Force System Resultants - Civil Engineering...

Engineering Mechanics: Statics

Force System Resultants

Chapter Objectives

To discuss the concept of the moment of a force and show how to calculate it in 2-D and 3-D systems.

Definition of the moment of a couple.

Chapter Objectives

• To present methods for determining the resultants of non-concurrent force systems in 2D and 3D systems.

• Reducing the given system of forces and couple moments into an equivalent force and couple moment at any point.

Chapter Outline

Moment of a Force

Force Couple

Principles of Moments

Moment of a Force

Carpenters often use a hammer in this way to pull a stubborn nail.

Through what sort of action does the force FH at the handle pull the

nail? How can you mathematically model the effect of force FH at

point O?

Moment of a force about a point (or an axis) is

a measure of the tendency of the force to

cause a body to rotate about the point or axis.

Moment of a Force 2D System

Case 1:

Fx horizontal and acts perpendicular to

the handle of the wrench and

is located dy from the point O

Fx tends to turn the pipe about the z axis

The larger the force or the distance dy the greater the turning effect



The larger the force ‘W’ or the distance ‘D’ the greater the turning effect at point ‘P’.

Note that:

Moment axis (z) is perpendicular to shaded plane (x-y). i.e. The remaining third axis: „z‟

Fx and dy lies on the shaded plane (x-y)

Moment axis (z) intersects

the plane at point O


Case 2:

Apply force Fz to the wrench Pipe does not rotate about z axis The pipe not actually rotates but Fz creates a tendency for rotation so causing (producing)moment along (Mo)x. Moment axis (x) is perpendicular to the shaded plane (y-z) Fz and dy lies on the shaded plane (y-z)


Case 3:

Apply force Fy to the wrench

No moment is produced about point O

Lack of tendency to rotate

as line of action passes

through O

Note that, Fy and dy both

lies on the same line (and not forming any

plane) hence No Moment is produced.


Moment of a force does not always cause rotation

Force F;

tends to rotate the beam clockwise about A with moment MA = dAF

tends to rotate the beam counterclockwise about B with moment MB = dBF


In 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 passing

through O and which is

perpendicular to the plane, is a vector quantity.


Moment MO is a vector having specified magnitude and direction.


M = F x d M = Magnitude of the moment

about point or axis [N.m]

F= Magnitude of the force [N]

d = perpendicular distance [m]

Direction is determined by using the right hand rule


Positive direction of the moment :

Anti clockwise


Positive moment

Negative moment


Magnitude:

Use simple multiplication;

For magnitude of MO: MO = F . d d = moment arm or perpendicular distance

from the axis at point O to its line of action

of the force.

F = Magnitude of the force

Units for moment is N.m, kN.m


Scalar Formulation


Scalar Formulation Direction:

Direction 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.

Direction:

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


Scalar Formulation

Direction MO is shown by a vector arrow with a curl

to distinguish it from force vector Fig b.

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

Arrowhead shows the sense of rotation caused by F.

Using the right hand rule, the direction and sense of the moment vector points out of the page.


Scalar Formulation

FrM

r Position vector which runs from the

moment reference point to any point

on the line of action of the force.

In some two dimensional problems and

most of the three dimensional problems,

it is convenient to use a vector approach

for moment calculations.

The MOMENT of a force about point A

may be represented by the cross product

expression.


Vector Formulation

Without using the right hand rule directly apply the equation through

cross product of vectors.

MO = d X F

The cross product directly gives the magnitude and the direction. Units for moment is N.m, kN.m


Vector Formulation

FrMo

Sarrus’ Rule

+k - k

)kFr)kFrMxyyxo

( (


Vector Formulation

Example:

For each case, determine the moment of the

force about point O


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 as a colored curl

o

o

(a)M (2m)(100N) 200.000 N.m (CW)

(b)M (0.75m)(50N) 37.500 N.m (CW)


Solution

o

o

o

(c) M (4m 2cos30 m)(40N) 229.282 N.m (CW)

(d) M (1sin 45 m)(60N) 42.426 N.m (CCW)

(e) M (4m 1m)(7kN) 21.000 kN.m (CCW)


Example:

Determine the moments of

the 800 N force acting on the

frame about points A, B, C

and D.


Solution

(Scalar Analysis)

Line of action of F passes through C

A

B

C

D

M = (2.5m)(800N) = 2000 N.m (CW)

M = (1.5m)(800N) = 1200 N.m (CW)

M = (0m)(800N)= 0 N.m

M = (0.5m)(800N) = 400 N.m (CCW)




Also known as Varignon‟s Theorem

This principle states that the moment of a force about a point is equal to the sum of moments of the force’s components about the point.


“Moment of a force about a point is equal to the sum of the moments of the forces‟ components about the point”


Solution Method 1: From trigonometry using triangle BCD, CB = d = 100cos45° = 70.7107mm

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

= 14142.136 N.mm (CCW)

= 14.142 N.m (CCW)

As a Cartesian vector,

MA = {14.142 k} N.m


Solution

Method 2:

Resolve 200 N force into x and y components

Principle of Moments

MA = ∑dF

MA =(200)(200sin45°) – (100)(200cos45°)

= 14142.136 N.mm (CCW)

= 14.142 N.m (CCW)


“Moment of a force about a point is equal to the sum of the moments of the forces‟ components about the point

m 898.275sin3 d

mkN 489.14

898.25

FdMO

mkN 489.14

30cos345sin530sin345cos5

xyyxO dFdFM

Example:

Example:

The force F acts at the end of the angle

bracket. Determine the moment of the force

about point O.


Solution

Method 1:

Resolve the given force into components and than apply the moment equation.

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

= -98.5641 N.m

As a Cartesian vector,

MO = {-98. 5641k} N.m


Solution

Method 2:

Express as Cartesian vector

r = {0.4i – 0.2j} m

F = {400sin30°i – 400cos30°j} N

= {200.000i – 346.410j}N

For moment,

O

i j k

M rXF 0.4 0.2 0

200.000 346.410 0

-98.564k N.m


Example (T):

Determine the moment of the 600 N force with respect to

point O in both scalar and vector product approaches.


Resultant Moment of

System of Coplanar Forces

Resultant moment MRo = addition of the moments

produced by all the forces

algebraically since all

moment forces are

collinear (for 2D case).

MRo = ∑F.d

taking counterclockwise (CCW), to be positive.

Resultant moment, MRo = addition of the moments produced by all the forces algebraically, since all moment forces are collinear (for 2D case).

MRo =M1 – M2 + M3

=∑dF= d1 F1 – d2 F2 + d3 F3

taking counterclockwise (CCW)

to be positive.

Resultant Moment of


Counterclockwise is positive

ORM Fd

Resultant Moment of


+

Example:

Determine the resultant moment of the four

forces acting on the rod about point O.

Resultant Moment of


Solution: (by scalar analysis)

Note that always positive moments acts in the +k

direction, CCW

)CW(m.N923.333

m.N923.333

)m30cos3m4)(N40(

)m30sin3)(N20()m0)(N60()m2)(N50(M

d.FM

Ro

Ro

Resultant Moment of


Solution: (by vector analysis)

(CW) or (-k) N.m333.923

(-k)] [263.923 (k)] [30 [0] k)]( [100

(-j)] 40 X )(i )3cos30(4 [

(i )] 20 X (-j) [3s in30(i )] 60 X [0(-j)] 50 X (i ) [2MRo

dXFMRo

Resultant Moment of


Moment (Revision)

Moment

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

MO = r X F where r represents position vector

from O to any point lying

on the line of action of F.

• Remember M = r (vector) X F (vector)

• Find the length ‘r’ vectorially for each

force ‘F’

• if not given, find the vectorial

representation of ‘F’ also.

Moment

In some two dimensional problems and

many three dimensional problems, it is

convenient to use a vector approach for

moment calculations. The MOMENT of

a force about point A may be

represented by the cross product

expression

FrM

Moment

FrM

r

Position vector which runs from the

moment reference point to any point

on the line of action of the force

A

F

1r

2r

3r

MA

Due to the principle of

transmissibility, can act at any point

along its line of action and still create

the same moment about point A.

F

FrFrFrM A

321

Moment

The Moment Vector

The result obtained from r X F doesn‟t depend on where the vector r intersects the line of action of F:

r = r’ + u

r F = (r’ + u) F = r’ F

because the cross product of the parallel vectors u and F is zero.

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

Moment of a Couple

A couple is defined as two

parallel forces with the same

magnitude but opposite in

direction separated by a

perpendicular distance d.

The moment of a couple is defined as:

MO = F . d (using a scalar analysis; right hand rule for direction),

MO = d X F (using vector analysis).

Moment of a Couple

The net external effect of a couple is zero since

the net force equals zero and the magnitude of

the net moment equals F.d

Moments due to couples can be added using

the same rules as adding any vectors.

The moment of a couple is a free vector.

It can be moved anywhere on the body

and have the same external effect on the

body.

O

a

d

F

F A

B

C

MO= F (a+d) – F a = F d

MO=MA=MB=MC

Moment of a couple has the same value for all

moment centers.

Moment of a Couple „2D‟

M M

M M

2D CCW couple

2D CW couple


The moment of a couple is a free vector. It can be moved

anywhere on the body and have the same external effect on

the body.

=



APPLICATIONS


APPLICATIONS

(continued)


Scalar Formulation Magnitude of couple moment

M = F.d

Direction and sense are determined by right hand rule

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


Vectorial Formulation

M = d X F

In all cases, M acts perpendicular

to plane containing the forces.


Example:

A couple acts on the gear teeth. Replace it

by an equivalent couple having a pair of

forces that act through points A and B.

=


Solution

Magnitude of couple

M = 24 N.m

Direction out of the page since

forces tend to rotate CCW

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 = F.d

24 N.m = F (0.2m)

F = 120.000 N


Example (T):

Two different couples are equivalent if they produce the same moment, (magnitude as well as direction).

Equivalent Couples


Two couples are equivalent if they produce the same moment

with magnitude and direction.

= = = =


Example (T):


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