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Top school in delhi ncr

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Top school in delhi ncr

  1. 1. Top School in Delhi NCR :By school.edhole.com
  2. 2. 2 Section IV Failures Due to Static Loading school.edhole.com
  3. 3. 3 Talking Points  Static Loading?  Failure Theories:  Maximum-Normal Stress Theory  Maximum-Shear Stress Theory  The Distortion-Energy Theory  Factor of Safety for each Failure Theory school.edhole.com
  4. 4. 4 ?Static Loading  Static Load is a stationary force or moment acting on a member.  Stationary means that the force or moment does not change in magnitude.  Sometime the load is assumed to be static when it is known that some variation to be expected. Such assumptions are made to simplify the design computations when variations in load are few or minor in nature. school.edhole.com
  5. 5. 5 Failure Theories i. Maximum-Normal Stress Theory  This theory states that failure occurs whenever the largest principal stress equals the strength.  Suppose the principal stresses are ordered as:  Then if yielding is the criteria of failure, then failure occurs whenever: where: tensile yield stress = yt S  If ultimate tensile strength is used (in case of brittle materials), failure occur whenever: where: Principal Element 1 2 3 s >s >s yt yc ³ S £ -S 1 3 s or s compressive yield stress = yc S ut uc ³ S £ -S 1 3 s or s tensile ultimate stress = compressive ultimate stress = ut S uc S Pure tension Pure torsion Note: For pure torsion s1 = t = s3 and s2 = 0. This means that the part fails in torsion when t = Sy. But experiments show that a part in torsion will deform at about 60% of the yield strength. This is one of the reasons the maximum-normal stress theory is not recommended. school.edhole.com
  6. 6. Failure Theories ii. Maximum-Shear Stress Theory  This theory is used only in the ductile materials.  It states that yielding begins whenever the maximum shear stress becomes equal to the maximum shear stress in a tension test specimen of the same material when that specimen begins to yield. 6 2 (for simple tension) t = s max 1 t s s ( ) 2 (for pure torsion) = - max 1 3  The maximum-shear stress theory predicts that failure will occur whenever: t s s s s max or  Shear yield strength is given by: y y S S = - 1 2 ³ - ³ 1 3 2 2 sy y S = 0.5S school.edhole.com
  7. 7. a) both volume change and angular distortion  For figure (a):  For unit cube, the work done in any of the principal (1 2 )[ 2 ( )] (2) 1 2 2 3 1 3 7 Failure Theories iii. The Distortion-Energy Theory  This theory is called the shear-energy theory.  It is used also for ductile materials.  Like the maximum-shear stress theory, this theory is employed to define the beginning of yield. It predicts that yielding will occur whenever the distortion-energy in a unit volume equals the distortion energy in the same volume when uniaxially stressed to the yield strength.  The distortion energy is obtained by subtracting equation (5) from equation (2): = 1 + ¢ 3 n s ¢ = - + - + - E  So yielding is predicted to occur when: y school.edsh¢ ³oS le.com  Shear yield strength is given by: b) volume change c) distortion without volume change directions is: , where n= 1,2,3  The total strain energy is: or U = E s 2 +s +s 2 - n s s +s s +s s 1 3  The average value savg:  The remaining stress, (s1-savg), (s2-savg), (s3-savg) will produce angular distortion, figure (c). Substituting (savg for s1, s2, s3) in equation (2) gives the amount of strain energy producing only volume change. s 2 = s +s +s 3 avg  Substituting in equation (4) gives: 1 2 3 s >s >s 2 (1) n n n U =s e 1 2 3 U =U +U +U [ 2( )] (5) = 1 - 2 n s +s +s + s s +s s +s s v 1 6 1 2 2 3 1 3 2 3 2 2 2 E U ( ) 3 (3) 1 2 3 s = s +s +s avg (3 2 2 )(1 2 ) (4) v U = s E - n avg [( ) ]2 1 2 3 2 2 where, [( ) 2 ( ) 2 ( ) 2 ] 2 1 2 2 3 3 1 2 d s s s s s s s U sy y S = 0.577S
  8. 8. 8 Factor of Safety for each Failure Theory  For maximum-normal stress theory n = S n = - S = = - s s or (for ductile materials) s y s y n S n S 1 3 s s or (for brittle materials) s u s u 1 3  For maximum-shear stress theory ( ) 1 3 = s -s s y n S  For the distortion-energy theory: = ¢ s y n S school.edhole.com s ¢ = - + - + - where, s [( s s ) 2 ( s s ) 2 ( s s )2 ] 2 1 2 2 3 3 1
  9. 9. 9 Example A material has a yield strength of 600 MPa. Compute the fator of safety for each of the failure theories for ductile materials. Use the following stress states: a)- s1 = 420 MPa: s2 = 420 MPa, s3 = 0 b)- s1 = 420 MPa, s2 = 180 MPa, s3 = 0 c)- s1 = 420 MPa, s2 = 0 MPa, s3 = -180 MPa d)- s1 = 0 MPa, s2 = -180 MPa, s3 = -420 MPa. school.edhole.com
  10. 10. 9 Example A material has a yield strength of 600 MPa. Compute the fator of safety for each of the failure theories for ductile materials. Use the following stress states: a)- s1 = 420 MPa: s2 = 420 MPa, s3 = 0 b)- s1 = 420 MPa, s2 = 180 MPa, s3 = 0 c)- s1 = 420 MPa, s2 = 0 MPa, s3 = -180 MPa d)- s1 = 0 MPa, s2 = -180 MPa, s3 = -420 MPa. school.edhole.com

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