10'5 Loads Distributed Uniformly Along Cables

1
10.5 Loads Distributed Uniformly Along Cables

• A cables own weight subjects it to a load that
  is distributed uniformly along its length
• If a cable is subjected to equal, parallel forces
  spaced uniformly along its length, the load on
  the cable can often be modeled as a load
  distributed uniformly along its length
• Suppose that a cable is acted on by a distributed
  load that subjects each element ds of its length
  to a force w ds, where w is constant
• The free-body diagram is obtained by cutting the
  cable at its lowest point at a point a distance
  s along its length

2
10.5 Loads Distributed Uniformly Along Cables

• The terms T0 T are the tensions at the lowest
  point at s respectively
• The distributed load exerts a downward force ws
• The origin of the coordinate system is located at
  the lowest point of the cable
• Let the function y(x) be the
  curve described by the cable
  in the x-y plane

3
10.5 Loads Distributed Uniformly Along Cables

• Shape of the Cable
• From the free-body diagram, we obtain the
  equilibrium equations
• T sin ? ws
  (10.13)
• T cos ? T0
  (10.14)
• Dividing the Eq. (10.13) by Eq. (10.14), we
  obtain
• (10.15)
• where
• (10.16)

4
10.5 Loads Distributed Uniformly Along Cables

• The slope of the cable dy/dx tan ?, so Eq.
  (10.15) can be written as
• The derivative of this equation with respect to x
  is
• (10.17)

5
10.5 Loads Distributed Uniformly Along Cables

• By using the relation ds2 dx2 dy2, we can
  write the derivative of s with respect to x as
• (10.18)
• where is the slope.
• Now, with Eq. (10.18), we can write Eq. (10.17)
  as
• The slope s 0 at x 0

6
10.5 Loads Distributed Uniformly Along Cables

• Integrating this equation yields
• and we obtain the slope as a function of x
• (10.19)
• Then integrating this equation with respect to x
  yields the curve described by the cable, which is
  called a catenary
• (10.20)

7
10.5 Loads Distributed Uniformly Along Cables

• Tension of the Cable
• Using Eq. (10.14) the relation dx cos ? ds,
  we obtain
• Substituting Eq. (10.18) into this expression
  using Eq. (10.19) yields the tension in the cable
  as a function of x
• (10.21)

8
10.5 Loads Distributed Uniformly Along Cables

• Length of the Cable
• From Eq. (10.15), the length s of the cable from
  the origin to the point at which the angle
  between the cable the x axis equals ? is
• Substituting Eq. (10.19) into this equation, we
  obtain an expression for the length s of the
  cable in the horizontal interval from its lowest
  point to x
• (10.22)

9
Example 10.8 Cable Loaded by its Own Weight

• The mass per unit length of the cable in Fig.
  10.26 is 1 kg/m. The tension at its lowest point
  is 50 N. Determine the distance h the maximum
  tension in the cable.

10
Example 10.8 Cable Loaded by its Own Weight

• Strategy
• The cable is subjected to a load w (9.81
  m/s2) (1 kg/m) 9.81 N/m distributed
  uniformly along its length. Since we know w T0,
  we can determine a w/T0. Then we can determine
  h from Eq. (10.20). Because the maximum tension
  occurs at the greatest distance from the lowest
  point of the cable, we can determine it by
  letting x 10 m in Eq. (10.21).

11
Example 10.8 Cable Loaded by its Own Weight

• Solution
• The parameter a is
• In terms of a coordinate system with its
  origin at the lowest point of the cable, the
  coordinates of the right attachment point are x
  10 m, y h.

12
Example 10.8 Cable Loaded by its Own Weight

• Solution
• From Eq. (10.20),
• From Eq. (10.21), the maximum tension is

13
Example 10.8 Cable Loaded by its Own Weight

• Critical Thinking
• In this example we specified the mass per unit
  length of the cable the tension T0 at the
  cables lowest point used them to determine the
  distance h in Fig. 10.26
• In a real application, you would be much more
  likely to know the mass per unit length of the
  cable the distance h need to determine the
  tension T0
• If h is given, the x y coordinates of the
  cables highest points relative to the lowest
  point are known

14
Example 10.8 Cable Loaded by its Own Weight

• Critical Thinking
• Substituting the coordinates of 1 of these points
  into Eq. (10.20) yields an equation for the
  coefficient a w/T0, from which T0 could be
  determined
• However, notice that the transcendental Eq.
  (10.20) would have to be solved numerically for
  the value of a

15
10.6 Discrete Loads

• Consider the case of an arbitrary number N of
  objects suspended from a cable
• Assume that the weight of the cable can be
  neglected in comparison to the suspended weights
  that the cable is sufficiently flexible that we
  can approximate its shape by a series of straight
  segments

16
10.6 Discrete Loads

• Determining the Configuration Tensions
• Suppose that the horizontal distances b1, b2,,
  bN1 are known that the vertical distance hN1
  specifying the cables right attachment point is
  known
• Determine the configuration (shape) of the cable
  by solving for the vertical distances h1, h2,,
  hN specifying the positions of the attachment
  points
• Determine the tensions in the segments 1, 2,,
  N1 of the cable

17
10.6 Discrete Loads

• Begin by drawing a free-body diagram, cutting the
  cable at its left attachment point just to the
  right of the weight W1
• We resolve the tension in the cable at the left
  attachment point into its horizontal vertical
  components Th Tv

18
10.6 Discrete Loads

• Summing moments about the attachment point A1, we
  obtain the equation
• S Mpoint A1 h1Th ? b1Tv 0
• The next step is to obtain a free-body diagram by
  cutting the cable at its left attachment point
  just to the right of the weight W2

19
10.6 Discrete Loads

• Summing moments about A2, we obtain
• S Mpoint A2 h2Th ? (b1 b2)Tv b2W1 0
• Proceeding in this way, cutting the cable just to
  the right of each N weights, we obtain N
  equations
• We can also draw a free-body diagram by cutting
  the cable at its left right attachment points
  sum moments about the right attachment point
• In this way, we obtain N1 equations in terms of
  N2 unknowns the components of the tension Th
  Tv the vertical positions of the attachment
  points h1, h2,, hN

20
10.6 Discrete Loads

• If the vertical position of just 1 attachment
  point is also specified, we can solve the system
  of equations for the vertical positions of the
  other attachment points, determining the
  configuration of the cable
• Once we know the configuration of the cable the
  force Th, we can determine the tension in any
  segment by cutting the cable at the left
  attachment point within the segment summing
  forces in the horizontal direction

21
10.6 Discrete Loads

• Comments on Continuous Discrete Models
• Consider a cable subjected to a horizontally
  distributed load w
• The total force exerted on it is wL
• Since the cable passes through
  the point x L/2, y
  L/2, we find from Eq.
  (10.10) that a 4/L, so
  the equation for the curve
  described by the cable is y
  (2/L)x2

22
10.6 Discrete Loads

• Compare the shape of the cable with the
  distributed load to that of a cable of negligible
  weight subjected to 3 discrete loads W wL/3
  with equal horizontal spacing (we chose the
  dimensions of the cable with discrete loads so
  that the heights of the2 cables would be equal at
  their midpoints)

23
10.6 Discrete Loads

• Compare the shape of the cable with the
  distributed load to that of a cable of negligible
  weight subjected to 5 discrete loads W wL/5
  with equal horizontal spacing

24
10.6 Discrete Loads

• Compare the tension in the cable subjected to the
  distributed load to those in the cable subjected
  to 3 5 discrete loads

25
10.6 Discrete Loads

• The shape the tension in the cable with a
  distributed load are approximated by the shapes
  tensions in cables with discrete loads
• Although the approximation of the tension is less
  impressive that the approximation of the shape,
  it is clear that the former can be improved by
  increasing the number of discrete loads

26
10.6 Discrete Loads

• This approach, approximating a continuous
  distribution by a discrete model, is very
  important in engineering
• It is the starting point of the finite difference
  finite element methods
• The opposite approach, modeling discrete systems
  by continuous models, is also widely used
• E.g. when the forces exerted on a bridge by
  traffic are modeled as a distributed load

27
Example 10.9 Cable Subjected to Discrete Loads

• 2 masses m1 10 kg m2 20 kg are suspended
• from the cable in Fig. 10.30.
• (a) Determine the vertical distance h2
• (b) Determine the tension in cable segment 2

28
Example 10.9 Cable Subjected to Discrete Loads

• Strategy
• We will obtain 3 free-body diagrams by cutting
  the cable at the left attachment point (1) just
  to the right of the mass m1 (2) just to the
  right of the mass m2 (3) at the right
  attachment point. By writing a moment equation
  for each free-body diagram, we will obtain 3
  equations in terms of the 2 components of the
  tension at the left attachment point unknown
  vertical distance h2. once the geometry of the
  cable is determined, we can use equilibrium to
  determine the tension in segment 2.

29
Example 10.9 Cable Subjected to Discrete Loads

• Solution
• (a) Begin by cutting the cable at the left
  attachment point just to the right of the mass
  m1 resolve the tension at the left attachment
  point into horizontal vertical components

30
Example 10.9 Cable Subjected to Discrete Loads

• Solution
• Summing moments about A1 yields
• S Mpoint A1 (1 m)Th ? (1 m)Tv 0
• We then cut the cable just to the right of the
  mass
• m2 sum moments about A2
• S Mpoint A2 h2Th ? (2 m)Tv (1 m) m1g 0

31
Example 10.9 Cable Subjected to Discrete Loads

• Solution
• The last step is to cut the cable at the right
• attachment point sum moments about A3
• S Mpoint A3 (?3 m)Tv (2 m) m1g (1 m) m2g
  0
• We have 3 equations in terms of the unknowns Th,
• Tv h2. Solving them yields Th Tv 131 N h2
  
• 1.25 m.

32
Example 10.9 Cable Subjected to Discrete Loads

• Solution
• (b) To determine the tension in segment 2, use
  the 1st free-body diagram. The angle between the
  force T2 the horizontal is
• arctan (h2 ? 1)/1 14.0
• Summing forces in the horizontal direction
  gives
• T2cos 14.0 ? Th 0
• Solving, we obtain

33
Example 10.9 Cable Subjected to Discrete Loads

• Critical Thinking
• The systematic solution procedure we applied to
  this cable system with 3 segments resulted in 3
  equations
• In addition to the 2 components of the tension at
  the left attachment point, we were able to
  determine the unknown vertical distance h2
• There was 1 excess equation with which to
  determine h2

34
Example 10.9 Cable Subjected to Discrete Loads

• Critical Thinking
• Instead of the height h2, some other parameter of
  the system could have been left unspecified, such
  as 1 of the masses or the horizontal position of
  1 of the masses
• In a cable system with N segments, N?2 parameters
  can be left unspecified

35
Computational Example 10.10

• As the 1st step in constructing a suspended
  pedestrian bridge, a cable is suspended across
  the span from attachment points of equal height
  (Fig. 10.31). The cable weighs 50 N/m is 42 m
  long. Determine the maximum tension in the cable
  the vertical distance from the attachment
  points to the cables lowest point.

36
Computational Example 10.10

• Strategy
• Eq. (10.22) gives the length s of the cable as
  a function of the horizontal distance x from the
  cables lowest point parameter a w/T0. The
  term w is the weight per unit length T0 is the
  tension in the cable at its lowest point. We know
  that the half-span of the cable is 20 m, so we
  can draw a graph of s as a function of a
  estimate the value of a for which s 21 m. Then
  we can determine the maximum tension from Eq.
  (10.21) the vertical distance to the cables
  lowest point from Eq. (10.20).

37
Computational Example 10.10

• Solution
• Setting x 20 m in Eq. (10.22),
• We compute s as a function of a

38
Computational Example 10.10

• Solution
• The length s 20 m when the parameter a is
  approximately 0.027 m?1.

By examining the computed results near a 0.027
m?1 We see that s is approximately 21 m when a
0.0272 m?1.
39
Computational Example 10.10

• Solution
• Therefore, the tension at the cables lowest
  point is
• and the maximum tension is

40
Computational Example 10.10

• Solution
• From Eq. (10.20), the vertical distance from
  the cables lowest point to the attachment points
  is

41
Computational Example 10.10

• Critical Thinking
• From this example you can see how computational
  results can be used in the design of a suspended
  cable system
• By using a as a parameter, we could determine the
  maximum tension the vertical distance to the
  cables lowest point for cables with a range of
  lengths
• By calculations of this kind, the design engineer
  can choose the properties the cable must have to
  satisfy the criteria of a particular application

42
10.7 Pressure the Center of Pressure

• A surface immersed in a gas or liquid is
  subjected to forces exerted by molecular impacts
• If the gas or liquid is stationary, the load can
  be described by a function p, the pressure,
  defined such that the normal force exerted on a
  differential element dA of the surface is p dA

43
10.7 Pressure the Center of Pressure

• Notice the parallel between the pressure a load
  w distributed along a line, which is defined such
  that the force on a differential element dx of
  the line is w dx
• The dimensions of p are (force)/(area)
• SI units newtons per square meter or pascals
  (Pa)
• In some applications, it is convenient to use the
  gage pressure
• pg p ? patm
  (10.23)
• where patm is the pressure of the atmosphere

44
10.7 Pressure the Center of Pressure

• Atmospheric pressure varies with location
  climatic conditions
• Sea level 1 105 Pa
• If the distributed force due to pressure on a
  surface is represented by an equivalent force,
  the point at which the line of action of force
  intersects the surface is called the center of
  pressure
• Consider a plane area A subjected to a pressure p
  introduce a coordinate system such that the
  area lies in the x-y plane

45
10.7 Pressure the Center of Pressure

• The normal force on each differential element of
  area dA is p dA, so the total normal force on A
  is
• (10.24)

46
10.7 Pressure the Center of Pressure

• Equating the moment of F about the origin to the
  total moment due to the pressure about the origin
  gives
• And using Eq. (10.24), we obtain
• (10.25)

47
10.7 Pressure the Center of Pressure

• These equations determine the position of the
  center of pressure when the pressure p is known
• If p is uniform, the total normal force F pA
  Eqs. (10.25) indicate that the center of pressure
  is the centroid of A
• The term p dA in Eq. (10.24) is equal
  to a differential element of dV of the
  volume between the surface defined
  by the pressure distribution the area
  A

48
10.7 Pressure the Center of Pressure

• The total force exerted by the pressure is
  therefore equal to this volume
• Substituting p dA dV into Eqs. (10.25), we
  obtain

49
10.7 Pressure the Center of Pressure

• The center of pressure coincides with the x y
  coordinates of the centroid of the volume

50
10.8 Pressure in a Stationary Liquid

• The pressure in a liquid at rest increases with
  depth
• Neglecting changes in the density of the liquid,
  we can determine the dependence of the pressure
  on depth by using a simple free-body diagram
• Introducing a coordinate system with its origin
  at the surface of the liquid the positive x
  axis downward, we draw a free-body diagram of a
  cylinder of liquid that extends from the surface
  to a depth x

51
10.8 Pressure in a Stationary Liquid

• The top of the cylinder is subjected to the
  pressure at the surface, which we call p0

52
10.8 Pressure in a Stationary Liquid

• The sides bottom of the cylinder are subjected
  to pressure by the surrounding liquid, which
  increases from p0 at the surfaces to a value p at
  the depth x
• The volume of the cylinder is Ax, where A is its
  cross-sectional area
• Therefore, its weight W ?Ax, where ? is the
  weight density of the liquid (? ?g)
• Since the liquid is stationary, the cylinder is
  in equilibrium

53
10.8 Pressure in a Stationary Liquid

• From the equilibrium equation
• S Fx p0A ? pA ?Ax 0
• We obtain a simple expression for the pressure p
  of the liquid at depth x
• (10.26)
• Thus, the pressure increases linearly with depth
  the derivation we have used illustrates why
  the pressure at a given depth literally holds up
  the liquid above that depth

54
10.8 Pressure in a Stationary Liquid

• If the surface of the liquid is open to the
  atmosphere, p0 patm we write Eq. (10.26) in
  terms of the gage pressure pg p ? patm as
• (10.27)
• In SI units, the density of water at sea level
  conditions is ? 1000 kg/m3, so its weight
  density is approximately ? ?g 9.81 kN/m3

55
10.8 Pressure in a Stationary Liquid

• The force moment due to the pressure on a
  submerged plane area can be determined in 2 ways
• 1.Integration integrate Eq. 910.26) or Eq.
  (10.27).
• 2.Volume analogy determine the total force by
  calculating the volume between the surface
  defined by the pressure distribution the area
  A. The center of pressure coincides with the x
  y coordinates of the centroid of the volume

56
Example 10.11 Pressure Force Center of Pressure

• An engineer making preliminary design studies
  for a canal lock needs to determine the total
  pressure force on a submerged rectangular plate
  (Fig. 10.37) the location of the center of
  pressure. The top of the plate is 6 m below the
  surface. Atmospheric pressure is patm 1 105 Pa
  the weight density of the water is ? 9.81
  kN/m3.

57
Example 10.11 Pressure Force Center of Pressure

• Strategy
• We will determine the pressure force on a
  differential element of area of the plate in the
  form of a horizontal strip integrate to
  determine the total force moment exerted by the
  pressure.

58
Example 10.11 Pressure Force Center of Pressure

• Solution
• In terms of a coordinate system with its
  origin at the surface the positive x axis
  downward, the
  pressure of the water is p patm ?x. The
  horizontal strip dA (8 m) dx.

59
Example 10.11 Pressure Force Center of Pressure

• Solution
• Therefore, the total force exerted on the face
  of the plate by the pressure is

60
Example 10.11 Pressure Force Center of Pressure

• Solution
• The moment about the y axis due to the
  pressure on the plate is

61
Example 10.11 Pressure Force Center of Pressure

• Solution
• The force F acting at the center of pressure
  exerts a moment about the y axis equal to M
• xp F M
• Therefore, the location of the center of
  pressure is

62
Example 10.11 Pressure Force Center of Pressure

• Critical Thinking
• Notice that the center of pressure does not
  coincide with the centroid of the area
• The center of pressure of a plane area generally
  coincides with the centroid of the area only when
  the pressure is uniformly distributed
• In this example, the pressure increases with
  depth as a result, the center of pressure is
  below the centroid

63
Example 10.12 Gate Loaded by a Pressure
Distribution

• The gate AB in Fig. 10.38 has water of 2-m
  depth on the right side. The width of the gate
  (the dimension into the page) is 3 m its weight
  is 1000 N. The weight density of the water is ?
  9.81 kN/m3.Determine the reactions on the gate at
  the supports at A B.

64
Example 10.12 Gate Loaded by a Pressure
Distribution

• Strategy
• The left face of the gate the right face
  above the level of the water are exposed to
  atmospheric pressure. From Eqs. (10.23)
  (10.26), the pressure in the water is the sum of
  atmospheric pressure the gage pressure pg ?x,
  where x is measured downward from the surface of
  the water. The effects of atmospheric pressure
  cancel, so we need to consider only the forces
  moments exerted on the gate by the
  gage pressure. We will determine
  them by integrating also by
  calculating the volume of the
  pressure distribution.

65
Example 10.12 Gate Loaded by a Pressure
Distribution

• Solution
• Integration
• In terms of the differential element
  dA, the force exerted on the gate
  by the gage pressure is
• and the moment about the y axis is

66
Example 10.12 Gate Loaded by a Pressure
Distribution

• Solution
• The position of the center of pressure is
• Volume Analogy
• The gage pressure at the bottom of
  the gate is pg (2 m)?, so the volume
  of the pressure distribution is

67
Example 10.12 Gate Loaded by a Pressure
Distribution

• Solution
• The x coordinate of the centroid of the
  triangular distribution, which is the center of
  pressure, is
• Determining the Reactions
• Draw the free-body diagram of the gate.

68
Example 10.12 Gate Loaded by a Pressure
Distribution

• Solution
• From the equilibrium equations
• we obtain Ax ?1000 N, Az 45.78 kN B 13.08
  kN.

69
Example 10.12 Gate Loaded by a Pressure
Distribution

• Critical Thinking
• The motivation for defining the gage pressure is
  demonstrated by this example
• In many applications it is the difference between
  the pressure atmospheric pressure, not the
  pressure itself, that is significant

70
Example 10.13 Determination of a Pressure Force

• The container in Fig. 10.39 is filled with a
  liquid with weight density ?. Determine the force
  exerted by the pressure of the liquid on the
  cylindrical wall AB.

71
Example 10.13 Determination of a Pressure Force

• Strategy
• The pressure of the liquid on the cylindrical
  wall varies with depth.
• It is the force exerted by this pressure
  distribution we want to determine. We could
  determine it by integrating over the cylindrical
  surface but we can avoid that by drawing a
  free-body diagram of the quarter-cylinder of
  liquid to the right of A.

72
Example 10.13 Determination of a Pressure Force

• Solution
• Draw the free-body diagram of the
  quarter-cylinder of liquid. The pressure
  distribution on the cylindrical surface of the
  liquid is the same one that acts on cylindrical
  wall. If we denote the force exerted on the
  liquid by this pressure distribution by Fp, the
  force exerted by the liquid on the cylindrical
  wall is ?Fp. The other forces parallel to the x-y
  plane that act on the quarter-cylinder of liquid
  are its weight, atmospheric pressure at the upper
  surface the pressure distribution of the liquid
  on the left side.

73
Example 10.13 Determination of a Pressure Force

• Solution
• The volume of liquid is , so the
  force exerted on the free-body diagram by the
  weight of the liquid is
• The force exerted on the upper surface by the
  atmospheric pressure is Rbpatmi.

74
Example 10.13 Determination of a Pressure Force

• Solution
• Integrate to determine the force exerted by
  the pressure on the left side of the free-body
  diagram. Its magnitude is
• From the equilibrium equation

75
Example 10.13 Determination of a Pressure Force

• Solution
• We obtain the force exerted on the wall AB by the
  
• pressure of the liquid
• Critical Thinking
• The need to integrate over a curved surface to
  calculate a pressure force can often be avoided
  by choosing a suitable free-body diagram as we
  have done in this example (see Problem 10.101)

76
Chapter Summary

• Beams
• The internal forces moment in abeam are
  expressed as the axial force P, shear force V
  bending moment M
• Their positive directions are defined as

77
Chapter Summary

• By cutting a beam at an arbitrary position x, the
  axial force P, shear force V bending moment M
  can be determined as functions of x
• Depending on the loading, it may be necessary to
  draw several free-body diagrams to determine the
  distributions over the entire beam
• The graphs of V M as functions of x are the
  shear force bending moment diagrams

78
Chapter Summary

• The distributed load, shear force bending
  moment in a portion of a beam subjected only to a
  distributed load satisfy the relations
• (10.4)
• (10.6)
• For segments of a beam that are unloaded or
  subjected to a distributed load, these equations
  can be integrated to determine V M as functions
  of x

79
Chapter Summary

• To obtain the complete shear force bending
  moment diagrams, forces couples must also be
  accounted for

80
Chapter Summary

• Cables
• Loads Distributed Uniformly Along a Straight
  Line
• If a suspended cable is subjected to a
  horizontally distributed load w, the curve is
  described by the cable is the parabola
• (10.10)
• where a w/T0 T0 is the tension in the cable
  at x 0

81
Chapter Summary

• The tension in the cable at a position x is
• (10.10)
• and the length of the cable in the horizontal
  interval from 0 to x
• (10.10)

82
Chapter Summary

• Loads Distributed Uniformly Along a Straight
  Line
• If a suspended cable is subjected to a load w
  distributed along its length, the curve described
  by the cable is the catenary
• (10.20)
• where a w/T0 T0 is the tension in the cable
  at x 0

83
Chapter Summary

• The tension in the cable at a position x is
• (10.21)
• and the length of the cable in the horizontal
  interval from 0 to x is
• (10.22)

84
Chapter Summary

• Discrete Loads
• If N known weights are suspended from a cable
  positions of the attachment points of the cable,
  the horizontal positions of the attachment points
  of the weights the vertical position of the
  attachment point of 1 of the weights are known,
  the configuration of the cable the tension in
  each of its segments

85
Chapter Summary

• Liquids Gases
• The pressure p on a surface is defined so that
  the normal force exerted on an element dA of the
  surface is p dA
• The total normal force exerted by pressure on a
  plane area A is
• (10.24)
• The center of pressure is the point on A at which
  F must be placed to be equivalent to the pressure
  on A

86
Chapter Summary

• The coordinates of the center of pressure are
• (10.25)
• The pressure in a stationary liquid is
• (10.26)
• where p0 is the pressure at the surface, ? is
  the weight density of the liquid x is depth
• If the surface of the liquid is open to the
  atmosphere, p0 patm, the atmospheric pressure