Technical University of Sofia
Branch Plovdiv
Theoretical Mechanics
STATICS
KINEMATICS
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Lecture 7
Distributed Loads
Sometimes, a body may be subjected to a loading that is distributed over its surface. For
example, the pressure of the wind on the face of a sign, the pressure of water within a tank, or
the weight of sand on the floor of a storage container, are all distributed loadings.
Lecture 7
Distributed Loads
Line Distribution.
When a force is distributed along a line, as in the continuous vertical load supported by a
suspended cable, the intensity w of the loading is expressed as force per unit length of line,
newtons per meter (N/m) or pounds per foot (Ib/ft).
Area Distribution.
When a force is distributed over an area, as with the hydraulic pressure of water against the inner
face of a section of dam, the intensity is expressed as force per unit area. This intensity is called
pressure for the action of fluid forces and stress for the internal distribution of forces in solids. The
basic unit for pressure or stress in SI is the newton per square meter (N/m2), which is also called
the pascal (Pa).
Lecture 7
Distributed Loads
Volume Distribution.
A force which is distributed over the volume of a body is called a body force. The most
common body force is the force of gravitational attraction, which acts on all elements of mass
in a body. The determination of the forces on the supports of the heavy cantilevered structure
would require accounting for the distribution of gravitational force throughout the structure. The
intensity of gravitational force is the specific weight rg, where r is the density (mass per unit
volume) and g is the acceleration due to gravity. The units for rg are (kg/m3)(m/s2) = N/m3 in SI
units.
Lecture 7
Centers of mass and centroids
Consider a three-dimensional body of any size and shape, having a mass m. If we suspend
the body from any point such as A, the body will be in equilibrium under the action of the
tension in the cord and the resultant W of the gravitational forces acting on all particles of the
body. This resultant is clearly collinear with the cord. Assume that we mark its position by
drilling a hypothetical hole of negligible size along its line of action. We repeat the experiment
by suspending the body from other points such as B and C, and in each instance we mark the
line of action of the resultant force. For all practical purposes these lines of action will be
concurrent at a single point G, which is called the center of gravity of the body.
Lecture 7
Centroids of common shapes of areas and lines
Lecture 7
Centroids of common shapes of areas and lines
Lecture 7
Centroids of common shapes of areas and lines
Lecture 7
Centroids of common volumes
Lecture 7
Centroids of common volumes
Lecture 7
Centroids of common volumes
Lecture 7
Centroids of common volumes
Lecture 7
Centroids of common volumes
Lecture 7
Composite plates and bodies
In many instances, a flat plate can be divided into rectangles, triangles, or the other common
shapes. The coordinates X and Y of its center of gravity G can be determined from the abscissas
x1, x2, . . . , xn and y1, y2, . . . , yn of the centers of gravity of the various parts by expressing that
the moment of the weight of the whole plate about the x and y axis is equal to the sum of the
moments of the weights of the various parts about the same axis.
XW  x1W1  x2W2  ...  xnWn
YW  y1W1  y2W2  ...  ynWn
x1W1  x2W2  ...  xnWn
W
y W  y2W2  ...  ynWn
Y  1 1
W
X
If the plate has a constant thickness t , density r and area A, its weight is W=rgtA. Then the
coordinate of G are expressed by the areas A1 , A2 ,…, An of the different components.
Lecture 7
Composite plates and bodies
Sample problem 1
For the plane area shown, determine the location of the centroid.
Solution
Components of Area.
The area is obtained by adding a rectangle, a triangle, and a semicircle and by then subtracting
a circle. Using the coordinate axes shown, the area and the coordinates of the centroid of each of
the component areas are determined and entered in the table below.
Lecture 7
Sample problem 1
The area of the circle is indicated as
negative, since it is to be subtracted from
the other areas. We note that the
coordinate y of the centroid of the triangle
is negative for the axes shown.
Lecture 7
Sample problem 1
Component
A, mm2
x, mm
y, mm
xA, mm3
yA, mm3
Rectangle
9.6x103
60
40
576x103
384x103
Triangle
3.6x103
40
-20
144x103
-72x103
Semicircle
5.655x103
60
105.46
339.3x103
596.4x103
Circle
-5.027x103
60
80
-301.6x103
-402.2x103
∑xA=757.7x103
∑yA=506.2x103
∑A=13.828x103
Location of Centroid.
Substituting the values given in the table into the equations defining the centroid of a composite
area, we obtain
X 13.828 103   757.7 103
X  54.8 mm
Y 13.828  103   506.2  103
Y  36.6 mm
Lecture 7
Distributed Loads on Beams
Consider a beam supporting a distributed load. This load may consist of the weight of materials
supported directly or indirectly by the beam, or it may be caused by wind or hydrostatic pressure.
The distributed load can be represented by plotting the load w supported per unit length. This
load is expressed in N/m. The magnitude of the force exerted on an element of beam of length dx
is dW = w dx, and the total load supported by the beam is
L
W   w dx
0
We observe that the product w dx is equal in magnitude to the
element of area dA. The load W is thus equal in magnitude to
the total area A under the load curve.
The point of application P of the equivalent concentrated load
W is obtained by expressing that the moment of W about point
O is equal to the sum of the moments of the elemental loads
dW about O:
L
L
xW   xw dx
0

x
 xw dx
0
W
Lecture 7
Distributed Loads on Beams
Three most common cases and the resultants of the distributed loads in each case.
Lecture 7
Sample problem 2
Solution
• The magnitude of the concentrated load is
equal to the total load or the area under the
curve.
• The line of action of the concentrated load
passes through the centroid of the area
under the curve.
A beam supports a distributed load as
shown. Determine the equivalent
concentrated load and the reactions at the
supports.
• Determine the support reactions by
summing moments about the beam ends.
Lecture 7
Sample problem 2
• The magnitude of the concentrated load is equal to the
total load or the area under the curve.
F  18.0 kN
• The line of action of the concentrated load passes through
the centroid of the area under the curve.
X 
63 kN  m
18 kN
X  3.5 m
Lecture 7
Sample problem 2
• Determine the support reactions by summing moments
about the beam ends.
 M A  0 : B y 6 m  18 kN 3.5 m  0
B y  10.5 kN
 M B  0 :  Ay 6 m  18 kN 6 m  3.5 m  0
Ay  7.5 kN