Chapter 1
Introduction
Concept of Stress
Road Map:
Statics Mechanics of Materials 
Elasticity  Plasticity
Fracture Mechanics
Fatigue
Creep
Mechanics of Materials is important foundation for:
1. Machine Design I & II
2. Advanced Mechanics Courses
3. Elasticity & Plasticity
4. Finite Element Methods
Statics: main concern: Equilibrium
Fx  0
Fy  0
M o  0
Mechanics of Materials:
1. Equilibrium
Stress
Strain
2. Deflection (Deformation)
3. Yielding or Failure
Forces
“Mechanics of Materials” is a branch of
Mechanics that develops relationships
between :
The external
Intensity of internal forces
loads
(stress / strain / deformation)
1.3 Stresses in the Members of a Structure
P

A
(1.5)
Assumptions:
1. Uniform distribution of stress
2. Uniform material properties
Units:
SI Units:
English Units:
(1.5)
psi = lb/in2
ksi = 103 psi
1.5 Axial Loading: Normal Stress
The normal stress:
P
  average stress 
A
The more general definition
of normal stress is:
F
  lim
A0 A
(1.6)
(1.5)
If the stress distribution is not uniform:
dF   dA
External force
internal force
1. In engineering practice we assume the stress is uniform
This is only true:
If the line of action of the concentrated loads P and P’
passes through the centroid of the section considered.
2. The distribution of the internal stress cannot be
uniform if the load is eccentric.
1.6 Shearing Stress
 ave
P

A
(1.7)
 ave
1.7
P F /2 F
 

A
A
2A
Bearing Stress in Connections
P P
b  
A td
(1.10)
1.9 Method of Problem Solution
To solve a problem, use the following procedures:
1. Draw FBDs
2. Apply Equations of Equilibrium
3. Determine , , and deformation (deflections)
4. Check your answers.
1.10 Numerical Accuracy
Accuracy Criteria:
1. the accuracy of the given data
2. the accuracy of the computations performed.
For eng. practice, an accuracy of 0.2% is acceptable.
1. Use 4 important figures to record numbers
with a “1”
2. Use 3 important figures in all other cases.
Examples:
A force of 40 lb. should be read 40.0 lb, and
A force of 15 lb. should be read 15.00 lb.
beginnings
1.11 Stress on an Oblique Plan under Axial
Loading
F  P cos
F

A
V  P sin
V

A
(1.13)
(1.12)
Since Ao = Acos, or A = Ao/cos
The max normal stress occurs at  = 0o
The max shear stress occurs at  = 45o
At  = 45o the normal stress is
1.12 Stress under General Loading Conditions:
Components of Stress
F x
  lim
A  0  A
Notation:
Vxz
 xy  lim
A  0
V yx
A
 xz
Vzx
 lim
A  0  A
The plane is ┹ to x-axis
The vector is // to z-direction
(1.18)
Notation:
The surface is ┹ to
x-axis
 xy
The direction of the
component: shear stress
is // to y-direction
All the forces in a system must fulfill the
equation of equilibrium:
Fx  0
Fy  0
M x '  0
M y '  0
Fz  0
Mz '  0
(1.19)
(1.20)
Applying equation of equilibrium
M z  0
 xy   yx
 yz   zy
(1.21)
 zx   xz
1
(1.22)
 xy   yx
Therefore, only six components are required to
uniquely define the stress state of a material.
 x ,  y ,  z ,  xy ,  yz , and  zx .
Max normal stress occurs at  = 0o
Max shear stress occurs at  = 45o
The cube at  = 45o is subjected to the
same magnitude of normal and shear
stresses at all four sides
1.13 Design Considerations
Concept of Factor of Safety:
Factor of Safety = F.S. =
ultimate load
allowable load
Factor of Safety = F.S. =
ultimate stress
allowable stress
End
 D PD   L PL   PU
 ij 
Pij
A