Engineering Mechanics: Statics

1
Engineering Mechanics Statics

• Chapter 9
• Center of Gravity and Centroid

2
Chapter Objectives

• To discuss the concept of the center of gravity,
  center of mass, and the centroid.
• To show how to determine the location of the
  center of gravity and centroid for a system of
  discrete particles and a body of arbitrary shape.
• To use the theorems of Pappus and Guldinus for
  finding the area and volume for a surface of
  revolution.
• To present a method for finding the resultant of
  a general distributed loading and show how it
  applies to finding the resultant of a fluid.

3
Chapter Outline

• Center of Gravity and Center of Mass for a System
  of Particles
• Center of Gravity and Center of Mass and Centroid
  for a Body
• Composite Bodies
• Theorems of Pappus and Guldinus
• Resultant of a General Distributed Loading
• Fluid Pressure

4
9.1 Center of Gravity and Center of Mass for a
System of Particles

• Center of Gravity
• Locates the resultant weight of a system of
  particles
• Consider system of n particles fixed within a
  region of space
• The weights of the particles
• comprise a system of parallel
• forces which can be replaced
• by a single (equivalent) resultant
• weight having defined point G
• of application

5
9.1 Center of Gravity and Center of Mass for a
System of Particles

• Center of Gravity
• Resultant weight total weight of n particles
• Sum of moments of weights of all the particles
  about x, y, z axes moment of resultant weight
  about these axes
• Summing moments about the x axis,
• Summing moments about y axis,

6
9.1 Center of Gravity and Center of Mass for a
System of Particles

• Center of Gravity
• Although the weights do not produce a moment
  about z axis, by rotating the coordinate system
  90 about x or y axis with the particles fixed in
  it and summing moments about the x axis,
• Generally,

7
9.1 Center of Gravity and Center of Mass for a
System of Particles

• Center of Gravity
• Where represent the coordinates of the
  center of gravity G of the system of particles,
• represent the coordinates of each particle in
  the system and represent the resultant sum
  of the weights of all the particles in the
  system.
• These equations represent a balance between the
  sum of the moments of the weights of each
  particle and the moment of resultant weight for
  the system.

8
9.1 Center of Gravity and Center of Mass for a
System of Particles

• Center Mass
• Provided acceleration due to gravity g for every
  particle is constant, then W mg
• By comparison, the location of the center of
  gravity coincides with that of center of mass
• Particles have weight only when under the
  influence of gravitational attraction, whereas
  center of mass is independent of gravity

9
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Center Mass
• A rigid body is composed of an infinite number of
  particles
• Consider arbitrary particle
• having a weight of dW

10
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Center Mass
• ? represents the specific weight and dW ?dV

11
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Center of Mass
• Density ?, or mass per unit volume, is related to
  ? by ? ?g, where g acceleration due to
  gravity
• Substitute this relationship into this equation
  to determine the bodys center of mass

12
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Centroid
• Defines the geometric center of object
• Its location can be determined from equations
  used to determine the bodys center of gravity or
  center of mass
• If the material composing a body is uniform or
  homogenous, the density or specific weight will
  be constant throughout the body
• The following formulas are independent of the
  bodys weight and depend on the bodys geometry

13
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Centroid
• Volume
• Consider an object subdivided into volume
  elements dV, for location of the centroid,

14
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Centroid
• Area
• For centroid for surface area of an object, such
  as plate and shell, subdivide the area into
  differential elements dA

15
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Centroid
• Line
• If the geometry of the object such as a thin rod
  or wire, takes the form of a line, the balance of
  moments of differential elements dL about each of
  the coordinate system yields

16
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Line
• Choose a coordinate system that simplifies as
  much as possible the equation used to describe
  the objects boundary
• Example
• Polar coordinates are appropriate for area with
  circular boundaries

17
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Symmetry
• The centroids of some shapes may be partially or
  completely specified by using conditions of
  symmetry
• In cases where the shape has an axis of symmetry,
  the centroid of the shape must lie along the line
• Example
• Centroid C must lie along the
• y axis since for every element
• length dL, it lies in the middle

18
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Symmetry
• For total moment of all the elements about the
  axis of symmetry will therefore be cancelled
• In cases where a shape has 2 or 3 axes of
  symmetry, the centroid lies at the intersection
  of these axes

19
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Procedure for Analysis
• Differential Element
• Select an appropriate coordinate system, specify
  the coordinate axes, and choose an differential
  element for integration
• For lines, the element dL is represented as a
  differential line segment
• For areas, the element dA is generally a
  rectangular having a finite length and
  differential width

20
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Procedure for Analysis
• Differential Element
• For volumes, the element dV is either a circular
  disk having a finite radius and differential
  thickness, or a shell having a finite length and
  radius and a differential thickness
• Locate the element at an arbitrary point (x, y,
  z) on the curve that defines the shape

21
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Procedure for Analysis
• Size and Moment Arms
• Express the length dL, area dA or volume dV of
  the element in terms of the curve used to define
  the geometric shape
• Determine the coordinates or moment arms for the
  centroid of the center of gravity of the element

22
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Procedure for Analysis
• Integrations
• Substitute the formations and dL, dA and dV into
  the appropriate equations and perform
  integrations
• Express the function in the integrand and in
  terms of the same variable as the differential
  thickness of the element
• The limits of integrals are defined from the two
  extreme locations of the elements differential
  thickness so that entire area is covered during
  integration

23
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Example 9.1
• Locate the centroid of
• the rod bent into the
• shape of a parabolic arc.

24
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Differential element
• Located on the curve at the arbitrary point (x,
  y)
• Area and Moment Arms
• For differential length of the element dL
• Since x y2 and then dx/dy 2y
• The centroid is located at

25
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

26
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Example 9.2
• Locate the centroid of
• the circular wire
• segment.

27
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Differential element
• A differential circular arc is selected
• This element intersects the curve at (R, ?)
• Length and Moment Arms
• For differential length of the element
• For centroid,

28
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

29
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Example 9.3
• Determine the distance
• from the x axis to the
• centroid of the area
• of the triangle

30
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Differential element
• Consider a rectangular element having thickness
  dy which intersects the boundary at (x, y)
• Length and Moment Arms
• For area of the element
• Centroid is located y distance from the x axis

31
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

32
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Example 9.4
• Locate the centroid for
• the area of a quarter
• circle.

33
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Method 1
• Differential element
• Use polar coordinates for circular boundary
• Triangular element intersects at point (R,?)
• Length and Moment Arms
• For area of the element
• Centroid is located y distance from the x axis

34
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

35
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Method 2
• Differential element
• Circular arc element having thickness of dr
• Element intersects the
• axes at point (r,0) and
• (r, p/2)

36
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Area and Moment Arms
• For area of the element
• Centroid is located y distance from the x axis

37
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

38
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Example 9.5
• Locate the centroid of
• the area.

39
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Method 1
• Differential element
• Differential element of thickness dx
• Element intersects curve at point (x, y), height
  y
• Area and Moment Arms
• For area of the element
• For centroid

40
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

41
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Method 2
• Differential element
• Differential element of thickness dy
• Element intersects curve at point (x, y)
• Length (1 x)

42
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Area and Moment Arms
• For area of the element
• Centroid is located y distance from the x axis

43
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

44
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Example 9.6
• Locate the centroid of
• the shaded are bounded
• by the two curves
• y x
• and y x2.

45
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Method 1
• Differential element
• Differential element of thickness dx
• Intersects curve at point (x1, y1) and (x2, y2),
  height y
• Area and Moment Arms
• For area of the element
• For centroid

46
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

47
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Method 2
• Differential element
• Differential element of thickness dy
• Element intersects curve at point (x1, y1) and
  (x2, y2)
• Length (x1 x2)

48
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Area and Moment Arms
• For area of the element
• Centroid is located y distance from the x axis

49
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

50
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Example 9.7
• Locate the centroid for the
• paraboloid of revolution,
• which is generated by
• revolving the shaded area
• about the y axis.

51
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Method 1
• Differential element
• Element in the shape of a thin disk, thickness
  dy, radius z
• dA is always perpendicular to the axis of
  revolution
• Intersects at point (0, y, z)
• Area and Moment Arms
• For volume of the element

52
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• For centroid
• Integrations

53
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Method 2
• Differential element
• Volume element in the form of thin cylindrical
  shell, thickness of dz
• dA is taken parallel to the axis of revolution
• Element intersects the
• axes at point (0, y, z) and
• radius r z

54
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Area and Moment Arms
• For area of the element
• Centroid is located y distance from the x axis

55
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Integrations

56
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Example 9.8
• Determine the location of
• the center of mass of the
• cylinder if its density
• varies directly with its
• distance from the base
• ? 200z kg/m3.

57
9.2 Center of Gravity and Center of Mass and
Centroid for a Body
View Free Body Diagram

• Solution
• For reasons of material symmetry
• Differential element
• Disk element of radius 0.5m and thickness dz
  since density is constant for given value of z
• Located along z axis at point (0, 0, z)
• Area and Moment Arms
• For volume of the element

58
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• For centroid
• Integrations

59
9.2 Center of Gravity and Center of Mass and
Centroid for a Body

• Solution
• Not possible to use a shell element for
  integration since the density of the material
  composing the shell would vary along the shells
  height and hence the location of the element
  cannot be specified

60
9.3 Composite Bodies

• Consists of a series of connected simpler
  shaped bodies, which may be rectangular,
  triangular or semicircular
• A body can be sectioned or divided into its
  composite parts
• Provided the weight and location of the center of
  gravity of each of these parts are known, the
  need for integration to determine the center of
  gravity for the entire body can be neglected

61
9.3 Composite Bodies

• Accounting for finite number of weights
• Where
• represent the coordinates of the center of
  gravity G of the composite body
• represent the coordinates of the center of
  gravity at each composite part of the body
• represent the sum of the weights of all the
  composite parts of the body or total weight

62
9.3 Composite Bodies

• When the body has a constant density or specified
  weight, the center of gravity coincides with the
  centroid of the body
• The centroid for composite lines, areas, and
  volumes can be found using the equation
• However, the Ws are replaced by Ls, As and
  Vs respectively

63
9.3 Composite Bodies

• Procedure for Analysis
• Composite Parts
• Using a sketch, divide the body or object into a
  finite number of composite parts that have
  simpler shapes
• If a composite part has a hole, or a geometric
  region having no material, consider it without
  the hole and treat the hole as an additional
  composite part having negative weight or size

64
9.3 Composite Bodies

• Procedure for Analysis
• Moment Arms
• Establish the coordinate axes on the sketch and
  determine the coordinates of the center of
  gravity or centroid of each part

65
9.3 Composite Bodies

• Procedure for Analysis
• Summations
• Determine the coordinates of the center of
  gravity by applying the center of gravity
  equations
• If an object is symmetrical about an axis, the
  centroid of the objects lies on the axis

66
9.3 Composite Bodies

• Example 9.9
• Locate the centroid of the wire.

67
9.3 Composite Bodies

• Solution
• Composite Parts
• Moment Arms
• Location of the centroid for each piece is
  determined and indicated in the diagram

68
9.3 Composite Bodies

• Solution
• Summations

Segment Segment L (mm) x (mm) y (mm) z (mm) xL (mm2) yL (mm2) zL (mm2)
1 1 188.5 60 -38.2 0 11 310 -7200 0
2 2 40 0 20 0 0 800 0
3 3 20 0 40 -10 0 800 -200
Sum 248.5 248.5 11 310 -5600 -200
69
9.3 Composite Bodies

• Solution
• Summations

70
9.3 Composite Bodies

• Example 9.10
• Locate the centroid of the plate area.

71
9.3 Composite Bodies

• Solution
• Composite Parts
• Plate divided into 3 segments
• Area of small rectangle considered negative

72
9.3 Composite Bodies

• Solution
• Moment Arm
• Location of the centroid for each piece is
  determined and indicated in the diagram

73
9.3 Composite Bodies

• Solution
• Summations

Segment Segment A (mm2) x (mm) y (mm) xA (mm3) yA (mm3)
1 1 4.5 1 1 4.5 4.5
2 2 9 -1.5 1.5 -13.5 13.5
3 3 -2 -2.5 2 5 -4
Sum 11.5 11.5 -4 14
74
9.3 Composite Bodies

• Solution
• Summations

75
9.3 Composite Bodies

• Example 9.11
• Locate the center of mass of the
• composite assembly. The conical
• frustum has a density of
• ?c 8Mg/m3 and the hemisphere
• has a density of ?h 4Mg/m3.
• There is a 25mm radius
• cylindrical hole in the center.

76
9.3 Composite Bodies
View Free Body Diagram

• Solution
• Composite Parts
• Assembly divided into 4 segments
• Area of 3 and 4 considered negative

77
9.3 Composite Bodies

• Solution
• Moment Arm
• Location of the centroid for each piece is
  determined and indicated in the diagram
• Summations
• Because of symmetry,

78
9.3 Composite Bodies

• Solution
• Summations

Segment Segment m (kg) z (mm) zm (kg.mm)
1 1 4.189 50 209.440
2 2 1.047 -18.75 -19.635
3 3 -0.524 125 -65.450
4 4 -1.571 50 -78.540
Sum 3.141 3.141 45.815
79
9.3 Composite Bodies

• Solution
• Summations

80
9.4 Theorems of Pappus and Guldinus

• A surface area of revolution is generated by
  revolving a plane curve about a non-intersecting
  fixed axis in the plane of the curve
• A volume of revolution is generated by revolving
  a plane area bout a nonintersecting fixed axis in
  the plane of area
• Example
• Line AB is rotated about
• fixed axis, it generates
• the surface area of a
• cone (less area of base)

81
9.4 Theorems of Pappus and Guldinus

• Example
• Triangular area ABC rotated
• about the axis would
• generate the volume of
• the cone
• The theorems of Pappus and Guldinus are used to
  find the surfaces area and volume of any object
  of revolution provided the generating curves and
  areas do not cross the axis they are rotated

82
9.4 Theorems of Pappus and Guldinus

• Surface Area
• Area of a surface of revolution product of
  length of the curve and distance traveled by the
  centroid in generating the surface area

83
9.4 Theorems of Pappus and Guldinus

• Volume
• Volume of a body of revolution product of
  generating area and distance traveled by the
  centroid in generating the volume

84
9.4 Theorems of Pappus and Guldinus

• Composite Shapes
• The above two mentioned theorems can be applied
  to lines or areas that may be composed of a
  series of composite parts
• Total surface area or volume generated is the
  addition of the surface areas or volumes
  generated by each of the composite parts

85
9.4 Theorems of Pappus and Guldinus

• Example 9.12
• Show that the surface area of a sphere is
• A 4pR2
• and its volume
• V 4/3 pR3

86
9.4 Theorems of Pappus and Guldinus
View Free Body Diagram

• Solution
• Surface Area
• Generated by rotating semi-arc about the x axis
• For centroid,
• For surface area,

87
9.4 Theorems of Pappus and Guldinus

• Solution
• Volume
• Generated by rotating semicircular area about the
  x axis
• For centroid,
• For volume,

88
9.5 Resultant of a General Distributed Loading

• Pressure Distribution over a Surface
• Consider the flat plate subjected to the loading
  function ? ?(x, y) Pa
• Determine the force dF acting on the differential
  area dA m2 of the plate, located at the
  differential point (x, y)
• dF ?(x, y) N/m2(d A m2)
• ?(x, y) d AN
• Entire loading represented as
• infinite parallel forces acting on
• separate differential area dA

89
9.5 Resultant of a General Distributed Loading

• Pressure Distribution over a Surface
• This system will be simplified to a single
  resultant force FR acting through a unique point
  on the plate

90
9.5 Resultant of a General Distributed Loading

• Pressure Distribution over a Surface
• Magnitude of Resultant Force
• To determine magnitude of FR, sum the
  differential forces dF acting over the plates
  entire surface area dA
• Magnitude of resultant
• force total volume under
• the distributed loading
• diagram

91
9.5 Resultant of a General Distributed Loading

• Pressure Distribution over a Surface
• Location of Resultant Force
• Line of action of action of the resultant force
  passes through the geometric center or centroid
  of the volume under the distributed loading
  diagram

92
9.6 Fluid Pressure

• According to Pascals law, a fluid at rest
  creates a pressure ? at a point that is the same
  in all directions
• Magnitude of ? measured as a force per unit area,
  depends on the specific weight ? or mass density
  ? of the fluid and the depth z of the point from
  the fluid surface
• ? ?z ?gz
• Valid for incompressible fluids
• Gas are compressible fluids and thus the above
  equation cannot be used

93
9.6 Fluid Pressure

• Consider the submerged plate
• 3 points have been specified

94
9.6 Fluid Pressure

• Since point B is at depth z1 from the liquid
  surface, the pressure at this point has a
  magnitude of ?1 ?z1
• Likewise, points C and D are both at depth z2 and
  hence ?2 ?z2
• In all cases, pressure acts normal to the surface
  area dA located at specified point
• Possible to determine the resultant force caused
  by a fluid distribution and specify its location
  on the surface of a submerged plate

95
9.6 Fluid Pressure

• Flat Plate of Constant Width
• Consider flat rectangular plate of constant width
  submerged in a liquid having a specific weight ?
• Plane of the plate makes an angle with the
  horizontal as shown

96
9.6 Fluid Pressure

• Flat Plate of Constant Width
• Since pressure varies linearly with depth,
• the distribution of pressure over the plates
• surface is represented by a trapezoidal
• volume having an
• intensity of ?1 ?z1
• at depth z1 and
• ?2 ?z2 at depth z2

97
9.6 Fluid Pressure

• Magnitude of the resultant force FR volume of
  this loading diagram and FR has a line of action
  that passes through the volumes centroid, C
• FR does not act at the centroid of the plate but
  at
• point P called the center of
• pressure
• Since plate has a constant
• width, the loading diagram
• can be viewed in 2D

98
9.6 Fluid Pressure

• Flat Plate of Constant Width
• Loading intensity is measured as force/length and
  varies linearly from
• w1 b?1 b?z1 to w 2 b?2 b?z2
• Magnitude of FR trapezoidal area
• FR has a line of action that passes through the
  areas centroid C

99
9.6 Fluid Pressure

• Curved Plate of Constant Width
• When the submerged plate is curved, the pressure
  acting normal to the plate continuously changes
  direction
• For 2D and 3D view of the loading distribution,
• Integration can be used to determine FR and
  location of center of centroid C or pressure P

100
9.6 Fluid Pressure

• Curved Plate of Constant Width
• Example
• Consider distributed loading acting on the curved
  plate DB

101
9.6 Fluid Pressure

• Curved Plate of Constant Width
• Example
• For equivalent loading

102
9.6 Fluid Pressure

• Curved Plate of Constant Width
• The plate supports the weight of the liquid Wf
  contained within the block BDA
• This force has a magnitude of
• Wf (?b)(areaBDA)
• and acts through the centroid of BDA
• Pressure distributions caused by the liquid
  acting along the vertical and horizontal sides of
  the block
• Along vertical side AD, force FADs magnitude
  area under trapezoid and acts through centroid
  CAD of this area

103
9.6 Fluid Pressure

• Curved Plate of Constant Width
• The distributed loading along horizontal side AB
  is constant since all points lying on this plane
  are at the same depth from the surface of the
  liquid
• Magnitude of FAB is simply the area of the
  rectangle
• This force acts through the area centroid CAB or
  the midpoint of AB
• Summing three forces,
• FR ?F FAB FAD Wf

104
9.6 Fluid Pressure

• Curved Plate of Constant Width
• Location of the center of pressure on the plate
  is determined by applying
• MRo ?MO
• which states that the moment of the resultant
  force about a convenient reference point O, such
  as D or B sum of the moments of the 3 forces
  about the same point

105
9.6 Fluid Pressure

• Flat Plate of Variable Width
• Consider the pressure distribution acting on the
  surface of a submerged plate having a variable
  width

106
9.6 Fluid Pressure

• Flat Plate of Variable Width
• Resultant force of this loading volume
  described by the plate area as its base and
  linearly varying pressure distribution as its
  altitude
• The shaded element may be used if integration is
  chosen to determine the volume
• Element consists of a rectangular strip of area
  dA x dy located at depth z below the liquid
  surface
• Since uniform pressure ? ?z (force/area) acts
  on dA, the magnitude of the differential force dF
• dF dV ? dA ?z(xdy)

107
9.6 Fluid Pressure

• Flat Plate of Variable Width
• Centroid V defines the point which FR acts
• The center of pressure which lies on the surface
  of the plate just below C has the coordinates P
  defined by the equations
• This point should not be mistaken for centroid of
  the plates area

108
9.6 Fluid Pressure

• Example 9.13
• Determine the magnitude and location of the
• resultant hydrostatic force acting on the
  submerged
• rectangular plate AB. The
• plate has a width of 1.5m
• ?w 1000kg/m3.

109
9.6 Fluid Pressure

• Solution
• The water pressures at depth A and B are
• Since the plate has constant
• width, distributed loading
• can be viewed in 2D
• For intensities of the load at
• A and B,

110
9.6 Fluid Pressure

• Solution
• For magnitude of the resultant force FR created
  by the distributed load
• This force acts through the
• centroid of the area
• measured upwards from B

111
9.6 Fluid Pressure

• Solution
• Same results can be obtained by considering two
  components of FR defined by the triangle and
  rectangle
• Each force acts through its associated centroid
  and has a magnitude of
• Hence

112
9.6 Fluid Pressure

• Solution
• Location of FR is determined by summing moments
  about B

113
9.6 Fluid Pressure

• Example 9.14
• Determine the magnitude of the resultant
• hydrostatic force acting on the surface of a
  seawall
• shaped in the form of a parabola. The wall is 5m
• long and ?w 1020kg/m2.

114
9.6 Fluid Pressure

• Solution
• The horizontal and vertical components of the
  resultant force will be calculated since
• Then
• Thus

115
9.6 Fluid Pressure

• Solution
• Area of the parabolic sector ABC can be
  determined
• For weight of the wafer within this region
• For resultant force

116
9.6 Fluid Pressure

• Example 9.15
• Determine the magnitude and location of
• the resultant force acting on the triangular
• end plates of the wafer of the water trough.
• ?w 1000 kg/m3

117
9.6 Fluid Pressure
View Free Body Diagram

• Solution
• Magnitude of the resultant force F volume of
  the loading distribution
• Choosing the differential volume element,
• For equation of line AB
• Integrating

118
9.6 Fluid Pressure

• Solution
• Resultant passes through the centroid of the
  volume
• Because of symmetry
• For volume element

119
Chapter Summary

• Center of Gravity and Centroid
• Center of gravity represents a point where the
  weight of the body can be considered concentrated
• The distance to this point can be determined by a
  balance of moments
• Moment of weight of all the particles of the body
  about some point moment of the entire body
  about the point
• Centroid is the location of the geometric center
  of the body

120
Chapter Summary

• Center of Gravity and Centroid
• Centroid is determined by the moment balance of
  geometric elements such as line, area and volume
  segments
• For body having a continuous shape, moments are
  summed using differential elements
• For composite of several shapes, each having a
  known location for centroid, the location is
  determined from discrete summation using its
  composite parts

121
Chapter Summary

• Theorems of Pappus and Guldinus
• Used to determine surface area and volume of a
  body of revolution
• Surface area product of length of the
  generating curve and distance traveled by the
  centroid of the curve to generate the area
• Volume product of the generating area and the
  distance traveled by the centroid to generate the
  volume

122
Chapter Summary

• Fluid Pressure
• Pressure developed by a fluid at a point on a
  submerged surface depends on the depth of the
  point and the density of the liquid according to
  Pascals law
• Pressure will create a linear distribution of
  loading on a flat vertical or inclined surface
• For horizontal surface, loading is uniform
• Resultants determined by volume or area under the
  loading curve

123
Chapter Summary

• Fluid Pressure
• Line of action of the resultant force passes
  through the centroid of the loading diagram

124
Chapter Review
125
Chapter Review
126
Chapter Review
127
Chapter Review
128
Chapter Review
129
Chapter Review