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Solid Mechanics
Dr. Imran LatifDepartment of Civil and Environmental Engineering
College of EngineeringUniversity of Nizwa (UoN)
1
Chapter 1: Tension, Compression and Shear
Why do we study Mechanics of Solids? 2
Anyone
concerned
with
the
strength
and
physical
performance
of
natural/man‐made structures should study Mechanics of Solids
Introduction to Mechanics of Solids 3
Definition: Mechanics of solids is a branch of applied mechanics
that deals with the behaviour of solid bodies subjected to various
types of loading
Compression
Tension
Bending
Torsion
Shearing(streteched)
(twisted)
Introduction to Mechanics of Solids 4
Fundamental concepts
• stress and strain
• deformation and displacement
• elasticity and inelasticity
• load‐carrying capacity
Design and analysis of mechanical and structural systems
Introduction to Mechanics of Solids 5
•
Examination of stresses and
strains inside real bodies of
finite dimensions that deform
under loads
•
In order to determine stresses
and strains we use:
–
Physical properties of
materials
–
Theoretical laws and
concepts
Problem solving 6
• 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 Diagram 7
Free Body Diagram
The unknowns: RA
, RB
, RC
Free Body Diagram 8
Free Body Diagram
The unknowns: Ax
, Ay
, CB
Normal stress and strain•
Most fundamental concepts in Mechanics of Materials are stress and strain
•
Prismatic bar: Straight structural member with the
same
cross‐section
throughout its length
•
Axial force: Load directed along the axis of the member
•
Axial force can be tensile
or compressive
•
Axial loads: Tension (+) and compression (−)
•
Type of loading for landing gear strut and for tow bar?
9
Structural members subjected to axial loads
Normal stress and strain 10
A truss bridge is a type of beam bridge with a skeletal structure. The
forces of tension, or pulling, are represented by red lines and the
forces of compression, or squeezing, are represented by green lines.
Normal stress 11
Continuously distributed stresses
acting over the entire cross‐section.
Axial force P is the resultant of those
stresses
Stress (σ)
has units of force per unit
area
If stresses acting on cross‐section
are uniformly distributed then:
Units of stress in USCS: pounds per square inch (psi) or kilopounds per square inch (ksi)SI units: newtons per square meter (N/m2) which is equal to Pa
FBD of a segment of the bar
Segment of the bar before loading
Segment of bar after loading
Normal stresses in the bar
Limitations 12
The loads P are transmitted to the bar by pins that pass through
the holes
High localized stresses are produced
around the holes !!
Stress concentrations
Steel eyebar subjected to tensile loads P
Normal strain 13
A prismatic bar will change in length when under a uniaxial tensile
force, and obviously it will become longer
Definition of elongation per unit length
or strain (ε)
If bar is in tension, strain is tensile
and if in compression the strain is
compressive
Strain is a dimensionless quantity
(i.e. no units)
FBD of a segment of the bar
Segment of the bar before loading
Segment of bar after loading
Normal stresses in the bar
Line of action of the axial forces for a uniform stress distribution
14
It can be demonstrated that in
order to have uniform tension or
compression in a prismatic bar,
the axial force must act through
the centroid of the cross‐
sectional area.
Example
A short post constructed from a
hollow circular tube of aluminum
supports a compressive load of 26
kips (26000 lb). The inner and outer
diameters of the tube are d1
= 4 in.
and d2
= 4.5 in. respectively, and its
length is 16 in. The shortening of the
post due to the load is measured as
0.012 in.
Determine:
(a) Compressive stress in the
post.
(b) Compressive strain in the
post.
15
16
Problem 17
Problem
A circular aluminum tube of length L = 400 mm is loaded in compression by forces P (see figure). The outside and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal
direction.
(a) If the measured strain is 550 x 10-6
, what is the shortening of the bar?
(b) If the compressive stress in the bar is intended to be 40 MPa, what should be the load P?
18
Problem
Two steel wires, AB and BC, support a lamp weighing 18 lb (see figure). Wire AB is at an angle α
= 34°
to the horizontal and wire BC is at an angle β
= 48°. Both wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals 0.001 in.)
Determine the tensile stresses AB and BC in the two wires.
Mechanical properties of materials 19
In order to understand the mechanical
behaviour of materials we need to perform
experimental testing in the lab
A tensile test machine is a typical equipment
of a mechanical testing lab
ASTM (American Society for Testing and
Materials)
Stress () –
strain () diagrams 20
–
diagrams 21
Stress‐strain diagram for a typical structural
steel in tension (not to scale)
Nominal stress and strain (in the calculations we use the initial
cross‐sectional area A)
True stress (in the calculations we use the cross‐sectional area A
when failure occurs)
True strain if we use a
strain gauge
Stress‐strain diagrams
contain important
information about
mechanical properties
and behaviour
–
diagrams 22
OA: Initial region which
is linear and
proportional (Slope of
OA is called modulus of
elasticity)
BC: Considerable
elongation occurs with
no noticeable increase
in stress (yielding)
CD: Strain hardening –
changes in crystalline structure (increased
resistance to further deformation)
DE: Further stretching leads to reduction in the applied load and fracture
OABCE’: True stress‐strain curve
23
Tensile coupon test specimens showing elongation
immediately prior to failure (G.J. Davies)
–
diagrams 24
Stress‐strain diagram for a typical
structural steel in tension (drawn to
scale)
The strains from zero to point A are so
small as compared to the strains from
point A to E and can not be seen (it is
a vertical line…)
Metals, such as structural steel, that
undergo permanent large strains
before failure are ductile
Ductile materials
absorb large
amounts of strain energy
Ductile materials: aluminium, copper,
magnesium, lead, molybdenum,
nickel, brass, nylon, teflon
Aluminum alloys 25
Typical stress‐strain diagram for an
aluminum alloy.
Although ductile, aluminium alloys
typically do not have a clearly
definable yield point
However, they have an initial
linear region with a recognizable
proportional limit
Structural alloys have proportional
limits in the range of 70‐410 MPa
and ultimate stresses in the range
of 140‐550 MPa
Offset method 26
When the yield point is not
obvious, like in the previous
case, and undergoes large
strains, an arbitrary yield stress
can be determined by the
offset method (e.g. 0.002 or
0.2%)
The intersection of the offset
line and the stress‐strain curve
(point A) defines the yield
stress
Brittle materials 27
Typical stress‐strain diagram for a brittle
material showing the proportional limit
(point A) and fracture stress (point B)
Brittle materials fail at relatively
low strains and little elongation
after the proportional limit
Brittle materials: concrete, marble,
glass, ceramics and metallic alloys
The reduction in the cross‐
sectional area until fracture (point
B) is insignificant and the fracture
stress (point B) is the same as the
ultimate stress
28
Problem
Three different materials, designated A, B,and C, are tested in tension using test specimens having diameters of 0.505 in. and gage lengths of
2.0 in. (see figure).
At failure, the distances between the gage marks are found to be
2.13, 2.48, and 2.78 in., respectively. Also, at the failure cross sections the diameters are found to be0.484, 0.398, and 0.253 in., respectively. Determine the percent elongation and percent reduction in area of each specimen, and then, using your own judgment, classify each material as brittle or ductile.
29
Problem 1.3-6
A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data listed in the accompanying table. Plot the stress-strain curve and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), and yield stress at 0.2% offset. Is the material ductile or brittle?
Solution to problem 30
Modulus of elasticity: 2.35 GPa
Proportional limit: 47 MPa
Yield stress: 52.5 MPa
Material is brittle, because the strain after the proportional limit is exceeded is relatively small.
Elasticity 31
Fig. 1. Elastic behavior
Fig. 2. Plastic behavior
What happens when the load is removed (i.e.
the material is unloaded)?
Tensile load is applied from O to A (Fig 1) and
when load is removed the material follows the
same curve back. This property is called
elasticity
If we load the same material from O to B (Fig
2) and then unloading occurs, the material
follows the line BC. Line OC represents the
residual
or permanent
strain. Line CD
represents the elastic recovery of the material.
During unloading the material is partially
elastic
Plasticity 32
“
Plasticity
is the characteristic of a material which undergoes inelastic
strains beyond the strain at the elastic limit ”
When large deformations occur in a ductile material loaded in the
plastic region, the material is undergoing plastic flow
33
If the material is in the elastic range, it can
be loaded, unloaded and loaded again
without significantly changing the behaviour
When loaded in the plastic range, the
internal structure of the material is altered
and the properties change
If the material is reloaded (fig 1‐19), CB is a
linearly elastic region with the same slope as
the slope of the tangent to the original
loading curve at origin OBy stretching steel or aluminium into the
plastic range, the properties of the material
are changed
Reloading of a material
Creep 34
When loaded for periods of time, some
materials develop additional strains and
are said to creep
Even though the load P remains constant
after
time t0, the bar gradually lengthens
Relaxation is a process at which, after time t0
,
the stress in the wire gradually diminishes and
eventually is reaching a constant value
Creep is more important at high temperatures
and has to be considered in the design of
engines and furnaces
Hooke’s law 35
Many structural materials such as metals, wood, plastics and ceramics
behave both elastically and linearly when first loaded and their
stress‐
strain curve begin with a straight line passing through origin (line OA)
Linear elastic materials are useful for designing structures and
machines
when permanent deformations, due to yielding, must be avoided
Hooke’s law 36
Robert Hooke
(1635‐1703)
The linear relationship between stress and strain for a bar in
simple tension or compression is expressed by:
σ
= E εσ
is axial stress
ε
is axial strain
E is modulus of elasticity
Hooke’s law
The above equation is a limited version of Hooke’s Law relating only the
longitudinal stresses and strains that are developed during the
uniaxial
loading of a prismatic bar
Robert Hooke
was an English inventor, microscopist, physicist, surveyor,
astronomer, biologist and artist, who played an important role in the
scientific revolution, through both theoretical and experimental
work.
Modulus of Elasticity 37
E is called modulus of elasticity or Young’s modulus and
is a constant
It is the slope of the stress – strain curve in the linearly
elastic region
Units of E are the same as the units of stress (i.e. psi or
Pa)
Thomas Young
was an English polymath, contributing to the scientific
understanding of vision, light, solid mechanics, energy, physiology, and
Egyptology.
When a prismatic bar is loaded in tension the
axial elongation is accompanied by lateral
contraction
The lateral strain ε’
at any point in a bar is
proportional to the axial strain ε
at the same
point if the material is linearly elastic
The ratio of the above two strains is known as Poisson’s ratio
(ν)
Poisson’s ratio 38
lateral
contractionlongitudinal
extension
ν
= ‐
(lateral strain / axial strain = ‐
(ε’
/ ε
)
Poisson’s ratio 39
Simeon Denis Poisson(1781‐1840)
Siméon‐Denis Poisson
was a French mathematician, geometer, and
physicist.
The minus sign in the equation is because the lateral
strain is negative (width of the bar decreases) and the
axial tensile strain is positive. Therefore, the Poisson’s
ratio will have a positive value.
When using the Poisson’s ratio equation we need to
know that it applies only to a prismatic bar in uni‐axial
stress
Poisson’s value of concrete = 0.1 – 0.2
Poisson’s value of rubber = 0.5
Limitations 40
Poisson’s ratio is constant in the
linearly elastic range
Material must be homogeneous
(same composition at every
point)
Materials having the same
properties in all directions are
called isotropic
If the properties differ in
various directions the materials
called anisotropic
Shear stress and shear strain 41
tangentialperpendicular
Shear stress
acts tangential to the surface
of the material and not perpendicular
Consider the bolted connection of Figure
where A is a flat bar, C a clevis and B a bolt
When load P is applied, the bar and clevis
will press against the bolt and bearing
stresses
will be developed
The bar and clevis tend to shear
the bolt
Shear stress and shear strain 42
If we have a closer look from the side
view (fig b) and draw a FBD (fig c)
bearing area
total bearing force
Bearing stresses exerted by the clevis against the
bolt appear on the left‐hand side (1 and 3)
Stresses from the bar are on the right‐hand side (2)
Based on the assumption of uniform stress
distribution we can calculate an average bearing
stress σb
Shear stress and shear strain 43
The
bearing
area
Ab
is
defined
as
the
projected
area
of
the
curved
bearing
surface.
For
example
(for
stresses
labeled
1)
the
projected
area
on
which
the stresses
act is a
rectangle
with
height
equal
to
the
thickness of the clevis and width equal to the diameter of the bolt
The bearing force Fb
(for stresses labeled 1) is equal to P/2
The same area and force apply for stresses labeled 3For
bearing
stresses
labeled
2
the
bearing
area
is
a
rectangle
with
height
equal
to
the
thickness
of
the
flat
bar
and
width
equal
to
the
diameter of the bolt. The force is equal to P
Shear stress and shear strain 44
The FBD (fig. c) shows that there is a tendency to shear the bolt along the
cross sections mn and pq
From the FBD (fig. d) of the portion mnqp of the bolt we see that the shear
forces V act over the cut surfaces of the bolt. There are two planes of shear
(plane mn and plane pq). Therefore, the bolt is in double shear
The shear stresses acting on the cross section mn are shown (fig. e)Shear stresses are denoted by τ
Single shear 45
The axial force P in the metal bar is
transmitted to the flange of the
steel column through a bolt
A cross section of the column (fig b)
shows more details
Fig c shows the assumed distribution
of the bearing stresses acting on the
bolt
Cutting through the bolt at section
mn (fig d) we see the shear force V
(equal to load P). V is the resultant
of the shear stresses that act over
the cross‐sectional area of the bolt
Single shear 46
Shear stress and strain 47
Discussing about bolted connections we disregard friction which is
produced by tightening the bolts
Average shear stress on the cross section of a bolt is obtained by dividing
the total shear force V by the area A of the cross section on which it acts:
Shear stresses have the same units as normal stresses
The two previous examples (double and single shear) are examples
of
direct shear
Direct shear arises in the design of bolts, pins, rivets, keys, welds and
glued joints
48
Consider a small rectangular parallelepiped
element
Assume that a shear stress τ1
is uniformly
distributed over the right‐hand side area bc
(τ1bc
)
For equilibrium in the y direction the τ1bc
must
be balanced by an equal and of opposite
direction shear force on the left‐hand side
The forces τ1bc
acting on the right‐hand and
left‐hand side faces form a couple having a
moment about z‐axis equal to τ1
.abc
(counterclockwise direction)
Small element of material
subjected to shear stresses
Equality of shear stresses on perpendicular planes 49
Similarly, in order to have equilibrium of
the element, we have a shear force τ2
.ac
and consequently a clockwise couple of
moment τ2
.abc
It is therefore evident that for moment
equilibrium we have:
τ1
= τ21.
Shear stresses on opposite and parallel faces of an element
are equal in magnitude and opposite in direction
2.
Shear stresses on adjacent and perpendicular faces of an
element are equal in magnitude and have directions such
that both stresses point toward, or both point away from,
the line of intersection of the faces
50
Shear stresses
acting on an element of
material (fig a) are accompanied by shear
strains
The lengths of the sides of the element do
not change but, the shear stresses
produce a change in the shape of the
element
Rectangular parallelepiped becomes
oblique parallelepiped. Front and rear
faces become rhomboids
The angle γ
(fig b) is a measure of
distortion of the element and is called
shear strain
Hooke’s law in shear 51
We can plot shear stress‐strain diagrams
Hooke’s Law in shearτ
= Gγ
shear modulus of elasticity
shear strainshear stress
G has the same units as E (Young’s modulus)
G and E are also related by:
G = E / (2(1+ν)) Poisson’s ratio
Allowable stresses & allowable loads 52
The principal design interest is
strength
Strength is the capacity of the object
to support or transmit loads
The actual strength of a structure
must exceed
the required strength
Factor of safety must be greater than
1 if failure is to be avoided
Factors of safety from slightly above 1
to as much as 10 are used
Factor of safety (n) = Actual strength / Required strength
Allowable stresses & allowable loads 53
Allowable stress = Yield strength / Factor of safety
σallow
= σY
/ n1 τallow
= τY
/ n2
tension shear
Allowable stress = Ultimate strength / Factor of safety
OR…
σallow
= σU
/ n3 τallow
= τU
/ n4
tension shear
Allowable loads 54
Allowable load is also called permissible load or
the safe load
Allowable load = (Allowable stress) (Area)
For bars in tension or compression:Pallow
= σallow
. A
For pins in direct shear:Pallow
= τallow
. A
Permissible load based upon bearing:Pallow
= σb
. A