State of Stress in Three Dimensions

Transcription

1 State of Stress in Three Dimensions Theories of failure Introduction: Due to large numbers of examples of compound stresses met with in engineering practice, the cause of failure or permanent set under such conditions has attracted considerable attention. Certain theories have been advanced to explain the cause of failure and many of theories have received considerable experimental investigation. No great uniformity of opinion has been reached and there is still room for a great deal of further experimental investigation. 1

2 State of stress in Three Dimensions The principal theories of failure are Maximum principal stress theory Maximum shear stress or stress difference theory Strain energy theory Shear strain energy theory Maximum principal strain theory Mohr s theory

3 State of stress in Three Dimensions In all above said theories: σ et, σ ec = Tensile stress at the elastic limit in simple tension and compression respectively. σ 1, σ, σ 3 = Principal stresses in any complex system (such that σ 1 > σ > σ 3 ) It may be assumed that the loading is greater or static (and there is no cyclic or impact loading). 3

4 1. Maximum principal stress theory: This theory is usually associated with Rankine, but also received considerable support from other writers. This theory is simplest and the oldest theory of failure. According to this theory, failure will occur when the maximum principal tensile stress (σ 1 ) in the complex system reaches the value of maximum stress at the elastic limit (σ et ) in simple tension or the minimum principal stress (that is the maximum principal compressive stress) reaches the elastic limit stress (σ ec ) in simple compression. i.e., σ 1 = σ et ( in simple tension) σ 3 = σ ec ( in simple compression) σ 3 means numerical value of σ 3. 4

5 1. Maximum principal stress theory: If the maximum principal stress is the design criterion, then maximum principal stress must not exceed the working stress σ for the material. Hence σ 1 σ. This theory disregards the affect of other principal stresses and of the shearing stress on other planes through the element. For brittle materials which do not fail by yielding but fail by brittle fracture, the maximum principal stress theory is considered to be reasonably satisfactory. This theory appears to be approximately correct for ordinary cast-irons and brittle metals. 5

6 1. Maximum Principal stress theory: The maximum Principal stress theory is contradicted in the following cases: On a mild steel specimen when simple tension test is carried out sliding occurs approximately 45 0 to the axis of the specimen; this shows that the failure in this case is due to maximum shear stress rather than direct tensile stress. It has been found that a material which is even though weak in simple compression yet can sustain hydrostatic pressure far in excess of the elastic limit in simple compression. 6

7 1. Maximum Principal stress theory - Problems Problem 1 In a metallic body the principal stresses are +35 MN/m and -95MN/m, the third principal stress being zero. The elastic limit stress in simple tension as well as in simple compression is equal and is 0 MN/m. Find the factor of safety based on the elastic limit if the criterion of failure for the material is the maximum principal stress theory. 7

8 1. Maximum principal stress theory - Problems Solution: The given principal stresses are: σ 1 = +35 MN/m σ = 0 σ 3 = -95 MN/m and σ et = σ ec = 0 MN/m where, σ et = Elastic limit stress in tension, and σ ec = Elastic limit stress in compression. 8

9 1. Maximum principal stress theory - Problems σ 1 =σ t (working stress in tension), σ 1 = σ et F.S. F. S. = σ et = 0 = 6.8 σ 1 35 σ 3 = σ c ( working stress in compression), σ 3 = σ ec F.S. 95 = σ ec F.S. F.S. = 0 95 =.3. So the material according to the maximum principal stress theory will fail due to compressive principal stress. F. S. =.3. 9

10 1. Maximum principal stress theory - Problems Problem In a cast-iron body the principal stresses are +40 MN/m and -100 MN/m the third principal stress being zero. The elastic limit stresses in simple tension and in simple compression are 80 MN/m and 400 MN/m respectively. Find the factor of safety based on the elastic limit if the criterion of failure is the maximum principal stress theory. 10

11 1. Maximum principal stress theory - Problems Solution: Given Principal stresses are: σ 1 =40 MN/m σ = 0 σ 3 = -100 MN/m σ et = 80 MN/m σ ec = 400 MN/m Now, σ 1 =σ t (working stress in tension) σ 1 = σ et F.S. F. S. = σ et = 80 = σ

12 1. Maximum principal stress theory - Problems Also σ 3 = σ c ( working stress in compression) or σ 3 = σ ec F.S. 100 = σ ec F.S. F.S. = = 4. So the material according to the maximum principal stress theory will fail due to tensile principal stress. F. S. = 1

13 . Maximum shear stress or stress difference theory It is also called Guest s or Tresca s theory. This theory implies that failure will occur when the maximum shear stress q max in the complex system reaches the value of the maximum shear stress in simple tension at the elastic limit, i.e., q max = σ 1 σ 3 or σ 1 -σ 3 = σ et. = σ et in simple tension 13

14 . Maximum shear stress or stress difference theory In actual design σ et in the above equation is replaced by the safe stress. In case of any of the three principal stresses is compressive then it must be taken as σ 3 and its proper sign taken in above equation. That is, the maximum stress difference is to be equal to σ et. In the case of two dimensional tensile stress system, third stress must be taken as zero and thus the maximum stress difference calculated to equate it to σ et. This theory has been found to give satisfactory results for ductile materials. 14

15 . Maximum shear stress or stress difference theory Worth noting points: The theory doesn t give accurate results for the state of stress of pure shear in which the maximum amount of shear is developed (i.e., torsion test) The theory is not applicable in the case where the state of stress consists of triaxial tensile stress of nearly equal magnitude reducing the shear stress to a small magnitude, so that failure would be by brittle fracture rather than by yielding. The theory doesn t give as close results as found by experiments on ductile materials. However it gives safe results. 15

16 . Maximum shear stress theory - Problems Problem 1 A mild steel shaft 10 mm diameter is subjected to a maximum torque of 0 knm and a maximum bending moment of 1 knm at a particular section. Find the factor of safety according to the maximum shear stress theory if the elastic limit in simple tension is 0 MN/m. 16

17 . Maximum shear stress theory - Problems Solution d = 10 mm = 0.1 m. T= 0 knm M= 1 knm σ et = 0 MN/m F. S. =? M = π 3 d3 σ b. σ b = 3 M πd 3 = πd MN/m = MN/m 17

18 . Maximum shear stress theory - Problems T = π 16 f sd 3 f s = 16T πd 3 = π(0.1) MN/m = N/m σ = σ b ± σ b + fs = ± = 35.57± = MN/m or MN/m 18

19 . Maximum shear stress theory - Problems According to maximum shear stress theory σ 1 σ 3 = σ t σ 1 = MN/m σ = 0 σ 3 = MN m = σ t σ t = MN/m F.S. = σ et σ t = =

20 . Maximum shear stress theory - Problems Problem A shaft is subjected to a maximum torque of 10 knm and a maximum bending moment of 7.5 knm at a particular section. If the allowable equivalent stress in simple tension is 160 Mn/m, find the diameter of the shaft according to the maximum shear stress theory. 0

21 . Maximum shear stress theory - Problems Solution: Maximum torque, T= 10 knm Maximum bending moment, M = 7.5 knm Allowable equivalent stress in simple tension is, σ t = 160 MN/m Diameter of the shaft is d M =σ b π 3 d3 σ b = 3 M πd 3 T = f s π 16 d3 f s = 16 T πd 3 1

22 . Maximum shear stress theory - Problems Principal stresses are given by, σ 1,3 = σ b ± σ b + fs = 1 σ b ± σ b + 4f s = 1 3M πd 3 ± 3M 3T + πd 3 πd 3 σ 1 = 16 πd 3 M + M + T σ = 0 σ 3 = 16 πd 3 M M + T = 16 πd 3 M ± M + T

23 . Maximum shear stress theory - Problems According to maximum shear stress theory, σ t =σ 1 σ 3 = 16 + T 16 πd 3 πd 3 M M + T = 3 πd 3 M + T d 3 = 3 πσ t M + T = π = d= m or 9.6 mm d = 9.6 mm 3

24 3. Strain energy theory This theory which has a thermodynamic analogy and a logical basis is due to Haigh. This theory states that the failure of a material occurs when the total strain energy in the material reaches the total strain energy of the material at the elastic limit in simple tension. In a three dimensional stress system, the strain energy per unit volume is given by U= 1 E σ 1 + σ + σ 3 m σ 1σ + σ σ 3 + σ 3 σ 1 where σ 1, σ and σ 3 are of the same sign. 4

25 3. Strain energy theory Hence at the point of failure, 1 E σ 1 + σ + σ 3 m σ 1σ + σ σ 3 + σ 3 σ 1 = σ e E σ 1 + σ + σ 3 m σ 1σ + σ σ 3 + σ 3 σ 1 = σ e In actual design σ e in the above equation is replaced by the allowable stress obtained by dividing factor of safety. Taking two dimensional case (σ 3 = 0) the equation reduces to σ 1 + σ m σ 1σ = σ e If σ is the working stress in the material, the design criteria may be stated as follows σ 1 + σ m σ 1σ σ 5

26 Worth noting points: 3. Strain energy theory The results of this theory are similar to the experimental results for ductile materials ( i.e., the materials which fail by general yielding, for which σ et = σ ec approximately. It may be noted that order of σ 1, σ and σ 3 is immaterial) The theory doesn t apply to materials for which σ et is quit different from σ ec. The theory doesn t give results exactly equal to the experimental results even for ductile materials, even though the results are close to the experiments. 6

27 3. Strain energy theory Problem 1 A shaft is subjected to a maximum torque of 10 knm and a maximum bending moment of 7.5 knm at a particular section. If the allowable equivalent stress in simple tension is 160 MN/m, find the diameter of the shaft according to the strain energy theory. Take poisson s ratio, 1 m =

28 3. Strain energy theory Solution σ 1 = 16 πd 3 M + M + T ; σ = 0 σ 3 = 16 πd 3 M M + T Now according to strain energy theory, σ t = σ 1 + σ + σ 3 m σ 1σ + σ σ 3 + σ 3 σ 1 = σ 1 + σ 3 m σ 1σ 3 (since σ = 0) = 16 πd 3 M + M + T m (M M T ) = 16 πd 3 4M + T (1 + 1 m ) 8

29 3. Strain energy theory σ t = 16 πd 3 4M m T = 3 πd 3 M m = 3 πd 3 M + 0.6T T d 3 = 3 πσ t M + 0.6T = π d = m = 88.5 mm d = 88.5 mm. 9

30 A bolt is under an axial thrust of 9.6 kn together with a transverse force of 4.8 kn. Calculate its diameter according to: (i) Maximum principal stress theory (ii) Maximum shear stress theory, and (iii) Strain energy theory. 3. Strain energy theory Given: Factor of safety =3, yield strength of material of bolt = 70 N/mm and Poisson s ratio =

31 3. Strain energy theory A solid circular shaft is subjected to a bending moment of 60 knm and a torque of 6 knm. Design the diameter of the shaft according to (i) Maximum principal stress theory (ii) Maximum shear stress theory (iii) Maximum strain energy theory Take μ = 0.5, stress at elastic limit 50 N/mm and factor of safety =.5. 31

32 4. Shear Strain (Distortion) energy theory Maximum shear strain energy theory (or) Maximum distortion energy theory: This theory also known as von-misses Henkey criteria of elastic failure of elastic bodies. According to this theory, part of strain energy causes only changes in volume of the material and rest it causes distortion. At failure, the energy causing distortion per unit volume is equal to the distortion energy per unit volume in uniaxial state of stress at elastic limit. To derive the condition, let us split the state of stress at a point in case (i) into the two cases (Case (ii) and case (iii)) 3

33 4. Shear Strain (Distortion) energy theory σ 3 σ σ 1 σ 1 σ σ 3 Case (i) σ av σ av σ 3 σ av σ σ av av σ av σ 1 σ av σ 1 σ av σ av σ av Case (ii) σ -σ av σ 3 σ av Case (iii) 33

34 4. Shear Strain (Distortion) energy theory σ av is average stress and is given by, σ av = σ 1+σ +σ 3. 3 Since, in case (ii) stress in three mutually perpendicular direction is identical, there will be uniform change in all directions. σ av σ av σ av σ av σ av σ av Case (ii) 34

35 In case (iii), δv 4. Shear Strain (Distortion) energy theory σ 3 σ av = e v 1 + e + e 3 σ -σ av σ σ av + σ 3 σ av + = σ 1 σ av μ E E σ σ av μ E E σ 3 σ av μ E E = σ 1+σ +σ 3 3σ av E = σ 1+σ +σ 3 3σ av E σ 3 σ av + σ 1 σ av + σ 1 σ av + σ σ av μ E σ 1 + σ + σ 3 3σ av 1 μ σ 1 σ av σ 3 σ av σ 1 σ av =0 ( since σ 1+σ +σ 3 = σ 3 av ). There is no change in volume in case (iii). It causes only distorsion. σ σ av Case (iii) 35

36 4. Shear Strain (Distortion) energy theory Strain energy in Case (i) = Strain energy in case (ii)+ strain energy in case (iii) =Strain energy in Case (ii) + distortion energy in case (iii) Distortion energy = Strain energy in case (i) Strain energy in case (ii) We know that Strain energy per unit volume = 1 E σ 1 + σ + σ 3 m σ 1σ + σ σ 3 + σ 3 σ 1 Apply the above equation to case (i) and case (ii), we get Distortion energy = 1 E σ 1 + σ + σ 3 m σ 1σ + σ σ 3 + σ 3 σ 1 3σ av (1 μ) E 36

37 4. Shear Strain (Distortion) energy theory Distortion energy = 1 E σ 1 + σ + σ 3 3σ av m 1 E σ 1σ + σ σ 3 + σ 3 σ 1 3σ av = 1 E m σ 1 + σ + σ 3 3σ av = 1 E (1 + 1 m ) σ 1 + σ + σ 3 3 σ 1 + σ + σ 3 3 = 1 E (1 + 1 m ) 3σ σ + 3σ 3 σ 1 + σ + σ 3 = 1 6E (1 + 1 m ) 3σ 1 + 3σ + 3σ 3 σ 1 + σ + σ 3 = 1 6E (1 + 1 m ) σ 1 + σ + σ 3 σ 1 σ σ σ 3 σ 3 σ 1 = 1 6E (1 + 1 m ) σ 1 σ + σ σ 3 + σ 3 σ 1 ---(i) 37

38 4. Shear Strain (Distortion) energy theory In uniaxial direction, the state of stress at elastic limit is σ 1 = σ e, σ = σ 3 = 0. Distortion energy at the elastic limit = 1 6E (1 + 1 m ) σ e σ e = 1 6E (1 + 1 m ) σ e (ii) Equating (i) and (ii), we get criteria for failure is, 1 6E (1 + 1 m ) σ 1 σ + σ σ 3 + σ 3 σ 1 = 1 6E (1 + 1 m ) σ e σ 1 σ + σ σ 3 + σ 3 σ 1 = σ e This theory is found to give best results for ductile materials. 38

39 4. Shear Strain (Distortion) energy theory - Problem Problem: In a steel member, at a point, the major principal stress is 00 MN/m and the minor principal stress is compression. If the tensile yield point of the steel is 35 MN/m ; find the value of minor principal stress at which yielding will commence, according to each of the following criteria of failure. (i) Maximum shear stress (ii) Maximum total strain energy (iii) Maximum shear strain energy. Take μ =

40 4. Shear Strain (Distortion) energy theory - Problem Solution: σ 1 = 00 MN/m σ =? σ t = 35 MN/m μ = 0.6 (i) Maximum shear stress criteria: σ 1 σ = σ e 00 σ = 35 σ = 35MN/m (ii) Maximum total strain energy criteria: σ 1 + σ + σ 3 m σ 1σ + σ σ 3 + σ 3 σ 1 = σ e Here σ 3 = 0 σ 1 + σ m σ 1σ = σ e 40

41 4. Shear Strain (Distortion) energy theory - Problem σ 1 + σ m σ 1σ = σ e 00 + σ 104σ 555 = 0 σ 104σ 155 = 0 σ = 104 ± = = 81.89MN/m (comp.) (iii) Maximum shear strain energy criteria: σ 1 σ + σ σ 3 + σ 3 σ 1 = σ e σ 3 = 0 σ 1 + σ σ 1 σ + σ + σ 1 = σ e σ 1 + σ σ 1 σ = σ e σ 1 + σ σ 1 σ = σ e 41

42 4. Shear Strain (Distortion) energy theory - Problem σ 1 + σ σ 1 σ = σ e 00 + σ 00σ = 35 σ 00σ 155 = 0 σ = 00 ± = 58.8MN/m (comp.) 4

43 Glimpse of theories of failure (i) Maximum principal stress theory, σ 1 = σ et F. S. σ 3 = σ ec F.S. (ii) Maimum shear stress ( stress difference) theory, σ 1 -σ 3 = σ et F.S. (iii) Maximum strain energy theory, σ 1 + σ + σ 3 m σ 1σ + σ σ 3 + σ 3 σ 1 = σ et F.S. (iv) Maximum shear strain energy (Distortion energy) theory, σ 1 σ + σ σ 3 + σ 3 σ 1 = σ et F.S. 43

44 Problems A thick steel cylinder with an internal diameter 00 mm has to withstand an internal fluid pressure of 30 N/mm, calculate the thickness of the metal by using, (i) Maximum principal stress theory (ii) Maximum shear stress theory The tensile stress at yield point is 50 N/mm use factor of safety of.5. 44

45 Answers: (i) Maximum principal stress theory: b=650000, a=35 r 0 = mm t=36.7 mm (ii) Maximum shear stress theory: b= , a=0 r 0 = mm t=58.11 mm Hence minimum thickness of metal required is mm. Provide 60 mm thickness. Problems 45

46 Problems A cylindrical shell 1. m diameter is to be made of mild steel plates. It is subjected to an internal pressure of 1.5 MN/m. If the material yields at 00 MN/m, calculate the thickness of the plate on the basis of following theories of failure assuming a factor of safety of 3 in each case. (i) Maximum principal stress theory (ii) Maximum shear stress theory (iii) Maximum shear strain energy theory. 46

47 Problems At a point, the major principal stress is 10 N/mm (tensile) and the minor principal stress is compressive. If the yield stress of steel is 50 N/mm. Find the value of minor principal stress at which yielding takes place according to each of following theories of failure. (i) Maximum shear stress theory (iii) Maximum principal stress theory 47

48 Problems A solid circular shaft is subjected to a bending moment of 60 knm and a torque of 6 knm. Design the diameter of the shaft according to (i) Maximum principal stress theory (ii) Maximum shear stress theory (iii) Maximum strain energy theory Take μ = 0.5, stress at elastic limit 50 N/mm and factor of safety =.5. 48

49 A.U. Question Paper Problems A mild steel shaft is subjected to an end thrust producing a stress of 10 MPa and the maximum shearing stress on the surface arising from torsion is 90 MPa. The yield point of the material in simple tension was found to be 450 MPa. Calculate the factor of safety of the shaft according to (i) Maximum shear stress theory and (ii) Maximum distortion energy theory. (Nov/ Dec 014) 49

50 A.U. Question Paper Problems The inside and outside diameters of a cast iron cylinder are 40 mm and 150 mm respectively. If the ultimate strength of a cast iron is 180 MN/m find, according to each of the following theoriesthe internal pressure which would cause rupture: (i) Maximum principal stress theory, (ii) Maximum strain theory and (iii) Maximum strain energy theory. Poisson s ratio =0.5. Assume no longitudinal stress in the cylinder. (May/ June 013) 50

51 A.U. Question Paper Problems State the Haigh s theory. Also explain the maximum strain energy theory. (Nov/ Dec 014) 51