1. THEORIES OF FAILURE THEORIES OF FAILURE FOR DUCTILE MATERIALS
Failure in a ductile material is specified by the
initiation of yielding.
Failure in a brittle material is specified by fracture.
THEORIES OF FAILURE FOR DUCTILE MATERIALS
Maximum Shear Stress Theory (Tresca Yield Criterion)
The most common cause of yielding of a ductile material such as steel is slipping, which occurs along the contact planes of randomly ordered crystals that make up the material.
FG09_32-09UNP95_96.TIF
Notes:
Problem 9-95/9-96
This slipping is due to shear stress :

2. THEORIES OF FAILURE
The maximum shear stress theory for plane stress can be expressed for any two in-plane principal stresses as s1 and s2 by the following criteria:
s1 and s2 have same sign
FG09_32-09UNP95_96.TIF
Notes:
Problem 9-95/9-96
s1 and s2 have opposite sign

3. THEORIES OF FAILURE
Maximum Distortion Energy Theory (von Mises Yield Criterion) s1 s3 s2
von Mises Stress for 3-Dimension s2 s1
von Mises Stress for 2-Dimension
FG09_32-09UNP95_96.TIF
Notes:
Problem 9-95/9-96
so = sy
Yield stress, sy , occurs when

4. THEORIES OF FAILURE Allowable Stress, sallow & Safety Factor, SF
In designing a component or structure, it is introduced what so-called Allowable Stress, which is defined as
where
FG09_32-09UNP95_96.TIF
Notes:
Problem 9-95/9-96

5. Failure of a brittle material Failure of a brittle material
THEORIES OF FAILURE
THEORIES OF FAILURE FOR BRITTLE MATERIALS
Failure of a brittle material
in tension
45o
Failure of a brittle material
in torsion
FG09_32-09UNP95_96.TIF
Notes:
Problem 9-95/9-96
If the material is subjected to plane stress, we require that

6. PROBLEM-1 Stress in Shafts Due to Axial Load and Torsion
An axial force of 900 N and a torque of 2.50 N.m are applied to the shaft as shown in the figure. If the shaft has a diameter of 40 mm and the safety factor is 5, determine the minimum yield stress of the material used.
(a)
(b)
(a)
Internal Loadings
The internal loadings consist of the torque and the axial load is shown in Fig.(b)

7. PROBLEM-1 Due to axial load Due to torsional load
Maximum stress Components
(a)
(b)
Due to axial load
Due to torsional load
198.9 kPa

8. PROBLEM-1
The state of stress at point P is defined by these two stress components
Principal Stresses:
We get s1 = kPa
s2 = – kPa
FG09_23c.TIF
Notes:
Example 9-12: solution
The orientation of the principal plane:
= – 29o

9. PROBLEM-1 von Mises equivalent stress for 2-D = 794.8 kPa
Yield Stress of the shaft material can be found from:
FG09_23c.TIF
Notes:
Example 9-12: solution
= (5)(794.8) kPa
= kPa
= MPa

10. PROBLEM-2
Solid shaft has a radius of 0.5 cm and made of steel having yield stress of Y = 360 MPa. Determine if the loadings cause the shaft to fail according to the maximum-shear-stress theory and the maximum-distortion-energy theory.

11. PROBLEM-2
State of stress in shaft caused by axial force and torque. Since maximum shear stress caused by torque occurs in material at outer surface, we have

12. PROBLEM-2
Stress components acting on an element of material at pt A. Rather than use Mohr’s circle, principal stresses are obtained using stress-transformation eqns 9-5:

13. PROBLEM-2 Maximum-shear-stress theory
Since principal stresses have opposite signs, absolute maximum shear stress occur in the plane, apply Eqn 10-27,
Thus, shear failure occurs by maximum-shear-stress theory.

14. PROBLEM-3 Stress in Shaft due to Bending Load and Torsion
A shaft has a diameter of 4 cm. The cutting section shows in the figure is subjected to a bending moment of 2 kNm and a torque of 2.5 kNm. T x y z
Determine:
The critical point of the section
The stress state of the critical point.
The principal stresses and its orientation.
Select the material if SF = 6
FG09_23c.TIF
Notes:
Example 9-12: solution

15. PROBLEM-3 Analysis to identify the critical point Due to the torque Tx y z
Maximum shear stresses occur at the peripheral of the section.
Due to the bending moment M
Maximum tensile stress occurs at the bottom point (A) of the section.
FG09_23c.TIF
Notes:
Example 9-12: solution A
Conclusion: the bottom point (A) is the critical point

16. PROBLEM-3 Stress components Due to the torque T 198.9 kPa Tx y z
198.9 kPa
Due to the bending moment M
318.3 kPa
FG09_23c.TIF
Notes:
Example 9-12: solution A

17. PROBLEM-3 Stress state at critical point A 198.9 kPa txy = 198.9 kPa
sx = kPa
Principal stresses
25.65o s1 s2
We get s1 = kPa
s2 = – kPa
FG09_14-04UNP08_09.TIF
Notes:
Problems 9-08/09
The orientation of the principal plane:
= o

18. PROBLEM-3 von Mises equivalent stress for 2-D s2 s1 = 469 kPa
Yield Stress of the shaft material can be found from:
FG09_14-04UNP08_09.TIF
Notes:
Problems 9-08/09
= (6)(469) kPa
= kPa
= MPa

19. PROBLEM-4
Steel pipe has inner diameter of 60 mm and outer diameter of 80 mm. If it is subjected to a torsional moment of 8 kN·m and a bending moment of 3.5 kN·m, determine if these loadings cause failure as defined by the maximum-distortion-energy theory. Yield stress for the steel found from a tension test is Y = 250 MPa.

20. PROBLEM-4
Investigate a pt on pipe that is subjected to a state of maximum critical stress.
Torsional and bending moments are uniform throughout the pipe’s length.
At arbitrary section a-a, loadings produce the stress distributions shown.

21. PROBLEM-4

22. PROBLEM-4

23. PROBLEM-4

24. PROBLEM-5 Stress in Shafts Due to Axial Load, Bending and Torsion
A shaft has a diameter of 4 cm. The cutting section shows in the figure is subjected to a compressive force of 2500 N, a bending moment of 800 Nm and a torque of 1500 Nm.
FG09_17c.TIF
Notes:
Procedure for Analysis
Determine: 1. The stress state of point A.
2. The principal stresses and its orientation
3. Determine the required yield stress if SF = 4.

25. PROBLEM-5 t s Analysis of the stress components at point A
Due to comprsv load:
Due to torsional load:
Due to bending load: t s
FG09_17c.TIF
Notes:
Procedure for Analysis
(compressive stress)
Stress state at point A
Shear stress: t = tA
Normal stress: s = sA’ + sA”