5. DISTRIBUTED FORCES. CENTRES OF GRAVITY
5.1 Distributed forces. In engineering problems we often have to deal with
loads distributed along a line, over an area or within a certain volume. Let us
examine some simple cases of distributed forces.
A system of distributed forces along a line can be characterized by the load
per unit length of the line of application, which is called the intensity q. The
dimension of intensity is newton per meter: N/m.
1) Forces uniformly distributed along a straight line (Fig.5.1). The intensity q of
such system is a constant quantity.
2) Forces distributed along a straight line according to a linear law (Fig.5.2). An
example of such a load is the pressure of water against a dam, which drops from
a maximum at the bottom to zero at the surface. For such forces the intensity
varies from zero to qm.
qm
q
Fig. 5.1
Fig. 5.2
3) Forces uniformly distributed along the arc of a circle (Fig.5.3). An example of
such forces is the hydrostatic pressure on the sides of a cylindrical vessel.
4) Forces uniformly distributed along a curved line. An example of such forces
are gravity forces of a wire lying in the xy plane (Fig.5.4).
z
y
R
x
0
Fig. 5.3
Fig. 5.4
CG-1
5.2 Centre of gravity of a two-dimensional body. Centroids of areas and
lines. We have assumed so far that the attraction exerted by the earth on rigid
body could be represented by a single force W. This force, called the weight of
the body, was to be applied at the centre of gravity of the body. Actually, the
earth exerts a force on each of the particles forming the body. The action of the
earth on a rigid body should thus be represented by a large number of small
forces distributed over the entire body. We shall see in this chapter, however,
that all these small forces may be replaced by a single equivalent force W. We
shall also learn to determine the centre of gravity, i.e., the point of application of
the resultant W, for various shapes of bodies.
Let us consider a flat horizontal plane (Fig.5.5). We may divide the plate into
n small elements. The forces exerted by the earth on elements of plate will be
denoted by ∆W1, ∆W2,…, ∆Wn, respectively. These forces are directed towards
the centre of the earth; however, for all practical purposes they may be assumed
parallel. Their resultant is therefore a single force in the same direction. The
magnitude W of this force is obtained by the adding the magnitudes of the
elementary weights,
∑ Fz :
W = ∆W1 + ∆W2 + … ∆Wn
z
(5.1)
z
y
y
x
y
0
xG
x
∆W
G
yG
0
Fig. 5.5
x
W
Fig. 5.6
To obtain the coordinates xG and yG of the point G where the resultant W should
be applied, we write that the moments of W about the y and x axes are equal to
the sum of the corresponding moments of the elementary weights,
∑My :
∑Mx :
xGW = x1∆W1 + x2∆W2 + … + xn∆Wn
yGW = y1∆W1 + y2∆W2 + … + yn∆Wn
CG-2
(5.2)
If we now increase the number of elements into which the plate is divided and
simultaneously decrease the size of each element, we obtain at the limit
following expressions:
W = ∫ dW
x G W = ∫ xdW
y G W = ∫ ydW
(5.3)
These equations define the weight W and the coordinates xG and yG of the centre
of gravity G of a flat plate. The same equations may be derived for a wire lying
in the xy plane (Fig.5.6). We shall observe, in the latter case, that the centre of
gravity G will generally not be located on the wire.
In the case of homogeneous plates of uniform thickness (Fig.5.7a) the centre
of gravity G coincides with the centroid C of the area A of the plate, the latter
being define as follows:
x C A = ∫ xdA
y C A = ∫ ydA
(5.4)
b) y
a) y
L
A
xC
xC
C
yC
C
yC
0
x
0
Fig. 5.7
The integral ∫ xdA is known as the first moment of the area A with respect to
the y axis. Similarly the integral
∫ ydA
defines the first moment of the area A
with respect to the x axis. It is seen from eqs. (5.4) that, if the centroid of an area
is located on a coordinate axis, the first moment of the area with respect to this
axis is zero.
Similarly the centre of gravity of the homogeneous wire of uniform cross
section coincides with the centroid C of the line L defining the shape of the line
L (Fig.5.7b). The coordinates xC and yC of the centroid of the line L are obtained
from the equations
CG-3
x C L = ∫ xdL
y C L = ∫ ydL
Centroids of common shapes of areas and lines are shown in Fig.5.8
CG-4
(5.5)
5.3 Centre of gravity of a three-dimensional body. Centroid of a volume.
The method of determination of the centre of gravity G of a three-dimensional
body is, in principle, similar to that applied to plates in previous section. The
body, however, is divided into small volume elements of weighs ∆W. Increasing
the number of elements and simultaneously decreasing the size of each element,
we obtain at the limit
x G W = ∫ xdW
y G W = ∫ ydW
z G W = ∫ zdW
(5.6)
If the body is made of a homogeneous material, the equations
x C V = ∫ xdV
y C V = ∫ ydV
z C V = ∫ zdV
(5.7)
defining the centroid C of the volume V of the body may be used to a
determination of its centre of gravity.
The integral ∫ xdV is known as the first moment of the volume with respect to
the yz plane. Similarly, the integrals ∫ ydV and ∫ zdV define the first moments
of the volume with respect to the zx plane and the xy plane, respectively. It is
seen from eqs. (5.7) that, if the centroid of a volume is located in a coordinate
plane, the first moment of the volume with respect to that plane is zero.
CG-5
A volume is said to be symmetrical with respect to a given plane if to every
point P of the volume corresponds a point P’ of the same volume, such that the
line PP’ is perpendicular to the given plane and divided into two equal parts by
that plane. The plane is said to be a plane of symmetry for a given volume. When
a volume V possesses two planes of symmetry, the centroid of the volume must
be located on the line of intersection of the two planes. Finally, when a volume
V possesses three planes of symmetry which intersect in a well-defined point
(i.e., not along a common line), the point of intersection of the three planes must
coincide with the centroid of the volume. This property enables us to determine
immediately the centroid of the volume of spheres, ellipsoides, cubes,
rectangular parallelepipeds, etc.
Centroids of unsymmetrical volumes or of volumes possessing only one or
two planes of symmetry should be determined by integration. Centroids of
common shapes of volumes are shown in Fig.5.9. It should be observed that the
centroid of a volume of revolution in general does not coincide of its cross
section. Thus, the centroid of hemisphere is different from that of a semi circular
area, and the centroid of a cone is different from that of a triangle.
CG-6
Fig. 5.9
5.4 Determination of gravity centres of composite plates, wires and bodies.
CG-7
In many instances a flat plate may be divided into rectangles, triangles, or other
common shapes shown in Fig.5.8. The abscissa xG of its centre of gravity G may
be determined from the abscissas x1, x2,… of the centres of gravity of the various
parts by expressing that the moment of the weight of the whole plate about y
axis is equal to the sum of the moments of the weights of the various parts about
the same axis (Fig.5.10). The ordinate yG of the centre of gravity of the plate is
found in a similar way by equating moments about the x axis. Thus, denoting by
W the magnitude of the weight of a plate and by W1, W2,…, Wn the magnitudes
of the weights of its parts, we obtain:
∑My :
x G W = ∑ x i Wi
(5.8a)
∑Mx :
y G W = ∑ y i Wi
(5.8b)
z
z
y
y
W3
∑W
xG
W1
G
yG
0
G1
x
0
W2
G2
G3
x
Fig. 5.10
If the plate is homogeneous and of uniform thickness, the centre of gravity
coincides with the centroid C of its area. The abscissa xC and ordinate yC of the
centroid of the area may be then determined by expressing that the first moment
of the composite area with respect to either y or x axis, respectively is equal to
the sum of the first moments of the elementary areas with respect to the same
axis. Thus, coordinates of the area centroid can be determined from the
following equations
∑My :
x C ∑ A = ∑ xiAi
(5.9a)
∑Mx :
yC ∑ A = ∑ yi Ai
(5.9b)
Care should be taken to record the moment of each area with the appropriate
sign. First moments of areas, just like moments of forces, may be positive and
CG-8
negative. For example, an area whose centroid is located to the left of the y axis
will have a negative first moment with respect to that axis. Also, the area of a
hole should be recorded with a negative sign (Fig.5.11)
y
A2
A1
x1
x2
x3
A3
xi
A
xiA
A1 Semicircle
-
+
-
A2 Full rectangle
+
+
+
A3 Circular hole
+
-
-
x
Fig. 5.11
Similarly, it is possible in many cases to determine the centre of gravity of a
composite wire/body or centroid of a composite line/volume by dividing the
wire/body or line/volume into simpler elements. Proceeding analogously as in
the case of plates, we obtain the following equations defining the coordinates xG,
yG and zG of the composite body gravity centre G :
x G W = ∑ x i Wi ,
y G W = ∑ y i Wi ,
z G W = ∑ z i Wi
(5.10)
5.5 Determination of centroids by integration. The centroid of an area
bounded by analytical curves (i.e., curves defined by algebraic equations) is
usually determined by computing the integrals in Eqs.5.4
x C A = ∫ xdA
y C A = ∫ ydA
(5.4)
If the element of area dA is chosen equal to a small square of sides dx and dy,
the determination of each of these integrals requires a double integration in x
and y. A double integration is also necessary if polar coordinates are used.
In most cases, however, it is possible to determine the coordinates of the
centroid of an area by performing a single integration. This is achieved by
choosing for dA a thin rectangle or strip, or a thin sector of pie-shaped element
(Fig.5.12); the centroid of the thin rectangle is located at its centre, and the
centroid of the thin sector at a distance
2
r
3
from its vertex (as for triangle). The
coordinates of the centroid of the area under consideration are then obtained by
CG-9
expressing that the first moment of the entire area with respect to each of the
coordinate axes is equal to the sum (or integral) of the corresponding moments
of the elements area. Denoting by xel and yel the coordinates of the centroid of
the element dA, we write
∑My :
x G A = ∫ x el dA
(5.11a)
∑Mx :
y G A = ∫ y el dA
(5.11b)
P(x,y)
Q(x,y)
x
y
y
y
x
x
dy
xel
y
y
r
2/3
yel
0
0
dx
yel
θ
x
x
R( θ,r)
r
yel
0
xel
x
xel
a
a)
b)
c)
a+x
2
xel = x
x el =
yel = y/2
yel = y
dA = ydx
dA = (a – x)dy
2r
cos θ
3
2r
y el = sin θ
3
1
dA = r 2 dθ
2
x el =
Fig.5. 12
If the area itself is not already known, it may also be computed from these
elements.
The coordinates xel and yel of the centroid of the element of area should be
expressed in terms of the coordinates of a point located on the curve bounding
the area under consideration. Also, the element of the area dA should be
expressed in terms of the coordinates of the point and their differentials. This
has been done in Fig.5.12 for three common types of elements; the pie-shaped
element of part c should be used when the equation of the curve bounding the
area is given in polar coordinates. The appropriate expressions should be
CG-10
substituted in formulae (5.11), and the equation of the curve should be used to
express one of the coordinates in terms of the other. The integration is thus
reduced to a single integration which may be performed according to the usual
rules of calculus.
A similar procedure may be applied to the determination of centroids of a
volume bounded by analytical surfaces.
Examples of this method of problem solution will be given in the course of
tutorials.
5.6 Theorems of Pappus –Guldinus. These theorems deal with surfaces and
bodies of revolution.
A surface of revolution is a surface, which may be generated by rotating a
plane curve about a fixed axis lying in the same plane. For example (Fig.5.13),
the surface of a sphere may be obtained by rotating a semicircular arc ABC
about the diameter AC; the surface of a cone by rotating a straight line AB about
an axis AC; the surface of a torus or ring by rotating the circumference of a
circle about nonintersecting axis.
B
B
C
A
Sphere
A
C
Cone
A
C
Torus
Fig. 5.13
A body of revolution is a body which may be generated by rotating a plane
area about a fixed axis. A solid sphere may be obtained by rotating a semicircular area, a cone by rotating triangular area, and a solid torus by rotating a
full circular area (Fig.5.14).
Sphere
Cone
CG-11
Torus
Fig. 5.14
Theorem I. The area of a surface of revolution is equal to the length of the
generating curve times the distance traveled by the centroid of the curve while
the surface is being generated, i.e.
A = 2πyCL
(5.12)
Theorem II. The volume of a body of revolution is equal to the generating
area times the distance traveled by the centroid of the area while the body is
being generated, i.e.
V = 2πyCA
(5.13)
Fig.5.15 shows most important relation between curve element dL and a surface
of revolution, while Fig.5.16 shows relation between area element dA and a
volume generated by this element. For proofs of both theorems please consult
Beer & Johnston p. 184.
The theorems of Pappus-Guldin offer a simple way for computing the area of
surfaces of revolution. They may also be used conversely to determine the
centroid of a plane curve when the area of the surface generated by the curve is
known or to determine the centroid of a plane area when the volume of the body
generated by the area is known.
5.7 Distributed loads on beams. The concept of centroid of an area may be
used to solve other problems besides those dealing with the weight of flat plates.
Consider, for example, a beam supporting a distributed load; this load may
consist of the weight of materials supported directly or indirectly by the beam,
CG-12
or it may be caused by wind or hydrostatic pressure. The distributed load may be
represented by plotting the load w supported per unit length (Fig.5.17); this load
will be expressed in N/m. The magnitude of the force exerted on an element of
beam of length dx is dW = wdx, and the total load supported by the beam is
a)
b)
dw
w
W
w
W=A
dw = dA
xC
w
x
0
C
x
0
P
dx
xP
x
L
L
L
W = ∫ wdx
(5.14)
0
Fig. 5.17
But the product wdx is equal in magnitude to the element of area dA shown in
Fig.5.17a, and W is thus equal in magnitude to the total area A under the load
curve,
W = ∫ dA = A
(5.15)
We shall now determine where a single concentrated load W, of the same
magnitude W as the total distributed load, should be applied on the beam if it is
to produce the same reactions at supports (Fig.5.17b). This concentrated load W,
which represents the resultant of the given distributed loading, should be
CG-13
equivalent to this loading as far as the free-body diagram of the entire beam is
concerned. The point of application P of the equivalent concentrated load W
will therefore be obtained by expressing that the moment of W about point O is
equal to the sum of the moments of the elementary loads dW about O :
x P W = ∫ xdW
(5.16)
or, since dW = wdx = dA and W = A
L
x P A = ∫ xdA
(5.17)
0
Since the integral represents the first moment with respect to the w axis of the
area under the load curve, it may be replaced by the product xC A. We have
therefore xP = xC , where xC is the distance from the w axis to the centroid C of
the area A (i.e. area representing the distributed load).
A distributed load on beam may be replaced by a concentrated load; the
magnitude of this single load is equal to the area under the load curve, and its
line of action passes through the centroid of that area. It should be noted,
however, that the concentrated load is equivalent to the given loading only as far
as external forces are concerned. It may be used to determine reactions but
should not be used to compute internal forces and deflections.
Bibliography
Beer F.P, Johnston E.R., Jr., Vector Mechanics for Engineers, McGraw-Hill
CG-14