Analysis: The Current Production In Manufacturing Industry

CHAPTER 1

INTRODUCTION

1.1Preamble:

The current production in manufacturing industries are distinguished based on the product range extension, high frequency in changing production programs, demand for constant improvement in the quality of products, production time shortening, need for constant increasing technological level of products and decreasing their manufacturing costs. With such a market demand and intensive development ofinformation, scienceand techniques, the level and trend in manufacturing industries are composed of factors such as type of blank, machining process, order of operations, machinery, operational and sequential procedures, tools, fixtures, measurements etc., by solving all these elements optimally , technological solutions

Few other journal papers in support of the present work includes,

J.Cecil‘s journal on “Review And Future Trends about computer aided fixture design” describes about some of the approaches made in solving the drawbacks of existing approaches in designing fixtures.

A.Y.C.Nee Et Al has written about “Advanced Fixture Design for flexible manufacturing system”, this journal is about solving the fixture design problems using two approachi.e. Variant fixture design and Generative fixture design schemes.

Fazlina Binti Mansor’s, “Designing and Evaluating of Jig for Holding Cylindrical Parts for Mass Production for Drilling Operation” is a thesis about designing a drilling fixture for cylindrical parts. The design has been made with design software tools along with analysis using ANSYS software.

Bi, Z.M, Zhang, and W.J: “Flexible Fixture Design and Automation”: Review, Issues, and Future Directions.

Nikhil. G. Lokhande, C.K. Tembhurkar’s “Advanced Fixture for Angular Drilling on Cylindrical Objects”: These are the journals referred to follow the steps of fixture

(3.2)

Integrating equation 3.2, we get:

EI dy/dx=C_1+43.93x^2+59.15(x-109)^2…………………… (3.3)

Integrating equation 3.3, we get:

EIy=C_2+C_1 x+14.64x^2+19.716(x-109)^3…………… (3.4)

Where C1 and C2 are the constants which are solved using boundary conditions. at x=0; y=0 ,C_2=0 at x=109; y=0 ,C_1=-173977.4

Substituting the above boundary conditions in equation 3.4, we get:

EIy_c=5422778613.9

Since for the material selected young’s modulus is 210 PA, and the section modulus is given by the equation,

E=210×〖10〗^3 N/〖mm〗^2

I=π/64 d^4,d=22mm

With all the data available, we get maximum deflection as;

EI=2414792940 N/m^2 y_c=0.2245 mm

Solving equation 3.1 with all the known parameters we get the maximum bending moment as;

〖 M〗_max=74409.825 Nmm,74.409 KN mm

The equation for torque to be produced in a shaft, polar modulus and maximum shear stress is given as;

Torque (T)=J/R qs ∶ qs=F/A ∶ J=π/32 D^3 …………