Upload
others
View
0
Download
0
Embed Size (px)
Citation preview
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Lecture Note for Solid Mechanics- Deflections in Beam-
Prof. Sung Ho YoonDepartment of Mechanical EngineeringKumoh National Institute of Technology
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Text book : Mechanics of Materials, 6th ed.,
W.F. Riley, L.D. Sturges, and D.H. Morris, 2007.
Prerequisite : Knowledge of Statics, Basic Physics, Mathematics, etc.
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Flexural Loading : Beam Deflections
Differential equation of the elastic curve
EIM
EIMc
ELc
cLL
1
)(xM
dxydEI
dxyd
dxdy
dxyd
2
2
2
2
232
22
1
1
Mdx
ydEI
EIM
dxd
dxyd
dxdLd
dxd
dxydand
dxdy
2
2
2
2
2
2
1
tan
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
Differential equation of the elastic curve
Boundary conditions
For (a) and (b) : y=0 at x=0
For (c) and (d) : y=0 and dy/dx=0 at x=0
4
4
3
3
2
2
dxydEI
dxdVwload
dxydEI
dxdMVshear
dxydEIMmoment
dxdyslope
ydeflection
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
2142
23
13
22
2
2
2
2
2424727
612247
021212
7
127
1270
212
CxCxwxwLxwLEIy
CxwxwLxwLdxdyEI
LxforxwxwLxwLxMdx
ydEI
wLwLRLwLwLLRM AAB
)(
)(
Using boundary conditions of y=0 at x=0 and y=0 at x=L
(Example) (a) Equation of elastic curve for interval between the supports
xLxLLxxEI
wxy 32234 37372
)(
(b) Deflection midway between supports
EIwL
EIwL
EIwLy
LLLLLLLEI
wLy
xLxLLxxEI
wxy
43434
32
234
32234
1081710817128
223
27
23
722
37372
)(.)(.
)(
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
(c) Point of maximum deflection between the supports
LxLxLLxxEI
wdxdy
006211272
3223
LxLxLxLxLLxx
3251541011620062112 3223
.,.,.
Lx 5410. to the right of the left support
(d) Maximum deflection in the interval between the supports
EIwL
EIwLy
xLxLLxxEI
wLxy
4343
32234
1088710887
37372
5410
)(.)(.
).(
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
Deflection by integration of shear force and load equations
4
4
3
3
2
2
dxydEI
dxdVload
dxydEI
dxdMshear
dxydEImoment
dxdyslope
ydeflection
)(
)(
)(
xwdx
ydEI
xVdx
ydEI
xMdx
ydEI
4
4
3
3
2
2
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
wxwdx
ydEI )(44
Using boundary conditions
(Example) (a) Equation of elastic curve for interval between the supports
43
2
2
3
1
4
32
2
1
3
21
2
2
2
13
3
2624
26
2
CxCxCxCwxEIy
CxCxCwxdxdyEI
CxCwxxMdx
ydEI
CwxxVdx
ydEI
)(
)(
(1) y=0 at x=0, (2) M=0 at x=0, (3) M=0 at L=0, (4) y=0 at x=L
)( xLLxxEI
wy 334 224
(b) Maximum deflection of the beam
EI
wLEI
wLLxy384384
5044
).(
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Singularity functions
When several intervals and several sets of matching conditions are required, need a special mathematical apparatus available
X=a
1
X=a
1
X=a
1
X=a
1
X=a
1
2 ax 1 ax
0ax 1ax 2ax
axwhenaxwhen
ax
axandnwhenaxandnwhenaxax
nn
01
000
0
)(
1
01
1
1
1
nwhenaxnaxdxd
nwhenaxn
dxax
nn
nn
0
1
12
axdxax
axdxax
Deflections by Singularity Functions
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
Examples of singularity function
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
23
02
11
23
02
11
1
1312
01
0
0322111
1
0322111
2
2
0
xxwxxMxxPxR
xxwxxMxxPxRxM
xxwxxMxxPxRxV
xxwxxMxxPxR
LxRxxwxxMxxPxRxq
AL
AL
AL
AL
LAL
)(
)(
)(
Application of singularity function
LxxxxwMxxPxRxM
xxxMxxPxRxMxxxxxPxRxM
xxxRxM
AL
AL
L
L
32
314
3213
2112
11
2
0
)()()(
)()()()(
)(
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
0
247
2
2wLLLwLRM LB )(
(Example) Determine the deflection at the left end of the beam
From moment equilibrium equation in (b)
021
22
2
2
220
83
222LxwLxwLLxwLxw
dxydEI )(
Bending moment expression
Boundary conditionsLxatyandxatoy 00
21
223
44
1
122
33
240
1622424
220
163
266
CxCLxwLxwLLxwLxwEIy
CLxwLxwLLxwLxwdxdyEI
)(
)(
31
41
4444
422
44
247
1285
161638481
24160
1285000
384240
wLCwLLCwLwLwLwL
wLCCwLwL
444338497
38497
1285
247000 wLywLwLLwLEIy )(
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
Deflections by superposition
Resultant effect of several loads acting on a member simultaneously is the sum of the contributions from each of the loads applied individually.
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
(Example) Determine the deflection at the right end of the beam
mmmmyyyy
mmmEI
wLy
mmmyymLy
radEI
PL
mEI
PLy
4739973
85410418501050010288
510578
143203214001607100160710
016071020080360
00803601050010282
310252
01607101050010282
310253
321
69
434
3
21
2
69
232
69
333
1
..)(
..))()()((
))((...).().(
.)(.
.))()()((
))((
.))()()((
))((
E=28 GPaI=500(10-6) m4
Deflection at concentrated load :
Additional deflection of unloaded beam :
Total deflection due to concentrated load :
Deflection due to distributed load :
Total deflection :
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
Deflections due to shearing stressFor short heavily loaded beams deflection produced by shearing stresses can be significantly.
Deflection of neutral surface dy due to shearing stresses in the interval dx
r
r
Vdxd
QGIt
dxGIt
QVdxG
dxd
since the shear is negative
AGwxdxd
AItQ
AV
ItQV rr
23
23
23
max
Relative deflection due to shearing stress
since the shear V is -wx
Deflection at the left end of the beam
AGwLxdx
AGwd
L
s 43
23 2
0
Total deflection at the left end of the beam
AGwL
EAdwL
sf 43
23 2
2
4
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
(Example) Structural steel (E=29,000psi and G=11,000psi) cantilever beam with rectangular cross section 2in wide and 4in deep supports a concentrated load of 1000lb at the end of 3-ft span. Determine the percent increase in deflection at the free end of the beam resulting from the shearing stresses
Deflection dy of the neutral surface due to shearing stresses
dxGIt
QVdxG
dxd r
For the beam with rectangular cross section PVandbhQbhI 8
,12
23
dxbhGPd
23
Deflection νs at free end of the beam due to shearing stresses
inbhGPLds 0006136.0)10)(11)(4)(2(2
)12)(3)(1000(323
6
Deflection yf at free end of the beam due to flexural stresses
inEbhPL
EIPL
f 05028.0)4)(2)(10(29)]12(3)[1000(44
3 363
3
33
Percent increase in deflection
%22.1)100(05028.0
0006136.0Increase
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
Deflections by energy methods – Castigliano’s theoryCastigliano’s theory is applicable to any structure for which the force-deformation relations are linear
Work done by 2
L
k dPW 0
Using Hooke’s law
dE
dandE
1
EALdE
ALU2
22
0
2
Elastic strain energy per unit volume
For shear loading, expression is identical except that is replaced by and E by G.
Work done is equal to strain energy
22
00dALdLAUWk ))((
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
Strain energy is equal to the work done by the forces P1 and P2
2211 21
21 PPU
Let P1 be increased by P1, ν1 and ν2 be the changes in deflection due to incremental load
22111121 PPPU
If the order of loading is reversed so that P1 is applied first, followed by P1 and P2
22111111 21
21
21 PPPPUU
221111 PPP
Strain energy must be independent of the order of loading
111 2
1
PU
Total strain energy
2211112211 21
21
21 PPPPPUU
(*)
0111
PasPU
1121 PU
221111 21
21 PPPU
111121 PPU (**)
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
Castigliano’s theorem for the concentrated loads
iiP
U
Castigliano’s theorem for bending moments
iiM
U
Total strain energy under uniaxial stress
dVE
UE
uV 22
22
Total strain energy under a beam subjected to pure bending
dxI
ME
dxdAyIM
E
dVI
MyE
UE
u
L
L
A
V
0
2
0
22
2
22
21
21
21
2
dxPM
IM
EPUy
i
L
ii
01
Leibnitz’s rule
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
(Example) Determine the deflection at the end of cantilever beam.
Moment equation is
LwxPxM6
3
Deflection is given by
dxPM
IM
EPUy
i
L
ii
01
303
6
66
43
0
42
0
3
0
wLPL
dxL
wxPx
dxxL
wxPx
dxPMMEIy
L
L
L
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
(continued)
(Example) Determine the deflection at the center of a simply supported beam.
Consider a dummy load at the center of the beam.
Moment equation is
1
12
22
2222
LxxPM
LxPPxwxwLxMMM Pw
Deflection is given bydx
PM
IM
EPUy
i
L
ii
01
3845
2344
2222
4
2
322
0
32
0
12
0
wL
dxxLxxLwdxxLxw
dxLxxwxwLx
dxPMMEIy
L
L
L
L
L
12
11
21
00
10
11
0
1
221
221)(
2)(
2)(
LxPwxxR
LxRLxPxwxRxM
LxRLxPxwxRxV
LxRLxPxwxRxq
A
CA
CA
CA
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Flexural Loading : Beam Deflections
Statically indeterminate beams
Beams that the equations of equilibrium are not sufficient to determine all the reactions
- Integration method
Additional constraint provides addition information on slopes or deflections
Moment equations would contain reactions or loads that can not be evaluated from the available equations of equilibrium
Extra boundary conditions will yield the necessary additional equations
- Superposition method
The slope or deflection due to several loads is the algebraic sum of the slopes or deflections due to each of the loads
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Flexural Loading : Beam Deflections
2)(
2
2
2 wxMVxxMdx
ydEI
(Example) Determine the reactions at A and B.
The given beam is statically indeterminate due to 3 unknowns (M, V, R) and 2 equations of equilibrium
Consider boundary conditions :
Lxatyxatyanddxdy
0;000
The elastic curve equation :
2
423
1
32
2426
32
CwxMxVxEIy
CwxMxVxdxdyEI
LxatywLMVL 0124 2
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Flexural Loading : Beam Deflections
(continued)
From solving simultaneous equations
04
564
4364
4364
3764
3764
764
7 22
wLVRFwLwLR
wLwLV
wLwLM
y
2153232
83
450
wLMVL
LLwMVLM R
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Flexural Loading : Beam Deflections
(Example) A steel beam with E=30,000-ksi, 20ft long, and IC=100in4 is supported.
The resulting deflection at midpoint
0 Rw yyy
(a) Determine the reactions A, B, and C.
The deflection yw at midpoint is
inRREILRy
inEI
wLy
CCC
R
w
)10(96)100)(10)(30(48
)12*20(48
48.0)100)(10)(30(384)12*20)(12400(5
3845
66
33
6
44
lblbRC 50005000
lblbRR RL 15001500]500020*400[21
The reactions comes from equilibrim equation and symmery conditions
Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수
Homework
(8-4), (8-14), (8-41), (8-60)