1) The document discusses concepts of stress and strain in materials including definitions of strain as the ratio of deformation to length, stress-strain diagrams, Hooke's law relating stress and strain, and Poisson's ratio relating lateral and axial strain. 2) An example problem is presented to calculate the deformation of a steel rod under different axial loads applied at each end using the equations discussed. 3) A second example problem determines the modulus of elasticity and Poisson's ratio of an unknown material based on experimental data measuring the change in length and diameter of a rod under an axial load.

1. MECHANICS OF MATERIALS
“Strength of Materials”
By:
Dr. Arkan Fawzi Saeed
PhD. In Mechanical Engineering
arkan_alhazeen@yahoo.com

2. Chapter (2)
2
Dr. Arkan F. Saeed

3. 2- Stress and Strain
3
Dr. Arkan F. Saeed

4. 2.1: Strain “”:
It is the ratio of the total deformation “” over the total length “L”.
4
Dr. Arkan F. Saeed
dx
d
x
L x



 





 0
lim

5. 2.2: Stress-Strain Diagram:
5
Dr. Arkan F. Saeed
Stress-Strain diagrams of various materials vary widely, and different tensile
tests conducted on the same material may yield different results, depending upon
the temperature of the specimen and the speed of loading.

6. 6
Dr. Arkan F. Saeed
The above two types are typical ductile materials.
 yield : yield stress.
 all : allowable stress.
 U : Ultimate stress.
 B : Breaking stress (Rupture).

7. 7
Dr. Arkan F. Saeed
 The deformation after a critical value (y) of the stress, has been reached,
the specimen undergoes a large deformation with a relatively small increase
in the applied loads. This deformation is caused by slipping of the material
along oblique surfaces and its due, therefore, primarily to shearing stresses.
 The rupture occurs along a cone-shaped surface which forms an angle of
approximately (45o) with the original surface of the specimen. This
indicates that shear is primarily responsible for the failure of ductile
materials, and confirms the fact that:
“under an axial load, shearing stresses are largest on surfaces forming
an angle of (45o) with the loads”.

8. 8
Dr. Arkan F. Saeed
 Brittle materials, which comprise cast iron, glass and stone, are
characterized by the fact that rupture occurs without any noticeable prior
change in the rate of elongation.
 The strain at the time of rupture is much smaller for brittle than for ductile
materials. And the rupture occurs along a surface perpendicular to the load.
We conclude from this observation that normal stresses are primarily
responsible for the failure of brittle materials.

9. 9
Dr. Arkan F. Saeed

 .
E

2.3: Hook’s Law: Modulus of Elasticity:
Where;
E = Young’s Modulus of Elasticity
... Hook’s Law

10. 2.4: Deformations of members under axial loading:
10
Dr. Arkan F. Saeed
)
1
...(
.
 E

)
2
...(
.E
A
P
E




)
3
...(
L

 

)
4
...(
.
.
.
E
A
L
P
L 

 

)
5
...(
.
.


i i
i
i
i
E
A
L
P
or  )
6
...(
.
.
.
.
.
0





L
E
A
dx
P
E
A
dx
P
dx
d





11.  If we have relative displacement of one of the member;
11
Dr. Arkan F. Saeed

12. Example(1): Determine the deformation of the steel rod shown in figure
under the given loads (E=29*106 psi).
12
Dr. Arkan F. Saeed
L1 = L2 = 12 in L3 = 16 in
A1 = A2 = 0.9 in2 A3 = 0.3 in2
P1 = 60 kips = 60*103 lb
P2 = -15 kips = -15*103 lb
P3 = 30 kips = 30*103 lb
Now applying the equation;
     
  
in
A
L
P
A
L
P
A
L
P
E
E
A
L
P
i i
i
i
i
3
3
3
3
6
3
3
3
2
2
2
1
1
1
10
*
9
.
75
3
.
0
16
10
*
30
9
.
0
12
10
*
15
9
.
0
12
10
*
60
10
*
29
1
.
.
.
1
.
.






























 



13. 13
Dr. Arkan F. Saeed
2.5: Poisson’s ratio ():
E
x
x

 
A
P
x 
 and
Since:
We also note that the normal stresses on faces
respectively perpendicular to the y and z axes are
zero: y = z = 0.
Then:
x
z
x
y
strain
axial
strain
lateral




 




Where;
E
E
x
y
z
x
x






.






14. Example(2): A 500 mm long, 16 mm diameter rod made of a homogenous,
isotropic material is observed to increase in length by 300 m, and to
decrease in diameter by 2.4 m when subjected to an axial 12 kN load.
Determine the modulus of elasticity and Poisson's ratio of the material.
14
Dr. Arkan F. Saeed
The cross-sectional area of the rod is:
A = .r2 = .(8*10-3)2 = 201*10-6 m2
6
3
6
3
6
3
10
*
150
16
10
*
4
.
2
10
*
600
500
10
*
300
7
.
59
10
*
201
10
*
12
















d
L
MPa
A
P
y
y
x
x
x





From Hook’s Law; x = E.x, we obtain: GPa
E
x
x
5
.
99
10
*
600
10
*
7
.
59
6
6


 


25
.
0
10
*
600
10
*
150
6
6





 

x
y




15.  The principle is that the effect of a given combined loading on a structure
may be obtained by determining separately the effect of the various loads
and combining the results obtained under the following conditions:
1. Each effect is linearly related to the load which produces it.
2. The deformation resulting from any given load is small and does not affect
the conditions of application of the other loads.
15
Dr. Arkan F. Saeed
2.6: Multi-axial loading: Generalized Hook’s law:
E
E
E
E
E
E
E
E
E
z
y
x
z
z
y
x
y
z
y
x
x






























.
.
.
.
.
.