IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Rigid Bodies:
Equivalent System of Forces
13.12.2015
Dr. Engin Aktaş
1
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
In this chapter we will study
• the effects of forces on a rigid body.
• to replace the system of forces with a
simpler equivalent system.
13.12.2015
Dr. Engin Aktaş
2
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
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.
F
R1
W
R2
• The Internal Forces
Are the forces which hold together the
particles forming the rigid body.
13.12.2015
Dr. Engin Aktaş
3
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Principle of Transmissibility
Equivalent 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
provided on the same line of action
=
13.12.2015
W
R2
F=F’
B
acting at a different point
F
R1
same magnitude and direction
F’
A
=
F
R1
Dr. Engin Aktaş
W
R2
4
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Limitation
P1
B
A
(a)
P2
P2
(d)
=
P1=-P2
P1
B
P’2
P1=-P2
B
A
A
P1
=
(c)
(b)
A
=
B
A
B
P’2
(e)
P1
B
A
=
(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.
13.12.2015
Dr. Engin Aktaş
5
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Vector Product of Vectors
V=PxQ
V
Q
q
Also
referred
as
Cross Product of P
and Q.
Line of action of V
P
The magnitude of V
V = P Q sin q
The sense of V
V
Right 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.
13.12.2015
Dr. Engin Aktaş
6
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Properties
V
The magnitude of V is the area of the parallelogram.
Q
Q’
V = P x Q = P x Q’
P
The vector products are not commutative
The distributive property holds
but Q x P = -(P x Q)
P x (Q1 + Q2) = P x Q1 + P x Q2
The associative property does not apply
13.12.2015
QxPPxQ
P x (Q x S)  (P x Q) x S
Dr. Engin Aktaş
7
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Vector Products Expressed in Terms of
Rectangular Components
y
j
i
z
x
k
ixi= 0
jxi= -k
kxi= j
ixj= k
jxj= 0
kxj= -i
ixk= -j
jxk= i
kxk= 0
j
k
13.12.2015
i
Dr. Engin Aktaş
8
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
V=PxQ
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
i
j
k
i
j
k
i
j
V  Px
Py
Pz
Px
Py
Pz
Px
Py
Qx
Qy
Qz
Qx
Qy
Qz Qx
Qy
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
13.12.2015
Dr. Engin Aktaş
9
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Moment of a Force about a Point
Mo
Moment of F about O
F
q
r
O
d
Mo = r x F
Mo
A
The magnitude of moment of F about O.
Mo = r F sin q = F d
The magnitude of Mo measures the tendency of the force F to make the rigid
body rotate about a fixed axis directed along Mo.
13.12.2015
Dr. Engin Aktaş
10
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Problems Involving Only Two Dimensions
F
F
Mo
13.12.2015
O
O
Mo
Counterclockwise
Clockwise Moment
Mo = + F d
Mo = - F d
points out of screen
points into the screen
Dr. Engin Aktaş
11
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Varignon’s Theorem
The 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. (Varignon (1654-1722))
y
F4
F5
F3
r
O
z
13.12.2015
x
F1
F2
r x ( F1+F2+ … ) = r x F1 + r x F2 + …
Dr. Engin Aktaş
12
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Rectangular Components of the Moment of a Force
y
r=xi+yj+zk
Fy j
F = Fx i + Fy j + Fz k
yj
Mo= r x F
Fx i
r
O
zk
Mo = Mx i + My j + Mz k
xi
x
Fz k
z
Mx = y Fz – z Fy
13.12.2015
My = z Fx – x Fz
Dr. Engin Aktaş
Mz= x Fy – y Fx
13
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
y
(yA-yB) j
MB= rA/B x F = (rA-rB) x F
Fy j
rA/B A
B
i
M B  xA/B
Fx
j
k
yA/B
zA/B
Fy
Fz
Fx i
(xA-xB) i
Fz k
x
O
z
xA/B = xA - xB
13.12.2015
yA/B = yA - yB
Dr. Engin Aktaş
zA/B = zA - zB
14
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
2-D problems
y
Fy j
F
A (x, y, 0)
z
A
Fx i
yj
xi
Fy j
B
x
O
MB = MB k
F
Fx i
(yA-yB) j
y
O
AR231 Fall12/13
(xA-xB) i
x
Mo = M z k
z
MB = (xA-xB) Fy - (yA-yB) Fx
Mo = (x Fy - y Fx) k
Mo = Mz = x F y - y F x
13.12.2015
Dr. Engin Aktaş
15
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problems
13.12.2015
Dr. Engin Aktaş
16
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
A
60o
O
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
300 N (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)
A
a) Moment about O
300 N
r
d = (0.5 m) cos 60o = 0.25 m
MO = F d =(300 N) (0.25 m) = 75.0 Nm
60o
O
d
MO
13.12.2015
or
r = (0.5 m cos 60o) i + (0.5 m sin 60o) j = (0.25 i + 0.433 j) m
F = - 300 j N
MO = r x F = (0.25 i +0.433 j) x (-300 N) j = -75.0 k Nm
Dr. Engin Aktaş
17
IZMIR INSTITUTE OF TECHNOLOGY
A
AR231 Fall12/13
b) Horizontal Force
d = (0.5 m) sin 60o = 0.433 m
r
d
F
Department of Architecture
MO = F d =(F) (0.433 m) = 75.0 Nm
O
MO
60o
F = 75.0 Nm / 0.433 m = 173.2 N
or
r = (0.5 m cos 60o) i + (0.5 m sin 60o) j = (0.25 i + 0.433 j) m
F= FiN
MO = r x F = (0.25 i +0.433 j) x (F ) i = -75.0 k Nm
- 0.433 F k Nm = -75.0 k Nm
F = 173.2 N
13.12.2015
Dr. Engin Aktaş
18
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
A
F
c) Smallest Force
Since Mo = F d, when d is max F is the smallest.
d = (0.5 m)
MO = F d
60o
O
75.0 Nm = F (0.5 m)
F = 75.0 Nm / 0.5 m = 150.0 N
MO
30o
F
A
d) 750 N vertical force
B
75 Nm = (750 N) d
750 N
60o
O
d
d = 0.1 m
OB cos 60o = d
OB = 0.1 m / cos 60o = 200 mm
MO
13.12.2015
Dr. Engin Aktaş
19
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
800 N
60o
160 mm
A
A force of 800 N acts on a bracket as
shown. Determine the moment of the
force about B. (Beer and Johnston)
B
200 mm
Fx = (693 N) j
A
MB = rA/B x F
800 N
rA/B = (- 0.2 i +0.16 j ) m
Fx = (400 N) i
rA/B
B
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
= (- 138.6 Nm) k - 64.0 Nm) k
= (- 202.6 Nm) k
13.12.2015
Dr. Engin Aktaş
MB = 203 Nm
20
IZMIR INSTITUTE OF TECHNOLOGY
y
60 mm
C
B
z
30o
Department of Architecture
AR231 Fall12/13
200 N
60o
x
A
A 200-N force is applied as shown to
the bracket ABC. Determine the
moment of the force about A. (Beer and
Johnston)
A(0,-50,0) C(60,25,0)
rC/A = (0.06 i +0.075 j ) m
MA = rC/A x F
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) = -10.39 k - 6 j + 7.5 i
y
MA = 7.5 i – 6 j –10.39 k
MAz
MA = 14.15 Nm
MAx
x
z
13.12.2015
MAy
MA
Dr. Engin Aktaş
21
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Scalar Product of Two Vectors
Q
scalar product
P•Q=
q
or
P Q cos q
dot product
P
The result is a scalar
Properties
Commutative
Distributive
X
Associative
13.12.2015
P•Q=Q•P
P • (Q1 + Q2) = P • Q1 + P • Q2
?
P • (Q • S) = (P • Q) • S
P • (Q • S) and (P • Q) • S does not have a meaning
Dr. Engin Aktaş
22
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Moment of a Force about a Given Axis
MO
y
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.
F
MOL = l • MO = l • (r x F)
L
C
•
l
z
13.12.2015
O
x A
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.
Dr. Engin Aktaş
23
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample
Problem
A cube of side a is acted upon by a force P as
C
D
A
a
B P
G
y
E
shown. Determine the moment of P
a) About A
b) About the edge AB
c) About the diagonal AG of the cube
F
a)Moment about A
C
D
A
E
j
O
k
z
13.12.2015
rF/A = ai – aj
B
i
G
F
x
P
P
j
2
M A  rF/A  P  a i  j  
Dr. Engin Aktaş
P
k
2
P
2
j  k  
aP
i  j  k 
2
24
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
b) Moment about AB
M
D
C
l
B P
A
AB
 iM A  i
λ 
13.12.2015
AG
 λ M A 
2
AG

2
ai  aj  ak
AG
F
M
i  j  k  
aP
c) Moment about diagonal AG.
G
E
aP
1
3
a 3
i  j  k  
aP

1
3
i  j  k  
2
Dr. Engin Aktaş
i  j  k 
aP
6
1  1  1  
aP
6
25
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Moment of a Couple
Two forces F and –F having the same magnitude, parallel
lines of action, and opposite sense are said to form a couple.
rA x F + rB x (- F) = (rA – rB ) F
B
y
-F
r q
M
rB
A
rA
O
z
rA – rB = r
d
F
M=rxF
The vector is called the moment of the couple;
it is a vector perpendicular to the plane
containing the forces and its magnitude
x
M = r F sin q
d = r sin q
M=Fd
Since 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.
13.12.2015
Dr. Engin Aktaş
26
IZMIR INSTITUTE OF TECHNOLOGY
y
M
y
x
200 N
y
150 mm
O
M
300 N
x
M
z
200 N
300 N
This property indicates that
when a couple acts on a
rigid body, it does not matter
300 N
where the two forces
forming the couple act, or
what
magnitude
and
x 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.
300 N
Two couples having the
same moment M are
equivalent, whether they are
contained in the same plane
or in parallel planes
O
z
13.12.2015
AR231 Fall12/13
Equivalent Couples
O
z
Department of Architecture
Dr. Engin Aktaş
27
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Addition of Couples
-R
- F1
Consider two intersecting planes P1 and P2.
- F2
M = r x R = r x (F1 + F2)
P2
B
r
P1
A
F1
R
M = r x F1 + r x F2
F2
M1
M
M = M1 + M2
P2
P1
13.12.2015
M2
Dr. Engin Aktaş
28
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Couples May Be Represented by Vectors
M
y
M=Fd
y
y
M
-F
O
z
d
F
x
= O
x
z
= O
x
z
y
My
O
=
Mx
x
Mz
z
13.12.2015
Dr. Engin Aktaş
29
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Resolution of a Given Force into a Force at
a Different Point and a Couple
F
F
F
A
O
F
A
r
=
MO
r
O
A
=
O
-F
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.
13.12.2015
Dr. Engin Aktaş
30
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
y
Sample Problem
(Beer and Johnston)
B
C
AR231 Fall12/13
Determine the components of
the single couple equivalent to
x the two couples shown
150 N
100 N
E
100 N
A
100 N
Mx = - (150 N) (0.360 m) = - 54.0 N.m
My = + (100 N) (0.240 m) = 24.0 N.m
z
D
100 N
Mz = + (100 N) (0.180 m) = 18.00 N.m
150 N
M = - (54.0 N.m) i +(24.0 N.m) j + (18.0 N.m) k
13.12.2015
Dr. Engin Aktaş
31
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem (Beer and Johnston)
B
60 mm
200 N
60o
400 N
200 N
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.
O
150 mm
F = -(400 N) j
=
-(24 N . m) k
O
-(24 N . m) k
150 mm
O
-(60 N . m) k
-(400 N) j
13.12.2015
Dr. Engin Aktaş
32
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
C
-(400 N) j
-(84 N . m) k
=
60o
O
O
-(400 N) j
-(84 N . m) k = OC x F = ( OC cos 60o i + OC sin 60o j ) x ( -400 N) j
-(84 N . m) k =
- OC cos 60o (400 N) k
84 N . m
OC =
=
0.42 m = 420 mm
cos 60o (400 N)
13.12.2015
Dr. Engin Aktaş
33
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Reduction of a System of Forces to One
Force and a Couple
F2
M3
F2
M2
A2
F1
A1
r
r1 O 2
r3
MO R
F3
F1
=
O
R
=
F3
O
A3
M1
R = SF
13.12.2015
MOR = SMO = S(r x F)
Dr. Engin Aktaş
34
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Equivalent System of Forces
Two 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.
SF = SF’
13.12.2015
SMO = SM’O
and
SFx = SF’x
SFy = SF’y
SMx = SM’x
SMy = SM’y
SFz = SF’z
SMz = SM’z
Dr. Engin Aktaş
35
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
Sample Problem
(Beer and Johnston)
250 N
150 N
100 N
600 N
A
B
1.6 m
1.2 m
2m
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 A
b) An equivalent force couple system at B
c) A single force or resultant
a) An equivalent force couple system at A
R = SF = (150 N)j – (600 N)j + (100 N)j –(250 N)j = -(600 N)j
MRA = S(r x F) = (1.6i) x (-600j) + (2.8i) x (100j) + (4.8i) x (-250j) = - (1880 N.m)k
- (600 N)j
- (1880 N)k
B
A
13.12.2015
Dr. Engin Aktaş
36
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
b) An equivalent force couple system at B
MRB = MRA + 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
- (600 N)j
- (1880 Nm)k
B
A
(2880 Nm)k
- (600 N)j
A
13.12.2015
B
Dr. Engin Aktaş
(1000 Nm)k
37
IZMIR INSTITUTE OF TECHNOLOGY
Department of Architecture
AR231 Fall12/13
c) A single force or resultant
x
- (600 N)j
B
A
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
13.12.2015
Dr. Engin Aktaş
38
IZMIR INSTITUTE OF TECHNOLOGY
60 kN
200 kN
40 kN
100 kN
A
3m
y
-(700 kNm)k
x A square foundation mat supports the four
columns shown. Determine the magnitude
2.5 m
and point of application of the resultant of the
four loads.
2.5 m
C
O
2m
B
R = SF
-(400 kN)j
O (600 kNm)i
x
MRO = S(r x F)
r, m
F, kN
r x F, kN•m
0
-200j
0
5i
-60j
-300k
5i + 2.5k
2i + 5k
z
13.12.2015
AR231 Fall12/13
Sample Problem (Beer and Johnston)
y
z
Department of Architecture
Dr. Engin Aktaş
-40j
-100j
100i – 200k
500i – 200k
R = -400j
MRO = 600i –700k
39
IZMIR INSTITUTE OF TECHNOLOGY
y
O
r x R = MO R
xi
x
(xi + zk) x (-400j) = 600i – 700k
- 400xk + 400zi = 600i – 700k
- 400x = – 700
x = 1.750 m
13.12.2015
AR231 Fall12/13
-(400 kN)j
zk
z
Department of Architecture
Dr. Engin Aktaş
400z = 600
z = 1. 500 m
40
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Equivalent System of Forces