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
• 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
Course textbook
Mechanics of Materials, 6th 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
4
3
2m
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?
0.01m
0.02m
Step 1: Free body diagram
4
3
2m
240kN.m
200kN
120N
160kN
160kN
120kN
Step 2: Calculate moment of inertia
I=1/12 x (0.024)- 1/12 x (0.014) m4
=1.25 x 10-8 m4
A=0.022-0.012 m2
=0.0003 m2
0.01m
0.02m
Step 3: Shear and moment diagrams
4
3
V
2m
120
200kN
x
M
x
-240
Step 4: Calculation of maximum tensile stress
• Stress due to axial loading
 axial
F
160
 
kPa  533 .33MPa
A 0.0003
• Stress due to bending
 bend
Mc 240  0.01


kPa  1920 MPa
8
I
1.25 10
ANS: Total stress greater than failure stress
therefore failure will occur
Key to success
Ask questions and seek help if you feel like it!!!