Download - 6161103 5.3 equations of equilibrium

5.3 Equations of Equilibrium5.3 Equations of Equilibrium

� For equilibrium of a rigid body in 2D,

∑Fx = 0; ∑Fy = 0; ∑MO = 0

� ∑Fx and ∑Fy represent the algebraic sums of the x and y components of all the forces acting on x and y components of all the forces acting on the body

� ∑MO represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to x-y plane and passing through arbitrary point O, which may lie on or off the body


Alternative Sets of Equilibrium Equations

� For coplanar equilibrium problems, ∑Fx = 0; ∑Fy

= 0; ∑MO = 0 can be used

� Two alternative sets of three independent � Two alternative sets of three independent equilibrium equations may also be used

∑Fa = 0; ∑MA = 0; ∑MB = 0

When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis



� Consider FBD of an arbitrarily shaped body

� All the forces on FBD may be

replaced by an equivalent replaced by an equivalent

resultant force

FR = ∑F acting at point A and a

resultant moment MRA = ∑MA

� If ∑MA = 0 is satisfied, MRA = 0



� If FR satisfies ∑Fa = 0, there is no

component along the a axis and

its line of axis is perpendicular its line of axis is perpendicular

to the a axis

� If ∑MB = 0 where B does not lies

on the line of action of FR, FR = 0

� Since ∑F = 0 and ∑MA = 0, the

body is in equilibrium


� A second set of alternative equations is

∑MA = 0; ∑MB = 0; ∑MC = 0

� Points A, B and C do not lie on the


� Points A, B and C do not lie on the

same line

� Consider FBD, if If ∑MA = 0, MRA = 0

� ∑MA = 0 is satisfied if line of action of FR passes through point B

� ∑MC = 0 where C does not lie on line AB

� FR = 0 and the body is in equilibrium


Procedure for AnalysisFree-Body Diagram

� Establish the x, y, z coordinates axes in any suitable orientationsuitable orientation

� Draw an outlined shape of the body

� Show all the forces and couple moments acting on the body

� Label all the loadings and specify their directions relative to the x, y axes


Procedure for AnalysisFree-Body Diagram

� The sense of a force or couple moment having an unknown magnitude but having an unknown magnitude but known line of action can be assumed

� Indicate the dimensions of the body necessary for computing the moments of forces


Procedure for Analysis

Equations of Equilibrium

� Apply the moment equation of equilibrium ∑M = 0 about a point O that lies on the ∑MO = 0 about a point O that lies on the intersection of the lines of action of the two unknown forces

� The moments of these unknowns are zero about O and a direct solution the third unknown can be obtained


Procedure for Analysis


� When applying the force equilibrium ∑Fx = 0 and ∑F = 0, orient the x and y axes along and ∑Fy = 0, orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components

� If the solution yields a negative result scalar, the sense is opposite to that was assumed on the FBD


Example 5.6

Determine the horizontal and vertical

components of reaction for the beam loaded.

Neglect the weight of the beam in the Neglect the weight of the beam in the

calculations.


Solution

FBD

� 600N force is represented by its x and y components

� 200N force acts on the beam at B and is � 200N force acts on the beam at B and is

independent of the

force components

Bx and By, which

represent the effect of

the pin on the beam


Solution


M B ;0=∑→+

� A direct solution of Ay can be obtained by applying ∑MB = 0 about point B

� Forces 200N, Bx and By all create zero moment about B

NB

BN

x

x

424

045cos600

=

=−o


Solution

NA

mAmNmNmN

M

y

B

319

0)7()2.0)(45cos600()5)(45sin600()2(100

;0

=

=−−+

=∑oo

NB

BNNNN

F

NA

y

y

y

y

405

020010045sin600319

;0

319

=

=+−−−

=∑↑+

=

o


Solution

Checking,

M A ;0=∑

NB

mBmN

mNmNmN

M

y

y

A

405

0)7()7)(200(

)5)(100()2.0)(45cos600()2)(45sin600(

;0

=

=+−

−−−

=∑oo


Example 5.7

The cord supports a force of 500N and wraps over

the frictionless pulley. Determine the tension in the

cord at C and the horizontal cord at C and the horizontal

and vertical components at

pin A.


Solution

FBD of the cord and pulley

� Principle of action: equal but opposite reaction observed in the FBDobserved in the FBD

� Cord exerts an unknown load

distribution p along part of

the pulley’s surface

� Pulley exerts an equal but

opposite effect on the cord


Solution

FBD of the cord and pulley

� Easier to combine the FBD of the pulley and

contracting portion of the cord so that the contracting portion of the cord so that the

distributed load becomes internal to the system

and is eliminated from the

analysis


Solution


mTmN

M A

0)2.0()2.0(500

;0

=+

=∑

Tension remains constant as cord

passes over the pulley (true for

any angle at which the cord is

directed and for any radius of

the pulley

NT

mTmN

500

0)2.0()2.0(500

=

=+


Solution

NA

Fx

030sin500

;0

=+−

=∑→+o

NA

NNA

F

NAx

NA

y

y

y

x

933

030cos500500

;0

250

030sin500

=

=−−

=∑↑+

=

=+−

o

o


Example 5.8

The link is pin-connected at a and rest a

smooth support at B. Compute the horizontal smooth support at B. Compute the horizontal

and vertical components of reactions at pin A


Solution

FBD

� Reaction NB is perpendicular to the link at BB

� Horizontal and vertical

components of reaction

are represented at A


Solution


M A

0)75.0()1(60.90

;0

=+−−

=∑

NAx

NA

F

NN

mNmNmN

x

x

B

B

100

030sin200

;0

200

0)75.0()1(60.90

=

=−

=∑→+

=

=+−−

o


Solution

NNA

Fy

030cos20060

;0

=−−

=∑↑+o

NA

NNA

y

y

233

030cos20060

=

=−−


Example 5.9

The box wrench is used to tighten the bolt at

A. If the wrench does not turn when the load

is applied to the handle, determine the is applied to the handle, determine the

torque or moment applied to the bolt and the

force of the wrench on the bolt.


Solution

FBD

� Bolt acts as a “fixed support” it exerts force � Bolt acts as a “fixed support” it exerts force components Ax and Ay and a torque MA on the wrench at A


Solution

Equations of EquilibriumM A

12

;0

=∑

NAx

NNA

F

mNM

mNmNM

x

x

A

A

00.5

060cos3013

552

;0

.6.32

0)7.0)(60sin30()3.0(13

1252

=

=+

−

=∑→+

=

=−

−

o

o


Solution

NNA

F

y

y

060sin3013

1252

;0

=−

−

=∑↑+

o

( ) ( )( )

NmM

mNmNM

MCCW

NA

NNA

A

A

y

y

y

6.32

07.060sin303.013

1252

;0

0.74

060sin3013

52

=

=°−

−

=∑↑

=

=−

−


Solution

� Point A was chosen for summing the moments as the lines of action of the unknown forces Ax and Ay pass through this point and these forces are Ay pass through this point and these forces are not included in the moment summation

� MA must be included

� Couple moment MA is a free vector and represents the twisting resistance of the bolt on the wrench


Solution

� By Newton’s third law, the wrench exerts an equal but opposite moment or torque on the boltbolt

� For resultant force on the wrench,

� For directional sense,

� FA acts in the opposite direction on the bolt

( ) ( )

o1.8600.5

0.74tan

1.740.7400.5

1

22

==

=+=

−

N

N

NFA

θ


Solution

� Checking,

;0=∑MC

0.8.51.6.32.2.19

0)7.0(0.74.6.32)4.0(1312

52

=−+

=−+

mNmNmN

mNmNmN

Example 5.10

Placement of concrete from the

truck is accomplished using the

chute. Determine the force that the


chute. Determine the force that the

hydraulic cylinder and the truck

frame exert on the chute to hold it in

position. The chute and the wet

concrete contained along its length

have a uniform weight of 560N/m.


Solution

� Idealized model of the chute

� Assume chute is pin connected to the frame at A and the hydraulic cylinder BC acts as a at A and the hydraulic cylinder BC acts as a short link


SolutionFBD� Since chute has a length of 4m, total supported

weight is (560N/m)(4m) = 2240N, which is assumed to act at its midpoint, Gassumed to act at its midpoint, G

� The hydraulic cylinder exerts a horizontal force FBC on the chute

Equations of Equilibrium� A direct solution of FBC is obtained by the

summation about the pin at A


Solution

mNmF

M

BC

A

)2(30cos2240)5.0(

;0

+−

=∑o

NAx

NA

F

NF

mN

x

x

BC

7900

07900

;0

7900

0)0625.0(30sin2240

=

=+−

=∑→+

=

=+ o


Solution

2240

02240

;0

=

=−

=∑↑+

NA

NA

F

y

y

Checking,

0)0625.0(30sin2240)1(30cos2240

)30cos1(2240)5.0(7900

;0

2240

=+

++−

=∑

=

mNmN

mNmN

M

NA

B

y

oo

o


Example 5.11

The uniform smooth rod is subjected to a force and

couple moment. If the rod is supported at A by a

smooth wall and at B and C either at the top or smooth wall and at B and C either at the top or

bottom by rollers, determine

the reactions at these supports.

Neglect the weight of the rod.


SolutionFBD� All the support reactions act normal to the

surface of contact since the contracting surfaces are smoothare smooth

� Reactions at B and C are acting in the positive y’ direction

� Assume only the rollers located on the bottom of the rod are used for support


Solution


;0=∑ AM

030cos30cos300

;0

030sin30sin

;0

0)8)(30cos300()6(.4000)2(

''

''

''

=++−

=∑↑+

=−+

=∑→+

=+−+−

oo

oo

o

yy

y

xyy

x

yy

BCN

F

ABC

F

mNmCmNmB


Solution

� Note that the line of action of the force component passes through point A and this force is not included in the moment equation force is not included in the moment equation

� Since By’ is negative scalar, the sense of By’ is opposite to shown in the FBD

kNNC

kNNB

y

y

35.14.1346

10.1000

'

'

==

−=−=


Solution

� Top roller at B serves as the support rather than the bottom one

NA

ANN

x

x

173

030sin0.100030sin4.1346

=

=−− oo


Example 5.12

The uniform truck ramp has a weight of

1600N ( ≈ 160kg ) and is pinned to the body 1600N ( ≈ 160kg ) and is pinned to the body

of the truck at each end and held in position

by two side cables.

Determine the tension

in the cables.


Solution

� Idealized model of the ramp

� Center of gravity located at the midpoint since the ramp is approximately uniformramp is approximately uniform

FBD of the Ramp


Solution


TmT

M A

)30cos2(20sin)30sin2(20cos

;0

+−

=∑oooo

By the principle of transmissibility, locate T at C

md

md

NT

N

TmT

0154.120sin

2

10sin

5985

0)30cos5.1(1600

)30cos2(20sin)30sin2(20cos

=

=

=

=+

+−

oo

o

oooo


Solution

Since there are two cables supporting the

ramp,

T’ = T/2 = 2992.5NT’ = T/2 = 2992.5N

Download - 6161103 5.3 equations of equilibrium

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