3.analysis cable and arches 1

3/23/2011

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Cables and Arches 1

Nottingham University

1

General Procedure for Analysis

When solving the problems, do the work as neatly as possible. Being neat generally stimulates clear and orderly thinking and vice versa.

2

Addition of a System of Coplanar Forces

For resultant of two or more forces:• Find the components of the forces in the specified

axes • Add them algebraically• Add them algebraically• Form the resultant

In this subject, we resolve each force into rectangular forces along the x and y axes.

yx FFF

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• Coplanar Force Resultants- Positive scalars = sense of direction along the positive coordinate axes

- Negative scalars = sense of direction along - Negative scalars = sense of direction along the negative coordinate axes

- Magnitude of FR can be found by Pythagorean Theorem

sin

cos

22

RRy

RRx

RyRxR

FF

FF

FFF

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• Coplanar Force Resultants- Direction angle ? (orientation of the force) can be found by trigonometry

Rx

Ry

F

F1tan

5


Example

Determine x and y components of F1 and F2

acting on the boom. Express each force as a

Cartesian vector Cartesian vector

sin

cos

22

RRy

RRx

RyRxR

FF

FF

FFF

Rx

Ry

F

F1tan

6


Solution

Scalar Notation

NNNF x 10010030sin2001

Hence, from the slope triangle

NNNF y

x

17317330cos2001

1

12

5tan1

7


Solution

Alt, by similar triangles

N

F x

13

12

2602

Similarly,

NNF

N

x 24013

12260

13260

2

NNF y 10013

52602

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Example

The end of the boom O is subjected to three

concurrent and coplanar forces. Determine

the magnitude and orientation of the the magnitude and orientation of the

resultant force.

sin

cos

22

RRy

RRx

RyRxR

FF

FF

FFF

Rx

Ry

F

F1tan

9


Solution

Scalar NotationFF xRx

4

:

View Free Body Diagram

N

NNF

FF

NN

NNNF

Ry

yRy

Rx

8.296

5

320045cos250

:

2.3832.383

5

420045sin250400

10

Solution

Resultant Force


NNFR 8.2962.383 22

From vector addition,

Direction angle ? is

N485

8.37

2.383

8.296tan 1

N

N

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Condition for the Equilibrium of a Particle

• Particle at equilibrium if

- At rest

- Moving at constant a constant velocity

• Newton’s first law of motion

? F = 0

where ?F is the vector sum of all the forces acting on the particle

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Condition for the Equilibrium of a Particle

• Newton’s second law of motion? F = ma

• When the force fulfill Newton's first law of motion, motion,

ma = 0a = 0

therefore, the particle is moving in constant velocity or at rest

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The Free-Body Diagram

• Cables and Pulley- Cables (or cords) are assumed to have negligible weight and they cannot stretch- A cable only support tension or pulling force- Tension always acts in the direction of the cable- Tension force in a continuouscable must have a constant magnitude for equilibrium

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• Cables and Pulley- For any angle ?, the cable is subjected to

a constant tension T

throughout its length

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Procedure for Drawing a FBD 1. Draw outlined shape

- Isolate particle from its surroundings

2. Show all the forces2. Show all the forces

- Indicate all the forces

- Active forces: set the particle in motion

- Reactive forces: result of constraints and supports that tend to prevent motion

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Procedure for Drawing a FBD 3. Identify each forces

- Known forces should be labeled with proper magnitude and directionmagnitude and direction

- Letters are used to represent magnitude and directions of unknown forces

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• A spool is having a weight W which is suspended from the crane bottom

• Consider FBD at A since • Consider FBD at A since these forces act on the ring

• Cables AD exert a resultant force of W on the ring

• Condition of equilibrium is used to obtained TB and TC

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• The bucket is held in equilibrium by the cable

• Force in the cable = weight of the bucketweight of the bucket

• Isolate the bucket for FBD• Two forces acting on the

bucket, weight W and force T of the cable

• Resultant of forces = 0W = T

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Example

The sphere has a mass of 6kg and is

supported. Draw a free-body diagram of the

sphere, the cord

CE and the knot at C.

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Solution

FBD at Sphere

• Two forces acting,

View Free Body Diagram

• Two forces acting, weight and the force on cord CE.

• Weight of 6kg (9.81m/s2) = 58.9N

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SolutionCord CE• Two forces acting, force of

the sphere and force of the knotthe sphere and force of the knot

• Newton’s Third Law: FCEis equal but opposite

• FCE and FEC pull the cord in tension

• For equilibrium, FCE = FEC

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Coplanar Systems

• A particle is subjected to coplanar forces in the x-y plane

• Resolve into i and j components for equilibrium equilibrium

? Fx = 0? Fy = 0

• Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal o zero

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Coplanar Systems

• Scalar Notation- Sense of direction = an algebraic sign that corresponds to the arrowhead direction of the component along each axiscomponent along each axis

- For unknown magnitude, assume arrowhead sense of the force

- Since magnitude of the force is always positive, if the scalar is negative, the force is acting in the opposite direction

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Coplanar Systems

ExampleConsider the free-body diagram of the particle subjected to two forces

• Assume unknown force F acts to the right for equilibrium

? Fx = 0 ; + F + 10N = 0F = -10N

• Force F acts towards the left for equilibrium

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Coplanar Systems

Example

Determine the tension in

cables AB and AD for cables AB and AD for

equilibrium of the 250kg

engine.

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Coplanar Systems

SolutionFBD at Point A- Initially, two forces acting, forces

of cables AB and AD- Engine Weight - Engine Weight

= (250kg)(9.81m/s2) = 2.452kN supported by cable CA

- Finally, three forces acting, forces TB and TD and engine weight on cable CA

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Coplanar Systems

Solution

+? ? Fx = 0; TBcos30 - TD = 0

+? ? Fy = 0; TBsin30 - 2.452kN = 0

Solving, Solving,

TB = 4.90kN

TD = 4.25kN

*Note: Neglect the weights of the cables since they are small compared to the weight of the engine

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Coplanar Systems

Example

If the sack at A has a weight

of 20N (˜ 2kg), determine

the weight of the sack at B the weight of the sack at B

and the force in each cord

needed to hold the system in

the equilibrium position

shown.

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Coplanar Systems

Solution

FBD at Point E

- Three forces acting, - Three forces acting, forces of cables EG and EC and the weight of the sack on cable EA

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Coplanar Systems

Solution

+? ? Fx = 0; TEGsin30 - TECcos45 = 0

+? ? Fy = 0; TEGcos30 - TECsin45 - 20N = 0

Solving, Solving,

TEC = 38.6kN

TEG = 54.6kN

*Note: use equilibrium at the ring to determine tension in CD and weight of B with TEC known

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Moment of a Force – Scalar Formation

In General• Consider the force F and the point O which lies in the

shaded plane

• The moment M about point O, • The moment MO about point O,

or about an axis passing

through O and perpendicular

to the plane, is a vector quantity

• Moment MO has its specified

magnitude and direction 32

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Moment of a Force –Scalar Formation

Magnitude• For magnitude of MO,

MO = FdO

where d = moment arm or perpendicular distance from the axis at point O to its line of action of the force

• Units for moment is N.m

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Direction• Direction of MO is specified by using

“right hand rule”“right hand rule”

- fingers of the right hand are curled to follow the sense of rotation when force rotates about point O

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Direction- Thumb points along the moment axis to give the direction and sense axis to give the direction and sense of the moment vector- Moment vector is upwards and perpendicular to the shaded plane

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DirectionMO is shown by a vector arrow

with a curl to distinguish it from

force vector

Example (Fig b)

• MO is represented by the counterclockwise curl, which indicates the action of F

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Direction• Arrowhead shows the sense of rotation

caused by F• Using the right hand rule, the direction • Using the right hand rule, the direction

and sense of the moment vector points out of the page

• In 2D problems, moment of the force is found about a point O

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Direction• Moment acts about an axis

perpendicular to the plane containing Fand dFand d

• Moment axis intersects the plane at point O

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Resultant Moment of a System of

Coplanar Forces• Resultant moment, MRo = addition of the moments • Resultant moment, MRo = addition of the moments

of all the forces algebraically since all moment forces are collinear

MRo = ?Fd

taking clockwise to be positive

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Resultant Moment of a System of Coplanar Forces• A clockwise curl is written along the equation to • A clockwise curl is written along the equation to

indicate that a positive moment if directed along the + z axis and negative along the – z axis

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• Moment of a force does not always cause rotation

• Force F tends to rotate the beam clockwise about A with moment

MA = FdAMA = FdA

• Force F tends to rotate the beam counterclockwise about B with moment

MB = FdB

• Hence support at A prevents

the rotation

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Example

For each case, determine the moment of the

force about point Oforce about point O

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Solution• Line of action is extended as a dashed line to establish

moment arm d

• Tendency to rotate is indicated and the orbit is shown • Tendency to rotate is indicated and the orbit is shown as a colored curl

)(.5.37)75.0)(50()(

)(.200)2)(100()(

CWmNmNMb

CWmNmNMa

o

o

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Solution

)(.0.21)14)(7()(

)(.4.42)45sin1)(60()(

)(.229)30cos24)(40()(

CCWmkNmmkNMe

CCWmNmNMd

CWmNmmNMc

o

o

)(.0.21)14)(7()( CCWmkNmmkNMe o

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Example

Determine the moments of

the 800N force acting on the

frame about points A, B, C

and D.

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SolutionScalar Analysis

)(.1200)5.1)(800(

)(.2000)5.2)(800(

CWmNmNM

CWmNmNM A

Line of action of F passes through C

)(.400)5.0)(800(

.0)0)(800(

)(.1200)5.1)(800(

CCWmNmNM

mkNmNM

CWmNmNM

D

C

B

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Free-Body Diagrams

• FBD is the best method to represent all the known and unknown forces in a system

• FBD is a sketch of the outlined shape of the body, which represents it being isolated from its which represents it being isolated from its surroundings

• Necessary to show all the forces and couple moments that the surroundings exert on the body so that these effects can be accounted for when equations of equilibrium are applied

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Free-Body Diagrams

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Free-Body Diagrams

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Free-Body Diagrams

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Free-Body Diagrams

Support Reactions• If the support prevents the translation of a body in a

given direction, then a force is developed on the body in that direction

• If rotation is prevented, a couple moment is exerted • If rotation is prevented, a couple moment is exerted on the body

• Consider the three ways a horizontal member, beam is supported at the end- roller, cylinder- pin- fixed support

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Free-Body Diagrams

Support ReactionsRoller or cylinder• Prevent the beam from

translating in the vertical translating in the vertical direction

• Roller can only exerts a force on the beam in the vertical direction

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Free-Body Diagrams

Support ReactionsPin• The pin passes through a hold in the beam and

two leaves that are fixed to the ground• Prevents translation of the beam in any direction

F• The pin exerts a force F on the beam in this

direction

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Free-Body Diagrams

Support ReactionsFixed Support

• This support prevents both translation and rotation of the beamand rotation of the beam

• A couple and moment must be developed on the beam at its point of connection

• Force is usually represented in x and y components

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Free-Body Diagrams

• Cable exerts a force on the bracket

• Type 1 connections

• Rocker support for this bridge girder allows horizontal movements so that the bridge is free to expand and contract due to temperature

• Type 5 connections55

Free-Body Diagrams

• Concrete Girder rest on the ledge that is assumed to act as a smooth contacting surface

• Type 6 connections• Type 6 connections

• Utility building is pin supported at the top of the column

• Type 8 connections

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Free-Body Diagrams

• Floor beams of this building are welded together and thus form fixed connections

• Type 10 connections• Type 10 connections

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Free-Body Diagrams

External and Internal Forces• A rigid body is a composition of particles, both

external and internal forces may act on it• For FBD, internal forces act between particles which

are contained within the boundary of the FBD, are not For FBD, internal forces act between particles which are contained within the boundary of the FBD, are not represented

• Particles outside this boundary exert external forces on the system and must be shown on FBD

• FBD for a system of connected bodies may be used for analysis

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5.2 Free-Body Diagrams

Idealized Models• Needed to perform a correct force analysis of

any object• Careful selection of supports, material,

behavior and dimensions for trusty results• Complex cases may require developing several

different models for analysis

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Free-Body Diagrams

Idealized Models• Consider a steel beam used to support the roof

joists of a building• For force analysis, reasonable to assume rigid

body since small deflections occur when beam is body since small deflections occur when beam is loaded

• Bolted connection at A will allow for slight rotation when load is applied => use Pin

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Free-Body Diagrams

Support at B offers no resistance to horizontal movement => use Roller

• Building code requirements used to specify the roof loading (calculations of the joist forces)

• Large roof loading forces account for extreme loading • Large roof loading forces account for extreme loading cases and for dynamic or vibration effects

• Weight is neglected when it is small compared to the load the beam supports

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Free-Body Diagrams

Procedure for Drawing a FBD1. Draw Outlined Shape

• Imagine body to be isolated or cut free from its constraintsconstraints

• Draw outline shape

2. Show All Forces and Couple Moments

• Identify all external forces and couple moments that act on the body

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Free-Body Diagrams

Procedure for Drawing a FBD• Usually due to

- applied loadings- reactions occurring at the supports or at points of contact with other bodycontact with other body- weight of the body

• To account for all the effects, trace over the boundary, noting each force and couple moment acting on it

3. Identify Each Loading and Give Dimensions• Indicate dimensions for calculation of forces

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Free-Body Diagrams

Procedure for Drawing a FBD• Known forces and couple moments should be

properly labeled with their magnitudes and directionsdirections

• Letters used to represent the magnitudes and direction angles of unknown forces and couple moments

• Establish x, y and coordinate system to identify unknowns

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Free-Body Diagrams

Example

Draw the free-body diagram of the uniform

beam. The beam has a mass of 100kg.beam. The beam has a mass of 100kg.

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Free-Body Diagrams

Solution

Free-Body Diagram

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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 the bodyy 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

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Alternative Sets of Equilibrium Equations• For coplanar equilibrium problems, ?Fx = 0; ?Fy = 0;

?MO = 0 can be used

• Two alternative sets of three independent equilibrium • 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

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Example 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.

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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 beam70


SolutionEquations of Equilibrium

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

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Solution

NA

mAmNmNmN

M

y

B

319

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

;0

NB

BNNNN

F

NA

y

y

y

y

405

020010045sin600319

;0

319

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SolutionChecking,

M A ;0

NB

mBmN

mNmNmN

M

y

y

A

405

0)7()7)(200(

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

;0

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3.analysis cable and arches 1

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