Kesme kuvveti-Kayma gerilmesi-Kayma akımı-Kayma merkezi

Shear Forces-Shear stressShear flow-Shear center

6 - 2

Introduction

dAyMdAF

dAzMVdAF

dAzyMdAF

xzxzz

xyxyy

xyxzxxx

0

0

00

• Distribution of normal and shearing stresses satisfies

• Transverse loading applied to a beam results in normal and shearing stresses in transverse sections.

• When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces

• Longitudinal shearing stresses must exist in any member subjected to transverse loading.

6 - 3

Shear flow on the Horizontal Face of a Beam Element• Consider prismatic beam• For equilibrium of beam element

A

CD

ADCx

dAyI

MMH

dAHF 0

xVxdx

dMMM

dAyQ

CD

A

• Note,

flowshearI

VQxHq

xI

VQH

• Substituting,

6 - 4

Shear flow on the Horizontal Face of a Beam Element

flowShearI

VQxHq :

Shear flow,

where

section cross full ofmoment Second :

above area ofmoment First :

'

2

1

AA

A

dAyI

ydAyQ

6 - 5

Shear on the Horizontal Face of a Beam Element

• Same result is found for lower area

HHQQ

qIQV

xHq

axis neutral respect toh moment witFirst :0

6 - 6

Shear Stress on the Horizontal Face of a Beam Element

IVQ

xHq

• Shear flow,

• Shear stress is found by dividing the shear flow q with bz.

zz IbVQ

bq

V

xy

zb

• Shear stress

Örnek: Şekildeki yükleme durumu ve kesiti görülen kiriş için;a) C noktasındaki asal gerilmeleri ve doğrultularını bulunuz. b) Kesitteki kayma gerilmesi dağılımını gösteriniz.

0.5 m 1 m 0.5 m

6 kN 6 kN

A D E B

20

6020

20

40

y

C

G

0.5 m 1 m 0.5 m

6 kN 6 kN

A D E B

Çözüm: KKD - EMD

6 kN 6 kN

6 kN

6 kN

(+)

(-)

(+) (+) (+)

3 kNm

20

6020

20

40

y

C

G

mmy 30

60206020106020506020

46

23

23

1036.1

3010602012

20603050602012

6020

mmI

I

z

z

Ağırlık merkezi ve Atalet momenti

NmmkNmM

NkNV

D

D

6

3

1033

1066

33

11

212

11

101640400

4002020 ve40203080

mmyAQ

mmAmmy

C

MPaxyC 53.3201036.11016106

6

33

MPaIyM C

xC 2.661036.130103

6

6

İç kuvvetler:

Birinci moment:

Kayma gerilmesi:

Normal gerilme:

C

C

CD

İnce Cidarlı KirişlerdeSimetrik Olmayan Yüklemeler:

Kesme Kuvveti-Kayma AkımıKayma Merkezi-Kayma Gerilmesi

Kesitlerdeki iç kuvvetler (normal kuvvetleri):

İNCE CİDARLI AÇIK KESİTLERDE KAYMA GERİLMELERİ VE KAYMA MERKEZİ

Denge denklemi:

Kesme kuvveti:

b – b kesitindeki ortalama gerilme:

Kayma Akımı:

Kayma Gerilmesi:

«U» şeklindeki kesitin kayma merkezi

Kesitteki Kayma Gerilmesi Değişimi:

zItbhe

41

22

Kanattaki kesme kuvveti:

Kayma merkezi:

Kanattaki kayma akımı ve kesme kuvveti:

ÖRNEK: Şekilde görülen profilin boyutları

b=100 mm, h=150 mm ve t=3 mm olup,

profil P=800 N’luk bir kesme kuvvetine

maruz kaldığına göre:

a) Kayma merkezinin yerini bulunuz.

b) Kesit çevresi boyunca kayma gerilmesi

dağılımını gösteriniz.

e

b

h

tO O’

AB

D E

P

Çözüm:

e

b

h

tO O’

AB

D E

b=100 mm, h=150 mm, t=3 mm

462

1022.41501006121503 mmI

hbthI

hbtbtthI

III flangeweb

6121

21212

121

2

2

233

462

33 1022.41502131003100

12121503

121 mmI

Veya kısaca

İhmal edilebilir

AB kolundaki kayma akımını bulmak için s uzunluğundaki bir eleman dikkate alınır.

sAB

D E

21hy

..ET

21hy ve stA 1

sthsthAyQ2211

sIhtVsht

IV

IQVq

22

Statik momenti:

Kayma akımı:

A1

AB kolundaki kesme kuvvetini hesaplamak için A’dan B’ye kadar integral almak gerekir.

IhbtVF

sdsIhtVdsqF

bb

4

22

00

s

AB

D E

..ET

F

F

A1 21hy

mme

Ibhte

40

1022.441001503

4 6

2222

O’ noktasına göre moment alınırsa kayma merkezi

VhFehFeVM O 0'

Ibhte

Vh

IhbtVe

44

222

şeklinde bulunur.e

b

h

tO O’

AB

D E

V

s

..ET

B

D

max

21hy ve tsA 1 tshAyQ

211 Statik momenti:

Kayma gerilmesi denkleminden:

şeklinde bulunur.

Kesit çevresi boyunca kayma gerilmesi dağılımı

sIhV

tIQV

tq

2

1001022.42

1508002 6

bIhV

bs

B

A-B ve E-D kesitindeki kayma gerilmesi değişimi

B

D

A1

21hy

MPaBD 422.1

33

4150

23150

42

4222211

1094.30100 mmbQ

ttbAyAyQhth

hhh

Statik momenti:

Kayma gerilmesi denkleminden:

bulunur.

Kesit çevresi boyunca kayma gerilmesi dağılımı

31022.41094.30800

6

3

tIQV

MPa955.1max

B-D kesitindeki kayma gerilmesi (maksimum kayma gerilmesi)Maksimum kayma gerilmesi T.E. üzerinde meydana gelir. T.Ü. deki alanın statik momenti

1A

2A

Örnek: Şekilde kesiti görülen kirişin

a) Kanatlarda oluşan iç kuvvetleri hesaplayınız.

b) Kayma merkezinin yerini bulunuz. t =6 mm t1 =4 mm t2 =5 mm h1 =60 mm h2 =40 mm b=50 mm P=800 N

t1 t2

b

Ph1 h2

t

Çözüm: Denge denklemleri

00 21 PVVFy PVV 21

00 2 bVePM AePbV 2

PfbV 1veya

t1 t2

e fb

Ph1

h2

V1 V2

A Omaxx

bfe

43333 10567.99405650604121 mmI x

4332 10667.26405

121 mmI

4331 1072604

121 mmI

Atalet momentleri

Tüm kesitin Atalet momenti

Başlıkların atalet momentleri t1 t2

e fb

Ph1

h2

V1 V2

A Omax

x

xxx IhPhth

tIP

tIQV

842

222

22

22

2max

Sağ başlıktaki maksimum kayma gerilmesi

MPaIhP

x

61.110567.998

408008 3

222

max

xxx IIPht

IPth

IhPthV 2

322

22

22

22max2 12832

32

kNIIPV

x

26.21410567.9910667.26800 3

32

2

kNIIPV

x

5.57810567.99

1072800 3

31

1

ePbV 2 800

26.214502

PVbe

mme 4.13

Sağ başlıktaki kesme kuvveti

Sol başlıktaki kesme kuvveti

Kayma merkezinin yeri:

b=100 mm

h=150 mm

t=3 mm

Example: For the channel section, and neglecting stress concentrations, determine the maximum shearing stress caused by a V=800-N vertical shear applied at centroid C of the section, which is located to the right of the center line of the web BD.x

t

V

h

b

x

A

C

B

ED

x

Solution:

V

x

A

C

B

ED

=

e

V

x

A

C

B

ED

=

e

TV

x

A

C

B

EDe x

A

C

B

EDe

+ T

mmx 29

105030000

1503310025031002

46233 10219.4753100310012121503

121 mmI x

331094.302

75375753100 mmQ

3)10219.4()1094.30(800

6

3

tI

QV

xV

V

x

A

C

B

EDe

V

BB

D

D

MPaV 956.1

mmI

tbhex

4010219.44

3.1501004 6

2222

x

A

C

B

EDe

TO VxeVOCT

4333 1015.33100215031

31 mmtbJ ii

MPatJT

T 57.5231015.3102.55

3

3

NmNmmT 2.55102.558002940 3

MPaTV 526.5457.52956.1max

The maximum shearing stress

EXAMPLE 6.16 (from Craig): A pipe conveying fluid over a narrow stream crossing

must act as a beam as well as a conduit, as indicated in Fig. 1.

Assuming that the ratio of mean diameter to wall thickness satisfies the requirement

d/t>>1, determine the shear flow and shear stress distribution at a section due to the

transverse shear force, V, at that section, Also, determine the maximum shear stress on

the cross section, and express τmax in terms of V/A.

Solution: Because this is a circular cross section, it is convenient to select a FBD that

is defined by radial cutting planes at angle θ either side of the xy-plane, as indicated in

Fig. 2. In Fig. 2b there are two surfaces that have equal shear forces ∆H which serve to

balance the net force (F2 — F1) due to the flexural stresses. Therefore, we get

where, from the definition of shear flow

The geometric properties for a thin ring and for a sector of a thin ring are

shown in Fig. 3. Using the formulas in Fig. 3b, we have, for t<< r,

Review the Solution: As a check on the expression that we obtained for q, Eq.

(6). we can see if the resultant of this distribution is V. as it is supposed to be. The

force on the elemental area highlighted in Fig. 4 is

The vertical component, as shown on the

insert in Fig. 4, is, therefore,

So, due to symmetry, we have

Since the distribution looks “reasonable” (the shear flow has its

maximum at the neutral axis and is zero on the plane of symmetry),

and since it gives the correct resultant, we can assume that our

answer is correct.

ÖRNEK: Şekilde görülen profil P=12 kN’luk bir kesme kuvvetine maruz kaldığına göre:a) Kayma merkezinin yerini bulunuz.b) Kesit çevresi boyunca kayma gerilmesi değişimini gösteriniz.

103105.85

18 MPa

6 - 45

Shearing Stresses in Thin-Walled Members• Consider a segment of a wide-flange

beam subjected to the vertical shear V.

• The longitudinal shear force on the element is

xI

VQH

ItVQ

xtH

xzzx

• The corresponding shear stress is

• NOTE: 0xy0xz

in the flangesin the web

• Previously found a similar expression for the shearing stress in the web

ItVQ

xy

6 - 46

Shearing Stresses in Thin-Walled Members

• The variation of shear flow across the section depends only on the variation of the first moment.

IVQtq

• For a box beam, q grows smoothly from zero at A to a maximum at C and C’ and then decreases back to zero at E.

• The sense of q in the horizontal portions of the section may be deduced from the sense in the vertical portions or the sense of the shear V.

6 - 47

Sample Problem 6.3

Knowing that the vertical shear is 50 kips in a W10x68 rolled-steel beam, determine the horizontal shearing stresses in the top flange at the points a and C.

SOLUTION:• First moment for the shaded area,

3in98.15

in815.4in770.0in31.4

Q

Q

• The shear stress at a,

in770.0in394

in98.15kips504

3

ItVQ

ksi63.2

• First moment for the area over point C,

3in4.42

in215.2in770.0in43.4in815.4in770.0in40.9

Q

Q

• The shear stress at C,

in770.0in394

in4.42kips504

3

It

VQ

ksi989.6

6 - 50

Unsymmetric Loading of Thin-Walled Members

• Beam loaded in a vertical plane of symmetry deforms in the symmetry plane without twisting.

ItVQ

IMy

avex

• Beam without a vertical plane of symmetry bends and twists under loading.

ItVQ

IMy

avex

6 - 51

• When the force P is applied at a distance e to the left of the web centerline, the member bends in a vertical plane without twisting.

Unsymmetric Loading of Thin-Walled Members

• If the shear load is applied such that the beam does not twist, then the shear stress distribution satisfies

FdsqdsqFdsqVIt

VQ E

D

B

A

D

Bave

• F and F’ indicate a couple Fh and the need for the application of a torque as well as the shear load.

VehF

6 - 52

Example 6.05Determine the location for the shear center of the channel section with b = 4 in., h = 6 in., and t = 0.15 in.

6 - 53

Example 6.05• Inertia moment: • b = 4 in., h = 6 in., and t = 0.15 in.

hbthI

hbtbtthI

III flangeweb

6121

21212

121

2

2

233

6 - 54

Solution

IhFe

• where

IVthbF

dshstIVds

IVQdsqF

b bb

4

22

0 00

• Combining,

.in43.in62

in.4

32

bh

be .in6.1e

Shear stress in flanges• Determine the shear stress distribution for

V = 2.5 kips.

ItVQ

tq

• Shearing stresses in the flanges,

ksi22.2

in6in46in6in15.0in4kips5.26

66

62

22

2121

B

B

B hbthVb

hbthVhb

sI

VhhstItV

ItVQ

6 - 56

Shear stress in web• Determine the shear stress distribution for

V = 2.5 kips.

ItVQ

tq

• Shearing stress in the web,

ksi06.3in6in66in6in15.02

in6in44kips5.236243

64

max

max

2121

81

max

hbthhbV

thbthhbhtV

ItVQ