moments and rigid body equilibrium - faculty...
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S2017abnMoments 1
Lecture 5
Elements of Architectural Structures
ARCH 614
ELEMENTS OF ARCHITECTURAL STRUCTURES:
FORM, BEHAVIOR, AND DESIGN
ARCH 614
DR. ANNE NICHOLS
SPRING 2017
five
moments and
rigid body equilibrium
lecture
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moments
• forces have the tendency to make a body rotate about an axis
– same translation but different rotationhttp://www.physics.umd.edu
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moments
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moments
• a force acting at a different point causes
a different moment:
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moments
• defined by magnitude and direction
• units: Nm, kft
• direction:
+ cw (!)
- ccw
• value found from Fand distance
• d also called “lever” or “moment” arm
F
A
C
B
dFM
d
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moments
• with same F:
21 dFMdFM AA (bigger)
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moments
• additive with sign convention
• can still move the force
along the line of action
=
-MA = Fd
MB = Fd
d
F
A
Bd
MA = Fd
MB = Fd
d
F
A
B
d
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moments
• Varignon’s Theorem
– resolve a force into components at a point
and finding perpendicular distances
– calculate sum of moments
– equivalent to original moment
• makes life easier!
– geometry
– when component runs through point, d=0
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moments of a Force
• moments of a force
– introduced in Physics as
“Torque Acting on a Particle”
– and used to satisfy rotational equilibrium
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Lecture 5
Elements of Architectural Structures
ARCH 614
Physics and Moments of a Force
• my Physics book (right hand rule):
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Lecture 5
Elements of Architectural Structures
ARCH 614
• 2 forces
– same size
– opposite direction
– distance d apart
– cw or ccw
– not dependant on point of application
Moment Couples
d
F
A
d1
d2
F
dFM
21 dFdFM
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moment Couples
• equivalent couples
– same magnitude and direction
– F & d may be different
100 mm
300 N
300 N
150 mm200 N
200 N
250 mm
120 N120 N
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moment Couples
• added just like moments caused by one
force
• can replace two couples with a single
couple
+=
100 mm
300 N
300 N
150 mm200 N
200 N
250 mm
240 N240 N
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moment Couples
• moment couples in structures
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Lecture 5
Elements of Architectural Structures
ARCH 614
Equivalent Force Systems
• two forces at a point is equivalent to the resultant at a point
• resultant is equivalent to two components at a point
• resultant of equal & opposite forces at a point is zero
• put equal & opposite forces at a point (sum to 0)
• transmission of a force along action line
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Lecture 5
Elements of Architectural Structures
ARCH 614
• single force causing a moment can be
replaced by the same force at a
different point by providing the moment
that force caused
• moments are shown as arched arrows
Force-Moment Systems
-F
FF
d
A
AF
A
A
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Lecture 5
Elements of Architectural Structures
ARCH 614
Force-Moment Systems
• a force-moment pair can be replaced by
a force at another point causing the
original moment
F
d
-F
F
A
AF
A
A
F M=Fd
A
A
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Lecture 5
Elements of Architectural Structures
ARCH 614
Parallel Force Systems
• forces are in the same direction
• can find resultant force
• need to find location for equivalent
moments
R=A+B
C Dx
AB
ab
aA
bB
xBA )(
C D
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Lecture 5
Elements of Architectural Structures
ARCH 614
Equilibrium
• rigid body
– doesn’t deform
– coplanar force systems
• static:
A
CB
0 xx FR
0 yy FR
0MM
( H)
( V)
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Lecture 5
Elements of Architectural Structures
ARCH 614
Free Body Diagram
• FBD (sketch)
• tool to see all forces on a body or a
point including
– external forces
– weights
– force reactions
– external moments
– moment reactions
– internal forces
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Lecture 5
Elements of Architectural Structures
ARCH 614
Free Body Diagram
• determine body
• FREE it from:
– ground
– supports & connections
• draw all external forcesacting ON the body
– reactions
– applied forces
– gravity
mg
+ weight
100 lb
100 lb
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Lecture 5
Elements of Architectural Structures
ARCH 614
Free Body Diagram• sketch FBD with relevant geometry
• resolve each force into components– known & unknown angles – name them
– known & unknown forces – name them
– known & unknown moments – name them
• are any forces related to other forces?
• for the unknowns
• write only as many equilibrium equations as needed
• solve up to 3 equations
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Lecture 5
Elements of Architectural Structures
ARCH 614
Free Body Diagram
• solve equations
– most times 1 unknown easily solved
– plug into other equation(s)
• common to have unknowns of
– force magnitudes
– force angles
– moment magnitudes
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Lecture 5
Elements of Architectural Structures
ARCH 614
Reactions on Rigid Bodies
• result of applying force
• unknown size
• connection or support type
– known direction
– related to motion prevented
no vertical motion no translation
no translation
no rotation
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Lecture 5
Elements of Architectural Structures
ARCH 614
Supports and Connections
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Lecture 5
Elements of Architectural Structures
ARCH 614
Supports and Connections
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Lecture 5
Elements of Architectural Structures
ARCH 614
Moment Equations
• sum moments at intersection where the
most forces intersect
• multiple moment equations may not be
useful
• combos:
0Fx 0Fy 0M
1
0F 0M1 0M
2
0M1 0M2 0M
3
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Lecture 5
Elements of Architectural Structures
ARCH 614
Concentrated Loads
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Lecture 5
Elements of Architectural Structures
ARCH 614
Distributed Loads
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Lecture 5
Elements of Architectural Structures
ARCH 614
Beam Supports
• statically determinate
• statically indeterminate
L L L
simply supported
(most common)
overhang cantilever
L
continuous
(most common case when L1=L2)
L
L L
Propped Restrained
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Lecture 5
Elements of Architectural Structures
ARCH 614
Equivalent Force Systems
• replace forces by resultant
• place resultant where M = 0
• using calculus and area centroids
dx
w(x)
x
L
loadingloading
L
0AdAwdxW
dx
y
x
elx
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Lecture 5
Elements of Architectural Structures
ARCH 614
Load Areas
• area is width x “height” of load
• w is load per unit length
• W is total load
x
x/2
W
x/2
x
2x/3
W/2
x/3
x
x/2
W
x/6 x/3
W/2
0
Wxw w w 22
Wxw
w
2w
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Lecture 5
Elements of Architectural Structures
ARCH 614
Method of Sections
• relies on internal forces being in
equilibrium on a section
• cut to expose 3 or less members
• coplanar forces M = 0 too
A B
C
P
F
E
D
P
.
A
By
AC
AB
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Lecture 5
Elements of Architectural Structures
ARCH 614
Method of Sections
• joints on or off the section are good to
sum moments
• quick for few members
• not always obvious where to cut or sum
A B
C
P
F
E
D
P
. A
By
AC
AB
B
.