CHAPTER ONE
STRESS
Mechanics – science describing the
behavior of bodies under the action of
forces.
Mechanics of Materials is a branch
of mechanics that studies the internal
effects of stress and strain in a solid body
that is subjected to an external loading.
Stress is associated with the strength of
material from which the body is made,
while strain is a measure of the
deformation of the body.
THREE FUNDAMENTAL AREAS OF
ENGINEERING MECHANICS
1. Statics
2. Dynamics
}
Devoted primarily to the
study of the external
effects of forces on rigid
bodies.
}
Deals
with
the
relations
between
externally
applied
loads
and
their
internal effects on
bodies
EXTERNAL EFFECTS OF A FORCE
3. Mechanics of
Deformable
Bodies
1. Development of
forces at surfaces of
contact between
bodies.
2. Change in state of
motion of the body.
INTERNAL EFFECTS OF A FORCE
2. Sustained Load - load that is constant
over a long period of time.
3. Impact Load - impulsive and rapid
application of loads.
4. Repeated Load - successive application
and removal of loads.
B. According to Distribution
1. Concentrated Load - point load.
2. Distributed Load - a load distributed
along a line or a surface.
C. According to Location and Method of
Application
1. Centric Load - load that pass through
the centroid of the resisting section.
2. Torsional Load - load that twists a
member.
3. Flexural / Bending Load - load that is
applied transversely to the longitudinal
axis of the member.
4. Combined Loading - any combination of
the first 3 above.
CONCEPT of STRESS
1. Stress - intensity of load/force per unit
area (P/A)
Units: MPa, kPa, Pa, psi, ksi
2. Normal Stress, σ = P/A - stresses acting
perpendicular to the surface of a cross
section.
3. Shear Stress, τ = V/A - stresses acting
parallel or tangent to the surface of a cross
section
1. Deformation
2. Development
of internal
stresses.
Rigid Body – a body that does not deform
under the action of forces.
Non – Rigid Body – a body that deforms
under application of loads.
Consider two bars of equal length,
supporting maximum axial loads, which
bar is stronger?
LOAD CLASSIFICATION
A. According to Time
1. Static Load - gradual application of
loads for which equilibrium is achieved at a
very short time.
Engr. Nikko Reymon R. Manito
Mechanics of Deformable Bodies P a g e | 1
TWO TYPES OF EXTERNAL LOADS
1. SURFACE FORCES
are caused by the
direct contact of one
body with the surface
of another. In all cases
these
forces
are
distributed over the
area
of
contact
between the bodies.
a. Concentrated force – a single
force which applied to a point of the body
since the area is so small in comparison
with the total surface area of the body.
b. Linear distributed load – is
applied along a narrow strip of area and is
measured as having an intensity of force
per length along the strip and is
represented graphically by a series of
arrows.
c. Support Reactions – is the
surface forces that develop at the supports
or points of contact between bodies.
d. The resultant force of a
distributed load is equivalent to the area
under the distributed loading curve and
this resultant acts through the centroid or
geometric center of this area.
2. A BODY FORCE is developed when one
body exerts a force on another body without
direct physical contact between bodies.
These forces are normally represented by
single concentrated force acting on a body.
FOUR TYPES OF RESULTANT LOADINGS
1. Normal force (N) – this force acts
perpendicular to the area. It is developed
whenever the external loads tend to push
or pull on the two segment of the body.
2. Shear force (V) – the shear force lies on
the plane of the area and it is developed
when the external loads tend to cause the
two segments of the body to slide over one
another.
3. Torsional moment or Torque (T) – this
effects are developed when the external
loads tend to twist one segment of the body
with respect to the other about an axis
perpendicular to the area.
4. Bending moment (M) – it is caused by
external loads that tend to bend the body
about an axis lying within the plane of the
area.
Simple Stresses
1. Normal Stress
2. Shear Stress
3. Bearing Stress
4. Thin-walled Pressure Vessel
Stress is defined as the strength of a
material per unit area or unit strength. It is
the force on a member divided by area,
which carries the force, formerly express in
psi, now in N/mm2 or MPa.
σ
=
𝑃
𝐴
where P is the applied normal load in
Newton and A is the area in mm2. The
maximum stress in tension or compression
occurs over a section normal to the load.
 Normal Stress
Normal stress is either tensile stress or
compressive stress. Member’s subject to
pure tension (or tensile force) is under
tensile stress, while compression members
(members subject to compressive force) are
under compressive stress. Compressive
force will tend to shorten the member.
Tension force on the other hand will tend to
lengthen the member.
Example No. 1
A hollow steel tube with
an inside diameter of 100
mm must carry a tensile
load
of
400
kN.
Determine the outside
diameter of the tube if the
stress is limited to 120
MN/m2.
Example No. 2
A composite bar consists of an aluminum
section rigidly fastened between a bronze
section and a steel section as shown. Axial
loads are applied at the positions indicated.
Determine the stress in each section.
Example No. 3
For the truss shown in figure, determine
the stress in member’s AC and BD. The
cross-sectional area of each member is 900
mm2.
STRESS
Engr. Nikko Reymon R. Manito
Mechanics of Deformable Bodies P a g e | 2
Example No. 4
The block weight (W) in figure hangs from
the pin at A. The bars AB and AC are
pinned to the support at B and C. The areas
are 800 mm2 for AB and 400mm2 for AC.
Neglecting the weights of the bars,
determine the maximum safe value of W if
the stress in AB is limited to 110 MPa and
that in AC to 120 MPa
ASSIGNMENT NO. 1
Solve problems #s 105,107,108,109,111,
112. 114 Pages 12-14, SINGER & Pytel
book.
*Long coupon bond, format will be posted
at a later date.
 Shearing Stress
Forces parallel to the area resisting the
force cause shearing stress. It differs to
tensile and compressive stresses, which are
caused by forces perpendicular to the area
on which they act. Shearing stress is also
known as tangential stress.
τ
=
Example No. 6
If the wood joint in figure has a width of 150
mm, determine the average shear stress
developed along shear planes a-a and b-b.
Example No. 7
What force is required to punch a 20 mm
diameter hole in a plate that is 25 mm
thick? The shear strength is 350 MPa.
Example No. 8
A 200-mm-diameter pulley is prevented
from rotating relative to 60-mm-diameter
shaft by a 70-mm-long key, as shown in
Fig. P-118. If a torque T = 2.2 kN-m is
applied to the shaft, determine the width b
if the allowable shearing stress in the key is
60 MPa.
𝑉
𝐴
where
is the resultant shearing force
which passes through the centroid of the
area
being sheared.
Examples of shear stress
SUPPLEMENTARY EXAMPLES
SE No. 1
The beam is supported by a pin of A and a
short link BC. Determine the maximum
magnitude P of the loads the beam will
support if the average shear stress in each
pin is not to exceed 80 MPa. All pins are in
double shear as shown, and each has a
diameter of 18 mm.
Example No.5
Determine the shear stress in the 20 mm
diameter pin at A and the 30 mm diameter
pin at B that support the beam shown.
Engr. Nikko Reymon R. Manito
Mechanics of Deformable Bodies P a g e | 3
SE No. 2
The block is subjected to a compressive
force of 2 KN. Determine the average
normal and average shear stress developed
in the wood fibers that are oriented along
section a-a at 30 with the axis of the block.
ASSIGNMENT NO. 2
Solve problems #s 115,116,119,120,123
Pages 16-18, SINGER & Pytel book.
*Long coupon bond, format will be posted
at a later date.
Bearing Stress
Example No. 11
Figure shows a
roof truss and the
detail of the riveted
connection at joint
B. Using allowable
stresses of τ = 70
MPa and σb= 140
MPa, how many
19-mm-diameter
rivets are required
to fasten member BC to the gusset plate?
Member BE? What is the largest average
tensile or compressive stress in BC and
BE?
Bearing stress is the contact pressure
between the separate bodies. It differs from
compressive stress, as it is an internal
stress caused by compressive forces.
𝑃𝑏
b =
𝐴𝑏
σ
Example No. 9
In figure, assume that a 20-mm-diameter
rivet joins the plates that are each 110 mm
wide. The allowable stresses are 120 MPa
for bearing in the plate material and 60
MPa for shearing of rivet. Determine (a) the
minimum thickness of each plate; and (b)
the largest average tensile stress in the
plates.
 Thin Walled Pressure Vessels
A tank or pipe carrying a fluid or gas under
a pressure is subjected to tensile forces,
which resist bursting, developed across
longitudinal and transverse sections.
 TANGENTIAL STRESS, σt
(Circumferential Stress)
Consider the tank shown being subjected
to an internal pressure . The length of the
tank is
and the wall thickness is .
Isolating the right half of the tank:
Example No. 10
The lap joint shown in figure is fastened by
four ¾-in.-diameter rivets. Calculate the
maximum safe load P that can be applied if
the shearing stress in the rivets is limited
to 14 ksi and the bearing stress in the
plates is limited to 18 ksi. Assume the
applied load is uniformly distributed
among the four rivets.
Engr. Nikko Reymon R. Manito
The forces acting are the total pressures
caused by the internal pressure and the
total tension in the walls .
Mechanics of Deformable Bodies P a g e | 4
If there exist an external pressure and an
internal pressure , the formula may be
expressed as:
 LONGITUDINAL STRESS,
Consider the free body diagram in the
transverse section of the tank:
The total force acting at the rear of the tank
must equal to the total longitudinal stress
on the wall
. Since is so small
compared to , the area of the wall is close
to
If there exist an external pressure and
an internal pressure , the formula may
be expressed as:
Example No. 12
A cylindrical steel pressure vessel 400 mm
in diameter with a wall thickness of 20 mm,
is subjected to an internal pressure of 4.5
MN/m2. (a) Calculate the tangential and
longitudinal stresses in the steel. (b) To
what value may the internal pressure be
increased if the stress in the steel is limited
to 120 MN/m2? (c) If the internal pressure
were increased until the vessel burst,
sketch the type of fracture that would
occur.
Example No. 13
The wall thickness of a 4-ft-diameter
spherical tank is 5/6 inch. Calculate the
allowable internal pressure if the stress is
limited to 8000 psi.
Example No. 14
Calculate the minimum wall thickness for
a cylindrical vessel that is to carry a gas at
a pressure of 1400 psi. The diameter of the
vessel is 2 ft, and the stress is limited to 12
ksi
Example No. 15
A cylindrical pressure vessel is fabricated
from steel plating that has a thickness of
20 mm. The diameter of the pressure vessel
is 450 mm and its length is 2.0 m.
Determine the maximum internal pressure
that can be applied if the longitudinal
stress is limited to 140 MPa, and the
circumferential stress is limited to 60 MPa.
ASSIGNMENT NO. 3
Solve problems #s 125,127,133,134,137,
138, 140, 142 Pages 20-21 and 28-29,
SINGER & Pytel book.
*Long coupon bond, format will be posted
at a later date.
It can be observed that the tangential
stress is twice that of the longitudinal
stress.
 SPHERICAL SHELL
If a spherical tank of
diameter and thickness
contains gas under a
pressure of , the stress
at the wall can be
expressed as:
Engr. Nikko Reymon R. Manito
Mechanics of Deformable Bodies P a g e | 5