Triaxial Stress State
description
Transcript of Triaxial Stress State
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Triaxial Stress State
(+ve sense shown)
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3D Principal – Triaxial Stress
max int min
123
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3D Stress – Principal Stresses
3 21 2 3 0I I I
The three principal stresses are obtained as the three real roots of the following equation:
where
1
2 2 22
2 2 23 2
x y z
x y x z y z xy xz yz
x y z xy xz yz x yz y xz z xy
I
I
I
I1, I2, and I3 are known as stress invariants as they do not change in value when the axes are rotated to new positions.
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Stress Invariants for Principal Stress
3213
3132212
3211
I
I
I
1
2 2 22
2 2 23 2
x y z
x y x z y z xy xz yz
x y z xy xz yz x yz y xz z xy
I
I
I
Zero shear stress on principal planes
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Mohr’s Circle?
• There is no Mohr’s circle solution for problems of triaxial stress state
• Solution for maximum principal stresses and maximum shear stress is analytical
• Either closed form solution or numerical solution (or computer program) are used to solve the eigenvalue problem.
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Maximum Shear Stresses
2 3max,1 2
1 3
max,2 2
1 2
max,3 2
Absolute max shear stress is the numerically larger of:
NormalStress,
31
y’z’, absmax
2
x’y’y’z’
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3D Mohr’s Circle – Plane StressA Case Study – The two principal stresses are of the same sign
31 2
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3D Mohr’s Circle – Plane StressA Case Study – The two principal stresses are of opposite sign
31 2
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For the following state of stress, find the principal and critical values.
ij
120 50 0
50 80 0
0 0 0
MPa
Tensor shows that: z = 0 and xz = yz = 0
80 MPa
120 MPa
50 MPa
y
x
Example:
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120 MPa
0 MPa
0 MPa
80 MPa
0 MPa
0 MPa
y
z
x
z
The other 2 faces:
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She
ar S
tres
s, M
Pa
-25 0 25 50 75 100 125 150 175-80
-60
-40
-20
0
20
40
60
80
Normal Stress (MPa)
V
H
max = 77 MPa
3-D Mohr’s Circles
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Example: triaxial stress state, not plane stress
• Determine the maximum principal stresses and the maximum shear stress for the following triaxial stress state. (+ve values as defined in slide 1)
102530
253040
304020
MPa
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Solution
zyzzx
zyyxy
zxxyx
102530
253040
304020
MPa
1
2 2 22
2 2 23 2
x y z
x y x z y z xy xz yz
x y z xy xz yz x yz y xz z xy
I
I
I
= 20 + 30 –10 = 40 MPa
= -3025 MPa
= 89500 MPaSolve
3 21 2 3 0I I I
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Solution to Example
-800000
-600000
-400000
-200000
0
200000
400000
600000
-100 -80 -60 -40 -20 0 20 40 60 80 100
Stress (MPa)
Sig
ma
(MP
a)
-51.8 MPa
65.3 MPa
26.5 MPa
Open the Exel Spreadsheet “triaxial stress.xls” for a template to solve the cubic eqn.
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Results
MPa
MPa
MPa
MPa
5.58
)8.513.65(2/1
8.51
5.26
3.65
max
1
2
3
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Mohr’s circles
NormalStress, MPa
3=63.51= -51.8
y’z’, absmax=58.5
2=26.5
Shear (MPa)
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Safety Factor?If the stress state was determined on a steel crankshaft, made of forged
SAE1045 steel with a yield strength of 300 MPa, what is the factor of safety against yield?
1. Tresca Criterion: max= 58.5 MPa
6.2)5.58(2
300
2 max
ySFS
2. Max Principal Stress Criterion: max= 63.5 MPa
6.43.65
300
2 max
ySFS
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3. Von Mises Criterion:
9.23.103
300
e
ySFS
2/1222
2/1213
232
221
))8.51(3.65()3.655.26()5.268.51(2
1
)()()(2
1
e
e
MPa
MPa
MPa
8.51
5.26
3.65
1
2
3
=103.31 MPa