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COLLEGE OF ENGINEERING DEPARTMENT OF CIVIL & ARCHITECTURAL
ENGINEERING
CVEN 220 : ANALYSIS OF STRUCTURES
WAEL I. ALNAHHAL, Ph. D., P. Eng
Spring, 2015
Ch.8: DEFLECTION PART 1
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Deflections
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IntroductionIntroduction
Calculation of deflections is an important part
ofCalculationofdeflectionsisanimportantpartofstructuralanalysis
Excessivebeamdeflectioncanbeseenasamodeoffailure.
Extensiveglassbreakageintallbuildingscanbeattributedtoexcessivedeflections
Largedeflectionsinbuildingsareunsightly(andunnerving)and can
cause cracks in ceilings and
wallsandcancausecracksinceilingsandwalls.
Deflectionsarelimitedtopreventundesirablevibrations
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Beam DeflectionBeamDeflection
Bending changes
theBendingchangestheinitiallystraightlongitudinalaxisofthebeamintoacurvethatiscalledthe
flDeflectionCurve orElasticCurve
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Beam DeflectionBeamDeflection
To determine the deflection curve:Todeterminethedeflectioncurve:
Drawshearandmomentdiagramforthebeam
Directlyunderthemomentdiagramdrawalineforthebeamandlabelallsupports
Atthesupportsdisplacementiszeroh h i i h d fl i i
Wherethemomentisnegative,thedeflectioncurveis
concavedownward. Where the moment is positive the deflection
curve
isWherethemomentispositivethedeflectioncurveisconcaveupward
WherethetwocurvemeetistheInflectionPoint
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DeflectedShape
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Example1Drawthedeflectedshapeforeachofthebeamsshown
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Example2Drawthedeflectedshapeforeachoftheframesshown
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DoubleIntegrationMethod
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ElasticBeamTheory
Consider a differential element
Consideradifferentialelementofabeamsubjectedtopurebending.g
Theradiusofcurvature
ismeasuredfromthecenterofcurvaturetotheneutralaxis
SincetheNAisunstretched,thedx=d
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ElasticBeam TheoryElastic BeamTheory
Applying Hookes law and the Flexure formula we ApplyingHooke
slawandtheFlexureformula,weobtain:
M1EI
=
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ElasticBeamTheory The product EI is referred to as the flexural
rigidity TheproductEIisreferredtoastheflexuralrigidity. Sincedx
=d,then
M )(SlopedxEIMd =
In most calculus booksIn most calculus books
[ ]dxvd= 3
22 /1
InmostcalculusbooksInmostcalculusbooks
( )[ ]l titdxvdM
dxdv+22
23
2
)(//1
( )[ ] solutionexactdxdvEI +=2
23
2)(
/1
EIM
dxvd =2
2
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The Double Integration
MethodTheDoubleIntegrationMethodRelateMomentstoDeflections
Md 2
EIM
dxvd =2
2
( )== dxEI xMddvx )()(DoNotIntegrationConstants xEIdx )(
( ) xMUseBoundaryConditionstoEvaluateIntegrationConstants( )=
2)()( dxxEI xMxv Constants
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AssumptionsandLimitations
Deflectionscausedbyshearingactionnegligiblysmallcomparedtobending
Deflectionsaresmallcomparedtothecrosssectionaldimensionsofthebeam
Allportionsofthebeamareactingintheelasticrange
Beamisstraightpriortotheapplicationofloads
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ExamplesLx
y
PxPLM +=x
P
PL
P
PxPLM +=
MdxydEI =2
2
yd 2@x PxPL
dxydEI +=2
Integratingonce 1
2
2cxPPLx
dxdyEI ++=
2dx
@x=0 ( ) ( ) ( ) 020000 11
2
=++== ccPPLEIdxdy
32PLIntegratingtwice 2
32
62cxPPLxEIy ++=
@x=0 ( ) ( ) ( ) 00000 223
2 =++== ccPPLEIy3
( ) ( )62 22
y
62
32 xPPLxEIy +=@x=Ly=ymax EI
PL3
3
max =
62y
EIPLyPLLPLPLEIy3662
3
max
332
max ==+=
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y( )2xLWM =W ( )2
xLM =
MdxydEI =2
2
2 WydL
xx
WL2
2WL
@x ( )22 2 xLW
dxydEI =WL2
Integratingonce( )
1
3
32cxLW
dxdyEI +=
@x=0 ( ) ( )63
02
003
11
3 WLccLWEIdxdy =+==
( )66
33 WLxLW
dxdyEI =
66dx
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Integrating twice( ) 34 cxWLxLWEIy +=Integratingtwice 2646 cxEIy
+=
@x=0 ( ) ( ) ( )24
064
06
004
22
34 WLccWLLWEIy =+==
( ) 434 WLxWLxLWEIy +=
( ) ( )24646 22
y
( )24624
xxLEIy +=
Max occurs @ x = LMax.occurs@x=L
EIWLyWLWLLWEIy88246
4
max
444
max ==+=
EIWL8
4
max = EI8
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Exampley x
L
x
WL WL2 2
22xWxxWLM =
22 xWLyd222xWxWL
dxydEI =
Integrating 1
32
3222cxWxWL
ddyEI +=g g 13222dx
Sincethebeamissymmetric 02
@ ==dxdyLx
32 LL
( ) +
== 132
222
20
2@ c
LW
LWLEILx
24
3
1WLc =
3
2464
332 WLxWxWL
dxdyEI =
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Integrating2
343
cxWLxWxWLEIy +=g g 2244634 cxEIy +
@x=0 y=0 ( ) ( ) ( ) ( ) 2343
0000 cWLWWLEI += 02 = c( ) ( ) 2244634 2
xWLxWxWLEIy3
43 = xxxEIy242412
=
Max occurs @ x = L /25 4WLEIy =Max.occurs@x=L/2384max
EIy =
WL5 4EI384max
=
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Exampley x P
L/2
x
P PL/2
f2 LPyd
xPMLx22
0for =
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Integrating2
23
cxPLxPEIy +=g g 21634 cxEIy +
@x=0 y=0 ( ) ( ) ( ) 223
000 cPLPEI += 02 = c( ) ( ) 21634 2
xPLxPEIy2
3 = xxEIy1612
Max occurs @ x = L /23PLEIy =Max.occurs@x=L/2
48maxEIy =
PL3EI48max
=
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Example
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Example5
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MomentAreaTheorems
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MomentAreaTheoremsTheorem 1: The change in slope between any two
points onthe elastic curve equal to the area of the bending
momentdiagram between these two points, divided by the product
EI.
2
2
d v M dvdx EI dxd M M
= =
B
d M Md dxdx EI EI
M
= = B
B AA
M dxEI
=
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dt xdM=
B B
Md dxEI
M M
= B A
A A
M Mt x dx x dxEI EI
= =
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MomentAreaTheoremsTheorem 2: The vertical distance of point A on
a elasticcurve from the tangent drawn to the curve at B is equal
tothe moment of the area under the M/EI diagram betweentwo points
(A and B) about point A .
B
A BA
Mt x dxEI
= A
B
A BA
Mt x dxEI
= A
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Example1
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Example2
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Example3
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Example4
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Example5
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Example6
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Example7
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M/EI
30/EI20/EI
( )20 1 10 22 1 2 2t = + ( )/3
2 1 2 22 3
53.33 0.00741
C Bt EI EI
kN m rad
= + = =
EI
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Another Solution
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Conjugate-Beam Method
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Conjugate-Beam Method
2
2
dV d Mw wdx dx
= =2
2
dx dxd M d v Mdx EI dx EI = =
dx EI dx EI
Integrating
V wdx M wdx dx
M M
= =
M Mdx v dx dxEI EI = =
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ts
p
p
o
r
t
m
S
u
p
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B
e
a
m
u
g
a
t
e
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C
o
n
j
u
C
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Example 1Example 1Find the Max. deflection Take E=200Gpa,
I=60(106)
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EIVBB
5.562' ==
EIEIM
EI
BB5.14062)25(5.562'
===
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Example 2Example 2Find the deflection at Point C
C
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EIEIEIMCC
162)3(63)1(27'===
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Example 3Example 3Find the deflection at Point D
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3600EI
360
EI720 EI
MDD3600
' ==
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Example 4Example 4Find the Rotation at A
10 ft
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3.33EIA
3.33=
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Example 5Example 5
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Copyright2009PearsonPrenticeHallInc.
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Example 6Example 6
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MomentDiagramsandEquationsforMaximumDeflectionMaximumDeflection
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Example 4Example 4 Find the Maximum deflection for the following
structure based onThe previous diagrams p g
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Slide Number 1
