Lecture Note for Solid Mechanics - KOCWcontents.kocw.net/KOCW/document/2014/kumoh/yoonsungho2/1.pdf · 2016. 9. 9. · Lecture Note for Solid Mechanics - Deflections in Beam-Prof

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


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.


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


(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


(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

)(.)(.

)(


(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

)(.)(.

).(


(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


(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

).(


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


(continued)

Examples of singularity function


(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

)()()(

)()()()(

)(


(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 )(


(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.


(continued)


(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 :


(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


(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


(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 ))((


(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 (**)


(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


(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


(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



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



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



(continued)

From solving simultaneous equations

04

564

4364

4364

3764

3764

764

7 22

wLVRFwLwLR

wLwLV

wLwLM

y

2153232

83

450

wLMVL

LLwMVLM R



(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


Homework

(8-4), (8-14), (8-41), (8-60)

Documents

Lecture Note for Solid Mechanics - KOCWcontents.kocw.net/KOCW/document/2014/kumoh/yoonsungho2/1.pdf · 2016. 9. 9. · Lecture Note for Solid Mechanics - Deflections in Beam-Prof