İnce cİdarli tÜplerİn burulmasi thin-walled tubes
DESCRIPTION
İNCE CİDARLI TÜPLERİN BURULMASI THIN-WALLED TUBES. Kapalı tüpler. Dairesel kesitli-ince cidarlı tüplerin burulması. - PowerPoint PPT PresentationTRANSCRIPT
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İNCE CİDARLI TÜPLERİN BURULMASI
THIN-WALLED TUBESKapalı tüpler
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Dairesel kesitli-ince cidarlı tüplerin burulması
İnce cidarlı ve kapalı tüplerin burulma problemleri, Coulomb teorileri
ile çözülebilen dairesel tüplerden elde edilen sonuçlardan yararlanarak
elemanter olarak çözülebilmektedir.
oR
ido RRR 21
id RRt
Ortalama yarıçap Cidar kalınlığı
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1) Daire kesitli tüpler: İçi boş daire kesitli millerin burulması dikkate alınarak ince cidarlı tüplerin burulma formülleri çıkarılabilir.
• İçi boş dairesel kesitlerde maksimum kayma gerilmesi kesitin dış yüzeyine yakın noktalarda meydana gelir:
J
RT ddis max
• İnce cidarlı dairesel kesitlerde ortalama kayma gerilmesi cidar boyunca sabit kabul edilir ve aşağıdaki gibi hesaplanır:
J
RT oort
(a)
(b)
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İçi boş milin polar atalet momenti:
İçi boş dairesel kesitin dış ve iç yarıçapları kullanılarak aşağıdaki gibi bulunur:
442 id RRJ
Dairesel tüplerin polar atalet momenti:
Yukarıdaki polar atalet momenti ifadesi aşağıdaki gibi çarpanlara ayrılır:
222
22222 ididididid RRRRRRRRRRJ
tRR id oid RRR 2 222 2 oid RRR
22 22 oo RRtJ
burada
şeklindedir. Buna göre polar atalet momenti aşağıdaki gibi olur:
tRRtRJ ooo23 22
tARJ o2
(a)
(b)
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t cidar kalınlığı R ortalama yarıçapı yanında çok küçükse J=2πR3t=2ARt polar
atalet momenti kullanılabilir. (b) şeklinde görüldüğü gibi ince cidarlı tüplerde
cidar kalınlığı boyunca kayma gerilmelerinin değişmediği kabul edilebilir. Bu
durumda
tAR
RT
J
RT
o
ooort 2
At
Tort 2 2
oRA
Birim dönme açısı ise:
o
o
o R
R
tRG
T
GJ
T
2
2
2 3
burada s ortalama çevre uzunluğudur. Buna göre tüpün toplam dönme açısı aşağıdaki gibi olur:
t
s
GA
T24
t
s
GA
TLL
24
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Örnek: Şekildeki dairesel kesitli mil T=6 kNm’lik bir burulma
momentine maruz bırakıldığına göre meydana gelen kayma gerilmesini
ve birim dönme açısını hesaplayınız (G=25 GPa).
Dış ve iç çaplar sırası ile D=128 mm ve d=122 mm olarak verilmektedir.
d D
T=6 kNm
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2322
10272.124
125
4mm
DA o
mmdD
RD oo 1252
122128
22
mmdD
t 32
122128
2
463
3 106.432
12522 mmmmmmtRJ o
Kesit özellikleri (Alan ve Polar atalet momenti):
Ortalama çap ve cidar kalınlığı:
4623 106.432
12510272.1222 mmmmmmmmtARJ o
veya
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MPa
mm
mm
Nmm
J
RT o
52.81
2
125
106.4
10646
6
mmmrd
mmMPa
Nmm
GJ
T
o /989.2180
/1017.52
106.41025
106
6
463
6
MPa
mmmm
Nmm
At
T
5.81
)3()10272.12(2
106
2 23
6
Cidarda oluşan ortalama kayma gerilmesi:
veya
Birim dönme (burulma) açısı:
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464444 10604.41221283232
mmdDJ
MPamm
mmNmm
J
DT4.83
)10604.4(2
)128()106(
2 46
6
mmrdmmMPa
Nmm
GJ
T/10213.5
10604.41025
106 5463
6
mmrd o/987.2180
/05213.0
İçi boş mil durumuna göre kayma gerilmesi ve birim dönme açısı:
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The stresses acting on the longitudinal faces a-b and cd produce forces Fb and
Fc (Fig. 3-40d). These forces are obtained by multiplying the stresses by the
areas on which they act:
FIG. 3-40 Thin-walled tube of arbitrary cross-sectional shape
in which tb and tc represent the
thicknesses of the tube at points b
and c, respectively (Fig. 3-40d).
dxtFdxtF cccbbb
Non-circular Thin-Walled Hollow Shafts
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ccbbbcb
x
ttFFFF
F
0
0
FIG. 3-40 Thin-walled tube of arbitrary cross-sectional shape
In addition, forces F1 and F1 are produced by the stresses acting on faces
b-c and a-d. From the equilibrium of the element in the longitudinal
direction (the x direction), we see that Fb = Fc , or
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Kayma Akımı (Shear Flow)
constant tq t
q (3-59)or
Because the locations of the longitudinal cuts a-b and c-d were selected
arbitrarily, it follows from the preceding equation that the product of
the shear stress τ and the thickness t of the tube is the same at every
point in the cross section.
This product is known as the shear flow and is denoted by the letter q:
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• This relationship shows that the largest shear stress occurs where
the thickness of the tube is smallest, and vice versa.
• Naturally, in regions where the thickness is constant, the shear
stress is constant.
• Note that shear flow is the shear force per unit distance along the
cross section.
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İnce cidarlı tüplerde burulma formülü(Torsion Formula for Thin-Walled Tubes)
• The next step in the analysis is to relate the
shear flow q (and hence the shear stress τ) to
the torque T acting on the tube. For that
purpose, let us examine the cross-section of the
tube, as pictured in Fig. 3-41.
FIG. 3-41 Cross section of thin-walled tube
dsq
median line
r
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• The median line (also called the centerline or the midline) of the wall
of the tube is shown as a dashed line in the figure.
• We consider an element of area of length ds (measured along the
median line) and thickness t.
• The distance s defining the location of the element is measured along
the median line from some arbitrarily chosen reference point.
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• The total shear force acting on the element
of area is qds, and the moment of this
force about any point O within the tube is
FIG. 3-41 Cross section of thin-walled tube
dsq
median line
r dsqrdT
in which r is the perpendicular distance
from point O to the line of action of the
force qds. The total torque T produced by
the shear stresses is obtained by integrating
along the median line of the cross section:
mLdsrqT
0
in which Lm denotes the length of the median line.
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• The integral above can be difficult to integrate by formal
mathematical means, but fortunately it can be evaluated easily by
giving it a simple geometric interpretation.
• The quantity rds represents twice the area of the shaded triangle
shown in Fig. 3-41.
• (Note that the triangle has base length ds and height equal to r.)
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• Therefore, the integral represents twice the area Am enclosed by the
median line of the cross section:
m
LAdsr
m
20
• Therefore the shear flow is
m
m
L
A
TqAqdsrqT
m
22
0
• Now we can obtain a torsion shear formula for thin-walled tubes:
tA
T
t
q
m2
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Dairesel olmayan ince cidarlı tüplerin burulması2) Herhangi bir biçimdeki tüp kesitli çubuklar:
Şekil (a) da görüldüğü gibi herhangi bir kesiti olan çubuk dikkate alalım. Bu çubuktan çok küçük parçayı büyütüp dengesini inceleyelim. Bu eleman dengede olduğundan, örnek olarak karşılıklı kesitlerde bulunan V3 ve V4 kesme kuvvetleri de dengededir.
43 VV dztVdztV 244133 ve
dztdzt 2413 2413 tt 43 qq
Kayma gerilmesi cidar kalınlığı çarpımına kayma akımı denir ve q ile gösterilir.
0zFve olur.
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Dik köşelerde, yani birbirine dik kesitlerde kayma gerilmelerinin eşit olması şartından
yazılabilir. Buna göre kayma akımları
Cidar eksen eğrisi s üzerinde alınan (t ds) alan elemanına etkiyen dV kesme kuvveti
veya
şeklindedir.
ve 4231
sbt 222111 qttq
dstdstdV dsqdV
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ss
dshtdshtdVhT
dshdA 21 veya dAdsh 2
Kesit eğrisi boyunca kayma akımlarının eşit olması (q1=q2=q3=…) şartından dV kesme kuvvetinin büyüklüğü de sabit kalır. Bu kesme kuvvetinin kesit düzlemi içerisindeki herhangi bir O noktasına göre momenti, kesite etkiyen T burulma momentine eşit olmalıdır. Bu durumda
yazılır.
Tüplerde Kayma Gerilmesinin Bulunması
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dAtT 2 tAT 2AdA
İntegral içindeki (h ds) terimi, şekildeki taralı üçgen (dA) alanının iki
katıdır. Buna göre burulma momenti
şeklinde olur. Burulma momenti ifadesinden kayma gerilmesi
çekilirse aşağıdaki gibi olur:
tA
T
2
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Buna göre kesitteki en büyük kayma gerilmesi, cidar kalınlığının en küçük olduğu noktada meydana geleceği açıktır. Bu durumda maksimum kayma gerilmesi
minmax 2 tA
T
gibi olur. Burada A kesit cidar orta hattının sınırladığı alandır. A
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Kesitin θ birim dönme açısını hesaplamak için şekil değiştirme enerjisinde
yararlanılabilir. T burulma momentinin yaptığı iş, dφ= θ dz olduğu bilinerek
dzTTddU 2
1
2
1
şeklinde olur. dz boyundaki parçada biriken enerji, τ/2G enerji yoğunluğu
kullanılarak
VdV
GdU
2
2
şeklinde olur. Burada, τ kayma gerilmesi, G kayma modülü ve dV hacim elemanıdır.
Tüplerde Burulma Açısının Bulunması
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t
ds
GA
TL
t
ds
GA
T22 44
Yukarıdaki iş ve enerji ifadesi birbirine eşitlenirse
(burada ) V
dVG
dzT
22
2
elde edilir. Burada dV hacim elemanı olup dV=t ds dz yukarıdaki
denklemde yerine konulursa
t
ds
GA
dzTdz
Tdzdst
tA
T
Gdz
Ts 2
2
22
2
82veya
42
1
2
elde edilir. Bu ifade düzenlenirse tüpün birim dönme (burulma) açısı
aşağıdaki gibi olur:
tA
T
2
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3 - 26
Thin-Walled Hollow Shafts
• Summing forces in the x-direction on AB,
shear stress varies inversely with thickness
flowshear
0
qttt
xtxtF
BBAA
BBAAx
tA
T
qAdAqdMT
dAqpdsqdstpdFpdM
2
22
2
0
0
• Compute the shaft torque from the integral of the moments due to shear stress
t
ds
GA
TL24
• Angle of twist (from Chapt 11)
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Örnek: Boyutları şekilde verilen tüp, T=50 kNm’lik burulma
momentine maruz bırakılıyor. Buna göre:
a) Kesitte meydana gelen en büyük kayma gerilmesini ve yerini
bulunuz.
b) Birim dönme açısını hesaplayınız (G=70 GPa)
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Example 3.10Extruded aluminum tubing with a rectangular cross-section has a torque loading of 24 kip-in. Determine the shearing stress in each of the four walls with
(a) uniform wall thickness of 0.160 in. and wall thicknesses of
(b) 0.120 in. on AB and CD and 0.200 in. on CD and BD.
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SOLUTION:
• Determine the shear flow through the tubing walls
• Find the corresponding shearing stress with each wall thickness
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3 - 32
SOLUTION:
• Determine the shear flow through the tubing walls
in.
kip335.1
in.986.82
in.-kip24
2
in.986.8in.34.2in.84.3
2
2
A
Tq
A
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• Find the corresponding shearing stress with each wall thickness
with a uniform wall thickness,
in.160.0
in.kip335.1
t
q ksi34.8
with a variable wall thickness
in.120.0
in.kip335.1ACAB
in.200.0
in.kip335.1CDBD
ksi13.11 BCAB
ksi68.6 CDBC
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Various thin-walled members
İnce Cidarlı Tüplerin Burulması
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Torsion of circular and rectangular members
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- Dikdörtgen kesitli miller- Açık tüpler
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3 - 37
Torsion of Noncircular Members
• For uniform rectangular cross-sections,
• Circular torsion formulas are not valid for
non-circular shafts.
• Planar cross-sections of noncircular shafts
do not remain planar and stress and strain
distribution do not vary linearly
Gabc
TL
abc
T3
22
1max
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• At large values of a/b, the maximum shear stress and angle of twist for other
open sections are the same as a rectangular bar.
231max3
31
ve ba
T
baG
T
31
2
31
110 c
c
b
a
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231max3
31
a)ts
T
tsG
T
2322
31max3
323
31 2
2
b)ta
T
tta
T
taG
T
ttaG
T
max322
3113
1maxmax322
3113
1 2
2 c) t
tbtb
TtG
tbtbG
T
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GJ
T
n
ii tbJ1
331
maxmaxmax tJ
TtG
3 max74.1r
tk
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Using τall =40 MPa, determine the largest torque that may be applied to
each of the brass bars. Note that the two solid bars have the same cross-
sectional area, and that the square bar and square tube have the same
outside dimensions.
SAMPLE PROBLEM 3.9
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tA
T
2
3. Square Tube. For a tube of thickness t, the shearing stress is given
by following equation
where A is the area bounded by the center
line of the cross section. We have
211563434 mmmmmmA
We substitute τ=τall =40 MPa and t = 6 mm and
solve for the allowable torque:
tA
T
2
mmmm
TMPa
61156240
23 NmT 5553
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Örnek: Ortalama yarıçapları R, cidar kalınlıkları t olan kapalı ve açık dairesel tüp kesitli çubuklar T burulma momentine maruz bırakılırsa τmax ve ϴ oranlarını hesaplayınız.
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Değişken kesitli kesitlerin burulması
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Örnek: Şekildeki profilin taşıyabileceği burulma momentini hesaplayınız.
G=80 Gpa
τem=70 Mpa
ϴem=0.22 rd/m
5 55
2
25
5
110
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Örnek: Boyutları şekilde verilen ‘L’ profil
kesitin imal edildiği malzemenin emniyet
gerilmesi 60 MPa dır. Birim dönme açısı
için konulan sınır 0.2 rad/m olduğuna
göre kesitin taşıyabileceği burulma
momentini hesaplayınız. G=80 GPa.
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43
33313
31
10873.5
5100480
mmJ
tbJ ii
Çözüm:
Kesitin polar atalet momenti:
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5
10873.560 3
max
maxmax
t
JT
tJ
T
em
em
NmNmmT 5.7070480
Emniyet gerilmesine göre burulma momentinin bulunması:
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333
10873.5108010
2.0
GJT
GJ
T
em
em
NmNmmT 97.9393970
Emniyetli birim dönme açısına göre burulma momentinin bulunması:
TT
alınır.
NmTTem 48.70
Buna göre olduğundan
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Örnek: Boyutları şekilde verilen ‘T’ profil kesitte 1 ve 2 parçaları, kayma
modülleri sırası ile G1=60 GPa ve G2=80 GPa olan farklı malzemelerden
imal edilmiştir. Buna göre bu profilin taşıyabileceği burulma momentini
hesaplayınız.
MPa
MPa
em
em
90
70
2
1
Kayma emniyet gerilmeleri
100 mm
120 mm
8
7
1
2
T
Kompozit profiller
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26433
11
4333111
108.6851043.111060
1043.1171003
1
3
1
NmmmmMPaJG
mmtbJ
26433
22
4333222
104.16381048.201080
1048.2081203
1
3
1
NmmmmMPaJG
mmtbJ
Çözüm:Kesitlerin burulma rijitlikleri:
![Page 67: İNCE CİDARLI TÜPLERİN BURULMASI THIN-WALLED TUBES](https://reader035.vdocuments.site/reader035/viewer/2022062221/56812ef3550346895d949493/html5/thumbnails/67.jpg)
21 TTT
Kesitlerde oluşan iç burulma momentleri-dış burulma momenti dengesi:
T
1T
2T
![Page 68: İNCE CİDARLI TÜPLERİN BURULMASI THIN-WALLED TUBES](https://reader035.vdocuments.site/reader035/viewer/2022062221/56812ef3550346895d949493/html5/thumbnails/68.jpg)
ve22
22
11
11 JG
T
JG
T
Kesitlerde oluşan burulma açıları birbirine eşittir (Uygunluk Şartı):
1
2
21
2211
21
221122
2
11
1
JGJG
TT
JGJG
T
JG
T
JG
T
TJGJG
JGTT
JGJG
JGT
2211
222
2211
111 e v
![Page 69: İNCE CİDARLI TÜPLERİN BURULMASI THIN-WALLED TUBES](https://reader035.vdocuments.site/reader035/viewer/2022062221/56812ef3550346895d949493/html5/thumbnails/69.jpg)
ve 22
221
1
11 t
J
Tt
J
T
Kesitlerdeki kayma gerilmeleri:
ve 22211
2221
2211
111 emem JGJG
tGT
JGJG
tGT
Birleşik kesitin taşıyabileceği emniyetli burulma momentleri:
veemtG
JGJGT 1
11
2211 emtG
JGJGT 2
22
2211
Son iki denklemden bulunacak en küçük burulma momenti emniyetli değer olarak alınır.
![Page 70: İNCE CİDARLI TÜPLERİN BURULMASI THIN-WALLED TUBES](https://reader035.vdocuments.site/reader035/viewer/2022062221/56812ef3550346895d949493/html5/thumbnails/70.jpg)
NmNmmT
MPammMPa
NmmNmmT
tG
JGJGT em
4.387104.387
7071060
104.1638108.685
3
3
2626
111
2211
Birinci kesite göre, birleşik kesitin taşıyabileceği emniyetli burulma yükü
aşağıdaki gibi bulunur:
![Page 71: İNCE CİDARLI TÜPLERİN BURULMASI THIN-WALLED TUBES](https://reader035.vdocuments.site/reader035/viewer/2022062221/56812ef3550346895d949493/html5/thumbnails/71.jpg)
NmNmmT
MPammMPa
NmmNmmT
tG
JGJGT em
8.326108.326
9081080
104.1638108.685
3
3
2626
222
2211
Buna göre, emniyetli burulma momenti küçük olan değerdir:
NmTem 8.326
Birinci kesite göre, birleşik kesitin taşıyabileceği emniyetli burulma yükü
aşağıdaki gibi bulunur:
![Page 72: İNCE CİDARLI TÜPLERİN BURULMASI THIN-WALLED TUBES](https://reader035.vdocuments.site/reader035/viewer/2022062221/56812ef3550346895d949493/html5/thumbnails/72.jpg)
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b=100 mmh=150 mmt=3 mm
Example: For the channel section, and neglecting stress concentrations,
determine the maximum shearing stress caused by a 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
![Page 86: İNCE CİDARLI TÜPLERİN BURULMASI THIN-WALLED TUBES](https://reader035.vdocuments.site/reader035/viewer/2022062221/56812ef3550346895d949493/html5/thumbnails/86.jpg)
mmx 29
1050
30000
150331002
5031002
46233 10219.4753100310012
121503
12
1mmI x
331094.302
75375753100 mmQ
MPatI
QV
xV 956.1
310219.4
1094.308006
3
V
x
A
C
B
EDe
V
BB
D
D
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x
A
C
B
EDe
T
mmI
tbhe
x
4010219.44
3.150100
4 6
2222
O
VxeVOCT
4333 1015.3310021503
1
3
1mmtbJ ii
MPatJ
TT 57.523
1015.3
102.553
3
NmNmmT 2.55102.558002940 3
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MPaTV 526.5457.52956.1max
The maximum shearing stress
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Bölmeli Tüplerin Burulması
(İleri Mukavemet)
![Page 90: İNCE CİDARLI TÜPLERİN BURULMASI THIN-WALLED TUBES](https://reader035.vdocuments.site/reader035/viewer/2022062221/56812ef3550346895d949493/html5/thumbnails/90.jpg)
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