Download 5 Equilibrium of a Rigid Body Chapter Objectives

5 Equilibrium of a Rigid Body
Chapter Objectives
• Develop the equations of equilibrium for a rigid body
• Concept of the free-body diagram for a rigid body
• Solve rigid-body equilibrium problems using the equations
of equilibrium
101
Chapter Outline
•
•
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Conditions for Rigid Equilibrium
Free-Body Diagrams
Equations of Equilibrium
Two and Three-Force Members
Free Body Diagrams
Equations of Equilibrium
Constraints and Statical Determinacy
102
5.1 Conditions for Rigid-Body Equilibrium
•
The equilibrium of a body is expressed as
FR   F  0
 MR O   MO  0
•
Consider summing moments about some other point, such
as point A, we require
M
A
 r  FR   MR O  0
103
5.2 Free Body Diagrams
Support Reactions
• If a support prevents the translation of a body in a given direction, then
a force is developed on the body in that direction.
• If rotation is prevented, a couple moment is exerted on the body.
104
5.2 Free Body Diagrams
105
5.2 Free Body Diagrams
106
5.2 Free Body Diagrams
Internal Forces
•
•
•
External and internal forces can act on a rigid body
For FBD, internal forces act between particles which are
contained within the boundary of the FBD, are not
represented
Particles outside this boundary exert external forces on the
system
107
5.2 Free Body Diagrams
Weight and Center of Gravity
•
•
•
Each particle has a specified weight
System can be represented by a single resultant force,
known as weight W of the body
Location of the force application is known as the center of
gravity
108
5.2 Free Body Diagrams
Procedure for Drawing a FBD
1. Draw Outlined Shape
• Imagine body to be isolated or cut free from its constraints
• Draw outline shape
2. Show All Forces and Couple Moments
• Identify all external forces and couple moments that act on the body
3. Identify Each Loading and Give Dimensions
• Indicate dimensions for calculation of forces
• Known forces and couple moments should be properly labeled with
their magnitudes and directions
109
Example 5.1
Draw the free-body diagram of the uniform beam. The beam has a mass of
100kg.
Solution
Free-Body Diagram
• Support at A is a fixed wall
• Two forces acting on the beam at A denoted as Ax, Ay, with moment MA
• Unknown magnitudes of these vectors
• For uniform beam,
Weight, W = 100(9.81) = 981N
acting through beam’s center of gravity, 3m from A
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5.3 Equations of Equilibrium
• For equilibrium of a rigid body in 2D,
∑Fx = 0; ∑Fy = 0; ∑MO = 0
• ∑Fx and ∑Fy represent sums of x and y components of all the forces
• ∑MO represents the sum of the couple moments and moments of the force
components
2 alternative sets of 3 independent equations,
∑Fx = 0; ∑MA = 0; ∑MB = 0
114
5.3 Equations of Equilibrium
Procedure for Analysis
Free-Body Diagram
• Force or couple moment having an unknown magnitude but known line
of action can be assumed
• Indicate the dimensions of the body necessary for computing the
moments of forces
Procedure for Analysis
Equations of Equilibrium
• Apply ∑MO = 0 about a point O
• Unknowns moments of are zero about O and a direct solution the third
unknown can be obtained
• Orient the x and y axes along the lines that will provide the simplest
resolution of the forces into their x and y components
• Negative result scalar is opposite to that was assumed on the FBD
115
Example 5.5
Determine the horizontal and vertical components of reaction
for the beam loaded. Neglect the weight of the beam in the
calculations.
Free Body Diagrams
600N represented by x and y components
200N force acts on the beam at B
116
Solution
Equations of Equilibrium
   M B  0;
600 cos 45 N  Bx  0  Bx  424 N
 M B  0;
100 N (2m)  (600 sin 45 N )(5m)  (600 cos 45 N )(0.2m)  Ay (7m)  0
Ay  319 N
   Fy  0;
319 N  600 sin 45 N  100 N  200 N  B y  0
B y  405 N
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5.4 Two- and Three-Force Members
Two-Force Members
• When forces are applied at only two points on a member, the member is
called a two-force member
• Only force magnitude must be determined
Three-Force Members
When subjected to three forces, the forces are concurrent or parallel
125
Example 5.13
The lever ABC is pin-supported at A and connected to a short link BD. If
the weight of the members are negligible, determine the force of the pin on
the lever at A.
Solution
Free Body Diagrams
• BD is a two-force member
• Lever ABC is a three-force member
Equations of Equilibrium
 0.7 

  60.3
 0.4 
   Fx  0; FA cos 60.3  F cos 45  400N  0
  tan 1 
   Fy  0;
Solving,
FA sin 60.3  F sin 45  0
FA  1.07 kN
F  1.32 kN
126
5.5 Free-Body Diagrams
Support Reactions
As in the two-dimensional case:
• A force is developed by a support
• A couple moment is developed when rotation of the attached
member is prevented
• The force’s orientation is defined by the coordinate angles ,
and
127
5.5 Free-Body Diagrams
128
5.5 Free-Body Diagrams
129
Example 5.14
Several examples of objects along with their associated free-body diagrams
are shown. In all cases, the x, y and z axes are established and the unknown
reaction components are indicated in the positive sense. The weight of the
objects is neglected.
130
5.6 Equations of Equilibrium
Vector Equations of Equilibrium
• For two conditions for equilibrium of a rigid body in vector
form,
∑F = 0 ∑MO = 0
Scalar Equations of Equilibrium
• If all external forces and couple moments are expressed in
Cartesian vector form
∑F = ∑Fxi + ∑Fyj + ∑Fzk = 0
∑MO = ∑Mxi + ∑Myj + ∑Mzk = 0
131
5.7 Constraints for a Rigid Body
Redundant Constraints
• More support than needed for equilibrium
• Statically indeterminate: more unknown
loadings than equations of equilibrium
132
5.7 Constraints for a Rigid Body
Improper Constraints
• Instability caused by the improper constraining by the
supports
• When all reactive forces are concurrent at this point, the body
is improperly constrained
133
5.7 Constraints for a Rigid Body
Procedure for Analysis
Free Body Diagram
• Draw an outlined shape of the body
• Show all the forces and couple moments acting on the body
• Show all the unknown components having a positive sense
• Indicate the dimensions of the body necessary for computing
the moments of forces
134
Example 5.15
The homogenous plate has a mass of 100kg and is subjected to
a force and couple moment along its edges. If it is supported in
the horizontal plane by means of a roller at A, a ball and socket
joint at N, and a cord at C, determine the components of
reactions at the supports.
135
Solution
Free Body Diagrams
• Five unknown reactions acting on the plate
• Each reaction assumed to act in a positive coordinate direction
Equations of Equilibrium
 Fx  0; Bx  0
 Fy  0; B y  0
 Fz  0; Az  Bz  TC  300 N  981N  0
 M x  0; TC (2m)  981N (1m)  BZ (2m)  0
 M y  0; 300 N (1.5m)  981N (1.5m)  Bz (3m)  Az (3m)  200 N .m  0
Az = 790N
Bz = -217N
TC = 707N
136
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139
6 Structural Analysis
Chapter Objectives
• Determine the forces in the members of a truss using the
method of joints and the method of sections
• Analyze forces acting on the members of frames and
machines composed of pin-connected members
Chapter Outline
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Simple Trusses
The Method of Joints
Zero-Force Members
The Method of Sections
Space Trusses
Frames and Machines
140
6.1 Simple Trusses
• A truss composed of slender members joined together at their end points
Planar Trusses
• The analysis of the forces developed in the truss members is 2D
• Similar to roof truss, the bridge truss loading is also coplanar
141
6.1 Simple Trusses
Assumptions for Design
•
•
“All loadings are applied at the joint”
- Weight of the members neglected
“The members are joined together by smooth pins”
- Assume connections provided the center lines of the
joining members are concurrent
142
6.1 Simple Trusses
Simple Truss
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•
Form of a truss must be rigid to prevent collapse
The simplest form that is rigid or stable is a triangle
143
6.2 The Method of Joints
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For truss, we need to know the force in each members
Forces in the members are internal forces
For external force members, equations of equilibrium can
be applied
Force system acting at each joint is coplanar and concurrent
∑Fx = 0 and ∑Fy = 0 must be satisfied for equilibrium
144
6.2 The Method of Joints
Procedure for Analysis
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Draw the FBD with at least 1 known and 2 unknown forces
Find the external reactions at the truss support
Determine the correct sense of the member
Orient the x and y axes
Apply ∑Fx = 0 and ∑Fy = 0
Use known force to analyze the unknown forces
145
Example 6.1
Determine the force in each member of the truss and indicate whether the
members are in tension or compression.
Solution
• 2 unknown member forces at joint B
• 1 unknown reaction force at joint C
• 2 unknown member forces and 2 unknown reaction forces at point A
For Joint B,
   Fx  0;
500N  FBC sin 45 N  0  FBC  707.1N (C )
   Fy  0;
FBC cos 45 N  FBA  0  FBA  500N (T )
146
Solution
For Joint C,
   Fx  0;
 FCA  707.1 cos 45 N  0  FCA  500 N (T )
   Fy  0;
C y  707.1sin 45 N  0  C y  500 N
For Joint A,
   Fx  0;
500N  Ax  0  Ax  500N
   Fy  0;
500N  Ay  0  Ay  500N
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6.3 Zero-Force Members
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•
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Method of joints is simplified using zero-force members
Zero-force members is supports with no loading
In general, when 3 members form a truss joint, the 3rd
member is a zero-force member provided no external force
or support reaction is applied to the joint
153
Example 6.4
Using the method of joints, determine all the zero-force
members of the Fink roof truss. Assume all joints are pin
connected.
Solution
For Joint G,
   Fy  0  FGC  0
GC is a zero-force member.
For Joint D,
 Fx  0  FDF  0
154
Solution
For Joint F,
   Fy  0  FFC cos   0
  90 , FFC  0
For Joint B,
   Fy  0  FFC cos   0
  90 , FFC  0
FHC satisfy ∑Fy = 0 and therefore HC is not a zeroforce member.
155
6.4 The Method of Sections
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•
•
Used to determine the loadings within a body
If a body is in equilibrium, any part of the body is in
equilibrium
To find forces within members, an imaginary section is
used to cut each member into 2 and expose each internal
force as external
156
6.4 The Method of Sections
•
•
Consider the truss and section a-a as shown
Member forces are equal and opposite to those acting on the
other part – Newton’s Law
157
6.4 The Method of Sections
Procedure for Analysis
Free-Body Diagram
• Decide the section of the truss
• Determine the truss’s external reactions
• Use equilibrium equations to solve member forces at the cut session
• Draw FBD of the sectioned truss which has the least number of
forces acting on it
• Find the sense of an unknown member force
Equations of Equilibrium
• Summed moments about a point
• Find the 3rd unknown force from moment equation
158
Example 6.5
Determine the force in members GE, GC, and BC of the truss. Indicate
whether the members are in tension or compression.
Solution
Choose section a-a since it cuts through the three members
Draw FBD of the entire truss
   Fx  0;
 M A  0;
   Fy  0;
400 N  Ax  0  Ax  400 N
 1200 N (8m)  400 N (3m)  D y (12m)  0  D y  900 N
Ay  1200 N  900 N  0  Ay  300 N
159
Solution
• Draw FBD for the section portion
 M G  0;
 300N (4m)  400N (3m)  FBC (3m)  0  FBC  800N (T )
 M C  0;
 300N (8m)  FGE (3m)  0  FGE  800N (C )
   Fy  0;
3
300N  FGC  0  FGC  500N (T )
5
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6.5 Space Trusses
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•
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Consists of members joined together at their ends to form
3D structure
The simplest space truss is a tetrahedron
Additional members would be redundant in supporting
force P
163
6.5 Space Trusses
Assumptions for Design
•
2 force members
Method of Joints
•
•
Solve ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 at each joint
Force analysis has at least 1 unknown force and 3 unknown
forces
Method of Sections
•
When imaginary section is passed through a truss it must
satisfied
∑Fx = 0, ∑Fy = 0, ∑Fz = 0
∑Mx = 0, ∑My = 0, ∑Mz = 0
164
Example 6.8
Determine the forces acting in the members of the space truss.
Indicate whether the members are in tension or compression.
Solution
For Joint A,
P  {4 j }kN , FAB  FAB j , FAC   FAC k
 rAE
FAE  FAE 
 rAE




 FAE (0.577i  0.577 j  0.577k )
 F  0;
P  FAB  FAC  FAE  0
4 j  FAB j  FAC k  0.577 FAE i  0.577 FAE j  0.577 FAE k  0
FAC  FAE  0, FAB  4kN
165
Solution
For Joint B,
 Fx  0; RB cos 45  0.707 FBE  0
 Fy  0;4  RB sin 45  0
 Fz  0;2  FBD  0.707 FBE  0
RB  FBE  5.66kN (T )
FBD  2kN (C )
To show,
FDE  FDC  FCE  0
166
6.6 Frames and Machines
•
•
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Composed of pin-connected multi-force members
Frames are stationary
Apply equations of equilibrium to each member to
determine the unknown forces
167
6.6 Frames and Machines
Free-Body Diagram
•
Isolate each part by drawing its outlined shape
–
–
–
–
–
show all forces and couple moments act on the part
identify each known and unknown force and couple moment
indicate any dimension
apply equations of equilibrium
assumed sense of
unknown force or moment
– draw FBD
168
Example 6.9
For the frame, draw the free-body diagram of (a) each member, (b) the pin at
B and (c) the two members connected together.
Solution
Part (a)
BA and BC are not two-force
AB is subjected to the resultant forces from the pins
169
Solution
Part (b)
• Pin at B is subjected to two forces, force of the member BC and AB on
the pin
• For equilibrium, forces and respective components must be equal but
opposite
• Bx and By shown equal and opposite on members AB
Part (c)
• Bx and By are not shown as they form equal but opposite internal forces
• Unknown force at A and C must act in the same sense
• Couple moment M is used to find reactions at A and C
170
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7 Internal Force
Chapter Objectives
• Method of sections for determining the internal loadings in a
member
• Develop procedure by formulating equations that describe
the internal shear and moment throughout a member
• Analyze the forces and study the geometry of cables
supporting a load
182
Chapter Outline
•
•
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•
Internal Forces Developed in Structural Members
Shear and Moment Equations and Diagrams
Relations between Distributed Load, Shear and
Moment
Cables
183
7.1 Internal Forces Developed in Structural Members
•
•
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•
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The design of any structural or mechanical member requires the
material to be used to be able to resist the loading acting on the member
These internal loadings can be determined by the method of sections
Force component N, acting normal to the beam at the cut session
V, acting tangent to the session are normal or axial force and the shear
force
Couple moment M is referred as the bending moment
184
7.1 Internal Forces Developed in Structural Members
•
•
•
For 3D, a general internal force and couple moment
resultant will act at the section
Ny is the normal force, and Vx and Vz are the shear
components
My is the torisonal or twisting moment, and Mx and Mz are
the bending moment components
185
7.1 Internal Forces Developed in Structural Members
Procedure for Analysis
Support Reactions
• Before cut, determine the member’s support reactions
• Equilibrium equations used to solve internal loadings
during sectioning
Free-Body Diagrams
• Keep all distributed loadings, couple moments and forces
acting on the member in their exact locations
• After session draw FBD of the segment having the least
loads
186
7.1 Internal Forces Developed in Structural Members
Procedure for Analysis
Free-Body Diagrams (Continue)
• Indicate the z, y, z components of the force, couple
moments and resultant couple moments on FBD
• Only N, V and M act at the section
• Determine the sense by inspection
Equations of Equilibrium
• Moments should be summed at the section
• If negative result, the sense is opposite
187
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189
Example 7.3
Determine the internal force, shear force and the bending
moment acting at point B of the two-member frame.
190
Solution
Support Reactions
FBD of each member
Member AC
∑ MA = 0;
-400kN(4m) + (3/5)FDC(8m)= 0
FDC = 333.3kN
+→∑ Fx = 0;
-Ax + (4/5)(333.3kN) = 0
Ax = 266.7kN
+↑∑ Fy = 0;
Ay – 400kN + 3/5(333.3kN) = 0
Ay = 200kN
191
Solution
Support Reactions
Member AB
+→∑ Fx = 0;
NB – 266.7kN = 0
NB = 266.7kN
+↑∑ Fy = 0;
200kN – 200kN – VB = 0
VB = 0
∑ MB = 0;
MB – 200kN(4m) – 200kN(2m) = 0
MB = 400kN.m
192
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194
7.2 Shear and Moment Equations and Diagrams
•
•
•
Beams – structural members designed to support loadings
perpendicular to their axes
A simply supported beam is pinned at one end and roller
supported at the other
A cantilevered beam is fixed at one end and free at the other
195
7.2 Shear and Moment Equations and Diagrams
Procedure for Analysis
Support Reactions
• Find all reactive forces and couple moments acting on the beam
• Resolve them into components
Shear and Moment Reactions
• Specify separate coordinates x
• Section the beam perpendicular to its axis
• V obtained by summing the forces perpendicular to the beam
• M obtained by summing moments about the sectioned end
• Plot (V versus x) and (M versus x)
• Convenient to plot the shear and the bending moment diagrams below
the FBD of the beam
196
197
Solution
Shear diagram
Internal shear force is always positive within the shaft AB.
Just to the right of B, the shear force
changes sign and remains at
constant value for segment BC.
Moment diagram
Starts at zero, increases linearly to
B and therefore decreases to zero.
198
7.3 Relations between Distributed Load, Shear and Moment
Distributed Load
• Consider beam AD subjected to an arbitrary load
w = w(x) and a series of concentrated forces and moments
• Distributed load assumed positive when loading acts
downwards
199
7.3 Relations between Distributed Load, Shear and Moment
Distributed Load
• A FBD diagram for a small segment of the beam having a
length ?x is chosen at point x along the beam which is not
subjected to a concentrated force or couple moment
• Any results obtained will not apply
at points of concentrated loadings
• The internal shear force and
bending moments assumed
positive sense
200
7.3 Relations between Distributed Load, Shear and Moment
Distributed Load
• Distributed loading has been replaced by a resultant force ∆F
= w(x) ∆x that acts at a fractional distance k (∆x) from the
right end, where 0 < k <1
   Fy  0;V  w( x)x  (V  V )  0
V   w( x)x
 M  0;Vx  M  w( x)xk x   ( M  M )  0
M  Vx  w( x)k (x) 2
201
7.3 Relations between Distributed Load, Shear and Moment
Distributed Load
Slope of the
shear diagram
dV
 w(x)
dx
Slope of
shear diagram
dM
V
dx
Shear moment diagram
M BC   Vdx
Area under
shear diagram
Change in shear
Change in moment
Negative of distributed
load intensity
VBC    w( x)dx
Area under
shear diagram
202
7.3 Relations between Distributed Load, Shear and Moment
Force and Couple Moment
• FBD of a small segment of the beam
• Change in shear is negative
• FBD of a small segment of the beam located at the couple
moment
• Change in moment is positive
203
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205
Example 7.9
Draw the shear and moment diagrams for the overhang beam.
The support reactions are shown.
Shear Diagram
Shear of –2 kN at end A of the beam
is at x = 0.
Positive jump of 10 kN at x = 4 m
due to the force.
Moment Diagram
M x4  M x0  M  0 24  8 kN m
206
207
7.4 Cables
• Cables and chains used to support and transmit loads from
one member to another
• In force analysis, weight of cables is neglected
• Assume cable is perfectly flexible and inextensible
• Due to its flexibility cables has no resistance to bending
• Length remains constant before and after loading
208
7.4 Cables
Cable Subjected to Concentrated Loads
• For a cable of negligible weight, it will subject to constant
tensile force
• Known: h, L1, L2, L3 and loads P1 and P2
• Form 2 equations of equilibrium
• Use Pythagorean Theorem to relate the three segmental
lengths
209
Example 7.11
Determine the tension in each segment of the cable.
FBD for the entire cable.
   Fx  0;
 Ax  E x  0
 M E  0;
 Ay (18m)  4kN (15m)  15kN (10m)  3kn(2m)  0
Ay  12kN
   Fy  0;
12kN  4kN  15kN  3kN  E y  0
E y  10kN
Consider leftmost section which cuts cable BC since sag yC = 12m.
 M C  0;
Ax (12m)  12kN (8m)  4kN (5m)  0
Ax  E x  6.33kN
   Fx  0;
TBC cos  BC  6.33kN  0
   Fy  0;
12kN  4kN  TBC sin  BC  0
 BC  51.6 , TBC  10.2kN

210
7.4 Cables
Cable Subjected to a Distributed Load
• Consider weightless cable subjected to a loading function w = w(x)
measured in the x direction
Cable Subjected to a Distributed Load
• For FBD of the cable having length ∆x
• Since the tensile force changes continuously, it is denoted on the FBD by
∆T
• Distributed load is represented by second integration,
y
1
FH
  w( x)dxdx
211
Example 7.12
The cable of a suspension bridge supports half of the uniform
road surface between the two columns at A and B. If this
distributed loading wo, determine the maximum force
developed in the cable and the cable’s required length. The
span length L and, sag h are known.
212
Solution
y
Note w(x) = wo 
1
FH
  w dx dx
o
y
Perform two integrations 
1
FH
Boundary Conditions at x = 0 
 wo x 2


 C1 x  C2 
 2

y  0, x  0, dy / dx  0
C1  C2  0
Therefore,
y 
Curve becomes
wo
x2
2 FH
This is the equation of a parabola
Boundary Condition at x = L/2 
For constant,
FH 
y h
wo L2
4h
and y  2 x 2
8h
L
Tension, T = FH/cosθ
Slope at point B 
Therefore
Tmax 
dy
dx
 tan  max 
xL / 2
wo
FH
xL / 2
w L
  max  tan 1  o 
 2 FH 
FH
cos( max )
Using triangular relationship
Tmax 
4 FH2  wo2 L2
2
213
Solution
For a differential segment of cable length ds,
ds 
dx  dy
2
2
2
 dy 
 1    dx
 dx 
Determine total length by integrating,
   ds  2 
L/2
0
2
 8h 
1   2 x  dx
L 
Integrating yields,
2

L
4h 
L

1  4h 


1     sinh  
2
4h
 L
 L 

214
7.4 Cables
Cable Subjected to its Own Weight
• When weight of the cable is considered, the loading function
becomes a function of the arc length s rather than length x
• FBD of a segment of the cable is shown
215
7.4 Cables
Cable Subjected to its Own Weight
• Apply equilibrium equations to the force system
T cos   FH T sin    w( s)ds
•
dy 1

w(s)ds
dx FH 
Replace dy/dx by ds/dx for direct integration
2
dy
 ds 
   1
dx
 dx 
ds  dx  dy
2
2
Cable Subjected to its Own Weight
• Therefore
ds 
1
 1  2
dx  FH
•


1/ 2

 w(s)ds 
2
Separating variables and integrating
x
ds

1
1  2
 FH


1/ 2

 w(s)ds 
2
216
Example 7.13
Determine the deflection curve, the length, and the maximum tension in the
uniform cable. The cable weights wo = 5N/m.
Solution
For symmetry, origin located at the center of the cable.
Deflection curve expressed as y = f(x)
x
ds

1  1 / F 2
H

 w ds 
2 1/ 2
o
x
ds
1  1 / F w s  C  
2
H
2 1/ 2
o
1
Substitute
du  ( wo / FH )ds
u  1 / FH wo s  C1 
Perform second integration
x
or

FH
sinh 1 u  C2
wo
x
FH
wo




1  1


sinh
w
s

C

C

o
1 
2

 FH



217
Solution
x


FH 
1  1
wo s  C1   C2 
sinh 
wo 
 FH


Evaluate constants
dy
1

dx FH
 wo ds

dy  1
 
wo s  C1 
dx  FH

dy/dx = 0 at s = 0, then C1 = 0
s=0 at x=0, then C2=0
solve for s
y
s
w
FH
sinh  o
wo
 FH

x 

w 
dy
 sinh  o x 
dx
 FH 
w 
FH
cosh o x   C3
wo
 FH 
FH
C


Boundary Condition y = 0 at x = 0  3
wo
For deflection curve,
y
 wo  
FH 
x   1
cosh 
wo 
F
 H  
This equations defines a catenary curve.
218
Solution
Boundary Condition y = h at x = L/2
h

 wo 
FH 


cosh
x

1


F

wo 
 H 

Since wo = 5N/m, h = 6m and L = 20m,
6m 
 50 N  
FH 

  1
cosh

5N / m 
F
 H  
FH  45.9 N
For deflection curve,
y  9.19cosh 0.109 x   1m
x = 10m, for half length of the cable

45.9
 5N / m
10 m   12.1m

sinh 
2 5N / m
 45.9 N

Hence
  24.2m
Maximum tension occurs when is maximum at
s = 12.1m
dy
5 N / m12.1m 
 tan  
 1.32
dx s 12.1m
max
45.9 N
 max  52.8
Tmax 
FH
45.9 N

 75.9 N
cos max cos 52.8
219
8 Friction
Chapter Objectives
• Introduce the concept of dry friction
• To present specific applications of frictional force analysis
on wedges, screws, belts, and bearings
• To investigate the concept of rolling resistance
220
Chapter Outline
•
•
•
•
•
•
•
•
Characteristics of Dry Friction
Problems Involving Dry Friction
Wedges
Frictional Forces on Screws
Frictional Forces on Flat Belts
Frictional Forces on Collar Bearings, Pivot Bearings, and
Disks
Frictional Forces on Journal Bearings
Rolling Resistance
221
8.1 Characteristics of Dry Friction
Friction
•
•
•
•
Force that resists the movement of two contacting surfaces
that slide relative to one another
Acts tangent to the surfaces at points of contact with other
body
Opposing possible or existing motion of the body relative to
points of contact
Two types of friction – Fluid and Coulomb Friction
222
8.1 Characteristics of Dry Friction
•
•
•
Fluid friction exist when the contacting surface are
separated by a film of fluid (gas or liquid)
Depends on velocity of the fluid and its ability to resist
shear force
Coulomb friction occurs
between contacting surfaces
of bodies in the absence of a
lubricating fluid
223
8.1 Characteristics of Dry Friction
Theory of Dry Friction
Impending Motion
• Constant of proportionality μs is known as the coefficient of
static friction
• Angle ϕs that Rs makes with N is called the angle of static
friction
 Fs 
1   s N 
1
s  tan    tan 

tan
s

N
 N 
1
224
8.1 Characteristics of Dry Friction
Theory of Dry Friction
Typical Values of μs
Contact Materials
Coefficient of Static Friction μs
Metal on ice
0.03 – 0.05
Wood on wood
0.30 – 0.70
Leather on wood
0.20 – 0.50
Leather on metal
0.30 – 0.60
Aluminum on aluminum
1.10 – 1.70
225
8.1 Characteristics of Dry Friction
Theory of Dry Friction
• Resultant frictional force Fk is directly proportional to the
magnitude of the resultant normal force N
Fk = μkN
• Constant of proportionality μk is coefficient of kinetic friction
•
μk are typically 25% smaller than μs
• Resultant Rk has a line of action defined by ϕk, angle of kinetic
friction
 Fk
N
k  tan 1 

1   k N 
1

tan


  tan  k

 N 
226
8.1 Characteristics of Dry Friction
Theory of Dry Friction
• F is a static frictional force if equilibrium is maintained
• F is a limiting static frictional force when it reaches a
maximum value needed to maintain equilibrium
• F is termed a kinetic frictional force when sliding occurs at
the contacting surface
227
8.1 Characteristics of Dry Friction
Characteristics of Dry Friction
• The frictional force acts tangent to the contacting surfaces
• The max static frictional force Fs is independent of the area
of contact
• The max static frictional force is greater than kinetic
frictional force
• When slipping, the max static frictional force is
proportional to the normal force and kinetic frictional force
is proportional to the normal force
228
8.2 Problems Involving Dry Friction
Types of Friction Problems
• In all cases, geometry and dimensions are assumed to be
known
• 3 types of mechanics problem involving dry friction
- Equilibrium
- Impending motion at all points
- Impending motion at some points
229
8.2 Problems Involving Dry Friction
Types of Friction Problems
Equilibrium
• Total number of unknowns = Total number of available
equilibrium equations
• Frictional forces must satisfy F ≤ μsN; otherwise, slipping
will occur and the body will not remain in equilibrium
• We must determine the frictional
forces at A and C to check
for equilibrium
230
8.2 Problems Involving Dry Friction
Equilibrium Versus Frictional Equations
• Frictional force always acts so as to oppose the relative
motion or impede the motion of the body over its contacting
surface
• Assume the sense of the frictional force that require F to be
an “equilibrium” force
• Correct sense is made after solving the equilibrium
equations
• If F is a negative scalar, the sense of F is the reverse of that
assumed
231
Example 8.1
The uniform crate has a mass of 20kg. If a force P = 80N is applied on to the
crate, determine if it remains in equilibrium. The coefficient of static friction
is = 0.3.
   Fx  0;
80 cos 30  N  F  0
   Fy  0;
 80 sin 30  N  N C  196 .2 N  0
 M O  0;
80 sin 30  N (0.4m)  80 cos 30  N (0.2m)  N C ( x)  0
F  69.3 N , N C  236 N
x  0.00908 m  9.08mm
232
Solution
Since x is negative, the resultant force acts (slightly) to the left
of the crate’s center line.
No tipping will occur since x ≤ 0.4m
Max frictional force which can be developed at the surface of
contact
Fmax = μsNC = 0.3(236N) = 70.8N
Since F = 69.3N < 70.8N, the crate will not slip thou it is close
to doing so.
233
234
235
236
237
238
8.3 Wedges
•
•
•
A simple machine used to transform an applied force into much larger
forces, directed at approximately right angles to the applied force
Used to give small displacements or adjustments to heavy load
Consider the wedge used to lift a block of weight W by applying a force
P to the wedge
•
FBD of the block and the wedge
•
Exclude the weight of the wedge since it is small compared to weight of
the block
239
Example 8.6
The uniform stone has a mass of 500kg and is held in place in the horizontal
position using a wedge at B. if the coefficient of static friction μs = 0.3, at the
surfaces of contact, determine the minimum force P needed to remove the
wedge. Is the wedge self-locking? Assume that the stone does not slip at A.
Solution
Minimum force P requires F = μs NA at the surfaces of contact with the wedge.
FBD of the stone and the wedge as below.
On the wedge, friction force opposes the motion and on the stone at A, FA ≤
μsNA, slipping does not occur.
240
Solution
5 unknowns, 3 equilibrium equations for the stone and 2 for the wedge.
 M A  0;
 4905 N (0.5m)  ( N B cos 7  N )(1m)  (0.3N B sin 7  N )(1m)  0
N B  2383.1N
   Fx  0;
2383.1sin 7   0.3(2383.1 cos 7  )  P  0.3N C  0
   Fy  0;
N C  2383.1 cos 7  N  0.3(2383.1sin 7  )  0
N C  2452.5 N
P  1154.9 N  1.15kN
Since P is positive, the wedge must be pulled out.
If P is zero, the wedge would remain in place (self-locking) and the
frictional forces developed at B and C would satisfy
FB < μsNB
FC < μsNC
241
8.4 Frictional Forces on Screws
• Screws used as fasteners
• Sometimes used to transmit power or motion from one part
of the machine to another
• A square-ended screw is commonly used for the latter
purpose, especially when large forces are applied along its
axis
• A screw is thought as an inclined plane or wedge wrapped
around a cylinder
242
8.4 Frictional Forces on Screws
• A nut initially at A on the screw will move up to B when
rotated 360° around the screw
• This rotation is equivalent to translating the nut up an
inclined plane of height l and length 2πr, where r is the mean
radius of the head
• Applying the force equations of equilibrium, we have
M  rW tan s   
243
8.4 Frictional Forces on Screws
Downward Screw Motion
• If the surface of the screw is very slippery, the screw may
rotate downward if the magnitude of the moment is reduced
to say M’ < M
• This causes the effect of M’ to become
M’ = Wr tan(θ – ϕs)
244
Example 8.7
The turnbuckle has a square thread with a mean radius of 5mm and a lead of
2mm. If the coefficient of static friction between the screw and the
turnbuckle is μs = 0.25, determine the moment M that must be applied to
draw the end screws closer together. Is the turnbuckle self-locking?
Solution
Since friction at two screws must be overcome, this requires
M  2Wr tan    
W  2000 N , r  5mm, s  tan 1  s  tan 1 0.25  14.04
  tan 1  / 2r   tan 1 2mm / 2 5mm   3.64
Solving


M  2 2000 N 5mm  tan 14.04  3.64
 6374.7 N .mm  6.37 N .m

When the moment is removed, the turnbuckle will be self-locking
245
8.5 Frictional Forces on Flat Belts
• It is necessary to determine the frictional forces developed
between the contacting surfaces
• Consider the flat belt which passes over a fixed curved
surface
• Obviously T2 > T1
• Consider FBD of the belt
segment in contact with the surface
• N and F vary both in
magnitude and direction
246
8.5 Frictional Forces on Flat Belts
•
•
•
Consider FBD of an element having a length ds
Assuming either impending motion or motion of the belt, the magnitude of the
frictional force
dF = μ dN
Applying equilibrium equations
 Fx  0;
 d 
 d 
T cos
  dN  (T  dT ) cos
0
2
2




 Fy  0;
•
 d 
 d 
dN  (T  dT ) sin 
  T sin 
0
 2 
 2 
We have
dT
 dN  dT dN  Td
  d
T
T  T1 ,   0, T  T2 ,   
T dT



T T 0 d
T

In 2   T2  Te
1
T1
2
1
247
Example 8.8
The maximum tension that can be developed In the cord is
500N. If the pulley at A is free to rotate and the coefficient of
static friction at fixed drums B and C is μs = 0.25, determine the
largest mass of cylinder that can be lifted by the cord. Assume
that the force T applied at the end of the cord is directed
vertically downward.
248
Solution
Weight of W = mg causes the cord to move CCW over the drums at B and C.
Max tension T2 in the cord occur at D where T2 = 500N
For section of the cord passing over the drum at B
180° = rad, angle of contact between drum and cord
= (135° /180° ) = 3/4 rad
T2  T1e  s  ;
500 N  T1e 0.253 / 4  
T1 
500 N
e 0.253 / 4  

500 N
 277.4 N
1.80
For section of the cord passing over the drum at C
W < 277.4N
T2  T1e   ;
s
277.4  We0.253 / 4  
W  153.9 N
m
W
153.9 N

 15.7 kg
g 9.81m / s 2
249
Engineering Mechanics:
Statics in SI Units, 12e
5
Equilibrium of a Rigid Body
250
4 Force System Resultants
Chapter Objectives
• Concept of moment of a force in two and three dimensions
• Method for finding the moment of a force about a specified
axis.
• Define the moment of a couple.
• Determine the resultants of non-concurrent force systems
• Reduce a simple distributed loading to a resultant force
having a specified location
251
Chapter Outline
•
•
•
•
•
•
•
•
•
Moment of a Force – Scalar Formation
Cross Product
Moment of Force – Vector Formulation
Principle of Moments
Moment of a Force about a Specified Axis
Moment of a Couple
Simplification of a Force and Couple System
Further Simplification of a Force and Couple System
Reduction of a Simple Distributed Loading
252
4.9 Reduction of a Simple Distributed Loading
• Large surface area of a body may be subjected to distributed
loadings
• Loadings on the surface is defined as pressure
• Pressure is measured in Pascal (Pa): 1 Pa = 1N/m2
Uniform Loading Along a Single Axis
• Most common type of distributed
loading is uniform along a
single axis
253
4.9 Reduction of a Simple Distributed Loading
Magnitude of Resultant Force
• Magnitude of dF is determined from differential area dA
under the loading curve.
• For length L,
FR   wx dx   dA  A
L
A
• Magnitude of the resultant force is equal to the total area A
under the loading diagram.
254
4.9 Reduction of a Simple Distributed Loading
Location of Resultant Force
• MR = ∑MO
• dF produces a moment of xdF = x w(x) dx about O
• For the entire plate,
x FR   xw( x)dx
M Ro  M O
L
• Solving for x
x
 xw( x)dx  xdA
L
 w( x)dx
L

A
 dA
A
255
Example 4.21
Determine the magnitude and location of the equivalent resultant force acting
on the shaft.
Solution
For the colored differential area element,
dA  wdx  60 x 2 dx
For resultant force
FR  F ;
2
FR   dA   60 x 2 dx
A
0
2
 x3 
 23 03 
 60    60   
 3 0
3 3
 160 N
For location of line of action,
2
 x4 
 24 04 
60
x
(60
x
)
dx
60
xdA
4
4  4
0
A
 0
  1.5m
x


 
160
160
160
 dA
2
2
A
256
257
9 Center of Gravity and Centroid
Chapter Objectives
• Concept of the center of gravity, center of mass, and the
centroid
• Determine the location of the center of gravity and centroid
for a system of discrete particles and a body of arbitrary
shape
• Theorems of Pappus and Guldinus
• Method for finding the resultant of a general distributed
loading
258
9.1 Center of Gravity and Center of Mass for
a System of Particles
Mass Center
~
xm
~
ym
~
zm
x
;y 
,z 
m
m
m
•
•
•
Particles have weight only when under the influence of gravitational
attraction, whereas center of mass is independent of gravity
A rigid body is composed of an infinite number of particles
Consider a particle having weight of dW
~
~
~
xdW
ydW
zdW



x
;y 
;z 
dW dW dW
259
9.1 Center of Gravity and Center of Mass for
a System of Particles
Centroid of a Volume
• Consider an object subdivided into volume elements dV, for location of
the centroid,
 ~xdV
 ~ydV
~zdV
dV
dV
dV
x V
;y V
V
;z  V
V
V
Centroid of an Area
 ~xdA
 ~ydA ~zdA
dA
dA
xA
;y  A
A
;z  A
dA
A
A
Centroid of a Line
 ~xdL
 ~ydL ~zdL
dL
dL
xL
L
;y  L
L
;z  L
dL
L
260
Example 9.1
Locate the centroid of the rod bent into the shape of a parabolic
arc.
261
Solution
Differential element
Located on the curve at the arbitrary point (x, y)
Area and Moment Arms
For differential length of the element dL
dL 
 dx    dy 
2
2
2
 dx 
    1 dy
 dy 
Since x = y2 and then dx/dy = 2y
2
dL   2 y   1 dy
The centroid is located at
 ~x dL 1 x 4 y 2  1 dy 1 y 2 4 y 2  1 dy
xL
 01
 01
2
 dL 0 4 y  1 dy 0 4 y 2  1 dy
L
0.6063
 0.410 m
1.479
 ~ydL 1 y 4 y 2  1 dy
yL
 01
 dL  4 y 2  1 dy

L

0
0.8484
 0.574 m
1.479
262
263
264
265
266
267
268
269
270
Example 9.10
9.2 Composite Bodies
Locate the centroid of the plate area.
Solution
Composite Parts
Plate divided into 3 segments.
Area of small rectangle considered “negative”.
Moment Arm
Location of the centroid for each piece is determined and indicated in the
diagram.
Summations
~
xA  4
x

 0.348 mm
 A 11 .5
~
y A 14
y

 1.22 mm
 A 11 .5
271
9.3 Surface Area and Volume
•
A surface area (volume) of revolution is generated by revolving a plane
curve (area) about a non-intersecting fixed axis in the plane of the curve
(area).
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
Volume
A r L
• Volume of a body of revolution = product of generating area and
distance traveled by the centroid in generating the volume

V  r A
272
Example 9.12
Show that the surface area of a sphere is A = 4πR2 and its volume V = 4/3πR3.
Solution
Surface Area
Generated by rotating semi-arc about the x axis
For centroid,
r  2R / 
For surface area,
A  ~
r L;
 2R 
2
A  2 
 R  4R
  
Volume
Generated by rotating semicircular area about the x axis
For centroid,
r  4 R / 3
For volume,
V  ~
r A;
4R   1
4 3

2
V  2 
  R   R
3


 2
 3
273
9.4 Resultant of a General Distributed
Loading
Pressure Distribution over a Surface
• Consider the flat plate subjected to the loading function ρ=ρ(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 = ρ[(x, y) N/m2](d A m2)
= ρ[(x, y) d A]N
Pressure Distribution over a Surface
• This system will be simplified to a single resultant force FR acting
through a unique point on the plate
FR    ( x, y)dA   dV  V
A
V
274
9.4 Resultant of a General Distributed
Loading
Magnitude of Resultant Force FR
• Magnitude of resultant force = total volume under the
distributed loading diagram
• Location of Resultant Force is
x ( x, y )dA  xdV

x

  ( x, y)dA  dV
y ( x, y )dA  ydV

y

  ( x, y)dA  dV
A
A
A
A
V
V
V
V
275
9.5 Fluid Pressure
• According to Pascal’s law, a fluid at rest creates a pressure at a
point that is the same in all directions
• Magnitude of depends on the specific weight or mass density
of the fluid and the depth z of the point from the fluid surface
p   z   gz
• Valid for incompressible fluids
• Gas are compressible fluids and the above equation cannot be
used
276
9.5 Fluid Pressure
Flat Plate of Constant Width
• Consider flat rectangular plate of constant width submerged in a liquid
having a specific weight
• As pressure varies linearly with depth, the distribution of pressure over
the plate’s surface is represented by a trapezoidal volume having an
intensity of w1  bp1  brz1 at depth z1 and w2  bp2  brz2 at depth z2
• Magnitude of the resultant force FR
= volume of this loading diagram
Curved Plate of Constant Width
277
Example 9.14
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.
Solution
The water pressures at depth A and B are
 A   w gz A  (1000 kg / m3 )(9.81m / s 2 )(2m)  19.62kPa
 B   w gzB  (1000 kg / m3 )(9.81m / s 2 )(5m)  49.05kPa
For intensities of the load at A and B,
wA  b A  (1.5m)(19.62kPa)  29.43kN / m
wB  b B  (1.5m)(49.05kPa)  73.58kN / m
FR  area of trapezoid

1
(3)(29 .4  73 .6)  154 .5 N
2
This force acts through the centroid
1  2(29.43)  73.58 
of the area,
h 
(3)  1.29 m
3  29.43  73.58 
measured upwards from B
278
A
279
4.9 Reduction of a Simple Distributed Loading
• Large surface area of a body may be subjected to distributed
loadings
• Loadings on the surface is defined as pressure
• Pressure is measured in Pascal (Pa): 1 Pa = 1N/m2
Uniform Loading Along a Single Axis
• Most common type of distributed
loading is uniform along a
single axis
280
4.9 Reduction of a Simple Distributed Loading
Magnitude of Resultant Force
• Magnitude of dF is determined from differential area dA
under the loading curve.
• For length L,
FR   wx dx   dA  A
L
A
• Magnitude of the resultant force is equal to the total area A
under the loading diagram.
281
4.9 Reduction of a Simple Distributed Loading
Location of Resultant Force
• MR = ∑MO
• dF produces a moment of xdF = x w(x) dx about O
• For the entire plate,
x FR   xw( x)dx
M Ro  M O
L
• Solving for x
x
 xw( x)dx  xdA
L
 w( x)dx
L

A
 dA
A
282
Example 4.21
Determine the magnitude and location of the equivalent
resultant force acting on the shaft.
283
Solution
For the colored differential area element,
dA  wdx  60 x 2 dx
For resultant force
FR  F ;
2
FR   dA   60 x 2 dx
A
0
2
x 
 23 03 
 60    60   
 3 0
3 3
 160 N
3
284
Solution
For location of line of action,
2
4


x
 24 04 
2
A xdA 0 x(60 x )dx 60 4  0 60 4  4 
x



160
160
160
 dA
2
A
 1.5m
Checking,
ab 2m(240 N / m)
A

 160
3
3
3
3
x  a  (2m)  1.5m
4
4
285
286
287
288
289
10 Moments of Inertia
Chapter Objectives
• Method for determining the moment of inertia for an area
• Introduce product of inertia and show determine the
maximum and minimum moments of inertia for an area
• Discuss the mass moment of inertia
290
Chapter Outline
•
•
•
•
•
•
•
•
Definitions of Moments of Inertia for Areas
Parallel-Axis Theorem for an Area
Radius of Gyration of an Area
Moments of Inertia for Composite Areas
Product of Inertia for an Area
Moments of Inertia for an Area about Inclined Axes
Mohr’s Circle for Moments of Inertia
Mass Moment of Inertia
291
10.1 Definition of Moments of Inertia for Areas
•
•
•
Centroid for an area is determined by the first moment of an
area about an axis
Second moment of an area is referred as the moment of
inertia
Moment of inertia of an area originates whenever one
relates the normal stress or force per unit area
292
10.1 Definition of Moments of Inertia for Areas
Moment of Inertia
• moments of inertia of the differential plane area dA
dIx  y2 dA dI y  x2 dA
Ix   y2 dA
A
I y   x2 dA
A
•
Formulate the second moment of dA about z axis
dJ O  r 2 dA
•
where r is perpendicular from the pole (z axis) to the element dA
Polar moment of inertia for entire area,
J O   r 2 dA  I x  I y
A
293
10.2 Parallel Axis Theorem for an Area
•
•
•
For moment of inertia of an area known about an axis
passing through its centroid, determine the moment of
inertia of area about a corresponding parallel axis using the
parallel axis theorem
Consider moment of inertia of the shaded area
A differential element dA is
located at an arbitrary distance y’
from the centroidal x’ axis
294
10.2 Parallel Axis Theorem for an Area
•
•
The fixed distance between the parallel x and x’ axes is defined as dy
For moment of inertia of dA about x axis
dI x   y ' d y  dA
2
•
For entire area
I x    y ' d y  dA
2
A
  y '2 dA  2d y  y ' dA  d y2  dA
A
 y ' dA  y  dA  0;
•
•
A
A
y  0  Ix  Ix  Ad y2 and I y  I y  Adx2
Similarly
For polar moment of inertia
JO  JC  Ad 2
295
10.3 Radius of Gyration of an Area
•
•
Radius of gyration of a planar area has units of length and is
a quantity used in the design of columns in structural
mechanics
For radii of gyration
kx 
Ix
A
ky 
Iy
A
kz 
JO
A
296
Example 10.1
Determine the moment of inertia for the rectangular area with respect to (a)
the centroidal x’ axis, (b) the axis xb passing through the base of the
rectangular, and (c) the pole or z’ axis perpendicular to the x’-y’ plane and
passing through the centroid C.
Solution
Part (a)
I x   y '2 dA  
A
h/2
h / 2
y '2 (bdy ' )  
h/2
h / 2
y '2 dy 
1 3
bh
12
By applying parallel axis theorem,
2
1
h 1
I xb  I x  Ad 2  bh 3  bh    bh 3
12
3
2
For polar moment of inertia about point C,
Iy ' 
1 3
1
hb and JC  Ix  I y '  bh(h2  b2 )
12
12
297
298
299
10.4 Moments of Inertia for Composite Areas
Example 10.4
Compute the moment of inertia of the composite area about the x axis.
Solution
Parallel Axis Theorem
Circle
I x  I x '  Ad y2
 
1
4
2
2
  25   25 75  11.4 106 mm4
4
Rectangle
I x  I x '  Ad y2

 
1
100 150 3  100 150 752  112.5 10 6 mm4
12
Summation
For moment of inertia for the composite area,
 
 
I x  11.4 106  112.5 106
 
 101 106 mm4
300
10.5 Product of Inertia for an Area
•
•
•
•
Moment of inertia for an area is different for every axis
about which it is computed
First, compute the product of the inertia for the area as well
as its moments of inertia for given x, y axes
Product of inertia for an element of area dA located at a
point (x, y) is defined as
dIxy = xydA
Thus for product of inertia,
I xy   xydA
A
301
10.5 Product of Inertia for an Area
Parallel Axis Theorem
• For the product of inertia of dA with respect to the x and y
axes
dI xy   x' d x  y ' d y dA
A
•
For the entire area,
dI xy    x' d x  y ' d y dA
A
  x' y ' dA  d x  y 'dA  d y  x'dA  d x d y  dA
A
•
A
A
A
Forth integral represent the total area A,
I xy  I x ' y '  Ad x d y
302
Example 10.6
Determine the product Ixy of the triangle.
Differential element has thickness dx and area dA = y dx
Using parallel axis theorem,
dI xy  dI xy  dA~
x~
y
~x , ~y  locates centroid of the element or origin of x’, y’ axes
Solution
Due to symmetry, dI  0 ~
x  x, ~
y  y/2
xy
2
 y h
  h  h 3
dI xy  0  ( ydx) x    xdx  x x   2 x dx
2 b
  2b  2b
Integrating we have
h2
I xy  2
2b

b
0
b2h2
x dx 
8
3
Differential element has thickness dy and
area dA = (b - x) dy.
For centroid,
~
x  x  (b  x) / 2  (b  x) / 2, ~
yy
For product of inertia of element
~
b x
dI xy  dI xy  dA~
x~
y  0  (b  x)dy
y
 2 
b   b  b / h  y 
1  2 b2 2 

  b  y dy 
y

y b  2 y dy

h  
2
2 
h


303
10.6 Moments of Inertia for an Area about Inclined Axes
x
u 
r    , r   
 y
v 
 x   cos 
r  A r ,    
 y   sin 
u   cos 
T

r  A r,    
 v    sin 
 sin   u 
cos    v 
sin    x 
cos    y 
y
Ixx   y2 dA
Iyy   x2 dA
v
Ixy   xydA
r
Consider moment of inertia matrix
 Ixx Ixy 
 Iuu Iuv 

I=
I


I

 uv Ivv 
 Ixy Iyy 
Then I  AT IA and I  A I AT
u
θ
0
x
304
10.6 Moments of Inertia for an Area about Inclined Axes
Principal Moments of Inertia
I
max
min

Ix  Iy
2
 Ix  Iy 
  I xy2
 
 2 
2
• Result can gives the max or min moment of inertia
for the area
• Or find the eigenvalue of I or I’ matrix
305
Example 10.8
Determine the principal moments of inertia for the beam’s cross-sectional
area with respect to an axis passing through the centroid.
Moment and product of inertia of the cross-sectional area,
 
 
I x  2.90 10 9 mm 4
I y  5.60 10 9 mm 4
 
I z  3.00 10 9 mm 4
Using the angles of inclination of principal axes u and v,
 
 
 
3.00 109
 Ixy
tan 2 p 

 2.22
 Ix  I y  / 2 2.90 109  5.60 109  / 2
  p1  32.9 , p 2  57.1
For principal of inertia with respect to the u and v axes
Ix  I y
 Ix  I y 
2

 
  Ixy
2
2


2
max
min
I
 
 
 Imax  7.54 109 mm4 , Imin  0.960 109 mm4
306
Solution
 2.9 3 
I

 3 5.6 
( -2.9)( -5.6)+9 =0
 2  8.5  7.24  0
eigenvalue of  (I)  0.96 or 7.54
3
3 
7.54  2.9

 4.64
 3 
eigenvector I x   x, 
x

x

0,
x


 3 1.94 
 4.64 
3
7.54

5.6






307
10.7 Mohr’s Circle for Moments of Inertia
• The circle constructed is known as a Mohr’s circle with
radius
 Ix  Iy 
  I xy2
R  
 2 
2
and center at (a, 0) where
a  I x  I y  / 2
308
10.7 Mohr’s Circle for Moments of Inertia
Determine Ix, Iy and Ixy
• Establish the x, y axes for the area, with the origin located at point P of
interest and determine Ix, Iy and Ixy
Principal of Moments of Inertia
• Points where the circle intersects the abscissa give the values of the
principle moments of inertia Imin and Imax
• Product of inertia will be zero at these points
Principle Axes
• This angle represent twice the angle from the x axis to the area in question
to the axis of maximum moment of inertia Imax
• The axis for the minimum moment of inertia Imin is perpendicular to 309
the
axis for Imax
Example 10.9
Using Mohr’s circle, determine the principle moments of the beam’s crosssectional area with respect to an axis
passing through the centroid.
Solution
Determine Ix, Iy and Ixy
Moments of inertia
 
I x  2.90 10 9 mm 4
 
I y  5.60 109 mm 4
 
I xy  3.00 10 9 mm 4
Center of circle
I
x
 I y  / 2  (2.90  5.60) / 2  4.25
Principal Moments of Inertia
 
 0.96010 mm
I max  7.54 109 mm 4
I min
9
4
310
10.8 Mass Moment of Inertia
• Mass moment of inertia is defined as the integral of the second moment
about an axis of all the elements of mass dm which compose the body
• For body’s moment of inertia
about the z axis,
I   r 2 dm
m
• The axis that is generally chosen
for analysis, passes through the
body’s mass center G
• When ρ being a constant,
I    r 2 dV
V
311
Example 10.10
Determine the mass moment of inertia of the cylinder about the z axis. The density
of the material is constant.
Shell Element
For volume of the element,
dV   2 r  h  dr
For mass,
dm   dV    2 rh dr 
dIz  r2 dm   2 hr3dr
R
I z   r 2 dm   2h  r 3dr 
m
0

2
R 4h
For the mass of the cylinder
R
m   dm   2h  rdr  hR 2
m
0
So that
Iz 
1
mR 2
2
312
313
10.8 Mass Moment of Inertia
Parallel Axis Theorem
For moment of inertia of body
about the z axis,


I   r 2 dm   d  x'  y '2 dm
m
m
2
  x'  y ' dm  2d  x' dm  d  dm
2
m
2
2
m
m
Parallel Axis Theorem
For moment of inertia about the z axis,
I = IG + md2
Radius of Gyration
I  mk
2
I
or k 
m
314
315