ME311 Machine Design

Transcription

1 ME311 Machine Design Lecture : Materials; Stress & Strain; Power Transmission W Dornfeld 13Sep018 airfield University School of Engineering Stress-Strain Curve for Ductile Material Ultimate Tensile racture Yield Elastic Limit Proportional Limit σ ε E σ E ε Slope Usually 0.% Offset. ε 0.00 l ε l 1

2 Effects of Hardness; Brittleness Hardening by Heat Treating or Cold Working increases the Yield Strength of materials. Strength, kpsi This slope does NOT change because the Modulus is the same. Terminology: Strength Material Property; Stress Applied Loading Manufacturing/Metalworking Metal fabrication, where sheets and bars are bent and formed, obviously depends on going beyond yield into the plastic forming range. It is common for highly-formed metals to require annealing to reduce their yield strengths for further forming.

3 Brittle versus Ductile Ductility is related to the amount of Plastic Deformation (Strain) at fracture. The strain at fracture % Elongation l % El fracture l 0 l 0 Ductile if %El > 5% Brittle if %El < 5% 100% Hamrock ig. 3.7 How to Crack Sheet Metal Bend it in a real tight radius. Why? θ Tension Neutral Compression R t or any wrap angle θ, the circumferential wrap is rθ. The Neutral length is (R+t/)θ, and the length in Tension is (R+t)θ. The strain is: l ( Tension Neutral) l Neutral ( R + t) θ ( R + t / ) θ t / ( R + t / ) θ ( R + t / ) t t (R + t) D + t 3

4 How to Crack Sheet Metal Example: Aluminum Dogbone Material is 6061-T6 Aluminum Inner Bend Radius 0.10 Thickness Calculate the strain. rom MatWeb.com: t ε R+ t Should it have cracked? What do you think the stress was? l t l R + t % See Example 3.1, Hamrock What s nu? Poisson s Ratio ν (nu) ε AXIAL ε LATERAL -ν ε AXIAL Question: Is there a resulting lateral stress? So, could you say there is strain without stress? Can you think of another way to get strain without stress? Poisson s ratio is around 0.3 for most metals. Lowest is 0. for Cast Iron; Highest is 0.44 for Lead. Shear modulus ( stiffness in twisting ) is Because most metals have a ν of about 0.3, this means that for most metals, G what percent of E? E G (1 + ν ) Hamrock Eqn

5 Hardness Tests Cone Pyramid Sphere UTS vs. Hardness Sut (ksi) Brinell Hardness 5

6 Ashby Charts Look them over. Understand what they are and how they represent differences among materials. My favorite is ig. 3.19, comparing Modulus of Elasticity and Density. Incr. Speed of Sound Beryllium Most Common Metals The natural frequency of a cantilever beam is nearly the same for steel or aluminum, but for Beryllium it is almost 3x higher! What is E ρ for Aluminum and for Steel? What are the units? Stresses in Straight Beams A uniform beam in pure bending M r Mc The maximum stress is σ max I M where I is the Area Moment of Inertia about the Centroid, and c is the distance from the Neutral Axis, or Centroid. We need to know these properties of the beam cross section to be able to calculate bending stress, one of the most common large stresses on structures. 6

7 Hamrock Chapter 4: Area Moment of Inertia Definitions: Centroid of an Area Area Moment of Inertia Parallel Axis Theorem x I I x A A I xda A y ' x x CENTROID + da Ad This is NOT same as J Polar Moment of Inertia, or as Mass Moment of Inertia, I m, (which has units Lb.In.Sec, and sets how fast a torque can rotationally accelerate a device.) Hamrock Section 4. Parallel Axis Theorem Y Y O O d Ix' Ix + Ad bh 3 /1 + bhd where: Ix' moment of inertia about an axis that IS NOT through the center Ix moment of inertia about an axis that is through the centroid bh 3 /1 d distance between the x axes A area of the cross section bh Note: the axes must be parallel b h X X 7

8 Moment of Inertia Procedure 1. ind neutral axis (centroid) of total area. Why? Because we will use I to calculate bending stress about the neutral axis. ( A 1 + A + A 3 ) r tot A 1 r 1 + A r + A 3 r 3 Solve for r tot. Cut-outs have negative areas. Pick a convenient axis to measure r s from. If you pick an outer edge, you will always know which direction r tot is!. Compute Moment of Inertia about the centroid of each sub-piece. r tot A 1 A A 3 r 1 r,3 or example: b h 3 bh I CG 1 4 πd I CG Use Parallel Axis Theorem to translate to about the neutral axis I X I CG + Ad where d is the distance from the sub-piece s centroid to the total area centroid, and A is the area of the sub-piece. 4. Add up the translated moments of inertia of all the pieces, subtracting moments if they are cut-outs. Hamrock Section 4..3 Elementary Load Building Blocks P σ A P τ A STRESS My σ I Tr τ J Hamrock Eqn. 4. Eqn Eqn. 4.3 rom DAWright at U of Western Australia 8

9 Stresses in Curved Beams Bending Stresses in a curved beam are not linearly distributed across the beam section, but have a hyperbolic distribution that is higher at the inside surface than for a straight beam. Stresses in Curved Beams 1) Draw a very good picture. Show the applied orce, Show ri, ro, Area ) Calculate the centroidal radius, R, based on the section type. (Hamrock 4.5.3) r Rectangular: i + r R o Circular: R r i + rc 3) Compute the neutral radius, r n, based on the section type. (Hamrock 4.5.3) Rectangular: ro ri rn ln( r r ) o i Circular: R + r n R r c b r o R r c R R r n r o r i e r i r i r n Rectangular ro ri rn ln( r r ) o Circular R + r n L i R rc Neutral radius, r n Centroidal radius, R 9

10 Stresses in Curved Beams () 4) Compute the eccentricity, e R - r n b e 5) Compute the moment about the centroidal radius, R. Here M x L, not x R, because the force is not through the center of curvature. r o R r c r n r i r i 6) Calculate the distances from the neutral axis to the inner and outer surfaces: c i r n r i and c o r o r n. R r i L 7) Calculate the stresses at the inside and outside surfaces: Mc i Mc o σ i and σ o, Aer Aer i where A the section area o R r o r n Neutral radius, r n Centroidal radius, R Stresses in Curved Beams (3) b e 8) Add or subtract any P/A stresses in this section (using superposition). r o R r c r n r i r i Rectangular: ( r r b Circular: σ π σ o i ) r c R L 9) As a check, compare the answer to MC / I for a straight beam with neutral axis on the centroid and see if it makes sense. R r i r n Neutral radius, r n Centroidal radius, R r o 10

11 Stresses in Curved Beams (4) rom DAWright at U of Western Australia The stress distribution: 1) is hyperbolic, rather than linear for a straight beam, ) passes through zero at the neutral axis rather than at the centroid, 3) has the highest stress at the inside radius. Stresses in Curved Beams Calculate and compare the stresses on a diameter bar with an applied bending moment of 1000 in-lb for these cases: 1) A straight bar ) A curved bar with R 3 in. M1000 in-lb M 1 1 R R o R i M M I π 4 4 r R + r n R r c Mci σ i Aer i Mco σ o Aer o 11

12 Transverse Shear Hamrock ig shows difference between behavior of boards not glued together (top) and boards glued together (bottom). Solid beams behave like boards glued together, and the interior layers see shear stress along them. The outside surfaces have nothing to constrain them, and therefore see no shear stress. Hamrock Section 4.6 Transverse Shear Even though we usually write transverse shear as τ P/A, that is just the average shear. In reality, it has a distribution that depends on the actual beam shape: As Hamrock says below Eqn. 4.66: In all cases, shear stress is zero on extreme fibers, is maximum on the neutral axis, and has a parabolic distribution through the thickness. rom 1

13 Notes on Power Power is the rate at which work gets done, so has units of (orce x Distance) / Time Work / Time. Example: 1 HP 550 Lb x t / Sec This can also be looked at as orce x (Distance / Time) orce x Velocity. If a motor is involved, it can also be viewed as Torque x Rotational Velocity (orce x Distance) x (Radians / Time) t Lb 1In 60Sec 1Re v 1HP 550 In Lb RPM Sec t Rad 6305 min π In the SI system, it is simply Power (Watts) Tω Torque (Nm) x Rotational Speed (Radians/Sec) or conversion, 1 HP Watts Note the orderly unit conversion. Hamrock Section 4.4. Power Example Horsepower: 1-1/ rame: 56 Shaft Diameter: 5/8" Volts: 115 volt ull load amps: 14. Phase: single Enclosure: Open dripproof No load speed: 3600 RPM Reversible: yes Service factor: 1.0 Weight: 9 lbs. A. If this motor didn t slow down when delivering its full rated power (it does), what torque would it be delivering? B. What would the maximum stress in the shaft be then? C. What is the efficiency (mechanical power/electrical power) of the motor? 13

14 Motor Calculations Torque Power / Speed Tr π τ, J r J 4 Electrical power (W) Volts x Amps ascinating Motor Notes or a NEMA D rame motor, the frame number is how many 1/16 th s of an inch from the shaft center to base (the D dimension). A size 56 frame is 56/ The standard controls mostly the motor mounting details, not the motor body dimensions 14

15 Simple Stress Distributions Axial load on a uniform bar σ P A P A Describe the maximum stress: 1. What position along the beam? Answer: Everywhere along its length.. What location in the cross section? Answer: It is uniform over the whole section. 3. What is the value? Answer: σ P/A Note that all material is equally stressed. What about E? Simple Stress Distributions Torsional load (torque) on a uniform round bar Tr τ max J T J π 4 r Describe the maximum stress: 1. What position along the beam? Answer: Everywhere along its length.. What location in the cross section? Answer: It is maximum at the outer surface. 3. What is the value? Answer: τ Tr/J Note that the material in the center of the bar (along its axis) isn t loaded very much. 15

16 Simple Stress Distributions Torsional load (torque) on a uniform tube Tr τ o max J T 4 J π ( r o r i 4 ) Describe the maximum stress: 1. What position along the beam? Answer: Everywhere along its length.. What location in the cross section? Answer: It is maximum at the outer surface. 3. What is the value? Answer: τ Tr/J Removing the material in the center of the bar (along its axis) saves weight and $ without too much loss. Simple Stress Distributions or a uniform beam in pure bending M r Describe the maximum stress: Mc σ max I 1. What position along the beam? Answer: Everywhere along its length.. What location in the cross section? Answer: It is maximum at the top & bottom surfaces. 3. What is the value? Answer: σ Mc/I M Note that the material in the center of the bar (along its axis) isn t loaded very much. 16

17 Simple Stress Distributions or an I beam in pure bending M Describe the maximum stress: 1. What position along the beam? Answer: Everywhere along its length.. What location in the cross section? Answer: It is maximum at the top & bottom surfaces. 3. What is the value? Answer: σ Mc/I M Mc σ max I Removing the material in the center of the bar (along its axis) saves weight and $ without too much loss. Non-Simple Stress Distributions or a cantilever beam with an end load Mc σ max I ML Describe the maximum bending stress: 1. What position along the beam? Answer: At the left end of the beam.. What location in the cross section? Answer: It is maximum at the top and bottom faces. 3. What is the value? Answer: σ Mc/I What about the transverse shear? 17

18 Non-Simple Stress Distributions It is best to map out the Shear, V, and Moment, M, distribution when they vary along the beam. ML The Transverse Shear, for this rectangular cross section beam, is parabolic, with a max of 1.5 x τ avg : This is why max bending stress is at the left end. Non-Simple Stress Distributions or a cantilever beam with an end load ML Describe the maximum transverse shear stress: 1. What position along the beam? Answer: Everywhere along its length.. What location in the cross section? Answer: It is maximum on the neutral axis (and zero at the top and bottom faces). 3. What is the value? Answer: τ 1.5 V/A How do I (or do I) combine the bending and the transverse shear? 18

19 Dealing with Several Stresses Considering: 1. Superposition. Mohr s circle propose how to handle the bending plus shear stresses. ML Summary 19

20 0 inch long beam with two 100lb loads Example Beam Loading Beam cross section is 1 inch square. Shear, V (Lb) dv w dx Moment, M (In.Lb.) dm V dx What stresses are where? Calculate Amongst Yourselves 1. Show the location (along the beam) of the max stresses.. Show the position (in the cross section) of the max stresses. 3. Calculate the max stress values. 4. Do they combine in any way? Describe. 0

21 Stresses in a Crank Arm Break it into pieces and look at the BD. 300 LB 3/4 in 1800 in.lb Moment 300 lb 100 in.lb Torque 6 in 4 in 300 lb This piece has a bending Moment and a twisting Torque. 6 in Rod 100 in.lb Torque 300 lb 100 in.lb Moment Note that the moment turns into a torque where the arm bends the corner. 300 lb 4 in Rod This piece has a bending Moment. Stresses in a Crank Arm We already know how to do a tip-loaded cantilever. Let s take a look at the piece that bends and twists. Let s slice off a thin section near the wall: 100 in lb Torque B 300 lb Shear A 1800 in.lb Moment 100 in lb Torque 1800 in.lb Moment 300 lb 100 in.lb Torque 6 in Rod 100 in.lb Torque 300 lb 300 lb Shear 1

22 Stresses in a Crank Arm 100 in lb Torque Now let s get the stresses. B 300 lb Shear A 1800 in.lb Moment 300 lb Shear C 100 in lb Torque B A y z x At "A": Bending M 1800 in.lb. Torsion T 100 in.lb. [Shear V 300 lb, but is at neutral axis, not on top surface.] Mc (1800)(0.375) 675 σ x 43,460 psi 4 I π (0.75) Tr (100)(0.375) 450 τ xz 14,487 psi 4 J π (0.75) V (4)(300) 100 τ BENDING 905 psi 3A 3π (0.75) τ TOTAL τ xz τ BENDING 13,58 psi Pure Shear At B": Torsion T 100 in.lb. Shear V 300 lb [Bending is at top and bottom only, not at neutral axis.] Note that at location C, the two shears would ADD to 15,39 psi. Stresses in a Crank Arm These stresses occur at the same point, so can use Mohr. or Xstress 43.5, Ystress 0.0, XYShear -14.5, Angle , Stress , Stress 0.00, Stress3-4.39, ShearMax z A y x Mc σ x 43,460 psi I Tr τ xz 14,487 psi J Shear Stress Y,T The Mohr circle gives: σ ksi σ ksi σ ksi τ Max ksi ace Stresses X,-T