) supports to maintain it in equilibrium.

ResultThe object will move when loaded.

SD/SI

See you in Dynamics

95

95


The supports exertparallel forces only.

45

The supports exert only concurrent forces.

SD/SI

96

96


SD/SI

97

97


SD/SI

98

98


SD/SI

99

99

Equilibrium

Here ends the most important chapter of the subject.

100

100

ConceptsWhen a body is in equilibrium, the resultant force and couple about any point O are both zero. Problems can be analyzed by drawing free body diagrams (FBDs) and substitute information in the FBD under consideration into the equilibrium equations.

Statically determinate (SD) problems can be solved using the equilibrium conditions alone.The statically indeterminate (SI) problems cannot due to too few or too many supports or constraints.

Review

Documents

2. Objects in Equilibrium - Chulalongkorn University: …pioneer.netserv.chula.ac.th/~mkuntine/42-111/files/ch2... · · 2012-01-272 2 Objectives Students must be able to #1 Course