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31-Oct-11 FAILURE THOERIES FOR STATIC LOADING What is Failure? • Failure – any change in a machine part which makes it unable to perform its intended function. • We will normally use a yield failure criteria for ductile materials. The ductile failure theories presented are based on yield. 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 14 1 31-Oct-11 Failure Theories • Static failure – Ductile – Brittle – Stress concentration • Recall – Ductile • Significant plastic deformation between yield and fracture – Brittle • Yield ~= fracture 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 15 Tensile Test • 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 16 2 31-Oct-11 Mohr’s Circle for Tensile Test • 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 17 Static Ductile Failure • Two primary theories for static ductile failure – Von Mises criterion • Maximum Distortion-energy Theory • MDE – Maximum Shear Stress criterion • MSS (TRESCA) 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 18 3 31-Oct-11 Static Ductile Failure • Max Shear Stress criterion – Material yields (fails) when: Sy 1) max 2 or 2) 1 3 S y Sy Sy – Factor of Safety: F .S 1 3 2 max 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 19 Maximum Shear Stress Criteria • 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 20 4 31-Oct-11 Static Ductile Failure • Von Mises criterion – Let the Mises stress (e, equivalent stress) be: e 1 2 1 2 2 2 3 2 1 3 2 – Then failure (yield) occurs when: e Sy Sy – Factor of Safety: F .S e • Typically, 1.25 F .S 4 • Want a margin of error but not completely overdesigned 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 21 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! 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 22 5 31-Oct-11 Comparison of MDE, MSS,MNS • 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 23 Ductile failure theory example • Given: – Bar is AISI 1020 hot-rolled steel • A DUCTILE material – F = 0.55 kN – P = 8.0 kN – T = 30 Nm • Find: – Factor of safety () • Two areas of interest: – A • Top – where max normal stress is seen (bending!) – B • Side – where max shear stress is seen 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 24 6 31-Oct-11 Element A • Consider the types of loading we have • Axial? – Yes – due to P • Bending? – Recall that bending produces and , depending on the element of interest – Yes – due to M ( at A, at B) – • Torsion? – Yes – due to T 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 25 Element A • Calculate stresses due to each load • Axial: P P 4P x A D D 2 2 4 • Bending: FL D x My 2 32 FL I D 4 D 3 64 • Shear: xy 0 • Torsion: T D xz Tc 2 16T J D 4 D 3 32 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 26 7 31-Oct-11 Element A • Look at a stress element • Sum up stresses due to all the loads • 4 P 32 FL 4 PD 32 FL x D 2 D3 D3 • 16T xz D 3 • x = 95.5 MPa • xz = 19.1 MPa 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 27 Element A • Draw Mohr’s Circle with the stresses that we calculated – x = 95.5 MPa – xz = 19.1 MPa – x at (x, xz) • (95.5, 19.1) – y at (y, zx) • (y, -xz) • (0, -19.1) – Find C • x y 95.5 0 2 ,0 2 ,0 47.8,0 – Find radius • R x Cx 2 xz 2 95.5 47.82 19.12 51.4 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 28 8 31-Oct-11 Out of Plane Maximum Shear for Biaxial State of Stress • Case 1 Case 2 Case 3 – 1,2 > 0 2,3 < 0 1 > 0, 3 < 0 1 = 0 2 = 0 – 3 = 0 3 – max 1 max 3 max 1 2 2 2 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 29 Element A • Find principal stresses – 1 = C + R • 99.2 MPa – 2 = C - R • -3.63 MPa – Think about 3-D Mohr’s Circle! – This is Case #3… – We want 1 > 2 > 3 – Assign 2 = 0 and 3 = -3.63 MPa • No failure theory was given, so use MDE 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 30 9 31-Oct-11 Element A • Find the von Mises stress (e) e 1 2 1 2 2 2 3 2 1 3 2 e 1 2 99.2 02 0 3.632 99.2 3.632 e 101MPa • Sy for our material = 331 MPa • Calculate the factor of safety – Sy 331 3.28 For yield e 101 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 31 Element B • Consider the types of loading we have • Axial? – Yes – due to P • Bending? – Recall that bending produces and , depending on the element of interest – Yes – due to M ( at A, at B) • Torsion? – Yes – due to T 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 32 10 31-Oct-11 Element B • Calculate stresses due to each load • Axial: P P 4P x A D 2 D 2 4 • Bending: – Use equation for round solid cross-section xy VQ 4V 4 F 16F Ib 3 A D 2 3D 2 4 3 • Shear: xy 0 T D • Torsion: xy Tc 2 16T J D 4 D 3 32 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 33 Element B • Look at a stress element • Sum up stresses due to all the loads • 4P x • D 2 16F 16T xy 19.1 .002 3D 2 D 3 • x = 25.5 MPa • xy = 19.1 MPa – Note small contribution of shear stress due to bending 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 34 11 31-Oct-11 Element B • Draw Mohr’s Circle with the stresses that we calculated – x = 25.5 MPa – xy = 19.1 MPa – x at (x, xy) • (25.5, 19.1) – y at (y, yx) • (y, -xy) • (0, -19.1) – Find C • x y 25.5 0 2 ,0 2 ,0 12.8,0 – Find radius • R x Cx 2 xz 2 25.5 12.82 19.12 22.96 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 35 Out of Plane Maximum Shear for Biaxial State of Stress • Case 1 Case 2 Case 3 – 1,2 > 0 2,3 < 0 1 > 0, 3 < 0 1 = 0 2 = 0 – 3 = 0 3 – max 1 max 3 max 1 2 2 2 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 36 12 31-Oct-11 Element B • Find principal stresses – 1 = C + R • 35.8 MPa – 2 = C - R • -10.2 MPa – Think about 3-D Mohr’s Circle! – This is Case #3… – We want 1 > 2 > 3 – Assign 2 = 0 and 3 = -10.2 MPa • No failure theory was given, so again use MDE 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 37 Element B • Find the von Mises stress (e) e 1 2 1 2 2 2 3 2 1 3 2 e 1 2 35.8 02 0 10.22 35.8 10.22 e 41.8MPa • Sy for our material = 331 MPa • Calculate the factor of safety – Sy 331 7.91 For yield e 41.8 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 38 13 31-Oct-11 Example, concluded • We found the factors of safety relative to each element, A and B – A – 3.28 – B – 7.91 • A is the limiting factor of safety – F.S = 3.3 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 39 Quiz • Given: – Shaft of st50 subject to loading shown • Find: – For a factor of safety of F.S = 2.8, what should the diameter of the shaft (d) be? 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 40 14 31-Oct-11 Solution • First, we need to find the forces acting on the shaft – Torque on shaft from pulley at B • TB = (300-50)(4) = 1000 in·lb – Torque on shaft from pulley at C • TC = (360-27)(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 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 41 Equilibrium • Use statics to solve for reactions forces – RAy = 222 lb – RAz = 106 lb – RDy = 127 lb – RDz = 281 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 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 42 15 31-Oct-11 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) 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 43 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 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 44 16 31-Oct-11 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 2 2 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 45 Calculate the max moment • We found the following: – MB x-y = 1780 in·lb – MB x-z = 848 in·lb – MC x-y = 762 in·lb – MC x-z = 1690 in·lb • M M xy M xz 2 2 Calculating the magnitudes with – MB = 1971.7 in·lb – MC = 1853.8 in·lb • 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. 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 46 17 31-Oct-11 Calculate the stresses at B • Bending stress ( and ) – 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 My • M = 1971 I • Plug in known values y d • max = (20x103)/d3 2 d 4 • Torsional stress Tc I 64 – T = 1000 J d c – = (5.1x103)/d3 2 d 4 J 32 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 47 Mohr’s Circle • Let’s look at our stress element • Now construct Mohr’s circle – C at (10 x 103)/d3 – R = (11.2 x 103)/d3 – 1 = (21.2 x 103)/d3 – 3 = (-1.2 x 103)/d3 • Use TRESCA to find the diameter. 31-Oct-11 Machine Design, Eng. Mahmoud Refaat 48 18

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