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

1

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

2

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 −

−==

3

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

4

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

−⇒

6

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

7

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

8

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

9

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

10

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

11

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.

12

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

13

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

14

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

15

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