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

2. General info M, W, F 8:00-8:50 A.M. at Room G-83 ESB Office: Room G-19 ESB E-mail: kostas.sierros@mail.wvu.edu Tel: 304-293-3111 ext.2310 Course notes: http://www.mae.wvu.edu/~cairns/teaching.html 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

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

4. 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

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

6. Structural integrity of materials is important…

7. 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

8. 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

9. 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

10. 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

11. Free body diagrams I

12. Free body diagrams II

13. 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

14. 200kN 3 4 2m 160kN 120kN 120N 160kN 240kN.m Step 1: Free body diagram

15. Step 2: Calculate moment of inertia 0.02m 0.01m I=1/12 x (0.02 4 )- 1/12 x (0.01 4 ) m 4 =1.25 x 10 -8 m 4 A=0.02 2 -0.01 2 m 2 =0.0003 m 2

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

17. 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

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