02 force systems

8/11/2019 02 Force Systems

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2 Force Systems

K L 2 1 0 3 , C L A S S 0 2

S E M E S T E R I 2 0 1 2 / 2 0 1 3

Force

Force is a vector quantity. Its effect on a body or a

direction of the force.

Generally, a complete description of a force must

include its magnitude, direction, andpoint ofapplication.


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External and Internal Effects

The effects of a force on a body (or structure) can bese arated into external and internal effects.

External effects: reaction forces at the supports sothat the whole structure is in equilibrium (more onthis in Chapter 3).

Internal effects: internal forces (will be discussed inCha ter and ) stress and deformation in the

structure (will be discussed in Mechanics ofMaterials).

Principle of Transmissibility

A force may be applied at any point onits line of action without altering the

resultant effects of the force external tothe body on which it acts.


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Components of a Force

Example 1

The forces F1, F2, and F3act on pointA of the

rac et as s own.

Determine thexand y

components of each ofthe three forces.


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o

1

o

1

600 cos 35 491 N

600sin 35 344 N

x

y

F

F

4 2

2

5

3500 300 N

5

x

yF

0.2800 358 NF

2 2

32 2

0.2 0.4

0.4800 716 N

0.2 0.4

x

yF

Resultant of Forces

1 2

1 2

1 2

x x x x

y y y y

R F F F

R F F F


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Example 2

Combine the twoforces P and T into asingle equivalent forceR.

o

o

o

6sin60tan 0.866 40.9

3 6cos60

BD

AD

o

o

800 600cos 40.9 346 lb

600sin 40.9 393 lb

x xR F

R F

2 2

1 o

524 lb

tan 48.6

x y

y

x

R R R

R

R

Graphical solution:

o

525 lb

49

R


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Moment about a Point

Definition: tendency of the force torotate the body about the point.

The magnitude of the moment isproportional to the magnitude ofthe force and to the momentarm d:

M r F

Varignons Theorem

The moment of a force about any point is equal to

force about the same point.

OM Rd Pp Qq


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Example 3

Calculate themagnitude of themoment about the

base point O of the600-N force indifferent ways.

(i) Moment arm: o o4cos 40 2sin 40

4.35 m

d

600 4.35 2610 N-m CWOM

o

1

o

2

600 cos 40 460 N

600sin 40 386 N

F

F

(ii) Components ofF:

460 4 386 2 2610 N-m CWOM


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(iii)

By principle of transmissibility,Fcan be moved to pointB or Cto eliminate the moment of componentF2 orF1, respectively.

o

14 2tan 40 5.68 md

386 6.77 2610 N-mOM

(iv)o

22 4cot 40 6.77 md

. -O

o o2 4 600 cos40 sin 402610 N-m

O

M r F i j i j

k

(v) Using vector formulation:

Couple

Definition: the momentM F d ,

opposite, andnoncollinear forces.

The moment of a couplehas the same value atall points.


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Equivalent Couples

Changing the value ofFand ddoes not change acou le as lon as the roductFdremains the same.

Likewise, a couple is not affected if the forces act in adifferent but parallel plane.

In these four cases, the couples are equivalent:

Example 4

The rigid structuralmember is subjected to

the two 100-N forces.Replace this couple by

an equivalent coupleconsisting of the twoforces P and P, eachof which has amagnitude of 400 N.Determine the properangle .


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The original couple is counterclockwise with magnitude of:

100 0.1 10 N-mM

The forces P and P produce a counterclockwise couple of:

.

10 16cos

Solving for :

1 o10cos 51.316

Force Couple System

Effect of a force acting on a body: Translation in the direction of the force.

Rotation about any axis which does not intersect the line of theforce.

Both effects can be represented by replacing a forceby a force and a couple.

Reversing the process, we can combine a force couple system into a single, equivalent force.


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Equivalent Force Systems

A group or a system of forces can be replaced by theresultant.

The resultant usually represents the simplest force couple combination without altering the externaleffect on the body.

x xR F R = F

y y

R F

M M Fd


Two force systems are equivalent if the force

systems are the same.

1 2 1 2x xR R

R = R

1 2

1 2

y y

M M


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These four systemsare e uivalent:

Example 5

Replace the horizontal 80-lb

equivalent system consisting ofa force at O and a couple.


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Apply two equal and opposite 80-lb forces at O andidentify the counterclockwise couple:

o80 9sin 60 624 lb-in.M

Example 6

Determine the

four forces and onecouple which act

on the plate shown.


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(i) Equivalent force-couple at O:o o

o o

: 40 80cos30 60cos45 66.9 N

: 50 80sin30 60sin 45 132.4 N

x x x

y y y

R F RR F R

2 2

66.9 132.4 148.3 NR

1 o132.4tan 63.2

66.9

o o: 140 50 5 60cos 45 4 60sin 45 7

237 N-m CW

O OM Fd M

(ii) Single equivalent force

148.3 237

1.60 m

ORd M d

d

237 1.792 m132.4

y OR b M b

In other form:

02 force systems

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