entc 376 chapter 9 lecture notes-iii-stress transformation
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Chapter 9: Stress Transformation
Finish Chapter 9Work Problems
Chapter 9: Stress Transformation
9.5 Stress in Shafts Due to Axial Load and Torsion
Example: An axial force of 900 N and a torque of 2.5 N∙m are applied to the shaft as shown. If the shaft has diameter of 40mm, determine the principal stresses at a point P on its surface.
Initial stress element at P
1
2x
x
kPa7.409
)9.198()2
2.7160()2
(R
kPa1.3582
kPa2.71602
2222y
yavg
xy
=
+−
=τ+σ−σ
=
=+
=σ+σ
=σ
x
x
3
Combined loadings (Ch. 8)C (σavg, 0)C (358.1,0)
A (σx , τxy)A (0, 198.9)
σ1 = 767.7 kPaσ2 = -51.5 kPa
Clockwise angle = 14.5o
= Tc/J
=N/A
9.6 Stress Variations Throughout a prismatic Beam
Direction (orientation) contours which give the directions of principal stresses of equal magnitude
Tensile principal stress contours which give the locations of identical tensile principal stresses
Stress Trajectories & Contours
σ2= σx =σLongitudinal
σ1= σy= σHoop
25° CCW°
°
Example A tank is to be made by rolling flat sheet of AISI 1040 cold‐drawn steel into thespiral shape as shown, where the spiral makes an angle of 65° with the horizontal axis of the tank. P=1.75 MPa, Di=900 mm.
(a) Specify a thickness, t, to provide N=4 based on sy or N=6 based on su.(b) Determine the stress condition on an element aligned with the weld.
Initial Stress element =Principal Stress Element
σ2= σx =σLongitudinal
σ1= σy= σHoop
Stress Element along Weld Line
25°
τvw σw
σvτwv
True 3‐D View of Initial Stress Element (not used in this example)
σ2= σx =σLongitudinal
σ1= σy= σHoop
σ3= σz= σRadial
X‐axis
)(205.1138
908)(9088900
8,
06.7)5.111(2
)900)(75.1(222
5.1116
6696
3.1414
5654
900nominal&
669,565:104075.1
.son based 6Nor son based 4N provide to t,, thicknessaSpecify (a)
min
uy
donecorrectisassumptionwallthint
DmeanmmtDD
assumptionwallthinofvaliditycheckmmtspecifysizepreferredFor
mmMPa
mmMPapDpDtt
pD
twotheofsmallerMPas
MPas
mmDgivenDcylinderwalledthinAssume
MPasMPassteelCDAISIMPap
m
im
allowable
m
Hoop
mmHoop
allowableu
d
yd
im
uy
−⇒>==
=+=+=−⇒=
====∴=
=====
===
==−
===
==
φ
σσσ
σσ
σ
Q
Cont’d
MPa66.492
MPa31.99)mm8(2
)mm908)(MPa75.1(t2
pD
)b(
1alLongitudinx2
mTangentialorHoopy1
=σ
=σ=σ=σ
===σ=σ=σ
elementstress principalthe as same the is elementstress initialthe case this In :s)coordinate Y-Xthe on based element(stress elementstress initialthe Construct
weld.the with aligned element an on conditionstress the Determine
σ2= σx =σLongitudinal= 49.66MPa
σ1= σy= σHoop= 99.31MPa
Initial Stress Element = Principal Stress Element,
and a special case in Mohr’s circle
Cont’d
X‐axis25° CCW
σv= 90.44 MPa
σw= 58.52MPaτwv
τvw = 19.02 MPa
Stress Element along Weld Line
τ (CW)
+σσ1σ2O=σavg
X‐axis
50° CCW
σv
σw
τvw
τwv •
•
• •0
R
(σv, τvw)
(σw, τwv)
MPa44.90)50cos(83.2448.74)50cos(RMPa02.19)50sin(83.24)50sin(R
MPa52.58)50cos(83.2448.74)50cos(R
.weldthealongconditionstresstherepresentthatcircletheonsintpothefindto5025x2byCCWRotate
MPa83.242/)66.4931.99(2/)(RRadiusMPa48.742/)66.4931.99(2/)(Center
avgv
wv
avgw
21
21avg
=°+=°+σ=σ=°=°=τ
=°−=°−σ=σ
°=°
=−=σ−σ==
=+=σ+σ=σ=
σ2= σx =σLongitudinal
σ1= σy= σHoop
25°
τvw σw
σv
τwv
Initial Stress element =Principal Stress Element
Stress Element along Weld Line
σ2= σx =σLongitudinal= 49.66MPa
σ1= σy= σHoop= 99.31MPa
25° CCW
σv= 90.44 MPa
σw= 58.52MPaτwv
τvw = 19.02 MPa
Summary
X‐axis
9.7 Absolute Maximum Shear Stress
General 3D state of stress Thru 3D stress transformation σ
and τ acting on any skewed plane can be determined.
There is an orientation having only principal stresses (σmin, σint, σmax)acting on the element – Triaxial Stress.
At another orientation the absolute maximum shear stress τabs max will occur.
2
2minmax
avg
minmax
σ+σ=σ
σ−σ=τ maxabs
How to find τabs max
←Circle with the largest R
Need to take a 3D view and find τabs max when the in‐plane principal stresses have the same sign – both tensile or both compressive
A plane stress case with principal stresses having the same sign
220)( maxmax
max'z'xσ
=−σ
=τ=τ maxabs
Given plane stress case
Both tensile
2)( minmax
max'y'xσ−σ
=τ=τ maxabs
No need to take a 3D view when the in‐plane principal stresses have the opposite – one tensile and one compressive
Opposite sign
Given plane stress case
A plane stress case with principal stresses having the opposite sign
τ (cw)
+σσ1=0
σ2σ3
+σ
τ (cw)
σ1
σ2σ3=0
Key: τmax is the radius of the largest circle.
Special Case 1 in TextbookSpecial Case 2 in Textbook
Initial Stress Elements
Principal Stress Elements
+σ
τ (cw)
σ1σ2=0σ3
σ1 > σ2 = 0 > σ3 σ1 > σ2 > σ3 = 0σ1 = 0 > σ2 > σ3
Need 3D View Need 3D View
Review: Mohr’s Circle
Example: Due to the applied loading, the element at the point on the frame is subjected to the state of stress shown. Determine the principal stresses and the absolute maximum shear stress at the point.
Opposite sign
σmax = -10+41.2=31.2σmin = -10-41.2 = -51.2θ = 38.0o
Example: The point on the surface of the cylindrical pressure vessel is subjected to the state of stress. Determine the absolute maximum shear stress at the point.
An orientation of an element 45° within the plane containing σmax=32MPa and σmin=0yields the state of absolute maximum shear stress and the associated average normal stress.
This is a case that principal stresses have the same sign, so need to think the stress state in 3D.
MPa162
0322
MPa162
0322minmax
avg
minmax
=+
=σ+σ
=σ
=−
=σ−σ
=τ maxabs
Homework # 10Due Thursday April 16
9-699-739-759-889-97
Exam 2: Thursday 4/23/09 (not next Thursday)
Lets work some problems
Materials after this slide are extra references
Initial State of Stress
Initial Stress Element(based on xyz coordinates)
Stress Analysis
Design Project
Min Principal Stress
Max Shear Stress
Chapter 11: The General Case of Combined Stress and Mohr’s Circle
Goals:
Combined Stress
σxAτxy
τyx
σy
x
y
Mohr’s Circle
σdirect or uniaxial
τdirectτtorsionalσbending
τv(beam)
Max Principal Stress
More about Initial Stress Element for a General Case of Combined Stresses
Direct Normaland/or
Bending Stress
Torsional Shearand/or
Vertical Shear
σx
Pointof
Interestτxy
τyx
σy
σy
σx
τxy
τyx
(Combined) Normal Stress (Combined) Shear Stress
What are the “resultant” σ’s (called max principal stress and min principal stress) and τmax (maximum shear) due to all stresses combined?
x
y
Plane Stress
Stress Transformation: Uniaxial Stress
θθσ=θθ=
θ
θ==τ
θσ=θ=
θ
θ==σ
θθ
θθ
cossincossinAP
cosA
sinPAV
coscosAP
cosA
cosPAN
x
2x
2
A
θ=θ=
sinPVcosPN
Forces
θ
θ=θ cos
AA
PAP
x =σ
Pθ
V
N
S = P
Pθ
σθ
τθ
x
y
xmax
xmax
21,45
,0
σ=τ=τ°=θ
σ=σ=σ°=θ
θ
θ
When
When
Stress Transformation: Biaxial Stress (no shear stress)
θθσ−σ=τ
θθσ−θσ=θττ
θσ+θσ=σ
θθσ+θσ=θσσ
θ
θ
θ
θ
θ
θ
cossin)(cos)tanA(sin)A()secA(
:sincos
sin)tanA(cos)A()secA(:
yx
yx
2y
2x
yx
of directionthe inm equilibriuForce
of directionthe inm equilibriuForce
θθ
θθ
τ=τ
σ+σ=σ+σ'
yx'
θσ−σ−=τ
θσ−σ−σ+σ=σ
θπ
+θτσ
θ
θ
θθ
2sin)(21
2cos)(21)(
21
:
yx'
yxyx'
'' for 2
substitute , and find To
θσ−σ+σ+σ= 2cos)(21)(
21
yxyx
θσ−σ= 2sin)(21
yx
stresses principalcalledare stress normal ofvalues min and max Such
smallest.the is otherthe and largestthe is one stressesthese Among ;σ to from varies σ yxθ σ
θσ−σ+σ+σ=σθ 2cos)(21)(
21
yxyxQ
σx
σy
σx
σy
σx
σy
τθ
σθ
τθ
σθ
σθ
τθ
τθ’
σθ’
σθ’
τθ’
A
Stress Transformation: General Case of Plane Stress
θτ−θσ−σ−=τ
θτ−θσ−σ+σ+σ=σ
θ
θ
2cos2sin)(21
2sin2cos)(21)(
21
xyyx
xyyxyx
:equationsstress shear and normal General
planes)principaltheonstressesshearnoareThere
stressprincipalmin
stressprincipalmax
.,e.i(0
)](21[)(
21
)](21[)(
210
dd
2xy
2yxyx2min
2xy
2yxyx1max
=τ
τ+σ−σ−σ+σ=σ==σ
τ+σ−σ+σ+σ=σ==σ⇒=θσ
θ
θ
θθ
))(21
()](21[
dd
avgmaxavgyx
2xy
2yxmax
στσ=σ+σ=σ
°τ+σ−σ=τ
=θτ
θ
θ
withdaccompanieis (i.e.,
planes)principaltheto45atoccur
0 Remember the τmax=((σx/2)2 + τxy2)½ in Ch. 10?
0
)2
()
avg
2xy
2yx22avg
=τσ=σ
τ+σ−σ
=τ+σ−σ
τσ
θθ
θθ
θθ
and at center withcircle a of equation anis This
(
:gives and Combine
σx
σy
τθ
σθ
τxyτyx
σx
σy
σx
σy
τxy
τyxτxy
τyx
Mohr’s Circle
Special Cases: Both Principal Stresses Have the Same Sign
σ1 > σ2 > σ3 =0 (Both are tensile)
σ1=0 > σ2 > σ3 (Both are compressive)
.
)no(
)(21
max
StressincipalPrMinStressincipalPrMaxmax
τ
τ
σ−σ=τ
thefindtowantwe and elementstressinitialanonactingstresses principal min and maxwithcasestressuniaxialaasviewedbe canIt
Plane Stress
Case 1: σ1 > σ2 > σ3 =0 (Both Are Tensile)
2
StressincipalPrMinTrue0Circles'Mohrst1thefromStressincipalPrMinCircles'Mohrst1thefromStressincipalPrMax
1max
3
2
1
σ=τ
==σ=σ=σ
.stressshearimummaxtruetheis
MPa3.1082
6.2162
)(21
and,stressprincipalimummintruethenowis0.,e.i
.caseD3aconsidertoneedwe,Thus!signsamethehaveand
MPa4.1036.56160RMPa6.2166.56160R
MPa6.564040baR
MPa40b
MPa40)120200(21)(
21a
MPa160)120200(21)(
21o,Center
:Circles'MohrFirst
131max
3
21
avg2
avg1
2222max
xy
yx
yxavg
==σ
=σ−σ=τ
=σ−
σσ
=−=−σ=σ
=+=+σ=σ=+=+==τ
=τ=
=−=σ−σ=
=+=σ+σ=σ=
Q
Same Sign
σ1
σ2
x-y planey’-z plane
x’-z plane
x-y plane
x’y’
Case 2: σ1=0 > σ2 > σ3 (Both are Compressive)
2
circles'Mohrst1thefromStressincipalPrMincircles'Mohrst1thefromStressincipalPrMax
StressincipalPrMaxTrue0
3max
3
2
1
σ=τ
=σ=σ
==σ
.stressshearimummaxtruetheis
MPa3.932
6.1862
)(21
and)principalmintrue(MPa6.186and,MPa4.43
,stressprincipalimummaxtruethenowis0.,e.i.caseD3aconsidertoneedwe,Thus
!signsamethehaveand
MPa6.1866.71115RMPa4.436.71115R
MPa6.713065baR
MPa30b
MPa65)18050(21)(
21a
MPa115)18050(21)(
21o,Center
:Circles'MohrFirst
331max
32
1
21
avg2
avg1
2222max
xy
yx
yxavg
−=−
=σ
=σ−σ=τ
−=σ−=σ=σ
−σσ
−=−−=−σ=σ
−=+−=+σ=σ=+=+==τ
=τ=
=+−=σ−σ=
−=−−=σ+σ=σ=
Q
σ3=–186.6MPa
σ2 =–43.4MPa
τ (cw)
+σσ1=0
σ2σ3
+σ
τ (cw)
σ1
σ2σ3=0
Key: τmax is the radius of the largest circle.
Special Case 1 in TextbookSpecial Case 2 in Textbook
Initial Stress Elements
Principal Stress Elements
+σ
τ (cw)
σ1σ2=0σ3
σ1 > σ2 = 0 > σ3 σ1 > σ2 > σ3 = 0σ1 = 0 > σ2 > σ3
Need 3D View Need 3D View
Review: Mohr’s Circle