lecture 5-6 beam mechanics of materials laboratory sec. 3-4 nathan sniadecki
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Mechanics of Materials Lab. Lecture 5-6 Beam Mechanics of Materials Laboratory Sec. 3-4 Nathan Sniadecki University of Washington. $100. Answer: What is the moment of inertia (Second Moment of Area) with respect to the x axis. $100. - PowerPoint PPT PresentationTRANSCRIPT
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Lecture 5-6Beam
Mechanics of Materials Laboratory Sec. 3-4
Nathan SniadeckiUniversity of Washington
Mechanics of Materials Lab
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$100
2x
A
I y dA
Answer: Answer: What is the moment of inertia (Second What is the moment of inertia (Second Moment of Area) with respect to the Moment of Area) with respect to the xx axis axis
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$100
2y
A
I x dA
Answer: Answer: What is the moment of inertia (second moment What is the moment of inertia (second moment of area) with respect to the of area) with respect to the yy axis axis
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$200
3112
I bh
Answer: Answer: What is the moment of inertia for a rectangleWhat is the moment of inertia for a rectangle
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$200
414
I r
Answer: Answer: What is the moment of inertia for a circleWhat is the moment of inertia for a circle
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$300
• The straight line that defines a surface where x and x are zero
Answer: Answer: What is the neutral axisWhat is the neutral axis
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$300
yx
y
M zI
Answer: Answer: What is the bending stress in the What is the bending stress in the xx-direction at a -direction at a distance distance yy from the origin of the coordinate from the origin of the coordinate system due to the loading of a couple vectorsystem due to the loading of a couple vector M Mxx acting in acting in xx-direction-direction
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• The point on the stress-strain curve where the material no longer deforms elastically, but also plastically.
Answer: Answer: What is the proportional limitWhat is the proportional limit
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$400
• The theorem that expresses that the moment of inertia Ix of an area with respect to an arbitrary x axis is equal to the moment of inertia Ixc with respect to the centroidal x axis, plus the product Ad2 of the area A and the square of the distance d between the two axis?
Answer: Answer: What is the Parallel Axis Theorem What is the Parallel Axis Theorem IIxx = I = Ixcxc + Ad + Ad22
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• The principle that states that the effect of a given combined loading on a structure can be obtained by determining the effects of each load separately and then combining the results obtained together as long as 1) each effect is linearly related to the load that produces it and 2) the deformation resulting from any given load is small and does not affect the conditions of application of the other loads
Answer: Answer: What is the Principle of SuperpositionWhat is the Principle of Superposition
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• The principle that states that except in the immediate vicinity of the point of loading, the stress distribution may be assumed to be independent of the actual mode of loading, i.e. for axial loading, at a distance equal to or greater than the width of a member, the distribution of stress across a given section is the same.
Answer: Answer: What is Saint-Venant’s PrincipleWhat is Saint-Venant’s Principle
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FINAL JEOPARDYFINAL JEOPARDY
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FINAL JEOPARDY
arctan y z
z y
M IM I
Answer: Answer: What is the What is the angleangle of the neutral axis for an of the neutral axis for an asymmetrically loaded beamasymmetrically loaded beam
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Inclined Load
Notice the sign convention: positive Mz compress upper part, negative stress; positive My extend front part, positive stress!
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Inclined Load
z
z
y
yx I
yMI
zM
Stress
Neutral axis
0z
z
y
yx I
yMI
zM
yz
zy
IMIM
zy tan
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Example
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Stress Distribution
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The centroid of the area A is defined as the point C with coordinates (yc, zc) which satisfies
Asymmetrical Beam
If the origin of y and z axes is placed at centroid C
(orientation is arbitrary.)
cA
ydA Ay cA
zdA Az
0A
ydA 0A
zdA
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( is inverse of )
x
x
x x
L y
L L L y yL y y
Ly
E yE
• Consider of beam segment AB of length L• After deformation, length of neutral surface
DE remains L, but JK becomes
Pure Bending
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Asymmetric Beam
zy
yxz
EI
dAyEydAM
2
yzy
yxy
EI
yzdAEzdAM
If z is a principal axis (symmetry), the product of inertia Iyz is zero My = 0, bending in x-y plane, analogous to a symmetric beam
z
yz
z
y
II
MM
When z axis is the neutral axis;
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Asymmetric Beam
yzz
zxz
EIyzdAEydAM
yz
zxy
EI
dAzEzdAM
2
y
yz
y
zII
MM
If y is a principal axis, the product of inertia Iyz is zero Mz = 0, bending in x-z plane, analogous to a symmetric beam
When y axis is the neutral axis;
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Asymmetric Beam
• When an asymmetric beam is in pure bending, the plane in which the bending moments acts is perpendicular to the neutral surface if and only if (iff) the y and z axes are principle centroidal axes and the bending moments act in one of the two principle planes. In such a case, the principle plane in which bending moment acts becomes the plane of bending and the usual bending theory is valid
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Analysis of Asymmetric Beam• Locating the centroid, and constructing a set of
principal axes• Resolving bending moment into My and Mz
• Superposition
z
z
y
yx I
yMI
zM
tantany
z
yz
zy
II
IMIM
zy
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Principle Axes
dAyIx2 dAxI y
2
xydAIxy
1 1, cos2 sin 22 2
x y x yx y xy
I I I II I I
2cos2sin211 xy
yxyx I
III
yx
xyp II
I
2
2tan
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Analysis of Asymmetric Beam
A channel section C 10 15.3
M = 15 kips-inIy=2.28 in4, Iz=67.4 in4
Location of Point C c=0.634 in
Location of Point A yA=5.00 in zA=-2.6+0.634=-1.966 in
Calculate bending stress
Locate neutral axis
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Analysis of Asymmetric Beam
ink605.2sin MM y
ink77.14cos MM z
psi3340z
z
y
yx I
yMI
zM
o
y
zII
zy
1.79
212.5tantan
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Normal Stress in Beam
y y
E y
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Curved Beams
• What if the beam is already ‘bent’?
• Where will the beam likely fail?
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Bending Stress for Curved Beam
• #1: Neutral surface remains constant: • #2: Deformation at JK:
R R
L r rR y R y
y y
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Bending Stress for Curved Beam
• #3: Strain:
• #4: Stress:
x
yR y
x
E yR y
E R rr
r R yr R y
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Bending Stress for Curved Beam
• #5: Neutral Axis:
• #6 Centroid:
0 1x
AdA RdA
r
1r rdA
A
e r R • #7: N.A. Location:
xy dA M
E MA r R
Since > 0 for M > 0R r
Aside: R = rn
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Location of N.A. in Curved Beam
• Cross-sectional dimensions define neutral axis location for a curve beam about C
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Curved Beams
Positive M decreases curvature
i
ii Aer
Mc
o
oo Aer
Mc
1/
( )
n
n
r A dAr
MyAe r y
Neutral axis is no longer the centroidal axis
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Curved Beam
1/
3.64in
n
o
i
bhr A dAbr drr
hrlnr
( )n
F MyA Ae r y
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Curved Beams
rr
IMsAr
Ie
yrAeMy
c
c
n
)(
rrs c
Curvature is small, e is small, rn is close to rc
Recover to straight beam
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Curved Beam
sbsrbrc
srsb
sbsr
s
e
c
c
Pay attention to the sign of s
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Curved Beam
srsb
sbsr
s
e
c
c
222 sRb
Pay attention to the sign of s
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• Read Mechanics of Materials Lab Sec. 4
• 4.26(e), 4.72 posted online
Assignment