Week 16 ꆱ 9.4 Deflection by integration of the shear force...
Transcript of Week 16 ꆱ 9.4 Deflection by integration of the shear force...
Week 16
§ 9.4 Deflection by integration of the shear force and
load equations
例 9-4
Fig. 9-14 Example 9-4. Deflections of a cantilever beam with a triangular load.
Differential equation of the deflection curve
LXLq
q)(0 −
=
LXLq
qEIV)(0''' −
−=−=
Shear force un the beam
120''' )(
2CXL
Lq
EIV +−=
B. C. C0)(''' =LV 1=0
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Week 16
20''' )(2
XLL
qEIV −=
20 )(2
XLL
qV −=⇒
Bending moment in the beam
230'' )(
6CXL
Lq
EIV +−−=
B. C. V C0)(''' =L 2=0
30'' )(6
XLL
qEIVM −−==
Slope and deflection of the beam
340' )(
24CXL
Lq
EIV +−=
4350 )(
120CXCXL
Lq
EIV ++−−=
B. C. V ,V 0)0(' = 0)0( =
LLqC
24
30
3 −= , L
LqC120
40
4 −=
)464(24
32230' XLXXLLLEIXq
V −+−=
)51010(120
32232
0 XLXXLLLEIXqV −+−−=
Angles of rotation and deflection at the free end
EILqLVB 24
)(3
0' −=−=θ
EILqLVB 30
)(4
0−=−=δ
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Week 16
例 9-5
Fig. 9-15 Example 9-5.
Deflections of a beam with an overhang.
Differential equation of the deflection curve
0<X<L, 2
''' PEIV −=
0<X<2
3L, PEIV −='''
Bending moments in the beam
0≦X≦L, 1''
2CPXEIVM +−==
0≦X≦2L3, 2
'' CPXEIVM +==
B. C. V ,0)0('' = 0)2
3( =LV
01 =C ,23
2PLC −
=
0≦X≦L, 2
'' PXEIVM −==
0≦X≦2
3L,
2)23('' XLPEIVM −
−==
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Week 16
Slopes and deflection of the beam
0≦X≦L, 3'
4CPXEIV +−=
0≦X≦2L3, 4
'
2)3( CXLPXEIV +
−−=
在 X=L處,Slope 應相同
42
3
2
4CPLCPL
+−=+−
43 2
34PLCC +=
0≦X≦L, 53
3'
12CXCPXEIV ++−=
0≦X≦2L3, 64
2'
12)29( CXCXLPXEIV ++
−−=
B. C. V ,V 0)0( = 0)( =L
(part AB)
05 =C ,12
2
3PLC = ,
65 2
4PLC =
B. C. V 0)( =L
(part BC)
4
3
6PLC −=
0≦X≦L, )(12
22 XLEI
PXV −=
0≦X≦2L3, )29103(
12323 XLXXLL
EIPV −+−−=
Deflection at the end of the over hang
EIPLLVC 8
)2
3(3
=−=δ
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Week 16
§ 9.5 Method of Superposition
Fig. 9-16 Simple beam with two loads.
“加成”
1. a uniform load of intensity q 2. a concentrated load P
1. uniform load q
EIqL
C 3845)(
4
1 =δ ,EI
qLBA 24
)()(3
11 == θθ
2. concentrated load P
EIPL
C 48)(
3
2 =δ ,EI
PLBA 16
)()(2
22 == θθ
EIPL
EIqL
EIPL
EIqL
AABA
CCC
1624)()(
483845)()(
23
21
34
21
+=+==
+=+=⇒
θθθθ
δδδ
Table of Beam Deflection Appendix G 5
Week 16
Fig. 9-17 Simple beam with a
triangular load. ©
2
0
0
1
B
r
(Deflection)
視 qdx為 concentrated load 查表 Table G-2,case 5, Appendix G
)43(48
22 aLEI
Pa−
代 qdx P
X a
)43(48
))(( 22 XLEI
xPdx−⇒
the intensity of the uniform load
LXq
q 02=
dxXLLEI
PX )43(48
222
−⇒
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Week 16
EILq
dxXXLLEIq
dxXLLEIXq
L
L
C
240)43(
24
)43(24
40222
0
20
2220
20
=−=
−=
∫
∫δ
(Angle of rotation)
查表 Table G-2,case 5
LEIbLPab
6)( +
代 PLEIXdxq
→6
2 0
X a
L-X b
dxXXLXLEIL
q 220 )2)((
3−−⇒
EILq
dxXXLXLEIL
qL
A
288041
)2)((33
0
220 2
0
=
−−= ∫θ
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Week 16
例 9-6
Fig. 9-18 Example 9-6.
Cantilever beam with a uniform load and a concentrated load.
Uniform load q
Case 2,Table G-1
)4(24
)(3
1 aLEI
qaB −=δ ,
EIqa
B 6)(
3
2 =θ
concentrated load P
EIPL
B 3)(
3
2 =δ ,EI
PLB 2
)(2
2 =θ
EIPLaL
EIqa
BBB 3)4(
24)()(
33
21 +−=+=⇒ δδδ
EIPL
EIqa
BBB 26)()(
23
21 +=+= θθθ
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Week 16
例 9-7
Fig. 9-19 Example 9-7. Cantilever beam with a uniform load acting on the right hand half of the beam.
Case 5,Table G-1
代 qdx P
X a
EIXLxqdxd B 6
)3)()(( 2 −=δ
EIxqdxd B 2
))(( 2
=θ
EIqL
XLXEIqd
LLBB
38441
)3(6
42
2
=
−==⇒ ∫∫ δδ
∫∫ ===LLBB EI
qLdxXEIqd
2
32
487
2θθ
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Week 16
例 9-8
Fig. 9-20 Example 9-8. Compound beam with a hinge.
©
2
0
0
1
(Deflection)
Two force on the cantilever beam-a uniform load + a concentrated load
Case 1 and 4,Table G-1
EIFb
EIqb
B 38
34
+=δ
代入 3
2PF =
EIPb
EIqb
B 92
8
34
+=δ
(Angle of Rotation)θA
1. an angle BAB’ by deflection 2. an additional angle by the bending of beam AB
aEIPb
aEIqb
aB
A 93
8)(
34
1 +==δ
θ
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Week 16
Case 5,Table G-2
LEIbLPab
6)( +
EIPa
aEIPb
aEIqb
aEI
aaaaP
AAA
A
814
92
8)()(
6
)3
)(3
)(3
2()(
234
21
2
++=+=⇒
+=
θθθ
θ
例 9-9
Fig. 9-21 Example 9-9. Simple beam with an overhang.
Deflection δ1-caused by the angle of rotation θB
(Fig.9-21b) simple beam AB q MB P
Case 1&7 of Table G-2
EILaqL
EILqa
EIqL
EILM
EIqL B
B 24)4(
624324
22233 −=+−=+=θ
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Week 16
EILaqaLa B 24
)4( 22
1−
== θδ
Deflection δ2-cantilever beam subjected to q Case 1,Table G-1
EIqa8
4
2 =δ
Deflection δC
)3)((24824
)4( 22422
21 LaLaLaEI
qaEI
qaEI
LaqaLC −++=+
−=+= δδδ
§ 9.7 Non-prismatic Beam
Fig. 9-27 Beams with varying moments of inertia (see also Fig. 5-23).
©
2
0
0
例 9-13
©
2
0
0
1
Fig. 9-28 Example 9-13. Simple beam with two different moments of inertia.
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Week 16
Differential equations of the deflection curve
2PxM = )
20 Lx ≤≤(
2'' PxEIV = )
40( Lx ≤≤
2)2( '' PxVIE = )
24( LxL
≤≤
B. C.
1. V(0)=0
2. 0)2
(' =LV
3. BCABLVLV )4
()4
( '' =
4. ABABLVLV )4
()4
( =
Slopes of the beam
1
2'
4C
EIPxV += )
40 Lx ≤≤(
2
2'
8C
EIPxV += )
24LxL
≤≤(
B. C. 2.
0)2
(' =LV
EIPL
32
2
2−
=C
)4(32
22' XLEI
PV −−= )24
( LxL≤≤
EIPLLV
1283)
4(
2' −=
B. C. 3
EIPLC
EIPLCL
EIP
1285
1283)
4(
4
2
1
2
12 −=⇒−=+
)325(128
22' XLEI
PV −−= )4
0( Lx ≤≤
EIPLVA 128
5)0(2
' −=−=θ
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Week 16
B. C. 4
EIPLC
EIPLCLLL
EIP
768
153613])
4(
34)
4([
323
4
3
432
−=⇒
−=+−−
)3224(768
323 XXLLEI
PV −+−= )24
( LxL≤≤
EIPLLVC 256
3)2
(3
−=−=δ
例 9-14
©
20
01
Br
Fig. 9-29 Example 9-14. Cantilever beam with two different moments of inertia.
Method of superposition
Deflection due to bending of part AC
Case 4,Table G-1
EIPL
EI
LP
243
)2
( 33
1 ==δ
Deflection due to bending of part CB
Case 4&6,Table G-1
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Week 16
EIPL
EI
LPL
EI
LPC 96
5)2(2
)2
)(2
(
3
)2
( 33
=+=δ
EIPL
EI
LPL
EI
LPC 16
32
)2
)(2
(
)2(2
)2
( 22
=+=θ
EIPL
EIPL
EIPLL
CC 487
163
965)
2(
323
2 =+=+= θδδ
Total deflection
EIPL
EIPL
EIPL
A 163
487
24
323
21 =+=+= δδδ
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Week 11
Homework
9.4-2 9.5-2 9.5-12 9.7-2
9.4-4 9.5-4 9.5-14
9.4-6 9.5-6 9.5-16
9.4-8 9.5-8 9.5-18
9.5-10
16