module 24 - static failure theories
DESCRIPTION
Failure TheoriesTRANSCRIPT
Fracture Mechanics and Static Failure Theories
Course Notes
Lesson
FAILURE OF DUCTILE MATERIALS UNDER STATIC LOADING
The von Mises-Hencky or Distortion-Energy Theory
1U =
2
1 1 2 2 3 3 1
U =2
1 2 3
1v v
E 1
2 1 3
1v v
E 2
3 1 2
1v v
E 3
2 2 21 2 3 1 2 2 3 1 32v
1U =
2E
COMPONENTS OF STRAIN ENERGY
h dU U U
1 1dh
2 2dh
3 3dh
1 2 3 1 2 3d d dh h h
1 2 3 1 2 33d d dh
Contd..
1 2 3 1 2 33d d dh
1 2 3
3h
2 2 2 2h h h h h h h h h hU v 1
2E
2 213 2 3
2 h hvE
Contd..
21 23
2h h
vU
E
2
1 2 31 23
2 3h
vU
E
2 2 21 2 3 1 2 2 3 1 3
1 22
6
v
E
DISTORTION ENERGY
d hU U U
2 2 21 2 3 1 2 2 3 1 32v
1
2E
2 2 21 2 3 1 2 2 3 1 3
1 22
6
v
E
2 2 21 2 3 1 2 2 3 1 3
1
3d
vU
E
21
3d y
vU S
E
Contd..
2 2 2 21 2 3 1 2 2 3 1 3
1 1
3 3y d
v vS U
E E
2 2 2 21 2 3 1 2 2 3 1 3yS
2 2 21 2 3 1 2 2 3 1 3yS
2 21 1 3 3yS
VON MISES EFFECTIVE STRESS
' 2 2 21 2 3 1 2 2 3 1 3
2 2 2 2 2 2
'6
2
x y y z z x xy yz zx
' 2 21 1 3 3
' 2 2 23x y x y xy
SAFETY FACTOR
'
ySN
2 2 21 2 3 1 2 2 3 1 3
yS
N
2 21 1 3 3
yS
N
PURE SHEAR
2 2 2 2 21 1 1 1 1 max3 3yS
1 max0.577 3
yy
SS
0.577ys yS S
MAXIMUM SHEAR STRESS THEORY
0.50ys yS S
max max 1 3 1 3
0.50 / 2
2ys y y yS S S S
N
MAXIMUM NORMAL-STRESS THEORY
Example 5-1
Problem :
Determine the safety factors for the bracket rod shown infigure 5-7 based on both the distortion-energy theory and the maximum shear theory and compare them.
Given :
The material is 2024-T4 aluminum with a yield strength of47 000 psi. The rod length l=6 in and arm a=8 in. The rod outsidediameter d=1.5 in. Load F=1 000 lb.
Assumptions :
The load is static and the assembly is at room temperature.Consider shear due to transverse loading as well as other stresses.
Solution
1 000 6 0.7518 108
0.249x
Fl cMCpsi
I I
1 000 8 0.7512 072
0.497xz
Fa rTrpsi
J J
1.
2 22 2
max
18 108 012 072 15 090
2 2x z
xz psi
1 max
18 10815 090 24 144
2 2x z psi
2 0
3 max
18 10815 090 6 036
2 2x z psi
Contd..
2.
Contd..
3. ' 2 21 1 3 3
2' 224 144 24 144 6 036 6 036 27 661psi
4.'
47 0001.7
27 661ySN
5. max
0.50 0.50 47 0001.6
15 090ySN
Stress Elements at Points A and B within Cross Section of RodFor Example 4-9
Contd..
7.
4 1 0004755
3 3 1.767bending
Vpsi
A
max 12 072 755 12 827torsion bending psi
8.
max
0.577 0.577 47 0002.1
12 827ySN
max
0.50 0.50 47 0001.8
12 827ySN
COULOMB-MOHR THEORY
THE MODIFIED-MOHR THEORY
1
utSN
1 1 3
ut uc
ut ut
S SN
S S
1 1 2 1 2
21
2uc ut
uc
S SC
S
Contd..
2 2 3 2 3
21
2uc ut
uc
S SC
S
3 3 1 3 1
21
2uc ut
uc
S SC
S
1 2 3 1 2 3, , , , ,MAX C C C
0 0if MAX
utSN
Example 5-2
Failure of Brittle Materials Under Static Loading
Problem :
Determine the safety factors for the bracket rod shown in Figure 5-7 based on the modified-Mohr theory.
Given :
The material is class 50 gray cast iron with Sut=52 500 psiand Suc=-164 000 psi. The rod length l=6 in and arm a=8 in. The rodoutside diameter d=1.5 in. Load F=1 000 lb.
Assumptions :
The load is static and the assembly is at room temperature.Consider shear due to transverse loading as well as other stresses.
1 000 6 0.7518 108
0.249x
Fl cMCpsi
I I
1 000 8 0.7512 072
0.497xz
Fa rTrpsi
J J
1.
Solution
2 22 2
max
18 108 012 072 15 090
2 2x z
xz psi
1 max
18 10815 090 24 144
2 2x z psi
2 0
3 max
18 10815 090 6 036
2 2x z psi
Contd..
2.
Contd..
3.1
52 4002.2
24 144 utSN =
4. 1 1 2 1 2
21
2uc ut
uc
S SC
S
164 000+2 52 5001
24144 0 24144 0 16 4152 164 000
psi
Contd..
2 2 3 2 3
21
2uc ut
uc
S SC
S
3 3 1 3 1
21
2uc ut
uc
S SC
S
164 000+2 52 5001
0 6 036 0 6 036 19322 164 000
psi
164 000+2 52 5001
24144 6 036 24144 6 036 18 3482 164 000
psi
Contd..
5. 1 2 3 1 2 3, , , , ,MAX C C C
16 415, 1932, 18 348, 24144, 0, 6 036 24144MAX
6.1
52 4002.2
24 144 utSN =
7.
4 1 0004755
3 3 1.767bending
Vpsi
A
max 12 072 755 12 827torsion bending psi
Contd..
8. 1 max 12 827 psi
2 0
3 max 12 827 psi
9.1
52 4004.1
12 827utS
N
10.1 8 721C psi
2 4106C psi
3 12 827C psi
Contd..
11. 12 827 psi
12. 52 5004.1
12 827utS
N