ch. 7 flexural loading : stresses in beamcontents.kocw.or.kr/document/1.solid mechanics -...

1

Ch. 7 Flexural Loading : Stresses in Beam

Ø Beam : member subjected to loads applied transverse to the long dimension, which cause the member to bend.

Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수

Ø Beams are classified on their supports or reactions.

2

7-1. Introduction

(a) Simple beam- supported by a roller- having on span- normal reaction and no couple

(b) Simple beam with overhang

(c) Continuous beam- with more than two simple supports

(d) Cantilever beam


(d) Cantilever beam- one end is built into a wall- neither move transversely nor rotate

(e) Beam fixed at left end and simply supported near the other end

(f) Beam fixed at both ends

3

7-1. (continued)

:rV Resisting shear at the section:rM Resisting moment at the section

:R Reaction at the support

å = 0yF

l From a free body diagram of entire beam

Rl From


o is any axis perpendicular to xy plane

rV

å = 0oMl From

rM

òò-=

-=

A xr

A xyr

dAyM

dAV

s

t

4

7-2. Flexural strains

l Neutral surface : surface of which longitudinal elements undergo no change in length

l Plane section before bending remains a plane after bending

l Neutral axis : intersection of neutral surface with any cross section

Þ-

==i

ifx L

LLLde


Þ-

==i

ifx L

LLLde

yyx

xxx rqr

qrqre 1-=

DD-D-

=D

D-D=

)()())(('

Strain developed in a fiber is directly proportional to the distance of the fiber from the neutral surface of the beam

5

7-3. Flexural stresses

l For linear elastic action

Normal stress on transverse cross section of the beam varies linearly with distance y from the neutral surface

yEE xx res -==

òò -=-=A xAr dAydFyM s

l Equilibrium equation for pure bending


0

0

=-=-=

÷÷ø

öççè

æ-=

===

ò

òò

òòå

AyEdAyE

dAyEdA

dAdFF

cA

AA x

A xAx

rr

rs

s

Yc is the distance from the neutral axis to the centroidal axis c-c of the cross section

Neutral axis passes through the centroid of the cross section

6

7-3. (continued)

where c is the distance to the surface of the beam farthest from the neutral surface

cEr

s -=max

cx cy

cyyE ss

rs ==-= max

yEE xx res -==by using

ycE xssr

==- max


òò -=-=A

cA xr dAy

cdAyM 2ss

cx cy

cyyE ss

rs ==-= max

Second moments of more complex areas can be derived from combinations of simple shapes

: second moment of area (Izz)

7

7-4. Elastic flexural formula

l For elastic flexural formula

yIMr

x -=s

l For the section about y-axis

where S is the section modulus of the beam

SM

IcM rr -=-=maxs


l For the section about y-axis

0==== òòò yzc

Ac

Ac

A x Ic

zydAc

dAyc

zdAz ssss

where Iyz is the mixed second moment of cross sectional area with respect to the centroidal y- and z-axes.

By considering equilibrium equation, å = 0yM

8

7-4. (continued)

yIM

IyM x

rr

xss -=Û-=

(Example 7-2) The maximum flexural stress at a given is 15 MPa. Determine (a) the resisting moment developed at the section, (b) percentage decrease in resisting moment if the dotted portion of the cross section is removed

(a) By using

464633

10676610676612

20010012

mmmbhI )(.)(. -==´

==

mkNmNcIM x

r -=-=== -

-

10101010100

1067661015 33

66

)()()(.)(s


mkNmNcIM x

r -=-=== -

-

10101010100

1067661015 33

66

)()()(.)(s

(b) 4646333

106521065212

1505012

20010012

mmmbhI )(.)(. -==´

-´

==

mkNmNcIM x

r -=-=== -

-

8971089710100

106521015 33

66

.)(.)()(.)(s

%.. 12110010

89710=´

-== Ddecreasepercent

9

7-4. (continued)

(Example 7-3) On a section where resisting moment is -75kN-m, determine (a) maximum tensile flexural stress, (b) maximum compressive flexural stress.


AMyAydAy A

ccA=Û=ò

The neutral axis is horizontal and passes through the centroid of the cross section

mmAMy

mmMmmA

Ac

A

9550012

5001871500187120075501002515051225200

50012755025150252003

2

===

=++=

=++=

,,,

,,))(())(().)((,)()()(

10

7-4. (continued)

The second moment of area of the part w.r.t. horizontal centroidal axis of the total cross section


For part A :

2CxCx yAII +='

4623

2 1013421055075125075 mmyAII CAAxCAAx )(.))(()(

' =+=+=

Similarly

4623

2

4623

2

1029345822520012

25200

101375150251215025

mmyAII

mmyAII

CCBxCCCx

CBBxCBBx

)(.).)(()(

)(.))(()(

'

'

=-+=+=

=+=+=

4646 105583105583 mmmIIII CxBxAxx )(.)(.''''-==++=

(Parallel axis theorem)

11

7-4. (continued)

(a) Since resisting moment is negative, maximum tensile flexural stress occurs at the top of the beam


MPamNIyM tr 7116107116

105583101301075 26

6

33

./)(.)(.

))()()((max ==

--=-= -

-

s

(a) Since resisting moment is negative, maximum tensile flexural stress occurs at the top of the beam

(b) Maximum compressive flexural stress occurs at the bottom of the beam

MPamNIyM br 385102885

10558310951075 26

6

33

./)(.)(.

))()()((max -=-=

---=-= -

-

s

12

7-5. Shear forces and bending moments

l Force equilibrium equation,

rr VVorVPwxR ==--

where V is the resultant of the external transverse forces called as transverse shear at the section

l Moment equilibrium equation,

rr MMorMhxPwxRx ==--- )(2

2

where M is the algebraic sum of the moments of the external forces called as bending moment at the section

å = 0yF

å = 0oM


where M is the algebraic sum of the moments of the external forces called as bending moment at the section

13

7-5. (continued)

(Example 7-4) Write equations for the shear force V and the bending moment M (a) in the interval AB, (b) in the interval BC, (c) in the interval CD.

åå

=--=

=--=

062000564008

022000364008

2

1

)())(()(

)())(()(

RM

RM

A

D

lbRandlbR 30001400 21 ==


14

7-5. (continued)

(a)

)( ftxlbVVFy

201400

01400

<<=

=-=å

)( ftxlbftxMMxMo

201400

01400

<<-=

=+-=å

(b)

)()(

ftxlbxVVxFy

622200400

024001400

<<+-=

=---=å

)(

)(

ftxlbftxxM

MxxxMo

628002200200

02

224001400

2 <<--+-=

=+÷øö

çèæ -

-+-=å


)(

)(

ftxlbftxxM

MxxxMo

628002200200

02

224001400

2 <<--+-=

=+÷øö

çèæ -

-+-=å

(c)

)()(

ftxlbxVVxFy

86200400

0200024001400

<<+-=

=----=å

)(

)()(

ftxlbftxxM

MxxxxMo

8611200200200

0620002

224001400

2 <<-++-=

=+-+÷øö

çèæ -

-+-=å

15

7-5. (continued)

(c)

)()(

ftxlbxVVxFy

86200400

0200024001400

<<+-=

=----=å

)(

)()(

ftxlbftxxM

MxxxxMo

8611200200200

0620002

224001400

2 <<-++-=

=+-+÷øö

çèæ -

-+-=å


)(

)()(

ftxlbftxxM

MxxxxMo

8611200200200

0620002

224001400

2 <<-++-=

=+-+÷øö

çèæ -

-+-=å

similarly

)()(

ftxlbxVxVFy

86200400

030008400

<<+-=

=+--=å

)(

)()(

ftxlbftxxM

xxxMMo

8611200200200

0830002

88400

2 <<-++-=

=-+÷øö

çèæ -

---=å

16

7-5. (continued)

(Example 7-5) On a section 3m to the right of A, determine (a) flexural stress at a point 25mm below the top of the beam, (b) maximum flexural stress on the section.


kNRRM

A

AB

62

081210768220

=

=-++=å )()())(()(

S152X19

mkNMMMo

-=

=+-+=å16

07103625258 )()().)((

mdmSmI

1524.0)10(121)10(20.9

36

46

==

=-

-

17

7-5. (continued)

(a) At a point 25mm below the top of the beam,

MPamNIyM

mmmdy

08910048910209

051201016

05120251252

4152252

266

3

./)(.)(.).)((

...

-=-=-=-=

==-=-=

-s

(b) Maximum flexural stress

)(.)(.

/)(.)()(

max

CMPaTMPa

mNSM

IMc

2313223132

1023132101211016 26

6

3

==

==== -s


)(.)(.

/)(.)()(

max

CMPaTMPa

mNSM

IMc

2313223132

1023132101211016 26

6

3

==

==== -s

on the bottomon the top

18

7-6. Load, Shear force and Bending moment

l Mathematical relationships between loads, shear forces, and bending moments

å = 0yF

l From force equilibrium

xwPVVVPxwV

avg

LavgL

D+=D

=D+-+D+ 0)(

(1) If P=0 and w=0, shear force is constant

RL VVV =Þ=D 0

(2) If , shear force jumps by the concentrated load as

0¹P0®Dx


PVVPV LR +=Þ=D

(3) If , the slope of shear force graph is equal to the intensity of loading as

wdxdV

xVxwV

xavg ==DD

ÞD=D®D 0lim

òò ==-Þ==DD

®D

2

1

2

1120

x

x

V

VxwdxdVVVw

dxdV

xVlim

Change in shear between sections is equal to the area under the load diagram

0=P0®Dx

19

7-6. (continued)

å = 0centerM

l From moment equilibrium

022

=D+D

D+-D

---D+ )()()( xwaxVVxVCMMM avgLLLL

wwandaxas

xaxforxwaxVxVCM

avg

avgL

®®®D

D<<

D-D-

DD+D+=D

00222

,

)(

(1) If , bending moment jumps by C as 0¹C 0®Dx


(1) If , bending moment jumps by C as

CMMCM LR +=Þ=D

(2) If , the slope of bending moment is equal to the value of shear force at that section as

)( xwaxVxVM avgL D-D

D+D=D2

0¹C

00 == PC ,

VdxdM

xM

x==

DD

®D 0lim

òò ==-2

1

2

112

x

x

M

MVdxdMMM

0®Dx

20

7-6. (continued)

l Shear and bending moment diagrams provide a convenient method for obtaining maximum values of shear and bending moment.

(2) algebraic equations in case of uniformly distributed or varies according to a known equation

(3) drawing shear diagram from the load diagram and bending moment diagram from shear diagram

(1) calculating values of shear and bending moment at various sections


21

7-6. (continued)

(Example 7-6) (a) write equations for shear and bending moment in AB, (b) write equations for shear and bending moment in BC, (c) draw complete shear and bending moment diagrams for the beam

(a) Section in AB

lbftMandlbVMM

VF

CC

CC

Cy

×-=-=Þ

=++-=

=---=

åå

86001700

03620010500

06200500

))(()(

)(

:0=å yFftxforlbV

VFy

40500

0500

<<-=Þ

=--=å


:0=å yFftxforlbV

VFy

40500

0500

<<-=Þ

=--=å

:0=å OMftxforlbftxM

MxMO

40500

0500

<<×-=Þ

=+=å )(

(b) Section in BC

:0=å yFftxforlbxV

VxFy

104200300

04200500

<<-=Þ

=----=å )(

:0=å OM

ftxforlbftxxM

MxxxMO

1041600300100

02442005002 <<×-+-=Þ

=+--+=å /))(()(

22

7-7. Shear stresses in beams

l Beam by stacking flat slabs one on top of another without fastening them together

Relative motion of the ends of the cards with respect to each otherSolid beam does not exhibit this relative movementIndication of presence of shearing stress on longitudinal planes


23

stressflexuralI

MywheredAF

dytdAdF

:-==

==

ò ss

ss

òò

òòD+

-=D+

-=

-=-=

c

yA

c

yA

dytyI

MMdAyI

MMF

dytyIMdAy

IMF

1

1

2

1

)()()(

)(

l Differential force on area dA

l Resultant normal forces

7-7. (continued)


òD

-=-=c

yH dytyIMFFV

112 )(

òò

ò

÷ø

öçè

æ-=÷

ø

öçè

æ-

DD

=

DD

-==

®D

c

y

c

yx

c

ys

Havg

dytytIdx

dMdytytIx

M

dytyxtI

MAV

11

1

110

)()(lim

)(

t

t

tIQV

H -=t

l Summation in horizontal direction

: Q is first moment of area

24

úúû

ù

êêë

é-÷

øö

çèæ==

==

ò

ò

21

22

221

1

yhIVdyty

tIV

dytytIV

tIQV

h

y

c

yt

l Transverse shear stress at a point of section

7-7. (continued)

( ) AV

htV

hthV

IhV

23

23

1288 3

22

max ====t

l This equation is used for rectangular cross sectionl Maximum shear stress is 1.5 times average shear stress

l This equation is worthless for I-beam or T-sections

For rectangular cross sections3% error : beam with depth having twice width

12% error : beam with square cross section100% error : beam with width having four times depth


l This equation is used for rectangular cross sectionl Maximum shear stress is 1.5 times average shear stress

l This equation is worthless for I-beam or T-sections

For rectangular cross sections3% error : beam with depth having twice width

12% error : beam with square cross section100% error : beam with width having four times depth

25

inyC 4)2(10)10(2)1)(2(10)7)(10(2=

++

=

7-7. (continued)

l Transverse shear stress at a point of section

( ) ( ) )( 82822 1

21

221

21

1

1

<<--=-=

== ò

yyI

VycI

V

dytytI

VtIQV c

yt

l Location of centroid

AMy A

c =Ü

l Variations of shearing stresses on an inverted T-shaped transverse section


( ) ( ) )( 82822 1

21

221

21

1

1

<<--=-=

== ò

yyI

VycI

V

dytytI

VtIQV c

yt

l Discontinuity at a junction of flange and stem because of abrupt change in thickness

l Maximum longitudinal and transverse shear stresses occur at the neutral surface at a section where transverse shear V is maximum.

( ) ( ) )( 24422 1

21

221

22

2

1

-<<--=-=

== ò

yyI

VycI

V

dytytI

VtIQV c

yt

26

7-7. (continued)

Let shear V be 37.5 kN for W203 X 22 section (I=20 X 106mm4)

at the neutral axis

MPamNtIQV

mmmQ

w

NANA

NA

9.32/)10(89.32

)10)(2.6)(10(0.20)10)(76.108)(10(5.37)10(76.108)10(76.108

)5.47)(2.6(95)99)(8(102

26

36

63

3633

@=

==

==

+=

--

-

-

t

l Detremination of the shearing stress in an I-beam


MPamNtIQV

mmmQ

w

NANA

NA

9.32/)10(89.32

)10)(2.6)(10(0.20)10)(76.108)(10(5.37)10(76.108)10(76.108

)5.47)(2.6(95)99)(8(102

26

36

63

3633

@=

==

==

+=

--

-

-

t

at the junction between web and flange

MPamNtIQV

mmmQ

w

JJ

J

4.24/)10(43.24

)10)(2.6)(10(0.20)10)(78.80)(10(5.37

)10(78.80)10(78.80)99)(8(102

26

36

63

3633

@=

==

===

--

-

-

t

27

7-7. (continued)

- Shear force V causes shear stress, yxxy tt =

- Resolve this shear stress into n-t coordinate system

xnxtxy and ttt Þ

0== nxxn tt- Outside surface is a free surface

- Any shearing stress at point A must be tangential to the surface of the shaft and not in the direction of shear force

l Beam with a solid circular cross section subjected to shaer load V


- At neutral axis, shear stress in the direction of V is

AV

rrrrV

ItVQ

NA

NANA 3

424

3424

2

====)()())((

max ppptt

- Any shearing stress at point A must be tangential to the surface of the shaft and not in the direction of shear force

32

34

2

32 rrrQNA =´=Üp

p

28

7-7. (continued)

(Example 7-10) Determine

(a) average shearing stress on a horizontal plane 4” above the bottom of the beam and 6’ from the left support

Shear force V on 6’ from the left support is 900 lbFirst moment Q4 is

3444 48426 inAyQ C === ))((

Second moment of area about the neutral axis is

Average shearing stress on a horizontal plane 4” above is

42323 253332102101213102102

121 inINA .))(())(())(())(( =+++=


Average shearing stress on a horizontal plane 4” above is

42323 253332102101213102102

121 inINA .))(())(())(())(( =+++=

psitI

VQNA

54022533

489004

44 .

)(.)(

===t

(b) Maximum transverse shearing stress in the beam

Maximum shearing stress will occur at the neutral axis on the cross section occurring largest shear force V.

364824 inQNA == ))((

psitI

VQsNA

NA 05422533

64900 .)(.)(

max ===t

29

7-7. (continued)

(c) Average shearing stress in the joint between the flange and the stem at 6’ from the left support

psitI

VQinAyQ

sNA

FJ

FCFF

65022533

60900602103 3

.)(.)())((

===

===

t

(d) Force transmitted from the flange to the stem by the glue in a 12” length of the joint at a section 6’ from the left support

lbAV JJg 112152126350 .))((. === t


lbAV JJg 112152126350 .))((. === t

(e) Maximum tensile flexural stress in the beam

)(..))((max

max TpsiI

cM 112153533

8126750===s

30

7-8. Principal stresses in flexural members

l To find principal stresses and maximum shearing stresses at the selected points on the sections of maximum shear V and maximum bending moment M

l flexural stress is maximum

l transverse and longitudinal shearing stresses are zero

flexural stresses are principal stress

20-

= pstmax

l At point on the top and bottom edge of the section


20-

= pstmax

l transverse and longitudinal shearing stressesare maximum

l flexural stress is zero

Shearing stresses are maximum shearing stresses

l At point on neutral axis

31

7-8. (continued)

(Example 7-10) Cantilever beam carries a uniformly distributed load of 160kN/m on a span of 2.5m. Determine maximum normal and shearing stresses in the beam.

Maximum bending moment and shearing transverse shear are

kNwLV

mkNwLM

400)5.2(160

5002)5.2(160

2

22

===

×-=-=-=


W610x145I=1243x106 mm2

S=4079x103 mm3

32

7-8. (continued)

At neutral axis, flexural stress is zero and

MPamNtI

VQmmQNA

06110986001190101243

10255210400

102552614291112852957198304

266

33

36

./)(.).)(())(.)((

)(.).)(.(.))(.(.

@===

=+=

-

-

t

In the web at the junction with top flange

)(./)(.)(

).)()(( TMPamNIyM 71141068114

1012432851010500 26

6

3

@=-

-=-= -s


MPamNtI

VQmmQJ

94710894701190101243

10771110400

1077112957198304

266

33

36

./)(.).)(())(.)(()(.))(.(.

@===

==

-

-

t

)(./)(.)(

).)()(( TMPamNIyM 71141068114

1012432851010500 26

6

3

@=-

-=-= -s

At the top surface, transverse shearing stress is zero and

)(./)(.)()( TMPamN

SM

IyM 61221058122

10409710500 26

6

3

@=-=-=-= -s

33

7-8. (continued)

For the points at the junction of web and flange

deg..

).(tantan

..).(.)(....)(....

..

).(..

minmaxmax

,

9319068114

89472212

21

47441742

3717051322

3417341741743457

1132051324174345741743457

89472

0681142

068114

22

11

2

1

22

22

21

-=-

-=

-=

@=--

=-

=

@-=-=

@=+=±=

-+÷øö

çèæ -

±+

=

+÷÷ø

öççè

æ -±

+=

--

yx

xyp

p

p

xyyxyx

pp

MPaMPa

CMPaMPaTMPaMPa

sst

q

sst

s

s

tssss

s


deg..

).(tantan

..).(.)(....)(....

..

).(..

minmaxmax

,

9319068114

89472212

21

47441742

3717051322

3417341741743457

1132051324174345741743457

89472

0681142

068114

22

11

2

1

22

22

21

-=-

-=

-=

@=--

=-

=

@-=-=

@=+=±=

-+÷øö

çèæ -

±+

=

+÷÷ø

öççè

æ -±

+=

--

yx

xyp

p

p

xyyxyx

pp

MPaMPa

CMPaMPaTMPaMPa

sst

q

sst

s

s

tssss

s

Stress Top edge Junction Neutral axis

sp1 122.6 MPa (T) 132.1 MPa (T) 61.0 MPa (T)

sp2 0 17.37 MPa (C) 61.0 MPa (C)

tmax 61.3 MPa 74.7 MPa 61.0 MPa

34

7-9. Flexural stresses – Unsymmetric bending

zkyka 21 ++=s

l stress variation is

òò

ò

-=

=

==

Arz

Ary

A

dAyM

dAzM

dAR

s

s

s 0

l force and moment equilibrium are

òòòòòò

òòò

---=

++=

=++=

AAArz

AAAry

AAA

dAyzkdAykdAyaM

dAzkdAyzkdAzaM

zdAkydAkdAaR

22

1

221

21 0


òòòòòò

òòò

---=

++=

=++=

AAArz

AAAry

AAA

dAyzkdAykdAyaM

dAzkdAyzkdAzaM

zdAkydAkdAaR

22

1

221

21 0

l The origin of coord. is at the centroid of cross section

òòòòò

===

==

AyzAyAz

AA

dAyzIdAzIdAyI

dAzdAy

,, 22

0

Iy, Iz : second moment of areaIyz : mixed second moment of area

35

7-9. Flexural stresses – Unsymmetric bending

yzzrz

yyzry

IkIkMIkIkM

AaR

21

21

0

--=

+===

22

21

0

yzzy

yzrzzry

yzzy

yzryyrz

IIIIMIM

k

IIIIMIM

k

a

-+

-=

-+

-=

=

l elastic flexural formula for unsymmetric bending is

rzyzzy

yzyry

yzzy

yzz

yzzy

yzrzzry

yzzy

yzryyrz

MIII

zIyIM

IIIyIzI

zIII

IMIMy

IIIIMIM

úû

ùêë

é

-+-

+úû

ùêë

é

--

=

úû

ùêë

é

-+

+úû

ùêë

é

-+

-=

22

22s


rzyzzy

yzyry

yzzy

yzz

yzzy

yzrzzry

yzzy

yzryyrz

MIII

zIyIM

IIIyIzI

zIII

IMIMy

IIIIMIM

úû

ùêë

é

-+-

+úû

ùêë

é

--

=

úû

ùêë

é

-+

+úû

ùêë

é

-+

-=

22

22s

l orientation of neutral axis is

0=+++- zIMIMyIMIM yzrzzryyzryyrz )()(

zIMIMIMIM

yyzryyrz

yzrzzryúû

ùêë

é

++

=

l slope of neutral axis is

yzryyrz

yzrzzry

IMIMIMIM

dzdy

++

== btan

36

7-9. (continued)

(Example 7-12) (a) Flexural stress at A

)(781.0781.0)300()108()240(135)5.4(108)4(1352

2

22

Cksiksi

MIIIzIyI

MIIIzIyI

MIIIyIzI

A

rzyzzy

yzy

rzyzzy

yzyry

yzzy

yzz

=-=úû

ùêë

é-

+-=

úúû

ù

êêë

é

-

+-=

úúû

ù

êêë

é

-

+-+

úúû

ù

êêë

é

-

-=

s

s

(b) Orientation of neutral axis

deg7.38

800.0135108tan

=

===+

+=

b

by

yz

yzryyrz

yzrzzry

II

IMIMIMIM


deg7.38

800.0135108tan

=

===+

+=

b

by

yz

yzryyrz

yzrzzry

II

IMIMIMIM

(c) Maximum tensile and compressive flexural stresses

)(16.1016.10)300()108()240(135)5.1(108)4(135

)5.1,4(

)(16.1016.10)300()108()240(135)5.1(108)4(135

)5.1,4(

2

2

Tksiksi

CAt

Cksiksi

BAt

C

B

==úû

ùêë

é-

---=

-

=-=úû

ùêë

é-

---=

-

s

s

37

7-10. Stress concentrations under flexural loadings

l flexural stress in smooth member

IMy

-=s

l flexural stress in the vicinity of discontinuity

÷øö

çèæ-=

IMyKs K : stress concentration factor that depends on geometry of the member


38

7-10. (continued)

494333

1016510165122550

12

41252

10301030

mmmbhI

MPaFS

MPa

uall

u

)(.)(.)(.

-====

===

=

ss

s

(Example 7-13) Cantilever is made of SAE 4340 steel and is 50mm wide. Determine the maximum safe moment M if FS of 2.5 with respect to failure by fracture is specified.

(1) For r=5mm,5112025532575 .., =Þ==== tKhrhw

(2) For r=10mm


(2) For r=10mm

28140251032575 .., =Þ==== tKhrhw

(3) For r=15mm18160251532575 .., =Þ==== tKhrhw

ttt

a

KKyKIM 2146

105121016510412

3

96

=-

-=-= -

-

))(.())(.)((s

yI

KyIM

IMy

t

allsss -=-=Þ-=

(1) For r=5mm mNK

Mt

×=== 1421511

21462146.

using

ch. 7 flexural loading : stresses in beamcontents.kocw.or.kr/document/1.solid mechanics -...

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