Mechanics of Materials(ME-294)
Lecture 2
Statics and Strength of Materials
Statics is the study of forces acting in equilibrium
on rigid bodies
• “Bodies” are solid objects, like steel cables, gear teeth, timber
beams, and axle shafts (no liquids or gases);
• “rigid” means the bodies do not stretch, bend, or twist;
• “equilibrium” means the rigid bodies are not accelerating.
In Strength of Materials, we keep the assumptions
of bodies in equilibrium, but we drop the “rigid”
assumption.
 Real cables stretch under tension and real axle
shafts twist under torsional load.
Strength of Materials
Statics is the study of forces acting in equilibrium on
rigid bodies.
• “Bodies” are solid objects, like steel cables, gear teeth, timber
beams, and axle shafts (no liquids or gases);
• “rigid” means the bodies do not stretch, bend, or twist;
• “equilibrium” means the rigid bodies are not accelerating.
In Strength of Materials, we keep the assumptions of
bodies in equilibrium, but we drop the “rigid”
assumption.
▫ Real cables stretch under tension and real axle shafts
twist under torsional load.
▫ The most fundamental concepts in mechanics of
materials are stress and strain.
Stress and Strain
• The words “stress” and “strain” are used interchangeably: “I’m
feeling stressed” or “I’m under a lot of strain.”
• In engineering, these words have specific, technical meanings.
If you tie a steel wire to a hook in the
ceiling and hang a weight on the
lower end, the wire will stretch.
•Divide the change in length by the
original length, and you have the
strain in the wire.
•Divide the weight hanging from the
wire by the wire’s cross sectional
area, and you have the tensile stress
in the wire.
Stress and strain are ratios.
▫ The symbol for stress is σ, the lower case Greek
letter sigma. Stress has units of force per unit area
 When SI units are used, force is expressed in newtons
(N) and area in square meters (m2). Consequently,
stress has units of newtons per square meter (N/m2),
that is, pascals (Pa).
1N/m2 = 1 Pa

 However, the pascal is such a small unit of stress that
it is necessary to work with large multiples, usually
the megapascal (MPa).
1 MPa = 106 Pa = 106 N/ m2 = 1 N/mm2
The symbol for strain is ε, the lower case Greek
letter epsilon.
 Because normal strain is the ratio of two lengths, it is
a dimension- less quantity, that is, it has no units.
Therefore, strain is expressed simply as a number,
independent of any system of units.
 Numerical values of strain are usually very small,
because bars made of structural materials undergo
only small changes in length when loaded.
A prismatic bar is a straight structural member
having the same cross section throughout its
length, and an axial force is a load directed
along the axis of the member, resulting in either
tension or compression in the bar
we will consider the tow bar of Fig. 1-1 and
isolate a segment of it as a free body.
The internal actions in the bar are exposed if we
make an imaginary cut through the bar at
section mn (Fig. 1-2c). Because this section is
taken perpendicular to the longitudinal axis of
the bar, it is called a cross section.
This action consists of continuously distributed
stresses acting over the entire cross section,
and the axial force P acting at the cross section
is the resultant of those stresses. (The resultant
force is shown with a dashed line in Fig. 1-2d.)
The force per unit area, or intensity of the forces distributed
over a given section, is called the stress on that section.
Tensile and Compressive Stress and strain
• A load that pulls is called a tensile load
and stress induced is tensile stress.
• If the load pushes, we call it a
compressive load and stress is called
compressive stress.
• The equations are the same.
• Stresses act in a direction perpendicular
to Cross section, they are called normal
stresses.
• Normal stresses may be either tensile or
compressive.
• Another type of stress, called shear
stress, that acts parallel to the surface.
• sign convention
▫ tensile stresses as positive(+)
▫ compressive stresses as negative(-)
Stress-Strain Curve
stress-strain curve illustrates the elastic
and plastic zones.
•If you hang a light weight to the wire
hanging from the ceiling, the wire
stretches elastically; remove the
weight and the wire returns to its
original length.
•Apply a heavier weight to the wire,
and the wire will stretch beyond the
elastic limit and begins to plastically
deform, which means it stretches
permanently. Remove the weight and
the wire will be a little longer (and a
little skinnier) than it was originally.
•Hang a sufficiently heavy weight, and
the wire will break.
• Two stress values are important in
engineering design.
▫ The yield strength, σys, is the limit
of elastic deformation; beyond this
point, the material “yields,” or
permanently deforms.
▫ The ultimate tensile strength, σUTS
(also called tensile strength, σTS) is
the highest stress value on the
stress-strain curve.
• The rupture strength is the stress
at final fracture; this value is not
particularly useful, because once
the tensile strength is exceeded,
the metal will break soon after.
• Young’s modulus, E, is the slope of
the stress-strain curve before the
test specimen starts to yield.
• The strain when the test specimen
breaks is also called the
elongation.
Many manufacturing operations on metals are
performed at stress levels between the yield strength
and the tensile strength.
▫ Bending a steel wire into a paperclip, deep-drawing sheet
metal to make an aluminum can, or rolling steel into
wide-flange structural beams are three processes that
permanently deform the metal, so σYS<σApplied .
▫ During each forming operation, the metal must not be
stressed beyond its tensile strength, otherwise it would
break, so σYS<σApplied<σUTS .
▫ Manufacturers need to know the values of yield and
tensile strength in order to stay within these limits.
Expression for elongation
In this Strength of Materials course, almost all of the problems are
elastic, so there is a linear relationship between stress and strain
Take an aluminum rod of length L, cross-sectional
area A, and pull on it with a load P. The rod will
lengthen an amount δ. We can calculate δ in three
separate equations, or we can use algebra to find a
simple equation to calculate δ directly.
Direct equation for calculating the change in length of the rod.
Deformation of a body due to self weight
In Fig determine the total increase of length of a bar of constant cross section
hanging vertically and subject to its own weight as the only load.
The normal stress (tensile) over any horizontal cross section
is caused by the weight of the material below that section.
The elongation of the element of thickness dy shown is
where A denotes the cross-sectional area of the bar and γ
its specific weight (weight/unit volume).
Integrating, the total elongation of the bar is
where W denotes the total weight of the bar.
Note that the total elongation produced by the weight of the bar is equal to that
produced by a load of half its weight applied at the end.
Principle of Superposition
Assignment#1
1. In 1989, Jason, a research-type submersible with
remote TV monitoring capabilities and weighing 35
200 N, was lowered to a depth of 646 m in an effort
to send back to the attending surface vessel
photographs of a sunken Roman ship offshore from
Italy. The submersible was lowered at the end of a
hollow steel cable having an area of 452 × 10–6 m2
and E= 200 GPa. Determine the extension of the
steel cable. Due to the small volume of the entire
system, buoyancy may be neglected. (Note: Jason
was the system that took the first photographs of the
sunken Titanic in 1986.) weight of steel per unit
volume is 77 kN/m3.
2. Two prismatic bars are rigidly
fastened together and support a
vertical load of 45 kN, as shown in
Figure. The upper bar is steel
having length 10 m and crosssectional area 60 cm2. The lower
bar is brass having length 6 m and
cross-sectional area 50 cm2. For
steel E = 200 GPa, for brass E =
100 GPa. Determine the maximum
stress in each material. Specific
weights of brass and steel are
84kN and 77kN.
3. A 70 kN compressive load is applied to a 5 cm
diameter, 3 cm tall, steel cylinder. Calculate
stress, strain, and deflection.
4. What tensile stress is required to produce a
strain of 8×10−5 in aluminum? Report the
answer in MPa. Aluminum has a Young’s
modulus of E = 70 GPa.
5. Stress-strain curves for materials in compression
differ from those in tension. Explain.
Quiz#1
• Chapter 1 and 2 –SOM by R. S. Khurmi
• Assignment#1