Chapter 5 Equilibrium of a Rigid Body 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 Chapter 5 Outline • 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 43
Transcript
Chapter 5 Equilibrium of a Rigid Body
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
Chapter 5 Outline • 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 43
5.1 Conditions for Rigid-Body Equilibrium
• The equilibrium of a body is expressed as
• Consider summing moments about some other point, such
as point A, we require
0
0
R
R OO
F F
M M
0A R RO
M r F M
44
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.
45
5.2 Free Body Diagrams
46
5.2 Free Body Diagrams
47
5.2 Free Body Diagram
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
48
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 49
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
50
5.5 3D Free-Body Diagrams
51
5.5 Free-Body Diagrams
52
5.7 Constraints for a Rigid Body
Redundant Constraints
• More support than needed for equilibrium
• Statically indeterminate: more unknown
loadings than equations of equilibrium
53
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
54
Chapter 6 Structural Analysis
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
Outline • Simple Trusses
• The Method of Joints
• Zero-Force Members
• The Method of Sections
• Space Trusses
• Frames and Machines
55
6.1 Simple Truss • 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
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
56
6.1 Simple Truss Simple Truss
• Form of a truss must be rigid to prevent collapse
• The simplest form that is rigid or stable is a triangle
• 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
Method of Joints
57
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,
)(500045cos
;0
)(1.707045sin500
;0
TNFFNF
F
CNFNFN
F
BABABC
y
BCBC
x
58
Solution
For Joint C,
For Joint A,
NCNC
F
TNFNF
F
yy
y
CACA
x
500045sin1.707
;0
)(500045cos1.707
;0
NAAN
F
NAAN
F
yy
y
xx
x
5000500
;0
5000500
;0
59
6.3 Zero-Force Members
• 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
60
Example 6.4
Using the method of joints, determine all the zero-force
members of the roof truss. Assume all joints are pin connected.
Solution
For Joint G,
GC is a zero-force member.
For Joint D,
00 GCy FF
00 DFx FF61
Solution
For Joint F,
For Joint B,
2BHF kN
0,90
0cos0
FC
FCy
F
FF
FHC satisfy ∑Fy = 0 and therefore
HC is not a zero-force member.
62
6.4 Method of Sections
• 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
• 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
63
Example 6.5
Determine the force in members GE, GC, and BC of the truss.
Indicate whether the members are in tension or compression.
NANNAF
NDmDmNmNM
NAANF
yyy
yyA
xxx
30009001200 ;0
9000)12()3(400)8(1200 ;0
4000400 ;0
Solution
Draw FBD of the entire truss
64
Solution
• Draw FBD for the section portion
)(50005
3300 ;0
)(8000)3()8(300 ;0
)(8000)3()3(400)4(300 ;0
TNFFNF
CNFmFmNM
TNFmFmNmNM
GCGCy
GEGEC
BCBCG
65
Example 6.5
Reaction force
400N
300N
900N
x
y
y
A
A
D
D 1500N
1200N 900N 1500N
E
800N
900N
400N
C
800N
1200N
1200N
900N 500N
B
800N
0N
800N
A
400N
300N
500N
800N
G
500N
0N
800N
500N
66
6.6 Frames • Composed of pin-connected multi-force members
• Frames are stationary
• Apply equations of equilibrium to each member to determine
the unknown forces
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
67
6kN 12kN 0 4kN 4kN 32kNx y x y y AA A B B C M m 68
Chapter 7 Internal Force
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
Outline • Internal Forces Developed in Structural Members
• Shear and Moment Equations and Diagrams
• Relations between Distributed Load, Shear and Moment
• Cables 69
7.1 Internal Forces in Structural Members
• 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
70
71
Example 7.3 Determine the internal force, shear force and
the bending moment acting at point B of the
two-member frame.
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
Solution
72
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
73
7.2 Shear and Moment Equations
• 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
74
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
75
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
2)()(
0)()(;0
)(
0)()(;0
xkxwxVM
MMxkxxwMxVM
xxwV
VVxxwVFy
76
7.3 Relations between Distributed Load, Shear and Moment
Distributed Load
Vdx
dM
)(xwdx
dV
Slope of the shear diagram
Negative of distributed load intensity
Slope of shear diagram
Shear moment diagram
VdxM BC
dxxwVBC )(
Change in shear Area under
shear diagram
Change in moment Area under shear diagram
77
78
79
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
mkN 842004
MMMxx
80
81
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
82
Example 7.11 Determine the tension in each segment of the cable.
FBD for the entire cable.
0; 0
0;
(18 ) 4 (15 ) 15 (10 ) 3 (2 ) 0
12
0; 12 4 15 3 0
10
x x x
E
y
y
y y
y
F A E
M
A m kN m kN m kn m
A kN
F kN kN kN kN E
E kN
Consider leftmost section which cuts cable BC since sag yC = 12m.
0;
(12 ) 12 (8 ) 4 (5 ) 0
6.33
0; cos 6.33 0
0; 12 4 sin 0
51.6 , 10.2
C
x
x x
x BC BC
y BC BC
BC BC
M
A m kN m kN m
A E kN
F T kN
F kN kN T
T kN
83
7.4 Cables
Cable Subjected to a Distributed Load
• Consider weightless cable subjected to a load w = w(x)
• 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,
dxdxxwF
yH
)(1
84
Cable
cos ( )cos( ) 0
sin ( )sin( ) 0
1( ) cos sin 02
T T T
T w w T T
w x x T y T x
1 cos[ cos ( )cos ] 0 ( cos ) 0
1 ( sin )[ sin ( )sin( )] 0 0
1
2
T dT T T T
x x dx
d TT w x T T w
x dx
w x
cos sin 0 tany dy
T Tx dx
cos constant
sin
sin 1tan ( )
cos
H
H
T F
T wdx
Ty dx dx wdx dx
T F
85
Solution Note w(x) = wo
Perform two integrations
Boundary Conditions at x = 0
Therefore, Curve becomes
Boundary Condition at x = L/2
dxdxwF
y o
H
1
21
2
2
1CxC
xw
Fy o
H
0/,0,0 dxdyxy
021 CC 2
2x
F
wy
H
o
hy For constant,
Tension, T = FH/cosθ
Slope at point B
Therefore
Using triangular relationship
2
2
2 4 and
8x
L
hy
h
LwF o
H
1max max
/2
tan tan2
o
Hx L
dy w L
dx F
)cos( max
maxHF
T
2
4 222
max
LwFT
oH
86
Solution
For a differential segment of cable length ds,
Determine total length by integrating,
Integrating yields,
dxxL
hds
L
2/
0
2
2
812
L
h
h
L
L
hL 4sinh
4
41
2
1
2
dxdx
dydydxds
2
221
87
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
88
7.4 Cables Cable Subjected to its Own Weight
• Apply equilibrium equations to the force system
• Replace dy/dx by ds/dx for direct integration
1cos sin ( ) ( )H
H
dyT F T w s ds w s ds
dx F
2
2 2 1dy ds
ds dx dydx dx
2/1
2
2)(
11
dssw
Fdx
ds
H
2/1
2
2)(
11 dssw
F
dsx
H
cos ( )cos( ) 0
sin ( )sin( ) 0
1( ) cos sin 02
T T T
T w s T T
w s s T y T x
1cos constant
1( sin ) 0
sin 1tan
cos
H
H
T Fs
dT w
s ds
dywds
dx F
Therefore
89
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)
2/1
22/11 dswF
dsx
oH
2/12
1
2/11 CswF
dsx
oH
Substitute
Perform second integration
or
1/1 CswFu oH dsFwdu Ho )/(
2
1sinh Cuw
Fx
o
H
21
1 1sinh CCsw
Fw
Fx o
Ho
H
90
Solution
Evaluate constants
dy/dx = 0 at s = 0, then C1 = 0
s=0 at x=0, then C2=0
solve for s
dswFdx
dyo
H
1
1
1Csw
Fdx
dyo
H
x
F
w
w
Fs
H
o
o
H sinh
21
1 1sinh CCsw
Fw
Fx o
Ho
H
x
F
w
dx
dy
H
osinh 3cosh CxF
w
w
Fy
H
o
o
H
Boundary Condition y = 0 at x = 0
For deflection curve,
This equations defines a catenary curve.
o
H
w
FC 3
1cosh x
F
w
w
Fy
H
o
o
H
91
Solution Boundary Condition y = h at x = L/2
Since wo = 5N/m, h = 6m and L = 20m,
1cosh x
F
w
w
Fh
H
o
o
H
1
50cosh
/56
H
H
F
N
mN
Fm NFH 9.45
For deflection curve,
x = 10m, for half length of the cable
Hence
Maximum tension occurs when is maximum at s = 12.1m
9.19 cosh 0.109 1y x m
max max
12.1
max
max
5 / 12.1tan 1.32 , 52.8
45.9
45.975.9
cos cos52.8
s m
H
N m mdy
dx N
F NT N
45.9 5 /
sinh 10 12.12 5 / 45.9
N mm m
N m N
24.2m
92
Chapter 8 Friction
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
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
s
sss
N
N
N
F
111 tantantan
1.10 – 1.70 Aluminum on aluminum
0.30 – 0.60 Leather on metal
0.20 – 0.50 Leather on wood
0.30 – 0.70 Wood on wood
0.03 – 0.05 Metal on ice
Coefficient of Static Friction μs Contact Materials
Typical Values of μs
94
Chapter 9 Center of Gravity and Centroid
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
9.1 Center of Gravity and Center of Mass Mass Center
• Consider a particle having weight of dW
m
mzz
m
myy
m
mxx
~,
~;
~
dW
dWzz
dW
dWyy
dW
dWxx
~
;
~
;
~
95
Example 9.1 Locate the centroid of the rod bent into the shape of a parabolic arc.
Solution For differential length of the element dL
Since x = y2 and then dx/dy = 2y
The centroid is located at
2
2 2
2
1
2 1
dxdL dx dy dy
dy
dL y dy
1 12 2 2
0 0
1 12 2
0 0
12
0
12
0
4 1 4 1 0.60630.410
1.4794 1 4 1
4 1 0.84840.574
1.4794 1
L
L
L
L
xdLx y dy y y dy
x mdL y dy y dy
ydLy y dy
y mdL y dy
96
9.2 Composite Bodies
Example 9.10
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
mmA
Ayy
mmA
Axx
22.15.11
14~
348.05.11
4~
97
9.5 Fluid Pressure • Magnitude of depends on the specific weight or mass density
of the fluid and the depth z of the point from the fluid surface
Valid for incompressible fluids
p z gz
Flat Plate of Constant Width
• 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 at depth z1 and at depth z2
• Magnitude of the resultant force FR = volume of this loading diagram
Curved Plate of Constant Width
1 1 1w bp brz 2 2 2w bp brz
98
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
For intensities of the load at A and B,
This force acts through the centroid
of the area,
measured upwards from B
mkNkPambw
mkNkPambw
BB
AA
/58.73)05.49)(5.1(
/43.29)62.19)(5.1(
kPamsmmkggz
kPamsmmkggz
BwB
AwA
05.49)5)(/81.9)(/1000(
62.19)2)(/81.9)(/1000(
23
23
1(3)(29.4 73.6) 154.5
2RF N
mh 29.1)3(58.7343.29
58.73)43.29(2
3
1
99
Chapter 10 Moments of Inertia
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
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
100
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
• moments of inertia of the differential plane area dA
• Formulate the second moment of dA about z axis
where r is perpendicular from the pole (z axis) to the element dA
• Polar moment of inertia for entire area,
2 2
2 2
x y
x yA A
dI y dA dI x dA
I y dA I x dA
dArdJO
2
2z x y
AJ r dA I I
101
10.2 Parallel Axis Theorem for an Area
• Determine the moment of inertia of area about a corresponding parallel.
• For moment of inertia of dA about x axis
• For entire area
• For polar moment of inertia
Ay
Ay
A
Ayx
yx
dAddAyddAy
dAdyI
dAdydI
22
2
2
'2'
'
'
2 2
2
' 0; 0
and x x y y y x
z C
y dA y dA y
I I Ad I I Ad
J J Ad
102
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
yx z
x y z
II Jk k k
A A A
103
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
By applying parallel axis theorem,
For polar moment of inertia about point C,
32/
2/
22/
2/
22
12
1')'('' bhdyybdyydAyI
h
h
h
hAx
3
2
32
3
1
212
1bh
hbhbhAdII xxb
3 2 2' '
1 1 and ( )
12 12y C x yI hb J I I bh h b 104
10.5 Product of Inertia for an Area • Moment of inertia for an area is different for every axis
• Product of inertia for an element of area dA located at a
point (x, y) is defined as
xy
xyA
dI xydA
I xydA
Parallel Axis Theorem
• For the product of inertia of dA with respect to the x and y axes
• For the entire area,
• Forth integral represent the total area A,
dAdydxdIA
yxxy ''
Ayx
Ay
A Ax
Ayxxy
dAdddAxddAyddAyx
dAdydxdI
''''
''
yxyxxy dAdII ''
105
10.6 Moments of Inertia for an Area about Inclined Axes
T
2 2
,
cos sin,
sin cos
cos sin,
sin cos
I dA, I dA, I dA
Consider moment of inertia matrix
I I
I I
xx yy xy
xx xy
xy yy
x ur r
y v
x ur r
y v
u xr r
v y
y x xy
A
A
I =
T T
I I
I I
Then and
uu uv
uv vv
I
I A I A I A I A
x
y
u
v
θ 0
r
Principal Moments of Inertia
• Or find the eigenvalue of I or I’ matrix
106
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,
494949 1000.31060.51090.2 mmImmImmI zyx
Solution
2
2.9 3
3 5.6
( -2.9)( -5.6)+9 =0
8.5 7.24 0
eigenvalue of ( ) 0.96 or 7.54
0.96 2.9 3 1.94 3 3eigenvector , 0,
3 0.96 2.9 3 1.94 1.94
7.54 2.9 3 4.6
3 7.54 5.6
x x x x x
x
I
I
I
4 3 30,
3 1.94 4.64x x
107
10.7 Mohr’s Circle for Moments of Inertia • The circle constructed is known as a Mohr’s circle with radius
and center at (a, 0) where
2
2
2xy
yxI
IIR
2/yx IIa
108
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 the
axis for Imax
109
10.8 Mass Moment of Inertia • Mass moment of inertia is defined as the integral of the second moment about an axis of all
