High Tech High Top Hat Technicians. An Introduction to Solid Mechanics. Is that supposed to bend there?

Transcription

1 High Tech High Top Hat Technicians An Introduction to Solid Mechanics Or Is that supposed to bend there?

2 Why don't we fall through the floor? The power of any Spring is in the same proportion with the Tension thereof: That is, if one power stretch or bend it one space, two will bend it two, three will bend it three, and so forward. And this is the Rule or Law of Nature, upon which all manner of Restituent or Springing motion doth proceed. - Robert Hooke Around 1676, Robert Hooke concluded that not only must solids resist weights or other loads by pushing back on them, but also that 1. Every solid changes shape by stretching or contracting when subjected to load 2. This change in shape allows the solid to react the load

3 What is Solid Mechanics? Solid Mechanics is the study of the behavior of solid materials subjected to applied forces By understanding the effect of loads on materials, we can build fantastic structures Salisbury Cathedral Or keep machines from breaking or bending in places we do not wish Understanding the effect of loads on materials is fundamental to structural design Burj Kahlifa m Shanghai Lupu Bridge

4 What kind of load: Tension Tension Tensile loads act to pull the material apart Force Force Tension is a normal load

5 What kind of load: Compression Compression Compressive loads act to push the material together Force Force Compression is a normal load

6 What kind of load: Shear Shear Shear loads act in the plane of the material to skew the material Force Force Shear is an in plane load Single shear Double shear

7 Beams in Bending Beams are structural members that offer resistance to bending caused by applied loads F F F Simple Continuous F F Cantilever End supported Cantilever Statically determinate Statically indeterminate Beam cross sections M Upper edge is in tension Lower edge is in compression Material between is in shear M H beams put the material where the forces are in bending plus more

8 Torsion in Constant Section Bars When subjected to torque, a constant section bar twists, with each section rotating about the longitudinal axis Each section of the bar twists relative to the adjoining sections resulting in shear force in the plane of the section as well as equal longitudinal shear force in a radial plane and tensile and compressive stresses at 45º Twisting of the bar causes the bar to lengthen a small amount which, in turn, causes a small longitudinal load. This load is negligible compared to the shear forces. The shafts in gearboxes and motors operate in torsion

9 Stress Stress in a solid is like pressure in a liquid or a gas. Stress is a measure of how hard the atoms and molecules that make up the material are being pushed together or pulled apart by external forces. We express stress as the load applied at a given point divided by the area of application Stress = s = load P = area A Force P Area A Stress in a bar in tension (compressive stress is exactly analogous) Force P As a consequence, stress and pressure have similar units SI: meganewtons per square meter (MN/m2) Imperial: pounds per square inch (psi) or, frequently, thousands of pounds per square inch (ksi) Stress and strain are born from the assumption of elastic conditions in a continuous material at any specified point inside that material

10 Strain Just as stress tells us how hard a material is being loaded, strain tells us how much the material is deformed by the load Force P Origina l le change of length l Strain = e = original length = L ngth, L Force P l Increase of length due to load P Strain in a bar in tension (compressive strain is exactly analogous) If a bar of original length, L, deforms due to the application of external load by the amount, l, then the strain in the bar, e, is the change in length divided by the original length Strains in engineering materials are typically very small and, therefore, frequently expressed as percentages

11 Stiffness: Young's Modulus When we plot stress versus strain for many materials, we see that, for moderate stresses, the plot is a straight line We say the material obeys Hooke's law in this region or that the deformation of the material under load is linearly proportional to the load The slope of the stress strain plot for a material in this range is Young's modulus of elasticity for this material Stress Rubber Unreinforced plastics Fresh bone Aluminum alloys Iron and Steel Diamond Strain Approximate Young's Moduli of Various Materials Material Rubber Unreinforced plastics In the elastic range, the deformed material returns to its original shape upon removal of the load 7 1,400 21,000 Aluminum alloys 70,000 Diamond Young's modulus is also called the elastic modulus, or E, and is commonly referred to as the stiffness of the material Young's Modulus (MN/m2) Fresh bone Iron and steel 210,000 1,200,000

12 Strength The ultimate strength of a material is the maximum stress the material withstands prior to breaking The limit or yield strength of a material is the stress at which the material ceases to behave elastically and begins to deform plastically Brittle material. Strong. Sudden, dramatic failure. Strong material. Non-ductile. C D Ductile material B A Stress A. Proportional limit B. Elastic limit C. Ultimate stress D. Fracture point Plastic material Strain Approximate Ultimate Tensile Strength of Various Materials Material Tensile Strength (MPa) Density (g/cm3) Muscle Concrete Aluminum*: cast Tendon Fresh bone Spider's silk Cotton & Silk , 1.3 Steel*: mild Bronze* 100 to Aluminum*: wrought 140 to Carbon fiber 350 to 1, Titanium* 700 to 1, Steel*: high tensile 1, Steel* : piano wire 3, * Metal alloys Very brittle

13 Strength of Our Common Materials Material Tension Yield (MPa) Ultimate (MPa) Aluminum, 6061-T6 AMS-QQ-A-200/8 Extruded shape Aluminum, 6063-T Steel, ASTM-A , cold finished Acrylic 55 Compression Shear Yield (MPa) Yield* (MPa) 324 Bearing Yield e/d = 2 Ultimate e/d = 2 Young's Modulus (GPa) Density (g/cm3) * Shear yield stress calculated as tensile yield stress multiplied by the theoretical ratio between shear and tensile stress for homogeneous, isotropic materials,

14 Stress in a Plate in Tension Determine the stress in a flat plate in tension A =W t A = A =252.0 mm P σ= A σ= ( ) 1,000,000 σ=2.118 MPa Symbol Name Value P Load N W Width 79.4 mm L Length 150 mm t thickness A Area σ Stress σmax Yield strength of aluminum 241 MPa G Shear modulus of aluminum 78.0 GPa mm

15 Stress in a Plate in Tension with a Hole Estimate the stress in a flat plate in tension with a hole P σ nom= t (W D) σ nom= ( ) σ nom=3.308 MPa Calculate the stress concentration factor: 2 3 D D D k = ( )+3.66 ( ) 1.53 ( ) W W W k =2.276 σ=k σ nom σ=7.530 MPa Symbol Name Value P Load N W Width 79.4 mm L Length 150 mm t Thickness D Hole diameter mm r Hole radius mm A Area σ Stress σmax Yield strength of aluminum 241 MPa G Shear modulus of aluminum 78.0 GPa mm

16 Hex Shaft in Torsion Calculate the maximum torque capability and associated angular deflection of a six (6) inch long one half inch steel hex shaft τmax = Symbol 1.09 T 3 a T L Θ= 4 a G 213,000,000= θ= 1.09 T ( )3 a ; ( ) 78,000,000,000 Name Value τmax Shear yield strength of steel 213 MPa T Torque solve Θ Total angular deflection solve a Length of one side m L Length of the shaft m G Shear modulus of steel 78.0 GPa T =77.05 Nm ; Θ= radians or degrees Since a CIM motor can produce a maximum of Nm, this steel hex shaft allows only an approximately 31 to 1 gear ratio with zero safety margin

17 Stress in a Plate with Load at the Hole Estimate the stress in a flat plate with a load applied at a hole Assume that the areas on either side of the hole at the hole diameter are the only material reacting load. Symbol Name Value P Load W Width 79.4 mm L Length 150 mm t Thickness D Hole diameter c Distance from load Ix Area moment of inertia M Moment σ Stress σmax Yield strength of aluminum 241 MPa G Shear modulus of aluminum 78.0 GPa We then have a beam, constrained on both ends with a point load applied N mm mm 39.7 mm

18 Stress in a Plate with Load at the Hole First, we need to determine the moment acting in the beam due to the point load. Maximum positive moment in a beam that is fixed at both ends with the point load acting at L/2: M= P L 8 M =0.723 N m Maximum stress in a beam in bending: σ= M C Ix σ=227.5 MPa Symbol Name Value P Load 91.0 N W Width 79.4 mm L Length 150 mm t Thickness D Hole diameter c Distance from load Ix Area moment of inertia M Moment σ Stress σmax Yield strength of aluminum 241 MPa G Shear modulus of aluminum 78.0 GPa mm mm 39.7 mm

19 Questions?