CHAPTER 2 Failure/Fracture Criterion


 Sheila Ray
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1 (11) CHAPTER 2 Failure/Fracture Criterion
2 (12) Failure (Yield) Criteria for Ductile Materials under Plane Stress Designer engineer: 1 Analysis of loading (for simple geometry using what you learn here and for complex ones using finite element approach) and 2 Does failure occur under the predicted loading condition? For uniaxial stress: Θp =0 & Θs = ± 45 : shear is responsible for the failure of ductile materials (τ max)y = σ Y/2) Different criterions for ductile (Tresca & von Mises) and brittle conditions (Coulomb & Mohr) MaximumShearingStress Criterion for Ductile Materials (Tresca, 1868) τmax < τmax)y (at uniaxial test) τmax)y = σy/2 therefore τmax < σy/2 τmax = σa/2 Uniaxial Stress (σx<σy) Plane Stress (?) It is more accurate to use experimental data for τ y if principal stresses have the same signs τmax = σa σb /2 if principal stresses have opposite signs If the principal stresses σa and σb have the same sign σa < σy and σb < σy If they have opposite signs σa σb < σy Assumptions: only shear is considered (not sensitive to σa & σb ), σyt = σyc
3 (13) Example 4: Consider a spherical vessel with thickness of t, diameter of D, and yield stress of σy. At what value of internal pressure (P) will yielding occur according to Tresca s criterion? = = 2 = 2 0 = 2 4 = 2 = 2 = 4 Example 5: Calculate allowable P according to Tresca s criterion and considering a factor of safety of 2. = 4 =, = 4 = = 2 = τ ± ( ) +, = ± ( ) +, = 2 ± ( 2 ) + 16 = 2 = 260 = = = σ = = 2 = > 0 = < Example 6: Calculate Factor of Safety according to the Tresca s criterion (σy = 280 MPa) At point K: σmax = MPa σmin = 22.9 MPa, ) =. = 62.2 = = 280/2 = 2.25, = 62.2 = = 2.25
4 (14) TBR 4: P = 25 kn, σ y = 420 MPa, Factor of Safety = 4. Find maximal T based on Tresca s criterion. = 2 = = 1 2 = (30 )(40 ) = = 2 (600 )(2 ) = 2400 = = = ( ), = ± ( ) + =. ± (. ) + ( ) = 2 = ( ) + ( 2400 ) = = 2 ( ) + ( 2400 ) = 4 = =
5 (15) MaximumDistortionEnergy Criterion (von Mises criterion) Reminding points: ) : = ) : = + + If: e = 0 then + + =0 0 = strainenergy density Energy is a scalar quantity and thus for the sake of ease principal axes are considered = + + Replace strains using Hook s law = no volume change + + =0 = + = Experimental evidence show that materials can withstand very high uniform (hydrostatic) stresses without yielding. In 1904 Huber proposed that yielding in a ductile material occurs when distortion energy per unit volume of the material exceeds or equals the distortion energy per unit volume of the same material when it is subjected to yielding in a sample tension test. To obtain u d then substitute stresses with,, : For uniaxial tension test 0 : For general state of stress: = = ( ) + ( ) + ( ) = Since only differences of the stresses are involved adding a constant stress to each does not alter the yield condition 1) Same results for six points (uniaxial and = conditions) 2) Tresca s criteria is more conservative For torsion (maximum difference): σa = σb = ±0.5σY (Tresca) σa = σb = ±0.577σY (Mises) τy/σy =
6 (16) Example 7: Consider a cylindrical vessel with internal pressure of P, radius of r, and yield stress of σy. Find minimal thickness of vessel based on von Mises criterion. = = 2 + = ( ) ( ) ( ) +( ) = 2 = : = 0 = 2 2 = Example 8: The element with yield stress of σy is under pure shear loading as shown. Find factor of safety based on VonMises criterion. τ τ, = ± ( ) + = ± τ = + = τ 2  τ(τ)+ τ 2 = 3 = Based on Tresca criterion: = = = Example 9: The state of stress shown occurs in a machine component made of a brass for which σy = 160 MPa. Using the maximumdistortionenergy criterion, determine the range of values of σz for which yield does not occur., =, = 145, 25 ± ( ) + = σz is the 3 rd principal stress as there is no shear stress on face z ± ( ) ( ) + ( ) + ( ) = ( 25) + ( 25 ) + ( 145) < < < < SOLVE USING TRESCA s CRITERIA. Is there a solution?
7 (17) TBR 5: The stresses on the surface of a hard bronze component are shown in the figure. The yield strength of the bronze is σy = 345 MPa. (a) What is the factor of safety predicted by the maximumshearstress theory of failure for the stress state shown? Does the component fail according to this theory? (b) What is the value of the Mises equivalent stress for the given state of plane stress? (c) What is the factor of safety predicted by the failure criterion of the maximum distortion energy theory of failure? Does the component fail according to this theory?
8 (18) Example 10: The 60mmdiameter shaft is made of a grade of steel with a 300 MPa tensile yield stress. Using the maximumshearingstress criterion and maximumdistortionenergy criterion determine the factor of safety for magnitude of the torque T= 5 knm, P = 100 kn, and M = 2.3 knm. Based on Tresca s criterion (for critical point): = = (60 ) = 35.36, = = ( ) (30 ) 4 (30 ) = =, = = ( ) (30 ) 2 (30 ) = = ± ( ) + = 210, = 66.2 = 2 = 210 ( 66.2) 2 = = 300 = = 1.08 Based on von Mises criterion: = + = (210) ( ) + ( 66.2) = = = 1.2 = 3.2
9 (19) Example 11: A hollow structural steel flexural member is subjected to the load shown. The yield strength of the steel is σy = 320 MPa. (a) Determine the factors of safety predicted at point K by the maximumshearstress theory of failure. (b) Determine the Mises equivalent stresses at point K. (c) Determine the factors of safety at point K predicted by the maximumdistortionenergy theory. 225 kn V = 225 kn M = 225 kn 1m = 225 knm =, = 50, = (150 16)(250 16) = = ( )(50 ) = () = = ( ), = 16, = = (125 4)(8 150) (67 8) 2 = = = ( )( ) = 63.18, (. )(), = ± ( ) +, = (.) ± ( (.) ) = 17.17, ( ) = = ) 2 = = 1.28 = + = (17.17) ( ) + ( ) = 242 ) = = 1.32
10 Example 12: A steel shaft with an outside diameter of 20 mm is supported in flexible bearings at its ends. Two pulleys are keyed to the shaft, and the pulleys carry belt tensions as shown. The yield strength of the steel is 350 MPa.(a) Determine the factors of safety predicted at points H and K by the maximumshearstress theory of failure. (b) Determine the Mises equivalent stresses at points H and K. (c) Determine the factors of safety at points H and K predicted by the maximumdistortionenergy theory. (20)
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13 (23) TBR 6: If the A36 steel (σy = 250 MPa) pipe has an outer and inner diameter of 30 mm and 20 mm, respectively, determine the factor of safety against yielding of the material at point A according to the maximumdistortionenergy theory.
14 (24) TBR 7: A force P0 applied by a lever arm to the shaft produces stresses at the critical point A having the values shown. Determine the load PS = cs P0 that would cause the shaft to fail according to the maximumshearstress theory, and determine the load PD = cd P0 that would cause failure according to the maximumdistortionenergy theory. The shaft is made of steel with σy = 36 ksi.,, 14.14, 10,20 * The stresses at point A are proportional to load Tresca: 2, 36, c S = 1.2 Von Mises: = 2, 36 0 and c D = 1.36 OR = (20) 2 (10 20)+(10) 2 = 26.4 ksi = =
15 (25) TBR 8: A sign is supported by a pipe (σy = 250 MPa) having outer diameter 110 mm and inner diameter 90 mm. The dimensions of the sign are 2.0 m 1.0 m, and its lower edge is 3.0 m above the base. Note that the center of gravity of the sign is 1.05 m from the axis of the pipe. The wind pressure against the sign is 1.5 kpa. Determine factor of safety based on Tresca and von Mises (maximumdistortionenergy theory) criterions at points A, B, and C. Tresca: FS = (250/2)/76 = 1.64 Tresca: FS = (250/2)/19.94 = 6.26 Tresca: FS = (250/2)/23.7 = 5.27
16 TBR 9: A steel shaft with an outside diameter of 20 mm is supported in flexible bearings at its ends. Two pulleys are keyed to the shaft, and the pulleys carry belt tensions as shown. The yield strength of the steel is σy = 350 MPa. Determine the factors of safety predicted at points H and K by the maximumshearstress theory and by the maximumdistortionenergy theory. (26)
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18 (28) Failure (Fracture) Criteria for Brittle Materials under Plane Stress For uniaxial tensile test failure occurs when σ = σu (ultimate stress). For plane stress conditions a criterion needs to be defined MaximumNormalStress Criterion (Coulomb s or Rankine; Criterion) No yielding for brittle conditions so caution is required For Brittle materials: 1 fracture is due to normal stress and 2 < (e.g., for cast iron = ). Shortcoming: Not good if, < 0. But it shows good agreement with experimental tests if there exists a tensile principal stress or for a brittle material with = (this is seldom due to presence of cracks). σ UT < < = (, ) Mohr s and simplifies Mohr (CoulombMohr) Criterion Applicable when results of various tests on material are available. If both principal stresses are positive, the state of stress is safe as long as σ a<σ UT and σ b<σ UT; if both principal stresses are negative, the state of stress is safe as long as σ a < σ UC and σ b < σ UC. σ UT σ UT σ UT τ Failure Envelope C σuc σut σ C2 C3 C1 D Failure Envelope = + = = + Usually More conservative than Mohr s criterion Only when principal stresses have opposite sign For design, incorporating the factor of safety FS, divide all strengths by FS: = =, If principal stresses have the same sign: = (, ) = (, ) Mohr s criteria can be used in ductile conditions in which yield stress in tension and compression are very different as Tresca and von Mises criterions both assume that yield stress in tension and compression are equal.
19 (29) Example 13: For a certain point of a castiron machine frame the state of stress on an element is as shown. Find factor of safety based on Mohr s criterion (σut = 295 MPa and σuc = 970 MPa)., = ± ( ) + = ± ( ) + 50 = 159.6, = Compression test τ (MPa) Tension test = 1 = σ (MPa) How to calculate FS using Mohr s circle? State of Stress Example 14: The shaft of a femur can be approximated as a hollow cylindrical shaft. The loads that cause femur bones to fracture are axial torque and bending moments. During strenuous activities (e.g., skiing) the femur is subjected to torque of T = 100 Nm. Determine the maximal bending moment M that the bone can support without failure according to maximum normal stress criteria (D = 24 mm, Di = 16 mm, σut = 120 MPa, σuc = 240 MPa). = = = = = = =0;, = ± + = ± = =120 = Based on CoulombMohr: =?? =1
20 (30) Example 15: The castaluminum rod shown is made of an alloy for which σut = 60 MPa and σuc = 120 MPa. Using Mohr s criterion, determine the magnitude of the torque T for which failure should be expected. = = , T Nmm Nm τ (MPa) Compression test Envelope Failure Envelope σ (MPa) Tension test
21 (31) Comparison of Yield and Fracture Criteria with Test Data E.P. Popov (1990), page 529 Remark: Pure shear test Principal stress: = = 1 1 = + = 2
22 (32) TBR 10: The castaluminum rod shown is made of an alloy for which σut = psi and σuc = psi. Using Mohr s criterion, determine the maximum magnitude of the force F for which factor of safety is equal to 2 at point A (neglect shear stress due to F and only consider bending and torsion stresses at point A). = =, ( =
23 (33) TBR 11: The shown press is made of cast iron having ultimate strength in tension (σut = 170 MPa) and compression (σuc = 650 MPa). Calculate the allowable load P according to the MohrCoulomb criteria and based on a factor of safety equal to 2.5. N. A. 190 mm M P = =
24 TBR 12: A 1.25in.diameter solid shaft is subjected to an axial force of P = 7,000 lb, a horizontal force of V = 1,400 lb, and a concentrated torque of T = 220 lbft, acting in the directions shown. Assume L = 6.0 in. The ultimate failure strengths for this material are 36 ksi in tension and 50 ksi in compression. Use the Mohr failure criterion to evaluate the safety of this component at points H and K. (34)
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27 (37) TBR 13: The shaft of a femur can be approximated as a hollow cylindrical shaft. The loads that cause femur bones to fracture are axial torque and bending moments. During strenuous activities (e.g., skiing) the femur is subjected to M = 100 Nm. Determine the maximal torque T that the bone can support without failure according to CoulombMohr criterion and factor of safety of 1.2 (D = 24 mm, Di = 16 mm, σ ut = 120 MPa, σ uc = 240 MPa). = = = = = =0;, = =91.8 Based on CoulombMohr: ± + =45.9± =. = 1 1.2
28 (38) TBR 14: A pipe (σy = 95 MPa) with an outside diameter of 140 mm and a wall thickness of 5 mm is subjected to the loadings shown. The internal pressure in the pipe is 1,600 kpa. (a) Determine factor of safety at point K according to Tresca and Mises criterion. (b) if the structure is put in the brittle conditions find factor of safety at point H according to CoulombMohr criterion ( σut = 80 MPa, σuc = 160 MPa).
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30 TBR 15: A pipe (σy = 250 MPa) with an outside diameter of 95 mm and a wall thickness of 5 mm is subjected to the loadings shown. (a) Determine factor of safety at point K according to Tresca and Mises criterion. (b) if the structure is put in the brittle conditions find factor of safety at point H according to CoulombMohr criterion ( σut = 250 MPa, σuc = 400 MPa). (40)
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32 (42) TBR 16: For a given brittle material under torsion the maximal shear stress is equal to 80 MPa. Moreover, the material fails under loading condition (1). Find factor of safety for this material under loading condition (2) (use Mohr s criterion). :, : :,,.,....