rigid bodies: equivalent system of forces

IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13

19.04.23 Dr. Engin Aktaş 1

Rigid Bodies: Equivalent System of ForcesEquivalent System of Forces



• the effects of forces on a rigid body.

• to replace the system of forces with a simpler equivalent system.

In this chapter we will study



External and Internal Forces • The External Forces

represent the action of other bodies on the rigid body under consideration. They control the external behavior of the body.

• The Internal Forces Are the forces which hold together the

particles forming the rigid body.

F

R2R1 W



Principle of TransmissibilityEquivalent Forces

The conditions of equilibrium or motion o a rigid body will remain unchanged if a force F acting at a given point of the rigid body is replaced by a force F’,

F

=F’A

B

same magnitude and direction

provided on the same line of action

acting at a different point

F=F’

F

R2R1 W=

F

R2R1 W



Limitation

= =

= =

P1=-P2

P1=-P2

A BP1 P2

(a)

A BP1

P’2

(b)

A B

(c)

A BP2 P1

(d)

A BP1P’2

(e)

A B

(f)

While the principle of transmissibility may be used freely to determine the conditions of motion or equilibrium of rigid bodies and to compute the external forces acting on these bodies, it should be avoided, or at least used with care, in determining internal forces and deformations.



Vector Product of Vectors

P

Q

V = P x Q

Line of action of V

The magnitude of V

V = P Q sin The sense of V

VRight Hand Rule

Right hand’s fingers show the direction of P and when you curl your fingers toward Q the direction of your thumb is the sense of V.

VAlso referred as Cross Product of P and Q.



Properties

P

QQ’

V

V = P x Q = P x Q’

The magnitude of V is the area of the parallelogram.

The vector products are not commutative Q x P P x Q

but Q x P = -(P x Q)

The distributive property holds P x (Q1 + Q2) = P x Q1 + P x Q2

The associative property does not apply P x (Q x S) (P x Q) x S



x

y

z

Vector Products Expressed in Terms of Rectangular Components

i

j

k

0

k

- j

- k

0

i

j

- i

i x i =

i x j =

i x k =

j x i =

j x j =

j x k =

k x i =

k x j =

k x k = 0

i

jk



V = P x Q

V = P x Q = (Px i + Py j + Pz k) x (Qx i + Qy j + Qz k)

V = (Py Qz - Pz Qy ) i + (Pz Qx - Px Qz) j + (Px Qy - Py Qx ) k

zyx

zyx

QQQ

PPP

kji

V

yx

yx

zyx

zyx

QQ

PP

QQQ

PPP

jikji

Repeat first and second columns to the right. The sum of the products obtained along the red line is then subtracted from the sum of the product obtained along the black lines



O

Moment of a Force about a Point

F

r

d A

Mo

Moment of F about O

Mo = r x F

Mo

Mo = r F sin = F d

The magnitude of moment of F about O.

The magnitude of Mo measures the tendency of the force F to make the rigid body rotate about a fixed axis directed along Mo.



Problems Involving Only Two Dimensions

O

d

F

Mo

Counterclockwise

Mo = + F d

points out of screen

O

d

F

Mo

Clockwise Moment

Mo = - F d

points into the screen



Varignon’s TheoremThe moment about a given point O of the resultant of several concurrent forces is equal to the sum of the moments of the various forces about the same point O. (VarignonVarignon (1654-1722))

x

y

z

OF1

F2

F3

F4

F5

r

r x ( F1+F2+ … ) = r x F1 + r x F2 + …



Rectangular Components of the Moment of a Force

x

y

z

O

Fy j

Fx i

Fz k

y j

z kx i

r = x i + y j + z k

r

F = Fx i + Fy j + Fz k

Mo= r x F

Mo = Mx i + My j + Mz k

Mx = y Fz – z Fy My = z Fx – x Fz Mz= x Fy – y Fx



x

y

z

O

Fy j

Fz k

Fx i

(yA-yB) j

rA/B AB

(z A-z B

) k (xA-xB) i

MB= rA/B x F = (rA-rB) x F

zyx

BABABA

FFF

zyx ///

kji

MB

xA/B = xA - xByA/B = yA - yB

zA/B = zA - zB



2-D problems

x

y

z

O

F

Fx i

Fy j

A (x, y, 0)

r

x i

y j

Mo = Mz kx

y

z

O

F

Fx i

Fy j

A

r A/B

MB = MB k

B(xA-xB) i

(yA-y

B)

j

Mo = (x Fy - y Fx) k

Mo = Mz = x Fy - y Fx

MB = (xA-xB) Fy - (yA-yB) Fx



Sample Problems



A 300-N vertical force is applied to the end of a lever which is attached to a shaft at O. Determine (a) the moment of the 300-N force about O(b) the magnitude of the horizontal force applied at A which creates the same moment about O.(c) the smallest force applied at A which creates the same moment about O.(d) how far from he shaft a 750-N vertical force must act to create the same moment about O.(Beer and Johnston)

O

300 N

60o

500

mm

A

A

60o

500

mm

dO

MO

a) Moment about O

d = (0.5 m) cos 60o = 0.25 m

MO = F d =(300 N) (0.25 m) = 75.0 Nmor

r

r = (0.5 m cos 60o) i + (0.5 m sin 60o) j = (0.25 i + 0.433 j) m

300 N

F = - 300 j N

MO = r x F = (0.25 i +0.433 j) x (-300 N) j = -75.0 k Nm



A

60o

500

mm

d

O

MO

b) Horizontal Force

d = (0.5 m) sin 60o = 0.433 m

MO = F d =(F) (0.433 m) = 75.0 Nm

or

r

r = (0.5 m cos 60o) i + (0.5 m sin 60o) j = (0.25 i + 0.433 j) m

F

F = F i N

MO = r x F = (0.25 i +0.433 j) x (F ) i = -75.0 k Nm

F = 75.0 Nm / 0.433 m = 173.2 N

- 0.433 F k Nm = -75.0 k Nm

F = 173.2 N



A

60o

500

mm

O

MO

c) Smallest Force

d = (0.5 m)

MO = F d

F

F = 75.0 Nm / 0.5 m = 150.0 N

Since Mo = F d, when d is max F is the smallest.

75.0 Nm = F (0.5 m)

F

30o

d) 750 N vertical force

75 Nm = (750 N) d

A

60o

500

mm

O

MO

750 N

d

B

OB cos 60o = d

d = 0.1 m

OB = 0.1 m / cos 60o = 200 mm



A force of 800 N acts on a bracket as shown. Determine the moment of the force about B. (Beer and Johnston)

60o

B

A

200 mm

160

mm

800 N

MB = rA/B x F

800 N

B

Fx = (400 N) i

Fx = (693 N) j

A

rA/B

rA/B = (- 0.2 i +0.16 j ) m

F = (800 N) cos 60o i + (800 N) sin 60o j

= 400 i + 693 j

MB = (-0.2 i + 0.16 j) x 400i + 693 j

64.0 Nm) k = (- 138.6 Nm) k -

= (- 202.6 Nm) k MB = 203 Nm



A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A. (Beer and Johnston)

MA = rC/A x F

rC/A = (0.06 i +0.075 j ) m F = -(200 N) cos 30o j + (200 N) cos 60o k

= -173.2 j + 100 k

MA = (0.06 i + 0.075 j) x (-173.2 j + 100 k) - 6 j = -10.39 k

MA = 14.15 Nm

MA = 7.5 i – 6 j –10.39 k

x

y

z

60 mm

25 mm

50 mmA

B C

30o

60o

200 N

A(0,-50,0) C(60,25,0)

+ 7.5 i

x

y

z

MAx

MAz

MAyMA



Scalar Product of Two Vectors

P

Q

P • Q = P Q cos

scalar product

or

dot product

The result is a scalar

PropertiesCommutative P • Q = Q • P

Distributive P • (Q1 + Q2) = P • Q1 + P • Q2

Associative P • (Q • S) = (P • Q) • S?

P • (Q • S) and (P • Q) • S does not have a meaningX



Moment of a Force about a Given Axis

x

y

zO

L

•

A

FMO

C

Let OL be an axis through O; we define the moment MOL of F about OL as the projection OC of the moment on the axis OL.

MOL = • MO = • (r x F)

the moment MOL of F about OL measures the tendency of the force F to impart to the rigid body a motion of rotation about a fixed axis OL.



Sample Problem

A B

C

F

D

E

PG

a

A cube of side a is acted upon by a force P as shown. Determine the moment of P

a) About Ab) About the edge ABc) About the diagonal AG of the cube

A B

C

F

D

E

G x

y

z

O ij

k

a)Moment about A

rF/A = ai – aj

kjP22

PP

kjikjjiPrM F/AA 22

PaPa



b) Moment about AB

22

PaPaM AB kjiiMi A

A B

C

F

D

E

PG

c) Moment about diagonal AG.

kjikjiAG

λ

3

1

3a

aaa

AG

6

111623

1 PaPaPaM AG kjikjiMλ A



Two forces F and –F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple.

Moment of a Couple

F-F

B

A

O x

y

z

rA

rB

rA x F + rB x (- F) = (rA – rB ) F

rA – rB = r

r M = r x F

The vector is called the moment of the couple; it is a vector perpendicular to the plane containing the forces and its magnitude

M = r F sin d = r sin

d

M = F dSince the r is independent of the choice of origin O, taking moment about another point would not change the result. Thus, the M of a couple is a free vector which may be applied at any point.

M



Equivalent Couples

O x

y

z

10

0 m

m100 mm

M

150 mm

200 N

200 N100 mm

O x

y

z

M

300 N

300 N

O x

y

z

M

100 mm

300 N

300 NTwo couples having the same moment M are equivalent, whether they are contained in the same plane or in parallel planes

This property indicates that when a couple acts on a rigid body, it does not matter where the two forces forming the couple act, or what magnitude and direction they have. The only thing which counts is the moment of the couple. Couples with the same moment will have the same effect on the rigid body.



P1

P2

Addition of Couples

P1

P2

Consider two intersecting planes P1 and P2.

F1

- F1

A

B

F2

- F2

r

R

-RM = r x R = r x (F1 + F2)

M = r x F1 + r x F2

M = M1 + M2

M1

M2

M



Couples May Be Represented by Vectors

O x

y

z

dF

-F

O x

y

z

=

M

M = F d

= O x

y

z

M

O x

y

z

= Mx

My

Mz



Resolution of a Given Force into a Force at a Different Point and a Couple

F

Or

A

=

F

-F

=

O

MO

F

F

O

r

A

A

A force-couple system

The force-couple system obtained by transferring a force F from a point A to a point O consists of F and a couple vector MO perpendicular to F.

Conversely, any force-couple system consisting of a force F and a couple MO which are mutually perpendicular may be replaced by a single equivalent force.



Sample Problem (Beer and Johnston)

x

y

z

100 N

100 N

150 N

150 N

240 mm

140 mm18

0 m

m18

0 m

m

A

D

C

E

BDetermine the components of the single couple equivalent to the two couples shown

100 N100 N

Mx = - (150 N) (0.360 m) = - 54.0 N.m

My = + (100 N) (0.240 m) = 24.0 N.m

Mz = + (100 N) (0.180 m) = 18.00 N.m

M = - (54.0 N.m) i +(24.0 N.m) j + (18.0 N.m) k




O

150 mm

300

mm

60 mm

400 N200 N

200 N

60o

B Replace the couple and force shown by an equivalent single force applied to the lever. Determine the distance from the shaft to the point of application of this equivalent force.

260

mm

O150 mm-(24 N . m) k

F = -(400 N) j

= O-(24 N . m) k

-(400 N) j

-(60 N . m) k



O

60o

O

-(400 N) j

-(84 N . m) k

-(400 N) j

-(84 N . m) k = OC x F

=

C

= ( OC cos 60o i + OC sin 60o j ) x ( -400 N) j

-(84 N . m) k =

cos 60o (400 N) OC =

84 N . m

- OC cos 60o (400 N) k

= 0.42 m = 420 mm



Reduction of a System of Forces to One Force and a Couple

F2

O r2

A2

A3

A1

F3

F1

r1

r3

=

F2

O

F3F1

M1

M2

M3

=

O

RMO

R

R = F MOR = MO = (r x F)



Equivalent System of ForcesTwo systems of forces F1, F2, F3 , etc., and F’1, F’2, F’3, etc., are equivalent if, and only if, the sums of the forces and the sums of the moments about a given point O of the forces of the two systems are, respectively, equal.

F = F’ and MO = M’O

Fx = F’x Fy = F’y Fz = F’z

Mx = M’x My = M’y Mz = M’z




A 4.8 m beam is subjected to the forces shown. Reduce the given system of forces to

a) An equivalent force couple system at Ab) An equivalent force couple system at Bc) A single force or resultant

a) An equivalent force couple system at A

R = F = (150 N)j – (600 N)j + (100 N)j –(250 N)j = -(600 N)j

150 N

600 N 100 N250 N

1.6 m 1.2 m 2 m

A B

MRA = (r x F) = (1.6i) x (-600j) + (2.8i) x (100j) + (4.8i) x (-250j) = - (1880 N.m)k

A B

- (600 N)j

- (1880 N)k



b) An equivalent force couple system at B

MRB = MR

A + BA x R = - (1880 N.m)k + (- 4.8m)i x (-600 N)j

= - (1880 N.m)k + (2880 N.m)k = (1000 N.m)k

A B

- (600 N)j

- (1880 Nm)k (2880 Nm)k

A B

- (600 N)j

(1000 Nm)k



c) A single force or resultant

A B

- (600 N)jx

r x R = MRA

(x)i x (-600 N)j = - (1880 N.m)k

- x(600 N)k = - (1880 N.m)k

x = 3.13 m




R = F MRO = (r x F)

A square foundation mat supports the four columns shown. Determine the magnitude and point of application of the resultant of the four loads.

r, m F, kN r x F, kN•my

x

z

O -200j

2.5 m

z

y

x

A

B

CO

2 m 3 m

2.5 m

200 kN 60 kN

40 kN100 kN

0 0

-60j5i -300k

-40j5i + 2.5k 100i – 200k-100j2i + 5k 500i – 200k

R = -400j MRO = 600i –700k

-(400 kN)j

(600 kNm)i-(700 kNm)k



y

x

z

O

-(400 kN)j

xi

zk

r x R = MOR

(xi + zk) x (-400j) = 600i – 700k

- 400xk + 400zi = 600i – 700k

- 400x = – 700

x = 1.750 m

400z = 600

z = 1. 500 m

rigid bodies: equivalent system of forces

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