Path to static failure of machine components

Transcription

1 Pat to static failure of macine components Load Stress Discussed last week (w) Ductile material Yield Strain Brittle material Fracture Fracture Dr. P. Buyung Kosasi,Spring 008 Name some of ductile and brittle materials Commonly used in engineering applications DUCTILE BRITTLE (S y close to S u ) Steel Plastics Rubbers Cast iron Hardened tool steel Concrete Woods Ceramics

2 Lecture 4 Failure under static load Failure of ductile material (C. 5 p. 9 5) Failure of brittle material (C. 5 p.54-6) Most common test to determine material strengt is Tensile test. Wat do we know from tensile test? Yield tensile stress (S y ) Ultimate tensile stress (S u ) 4

3 Mecanical properties of some material are listed in Appendi C (Norton). 5 How te yield tensile stress related to failure of part? F F y y Stress element of Tensile test specimen (uni-aial test) τ y y τ y τ y τ y y D stress element D stress element 6

4 Te answer is : 7 Using For ductile materials a) Te von Mises-Hencky teory (distortion energy teory) a) Te maimum Sear-Stress teory a) Maimum Normal-Stress teory For brittle materials b) Maimum Normal-Stress teory b) Te Coulomb-Mor teory b) Te Modified-Mor teory a) Te von Mises-Hencky or Distortion- Energy Teory Te total strain energy stored per unit volume in te material U (½) ε (J/m ) Etending to D stress state U ½ ( ε ε ε ) 8 In terms of principle normal stresses: [ ν ( )] U E 4

5 5 9 Component of strain energy ydrostatic distortion components U U U d d d d It is sown tat te ydrostatic stress is (page 4) ( ) / 0 Were U is due to ydrostatic loading i.e. equal principal stresses (volume cange witout canging sape no sear) [ ] υ E U U U U d d [ ] [ ] ) ( 6 ) ( υ υ E U E U U d is due to (angular) distortion (sape cange sear)

6 From tensile test, te uni-aial stress state at yield gives S y, 0. So, U yield υ S E y Comparing tis wit general stress state, so to prevent failure U > U yield d S y > or ) von Mises stress or effective stress or equivalent stress ` Te von Mises (effective) stress, `, can also be epressed in terms of applied stresses. > ( ) ( ) ( ) 6( τ τ τ ) y y z z y yz z For D case: y y τ y Factor of safety : N / S y 6

7 D visualization of te distortion energy failure teory. /S y Safe region (Distortion - Energy teory) /S y 0 /S y D visualization of te distortion energy failure teory. 4 7

8 a) Te maimum sear stress teory Te teory states tat failure occurs wen te maimum sear stress in a part eceeds te sear stress in a tensile specimen at yield (one-alf of te tensile yield strengt). i.e. part fails wen S ys 0.5 S y τ ma > S ys Factor of safety : /τ ma N S ys 5 D visualization of te maimum sear stress failure teory. /S y /S y -0.5 Safe region (Maimum Sear Stress teory)

9 a) Maimum normal stress teory Te teory states tat failure will occur wen te normal stress in te specimen reaces some limit on normal strengt suc as tensile yield strengt or ultimate tensile strengt. i.e. part fails wen {,, } > S y Factor of safety : N S y / ma{,, } 7 Safe region (Distortion - Energy teory) D visualization of te tree (ductile) failure teories. /S y Safe region (Maimum Normal Stress teory) 0 /S y Safe region (Maimum Sear Stress teory)

10 Wic teory sould we coose? 9 Application: For te loaded cantilever beam made of a material wit tensile yield strengt of 0 MPa. Wat is te maimum P tat te beam can support witout permanent yield? Te most critical spot P (N) m r 50 mm I πr 4 /4 M M ma P N.m m 4 A m V P N 0 0

11 Element y 0 τ y 04P 0 04P 0 04P 04 P τ y y τ y 0 M r/i 04 P (Pa) 04P 0 04P 0 0 τ y y 0 τ ma 00 P Based on te distortion energy teory Te von Mises stress at point is 04P For 04 P S y > 00 6 P < 4.7 kn > 04P

12 Based on te maimum sear stress teory S ys 0.5 S y 5 Mpa For τ ma 00 P < S ys 50 6 P < 4.7 kn Based on te maimum normal stress teory P < 4.7 kn (verify yourself) Element y 0 τ y ( 68.8 P) P 0 τ y y τ y 4V/A 68.8 P (Pa) ( 68.8 P) P τ y y 0 τ ma 68.8 P 4

13 Based on te distortion energy teory Te von Mises stress at point is 9.4 P For 9.4 P S y > 00 6 P < 08kN > 9.4 P 5 So te answer to te question Wat is te maimum P tat te beam can support witout permanent yielding? P <4. 7 kn 6

14 Failure mecanisms of brittle materials Ductile materials do not fracture on compression On tension fracture is due to normal stress alone On compression fracture is due to combination of normal stress and sear stress. 7 Caracteristics of brittle materials Failure mecanisms. In tension : due to normal stress alone In compression : due to combination of normal and sear stress Teir yield strengt (S y ) and ultimate strengt (S u ) are almost identical. So failure of brittle materials is normally associated wit fracture rater tan yield. And strengt refers to S u. S ut may be equal or not to S uc. Wen S uc > S ut, te materials are said to be uneven materials, or else tey are known as even materials. Teir sear strengt (S us ) can be greater tan teir tensile strengt (S ut ) unlike ductile materials were S us 0.5 S ut 8 4

15 Failure teories for brittle materials Even materials: b) Maimum Normal-Stress Teory Uneven materials: b) Te Coulomb-Mor Teory b) Te Modified-Mor Teory 9 b) Maimum Normal-Stress Teory for even materials.5 or /S uc Safe region (Maimum Normal Stress teory) or /S uc

16 b) Te Coulomb-Mor teory for uneven materials.0 (S ut, S ut ) Safe region (Coulomb-Mor teory) (-S uc, -S uc ) b) Modified-Mor teory for uneven materials Safe region (Modified-Mor teory) (S ut, S ut ) (-S uc, -S uc )

17 Wic teory sould we use for uneven materials? Calculation of factor of safety Safe region (Modified-Mor teory) N S ut / N S ut / / N S uc N S uc S ut S uc S ( ) ut 4 7

18 Modified-Mor effective stress by Dowling (for general D case and usually programmed) ~ ma( C, C C C 5, C,,, S N ~ut S ut Suc ( ) Suc Sut Suc ( ) Suc Sut Suc C ( ) Suc ~ 0 if ma < 0 ) Summary Lecture Static loading analysis Lecture Stress analysis of statically loaded parts Lecture 4 Failure analysis of stressed parts 6 8

19 Type of loadings Statically loaded parts Dynamically loaded parts Lecture 5 : Fatigue failure 7 9