1Unsymmetric Bending

Eğik EğilmeUnsymmetric Bending

Unsymmetric Bending 2

Chapter Outline

• Pure Bending• Unsymmetric Bending• Inertia moments• Maximum and minimum stresses

Unsymmetric Bending 3

• Analysis of pure bending has been limited to members subjected to bending couples acting in a plane of symmetry.

• The neutral axis of the cross section coincides with the axis of the couple

• Members remain symmetric and bend in the plane of symmetry.

Pure Bending / Basit Eğilme

Unsymmetric Bending 4

• Will now consider situations in which the bending couples do not act in a plane of symmetry.

• Cannot assume that the member will bend in the plane of the couples.

Unsymmetric Bending for Symmetric Cross-section Simetrik Kesitli Kirişlerde Eğik Eğilme

Unsymmetric Bending for symmetric cross-sectionWish to determine the conditions under which the neutral axis of a cross section of arbitrary shape coincides with the axis of the couple as shown.

• The resultant force and moment from the distribution of elementary forces in the section must satisfy

coupleappliedMMMF zyx 0,0

neutral axis passes through centroid

dAydA

cydAF mxx 0or 0

6

couple vector must be directed along a principal centroidal axis

inertiaofproductIdAyz

dAcyzdAzM

yz

mxy

0or

0

defines stress distribution

inertiaofmomentIIc

Iσz

m M

dA

cyyMM mz or

Unsymmetric Bending 4 - 7

Unsymmetric Bending for symmetric cross-section

Superposition is applied to determine stresses in the most general case of unsymmetrical bending.

Resolve the couple vector into components along the principle centroidal axes.

sincos MMMM yz kMjMM zy

8

Superpose the component stress distributions

zI

M

y

yMx

y yI

M

z

zMx

z

Combined stress, zy Mx

Mxx

zI

MyI

MzI

My

IM

yzy

y

z

zx

sincos

Unsymmetric Bending 9

Along the neutral axis, normal stress must be zero. Thus

tantan

0sincos

y

z

yzx

y

y

z

zx

II

zy

zI

MyI

M

zI

My

IM

Unsymmetric Bending 10

Example 4.08

A 1600 lb-in couple is applied to a rectangular wooden beam in a plane forming an angle of 30 deg. with the vertical. Determine

(a) the maximum stress in the beam,

(b) the angle that the neutral axis forms with the horizontal plane.

z

y

Unsymmetric Bending 11

• Resolve the couple vector into components along the principle centroidal axes and calculate the corresponding maximum stresses.

sincos MMMM yz

• Combine the stresses from the component stress distributions.

zI

My

IM

y

y

z

zx

Solution

z

y

MM

tan

Unsymmetric Bending 12

• Determine the angle of the neutral axis.

zIIy

MM

zMM

IIy

y

z

z

y

z

y

y

z tantan

zyzy tanor tan

0 zI

My

IM

y

y

z

zx

orzy

II

y

z tantan the angle of the neutral axis

Unsymmetric Bending 4 - 13

Resolve the couple vector into components and calculate the corresponding maximum stresses.

inlb80030sininlb1600

inlb138630cosinlb1600

y

z

MM

43

121

43121

in9844.0in5.1in5.3

in359.5in5.3in5.1

y

z

I

I

Unsymmetric Bending 14

• Determine the angle of the neutral axis.

o

y

z

z

y

IIMM

35.72

142.35772.09844.0359.5tantan

5772.01386800tan

Unsymmetric Bending 4 - 15

.75.0;75.1at occurs todue

stress nsilelargest te The

AM

max

44

psi1062

in75.0in9844.0inlb800in75.1

in359.5inlb1386

A

A

Ay

yA

z

zA z

IM

yI

M

Unsymmetric Bending 4 - 16

.75.0;75.1at occurs todue

stress normalsmallest The

EM

min

44

psi1062

in75.0in9844.0inlb800in75.1

in359.5inlb1386

E

E

Ey

yE

z

zE z

IM

yI

M

Unsymmetric Bending 4 - 17

• Distribution of the streeses in the cross-section.

z

x

y

b=60

mm

h=80

mm

L=1 m

F=10 kN

A

B

Örnek: Şekildeki kirişin A ve B noktalarında meydana gelen gerilmeleri hesaplayınız.

Çözüm: Kesit özellikleri (atalet momentleri):

46

31213

121

1056.2

8060

mmI

hbI

x

x

A b=60 mm

x

y

h=80

mm

F=10 kN

M

B

46

31213

121

1044.1

6080

mmI

bhI

y

y

Unsymmetric Bending 20

b=60 mm

x

y

h=80

mm

F=10 kN

M

Eğilme momenti bileşenleri:

NmmM

kNmmkNLFM61010

10110

siyo

o

87.36

87.368060tan

NmmkNmkNmMM

NmmkNmkNmMM

y

x

6

6

106687.36sin10sin

108887.36cos10cos

A

B

Unsymmetric Bending 21

o

y

x

II

13.53333,1tan

87.36tan44.156.2tantan

b=60 mm

x

y

h=80

mm

F=10 kN

M

A

B

TETarafsız Eksen (TE)

Unsymmetric Bending 22

b=60 mm

x

y

h=80

mm

F=10 kN

M

A

B

TEAy

yA

x

xA x

IM

yI

M

)40;30();( AyxA AA

MPaA

A

250

301044.110640

1056.2108

6

6

6

6

A ve B noktalarındaki gerilmeler:

Unsymmetric Bending 23

b=60 mm

x

y

h=80

mm

F=10 kN

M

A

B

TE

A ve B noktalarındaki gerilmeler:

MPa

xI

My

IM

B

B

By

yB

x

xB

250

301044.110640

1056.2108

6

6

6

6

)40;30();( ByxB BB

Unsymmetric Bending 24

• In general, the neutral axis of the section will not coincide with the axis of the couple.

Unsymmetric Bending

25

Simetrik Olmayan Kirişlerde Eğilme

)1( ve A zyA zx dAxMdAyM

Unsymmetric Bending 26

(2) 21 xCyCzzz

(3) 21 xCyCEE zz

(4) 21 xCyCz

Unsymmetric Bending 27

(5) - ve 2121 AyAx dAxCyCxMdAxCyCyM

(6)

212

21

2122

1

yxyAAy

xyxAAx

ICICdAxCxydACM

ICICdAxyCdAyCM

Unsymmetric Bending 28

22xyyx

xyxxy

IIIIMIM

C

21

xyyx

xyyyx

IIIIMIM

C

ve

7 2xyyx

xyxxyxyyyxz III

xIMIMyIMIM

Unsymmetric Bending 29

8 tan veya xyxIMIMIMIM

yxyyyx

xyxxy

xyyyx

xyxxy

IMIMIMIM

tan

Tarafsız Eksenin x ekseni ile yaptığı açı:

xyyyx

xyxxy

IMIMIMIM

1tan

σz gerilme fonksiyonu sıfıra eşitlenirse Tarafsız Eksen (TE) denklemi ve Tarafsız Eksenin açısı bulunur.

TE

Unsymmetric Bending 30

Örnek:

Şekilde kesiti ve yükleme durumu verilen kiriş için:a) Ağırlık merkezinin yerini belirleyiniz.b) Ağırlık merkezinden geçen eksenlere göre

atalet momentlerini bulunuz.c) Kesitteki eğilme momentinin değeri M=20

kNm olduğuna göre bileşenlerini bulunuz.d) Tarafsız Ekseni belirleyiniz.e) A, B ve C noktalarındaki gerilmeleri

hesaplayınız.f) Kiriş kesitindeki ekstremum gerilmeleri

belirleyerek emniyetli olup olmadığını irdeleyiniz.

M

A

B C

G

MPaem 300o30

Unsymmetric Bending 31

x

Kesitin ağırlık merkezinin yeri:

Unsymmetric Bending 32

Atalet Momentleri: Paralel Eksen Teoremi

X

Y

C

y

x

O

Ci

x

yyi

xi

2iyiixx dAII

2ixiiyy dAII

ii yxiixyxy ddAII

xxd

yyd

ix

iy

i

i

Unsymmetric Bending

Ağırlık merkezinden geçen x ve y eksenlerine kesitin atalet momentleri:

Paralel Eksen Teoremi ile

Y

X

mmymmy

1590

2

1

mmxmmx

7515

2

1

mmxxd

mmxxd

x

x

403575

203515

2

1

2

1

mmxxd

mmyyd

y

y

506515

256590

2

1

2

1

xI

46101.850403090252018030 mmI xy

Unsymmetric Bending 34

M

A

B C

GxM

yM

kNmMM ox 32.1730cos20cos

kNmMM oy 1030sin20sin

Eğilme momentinin bileşenleri:

Unsymmetric Bending 35

yIII

IMIMx

IIIIMIM

xyyx

xyyyx

xyyx

xyxyxz 22

Gerilme Denklemi:

66

66

22 10101010

1.8708.891.241.810708.832.17

1.8708.891.2491.24101.832.17

yxz

yxz 3.15182.231

3.1514.389

yxz 532.1574.2

Unsymmetric Bending 36

Tarafsız Eksen Denklemi:

0532.1574.2 yxz

Tarafsız Eksen Denklemi, Gerilme denklemi sıfıra eşitlenerek bulunur:

xxy tan68.1

Tarafsız Eksenin eğim açısı :

o24.5968.1tan

T.E.

37

T.E.

Kesitteki gerilmeler:

A

B C

A(-35;115) noktası MPaA 3.266115532.135574.2

MPaB 49.965532.135574.2

MPaC 4.31865532.185574.2

MPaMPa emC 3004.318max

Olduğundan kiriş emniyetsizdir.

B(-35;-65) noktası

C(85;-65) noktası

Kesit kontrolü

Unsymmetric Bending 38

Örnek: şekildeki ‘L’ kesitli kirişin C noktasına P=4 kN luk eğik bir

kuvvet uygulanmaktadır. Ҩ=60 olduğuna göre:

a) Kesitin ağırlık merkezinin yerini belirleyiniz.

b) Atalet momentlerini hesaplayınız.

c) T.E. Tarafsız ekseni bulunuz.

d) Maximum çeki ve bası gerilmelerini hesaplayınız.

xyyx III ,,

Unsymmetric Bending 39

Çözüm:

Ağırlık merkezinin yeri

Atalet momentleri

Ix =2.783x106 mm4

Iy =1.003x106 mm4

Ixy =-0.973x106 mm4

mmy

mmx

74.398001100

5800651100

74.198001100

4080051100

Unsymmetric Bending 40

Momentin bileşenleri

kNmRMkNR

PRM

A

A

AB

8.44.2224.2

0350

max

kNmM

MMkNmM

MM

y

y

x

x

4.2

60cos8.4cos157.4

60sin8.4sin

max

max

Unsymmetric Bending 41

Gerilme fonksiyonu

xyz 428.19945.0

66

662 10101010973.0003.1783.2

973.0157.4783.24.2973.04.2003.1157.4

xyz

2xyyx

xyxxyxyyyxz III

xIMIMyIMIM

Unsymmetric Bending 42

Tarafsız Eksen Denklemi ve Açısı

xyxyz

4362.10428.19945.0

oxy 15.55tan

Unsymmetric Bending 43

Gerilmeler

yxz 9945.0428.1

MPaA 6.125)74.39(9945.0)26.60(428.1

MPaB 108)26.80(9945.0)74.19(428.1

T.E. den en uzak noktalar:

26.80;74.19ve

74.39;26.60

B

A

Unsymmetric Bending 44

Örnek: Yükleme durumu ve kesiti görülen kirişin A, B ve C noktalarında meydana gelen gerilmeleri hesaplayınız ve σem=200 MPa için kirişin kontrolünü yapınız.

Fya b

z

y10

40

10

60 mm10

y

xC

F

A

B

C

mbakNF 1 2

45

Unsymmetric Bending 46

Unsymmetric Bending 47

Unsymmetric Bending 48

Unsymmetric Bending 49

Unsymmetric Bending 50

Unsymmetric Bending 51

Unsymmetric Bending 52

Unsymmetric Bending 4 - 53

General Case of Eccentric Axial Loading• Consider a straight member subject to equal

and opposite eccentric forces.

• The eccentric force is equivalent to the system of a centric force and two couples.

PbMPaMP

zy force centric

• By the principle of superposition, the combined stress distribution is

y

y

z

zx I

zMI

yMAP

• If the neutral axis lies on the section, it may be found from

APz

IM

yI

M

y

y

z

z