Download chapter 5 - Portal UniMAP

ENT151/4
STATICS
CHAPTER 4
FORCE SYSTEM RESULTANTS
4.1 Moment of a force – Scalar formulation
Moment - provides a measure of the tendency of the force to cause a body to rotate
about the point of axis.
This force tends to cause the pipe to turn about the z-axis.
Mo = Fx x d
Mo - moment of a force / torque
d - moment arm
Direction of moment - use right hand rule.
Resultant moment of a system of coplanar forces
MR = ∑F.d
1
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
The moment of any force will be +ve along the +ve z axis and –ve moment along the –ve
z axis.
EXAMPLES
Example 4-1 (pp 116)
Example 4-2 (pp 117)
Example 4-3 (pp 118)
4.2 Cross product
Magnitude and direction :
The cross product of two vectors A and B yields the vector C;
C = AxB;
(Eq. 1)
and is read “C equals A cross B”.
The magnitude of C is defined as the product of the magnitudes of A and B and the sine
of the angle θ between theirs tails.
Thus,
C = AB Sin θ
(0o ≤ θ ≤ 180o)
(Eq. 2)
Vector C has a direction that is perpendicular to the plane containing A and B such that C
is specified by the right-hand rule.
2
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Referring to the Figure (a);
Curling the fingers of the right hand from vector A (cross) to vector B, the thumb then
points in the direction of C.
Referring to the Figure (b);
From Eq. 1 and Eq. 2,
C = A x B = (AB Sin θ) uc
(Eq. 3)
where the scalar AB Sin θ defines the magnitude C and unit vector uc defines the
direction of C.
Law of operation :
There are 3 laws of operation in this topic;
1. The commutative law;
AxB ≠ BxA ;
AxB = -BxA
Referring to the Figures (c) i and (c) ii;
The cross product B x A yields a vector that acts in the opposite direction to C;
i.e., B x A = - C.
2. Multiplication by a scalar;
a ( A x B ) = ( aA ) x B = A x ( aB )
= (AxB)a
3. The distributive law;
Ax(B+D) = (AxD)+(AxD)
3
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Cartesian Vector Formulation :
From Eq. 3; it can be used to find the cross product of a pair of Cartesian unit vectors.
As example; to find i x j, the magnitude of the resultant vector is (i)(j)(sin 90o) = (1) (1)
(1) = 1. Its direction is determined using the right-hand rule. The resultant vector points
in the +k direction. (refer to the Figure (d))
Thus, i x j = (1)k.
In a similar manner,
ixj = k
ixk = -j
jxk = i
jxi = -k
kxi = j
kxj = -i
ixi = 0
jxj = 0
kxk = 0
Note:
Figure (e) shows a simple scheme for obtaining the same results when the need arises.
If the circle is constructed as shown, then “crossing” two unit vectors in a
counterclockwise fashion around the circle yield the positive third unit vector; e.g., k x i
= j.
4
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Moving clockwise, a negative unit vector is obtained; e.g., i x k = - j.
Now, Let us express the cross product of two general vectors A and B in a Cartesian
vector form;
A x B = ( Ax i + Ay j + Az k ) x ( Bx i + By j + Bz k )
= AxBx(i x i) + AxBy(i x j) + AxBz(i x k)
+ AyBx(j x i) + AyBy(j x j) + AyBz(j x k)
+ AzBx(k x i) + AzBy(k x j) + AzBz(k x k)
Then, carrying out the cross product operations and combining terms yields;
A x B = ( Ay Bz – Az By ) i - ( Ax Bz – Az Bx ) j + ( Ax By – Ay Bx ) k
(Eq. 4)
or this equation can be written in a compact determinant form;
AxB
=
i
Ax
Bx
j
Ay
By
k
Az
Bz
(Eq. 5)
Note:
To find the cross product of any two Cartesian vectors A and B, it is necessary to expand
a determinant whose first row of elements consists of the unit vectors i, j and k and
whose second and third rows represent the x, y, z components of the two vectors A and
B, respectively.
4.3 Moment of a Force – vector formulation
Mo = r x F
= r . F Sin θ
= F ( r . Sin θ )
where : r - position vector
5
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
= F.d
Cartesian vector :
Mo = ( ry Fz – rz Fy ) i - ( rx Fz – rz Fx ) j + ( rx Fy – ry Fx ) k
Mo = r x F
=
│
│
│
i
rx
Fx
j
ry
Fy
k
rz
Fz
│
│
│
Resultant moment :
MRo = ∑( r x F )
EXAMPLES
Example 4-4 (pp 124)
4.4
Example 4-5 (pp 125)
Principle of Moments
The moments of a force about a point is equal to the sum of the moments of the force’s
components about the point.
6
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Mo = r x F1 + r x F2
= r x ( F1 + F2 )
= r x F
EXAMPLES
Example 4-6 (pp 127)
Example 4-7 (pp 128)
4.5 Moment about specified axis
To find the components of the moment along a specified axis, Ma = F . da
If the line of action of a force F is perpendicular to any specify axis, the magnitud of the
moment of F about the axis can be determine by : Ma = F . da
Where : da – the perpendicular / shortest distance between the force line to the axis.
By vector analysis : Ma = ua ( r x F )
7
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
EXAMPLES
Example 4-8 (pp 142)
4.6
Example 4-9 (pp 143)
Moment of a couple
A couple is defined as two parallel forces that have the same magnitude, have opposite
directions and are separated by a perpendicular distance, d.
Moment couple = sum of the moments of both couple forces about any arbitrary point.
It is simpler to take moments about a point lying on the line of action of one of the forces.
e.g
At point A:
the moment of – F about A is zero, M = r x f
M depends only upon the position vector, r directed between the forces.
M = F.d
8
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
MR = M1 + M2
= ∑( r x F )
EXAMPLES
Example 4-10 (pp 150)
Example 4-11 (pp 151)
Example 4-12 (pp 152)
4.7 Equivalent system
Equivalent system occurs when the force and couple moment system produce the same
“external” effects translation and rotation of the body as their resultants.
In this section, you will learn how to maintain this equivalency when a single force is
applied to a specific point on a body and when it is located at another point O.
Point O Is On the Line of Action of the Force.
Referring to the Figure (a)i, which is subjected to the force F applied to point A.
In order to apply the force to point O without altering the external effects on the body,
equal but opposite forces F and –F are applied at O. (refer to Figure (a)ii)
The two forces indicated by the slash across them can be cancelled, leaving the force at
point O as required.(refer to Figure (a)iii). By using this construction procedure, an
equivalent system has been maintained between each of the diagrams, as shown by the
equal signs.
9
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Note :
The force has simply been “transmitted” along its line of action, from point A (Figure
(a)i) to point O (Figure (a)iii). The force can be considered as a sliding vector since it
can act at any point O along its line of action.
It is important to realize that only the external effects, such as the body’s motion or the
forces needed to support the body if it is stationary, remain unchanged after F is moved.
Certainly the internal effects depend on where F is located. For example, when F is acts
at A, the internal forces in the body have a high intensity around A; whereas movement
of F away from this point will cause these internal forces to decrease.
Point O Is Not On the Line of Action of the Force.
Referring to the Figure (b)i, where F is to be moved to point O without altering the
external effects on the body. Following the same procedure as before, equal but opposite
forces F and –F are applied at O. (refer to Figure (b)ii). Here the two forces indicated by
a slash across them form a couple which has a moment that is perpendicular to F and is
defined by the cross product M = r x F.
Since the couple moment is a free vector, it may be applied at any point P on the body as
shown in Figure (b)iii. In addition to this couple moment, F now acts at any point O as
required.
Summarization of these concepts:
When the point on the body is on the line of action of the force, simply transmit or slide
the force along its line of action to the point. When the point is not on the line of action
of the force, then move the force to the point and add a couple moment anywhere to the
body. This couple moment is found by taking the moment of the force about the point.
When these rules are carried out, equivalent external effects will be produced.
10
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
4.8 Resultant of a Force and couple system
When a rigid body is subjected to a system of forces and couple moments, it is often
simpler to study the external effects on the body by replacing the system by an equivalent
single resultant force acting at a specified point O and a resultant couple moment.
Referring to the Figure (c)i,
Since point O is not on the line of action of the forces, an equivalent effect is produced if
the forces are moved to point O and the corresponding couple moments M1 = r1 x F1
and M2 = r2 x F2 are applied to the body. The couple moment Mc is simply move to
point O since it is a free vector. (refer to Figure (c)ii)
By vector addition, the resultant force is FR = F1 + F2 and the resultant couple moment
is MRo = Mc + M1 + M2. (refer to Figure (c)iii)
=
=
Since equivalency is maintained between the diagrams in Figure (c)i, (c)ii and (c)iii, each
force and couple system will cause the same external effects (the same translation and
rotation of the body).
Note :
Both the magnitude and direction of FR are independent of the location of point O.
However, MRo depends upon this location since the moments M1 and M2 are determined
using the position vectors r1 and r2.
MRo is a free vector and can act at any point on the body, although point O is generally
chosen as its point of application.
Generally, the method of simplifying any force and couple moment system to a resultant
force acting at point O and a resultant couple moment can be represented by these
equations;
11
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
FR = ∑ F
and
(Eq. 1)
MRo = ∑ Mc + ∑ Mo
The first equation states that the resultant force of the system is equivalent to the sum of
all the forces. The second equation states that the resultant couple moment of the system
is equivalent to the sum of all the couple moments ∑ Mc, plus the moments about point O
of all the forces ∑ Mo.
If the force system lies in the x-y plane and any couple moments are perpendicular to this
plane, that is along the z axis, then the above equations reduce to the following three
scalar equations;
FRx = ∑ Fx
FRy = ∑ Fy
;
;
(Eq. 2)
MRo = ∑ Mc + ∑ Mo ;
Note :
The resultant force FR is equivalent to the vector sum of its two components FRx and FRy.
Procedure for Analysis :
Step 1:
Establish the coordinate axes with the origin located at the point O and the axes having a
selected orientation.
Force Summation
Step 2:
If the force system is coplanar, resolve each force into its x and y components. If a
component is directed along the positive x or y axis, it represents a positive scalar;
whereas if it is directed along the negative x or y axis, it is a negative scalar.
Step 3:
In three dimensions, represent each force as a Cartesian vector before summing the
forces.
12
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Moment Summation
Step 4:
When determining the moments of a coplanar force system about point O, it is generally
advantageous to use the principle of moments (determine the moment of the components
of each force rather than the moment of the force itself).
Step 5:
In three dimensions, use the vector cross product to determine the moment of each force
about the point. Here the position vectors extent from point O to any point on the line of
action of each force.
EXAMPLES
Example 4-14 (pp 164)
Example 4-15 (pp 165)
4.9 Reduction of a Force and Couple System
If the force and couple moment system is reduce to a resultant system at point P, only the
force resultant will have to be applied to the body.
EXAMPLES
Example 4-16 (pp 170)
Example 4-17 (pp 171)
13
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Example 4-18 (pp 172)
Example 4-19 (pp 173)
4.10 Reduction of a simple Distributed Loading
In this section, you will learn how to solve the reduction of a simple distributed loading
problem.
The intensity of loadings at each point on a surface is defined as the pressure p (force per
unit area), which can be measured in units of N/m2 or pascals (Pa), where 1 Pa = 1 N/m2.
Figure shows an example of a distributed pressure loading, which is uniform along one
axis of a flat rectangular body upon which the loading is applied.
The direction of the intensity of the pressure load is indicated by arrows shown on the
load-intensity diagram.
The entire loading on the plate is therefore a system of parallel forces, infinite in number
and each act on a separate differential area of the plate.
The loading function, p = p(x) Pa, is only a function of x since the pressure is uniform
along the y axis.
Multiply p = p(x) by the width a m of the plate;
Obtain w = [p(x) N/m2]a m = w(x) N/m
Figure shows the loading function (a measure of load distribution along the line y = 0
which is in the plane of symmetry of the loading.
14
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Note:
The loading function is measured as force per unit length, rather than a force per unit
area. The load-intensity diagram for w = w(x) can be represented by a system of coplanar
parallel forces as shown in Figure.
Magnitude of resultant force
From equation FR = ∑F, the magnitude of FR is equivalent to the sum of all the forces in
the system.
In this case integration must be used since there is an infinite number of parallel forces
dF acting along the plate (Figure (b)).
Since dF acting on an element of length dx and w(x) is a force per unit length, then at the
location x, dF = w(x) dx = dA.
In other words, the magnitude of dF is determined from the coloured differential area dA
under the loading curve.
For the entire plate length;
+ FR = ∑ F;
FR = ∫L w (x) dx
= ∫L dA
= A
(Eq. 1)
Magnitude of the resultant force is equal to the total area A under the loading diagram w
= w(x) as shown in Figure below.
15
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
Location of resultant force
Applying equation MRo = ∑ Mo, the location x of the line action of FR can be determined
by equating the moments of the force resultant and the force distribution about point O
(the y axis).
Since dF produces a moment of x dF = x w(x) dx about O (Figure a), then for the entire
plate (Figure b);
+ MRo = ∑ Mo
;
x FR = ∫L x . w(x) dx
Using Eq. 1 to solve x,
x = ∫L x w(x) dx
∫L w(x) dx
(Eq. 2)
x = ∫L x dA
∫L dA
This equation represents the x coordinate for the geometric centre or centroid of the area
under the distributed-loading diagram w(x).
The resultant force has a line of action which passed through the centroid C (geometric
centre) of the area defined by the distributed-loading diagram w(x). Once x is determined,
FR by symmetry passes through point (x, 0) on the surface of the plate.
Conclusion:
In the three-dimensional pressure loading p(x)(Figure (a)), the resultant force has a
magnitude equal to the volume under the distributed-loading curve p = p(x) and a line of
action which passes through the centroid (geometric centre) of this volume.
EXAMPLES
Example 4-20 (pp 182)
Example 4-21 (pp 183)
Example 44-22 (pp 184)
16
Zol Bahri Razali
School of Mechatronics, KUKUM
ENT151/4
STATICS
EXERCISES
Exercise 4 - 9
Exercise 4 - 14
Exercise 4 - 22
Exercise 4 - 36
Exercise 4 - 53
Exercise 4 - 63
Exercise 4 - 70
Exercise 4 - 82
Exercise 4 - 105
Exercise 4 - 121
Exercise 4 - 145
17
Zol Bahri Razali
School of Mechatronics, KUKUM