mechanics of materials - university of pittsburghpitt.edu/~qiw4/academic/engr0135/chapter5-1.pdf ·...

Department of Mechanical Engineering

ENGR 0135

Chapter 5 -1Moments

Department of Mechanical Engineeri

Concept of moment Introduction to moment Moment about a pointMoment about a lineCouplesSome examples


Characteristics of a moment O = moment center d = moment arm (perpendicular to the line of

action of P) A-A = axis of the moment A-A is perpendicular to the plane made by the

force and the distance vectors Mo = moment at O

Mo = |Mo| = |P|d

Arrow of the moment indicates the sense (based on convention)– Counterclockwise Positive– Clockwise Negative

Or use right-hand rule; the thumb tells the “direction” of the moment

Moment has magnitude and direction so moment is a vector

O

O

d

A

A

Mo

Moment

P

P

(scalar representation, tells the magnitude)


Principle of Moment –Varignon’s Theorem

If R = A + B, then Mo(R) = Mo(A) + Mo(B) Mo(R)= Moment at O due to force R Mo(A)= Moment at O due to force A Mo(B)= Moment at O due to force B

The theorem can be used to either decompose or combine moments produced by forces


Example – finding the magnitude of momenta) The moment of force F3 about point Cb) The moment of force F2 about point Bc) The moment of force F1 about point Bd) The moment of force F3 about point E

53.13o

15” 15” sin 53.13o

inlbLFM BCC .6000154003 =×==

inlbLFM oBCB .360013.53sin153002 =×==

inlbLFM

oDBB

.750)130.53sin515(2501

=−×=

=

inlb

LFMo

EBE

.400)130.53cos1510(400

3

=−×=

=

LDB

LEB


Example

F = 300 N

30o

250 mm

x

150 mm 200 mm

B

MB = ?


Fx = F cos 30o

250 mm

x

150 mm 200 mm

B

Fy = F sin 30o

MB = - Fx 0.250 – Fy 0.2 = -95.0 N.m

= 95.0 N.m counterclockwise

Apply Varignon’s theorem:Break up the Force into its XYZ components and take the moments from each component

A

MA = ?


Vector representation of a moment

Useful for 3D cases, especially when d is not easy to measure

O

dLine of action of force P Pr

α

r = position vector of AA = any point in the line of action of P

A

eFrM

PrM

o

o

)sin|~||~(|~

~~~

α=

×=

magnitudes

e= A unit vector describing the moment

Perpendicular to the plane of the moment

Parallel to the line of action

α

Note that d = r sin α


Position vectorAny point along the line of action is a valid

point to get the r


The direction of vector s

O

A

s

The curling direction of the fingers other than the thumb.The thumb points in the direction of positive M.

FrM o~~~ ×=


Cross Product of unit vectors - review

0~~~~~~~~

~~~0~~~~~

~~~~~~0~~

=×=×−=×

−=×=×=×

=×−=×=×

kkikjjki

ijkjjkji

jikkijii

Must be memorized !!!


Cartesian representation of Moment Recall: the cartesian components of a vector

applied to vector position and force

kFjFiFF

krjrirr

zyx

zyx~~~~

~~~~

++=

++=

zyx

zyx

FFFrrrkji

Fr

~~~~~ =× bejagicfhceibghafj

jihgfecba

−−−++=

Definition of the determinant

The cross product can be written as determinant of a matrix containing the unit vectors, and the components of r and F. Extra-point take-home quiz: Show it.


zyx

zyxo

FFFrrrkji

FrM

~~~~~~ =×=

Mo = The moment Mo about the origin of coordinates O produced by a force F


kFjFiFF

krjrirr

zyx

zyx~~~~

~~~~

++=

++=

kFrFrjFrFriFrFr

kMjMiMFFFrrrkji

Fr

xyyxzxxzyzzy

ozoyox

zyx

zyx

~)(~)(~)(

~~~~~~

~~

−+−+−=

++==×=oM

Two ways of representing the moment vector:

1. Using i, j, and k unit vectors = Cartesian vector form


2. Using direction cosines = using a unit vector

kjie

eMFr

zyx

o~cos~cos~cos~

~~~

θθθ ++=

=×=oM

)(

)(

)(

xyyxoz

zxxzoy

yzzyxo

FrFrMFrFrMFrFrM

−=

−=

−=

o

ozz

o

oyy

o

oxx M

MMM

MM

~cos~cos~cos === θθθ

222~ozoyoxoo MMMMM ++=== oM

The scalar components of the moment


Moment analysis Scalar Analysis

– Get the Cartesian components of the forces– Get the distances for all of the force(s) components relative to the

point of interest– Multiply each pair of the force and distance (scalar multiplication)

and determine the direction of each moment using right-hand rule– Sum up the moments according to their direction

Vector Analysis– Get the vector position of the point about which the moment is

calculated relative to any point in the line of application – Perform cross product of the force and position vectors


Problem 1

The 15-ft boom AB has a fixedend A. A steel cable is stretchedfrom the free end B of the boomto a point C located on thevertical wall. If the tension in thecable is 570 lb, determine themoment about A of the forceexerted by the cable at B.

C

B

A

x

y

z15 ft

6 ft10 ft


1. Determine the rectangular components of a force defined byits magnitude and direction. If the direction of the force isdefined by two points located on its line of action, the force canbe expressed by:

F = (dx i + dy j + dz k)Fd

C

B

A

x

y

z15 ft

6 ft10 ft Solving Problems on Your OwnThe 15-ft boom AB has a fixedend A. A steel cable is stretchedfrom the free end B of the boomto a point C located on thevertical wall. If the tension in thecable is 570 lb, determine themoment about A of the forceexerted by the cable at B.

Problem 1


2. Compute the moment of a force in three dimensions. If r is aposition vector and F is the force the moment M is given by:

M = r x F

C

B

A

x

y

z15 ft

6 ft10 ft


Problem 1 Solution

3. Determine the rectangularcomponents of a force definedby its magnitude and direction.

First note:

dBC = (_15)2 + (6) 2 + (_10) 2

dBC = 19 ftThen:

TBC = (_15 i + 6 j _ 10 k) = _ (450 lb) i + (180 lb) j _ (300 lb)k570 lb19

570 N

C

B

A

x

y

z15 ft

6 ft10 ft


Problem 1 Solution

Compute the moment of aforce in three dimensions.

Have:MA = rB/A x TBC

Where: rB/A = (15 ft) i

Then:

MA = 15 i x (_ 450 i + 180 j _ 300 k)

MA = (4500 lb.ft) j + (2700 lb.ft) k

570 N

C

B

A

x

y

z15 ft

6 ft10 ft


Moment of a force about a lineBasically, it means the rectangular vector component of a moment (about a point that is passed by the line) that acts along the line.

F

AO

B

e

r

Mo = r x F Moment about point OMOB = [(r x F) . e] e Moment about line OB

e is the unit vector along OBO is any point on the line OB


CouplesTwo equal, noncollinear, parallel of

opposite senseThe sum of these forces = zeroTends only to rotate a body a couple = a moment sum of the moments of the pair of forces

that comprise the couple


Characteristic 1

The magnitude of the moment of a couple about a point in the plane of a couple is equal to the forces multiplied by the perpendicular distance between the forces

a

b

O

F

F

Mo = F. b + F. a = F. (a + b) = F. d

d

The distances “a” and “b” are missing from the equation.

O can be anywhere as long as on the plane of the couple


Characteristic 2

Moment of a couple does not depend on the location of the moment center O

F1

F2 OrB

rA

12 FrFrM BAO ×+×=but F1 = - F2 = (say) F

so

eαsin

)(

/

/

Fr

FrFrrM

BA

BA

BAO

=

×=×−=

rA/B α

A couple is a Free Vector


Transformation of a coupleA couple can be rotated in its planeA couple can be translated to a parallel

position in its plane or to any parallel planeThe magnitude of the forces and the

distance between them can be changedprovided that the magnitude of the couple is maintained


Characteristics determining the external effect of a couple

Magnitude of the moment of the coupleSense (rotation) of the coupleAspect of the plane of the couple (the slope

or direction of the plane of the couple)

A couple can be represented as a vector


Couples ~ Vector characteristics Couples can be “summed up (vectorially)” to yield

a resultant couple A couple can be decomposed into its scalar

cartesian components Scalar components of the same directions can be

combined to produce a couple vector in that direction

Direction cosines can be computed in a similar method as for the forces and moments


Examples The magnitude of the force

F in is 595 N. – Determine the scalar

component of the moment at point O about line OC.


Example

mechanics of materials - university of pittsburghpitt.edu/~qiw4/academic/engr0135/chapter5-1.pdf ·...

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