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9966001305 Aeronautical Structures and Vibrations
Pablo Morata, [email protected]
February 17, 2014
Structural Analysis.
Beam Theory (I).
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Internal Forces
A deformable body is in equilibrium with the
externally applied loads (free-body diagram)
Cut by a section will create 2 solids each of one has
to be in equilibrium (external loads and
internal loads)Internal loads can be reduce to a force and a
moment applied at some point O
These internal forces have components in a given
coordinate system: consider the cut plane and
its normal vector to define that coordinate
system
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Internal Forces in a prismatic member
We define a prismatic member as the solid(elastic) created by the translation along a
curve known as directrix of a cross section
The cross section may vary along the directrix,
however in most of our applications will not
change
The directrix curvature should change smoothly
for this theory to be applicable. In most of
our applications the directrix will be a
straight line
We choose a coordinate system whosexaxis isalong directrix andyzplane is the cross
section
y
x
z
directrixCross
section
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Internal Forces in a prismatic member
At any section of the prismatic memberwe can reduce the internal forces to
a force and a moment:
Axial force, Shear forces
Torsional moment (torque, twisting
moment), bending moments
Most of our problems deal with prismatic members in
2D, that means only QyandMzare considered
zy
zy
MMT
QQN
,,
,,
y
x
z
yQ
N
zQ
yM
T
zM yQzM
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Sign ConventionsFor the external loads we consider them positive along the
positive sense of the coordinate axes defined.
For the internal forces the sign convention is the following:
Positive sense according to positive sense of axes defined in the
left hand side of the section. this means:
Mzis positive when it creates tension in the lower fiberand compression in the upper fiber
Qyis positive when it acts counterclockwise against the
material
yQyQ
z z
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Support reactionsBoundary conditions
The surface forces that develop at the supports or points of
contact between bodies are called reactions.
As a general rule, if the support prevents translationin a given
direction, then aforcemust be developed on the member in
that direction. Likewise, if rotationis prevented, a couple
momentmust be exerted on the member.This is what we call boundary conditions. For the usual 2D cases:
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Externally applied loadsBoundary conditions
Surface forces are caused by the direct contact of one body with the surface of
another.In all cases these forces are distributed over the area of contact between the
bodies.
If this area is small in comparison with the total surface area of the body, then
the surface force can be idealized as a single concentrated force, which is
applied to a point on the body,P. For example, the force of the ground on the wheels of a bicycle canbe considered as a concentrated force.If the surface loading is applied along a narrow strip of area, the loading can
be idealized as a linear distributed load, q
The resultant force of q is equivalent to the
area under the distributed loading curve, and
this resultant acts through the centroid Corgeometric center of this area. The loading along thelength of a beam is a typical example of where this idealization is often
applied.
A body force is developed when one body exerts a
force on another body without direct physical
contact between the bodies. Examples include the effectscaused by the earths gravitation or its electromagnetic field
q
RF
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Externally applied loadsBeams
Types of loading)(xq
)(xm
P
M
can be constant, linear or
any shape
Point force
Point Moment
10)( xxxxfq
Nm/mN][][
N/m]/[][][
Fm
LFq
Distributed force or moment Units:
Units:
Units:
N][][ FP
mN][][][ LFM
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Analysis of Statically determinate beams
First step is to determine the reactions. If the beam is statically determinate
there will be as many equations (3 equations in 2D analysis) as unknowns(reactions at supports)
P
l
a
Ox
y
SIMPLY SUPPORTED BEAM
Reactions in blue
ByROyR
OxR
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Analysis of Statically determinate beamsReactions
We consider the free-body diagram to set the equilibrium equations
P
l
a
Ox
y
EQUILIBRIUM EQUATIONS: 3EQ., 3 UNKNOWNSSTATICALLY
DETERMINATE
0:0 ByOyy RRPF
ByROyR
OxR
0:0 Oxx RF
0:0 lRPaM ByO
l
aPRBy
l
bP
l
alP
l
aPROy
)1(
b
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Analysis of Statically determinate beamsShear Forces and Bending Moments
We consider the free-body diagram to set the equilibrium equations after
applying the method of sections cutting at a generic location x
P
la
Ox
y
EQUILIBRIUM EQUATIONS APPLIED TO ANY OF THE 2 PORTIONS.
THE STANDARD IS TO KEEP LEFT HAND SIDE WHERE WE
DEFINED THE POSITIVE SIGN CONVENTION. Moments are taken
about the cut section, so for x>a
0:0 QRPF Oyy
ByR
OyR
OxR
0:0 Oxx RF
0)(:0 xRaxPMM Oyxat
b
z
M
yQ yQ
z
Mx
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Analysis of Statically determinate beamsShear Forces and Bending Moments Diagrams
Now we have obtained M(x)and Q(x)we can draw these functions of x. This is
what we call shear forces and bending moment diagrams x
x
y
EQUILIBRIUM EQUATIONS APPLIED TO ANY OF THE 2 PORTIONS.
THE STANDARD IS TO KEEP LEFT HAND SIDE WHERE WE
DEFINED THE POSITIVE SIGN CONVENTION. Moments are taken
about the cut section
l
aPQ
xl
b
PaxPM
)(
a b
l
bPQ
xl
b
PM
ax ax
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021100:0 ByO RM
Analysis of Statically determinate beamsShear Forces and Bending Moments Diagrams
Example withP=-100 kN, a=1 m, b=1 m,l= 2 m
x
y
1 1
50OyR 50ByR
100
0100:0 ByOyy RRF
50ByR 5050100 OyR
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Analysis of Statically determinate beamsShear Forces and Bending Moments Diagrams
Example withP=-100 kN, a=1 m, b=1 m,l= 2 m
x
y
50 Q
1 1
50 Q 1x1x
50
50
050100:0 QFy050:0 QFy
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Analysis of Statically determinate beamsShear Forces and Bending Moments Diagrams
Example withP=-100 kN, a=1 m, b=1 m,l= 2 m
x
y
xM 50100
1 1
xM 50 1x1x
50
050)1(100:0 xxMM xat050:0 xMM xat
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Analysis of Statically determinate beams
First step is to determine the reactions. If the beam is statically determinate
there will be as many equations (3 equations in 2D analysis) as unknowns(reactions at supports)
P
l
a
Ox
y
CLAMPED BEAM
Reactions in blue
OyR
OxR
OzM
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Analysis of Statically determinate beamsReactions, clamped beam
We consider the free-body diagram to set the equilibrium equations
P
l
a
Ox
y
EQUILIBRIUM EQUATIONS: 3EQ., 3 UNKNOWNSSTATICALLY
DETERMINATE
0:0 Oyy RPF
OyR
OxR
0:0 Oxx RF
0:0 OzO MPaM
PROy PaMOz
OzM
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Analysis of Statically determinate beamsShear Forces and Bending Moments, clamped beam
We consider the free-body diagram to set the equilibrium equations after
applying the method of sections cutting at a generic location x
P
l
a
Ox
y
EQUILIBRIUM EQUATIONS APPLIED TO ANY OF THE 2 PORTIONS.
THE STANDARD IS TO KEEP LEFT HAND SIDE WHERE WE
DEFINED THE POSITIVE SIGN CONVENTION. Moments are taken
about the cut section, so for xa shear force and bending
moment are zero):
0:0 QRF Oyy
OyR
OxR
0:0 Oxx RF
0:0 xRMMM OyOzxat
z
M
yQ yQ
z
Mx
OzM
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Analysis of Statically determinate beamsShear Forces and Bending Moments Diagrams, clamped beam
Now we have obtained M(x)and Q(x)we can draw these functions of x. This is
what we call shear forces and bending moment diagrams x
x
y
EQUILIBRIUM EQUATIONS APPLIED TO ANY OF THE 2 PORTIONS.
THE STANDARD IS TO KEEP LEFT HAND SIDE WHERE WE
DEFINED THE POSITIVE SIGN CONVENTION. Moments are taken
about the cut section
0 Q
0
M
a al
PQ
)( axPM
ax ax
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Relationships between Loads, Shear
Forces and Bending Moments
We consider a slice of a beam of length dx. We will see what is the relationshipbetween q, Q(x) and M(x).
)(xq
yy dQxQ )(
zz dMxM )(
)(xQy
)(xMz
Equilibrium (vertical forces):
0)())(()( dxxqdQxQxQ yyy
0)( dxxqdQy
)(xqdx
dQy
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Relationships between Loads, Shear
Forces and Bending Moments
We consider a slice of a beam of length dx. We will see what is the relationshipbetween q, Q(x) and M(x).
)(xq
yy dQxQ )(
zz dMxM )(
)(xQy
)(xMz
Equilibrium (of moments):
0))(()( kdxqdxdxQdMxMxM yzzz
0 kdxqdxdxQdM yz
kis a factor that identifies
the point of application
of the resultant of q(x)distribution. If q(x) is
constant k=1/2
As is a second order
differential we neglect it compared
to other terms resulting
yz Q
dx
dM
0 kdxqdx
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Relationships between Loads, Shear
Forces and Bending Moments
The signs in this formulas are due to our sign convention, a different signconvention would give different signs so be careful when using them.
yz Q
dx
dM)(xq
dx
dQy
These relationships are very helpful in drawing the shear force and bending
moment diagram:
because if the shear force iszero then the bending moment diagram
curve has a maximum or minimum (dM/dx=0) also the slopeof the bending moment diagram has the opposite sign(due
to sign convention) of the shear force diagram (dM/dx=-Q)
The curvature of the bending moment curve (convex or concave) is given
by the sign of qsince d2M/dx2=-dQ/dx=q (second derivative)
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Assignment #4aDraw the shear and bending moment diagrams of the following
beams
4a.1
4a.2
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Bibliography
Hibbeler, Mechanics of Materials
Ortiz Berrocal, Resistencia de Materiales
Timoshenko, Strength of Materials
Available at UEM Libraryhttp://biblioteca.uem.es/
http://biblioteca.uem.es/http://biblioteca.uem.es/Top Related