MECN2014 Introduction to Mechanical Engineering Design Recap Applied Mechanics A Ms T Mangera 2021/07/28 MECN2014 – Mechanical Engineering Design I 1 Grabcad.com Objectives • Understand component failures • What is the root cause of failures? • Acknowledge various materials and their unique properties • Determine whether a component fails • Understand the different static failure theories applicable to ductile and brittle materials 2021/07/28 2 Component Failure • • • • Separated into two or more pieces (fracture) Permanent Distortion Reliability Downgrade Compromised function Focus of predictability of permanent distortion or separation 2021/07/28 3 Component Failure (a) Failure of a truck drive-shaft spline due to corrosion fatigue. (b) Direct end view of failure. (Larry D. Mitchell, 1983.) 2021/07/28 4 Component Failure Impact failure of a lawn-mower blade driver hub. The blade impacted a surveying pipe marker. 2021/07/28 5 Component Failure Chain test fixture that failed in one cycle. To alleviate complaints of excessive wear, the manufacturer decided to case-harden the material. (a) Two halves showing fracture; this is an excellent example of brittle fracture initiated by stress concentration. (b) Enlarged view of one portion to show cracks induced by stress concentration at the support-pin holes. 2021/07/28 6 Component Failure Failure of an overhead-pulley retaining bolt on a weightlifting machine. A manufacturing error caused a gap that forced the bolt to take the entire moment load. 2021/07/28 7 Analysing Beams – General Approach • • • • External Loads Engineering Structure • Contains beam elements Generates Internal Forces & Moments Stresses Strains Structural Performance 2021/07/28 Axial Torsion Bending Combined • • • Deformation/deflections Yield/fracture Success/failure 8 Analysing Beams – General Approach Generates Internal Forces & Moments Stresses Strains Structural Performance Increasing or decreasing internal loads will affect the stresses and strains induced When we analyse a beams performance, we must do so at the “most stressed” section. The stress and strain behaviour affects the beam’s performance HOW? Graphical Tool: Shear Force (SF) & Bending Moment (BM) Diagrams 2021/07/28 9 SF and BM Diagrams Three methods: 1. Method of sections 2021/07/28 10 SF and BM Diagrams Three methods: 2. Graphical Integration Method 2021/07/28 11 SF and BM Diagrams Three methods: 3. Numerical Integration Method 2021/07/28 12 Graphical Integration Method Shear Force Diagram: 1. 2. 3. 4. 5. 6. 7. Draw SF diagram using the FBD Upward point loads create an increase in SF Downward point loads create a decrease in SF Graphically integrate distributed loads to find the change in SF, positive distributed loads create a decrease in SF Indicate the polynomial order for each segment Integrate complicated loads in their simple forms to account for the lower order polynomial coefficients (we can only handle simple geometries) Ensure that the SF diagram closes 2021/07/28 13 Graphical Integration Method Bending Moment Diagram: 1. 2. 3. 4. 5. Graphically integrate the SF w.r.t. x Integrate complicated areas in their simple forms to account for the lower order polynomial coefficients (we can only handle simple geometries) Clockwise moments create an increase in BM Anti-clockwise moments create a decrease in BM Ensure that the BM diagrams closes 2021/07/28 14 Area Formulae 2021/07/28 15 Class Example 1 2021/07/28 16 3500 + 2000 -1000 + 4500 -1000 5500 – 1000 – 1500 -1000 -1500 17 5500 3500 3000 -1000 πππ 2 0 -1000 -1500 πππ 3 3000×2 2 1500 × -2000+3500(2) 1000 3 2 × 2 -3000 + 3000(1) -1000(2) -11000+6000+1000+1000 5000-23000 -18000+3500(2) 18 5500 3500 3000 -1000 πππ 2 0 -1000 -1500 πππ 3 3000×2 2 1500 × 5000 1000 3 2 × 2 0 -2000 -3000 -11000 19 -180000 5500 3500 3000 Which location would you analyse for failure? -1000 πππ 2 0 -1000 -1500 πππ 3 3000×2 2 1500 × 5000 1000 3 2 × 2 0 -2000 -3000 -11000 20 -180000 Analysing Beams – General Approach • • • • External Loads Engineering Structure • Contains beam elements Generates Internal Forces & Moments Stresses Strains Structural Performance 2021/07/28 Axial Torsion Bending Combined • • • Deformation/deflections Yield/fracture Success/failure 21 Axial Stress or Direct Stress, ππππ πΉπΉ ππππ = π΄π΄ Normal stress: • Force per unit area • Force is the normal force • Cross sectional area (cut surface area) • Uniformly distributed ππππ ππππ F N Sign convention: Tension (+) Compression (-) 2021/07/28 +ππππ −ππππ 22 Transverse Shear Stress, ππππ • Type of shear stress • Maximum stress occurs at the Neutral Axis (NA) • Zero stress at the furthest points away from the NA Transverse Cut: F F Neutral Axis V 2021/07/28 ππππ 23 Bending Stress, ππππ F Sign convention: Tension (+) Compression (-) Compression −ππππ Neutral Axis R 2021/07/28 M ππππ Tension 24 • • • • • • Bending Stress, ππππ Type of axial or normal stress Top surface is under compression Bottom surface is under tension Maximum stress is at furthest point away from the NA Zero stress along the NA Axis for I is in the same direction of the M −ππππ Neutral Axis R 2021/07/28 M ππππ ππππ = πΌπΌ πΌπΌπ π π π π π π π π ππππ,ππππππ ππππ 4 = 64 32ππ = ππππ 3 ππππ 25 • • • • Torsional Shear Stress, πππ‘π‘ Type of shear stress Maximum stress is at furthest point away from the NA Zero stress along the NA Axis for J is in the same direction of the T T NA πππ‘π‘ πππ‘π‘,ππππππ Sign convention: Clockwise (+) Anticlockwise (-) 2021/07/28 16ππ = ππππ 3 ππππ πππ‘π‘ = π½π½ π½π½π π π π π π π π π ππππ 4 = 32 πππ‘π‘ T πππ‘π‘ πππ‘π‘ 26 Direct Shear Stress, ππππππππ ππππππππ 2021/07/28 ππ = π΄π΄ 27 Stress Distribution Summary NB! Draw stress distributions using actual stress directions. 2021/07/28 28 Recap stresses Axial Normal Stress Related Force Bending Normal Force, N Bending Moment, M ππππ ππππ Notation Distribution Highest @ furthest point AWAY from NA Uniform Shear Stress Transverse Shear Torsional Shear Direct Shear Related Force Shear Force, V Torque, T Shear Force, V ππππ πππ‘π‘ ππππππππ Notation Distribution 2021/07/28 Highest @ NA Highest @ furthest point AWAY from NA (caused by ⊥ force) Uniform 29 Stress Elements We’ve described the stress distribution for the cross section or cutting plane, but we know that this is not always uniform (e.g. bending stress) 2021/07/28 30 Class Example 2 Fy Fx M d 2021/07/28 = 5000 N = 2500 N = 2000 N.m = 25 - 50 mm 31 Section A-A ππππ,ππππππ ππππ,ππππππ ππππ,ππππππ 2021/07/28 16ππ 16 1071 = = 3ππππ 2 3ππ 0.025 32ππ 32 375 = = ππππ 3 ππ 0.025 4πΉπΉ 4 2500 = = ππππ 2 ππ 0.025 2 3 2 = 2.9 MPa = 244.5 MPa = 5.1 MPa 32 Section A-A 2021/07/28 33 Section B-B ππππ,ππππππ ππππ,ππππππ ππππ,ππππππ 2021/07/28 16ππ 16 1071 = = = 0.7 MPa 3ππππ 2 3ππ 0.05 2 32ππ 32 750 = = = 61.1 MPa ππππ 3 ππ 0.05 3 4πΉπΉ 4 2500 = = = 1.3 MPa ππππ 2 ππ 0.05 2 34 Section B-B 2021/07/28 35 Section C-C ππππ,ππππππ ππππ,ππππππ ππππ,ππππππ 2021/07/28 16ππ 16 3929 = = = 2.7 MPa 3ππππ 2 3ππ 0.05 2 32ππ 32 2750 = = = 224.1 MPa ππππ 3 ππ 0.05 3 4πΉπΉ 4 0 = = ππππ 2 ππ 0.05 2 = 0 MPa 36 Section C-C 2021/07/28 37 Stresses on inclined planes Why do we need to know the stresses on an inclined plane? Non - Transverse Cut: N F F V ππ The maximum normal or shear stress may not be on the transverse plane – it may be on an inclined plane We want to avoid failure so we must find the maximum stress! 2021/07/28 38 Principle Stresses and Principle Planes Principle planes: Plane at which the maximum and minimum NORMAL stresses occur. The shear stress is zero on these planes Principle stresses: Stresses that occur at the principle plane We want to locate the planes that the maximum values for πππ₯π₯ ′ and πππ₯π₯ ′π¦π¦′ occur at, and we define these planes by their angle of inclination, ππ. This tells us where failure is likely to occur. 2021/07/28 39 Recap ππ1,2 πππ₯π₯ + πππ¦π¦ = ± 2 ππππππππ = πππ₯π₯ − πππ¦π¦ 2 πππ₯π₯ − πππ¦π¦ 2 2 + πππ₯π₯π₯π₯ 2 2 + ππ π₯π₯π₯π₯ 2 2πππ₯π₯π₯π₯ π‘π‘π‘π‘π‘π‘π‘ππππ = πππ₯π₯ − πππ¦π¦ πππ₯π₯ − πππ¦π¦ π‘π‘π‘π‘π‘π‘π‘πππ π = − 2πππ₯π₯π₯π₯ We also found that the plane of the maximum shear stress is 45° from the principle planes ππ1 − ππ2 ππππππππ = ππ1 + ππ2 = πππ₯π₯ + πππ¦π¦ 2 2021/07/28 40 Mohr’s Circle For Mohr’s circle: Radius: Centre: πππ₯π₯ − πππ¦π¦ 2 2 + πππ₯π₯π₯π₯ 2 = ππππππππ πππ₯π₯ + πππ¦π¦ , 0 = ππππππππ , 0 2 Note: The stress transformation equations are based on an angle of 2ππ but the rotated stress element is based on ππ 2021/07/28 41 Principle Stresses/Planes for 3D stress Recall: We can use a 3D stress element to describe the state of stress at a point. For the 3D case we have a third principle stress, ππ3 , which is an out of plane principle stress, which is equal to zero. 2021/07/28 42 Principle Stresses/Planes for 3D stress The Mohr’s circle we have looked at represents all possible stress states for rotations about the z axis. We can look at rotation about the x and y axes and represent the stress states using Mohr’s circle in the yz and xz planes. 2021/07/28 43 Principle Stresses/Planes for 3D stress We plot all three rotations on the same axis to create the “3D Mohr’s circle”. ππ xz plane ππ1 ≥ ππ2 ≥ ππ3 yz plane ππππππππ ππ1 − ππ3 = 2 xy plane ππ3 will be the smallest stress so be careful: −ππ If the signs for the plane stress circle are different 2021/07/28 44 Principle Stresses/Planes for 3D stress ππ3 will be the smallest stress so be careful: If the signs of the principle stresses for the plane stress circle are different, ππ3 will not be zero, rather ππ2 will be zero and ππππππππ stays the same ππππππππ −ππ ππ1 ≥ ππ2 ≥ ππ3 yz plane σ2 σ3 2021/07/28 ππ σ1 xz plane ππππππππ ππ1 − ππ3 = 2 xy plane 45 Class Example 3 Draw the Mohr Circle, determine the principal normal and shear stresses, and draw the principal normal and shear stress elements for each stress element. 2021/07/28 46 ππ1,2 −100 + ππ1 = 2 −100 − ππ3 = 2 −100 2 −100 2 2 2 2πππ₯π₯π₯π₯ π‘π‘π‘π‘π‘π‘π‘ππππ = πππ₯π₯ − πππ¦π¦ πππ₯π₯ − πππ¦π¦ 2 2 + ππ π₯π₯π₯π₯ 2 + (−50)2 = 20.71 ππππππ + −50 2(−50) ; 2ππππ = +45° π‘π‘π‘π‘π‘π‘π‘ππππ = −100 2021/07/28 πππ₯π₯ + πππ¦π¦ ± = 2 2 = −120.71 ππππππ < 0 47 ππ1,2 = ± ππ1 = ππ2 = − −100 2 2 πππ₯π₯ − πππ¦π¦ 2 2 πππ₯π₯ − πππ¦π¦ π‘π‘π‘π‘π‘π‘π‘πππ π = − 2πππ₯π₯π₯π₯ + (−50)2 = 70.71 ππππππ −100 2 2 + −50 2 ππππππππ + πππ₯π₯π₯π₯ 2 πππ₯π₯ + πππ¦π¦ = 2 = −70.71 ππππππ −100 = −45° π‘π‘π‘π‘π‘π‘π‘πππ π = − 2 −50 −100 = −50 ππππππ ππππππππ = 2 2021/07/28 48 2021/07/28 49 2021/07/28 50 2021/07/28 51 2021/07/28 52
