Structure Analysis I - الصفحات...
Transcript of Structure Analysis I - الصفحات...
Structure Analysis IStructure Analysis IChapter 8Chapter 8
Deflections
IntroductionIntroduction
• Calculation of deflections is an important part ofCalculation of deflections is an important part of structural analysis
• Excessive beam deflection can be seen as a mode of failure.– Extensive glass breakage in tall buildings can be attributed to excessive deflections
– Large deflections in buildings are unsightly (and unnerving) and can cause cracks in ceilings and wallsand can cause cracks in ceilings and walls.
– Deflections are limited to prevent undesirable vibrations
Beam DeflectionBeam Deflection
• Bending changes theBending changes the initially straight longitudinal axis of the beam into a curve that is called the
flDeflection Curve or Elastic Curve
Beam DeflectionBeam Deflection
• To determine the deflection curve:To determine the deflection curve:– Draw shear and moment diagram for the beam
– Directly under the moment diagram draw a line for the beam and label all supports
– At the supports displacement is zero
h h i i h d fl i i– Where the moment is negative, the deflection curve is concave downward.
– Where the moment is positive the deflection curve isWhere the moment is positive the deflection curve is concave upward
– Where the two curve meet is the Inflection Point
Deflected Shape
Example 1Draw the deflected shape for each of the beams shown
Example 2Draw the deflected shape for each of the frames shown
Double Integration Method
Elastic‐Beam Theory
• Consider a differential element• Consider a differential element of a beam subjected to pure bending.g
• The radius of curvature ρ is measured from the center of curvature to the neutral axis
• Since the NA is unstretched, the dx=ρdθ
Elastic‐Beam TheoryElastic Beam Theory
• Applying Hooke’s law and the Flexure formula we• Applying Hooke s law and the Flexure formula, we obtain:
M1EI
=ρ
Elastic‐Beam Theory• The product EI is referred to as the flexural rigidity• The product EI is referred to as the flexural rigidity.• Since dx = ρdθ, then
M )(SlopedxEIMd =θ
In most calculus booksIn most calculus books
[ ]dxvd
= 3
22 /1
In most calculus books In most calculus books
( )[ ]l titdxvdM
dxdv+22
23
2
)(//1ρ
( )[ ]solutionexact
dxdvEI+
=
2
23
2)(
/1
EIM
dxvd=2
2
The Double Integration MethodThe Double Integration MethodRelate Moments to Deflections
Md 2
EIM
dxvd=2
2
( )∫== dx
EIxM
ddvx
)()(θ
Do NotIntegration Constants∫ xEIdx )(
( )∫∫
xMUse Boundary Conditions to Evaluate Integration Constants( )
∫∫= 2
)()( dx
xEIxMxv
Constants
Assumptions and Limitations
Deflections caused by shearing action negligibly small compared to bending
Deflections are small compared to the cross‐sectional dimensions of the beam
All portions of the beam are acting in the elastic range
Beam is straight prior to the application of loads
ExamplesLx
y
PxPLM +−=x
P
PL
P
PxPLM +=
Mdx
ydEI =2
2
yd 2
@ x PxPLdx
ydEI +−=2
Integrating once 1
2
2cxPPLx
dxdyEI ++−=
2dx
@ x = 0 ( ) ( ) ( ) 020000 11
2
=⇒++−=⇒= ccPPLEIdxdy
32PLIntegrating twice 2
32
62cxPPLxEIy ++−=
@ x = 0 ( ) ( ) ( ) 00000 22
32 =⇒++−=⇒= ccPPLEIy
3( ) ( )
62 22y
62
32 xPPLxEIy +−=
@ x = L y = ymax EIPL3
3
max =∆
62y
EIPLyPLLPLPLEIy3662
3
max
332
max −=⇒−=+−=
y
( )2xLWM −−=W
( )2
xLM =
Mdx
ydEI =2
2
2 WydL
xx
WL2
2WL
@ x ( )22 2xLW
dxydEI −−=WL2
Integrating once( )
1
3
32cxLW
dxdyEI +
−=
@ x = 0 ( ) ( )63
02
003
11
3 WLccLWEIdxdy
−=⇒+−
=⇒=
( )66
33 WLxLW
dxdyEI −−=∴
66dx
Integrating twice( ) 34
cxWLxLWEIy +−
=Integrating twice 2646cxEIy +−−=
@ x = 0 ( ) ( ) ( )24
064
06
004
22
34 WLccWLLWEIy =⇒+−−
−=⇒=
( )43
4 WLxWLxLWEIy +−−−=
( ) ( )24646 22y
( )24624
xxLEIy +=
Max occurs @ x = LMax. occurs @ x = L
EIWLyWLWLLWEIy88246
4
max
444
max −=⇒−=+−=
EIWL8
4
max =∆EI8
Exampley x
L
x
WL WL2 2
22xWxxWLM −=
22 xWLyd222
xWxWLdx
ydEI −=
Integrating 1
32
3222cxWxWL
ddyEI +−=g g 13222dx
Since the beam is symmetric 02
@ ==dxdyLx
⎞⎜⎛⎞
⎜⎛
32 LL
( ) ⇒+⎠⎞
⎜⎝⎛
−⎠⎞
⎜⎝⎛
== 132
222
20
2@ c
LW
LWLEILx
24
3
1WLc −=
3
2464
332 WLxWxWL
dxdyEI −−=∴
Integrating2
343
cxWLxWxWLEIy +−−=g g2244634
cxEIy +
@ x = 0 y = 0 ( ) ( ) ( ) ( ) 2
343
0000 cWLWWLEI +−−=⇒ 02 =⇒ c( ) ( ) 2244634 2
xWLxWxWLEIy3
43 −−=∴ xxxEIy242412
=∴
Max occurs @ x = L /25 4WLEIy =Max. occurs @ x = L /2 384maxEIy −=
WL5 4
∆EI384max =∆
Exampley x P
L/2
x
P PL/2
f2 LPyd
xPMLx22
0for =<<2 2
20for
22
LxxPdx
ydEI <<=
Integrating 1
2
22cxP
ddyEI +=g g 122dx
Since the beam is symmetric 02
@ ==dxdyLx
⎞⎜⎛
2L
( ) ⇒+⎠⎞
⎜⎝⎛
== 122
20
2@ c
LPEILx
16
2
1PLc −=
2
164
22 PLxP
dxdyEI −=∴
Integrating2
23
cxPLxPEIy +−=g g21634
cxEIy +
@ x = 0 y = 0 ( ) ( ) ( ) 2
23
000 cPLPEI +−=⇒ 02 =⇒ c( ) ( ) 21634 2
xPLxPEIy2
3 −=∴ xxEIy1612
∴
Max occurs @ x = L /23PLEIy =Max. occurs @ x = L /2
48maxEIy −=
PL3
∆EI48max =∆
Example
Example 5
Moment‐Area Theorems
Moment‐Area Theorems
Theorem 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
θ
θ
= ⇒ =
⎛ ⎞
B
d M Md dxdx EI EI
M
θ θ ⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠
B
B AA
M dxEI
θ = ∫
dt xdMθ=
⎛ ⎞
B B
Md dxEI
M M
θ ⎛ ⎞= ⎜ ⎟⎝ ⎠
B AA A
M Mt x dx x dxEI EI
= =∫ ∫
Moment‐Area Theorems
Theorem 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
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
M/EI
‐30/EI‐20/EI
‐
( )20 1 10 22 1 2 2t ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − ⋅ ⋅ + ⋅ − ⋅ ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟( )/
3
2 1 2 22 3
53.33 0.00741
C BtEI EI
kN m rad
= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − ⋅ =EI
Another Solution
Conjugate-Beam Method
Conjugate-Beam Method
2
2
dV d Mw wdx dx
= =
2
2
dx dxd M d v Mdx EI dx EIθ= =
dx EI dx EI
Integrating
V wdx M wdx dx
M M
⎡ ⎤= = ⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞
∫ ∫ ∫
∫ ∫ ∫M Mdx v dx dxEI EI
θ ⎡ ⎤⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∫ ∫ ∫
tspp
ort
m S
up-B
eam
ugat
e-Co
nju
C
Example 1Example 1Find the Max. deflection Take E=200Gpa, I=60(106)
EIVBB
5.562' −==θ
EIEIM
EI
BB5.14062)25(5.562
'−
===∆
Example 2Example 2Find the deflection at Point C
C
EIEIEIMCC
162)3(63)1(27'
−=−==∆
Example 3Example 3Find the deflection at Point D
3600∆
EI360
EI720 EI
M DD3600
' ==∆
Example 4Example 4Find the Rotation at A
10 ft
3.33θEIA
3.33=θ
Example 5Example 5
Copyright © 2009 Pearson Prentice Hall Inc.
Example 6Example 6
Moment Diagrams and Equations for Maximum DeflectionMaximum Deflection
Example 4Example 4 Find the Maximum deflection for the following structure based onThe previous diagrams p g