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# To select a material using a computer database called 'Cambridge Engineering Selector' (C.E.S).

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Introduction

Object To select a material using a computer database called 'Cambridge Engineering Selector' (C.E.S) Apparatus - Computer - 'Cambridge Engineering Selector' database program Theory The Problem Below is a brief description of the theory regarding this lab: - Oars are light, stiff beams. They must also have reasonable fracture toughness (KIC) and acceptable price per unit mass (Cm). The performance index for a light, stiff beam is: M1 = E1/2/? Where E is the Young's modulus and ? is the density. To select the best materials, perform two selection stages: (i) In stage 1, select materials with M1 > 7 (GPa) /(Mg/m) (ii) In stage 2, select materials with KIC > 1 MPa.m and Cm < 100 GBP/kg. CES Selector Materials for Oars: The solution The performance index for a light, stiff beam (M1) is plotted in stage 1. 'Density' is plotted on the x-axis and 'Young's Modulus' on the y-axis. A selection line of gradient 2, through the point (1.0, 49) is plotted. The constraints on adequate fracture toughness and price are plotted in stage 2. 'Fracture Toughness' is plotted on the x axis and 'Density' on the y axis. A selection box whose upper left corner is at (1.0, 100) is defined. In stage 1, the line representing the performance index is moved 'up' until only a small subset of records remains in the selection. Magnified views of the two selection charts are shown in figures M5.3.1 and M5.3.2 (results intersection and hide failed records on), and the materials passing both stages are shown in figure M5.3.3. ...read more.

Middle

CALCULATION OF THE GRADIENT FOR BOTH GRAPHS The gradient of the lines in both graphs were calculated using the performance index for the bending of rods, the formula used was: - E/P = Young's Modulus / Density In order to get the above equation into the correct term for a gradient or a curve (y=m x + c) both sides of the equation had to be logged: LOG E - LOG ? = LOG C Transpose for LOG E LOG E = LOG ? + LOG C The equation for a straight line is y = mx + c From the above it is fair to mention that: - Y = LOG E X = LOG ? M = 1 The performance used in this lab was E 1/2 / P = C If you take log on both sides of the equation above: 1/2 LOG E - LOG ? = LOG C Transpose for 1/2 LOG E: LOG C + LOG ? = 1/2 LOG E Multiply both sides by 2 to get LOG E LOG E = 2 LOG ? + 2 LOG C From the above it can be assumed that: Y = LOG E M = 2 X = LOG ? C = 2 LOG C M (The gradient) = 2 The gradient in the first graph of Density Vs. Young's Modulus is 2. If another performance index is used: K IC / p = c The log of both sides of the equation gives: - 2/3 LOG K ic = LOG ? ...read more.

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