ed.misa.donga.ac.kr/files/4_3_순수굽힘.pdfmechanics of materials beer • johnston • dewolf...

Copyright © 2011 by Tae Hyun Baek, Kunsan National University. All rights reserved.

MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 1

한개의대칭면을지닌부재의소성변형(Plastic Deformations of Members with a Single Plane of Symmetry)

• 수직으로대칭면을갖는완전한 소성변형보

• 요소의압축력과인장력력의합력인 R1과 R2에의하여우력(couple)이발생.

YY AARRσσ 21

21==

중립축은단면을똑같은면적으로분할

• 부재의소성모멘트

( )dAM Yp σ21=

• 중립축은단면의도심을통과한다고가정할수없음.

http://100.naver.com/100.nhn?type=image_list&docid=23498&dir_id=12160105�


MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 2

잔류응력(Residual Stresses)

• 굽힘모멘트가크게증가되면탄소성재료로만들어진부재에서소성영역이발생

• 선형관계는하중을제거하는동안에는부재의모든점에서작용할수있으므로부재가완전탄성을지닌다고가정하여해석한다.

• 잔류응력은잔류응력은굽힘모멘트M (elastoplastic deformation) 의작용에의한응력과부재의하중을제거하기위하여작용되는크기가같고방향이반대인굽힘모멘트 –M(elastic deformation) 에의한역응력을중첩의원리를적용하여구한다.

• 임의지점의최종값은일반적으로 0 이되지않는다.



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 3

예제 4.05, 4.06(Example 4.05, 4.06)

• 직사각형단면부재가 36.8 kN의굽힘모멘트를받고있다. 부재가 240 MPa의항복응력을지닌탄소성재료로만들어졌으며탄성계수가 200 GPa일때, (a) 탄성영역의두께와 (b) 중립면의곡률반지름을계산하시오.

• 하중이 0으로감소한후의 (c) 잔류응력분포와 (d) 곡률반지름을구하시오.



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 4

예제 4.05, 4.06 (Example 4.05, 4.06)

( )( )

( )( )

mkN 8.28

MPa240m10120

m10120

m1060m1050

36

36

233322

32

⋅=

×==

×=

××==

−

−

−−

YY cIM

bccI

σ

• 최대탄성모멘트:

• 탄성영역(elastic core)의두께:

( )

666.0mm60

1mkN28.8mkN8.36

1

2

2

31

23

2

2

31

23

==

−⋅=⋅

−=

YY

Y

YY

ycy

cy

cyMM

mm802 =Yy

• 곡률반지름

3

3

3

9

6

102.1m1040

102.1

Pa10200Pa10240

−

−

−

×

×==

=

×=

×

×==

Y

Y

YY

YY

y

y

E

ερ

ρε

σε

m3.33=ρ



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 5

예제 4.05, 4.06 (Example 4.05, 4.06)

• M = 36.8 kN-m

MPa240mm40

Y ==

σYy

• M = -36.8 kN-m

Y

36

2MPa7.306m10120mkN8.36

σ

σ

<=×

⋅==′

IMc

m

• M = 0

6

3

6

9

6

105.177m1040

105.177

Pa10200Pa105.35

core, elastic theof edge At the

−

−

−

×

×=−=

×−=

×

×−==

x

Y

xx

y

E

ερ

σε

m225=ρ



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 6

대칭면에 작용하는 편심축하중(Eccentric Axial Loading in a Plane of Symmetry)

• 비래한도(proportional limit) 이하에서유효하며, 굽힘에의한변형은치수에영향을거의미치지않고하중작용점에가까운응력이아니어야한다.

• 편심하중

PdMPF

==

• 편심하중에의한응력은중심축하중에의한균일한응력과순수굽힘에의한선형적인응력분포를중첩(superpose)시켜계산

( ) ( )

IMy

AP

xxx

−=

+= bendingcentric σσσ



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 7

예제 4.07(Example 4.07)

12 mm 지름의저탄소강봉을구부려서그림과같은링크를만들었다. 링크가 700 N의하중을받고있을때 (a) 링크의곧은부분에서발생하는가장큰인장및압축응력, (b) 단면의도심축과중립축사이의거리를구하시오.

풀이:

• 등가의중심축하중과굽힘모멘트를구한다.

• 중심축하중에의한균일한응력과굽힘모멘트에의한선형으로분포된응력을서로더한다(중첩).

• 중첩시킨응력분포로부터안쪽과바깥쪽끝의최대인장응력과압축응력을계산한다.

• 수직응력이 0 이되는지점을결정하여중립축을구한다.



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 8

Example 4.07

• 등가의중심축하중과굽힘모멘트

( )( )Nm2.11

m016.0N700N700

===

=PdM

P

( )

MPa2.6m101.113

N700mm1.113

mm6

260

2

22

=×

==

=

==

−AP

cA

σ

ππ

• 중심축하중에의한수직응력

( )

( )( )

MPa66m109.1017

m006.0Nm2.11mm9.1017

mm6

412-

4

4414

41

=×

==

=

==

IMc

cI

mσ

ππ

• 굽ㅎ미모멘트에의한수직응력



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 9

예제 4.07(Example 4.07)

• 최대인장및압축응력

662.6

662.6

0

0

−=−=+=+=

mc

mt

σσσ

σσσMPa2.72=tσ

MPa8.59−=cσ

• 중립축위치

( )Nm.211

m109.1017Pa102.6

0

4126

0

0

−××==

−=

MI

APy

IMy

AP

mm56.00 =y



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 10

견본문제 4.8 (Sample Problem 4.8)

주철링크의최대허용인장응력이 30 MPa이고압축허용응력이 120 MPa일때, 링크에작용할수있는가장큰힘 P를구하라.

풀이:

• 등가의중심축하중과굽힘모멘트를결정한다.

• 허용인장및압축응력에대한임계하중을구한다.

• 최대허용하중은두개의임계하중중작은값이다.

From Sample Problem 4.2,

49

23

m10868

m038.0m103

−

−

×=

=

×=

I

YA

• 중심축하중에의한응력과굽힘에의한응력을겹쳐놓는다(superpose).



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 11

견본문제 4.8 (Sample Problem 4.8)

• 등가의중심축하중과굽힘모멘트를결정한다

moment bending 028.0load centric

m028.0010.0038.0

====

=−=

PPdMPd

• 허용인장및압축응력에대한임계하중

kN0.77MPa1201559kN6.79MPa30377

=−=−===+=

PPPP

B

Aσσ

kN 0.77=P• 최대허용하중

• 중심축하중응력과굽힘응력을겹친다.( )( )

( )( ) PPPI

McAP

PPPI

McAP

AB

AA

155910868

022.0028.0103

37710868

022.0028.0103

93

93

−=×

−×

−=−−=

+=×

+×

−=+−=

−−

−−

σ

σ



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 12

비대칭 굽힘 (Unsymmetric Bending)

• 순수굽힘에대한해석은, 적어도하나의대칭면을갖으며그대칭평면에작용하는모멘트를받는부재들에제한

• 굽힘모멘트가부재의대칭면상에서작용하지않는경우를고려하자.

• 일반적으로단면의중립축은모멘트축(axis of couple)과일치하지않는다.

• 이경우, 부재가모멘트평면내에서구부러질것이라고가정할수없다.

• 횡단면의중립축은모멘트축과일치한다.

• 부재들은대칭적으로남아있고그평면내에서구부러진다.



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 13


임의모양을갖는횡단면의중립축이그단면에작용하는힘을나타내는모멘트M의축과일치하기위한엄밀한조건의결정

• 모멘트벡터 (couple vector) 는주도심축 (principal centroidal axis) 방향을향한다.

inertiaofproductIdAyz

dAcyzdAzM

yz

mxy

==∫=

∫

−=∫==

0or

0 σσ

• 단면의요소힘분포로부터모멘트와합력은다음식을만족

coupleappliedMMMF zyx ==== 0

• 중립축은도심을통과∫=

∫

−=∫==

dAy

dAcydAF mxx

0or

0 σσ

• 응력분포를정의

inertiaofmomentIIc

Iσ

dAcyyMM

zm

mz

===

∫

−−==

Mor

σ



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 14


비대칭굽힘의가장일반적인경우, 응력을결정할때중첩의원리를사용할수있다.

• 응력성분분포를중첩(superpose)한다.

y

y

z

zx I

yMI

yM+−=σ

• 모멘트벡터(couple vector)를주도심축에따른성분으로분해

θθ sincos MMMM yz ==

• 중립축에서는( ) ( )

θφ

θθσ

tantan

sincos0

y

z

yzy

y

z

zx

II

zy

IyM

IyM

IyM

IyM

==

+−=+−==



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 15

예제 4.08 (Example 4.08)

직사각형단면의나무보에 180 N-m의모멘트가수직에서 30o기울어진각도를이루는면에작용하고있다. (a) 보의최대응력, (b) 중립면이수평면과이루는각도를구하라.

풀이:

• 모멘트벡터를주도심축에따른성분으로분해하고상응하는최대응력을계산한다.

θθ sincos MMMM yz ==

• 응력성분분포로부터응력을조합(combine)하면,

y

y

z

zx I

zMI

yM+=σ

• 중립축각도를결정

θφ tantany

zII

zy==



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 16

Example 4.08

• 모멘트벡터를주도심축에따른성분으로분해, 상응하는최대응력을계산

( )( )( )( )( )( )

( )( )

3 -6 4112

3 -6 4112

1 -6 4

2

180 Nm cos30 155.9 Nm

180 Nm sin 30 90 Nm

0.04m 0.09m 2.43 10 m

0.09m 0.04m 0.48 10 m

155.9 Nm 0.045m

2.89MPa2.43 10 m

z

y

z

y

z

z

z

y

y

M

M

I

I

M AB

M yI

M AD

M z

σ

σ

= =

= =

= = ×

= = ×

= = =×

=

에의한가장큰인장응력은를따라발생

에의한가장큰인장응력은를따라발생

( )( )-6 4

90 Nm 0.02m3.75MPa

0.48 10 myI= =

×

• 조합하중으로인한가장큰응력은 A에서발생

75.389.221max +=+= σσσ MPa64.6max =σ



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 17

예제 4.08 (Example 4.08)

• 중립축의각도를결정

9.2

30tanm1048.0m1043.2tantan 46-

4-6

=

××

== θφy

z

II

o71=φ



MEC

HAN

ICS

OF M

ATERIA

LS

Beer •

Johnston • DeW

olf • Mazurek

5thEd.

Mechanics of M

aterials

5- 18

편심 축하중의 일반적인 경우 (General Case of Eccentric Axial Loading)

• 크기가같고반대방향으로작용하는편심축하중을받는곧은부재에고려

• 겹침의원리에의한조합응력분포

y

y

z

zx I

zMI

yMAP

+−=σ

• 중립축이단면에위치할경우, 다음식으로계산

APz

IM

yI

My

y

z

z =−

• 편심축하중은도심력(centric force)과두개의모멘트받는시스템과같다.

PbMPaMP

zy === force centric


ed.misa.donga.ac.kr/files/4_3_순수굽힘.pdfmechanics of materials beer • johnston • dewolf...

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