구조역학II 처짐 강의노트 보충02

(1)

II-02 / 처짐과 탄성변형 37 -Problem_02 다음 켄틸레버보에서 처짐각 와 처짐 을 구하여라. Solution ② 204 It EIEFJI.If ① BMD _a'

#¥

B'

pt

A . B _EI t " - BMD ' ' , to 8 EEZ EIkf8tF7H8t . pl -0

¥

•

-SED ③ OA = Sai = _O A' E 2198 HH EdHFI 018 , HEBE " O " 012k HE 01848140kt . ④ OB = SB' = VB' 71924101 EMILY SB' ( of 4th 23 VB') _E EV

-- ₀ E FEI FTE I 71ft 'Ect

. o: 013=513 ' ₌

I

. L -

1¥

=

27¥

C 2) ⑤ 8A = MA' = O A' E 2198 Htt 4271018 , " Ma' = o "

⑥

813 = MB' 24 A is . : % " _.'

=¥i÷÷•

_, ⇐

÷i¥=¥

(2)

II-02 / 처짐과 탄성변형 39 -Problem_03 다음 단순보에서 지점 A, B에서 처짐각 와 보의 중앙 C점에서 처짐각 , 보의 최대처짐 m ax 을 구하여라. Solution O C ① BMD _GE -- Me

.

e-←

¥74

*¥¥

- -Me --

E. I

- g.

E. ¥

¥

E-

:c

=

#

-

#

VA'

-9

qlz ② 204kt EItb8t%

÷¥¥⇐

A' C " B' ③ OA -- SA' =VA' A = f- bh

HEH EMMA El Italy OA = 7804801117 A' 23421929 Sa'

= 0204801117 A' 2128 VIM VA' * .

,÷÷%÷÷

.

I⇒"

_IF

* * Val =

-913

VB go _, _y , . et Elton H 212611229278 24ft I OA '- SA' = 2¥ VIEW 2714- HEA. ④ OB -- SB' t 8178371 at 72¥ KEEN-60101

safely

.

n

A' El- B 'T 22423 INE 'feet . ① SB' :. 013=5,3 ' ₌

#

CG) 24EI SED ↳ -7142-3471010-2 C SFD Z₎

I

Amok

_.

¢

0lM

(3)

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⇐

⇒

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.

⇒

em

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i

Eti

.

-i==uaf¥

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⑤

Oc -- So' a "

÷i

.

.÷÷

. . c ' . Oc -- O I ItHo7401 ' ' O' ' 0kt . Z 718717k EFFORT .

#

42 ⑥ 8mm

Et# KOHN Smay CKuen ItHotI 247621501 " _O"

14121

F, It Hobo FIFE KOLE 4121011kt FITFoley.

T

. 8mm , = Sci = Me' 0=094121 =

Mate

.

I

-

F.

E. ¥E⇐⇒=

-

E

--

8¥84

-- eh 8 48 EI 3845L 3842-2 384 EI

-€18615123

) 714 24 EI a

¥!

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mantis

_,

test

. I

#

42

(5)

Ei

¥?¥¥±e÷÷

(6)

II-02 / 처짐과 탄성변형 41 -Problem_05 다음 켄틸레버보에서 처짐각 와 처짐 을 구하여라. Solution

O

' '70kg f = noon -- no kgf section

¥

|#m

" mm

÷##

1. 5M i. sm E = 170014Pa mm ① BMD ② FOYE t EIHo8t7f

AIB

A' _B'

#

Ma= - (1700×1.5) = - 1050N .M

¥4

1050N 'M Me = _O 1050 ET

③

Oc , OB

;)

Oc -- Si = '

¥0

.kz. 1.5 =

¥I

= 0.0225rad _a HEHE radian there Either

. = 1.290 ( 0,0225rad _x

¥

= 1.290) 9 , 000 Ni M2 O . EI

-thx

= 35

↳

_too _Gpa = 1700×109×1%2 -Eta

#

TT) _OB -- SB ' = VB' =

¥-0

.-12 . I-5 = ME = 0.0225rad = 1.290 35,000 ④ Sc , SB T) 8c= Me' = .

I

. 1.5×32' I-5 = =

o

= _0.0225M = 2.25cm Cb) IT) 813=1413' =

¥0.2

. I

-5×(71.5+1.5)

=

I=÷;⇐=

0.0563M = 5.63cm Cb) → In A'

.

B' o

t÷¥÷÷!÷

÷

.

fEe¥¥µ

. . . m Oct _-290 8c= 2.25cm B -- 1.290

*

BE

viii.

on

¥④

,as

¥I

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or

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-9E⑤T

-

-0

00 "

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:

TO

0

00

-\ Ah Bl ₄₂₄ wrath my Tu -* •

(7)

He -441 ① OH -441

;)

Oc -- Sc' = '

¥0

.

Z

. 1.5

¥Ij

= = 0.0225rad - -t

_I

. M = = Elo . Yuet there 7¥71

EEE Eat DI EMF

N/m2- MY N 'M

'

I _gave radian HEINI 88

T) do= Mc' =

¥40

.

I

. 1.5×-32. I-5 = 2.25cm Cb)

t⇐§

,

mF=÷mm÷

-- metacenter #.

(8)

II-02 / 처짐과 탄성변형 42 -Problem_06 다음 켄틸레버보에서 처짐각 와 처짐 을 구하여라. Solution ∙ 탄성범위 내에서 구조물 해석이 이루어지므로 다음 그림과 같이 하중조건을 분리하여 처 짐과 처짐각을 구한 후 그 값을 더해주면 된다. p =L 00kW _, 4=20 kN/m , A -_ _2M , balm _, E- 204Pa 200×500 MM section

④

His 'VE

:p

③ I) OB , p =

#

= 100×10-00×-32 = 0.0108rad ( = 0.620 ) ZEI _{2×41,666,667}

•

E I = 20×109 _A = 41,666,667 N' m' 11 _P

f-

) 8,3. p =

¥

= = o. 02117M = 21.17mm Cb) ①

%e-=↳B

-3 EI

3×41,666,667

+ ② EEE 'VE if ②

%l##t

. BMD - Ma - -ga .

E

= -

Ef

B 20×103×22 A e- a -1 by = - - = 80,000 _Nim MA 2 1- 2M -1 A' - K , • 2004 Et EINEN -To

¥fT¥t

B' T)

OB

.q = SB_' _EZ _'

43.274

SB' = _-13

-T¥%f

2=0

. . 00128rad (2 ) ( = 0.017330 ) o: Omg = 0.00128rad C 2)

→ 21-427 Size E 131 HEI TIES UBI It 7-7171. Heat

⇒ _EV '

- O OTI EIZHHVBIEZTFOEI 1719- 'Eet

. TT) 813 -of = MB' MB' ₌

j

- . 2 .

(3-4.2+1)=0.0032

M % 8,3,go = 010032M = 3.2mm Cb) 41,666,665 ③ T) OB = 0,010ft 0.00128 = 0.01208rad C 2) ( = 0.6920 ) TT₎ 813=21.171-3.2 = 24.9mm Cf)

Er

"¥EI

. . a ⇐

2oxjq@imOiia-2oxioIaoymm.2oxo3xyTxswID.A

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.

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I

④

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g. a

t.EE

Alfa

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b -B A

¥i-b - _i Ya= VA =

wia-q.ae

+

sit

-1mm _g.

←

¥i¥T÷E

,¥

streets

Cabot 714712

" o' ' Eh ? 5h bet Cuza og Oc

¥f

= So =

#

13µg , _zzy ° " 42M£ . . " ''

j

_Mp = _o , Me - _o , Ma -- of. ax -8 = g , Mi --

¥23

.

Ea

=

's

o, = so. = Up.

E¥m

co

_..

--

EE

:

÷÷n*¥±

Ea. GEE

(10)

II-02 / 처짐과 탄성변형 43 -Problem_07 다음 켄틸레버보에서 처짐각 와 처짐 을 구하여라. Solution ② BOY Et Effertz ① BMD B'

¥itt÷E÷

:*

Iii

I _II t

③

OB = SBI = VB'

• Sis' ₌ Ait Azt As = .

⇐

¥1

E)

. t

(

III. E)

e

( I

.EEE.

⇒

=¥¥*¥i+¥=¥

' .: Ob = 2) 48 EI ④ SB = MB' . MB' = Aix _, t Aux . + As.X , =

(

I

.

⇒

t⇐

(

E. It

E)

t

¥3

(

E.

It

⇒

=

E.it#i-5aE.ii

=

G3t8-w

₌

¥4

3- 84 EI 384 EI .: OB --

Et

Cb) 384 EI is ÷

in

'

:

:*

Eaten

^¥④]

,

piano

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HE

-OO

(11)

k¥07

c-Enter

⇒

÷

.

*

¥

EL

Sathi

-117

④

t

(

Ix

+

I

Fx

=

¥

MEM.

Iii

.⇒

. Va.-85=0. I. VA -- 0 .

#

SGD _8B= _Mrs'

¥±¥÷i÷*

. . . .

em

A C B

f-1-1

4-t-zfE.TW

(12)

II-02 / 처짐과 탄성변형 44 -Problem_08 다음 켄틸레버보에서 처짐 을 구하여라. (단, Mo=0.4tf.m, P=1.6tf, L=2.4m, EI=600tf.m2) Solution *

stain

. . ③ P ① SB .p =

31¥

= = 0.01229 m = 12.3mm Cb) ② SB- Mo . BMD ° TOY Et

EII-ftotzz.EE

in .

#

¥e¥#

B ' ° 013 - Mo = MB' = -

¥0

.

t

.

¥1

ETL = -

¥j

= -= - 0.005776M =

-5.176mm

(9)

&

_Eez _It _Hotta

's !! ③ 813 = SB -p t 8 B- Mo = 12.3 - 5.176 ₌ 6.54mm Cb)

Liii

::

@

!.g②

(13)

¥ 27N 7 HENK . Ekta .

Weinand

fair

.mn#ni-n-.':i::i:::::

v.

za

: them 9Th . b

010-9386

h , Ehrman . 1221233120kV . ctfu .-01 237106 . Eam . Keil . .

totem

)

(

_animation

_!

- 01244

(14)

- i

OB.ch#--I12lo2tHeELH0dKollHA7toIFHItHo8A=

? @ ( Ed, E = 2.0×106 kgfloma , I = 16,000cm 't ) D

①

BMD _{② Both} _Et _EIk68tE totem = 2.5×4

I

o¥

#

A B C _g , cc B' _II A' . 104¥71 . HEUTE 'rbHE Eid 0¥

EHTLHEH.FI

TYE It 2-117181-2,42801 _ELIEL • ¥21514121011 t IEHVHHEIEE HEHE _-01-2%2 '

IEEE

:

.IE?a7amoaB#Hde.eiEEatatont-soioa

. I 4 so ③ Sa = 204 ENVY Ma ' 0112

a-

I

' ' 4= EL so A' b i. Set = Ma' = × 6

q-pg.jp?Ejiij,?E4-utH2ZEEEoI#

20×6 =

-tf , m -4.412 Eod (2.0×1017)×116,000×10 - f ) _① →

f

¥0 * ol E EIEKOEIH GIEL # VB

's

= 0.0375M = _3.175cm Cb)

_§

- Vis' -- SB' -- OB A' 1- _Lais - I IE} Ma' = 8A =

¥

x LAB Ii El EII toff !! -- Gm .; 8A = SB't LAB ② = OBXLAB - set

Galibi

f

" ⑦ 'm 't' as

*t

- d-B C f 2 2 * _LHDLEEI AB FEMI 81-301 _O , = =

,f¥

=

b-• 013=513' = VB' ofieff 'T BCEEEI 247676451 → ⇐

. 8A= Ma' = Sis AINU

'x

the 'EEEt ¥481. et

is8*-013×6

=¥xb=¥

= Ups

'x6

⇒

7748A

-013×714,0-17456019-1

! =

#

(2.0×107)/16000×10 -B ) = 0.03175M = _3.175cm Meo , Mb = co ,

-O

" " "

÷

. * _. ' _Mio . • o m = r - 3- h = _4-3 - I _so -2bar

-¥⑧&

¥

" .

o

.

*

o

(15)

-wtf _D _ptn been → ' In g-wtf g-

¥99

0¥

,

Ism

-725+2.5-4

* am # ref ftp.EZ?hZt 1- 10 Va=tet set

z5Ea④_d

. b

*

* . f-_tf OB'

Tozz

ma. ④

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s±-

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us,

ai-7iIi.e-EMii@Da.y

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Ma'

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&

9) -- da

(16)

II-02 / 처짐과 탄성변형 45

-2.3 하중재하 조건에 따른 처짐과 처짐각 정리

BBQ

:#

⑨ n - m --- * '

EE

ion

⇒

o¥→

. a

:÷÷⇒

Oreo

:¥

I

&

*

d-•

geo