kesme kuvveti-kayma gerilmesi-kayma akımı-kayma merkezi
DESCRIPTION
Kesme kuvveti-Kayma gerilmesi-Kayma akımı-Kayma merkezi. Shear Forces-Shear stress Shear flow-Shear center. Introduction. Transverse loading applied to a beam results in normal and shearing stresses in transverse sections. Distribution of normal and shearing stresses satisfies. - PowerPoint PPT PresentationTRANSCRIPT
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Kesme kuvveti-Kayma gerilmesi-Kayma akımı-Kayma merkezi
Shear Forces-Shear stressShear flow-Shear center
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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.
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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,
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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
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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
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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
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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
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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
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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
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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
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İnce Cidarlı KirişlerdeSimetrik Olmayan Yüklemeler:
Kesme Kuvveti-Kayma AkımıKayma Merkezi-Kayma Gerilmesi
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Kesitlerdeki iç kuvvetler (normal kuvvetleri):
İNCE CİDARLI AÇIK KESİTLERDE KAYMA GERİLMELERİ VE KAYMA MERKEZİ
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Denge denklemi:
Kesme kuvveti:
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b – b kesitindeki ortalama gerilme:
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Kayma Akımı:
Kayma Gerilmesi:
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«U» şeklindeki kesitin kayma merkezi
Kesitteki Kayma Gerilmesi Değişimi:
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zItbhe
41
22
Kanattaki kesme kuvveti:
Kayma merkezi:
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Kanattaki kayma akımı ve kesme kuvveti:
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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xxx IhPhth
tIP
tIQV
842
222
22
22
2max
Sağ başlıktaki maksimum kayma gerilmesi
MPaIhP
x
61.110567.998
408008 3
222
max
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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:
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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
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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
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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
![Page 34: Kesme kuvveti-Kayma gerilmesi-Kayma akımı-Kayma merkezi](https://reader030.vdocument.in/reader030/viewer/2022012818/5681610c550346895dd05e2c/html5/thumbnails/34.jpg)
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
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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.
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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
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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,
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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
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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.
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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.
![Page 43: Kesme kuvveti-Kayma gerilmesi-Kayma akımı-Kayma merkezi](https://reader030.vdocument.in/reader030/viewer/2022012818/5681610c550346895dd05e2c/html5/thumbnails/43.jpg)
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103105.85
18 MPa
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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
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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.
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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.
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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
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• 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
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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
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• 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
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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.
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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
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Solution
IhFe
• where
IVthbF
dshstIVds
IVQdsqF
b bb
4
22
0 00
• Combining,
.in43.in62
in.4
32
bh
be .in6.1e
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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
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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
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