ME 330 Engineering Materials Lecture 3 Tension/Bending/Torsion/Material Selection Bending Torsion Material Selection Techniques Please read Chapter 6
Next Lecture... Chemistry Review Crystals and crystallographic planes Metallic structures (BCC,FCC,HCP) Single vs. Polycrystalline Crystal Structure Theoretical Strength Point Defects Linear Defects Planar Defects Volume Defects Microscopy Please read Chapters 2 & 3Please read Chapters 2 & 3
Basic Mechanical Tests Tension Most common mechanical test Gage section reduced to ensure deflection here Load cell measures applied load Extensometer ensures l measured from gage region Compression Similar to tensile test Good for brittle specimens … hard to grip Often much different properties in compression Torsion Test of pure shear Member twisted by angle , calculate shear strain Measure applied torque, calculate shear stress Bending In all cases, a displacement is applied and you measure load Calculate stress from measured load Calculate strain from change in gage length
Bending From: Galileo, "Discorsi E Dimostrazioni Matematiche," English translation by H. Crew & A. de Salvio, Macmillan., Cantilever Bending Four Point Bending Can calculate both shear and normal stresses if the beam remains elastic. Shear accommodates changes in moment.
Bending A linear distribution of strain is assumed and related to the beam curvature, . Also assume plane sections remain plane. From: S.H. Crandall, N.C. Dahl & T.J. Lardner,, An Introduction to the Mechanics of Solids (2nd Ed.), McGraw-Hill, 1972, p In general, the curvature of a beam subject to bending loads is,
Bending At L/2, the angle = 0 due to symmetry. With the crude assumption that there is no curvature in the end sections, For small deformations, one could drop the trigonometry terms, but remember angles must have units of radians. For Four point bending, the curvature is constant (so is the moment) between the inner supports.
Bending If a homogeneous isotropic linear elastic material is assumed, Where the maximum stress is on the “outer fiber”, or at y = h. Where P is the load.
Bending Clearly for a 1045 steel, the response is not elastic, as evidenced by the permanent curvature of the beam upon unloading.
Torsion For linear elastic analysis the following apply, Again a linear distribution of strain, but this time with a shear strain.
Torsion For Mohr space, y = y /2 and for effective or von Mises y = y /. Plastic deformation may occur making calculation of stresses more difficult. But yield quantities may be approximated using elastic formulae.
Torsion Use the initial portion of the curve to calculate G. Units as shown are MPa.
Material Selection
Stiffness - Density
Strength - Density
Stiffness - Strength
Case Study: Shaft in Torsion Very common situation Rotating machinery –Automotive Engine Power transmission –Agriculture –Electrical motor –Machine tool –Etc. Choose material –Adequate strength –Minimize weight
Case Study: Shaft in Torsion Important equations: –Shear stress –Polar moment of inertia –Mass Objective: Lightest shaft with adequate strength
Selection Methodology
Performance Metric? Adequate strength when shear stress is lower than shear stress limit divided by safety factor Minimize mass which is related to dimensions and density Substitute r into shear stress equation Solve for mass, m, which is our performance metric
Performance Index? Minimize mass by maximizing material index Where material index
Materials selection plots (Ashby Plots) –M.F. Ashby, Materials Selection in Mechanical Design, Butterworth- Heinemann, UK, For our material index, look at density versus strength Assume as minimum –M i = 10 MPa 2/3 m 3 /Mg Strength - Density M i =10
Strength - Density M i =10 Design considerations: –Why aren’t crankshafts made of wood? Plastic? –Why not ceramic? –Why not carbon fiber composites? To complete picture: –Fatigue –Creep –Thermal issues –Cost Material Manufacturing Maintenance –ETC!!
Strength of Carbon Graphite 10 – 15 MPa Diamond10 GPa Nanotube> 100 GPa Why ?