Theories of Failure

1. Theories of Failure Failure of a member is defined as one of two conditions. 1. Fracture of the material of which the member is made. This type of failure is the characteristic of brittle materials. 2. Initiation of inelastic (Plastic) behavior in the material. This type of failure is the one generally exhibited by ductile materials. When an engineer is faced with the problem of design using a specific material, it becomes important to place an upper limit on the state of stress that defines the material's failure. If the material is ductile, failure is usually specified by the initiation of yielding, whereas if the material is brittle it isspecified by fracture.

2. These modes of failure are readily defined if the member is subjected to a uniaxial state of stress, as in the case of simple tension however, if the member is subjected to biaxial or triaxial stress, the criteria for failure becomes more difficult to establish. Inthis section we will discuss four theories that are often used in engineering practice to predict the failure of a material subjected to a multiaxial state of stress. A failure theory is a criterion that is used in an effort to predict the failure of a given material when subjected to a complex stress condition.

3. Several theories are available however, only four important theories are discussed here. • Maximum shear stress (Tresca) theory for ductile materials. • Maximum principal stress (Rankine) theory. • Maximum normal strain (Saint Venan’s) theory. • Maximum shear strain (Distortion Energy) theory.

4. Maximum shear stress theory for Ductile Materials The French engineer Tresca proposed this theory. It states that a member subjected to any state of stress fails (yields) when the maximum shearing stress (τmax)in the member becomes equal to the yield point stress (τy)in a simple tension or compression test (Uniaxial test). Since the maximum shear stress in a material under uniaxial stress condition is one half the value of normal stress and the maximum normal stress (maximum principal stress) is max, then from Mohr’s circle.

6. In case of Biaxial stress state

7. Problem 01:- The solid circular shaft in Fig. (a) is subject to belt pulls at each end and is simply supported at the two bearings. The material has a yield point of 36,000 Ib/in2• Determine the required diameter of the shaft using the maximum shear stress theory together with a safety factor of 3.

8. 400 + 200 lb 200 + 500 lb

14. Maximum Principal Stress theory or (Rankine Theory) According to this theory, it is assumed that when a member is subjected to any state of stress, fails (fracture of brittle material or yielding of ductile material) when the principal stress of largest magnitude. (1) in the member reaches to a limiting value that is equal to the ultimate stress,

15. Problem 02:- The solid circular shaft in Fig. 1 (a) is subject to belt pulls at each end and is simply supported at the two bearings. The material has a yield point of 36,000 Ib/in2• Determine the required diameter of the shaft using the maximum Principal stress theory together with a safety factor of 3.

16. 400 + 200 lb 200 + 500 lb

19. According to maximum normal stress theory .

21. Example 03 The solid cast-iron shaft shown in Fig. is subjected to a torque of T = 400 Ib . ft. Determine its smallest radius so that it does not fail according to the maximum-Principal-stress theory. A specimen of cast iron, tested in tension, has an ultimate stress of (σult)t = 20 ksi. Solution The maximum or critical stress occurs at a point located on the surface of the shaft. Assuming the shaft to have a radius r, the shear stress is

24. Mohr's circle for this state of stress (pure shear) is shown in Fig. . Since R = max, then The maximum-Principal-stress theory,, requires |1| ≤ ult Thus, the smallest radius of the shaft is determined from

25. Maximum Normal Strain or Saint Venant’s Criterion In this theory, it is assumed that a member subjected to any state of stress fails (yields) when the maximum normal strain at any point equals, the yield point strain obtained from a simple tension or compression test (y = σy/E). Principal strain of largest magnitude |max| could be one of two principal strain 1 and 2 depending upon the stress conditions acting in the member . Thus the maximum Principal strain theory may be represented by the following equation.

26. As stress in one direction produces the lateral deformation in the other two perpendicular directions and using law of superposition, we find three principal strains of the element. x= y= z= x= y= z= x= y= z= σx σx / E -μσx / E -μσx / E σy -μσy / E σy / E -μσy / E σz -μσz / E -μσz / E σz / E

27. x = σx / E -μσy / E -μσz / E = σx / E -μ / E (σy + σz) y = σy / E -μσx / E -μσz / E = σy / E -μ / E (σx + σz) z= σz / E -μσy / E -μσx / E = σz / E -μ / E (σx + σy) (2) Thus

28. Also

29. Maximum Shear Strain Energy (Distortion Energy Criterion (Von MISES Criterion) According to this theory when a member is subjected to any state of stress fails (yields) when the distortion energy per unit volume at a point becomes equal to the strain energy of distortion per unit volume at failure (yielding). The distortion strain energy is that energy associated with a change in the shape of the body. The total strain energy per unit volume also called strain energy density is the energy in a body stored internally throughout its volume due to deformation produced by external loading. If the axial stress

30. Distortion energy per unit volume is given by Strain energy due to distortion per unit volume for biaxial stress system According to distortion energy theory So, equation becomes