Failure from static loading

Transcription

1 Failure from static loading Topics Quiz /1/07 Failures from static loading Reading Chapter 5 Homework HW 3 due /1 HW 4 due /8

2 What is Failure? Failure any change in a machine part which makes it unable to perform its intended function.(from Spotts M. F. and Shoup T. E.) We will normally use a yield failure criteria for ductile materials. The ductile failure theories presented are based on yield.

3 Failure Theories [Note Theories] Static failure Recall Ductile Brittle Stress concentration Ductile Significant plastic deformation between yield and fracture Brittle Yield fracture

4 Tensile Test

5 Linear Stress Strain Plot

6 Failure Theory Problem Statement Given: Stress-strain data for simple uniaxial tension Find: When failure occurs for general state of stress

7 Mohr s Circle for Tensile Test

8 Static Ductile Failure Two primary theories for static ductile failure Von Mises criterion Maximum Distortion-energy Theory MDE Maximum Shear Stress criterion MSS

9 Static Ductile Failure Max Shear Stress criterion Material yields (fails) when: 1) ) τ max S ( 1 3) S y y or Factor of Safety: η S y S y ( 1 3) τ max

10 Safety Factors

11 Maximum Shear Stress Criteria

12 Static Ductile Failure Von Mises criterion Let the Mises stress ( e, equivalent stress) be: e 1 [( ) ( ) ( ) ] + + Then failure (yield) occurs when: e S y Factor of Safety: η S y e Typically, 1.5 η 4 Want a margin of error but not completely overdesigned

13 Comparison of MDE, MSS,MNS

14 Which theory to use? Look at a plot of the principal stresses B vs. A The non-zero principal stresses Failure occurs when the principal stresses lie outside the enclosed area Shape of area depends on the failure theory Data points are experimental results MSS Slightly more conservative Easier to calculate MDE More accurate If not specified, use this one!

15 Hydrostatic Stress State Diagonal

16 Ductile failure theory example Given: Bar is AISI 100 hot-rolled steel A DUCTILE material F 0.55 kn P 8.0 kn T 30 Nm Find: Factor of safety (η) Two areas of interest: A Top where max normal stress is seen (bending!) B Side where max shear stress is seen

17 Element A Consider the types of loading we have Axial? Yes due to P Bending? Recall that bending produces and τ, depending on the element of interest Yes due to M ( at A, τ at B) Torsion? Yes due to T

18 Element A Calculate stresses due to each load Axial: P P 4P x A πd πd Bending (A has normal Stresses): 4 Normal D ( FL) My 3FL x 4 3 I πd πd 64 Shear: τ xy 0 D ( T ) Torsion: Tc 16T τ xz 4 3 J πd πd 3

19 Element A Look at a stress element Sum up stresses due to all the loads x 4P πd + 3FL 3 πd 4PD + 3FL 3 πd τ xz 16T πd 3 x 95.5 MPa τ xz 19.1 MPa

20 Element A Draw Mohr s Circle with the stresses that we calculated x 95.5 MPa τ xz 19.1 MPa x at ( x, τ xz ) (95.5, 19.1) y at ( y, τ zx ) ( y, -τ xz ) (0, -19.1) Find C C x + y, ,0 ( 47.8,0) Find radius, R R ( ) + τ ( ) x C x xz

21 Out of Plane Maximum Shear for Biaxial State of Stress Max. Shear: 1 3 Case 1 1, > τ max τ Case,3 < τ max max Case 3 1 >0, 3 < τ max

22 Element A Find principal stresses (D) 1 C + R 99. MPa C - R MPa Think about 3-D Mohr s Circle! This is Case #3 We want 1 > > 3 Assign 0 and MPa No failure theory was given, so use MDE

23 Element A Find the von Mises stress ( e ) MSM Theory: S y η ( ) e e e MPa [( ) + ( ) + ( ) ] 1 3 [( 99. 0) + ( ) + ( ) ] 1 3 S y for our material 331 MPa Calculate the factor of safety η S y e For yield

24 Element B Consider the types of loading we have Axial? Yes due to P Bending? Recall that bending produces and τ, depending on the element of interest Yes due to M ( at A, τ at B) Torsion? Yes due to T

25 Element B Calculate stresses due to each load Axial: P P 4P x A πd πd 4 Bending (B has Shear Stresses due to bending): No Normal Stresses at B: x 0 Shear Stresses at B: Use equation for round solid cross-section ( 4F ) τ VQ 4V xy Ib 3A πd 3 4 D ( T ) Tc 16T Torsion: τ xy 4 3 J πd πd F πd

26 Element B Look at a stress element Sum up stresses due to all the loads x τ xy x 5.5 MPa τ xy 19.1 MPa 4P πd 16F 16T πD πd Note small contribution of shear stress due to bending

27 Element B Draw Mohr s Circle with the stresses that we calculated x 5.5 MPa τ xy 19.1 MPa x at ( x, τ xy ) (5.5, 19.1) y at ( y, τ yx ) ( y, -τ xy ) (0, -19.1) Find C x + y, ,0 ( 1.8,0) Find radius R ( ) + τ ( ) x C x xz

28 Out of Plane Maximum Shear for Biaxial State of Stress Case 1 1, > τ max Case,3 < τ max Case 3 1 >0, 3 < 0 0 τ max 1 3

29 Element B Find principal stresses (D) 1 C + R 35.8 MPa C - R -10. MPa Think about 3-D Mohr s Circle! This is Case #3 We want 1 > > 3 Assign 0 and MPa No failure theory was given, so again use MDE

30 Element B Find the von Mises stress ( e ) e e e 1 [( ) ( ) ( ) ] MPa [( ) ( 0 10.) ( ) ] S y for our material 331 MPa Calculate the factor of safety η S y e For yield

31 Example, concluded We found the factors of safety relative to each element, A and B A 3.8 B 7.91 A is the limiting factor of safety η 3.3

32 Static Brittle Failure Three primary theories for static brittle failure Maximum Normal Stress (MNS) Coulomb-Mohr Theory Modified-Mohr Theory

33 Mohr s Circle for MNS

34 Static Brittle Failure Maximum Normal Stress (MNS) Oldest failure hypothesis, attributed to Rankine Failure occurs whenever one of the three principal stresses equals the yield strength Say 1 > > 3 (our convention) Failure occurs when either 1 S t or 3 -S c Note brittle materials have both a tensile and compressive strength η S t / 1 or η -S c / 3 Plot of B vs. A

35 Static Brittle Failure Coulomb-Mohr Theory (AKA Internal Friction) S ut S ut Stress Region Mohr s Circle Failure Factor of Safety S uc A,B >0 A S ut η Sut 1 A > 0, B < 0 A S ut B S uc 1 1 A η S ut B S uc S uc A,B 0 B S uc η S uc B Use table, or look at load line

36 The Load Line Coulomb-Mohr B r A Equation of line from origin to point ( A, B ) Then, η A S ( S rs ) uc uc S ut ut Note: Strength of a part can be considered the stress necessary to cause failure. To find a part s strength at onset of failure, use η 1.

37 Modified Mohr Failure Theory

38 Static Brittle Failure Modified Mohr Theory Stress Region Mohr s Circle Failure Factor of Safety A,B >0 A S ut η S ut A B S ut A,B 0 B S uc η S uc B A > 0 -S ut < B B < 0 S ut A S ut η S ut A A > 0 B < -S ut S ut See Equation A See Equation B Which to use? (C-M or Mod-M) In general, Mod-M is more accurate A: SucSut 1 S B: ( S S ) uc ut A ut B η S uc S ut ( Suc Sut ) A Sut B

39 The Load Line Modified Mohr B r A Equation of line from origin to point ( A, B ) Then, η A ( S ( 1+ r) S ) uc S uc S ut ut Note: Again, strength of a part can be considered the stress necessary to cause failure. To find a part s strength at onset of failure, use η 1. For both Coulomb-Mohr and Modified Mohr, you can use either the table equations or the load line equations.

40 Brittle Failure example Given: Shaft of ASTM G5 cast iron subject to loading shown From Table A-4 S ut 6 kpsi S uc 97 kpsi Find: For a factor of safety of η.8, what should the diameter of the shaft (d) be?

41 Brittle Failure example First, we need to find the forces acting on the shaft Torque on shaft from pulley at B T B (300-50)(4) 1000 in lb Torque on shaft from pulley at C T C (360-7)(3) 1000 in lb Shaft is in static equilibrium Note that shaft is free to move along the x-axis (bearings) Draw a FBD Reaction forces at points of attachment to show constrained motion

42 Equilibrium Use statics to solve for reactions forces R Ay lb R Az 106 lb R Dy 17 lb R Dz 81 lb OK, now we know all the forces. The problem gives us a factor of safety, but unlike our last example, we aren t told specific places (elements) at which to look for failure! We are going to have to calculate stresses What do we need? Axial forces, bending moments, and torques We need to find our moments HOW? Shear-Moment diagrams will give us the forces and moments along the shaft. Failure will likely occur where the max values are seen

43 Torsion and moment diagrams Let s look at torsion and how it varies across the shaft We calculated the torques at B and C to be 1000 in lb each Plot that along the shaft and we see that max torque occurs at B and C (and all points between)

44 Torsion and moment diagrams Now let s look at the moments We have a 3-D loading How are we going to do the V-M diagrams? Look at one plane at a time Moment in the x-y plane From geometry you can calculate the values of the moment at B and C

45 Torsion and moment diagrams Moment in the x-z plane Failure is going to occur at either B or C, since these are locations where maximum moments are seen But we have moments in both planes To find the max bending stresses, we must find the total maximum moment Just as we would vectorally add the two force components to find the force magnitude, we can vectorally add the two moment components to find the moment magnitude M M xy + M xz

46 Calculate the max moment We found the following: M B x-y 1780 in lb M B x-z 848 in lb M C x-y 76 in lb M C x-z 1690 in lb Calculating the magnitudes with M B in lb M C in lb M M xy + M xz Since the overall max moment is at B, we will expect failure there, and use M B in our stress calculations If we had been told the location of interest, we would essentially start here.

47 Calculate the stresses at B Bending stress ( and τ) We know from experience that is the predominant stress, so essentially we will look for failure at an element at the top of the shaft M 1971 Plug in known values max (0x10 3 )/d 3 Torsional stress T 1000 τ (5.1x10 3 )/d 3 Tc τ J d c πd J 3 4 My I d y 4 πd I 64

48 Mohr s Circle Let s look at our stress element Now construct Mohr s circle C at (10 x 10 3 )/d 3 R (11. x 10 3 )/d 3 1 (1. x 10 3 )/d 3 3 (-1. x 10 3 )/d 3 Use Coulomb-Mohr theory for brittle failure 1 S ut 3 S uc 1 η d 97d d 1.3" If making a design recommendation, you would recommend the next largest standard dimension (16 th s) d in

49 Stress concentration A stress concentration is any geometric discontinuity in an element that is subjected to stress Aside from reducing the crosssectional area, these stress concentrations do not significantly affect static ductile failure Stress concentrations DO, however, have a significant influence on brittle failure

50 Analytical approach to stress concentrations max k t nom τ max k ts τ nom k t, k ts are stress concentration (SC) factors nom, τ nom are nominal stresses Nominal those stresses that are calculated before taking the SC s into account SC factors are given in the text on page Equations for the nominal stresses (taking into account geometry change due to the SC s) are given in the same charts

51 Stresses at a Hole in an Infinite Plate

52 Hoop Stress at a Hole in an Infinite Plate

53 Radial Stress at a Hole in an Infinite Plate

54 Stress Concentration in Ductile Material

55 Brittle failure example Given: Find: ASTM 30 cast iron S ut 31 ksi S uc 109 ksi How much torque before failure with and without the stress concentration? Note asked to find failure (not given a safety factor) Use η 1 to find onset of failure

56 Brittle failure example Without the SC Tc τ J c 0.5 J πd 3 4 π 0.098in 3 τ (5.1)T A 5.1 T, B -5.1 T Use Coulomb-Mohr (easier) A B 1 S S ut 5.1T T 3 uc 5.1T ( ) T 4730 in lb

57 Brittle failure example With the SC Refer to figure A-15-15, pg Picture shows us the loading and geometry Equation is given to calculate the nominal stress considering the geometry with the SC Axis and data labels tell us the quantities we need to calculate (using the figure as a guide)

58 Brittle failure example With the SC d D r 1 (.05) 0.95 D/d 1/ r/d 0.05/ k t ~ 1.8 τ max k t τ nom τ Tc 16T J πd nom T

59 Brittle failure example τ max k t τ nom τ Tc 16T J πd nom τ max k t τ nom (1.8)(5.94T) T Construct stress element and Mohr s Circle as before Use Coulomb-Mohr theory A B S S ut 10.69T T 3 uc T ( ) 1 1 T 58 in lb About half the load that could be withstood in the absence of the SC! T

60 Brittle failure example What if we consider a solid shaft (no SC s) with a diameter of d (0.95)? ( ) d Tc T τ 5.9T 4 J πd 3 Again, use Coulomb-Mohr A S ut B S 5.9T 5.9T T 4090in lb Note, this is a greater amount of torque than a shaft with larger diameter but with a SC uc 1 3 1

61 Design to avoid stress concentrations Avoid sudden changes in cross-section Avoid sharp inside corners Force-flow analogy Imagine flow of incompressible fluid through part Sudden curvature in streamlines High stress concentration!

62 Design to avoid stress concentrations