Mechanics of Materials – MAE 243 (Section 002) Spring 2008 Dr. Konstantinos A. Sierros

General info M, W, F 8:00-8:50 A.M. at Room G-83 ESB Office: Room G-19 ESB Tel: ext.2310 Course notes: USER NAME: cairns PASSWORD: materials Facebook : Konstantinos Sierros (using courses: Mechanics of Materials) Office hours: M, W 9:00-10:30 A.M. or by appointment

Course textbook Mechanics of Materials, 6 th edition, James M. Gere, Thomson, Brooks/Cole, 2006

Why do we study Mechanics of Materials? Anyone concerned with the strength and physical performance of natural/man-made structures should study Mechanics of Materials

Why do we study Mechanics of Materials? SAFETY and COST !!

Structural integrity of materials is important…

1.1: Introduction to Mechanics of Materials Definition: Mechanics of materials is a branch of applied mechanics that deals with the behaviour of solid bodies subjected to various types of loading Compression Tension (stretched) Bending Torsion (twisted) Shearing

1.1: Introduction to Mechanics of Materials Fundamental concepts stress and strain deformation and displacement elasticity and inelasticity load-carrying capacity Design and analysis of mechanical and structural systems

1.1: Introduction to Mechanics of Materials Examination of stresses and strains inside real bodies of finite dimensions that deform under loads In order to determine stresses and strains we use: 1.Physical properties of materials 2.Theoretical laws and concepts

Problem solving Draw the free-body diagram Check your diagram Calculate the unknowns Check your working Compute the problem Check your working Write the solution Check your working

Free body diagrams I

Free body diagrams II

Statics example 200kN A steel beam with a tensile strength of 700 MPA is loaded as shown. Assuming that the beam is made from hollow square tubing with the dimensions shown will the loading in the x direction exceed the failure stress? 3 4 2m 0.02m 0.01m

200kN 3 4 2m 160kN 120kN 120N 160kN 240kN.m Step 1: Free body diagram

Step 2: Calculate moment of inertia 0.02m 0.01m I=1/12 x ( )- 1/12 x ( ) m 4 =1.25 x m 4 A= m 2 = m 2

Step 3: Shear and moment diagrams 200kN 3 4 2mV x 120 M x -240

Stress due to axial loading Stress due to bending ANS: Total stress greater than failure stress therefore failure will occur Step 4: Calculation of maximum tensile stress

Key to success Ask questions and seek help if you feel like it!!!