chapter 7 bending deformation -...
TRANSCRIPT
2008 Architecture LJM 1
7.1 Introduction
7.2 Approximately Differential equation of deflection
curve
7.3 Integration method of determining the beam
deflections
7.4 Superposition method of determining the beam
deflections
7.5 Statically indeterminate beams
7.6 Stiffness criteria of beams; Optimum design of
beams for stiffness
Chapter 7 Bending DeformationChapter 7 Bending Deformation
2008 Architecture LJM 2
7.1 Introduction
7.2 Approximately Differential equation of deflection
curve
7.3 Integration methof of determining the beam
deflections
7.4 Superposition methof of determining the beam
deflections
7.5 Statically inderminate beams
7.6 Stiffness criteria of beams; Optimum design of
beams for stiffness
Chapter 7 Bending DeformationChapter 7 Bending Deformation
2008 Architecture LJM 3
Highlights::::Feflecion calculation of beams and plane frames
Objectives::::①①①①Stiffness check;;;;
②②②②Solution of Statically Inderminate Problems
7.1 Introduction
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⑴⑴⑴⑴ Deflection, w::::vertical displacement of centriod of the
cross-section.
⑵⑵⑵⑵ Rotational angle, θθθθ, ::::Angle rotated about neutral axis
2. Basic amounts measuring the beam deflection
3. Relation between w and θθθθ::::
1. Deflection curve:::: axis of the beam after deformation,
smooth ever-curve
dtan w'
d
w
xθ θ= = =
F
xθθθθ
C
w
C1
w
w =w (x)
7.1 Introduction
2008 Architecture LJM 5
7.1 Introduction
7.2 Approximately Differential equation of deflection
curve
7.3 Integration method of determining the beam
deflections
7.4 Superposition method of determining the beam
deflections
7.5 Statically indeterminate beams
7.6 Stiffness criteria of beams; Optimum design of
beams for stiffness
Chapter 7 Bending DeformationChapter 7 Bending Deformation
2008 Architecture LJM 6
z
z
EI
)x(M=
ρρρρ1
1. Differential equation of deflection curve
EI
)x(M)x("w ±=∴
挠曲线近似微分方程挠曲线近似微分方程挠曲线近似微分方程挠曲线近似微分方程。。。。
)x("w 'w
)x("w±≈
+±=
23
2 )1(
1
ρρρρ
小变形小变形小变形小变形
w
x
M>0
0>)x(''w
w
x
M<0
0<)x("w EI
)x(M)x("w =∴
7.2 Differential equation of deflection curve
2008 Architecture LJM 7
7.1 Introduction
7.2 Approximately Differential equation of deflection
curve
7.3 Integration method of determining the beam
deflections
7.4 Superposition method of determining the beam
deflections
7.5 Statically indeterminate beams
7.6 Stiffness criteria of beams; Optimum design of
beams for stiffness
Chapter 7 Bending DeformationChapter 7 Bending Deformation
2008 Architecture LJM 8
CxdEI
xMx'w +== ∫
)()(θθθθ
DCxxdxdEI
Mw ++= ∫ ∫ )(
1. Integrate the equation:
2.Boundary conditions FA BC
F
D
For uniform straight beams:
)()( xMx"EIw =
Equation of
rotational angle
Deflection equation
0=Aw
0=Bw
0=Dw0=Dθθθθ
7.3 Integration methof of determining
the beam deflections
2008 Architecture LJM 9
Discussions::::
①①①①Suitable to small deformation members, linear-elasticity
materials.
②②②②Advantage: Find deflection and rotational angle of
arbitrary sections; Weakness: Troublesome
Continuity condition:
Smooth condition::::
+− =CC
ww
+− =CC
θθθθθθθθP
A BC
wc
7.3 Integration methof of determining
the beam deflections
2008 Architecture LJM 10
FD
a a a
BA
C
MFa/2
Fa/2x
Fa/2Ex.7.1 Draw the deflection curve of the cantilever beam,,,,EI.
Solution::::1. Basic foundation
EI
xM"w
)(=
Determine curve,,,,Obey B.C and C.C
2. Draw sketch of the
deflection curve
DCGA
w
7.3 Integration method of determining
the beam deflections
2008 Architecture LJM 11
Ex.7.2 Determine the deflection curve |w|max |θ|max of the
cantilever beam .
�Bending equation
)()( lxFxM −=
�Differential equation and integration
)()( xlPxM"EIw −−==
CFlxxF
'EIw +−= 2
2
DCxxFl
xF
EIw ++−= 23
26
Solution:
F
l
xw
x
BA
Bθθθθ
7.3 Integration methof of determining
the beam deflections
2008 Architecture LJM 12
����Find constant from B.C.
X = 0: θA=0, C=0
wA=0, D=0
FlxxF
'EIw −= 2
223
26x
Flx
FEIw −=
Substitution B.C. into the above equations, gets:
④④④④Maximum deflection and rotational angle at B.
)(EI
FlwB ↓−=
3
3
EI
Fl'wBB
2
2
−==θθθθ ( )
F
l
w
wBx
BA
Bθθθθ
x
7.3 Integration methof of determining
the beam deflections
2008 Architecture LJM 13
FBC
Ex 7.3 Discuss the deformation of the simple-supported
beam shown.
x1
l
w
ASolution:
FA=Fb/l, FB=Fa/l
(2) List bending-moment equations
Portion AC :
)ax(l/FbxM ≤≤= 111 0
)()( 2222 lxaaxFl/FbxM ≤≤−−=
x2
Protion BC :
(1) Find constraint reactions
a bx
7.3 Integration methof of determining
the beam deflections
FA FB
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(3) List differential equations and then integrate
)lxa()ax(Fxl
FbM"EIw ≤≤−−== 22222
For portion CB:
)ax(xl
Fb)x(M"EIw ≤≤== 1111 0
111
3
11
1
2
1
6
2 1
DxCxl
Fb'EIw
Cxl
Fb'EIw
++=
+=
222
3
2
3
22
2
2
2
2
22
66
22
DxC)ax(l
Fx
l
FbEIw
C)ax(P
xl
Pb'EIf
++−−=
+−−=
For portion AC :
7.3 Integration methof of determining
the beam deflections
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(4) Determine integration constants by B.C and C.C
)ax(D:)(w ≤≤== 111 0000
:)a('w)a('w 21 =
2
22
1
2
222C)aa(
Fa
l
FbCa
l
Fb+−−=+
066
22
33
22 =++−= DlCbl
Fl
l
Fb)l(w
Continuity conditions:
22
33
11
3
21
666DaC)aa(
Fba
l
FbDaCa
l
Fb
:)a(w)a(w
++−−=++
=
(a)
(b)
(c)
(d)
Boundary conditions:
7.3 Integration methof of determining
the beam deflections
2008 Architecture LJM 16
)bl(l
FbCC
DD
22
21
21
6
0
−−==
==
Upon solving the four equations simultaneously, find
)xbl(xl
FbEIw
)xbl(l
Fb'EIw
2
1
22
11
2
1
22
1
6
36
−−−=
−−−=
])()[(6
])(3
)3[(6
3
22
2
2
22
2
2
2
2
2
22
2
axb
lxxbl
l
FbEIw
axb
lxbl
l
Fb'EIw
−+−−−=
−+−−−=
AC:
CB:
FBCA
7.3 Integration methof of determining
the beam deflections
2008 Architecture LJM 17
(5) Discussion
2 2
A
B
Fb Fabθ = - (l - b ) = - (l + b)
6EIl 6EIl
Fabθ = (l + a)
6EIl
When a>b, θB > θA
Maximum deflection:
When a>b, θA<0, θC >0. Thus, the point of
θ1(w1' )=0 occurs in the longer segment of the beam.
wmax
Maximum rotational angle:
2 2
0x = (l - b )/3
x0
EI
Flw,
la max
482
3
==
7.3 Integration methof of determining
the beam deflections
FBCA
)bl(EIl
Fab)bl(
EIl
Fbw
max+−=−−=
639
322
2008 Architecture LJM 18
7.1 Introduction
7.2 Approximately Differential equation of deflection
curve
7.3 Integration method of determining the beam
deflections
7.4 Superposition method of determining the beam
deflections
7.5 Statically indeterminate beams
7.6 Stiffness criteria of beams; Optimum design of
beams for stiffness
Chapter 7 Bending DeformationChapter 7 Bending Deformation
2008 Architecture LJM 19
1. Superposition by loads::::
∑=⋅⋅⋅ )()( 21 iin FF,F,F θθθθθθθθ
2. Superposiyion using analysis of
portion-by-portion
∑=⋅⋅⋅ )()( 21 iin FwF,F,Fw
Where F I is generalized force, including force and couple
7.4 Superposition method of determining
the beam deflections
2008 Architecture LJM 20
Ex.7.4 Find deflection at C and
rotational angle at A by
superposition.
Solution::::�Exerting load alone
�Deformation caused by one load
EI
Faw F,C
6
3
=EI
FaF,A
4
2
=θθθθ
EI
qaw q,C
24
54
=EI
qaq,A
3
3
=θθθθ
qF
A B
Ca a
�Superposition
EI
qaFq,AF,AA
12
43 +=+= θθθθθθθθθθθθ
3 4
C
F a 5 q aw = +
6 E I 2 4 E I
F =
A B
q
+
A B
7.4 Superposition method of finding deflections
2008 Architecture LJM 21
F
lBA
qMe
BA
Me
q
BA
F
BA
+
+
w = wMe+wq+wF
EI
Fl
EI
ql
EI
lMw e
B382
342
−−=
EI
Fl
EI
ql
EI
lM eB
262
232
−−=θθθθ
Ex.7.4 Find deflection and
rotational angle at B by
superposition.
( )
( )
7.4 Superposition method of determining
the beam deflections
Solution::::
2008 Architecture LJM 22
l
BA
Principle of portion-by-portion analysis
=+
Fl a
A
BC
CB
Fa
w2
EI
alFawwwC
3
)(2
21
+=+=
FM=Fa
)(3
2
1⇓==
EI
lFaaw
Bθθθθ
w1
Bθθθθ
Basic consideration::::Deflection equalization
Basic theory::::Force
transition
Basic results::::Application
directly
)(3
3
2⇓=
EI
Faw
7.4 Superposition method of finding deflections
2008 Architecture LJM 23
A Cq
B
F
a a/2 a/2
Aq
FAy
FBy
(a)
B
F
a/2 a/2
(b)
A CB
q,Aθθθθ
wB
Ex.7.5 Combined beam AC。。。。EI,,,,F=qa。。。。Find wB and θB.
解解解解2
qaFF ByAy ==
)(48
13
)2
3(6
)2(
324
33
↓=
−+⋅=
+=
EI
qa
aa
EI
/aF
EI
aqa
wwwF,BF,BB By
EI
qa
EI
qa
EI
qa
EI
/aF
EI
aF
BBB
q,BB
By
B
12
5
24
8
3
2
)2(
2
3
3
322
−=−−=
==
=+=
−+
−
+
ππππθθθθθθθθππππθθθθ∆∆∆∆
θθθθθθθθ
θθθθ
+Bθθθθq,B
θθθθ
7.4 Superposition method of finding deflections
2008 Architecture LJM 24
Ex.7.6 Determine wC .
����For infinitesimal portion dx
����From Table of deflection of beams.
����Superposition
EI
)bL(b)Fd(w dF,C
48
43 32 −−−−====
bdL
bqxd)x(qFd 02========
bdEI
)bL(qb
24
43 322 −−−−====
EI
qLbd
EIL
)bL(qbww
L.
dF,Cq,C24024
43 450
0
322
====−−−−
======== ∫∫∫∫∫∫∫∫
q0
0.5L 0.5L
x dx
b
x
f
C
7.4 Superposition method of finding deflections
Solution::::
2008 Architecture LJM 25
7.1 Introduction
7.2 Approximately Differential equation of deflection
curve
7.3 Integration method of determining the beam
deflections
7.4 Superposition method of determining the beam
deflections
7.5 Statically indeterminate beams
7.6 Stiffness criteria of beams; Optimum design of
beams for stiffness
Chapter 7 Bending DeformationChapter 7 Bending Deformation
2008 Architecture LJM 26
1. Take equalization system
2. List compatibility equation
3. Introduce physical law to get supplementary
4. Solve the equation to get redundant reaction
A
FB
A
FB
q q
Redundant reactions(多余支反力多余支反力多余支反力多余支反力)
FBy
Redundant restraints(多于约束多于约束多于约束多于约束)
7.5 Statically indeterminate beams
Solution::::
2008 Architecture LJM 27
A
F
l/2 l/2
A
FByF
B
MA
A
F
Solution::::
相当系统相当系统相当系统相当系统E.E
①①①①Take equalization equation
②②②②Compability condition
③③③③Fond redundant reaction
Change S.ID.P to S.D.P!!!!
0=Bw
Bw = 0
3 3By
B
F l 5Flw = - = 0
3EI 48I
)(16
5↓=
FFBy
Ex.7.7 Find reactions of the beam
④④④④Find other reactions A
FByFMA
FAy
16
3)(
16
11 FlM
FF AAy =↓= ( )
7.5 Statically indeterminate beams
2008 Architecture LJM 28
BC
A
F
D
E
Example7.8 Two cantilever beams of AD and BE are joined
by a steel rod CD. Determine the deflection of the cantilever
beam AD, at D due to a force F applied at E.
EI
EA
l
l
l
l
7.5 Statically indeterminate beams
2008 Architecture LJM 29
Solution:
(1) Set up a equivalent system
(2) Compatible condition
wD=wC -△△△△rod (a)
wC = (wC)p- (wC)FD
WhereEI
lFw D
D3
3
=
EI
Fl
EI
Flw FD
23)(
33
+=
EI
lFw D
FD D 3)(
3
=
wc
△△△△rod
B CF
E
Fig.b
FD
A
D
FD
wD
Fig.a
7.5 Statically indeterminate beams
2008 Architecture LJM 30
BC
A
F
D
E
FD
wc
△△△△rod
wD
Substitute wD, wC and △△△△rod
into Eq.(a):
(3) Find the deflection of D, wD
EA
lFD
rod=∆∆∆∆
A/Il
FlFD
64
52
2
+=
)()64(3
52
5
↓+
=A/IlEI
FlwD
7.5 Statically indeterminate beams
2008 Architecture LJM 31
=
q0
lA
B
l
MA
BA
Ex.7.9:::: Shown is the beam AB
of length l, EI, subjected to
uniform load q. Draw M-
diagram.
MA, FB--多余约束反力多余约束反力多余约束反力多余约束反力。。。。
q0
EI
q0
L FB
AB
7.5 Statically indeterminate beams
Solution::::
(1) Set up a equivalent system
Different Equivalent systems!
2008 Architecture LJM 32
�Compatible equation
0=+=BF,Bq,BB www
+
q0
L FB
AB
=
FB
AB
q0
AB
�physical relations
�Supplementary equation
EI
LFw
EI
qLw B
F,Bq,B B 3;
8
34
−==
038
34
=−EI
LF
EI
qL B
8
3qLF
B=∴
�M-diagram
7.5 Statically indeterminate beams
M
12823 2 /ql
82 /ql
2008 Architecture LJM 33
Solution::::����Set up E.SMA
Ex.7.9::::Plot Bending-moment
diagram of simple-supported
beam AB shown.
BA
0====++++==== q,AM,AA Aθθθθθθθθθθθθ
EI
lMA
M,A A 3====θθθθ
EI
qlq,A
24
3
====θθθθ
�Geometry equation
8
2qlM A −−−−====
�Reactions::::8
3
8
5 qlF,
qlF
BA========
�Bending-moment DiagramM 29 /128ql
82 /ql
㈠㈠㈠㈠
㈩㈩㈩㈩
lA
BEI
q0
FBFA
MA
BA
7.5 Statically indeterminate beams
Have::::
2008 Architecture LJM 34
Solution::::
相当相当相当相当系统系统系统系统E.S.
①①①①Equivalent system
②②②②Compatible equation
21 CC ww =
EI
lF
EI
lF
w RR
C243
)2
( 33
2 ========
Ex.7.10 Determine maximum deflection of the cantilever beam
AB shown. Suppose that EI for the two beams are equal.
④④④④Maximum deflection
A FR C
F
B
FR
A C
C BA
l/2 l/2
EI
lFFw R
C48
)25(3
1
−−−−====
EI
lFF
EI
lF RR
48
)25(
24
33 −=
4
5FFR =
33 3
RB
5F lFl 13Flw = - =
3EI 48EI 64EI
B
B'
w%
w= 61
F
7.5 Statically indeterminate beams
2008 Architecture LJM 35
BCF,Bq,BB LwwwB
∆∆∆∆=−=
=
Ex.7.11 Determine axial load in rod BC for the structure shown.
q0
FB
FB
A
+
q0
A
LBCEA
q0
LA B
C
EI
7.5 Statically inderminate beams
①①①①Equivalent system
②②②②Compatible equation
Solution::::
E.S.
2008 Architecture LJM 36
=
LBCEA
x
f
q0
L FB
A B
C
FB
AB
+
q0
AB
�Physical relations
�Supplementary
�Others((((Reactions, stresses,
deflections, and so on.))))
EI
LFw;
EI
qLw B
R,Bq,B B 38
34
−==
EA
LF
EI
LF
EI
qL BCBB =−38
34
)3(8 3
4
I/LA/LI
qLF
BC
B +=∴
EA
LFL BCB
BC =∆∆∆∆
7.5 Statically inderminate beams
2008 Architecture LJM 37
7.1 Introduction
7.2 Approximately Differential equation of deflection
curve
7.3 Integration method of determining the beam
deflections
7.4 Superposition method of determining the beam
deflections
7.5 Statically indeterminate beams
7.6 Stiffness criteria of beams; Optimum design of
beams for stiffness
Chapter 7 Bending DeformationChapter 7 Bending Deformation
2008 Architecture LJM 38
[ ] wmax
δδδδ≤ [ ]θθθθθθθθ ≤max
[w]: Allowable deflection,,,,
[θθθθ]: Allowable rotational angle
� Check the stiffness;
����、、、、Determine allowable loads.
1. Stiffness conditions of beams
Three types of stiffness calculation
����、、、、Design sections;
7.6 Stiffness criteria of beams;
Optimum design of beams for stiffness
2008 Architecture LJM 39
Strength:::: [ ] W
M
z
max σσσσσσσσ ≤=
zEI
XM''w
)(=Stiffness:
Methods::::
→3S are related to internal forces and
properties of the cross-section
Reducing Bending moment M
Enhancing Inertia moment I or Modulus of section W
Select materials rationally
Stability:::: l
EImimcr 2
2ππππσσσσ =
2. Methods of enhancing deformations of beams
7.6 Stiffness criteria of beams;
Optimum design of beams for stiffness
2008 Architecture LJM 40
Ex.7.12 The hollow circular rod AC of in-diameter d=40mm
and out-diameter D = 80mm. E = 210GPa,,,,[δ]= 10-5 m at C.
[θθθθ] = 0.001. at C, F1 = 1kN, F2 = 2kN. Check the stiffness of
the overhanging beam.
F2
B
l=400
P2
A
Ca=100
200 D
F1
B
+
F2 M
=
F1
+F2
7.6 Stiffness criteria of beams;
Optimum design of beams for stiffness
Analysis of deformations
2008 Architecture LJM 41
2
1C1 B1
F l aw = θ a =
16EI
2
1B1
F lθ =
16EI
2B3
F laθ = -
3EI2
2
C 3 B 3
F law = θ a = -
3E I
����From Table of deformations
02 =Bθθθθ
C
F aw
EI= −
3
22
3
F2
B C
++
=
( )
( )
( )
CF1
A BD
The overhang beam
C
F2
B
D
A
M
7.6 Stiffness criteria of beams;
Optimum design of beams for stiffness
Solution::::
2008 Architecture LJM 42
C
F l a F a F law
EI EI EI= − −
2 3 2
1 2 2
16 3 3
2
1 2B
F l F laθ = -
16EI 3EI
����Deformations by superposition
481244441018810)4080(
64
143)(
64m
. dDI
−− ×=×−=−=ππππ
[ ] 00101042304
..max =<×= − θθθθθθθθ
[ ] mm.w max
56 1010195 −− =<×= δδδδ
����Check the stiffness
)(104230)3
200
16
400(
1880210
40 4 弧度−×−=−×
= ..
Bθθθθ
m.wC
610195
−×−=
( )
7.6 Stiffness criteria of beams;
Optimum design of beams for stiffness