Idea Transcript
ME111 Instructor: Peter Pinsky Class #21 November 13, 2000 Today’s Topics
• Failure of ductile materials under static loading.
• The von Mises yield criterion.
2. Consider two designs of a lug wrench for an automobile: (a) single ended, (b) double ended. The distance between points A and B is 12 in. and the handle diameter is 0.625 in. What is the maximum force possible before yielding the handle if Sy = 45 kpsi?
3. A storage rack is to be designed to hold a roll of industrial paper. The weight of the roll is 53.9 kN, and the length of the mandrel is 1.615 m. Determine suitable dimensions for a and b to provide a factor of safety of 1.5 if: (a) The beam is a ductile material with S y = 300 Mpa (b) The beam is a brittle material with Sut = 150 Mpa, S uc = 570 Mpa.
• The maximum shear stress criterion.
Reading Assignment Juvinall, 6.5 – 6.8
Problem Set #7. Due Wednesday, November 22. 1. The stress (in kpsi) at a point is given bys 1 = 10, s 2 = 0, s 3 = − 20 Calculate the factor of safety against failure if the material is: (a) Brittle with Sut = 50, S uc = 90 and using modified-Mohr theory. (b) Ductile with S y = 40 using (i) the max. shear stress criterion, and (ii) von Mises criterion.
4. For the problem in Example 19.1 (torsion-bar spring), what diameter d will provide a factor of safety of N = 3 against yielding based on von Mises with S y = 150 MPa.
5. Repeat Example 21.2 (a) and (b) using the maximum shear stress criterion.
6. For the beam shown, determine the factor of safety for: (a) Ductile material with S y = 300 Mpa (b) Brittle material with S ut = 150 Mpa, S uc = 570 Mpa.
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21.1 Failure of Ductile Materials Under Static Loading Static Loads: • Brittle materials fail by cracking or crushing and are typically limited by their tensile strengths • Ductile materials fail by yielding and are typically limited by their shear strengths Recall Uniaxial Tension Test:
Problem 2
Fb
Fd 2
Fd 1
t
Fd 1
Problem 3
s 2, s 3
Fb s1
Problem 6
s
Fd 2
2
21.2 Three-Dimensional Stress States • It’s useful to think about 3-d stress states using principal stresses:
Recall Torsion Test:
z 3 sz
tt zy
tt zx
tt yz
tt xz sx
s3 s
y
y
tt yx
tt xy
x
Fd t
ss
2
2
1
Principal stresses
Fb
s
s1
3
−
I1ss 2 + I 2ss − I 3 = 0
I1 = s x + s y + s z Fd
I2 = s xs y + s ys z + s zs x − t 2xy − t
2 yz
2
− t zx
I3 = s xs ys z + 2t xyt yzt zx − s xt 2yz − s yt 2zx − s zt xy2 Fb s
2
s3
s
s
Maximum shear stresses
1
t 13 = t 21 = t 32 =
s1− s 3
2 s 2 − s1
2
tmax = max( t13 , t21 , t32 )
s3 −s2
2
3
21.3 Split of Stress into Mean and Distortional Components
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21.4 Yielding of Ductile Materials • A ductile material yields when a yield criterion is exceeded.
s3
• Two yield criteria are important: Given principal stress state s2
The von Mises Yield Theory
2
s1
• The strain energy per unit volume of an elastic body has the form: 1
3
sm sm = sm
2
sm
1 (s + s 2 + s 3 ) 3 1
Mean stress produces volumetric strain
U = Um + Ud 1 − 2n 2 Um = [s + s 22 + s 32 + 2(s 1s 2 + s 2s 3 + s 3s 1 )] 6E 1 1+ n 2 Ud = [s + s 22 + s 32 − s 1s 2 − s 2s 3 − s 3s 1 ] 3E 1 where U h is the strain energy per unit volume associated with pure volume change (dilation) due to the mean (hydrostatic) stress:
1 sm =
3
s3 −s
1 (s + s 2 + s 3 ) 3 1
and where U d is the strain energy per unit volume associated with pure distortion m
• Note that in a tensile test at yield : s
s1−sm 1
2
−sm
2
Produces distortional strain responsible for plastic yielding
s1 =
Sy , s 2 = s 3 = 0
and, in this special case,
Ud =
1+ n 2 S 3E y
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• The von Mises yield criterion predicts failure in a general 3-d stress state when the distortion energy per unit volumeU d is equal to the distortion energy per unit volume in the tensile test specimen at failure, i.e.
Ud =
Von Mises ellipse (plane stress) s s3
No yielding
Yield surface
1 + nn 2 S 3E y
Pure tension ss 1
1 + nn 2 [ss + ss 22 + ss 32 − ss 1ss 2 − ss 2ss 3 − ss 3ss 1 ] = 1 + nn S 2y 3E 1 3E
Pure shear (torsion)
or, tt
S y = s 12 + s 22 + s 32 − s 1s 2 − s 2 s 3 − s 3s 1 In terms of
Special cases:
x, y and z stresses
Pure tension 2
Sy =
2
2
(ss x − ss y ) + ( ss y − ss z ) + ( ss z − ss x ) + 6 (tt
2 xy
+ tt
2 yz
+ tt
2 zx
)
2
For plane stress,ss 2 = 0
Sy =
Note, this is rather arbitrary but we’ll work with this as the plane stress state. 2
2
ss 1 − ss 1ss 3 + ss 3
s
s s2 =s s3 =
0
S y = ss 1 tt
Pure shear (torsion) s s 1 = −s s 3 = tt max ,
In terms of x, y and z stresses
s s2 =
S y = 3 ss 1 = 3tt max
0 s
S y = ss x2 + ss y2 − ss xss y + 3tt 2xy
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Von Mises Effective Stress Convenient to introduce the von mises effective stressss ′ ss ′ =
2
2
Example 21.1 For the bracket shown, determine the von Mises stress and factor of safety against yielding at points (a) and (b) if S y =10,000.
2
ss 1 + ss 2 + ss 3 − ss 1ss 2 − ss 2ss 3 − ss 3ss 1
30
20 C
(The von Mises effective stress ss ′ is defined as the uniaxial tensile stress that would create the same distortion energyU d as is created by the the actual combination of applied stress)
z
Rz , M z
Ry , M y y
von Mises (distortion energy) Yield Criterion
A
4
x
The von Mises yield criterion predicts failure when:
3
5
Rx , M x
40
B
5
(a)
Sy = s ′
(b)
From previous analysis we found:
A At (a):
s s1 = s s′=
Factor of Safety Against Yielding
N= Sy N
271, ss 3 = − 7590
=s′
At (b):
s s 12 − s s 1s s3+s s 32 =
Sy
s s1 = ss ′ =
N=
10,000 = 1.3 7,729
=
s s′
7,729
5096, ss 3 = − 343 2
2
ss 1 − ss 1ss 3 + ss 3 =
Sy s s′
=
5,274
10,000 = 1.9 5,274
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21.5 Maximum Shear Stress Criterion (Tresca Condition)
Example 21.2 A thin -walled cylindrical pressure vessel is subject to an axial force and torque loads as shown:
• Another important criteria is based on the theory that shear stress controls yielding (in contrast to von Mises theory based on distortion energy). • This theory was developed before the von Mises criterion and in practice is slightly more conservative.
p
P
T
r t
• The theory is easy to use in an “analytical” setting but is not well suited to use in finite element codes because of the corners in the yield surface (see below). • The maximum shear stress theory states that yielding will occur when the maximum shear stress reaches the shear stress in a uniaxial test specimen at yield, i.e.
(a) Given:
max(
p = 15 MPa , r = 35 mm, t = 3 mm, T = 450 kN ⋅ mm, S y = 290 MPa Determine the range of values of the axial load P which will provide a factor of safety against yielding of at least 1.4 based on the von Mises criterion.
or
s1−s2
2
,
s 2 −s3
2
,
s 3 −s 1
2
)=
Sy 2
max( s 1 − s 2 , s 2 − s 3 , s 3 − s 1 ) = S y
For plane stress with s
2
=
0
max( s 1 , s 3 , s 3 − s 1 ) = S y
(b) Given:
p = 20 MPa, t = 5 mm, T = 800 kN ⋅ mm, P = 100 kN , S y = 290 MPa Determine the range of values of the radius r which will provide a factor of safety against yielding of at least 1.4 based on the von Mises criterion.
Factor of Safety Against Yielding
Sy = max( s 1 , s 3 , s 3 − s 1 ) N
Remark: von Mises is the preferred theory
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Example 21.3
Von Mises and Tresca for Plane Stress
s3= Stress states inside the yield surface have not yielded
For the bracket shown, determine the factor of safety against yielding using the maximum shear stress theory at points (a) and (b) if S y =10,000.
0
von Mises:
s1 =
Sy
Tresca:
s1 =
Sy
30
20 C
z
Rz , M z
s s3
Ry , M y y
Von Mises yield surface
A
Pure tension
4
x
ss 1
3
5
Rx , M x
40
B
5
(a)
Pure shear (torsion)
(b)
From previous analysis we found:
A
Tresca yield surface At (a): s 1 = −s 3 von Mises: s 1 = Tresca:
s1 =
s s1 =
N= 1 S 3 y
1 S 2 y
271, ss 3 = − 7590
Sy 10,000 = = 1.27 max( s 1 , s 3 , s 3 − s 1 ) 7,861 s s1 =
At (b):
N =
5096, ss 3 = − 343 Sy
max( s 1 , s 3 , s 3 − s 1 )
=
10 ,000 = 1 .84 5,439
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