ENT 152/4: Applied Mechanics RUSLIZAM DAUD [email protected]/013-4430827 Teaching Plan Final...

ENT 152/4: Applied MechanicsRUSLIZAM DAUD

[email protected]/013-4430827

ENT 152/4: Applied MechanicsRUSLIZAM DAUD

[email protected]/013-4430827

Teaching PlanFinal

(50%)

Tests (10%

)

Lab Report(20%)

Mini Project / Quiz

(20%)Tutorial

Force System Resultants1.Equilibrium of a Rigid Body FQ1

MQ1MQ2

L1: Resolution of a forceL2: Equilibrium of a beam (2 forces and 3 forces)

Q1 (5%) T1

3.Structure Analysis4.Internal Forces-----------------------------------------------5.Friction6.Center of Gravity and Centroid7.Moments of Inertia

FQ2

FQ3

MQ3MQ4---------

L3: Friction of inclined planeL4: Work done by a variable forces (tangential and vertical effort)

Mini Project:Report (6%)Oral presentation (4%)

T2

T3

8.Kinematics of a ParticleFQ4FQ5

L5: Mechanical conservation of energy Q2 (5%) T4

9.Planar Kinematic of a Rigid Body FQ6 L6: Projectile motion T5

1. R.C. Hibbeler “Engineering Mechanics STATICS” 11th Edition, 2007

2. R.C. Hibbeler “Engineering Mechanics DYNAMICS” 11th Edition, 2007

3. Beer, Johnston, Clausen “Vector Mechanics for Engineers” Statics,2007

4. Beer, Johnston, Clausen “Vector Mechanics for Engineers” Dynamics,2007

5. Kumar “Engineering Mechanics” 3rd Edition, 2003

Applied MechanicsApplied Mechanics

Chapter 1: Force System

Resultants

Chapter 1: Force System

Resultants

Chapter ObjectivesChapter Objectives

To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions.

To provide a method for finding the moment of a force about a specified axis.

To define the moment of a couple. To present methods for determining the

resultants of non-concurrent force systems. To indicate how to reduce a simple distributed

loading to a resultant force having a specified location.

Chapter OutlineChapter Outline

Moment of a Force – Scalar Formation

Cross ProductMoment of Force – Vector

FormulationPrinciple of MomentsMoment of a Force about a Specified

Axis

Chapter OutlineChapter Outline

Moment of a CoupleEquivalent SystemResultants of a Force and Couple

SystemFurther Reduction of a Force and

Couple SystemReduction of a Simple Distributed

Loading

Moment of a force about a point or axis – a

measure of the tendency of the force to cause a body to rotate about the point or

axisCase 1Consider horizontal force Fx,

which acts perpendicular to the handle of the wrench and is located dy from the point O

4.1 Moment of a Force – Scalar

Formation


Formation

Fx tends to turn the pipe about the z axisThe larger the force or the distance dy,

the greater the turning effectTorque – tendency of

rotation caused by Fx

or simple moment (Mo) z

4.1 Moment of a Force

– Scalar Formation



Moment axis (z) is perpendicular to shaded plane (x-y)

Fx and dy lies on the shaded plane (x-y)Moment axis (z) intersects

the plane at point O





Case 2Apply force Fz to the wrenchPipe does not rotate about z axisTendency to rotate about x axisThe pipe may not actually

rotate Fz creates tendency

for rotation so moment (Mo) x is produced





Case 2Moment axis (x) is perpendicular to

shaded plane (y-z)

Fz and dy lies on the shaded plane (y-z)








– Scalar FormationCase 3Apply force Fy to the wrenchNo moment is produced about point OLack of tendency to rotate

as line of action passes through O




– Scalar FormationIn General Consider the force F and the point O which lies

in the shaded plane The moment MO about point O,

or about an axis passingthrough O and perpendicularto the plane, is a vector quantity

Moment MO has its specified

magnitude and direction


Formation


FormationMagnitudeFor magnitude of MO,

MO = Fdwhere 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


Formation


FormationDirectionDirection of MO is specified by

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


Formation


FormationDirection

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


Formation


FormationDirectionMO is shown by a vector arrow

with a curl to distinguish it from force vectorExample (Fig b)MO is represented by the

counterclockwise curl, which indicates the action of F


Formation


FormationDirectionArrowhead shows the sense of

rotation caused by FUsing 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


Formation


FormationDirectionMoment acts about an axis

perpendicular to the plane containing F and d

Moment axis intersects the plane at point O


Formation


FormationResultant Moment of a System of Coplanar ForcesResultant moment, MRo = addition of the

moments of all the forces algebraically since all moment forces are collinear

MRo = ∑Fd

taking clockwise to be positive


Formation


FormationResultant Moment of a System of Coplanar ForcesA 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




– Scalar Formation Moment of a force does not always cause rotation

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

MA = FdA

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

MB = FdB

Hence support at A prevents the rotation


Formation


FormationExample 4.1For each case, determine the moment of

the force about point O


Formation


FormationSolution Line of action is extended as a dashed line

to establish moment arm d Tendency to rotate is indicated and the orbit

is shown as a colored curl mNmNMo .2002100 )(.5.37)75.0)(50()(

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

CWmNmNMb

CWmNmNMa

o

o


Formation


FormationSolution

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

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

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

CCWmkNmmkNMe

CCWmNmNMd

CWmNmmNMc

o

o

o


Formation


FormationExample 4.2Determine the moments of the 800N force acting on the frame about points A, B, C and D.

4.1 Moment of a Force – Scalar Formation


SolutionScalar Analysis

Line of action of F passes through C

)(.400)5.0)(800(

.0)0)(800(

)(.1200)5.1)(800(

)(.2000)5.2)(800(

CCWmNmNM

mkNmNM

CWmNmNM

CWmNmNM

D

C

B

A


Formation


FormationExample 4.3Determine the resultant moment of the

four forces acting on the rod about point O



SolutionAssume positive moments acts in the +k direction, CCW

)(.334

.334

)30cos34)(40(

)30sin3)(20()0)(60()2)(50(

CWmN

mN

mmN

mNmNmNM

FdM

Ro

Ro

4.2 Cross Product4.2 Cross Product

Cross product of two vectors A and B yields C, which is written as

C = A X BRead as “C equals A cross B”


Magnitude Magnitude of C is defined as the

product of the magnitudes of A and B and the sine of the angle θ between their tails

For angle θ, 0° ≤ θ ≤ 180°Therefore,

C = AB sinθ


Direction Vector C has a direction that is

perpendicular to the plane containing A and B such that C is specified by the right hand rule- Curling the fingers of the righthand form vector A (cross) to vector B- Thumb points in the direction of vector C


Expressing vector C when magnitude and direction are known

C = A X B = (AB sinθ)uC

where scalar AB sinθ defines the magnitude of vector C unit vector uC

defines the direction of vector C


Laws of Operations1. Commutative law is not valid

A X B ≠ B X ARather,

A X B = - B X A Shown by the right hand rule Cross product A X B yields a vector opposite

in direction to CB X A = -C


Laws of Operations2. Multiplication by a Scalar

a( A X B ) = (aA) X B = A X (aB) = ( A X B )a

3. Distributive Law A X ( B + D ) = ( A X B ) + ( A X D )

Proper order of the cross product must be maintained since they are not commutative


Cartesian Vector FormulationUse C = AB sinθ on pair of

Cartesian unit vectorsExampleFor i X j, (i)(j)(sin90°) = (1)(1)(1) = 1


Laws of Operations In a similar manner,

i X j = k i X k = -j i X i = 0j X k = i j X i = -k j X j = 0k X i = j k X j = -i k X k = 0

Use the circle for the results. Crossing CCW yield positive and CW yields negative results


Laws of Operations Consider cross product of vector A and B

A X B = (Axi + Ayj + Azk) X (Bxi + Byj + Bzk)

= AxBx (i X i) + AxBy (i X j) + AxBz (i X k) + AyBx (j X i) + AyBy (j X j) + AyBz (j X

k) + AzBx (k X i) +AzBy (k X j) +AzBz (k X k) = (AyBz – AzBy)i – (AxBz - AzBx)j + (AxBy –

AyBx)k


Laws of Operations In determinant form,

zyx

zyx

BBB

AAA

kji

BXA

4.3 Moment of Force - Vector Formulation4.3 Moment of Force - Vector Formulation

Moment of force F about point O can be expressed using cross product

MO = r X F

where r represents position vector from O to any pointlying on the line of action of F


MagnitudeFor magnitude of cross product,

MO = rF sinθwhere θ is the angle measured between tails of r and F

Treat r as a sliding vector. Since d = r sinθ,

MO = rF sinθ = F (rsinθ) = Fd


DirectionDirection and sense of MO are determined

by right-hand rule - Extend r to the dashed position - Curl fingers from r towards F- Direction of MO is the same

as the direction of the thumb


Direction*Note:

- “curl” of the fingers indicates the sense of rotation- Maintain proper order of r and F since cross product is not commutative


Principle of TransmissibilityFor force F applied at any point A,

moment created about O is MO = rA x F

F has the properties of a sliding vector and therefore act at any point along its line of action and still create the same moment about O


Cartesian Vector Formulation For force expressed in Cartesian

form,

where rx, ry, rz represent the x, y, zcomponents of the position vectorand Fx, Fy, Fz represent that of the force vector

zyx

zyxO

FFF

rrr

kji

FXrM


Cartesian Vector Formulation With the determinant expended,

MO = (ryFz – rzFy)i – (rxFz - rzFx)j + (rxFy – yFx)k MO is always perpendicular to

the plane containing r and F Computation of moment by cross

product is better than scalar for 3D problems


Cartesian Vector FormulationResultant moment of forces about

point O can be determined by vector addition

MRo = ∑(r x F)


Moment of force F about point A, pulling on cable BC at any point along its line of action, will remain constant

Given the perpendicular distance from A to cable is rd

MA = rdF In 3D problems,

MA = rBC x F


Example 4.4 The pole is subjected to a 60N force that

is directed from C to B. Determine the magnitude of the moment created by this force about the support at A.


Solution Either one of the two position vectors can

be used for the solution, since MA = rB x F or MA = rC x F

Position vectors are represented as rB = {1i + 3j + 2k} m and rC = {3i + 4j} m

Force F has magnitude 60N and is directed from C to B


Solution

Substitute into determinant formulation

kji

kji

FXrM

Nkji

kjiN

uNF

BA

F

)]40(3)20(1[)]40(2)40(1[)]20(2)40(3[

402040

231

402040

)2()1()2(

)092)493)31()60(

)60(

222


SolutionOr

Substitute into determinant formulation

For magnitude,

mN

M

mNkjiM

kji

kji

FXrM

A

A

CA

.224

)100()120()160(

.100120160

)]40(4)20(3[)]40(0)40(3[)]20(0)40(4[

402040

043

222


Example 4.5 Three forces act on the rod. Determine

the resultant moment they create about the flange at O and determine the coordinate direction angles of the moment axis.


Solution Position vectors are directed from point

O to each force rA = {5j} m and

rB = {4i + 5j - 2k} m For resultant moment about O,

mNkji

kjikjikji

FXrXFrFXrFXrM CBARo

.604030

304080

254

0500

050

204060

050

)( 321


SolutionFor magnitude

For unit vector defining the direction of moment axis,

kji

kji

M

Mu

mNM

Ro

Ro

Ro

76852.05121.03941.0

10.78

604030

.10.78)60()40()30( 222


SolutionFor the coordinate angles of the moment

axis,

8.39;7682.0cos

121;5121.0cos

4.67;3841.0cos

4.4 Principles of Moments4.4 Principles of Moments

Also known as Varignon’s Theorem“Moment of a force about a point is equal to the sum of the moments of the forces’ components about the point”

For F = F1 + F2,

MO = r X F1 + r X F2

= r X (F1 + F2)

= r X F


The guy cable exerts a force F on the pole and creates a moment about the base at A

MA = Fd If the force is replaced

by Fx and Fy at point B where the cable acts on the pole, the sum of moment about point A yields the same resultant moment


Fy create zero moment about A

MA = Fxh Apply principle of

transmissibility and slide the force where line of action intersects the ground at C, Fx

create zero moment about A

MA = Fyb


Example 4.6A 200N force acts on the bracket.

Determine the moment of the force about point A.


SolutionMethod 1:From trigonometry using triangle BCD,

CB = d = 100cos45° = 70.71mm = 0.07071m

Thus, MA = Fd = 200N(0.07071m)

= 14.1N.m (CCW)As a Cartesian vector,

MA = {14.1k}N.m


SolutionMethod 2: Resolve 200N force into x and y components Principle of Moments

MA = ∑Fd MA = (200sin45°N)(0.20m) – (200cos45°)(0.10m)

= 14.1 N.m (CCW)Thus,

MA = {14.1k}N.m


Example 4.6The force F acts at the end of the angle bracket. Determine the moment of the

force about point O.


SolutionMethod 1MO = 400sin30°N(0.2m)-400cos30°N(0.4m)

= -98.6N.m = 98.6N.m (CCW)

As a Cartesian vector, MO = {-98.6k}N.m


SolutionMethod 2: Express as Cartesian vector

r = {0.4i – 0.2j}NF = {400sin30°i – 400cos30°j}N = {200.0i – 346.4j}N

For moment,

mNk

kji

FXrMO

.6.98

04.3460.200

02.04.0

4.5 Moment of a Force about a Specified Axis


For moment of a force about a point, the moment and its axis is always perpendicular to the plane containing the force and the moment arm

A scalar or vector analysis is used to find the component of the moment along a specified axis that passes through the point



Scalar AnalysisExample Consider the pipe assembly that lies in the

horizontal plane and is subjected to the vertical force of F = 20N applied at point A.

For magnitude of moment, MO = (20N)(0.5m) = 10N.m

For direction of moment, apply right hand rule



Scalar Analysis Moment tends to turn pipe about the

Ob axis Determine the component of MO about

the y axis, My since this component tend to unscrew the pipe from the flange at O

For magnitude of My,

My = 3/5(10N.m) = 6N.m For direction of My, apply right hand

rule



Scalar Analysis If the line of action of a force F is

perpendicular to any specified axis aa, for the magnitude of the moment of F about the axis,

Ma = Fda

where da is the perpendicular or shortest distance from the force line of action to the axis

A force will not contribute a moment about a specified axis if the force line of action is parallel or passes through the axis



A horizontal force F applied to the handle of the flex-headed wrench, tends to turn the socket at A about the z-axis

Effect is caused by the moment of F about the z-axis

Maximum moment occurs when the wrench is in the horizontal plane so that full leverage from the handle is achieved

(MZ)max = Fd



For handle not in the horizontal position,

(MZ)max = Fd’where d’ is the perpendicular distance from the force line of action to the axis

Otherwise, for moment, MA = FdMZ = MAcosθ



Vector AnalysisExampleMO = rA X F = (0.3i +0.4j) X (-20k)

= {-8i + 6j}N.mSince unit vector for this axis is ua = j,

My = MO.ua

= (-8i + 6j)·j = 6N.m



Vector Analysis Consider body subjected to

force F acting at point A To determine moment, Ma,

- For moment of F about any arbitrary point O that lies on the aa’ axis

MO = r X F where r is directed from O to A- MO acts along the moment axis bb’, so projected MO on the aa’ axis is MA



Vector Analysis- For magnitude of MA,

MA = MOcosθ = MO·ua

where ua is a unit vector that defines the direction of aa’ axis

MA = ua·(r X F)

- In determinant form,

zyx

zyxazayaxa

FFF

rrr

kji

kujuiuM

)(



Vector Analysis- Or expressed as,

where uax, uay, uaz represent the x, y, z components of the unit vector defining the direction of aa’ axis and rx, ry, rz represent that of the position vector drawn from any point O on the aa’ axis and Fx, Fy, Fz represent that of the force vector

zyx

zyx

azayax

axa

FFF

rrr

uuu

FXruM )(



Vector Analysis The sign of the scalar indicates the

direction of Ma along the aa’ axis- If positive, Ma has the same sense as ua - If negative, Ma act opposite to ua

Express Ma as a Cartesian vector,Ma = Maua = [ua·(r X F)] ua

For magnitude of Ma, Ma = ∑[ua·(r X F)] = ua·∑(r X F)



Wind blowing on the sign causes a resultant force F that tends to tip the sign over due to moment MA created about the aa axis

MA = r X F For magnitude of projection

of moment along the axis whose direction is defined by unit vector uA is

MA = ua·(r X F)



Example 4.8 The force F = {-40i + 20j + 10k} N acts on the

point A. Determine the moments of this force about the x and a axes.



SolutionMethod 1

Negative sign indicates that sense of Mx is opposite to i

mN

FXriM

iu

mkjir

Ax

x

A

.80

102040

643

001

)(

}643{



SolutionWe can also compute Ma using rA as rA

extends from a point on the a axis to the force

mN

FXruM

jiu

AAa

A

.120

102040

643

054

53

)(

54

53



SolutionMethod 2 Only 10N and 20N forces contribute moments

about the x axis Line of action of the 40N is parallel to this axis

and thus, moment = 0 Using right hand rule

Mx = (10N)(4m) – (20N)(6m) = -80N.mMy = (10N)(3m) – (40N)(6m) = -210N.mMz = (40N)(4m) – (20N)(3m) = 100N.m



Solution



Example 4.9 The rod is supported by two brackets at A and

B. Determine the moment MAB produced by F = {-600i + 200j – 300k}N, which tends to

rotate the rod about the AB axis.



Solution Vector analysis chosen as moment arm

from line of action of F to the AB axis is hard to determine

For unit vector defining direction of AB axis of the rod,

For simplicity, choose rD

ji

jirr

u

FXruM

B

BB

BAB

447.0894.0

)2.0()4.0(

2.04.0

)(

22



Solution For force,

In determinant form,

Negative sign indicates MAB is opposite to uB

mN

FXruM

NkjiF

DBAB

.67.53

300200600

02.00

0447.0894.0

)(

}300200600{



Solution In Cartesian form,

*Note: if axis AB is defined using unit vector directed from B towards A, the above formulation –uB should be used.

MAB = MAB(-uB)

mNji

jimNuMM BABAB

.}0.240.48{

)447.0894.0)(.67.53(

4.6 Moment of a Couple4.6 Moment of a Couple

Couple - two parallel forces - same magnitude but opposite direction- separated by perpendicular distance d

Resultant force = 0 Tendency to rotate in specified direction Couple moment = sum of

moments of both couple forces about any arbitrary point


Example Position vectors rA and rA are directed

from O to A and B, lying on the line of action of F

and –F Couple moment about O

M = rA X (-F) + rA X (F) Couple moment about A

M = r X Fsince moment of –F about A = 0


A couple moment is a free vector- It can act at any point since M depends only on the position vector r directed between forces and not position vectors rA and rB, directed from O to the forces

- Unlike moment of force, it do not require a definite point or axis


Scalar FormulationMagnitude of couple

moment M = Fd

Direction and sense are determined by right hand rule

In all cases, M acts perpendicular to plane containing the forces


Vector Formulation For couple moment,

M = r X F If moments are taken about point

A, moment of –F is zero about this point

r is crossed with the force to which it is directed


Equivalent CouplesTwo couples are equivalent if they

produce the same momentSince moment produced by the

couple is always perpendicular to the plane containing the forces, forces of equal couples either lie on the same plane or plane parallel to one another


Resultant Couple MomentCouple moments are free vectors

and may be applied to any point P and added vectorially

For resultant moment of two couples at point P,

MR = M1 + M2

For more than 2 moments,MR = ∑(r X F)


Frictional forces (floor) on the blades of the machine creates a moment Mc that tends to turn it

An equal and opposite moment must be applied by the operator to prevent turning

Couple moment Mc = Fd is applied on the handle


Example 4.10A couple acts on the gear teeth. Replace it by an equivalent couple having a pair of forces that cat through points A and B.


Solution Magnitude of couple

M = Fd = (40)(0.6) = 24N.m Direction out of the page since

forces tend to rotate CW M is a free vector and can

be placed anywhere


Solution To preserve CCW motion,

vertical forces acting through points A and B must be directed as shown

For magnitude of each force, M = Fd24N.m = F(0.2m)F = 120N


Example 4.11Determine the moment of the couple

acting on the member.


Solution Resolve each force into horizontal and vertical

componentsFx = 4/5(150kN) = 120kNFy = 3/5(150kN) = 90kN

Principle of Moment about point D, M = 120kN(0m) – 90kN(2m)

+ 90kN(5m) + 120kN(1m) = 390kN (CCW)


Solution Principle of Moment about point A,

M = 90kN(3m) + 120kN(1m) = 390kN (CCW)

*Note: - Same results if take momentabout point B- Couple can be replaced by two couples as seen in figure


Solution- Same results for couple replaced by two couples - M is a free vector and acts on any point on the member-External effects such as support reactions on the member, will be the same if the member supports the couple or the couple moment


Example 4.12Determine the couple moment acting on

the pipe. Segment AB is directed 30° below the x-y plane.


SolutionMethod 1Take moment about point O,

M = rA X (-25k) + rB X (25k)= (8j) X (-25k) + (6cos30°i + 8j – 6sin30°k) X (25k)= {-130j}N.cm

Take moment about point A M = rAB X (25k)

= (6cos30°i – 6sin30°k) X (25k)= {-130j}N.cm


SolutionMethod 2Take moment about point A or B,

M = Fd = 25N(5.20cm) = 129.9N.cm

Apply right hand rule, M acts inthe –j direction M = {-130j}N.cm


Example 4.13Replace the two couples acting on the pipe column by a resultant couple moment.


Solution For couple moment developed by the

forces at A and B, M1 = Fd = 150N(0.4m) = 60N.m

By right hand rule, M1 acts in

the +i direction, M1 = {60i} N.m

Take moment about point D, M2 = rDC X FC


SolutionM2 = rDC X FC

= (0.3i) X [125(4/5)j – 125(3/5)k]= (0.3i) X [100j – 75k]= {22.5j + 30k} N.m

For resultant moment, MR = M1 + M2

= {60i + 22.5j + 30k} N.m

4.7 Equivalent System4.7 Equivalent System

A force has the effect of both translating and rotating a body

The extent of the effect depends on how and where the force is applied

We can simplify a system of forces and moments into a single resultant and moment acting at a specified point O

A system of forces and moments is then equivalent to the single resultant force and moment acting at a specified point O


Point O is on the Line of Action Consider body subjected to force F applied

to point A Apply force to point O without altering

external effects on body- Apply equal but opposite forces F and –F at O


Point O is on the Line of Action- Two forces indicated by the slash across them can be cancelled, leaving force at point O- An equivalent system has be maintained between each of the diagrams, shown by the equal signs


Point O is on the Line of Action- Force has been simply transmitted along its line of action from point A to point O- External effects remain unchanged after force is moved- Internal effects depend on location of F


Point O is Not on the Line of ActionF is to be moved to point ) without

altering the external effects on the bodyApply equal and opposite forces at point

OThe two forces indicated by a slash

across them, form a couple that has a moment perpendicular to F


Point O is Not on the Line of ActionThe moment is defined by cross product

M = r X FCouple moment is free vector and can be

applied to any point P on the body


Consider effects when a stick of negligible weight supports a force F at its end

When force is applied horizontally, same force Is felt at the grip regardless of where it is applied along line of action

This relates the Principle of Transmissibility


When force is applied vertically, it cause a downwards force F to be felt at the grip and a clockwise moment of M = Fd

Same effects felt when F is applied at the grip and M is applied anywhere on the stick

In both cases, the systems are equivalent

4.8 Resultants of a Force and Couple

System


System Consider a rigid body Since O does not lies on the line

of action, an equivalent effect is produced if the forces are moved to point O and the corresponding moments are

M1 = r1 X F1 and M2 = r2 X F2

For resultant forces and moments,

FR = F1 + F2 and MR = M1 + M2


System


System Equivalency is maintained thus

each force and couple system cause the same external effects

Both magnitude and direction of FR do not depend on the location of point O

MRo depends on location of point O since M1 and M2 are determined using position vectors r1 and r2

MRo is a free vector and can acts on any point on the body


System


System Simplifying any force and couple

system, FR = ∑FMR = ∑MC + ∑MO

If the force system lies on the x-y plane and any couple moments are perpendicular to this plane,

FRx = ∑Fx

FRy = ∑Fy

MRo = ∑MC + ∑MO


System


SystemProcedure for Analysis When applying the following equations,

FR = ∑FMR = ∑MC + ∑MO

FRx = ∑Fx

FRy = ∑Fy

MRo = ∑MC + ∑MO

Establish the coordinate axes with the origin located at the point O and the axes having a selected orientation

4.8 Resultants of a Force and Couple System


Procedure for Analysis Force Summation For coplanar force system, resolve each

force into x and y components If the component is directed along the

positive x or y axis, it represent a positive scalar

If the component is directed along the negative x or y axis, it represent a negative scalar

In 3D problems, represent forces as Cartesian vector before force summation



Procedure for Analysis Moment Summation For moment of coplanar force system

about point O, use Principle of Moment Determine the moments of each

components rather than of the force itself In 3D problems, use vector cross product

to determine moment of each force Position vectors extend from point O to any

point on the line of action of each force



Example 4.14 Replace the forces acting on the brace by

an equivalent resultant force and couple

moment acting at point A.



Solution Force Summation For x and y components of resultant force,

NN

NNF

FF

NN

NNF

FF

Ry

yRy

Rx

xRx

8.8828.882

54sin400600

;

8.3828.382

54cos400100

;



Solution For magnitude of resultant force

For direction of resultant force

6.66

8.382

8.882tantan

962

)8.882()8.382()()(

11

2222

Rx

Ry

RyRxR

F

F

N

FFF



Solution Moment Summation Summation of moments about point A,

When MRA and FR act on point A, they will produce the same external effect or reactions at the support

)(.551.551

)3.0)(54cos400(

)8.0)(54sin400()4.0(600)0(100

;

CWmNmN

mN

mNmNNM

MM

RA

ARA



Example 4.15 A structural member is subjected to a couple

moment M and forces F1 and F2.

Replace this system with an equivalent resultant force

and couple moment acting at its base, point O.



Solution Express the forces and

couple moments as Cartesian

vectors

mNkjkjM

Njiji

rr

NuNF

NkF

CB

CBCB

.}300400{53

50054

500

}4.1666.249{)1.0()15.0(

1.015.0300

)300()300(

}800{

22

2

1



Solution Force Summation

mNkji

kji

kXkkj

FXrFXrMMMM

Nkji

jikFFF

FF

BCOCRo

R

R

.}300650166{

04.1666.249

11.015.0)800()1()300400(

}8004.1666.249{

4.1666.249800

;

21

21

4.9 Further Reduction of a Force and Couple

System


System Consider special case where system of forces

and moments acting on a rigid body reduces at point O to a resultant force FR = ∑F and MR = ∑ MO, which are perpendicular to one another

Further simplify the system by moving FR to another point P either on or off the body


System


System Location of P, measured from point O, can be

determined provided FR and MR known P must lie on the axis bb, which is

perpendicular to the line of action at FR and the aa axis

Distance d satisfies MRo = Frd or d = MRo/Fr


System


System With FR located, it will produce the same

resultant effects on the body If the system of forces are concurrent, coplanar

or parallel, it can be reduced to a single resultant force acting

For simplified system, in each case, FR and MRo will always be perpendicular to each other


System


SystemConcurrent Force Systems All the forces act at a point for which

there is no resultant couple moment, so that point P is automatically specified


System


SystemCoplanar Force Systems May include couple moments directed

perpendicular to the plane of forces Can be reduced to a single resultant force When each force is moved to any point O in

the x-y plane, it produces a couple moment perpendicular to the plane


System


SystemCoplanar Force Systems For resultant moment,

MRo = ∑M + ∑(r X F) Resultant moment is perpendicular to

resultant force FR can be positioned a distance d from O to

create this same moment MRo about O


System


SystemParallel Force Systems Include couple moments that are perpendicular

to the forces Can be reduced to a single resultant force When each force is moved to any point O in the

x-y plane, it produces a couple moment components only x and y axes


System


SystemParallel Force Systems For resultant moment,

MRo = ∑MO + ∑(r X F) Resultant moment is perpendicular to resultant

force FR can be positioned a distance d from O to

create this same moment MRo about O


System


System Three parallel forces acting on the stick can

be replaced a single resultant force FR acting at a distance d from the grip

To be equivalent, FR = F1 + F2 + F3

To find distance d, FRd = F1d1 + F2d2 +F3d3

Procedure for Analysis Establish the x, y, z axes and locate the

resultant force an arbitrary distance away from the origin of the coordinates

Force Summation For coplanar force system, resolve each

force into x and y components If the component is directed along the

positive x or y axis, it represent a positive scalar

If the component is directed along the negative x or y axis, it represent a negative scalar


System


System

Procedure for Analysis Force Summation Resultant force = sum of all the forces in

the systemMoment Summation Moment of the resultant moment about

point O = sum of all the couple moment in the system plus the moments about point O of all the forces in the system

Moment condition is used to find location of resultant force from point O


System


System

Reduction to a Wrench In general, the force and couple moment

system acting on a body will reduce to a single resultant force and a couple moment at o that are not perpendicular

FR will act at an angle θ from MRo

MRo can be resolved into one perpendicular M┴ and the other M║ parallel to line of action of FR


System


System

Reduction to a Wrench M┴ can be eliminated by moving FR to point

P that lies on the axis bb, which is perpendicular to both MRo and FR

To maintain equivalency of loading, for distance from O to P, d = M┴/FR

When FR is applied at P, moment of FR tends to cause rotation in the same direction as M┴


System


System

Reduction to a Wrench Since M ║ is a free vector, it may be moved to P so

that it is collinear to FR

Combination of collinear force and couple moment is called a wrench or screw

Axis of wrench has the same line of action as the force

Wrench tends to cause a translation and rotation about this axis


System


System

Reduction to a Wrench A general force and couple moment

system acting on a body can be reduced to a wrench

Axis of the wrench and the point through which this axis passes can always be determined


System


System

Example 4.16 The beam AE is subjected to a system of coplanar forces. Determine the magnitude, direction and location on the beam of a resultant force which is equivalent to the given system of forces

measured from E


System


System


NN

NNF

FF

NN

NNF

FF

Ry

yRy

Rx

xRx

0.2330.233

20060sin500

;

0.3500.350

10060cos500

;


System


System



7.33

0.350

0.233tantan

5.420

)0.233()0.350()()(

11

2222

Rx

Ry

RyRxR

F

F

N

FFF


System


System

Solution - Moment Summation Summation of moments about point E,

md

mNmN

mNmNNdN

MM ERE

07.50.233

1.1182

)5.2)(200()5.0)(100(

)0)(60cos500()4)(60sin500()0(350)(0.233

;


System


System

Example 4.17 The jib crane is subjected to three

coplanar forces. Replace this loading by an

equivalent resultant force and specify where the resultant’s line of action intersects the column AB and boom BC.


System


System


NkN

kNNF

FF

kNkN

kNkNF

FF

Ry

yRy

Rx

xRx

60.260.2

6.054

5.2

;

25.325.3

75.153

5.2

;


System


System



7.38

25.360.2

tantan

16.4

)60.2()25.3()()(

11

2222

Rx

Ry

RyRxR

F

F

kN

FFF


System


System

Solution Moment SummationMethod 1 Summation of moments about

point A,

my

mkNmkN

mkNmkn

kNykN

MM ARA

458.0

)6.1(54

50.2)2.2(53

50.2

)6.0(6.0)1(75.1

)0(60.2)(25.3

;


System


System

Solution Principle of Transmissibility

mx

mkNmkN

mkNmkn

xkNmkN

MM ARA

177.2

)6.1(5

450.2)2.2(

5

350.2

)6.0(6.0)1(75.1

)(60.2)2.2(25.3

;


System


System

Solution Method 2 Take moments about point A,

49.160.225.3

)6.1(5

450.2)2.2(

5

350.2

)6.0(6.0)1(75.1

)(60.2)(25.3

;

xy

mkNmkN

mkNmkn

xkNykN

MM ARA


System


System

Solution To find points of intersection,

let x = 0 then y = 0.458m Along BC, set y = 2.2m then

x = 2.177m


System


System

Example 4.18 The slab is subjected to four parallel

forces. Determine the magnitude and direction

of the resultant force equivalent to the given

force system and locate its point of application

on the slab.


System


System


NN

NNNNF

FF

R

R

14001400

500400100600

;


System


System

Solution Moment Summation

mx

x

NNmNmNxN

MM

my

y

NmNmNNyN

MM

yRy

xRx

00.3

42001400

)0(500)0(400)6(100)8(600)(1400

;

50.2

35001400

)0(500)10(400)5(100)0(600)(1400

;


System


System

Solution A force of FR = 1400N placed at point P (3.00m,

2.50m) on the slab is equivalent to the parallel force system acting on the slab


System


System

Example 4.19 Three parallel bolting forces act on the rim of the circular cover plate. Determine the magnitude and direction of a resultant force equivalent to the given force system and locate its point of application, P on the cover plate.


System


System


Nk

kkkF

FF

R

R

}650{

150200300

;


System


System

Solution Moment Summation

Equating i and j components,

85.84160650

85.84240650

85.8485.84160240650650

)150()45cos8.045sin8.0(

)20()8.0()300()8.0()650()(

)150()200()300(

;

y

x

ijijjyjx

kXji

kXjkXikXjyix

kXrkXrkXrFXr

MM

CBAR

ORo


System


System

Solution Solving, x = 0.239m and y = -0.116m Negative value of y indicates that the +ve

direction is wrongly assumed Using right hand rule,

)45cos8.0(150)8.0(200650

;

)45sin8.0(150)8.0(300650

;

mNmNx

MM

mNmNx

MM

xRx

yRy


System


System

4.10 Reduction of a Simple Distributed

Loading


LoadingLarge surface area of a body may be

subjected to distributed loadings such as those caused by wind, fluids, or weight of material supported over body’s surface

Intensity of these loadings at each point on the surface is defined as the pressure p

Pressure is measured in pascals (Pa)1 Pa = 1N/m2


Loading


Loading Most common case of distributed pressure loading is uniform loading along one axis of a flat rectangular body

Direction of the intensity of the pressure load is indicated by arrows shown on the load-intensity diagram

Entire loading on the plate is a system of parallel forces, infinite in number, each acting on a separate differential area of the plate


Loading


Loading Loading function p = p(x) Pa, is a function of x since pressure is uniform along the y axis

Multiply the loading function by the width w = p(x)N/m2]a m = w(x) N/m

Loading function is a measure of load distribution along line y = 0, which is in the symmetry of the loading

Measured as force per unit length rather than per unit area


Loading


LoadingLoad-intensity diagram for w = w(x)

can be represented by a system of coplanar parallel

This system of forces can be simplified into a single resultant force FR and its location can be specified

Magnitude of Resultant Force FR = ∑F Integration is used for infinite number of

parallel forces dF acting along the plate For entire plate length,

Magnitude of resultant force is equal to the total area A under the loading diagram w = w(x)


Loading


Loading

AdAdxxwFFFL A

RR )(;


Loading


LoadingLocation of Resultant Force MR = ∑MO

Location of the line of action of FR can be determined by equating the moments of the force resultant and the force distribution about point O

dF produces a moment of xdF = x w(x) dx about O

For the entire plate,

x

L

RORo dxxxwFxMM )(;


Loading


LoadingLocation of Resultant Force Solving,

Resultant force has a line of action which passes through the centroid C (geometric center) of the area defined by the distributed loading diagram w(x)

A

A

L

L

dA

xdA

dxxw

dxxxw

x)(

)(


Loading


LoadingLocation of Resultant Force Consider 3D pressure loading p(x), the

resultant force has a magnitude equal to the volume under the distributed-loading curve p = p(x) and a line of action which passes through the centroid (geometric center) of this volume

Distribution diagram can be in any form of shapes such as rectangle, triangle etc


Loading


Loading Beam supporting this stack of lumber is

subjected to a uniform distributed loading, and so the load-intensity diagram has a rectangular shape

If the load-intensity is wo, resultant is determined from the are of the rectangle

FR = wob


Loading


Loading Line of action passes through the centroid

or center of the rectangle, = a + b/2 Resultant is equivalent to the distributed

load Both loadings produce same “external”

effects or support reactions on the beam

x


Loading


LoadingExample 4.20Determine the magnitude and location of

the equivalent resultant force acting on the

shaft


Loading


LoadingSolutionFor the colored differential area element,

For resultant force

N

x

dxxdAF

FF

dxxwdxdA

AR

R

160

30

32

603

60

60

;

60

332

0

3

2

0

2

2


Loading


LoadingSolutionFor location of line of action,

Checking,

mmax

mNmabA

m

xdxxx

dA

xdA

x

A

A

5.1)2(43

43

1603

)/240(23

5.1

160

40

42

60

160

460

160

)60(44

2

0

42

0

2


Loading


LoadingExample 4.21A distributed loading of p = 800x Pa

acts over the top surface of the beam.

Determine the magnitude and location of the

equivalent force.


Loading


LoadingSolution Loading function of p = 800x Pa indicates

that the load intensity varies uniformly from p = 0 at x = 0 to p = 7200Pa at x = 9m

For loading, w = (800x N/m2)(0.2m) = (160x) N/m

Magnitude of resultant force = area under the triangleFR = ½(9m)(1440N/m) = 6480 N = 6.48 kN


Loading


LoadingSolutionResultant force acts through the centroid

of the volume of the loading diagram p = p(x)

FR intersects the x-y plane at point (6m, 0)

Magnitude of resultant force = volume under the triangleFR = V = ½(7200N/m2)(0.2m)

= 6.48 kN


Loading


LoadingExample 4.22The granular material exerts the distributed loading on the beam. Determine the

magnitude and location of the equivalent resultant of

this load


Loading


LoadingSolution Area of loading diagram is trapezoid Magnitude of each force = associated area

F1 = ½(9m)(50kN/m) = 225kNF2 = ½(9m)(100kN/m) = 450kN

Line of these parallel forces act through the centroid of associated areas and insect beams at

mmxmmx 5.4)9(21

,3)9(31

21


Loading


LoadingSolution Two parallel Forces F1 and F2 can be

reduced to a single resultant force FR For magnitude of resultant force,

For location of resultant force,

mx

x

MM

kNxF

FF

ORo

R

R

4

)450(5.4)225(3)675(

;

675450225

;

Solution*Note: Trapezoidal area can be divided into two

triangular areas,F1 = ½(9m)(100kN/m) = 450kN

F2 = ½(9m)(50kN/m) = 225kN


Loading


Loading

mmxmmx 3)9(31

,3)9(31

21

Chapter SummaryChapter Summary

Moment of a Force A force produces a turning effect about the

point O that does not lie on its line of action

In scalar form, moment magnitude, MO = Fd, where d is the moment arm or perpendicular distance from point O to its line of action of the force

Direction of the moment is defined by right hand rule

For easy solving, - resolve the force components into x and y components


Moment of a Force- determine moment of each component about the point- sum the results

Vector cross product are used in 3D problems

MO = r X F

where r is a position vector that extends from point O to any point on the line of action of F


Moment about a Specified Axis Projection of the moment onto the axis is

obtained to determine the moment of a force about an arbitrary axis provided that the distance perpendicular to both its line of action and the axis can be determined

If distance is unknown, use vector triple product

Ma = ua·r X Fwhere ua is a unit vector that specifies the direction of the axis and r is the position vector that is directed from any point on the axis to any point on its line of action


Couple Moment A couple consists of two equal but opposite

forces that act a perpendicular distance d apart

Couple tend to produce rotation without translation

Moment of a couple is determined from M = Fd and direction is established using the right-hand rule

If vector cross product is used to determine the couple moment, M = r X F, r extends from any point on the line of action of one of the forces to any point on the line of action of the force F


Reduction of a Force and Couple System Any system of forces and couples can be

reduced to a single resultant force and a single resultant couple moment acting at a point

Resultant force = sum of all the forces in the system

Resultant couple moment = sum of all the forces and the couple moments about the point

Only concurrent, coplanar or parallel force system can be simplified into a single resultant force


Reduction of a Force and Couple System For concurrent, coplanar or parallel force

systems, - find the location of the resultant force about a point- equate the moment of the resultant force about the point to moment of the forces and couples in the system about the same point

Repeating the above steps for other force system will yield a wrench, which consists of resultant force and a resultant collinear moment


Distributed Loading A simple distributed loading can be

replaced by a resultant force, which is equivalent to the area under the loading curve

Resultant has a line of action that passes through the centroid or geometric center of the are or volume under the loading diagram

Chapter ReviewChapter Review

ENT 152/4: Applied Mechanics RUSLIZAM DAUD [email protected]/013-4430827 Teaching Plan Final...

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Transcript of ENT 152/4: Applied Mechanics RUSLIZAM DAUD [email protected]/013-4430827 Teaching Plan Final...