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Problem Set [Professor Video][Professor Note] [Faculty Video][Faculty Note]
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod. Knowing that, with an
axial load of 6000 N acting on the rod, the distance between the gage marks is 250.18 mm, determine the
modulus of elasticity of the aluminum used in the rod. [Example]
1.
Rod BD is made of steel (E = 29 X 106 psi) and is used to brace the axially compressed member ABC. The
maximum force that can be developed in member BD is 0.02P. If tIe stress must not exceed 18 ksi and the
maximum change in length of BD must not exceed 0.001 times the length of ABC, determine the smallest-
diameter rod that can he used for member BD.
2.
Each of the links AD and CD is made of aluminum (E = 75 GPa) and has a cross-sectional area of 125 mm2.
Knowing that they support the rigid member BC, determine the deflection of point E. [Example]
3.
Two cylindrical rods, one of steel and the other of brass, are joined at C and restrained by rigid supports at A
and E. For the loading shown and knowing that Es = 200 GPa and E
B = 105 GPa, determine (a) the reactions
at A and E. (b) the deflection of point C. [Example]
4.
Strength of Materials/ Unit 6/ Module 3 Strain, Hooke's Law
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Additional Problems:
For the steel truss (E = 200 GPa) and loading shown, determine the deformations of the members AB and
AD. knowing that their cross-sectional areas are 2400 mm2 and 1800 mm
2, respectively. [Example] (Ans:
)
1.
Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of
P is 4 kN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B.
[Example] (Ans: (a) 32.8 kN and (b) 0.0728 mm)
2.
The 4-mm-diameter cable BC is made of a steel with E = 200 GPa Knowing that the maximum stress in the
cable must not exceed 190 MPa and that the elongation of the cable must not exceed 6 mm, find the
maximum load P that can be applied as shown. (Ans: 1.988 kN)
3.
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The 1.5-m concrete post is reinforced with six steel bars, each with a 28-mm diameter. Knowing that Es = 200
GPa and Ec = 25 GPa, determine the normal stresses in the steel and in the concrete when a 1550-kN axial
centric force P is applied to the post. [Example] (Ans: )
4.
Examples
The rigid bar BDE is supported by two links AB and CD. Link
AB is made of aluminum (E = 70 GPa) and has a cross-
sectional area of 500 mm2; link CD is made of steel (E = 200
GPa) and has a cross-sectional area of 600mm2. For the 30-kN
force shown, determine the deflection (a) of B, (b) of D, (c) of E.
1.
Sol:
Free Body: Bar BDE
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a. Deflection of B. Since the internal force in link AB is
compressive, we have P = -60 kN
The negative sign indicates a contraction of member AB.
and, thus, an upward deflection of end B:
b. Deflection of D. Since in rod CD, P = 90 kN, we write
c. Deflection of E. We denote by and the displaced
positions of points B and D. Since the bar BDE is rigid, points ,
, and lie in a straight line and we write
The rigid castings A and B are connected by two -in.-diameter steel bolts CD and GH and are in
contact with the ends of 1.5-in.-diameter aluminum rod EF. Each bolt is single-threaded with a pitch of
0.1 in., and after being snugly fitted, the nuts at D and H are both tightened one-quarter of a turn.
Knowing that E is 29 X 106 psi for steel and 10.6 X 10
6 psi for aluminum, determine the normal stree in
the rod.
2.
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Sol:
Deformations:
Bolts CD and GH. Tightening the nuts causes tension in the
bolts. Because of symmetry, both are subjected to the same
internal force Pb and undergo the same deformation δ We have
(1)
Rod EF. The rod is in compression. Denoting by Pr the
magnitude of the force in the rod and by δr the deformation of the
rod, we write
(2)
Displacement of D Relative to B. Tightening the nuts
one-quarter of a turn causes ends D and H of the bolts to undergo
a displacement of (0.1 in.) relative to casting B. Considering end
D, we write
(3)
But , where δD
and δB
represent the
displacements of D and B. If we assume that casting A is held in a
fixed position while the nuts at D and H are being tightened, these
displacements are equal to the deformations of the bolts and of
the rod, respectively. We have, therefore,
(4)
Substituting from (1), (2), and (3) into (4), we obtain
0.025 in. = 1.405 X 10-6
Pb + 0.6406 X 10
-6
Pr (5)
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Free Body: Casting B
(6)
Forces in Bolts and Rod
Substituting for Pr from (6) into (5), we have
0.025 in. = 1.405 X 10-6
Pb + 0.6406 X 10
-6(2p
b)
Pb = 9.307 X 103 lb = 9.307 kips
Pr
= 2Pb = 2(9.307 kips) = 18.61 kips
Stress in Rod
A 500-mm-long, 16-mm-diameter rod made of a homogeneous, isotropic material is observed to
increase in length by 300 µm, and to decrease in diameter by 2.4 µm when subjected to an axial 12-kN
load. Determine the modulus of elasticity and Poisson's ratio of the material.
3.
Sol:
The crossp-sectional area of the rod is
Choosing the x axis along the axis of the rod (Fig. A), we
write
From Hooke's law, , we obtain
and, from Eq.(9),
Determine the value of the stress in portions AC and CB of the steel bar shown (Fig. B) when the
temperature of the bar is -500F, knowing that a close fit exists at both of the rigid supports when
temperature is +750F. Use the values E = 29 X 106 psi and α = 6.5 X 106/ 0F for steel.
4.
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Sol:
We first determine the reactions at the supports. Since the problem is
statically indeterminate, we detach the bar from its supports at B and let it
undergo the temperature change
The corresponding deformation (Fig. C) is
Applying now the unknown force RB
at end B (Fig. C(c)), we use to express
the corresponding deformation δR
. Substituting
L1 = L
2 = 12 in.
A1 = 0.6 in
2 A
2 = 1.2 in
2
P1 = P
2 = R
B E = 29 X 10
6 psi
into below Eq., we write
Expressing that the total deformation of the bar must be zero as a result of the imposed constraints, we write
δ = δr + δR = 0
= -19.50 X 10-3 in. + (1.0345 X 10-6
in./lb) RB = 0 from which we obtain
RB = -18.85 X 103 lb = 18.85 kips The reaction at A is equal and opposite.
Noting that the forces in the two portions of the bar are P1 = P
2 = 18.85 kips, we obtain the following values of
the stress in portions AC and CB of the bar:
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We cannot emphasize too strongly the fact that, while the total deformation of the bar must be zero, the
deformations of the portions AC and CB are not zero. A solution of the problem based on the assumption that these
deformations are zero would therefore he wrong. Neither can the values of the strain in AC or CB be assumed
equal to zero. To amplify this point, let us determine the strain in portion AC of the bar. The strain can be
divided into two component parts: one is the thermal strain produced in the unrestrained bar by the temperature
change ΔT (Fig. C(b)). From Eq. 10 we write
The other component of is associated with the stress due to the force RB
applied to the bar (Fig.
C(c)). From Hooke's law, we express this component of the strain as
Adding the two components of the strain in AC, we obtain
A similar computation yields the strain in portion CB of the bar:
The deformations δAC and δCB of the two portions of the bar are expressed respectively as
We thus check that, while the sum of the two deformations is zero, neither of the
deformations is zero.
Faculty Notes
1. Normal Strain
A solid body subjected to a change of temperature or to an external load
deforms. For example, while a specimen is being subjected to an increasing
force P as shown in Fig 1, a change in length of the specimen occurs between
any two points, such as A and B. Initially, two such points can be selected an
arbitrary distance apart. Thus, depending on the test, either 1-, 2-, 4-, or 8-in
lengths are commonly used. This initial distance between the two points is
called a gage length. In an experiment, the change in the length of this distance
is measured. Mechanical dial gages, such as shown in Fig. 1, have been largely
replaced by electronic extensometers for measuring these deformations. An
example of small clip-on extensometer is shown in Fig. 2.
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During an experiment, the change in gage length is noted as a function of
the applied force. With the same load and a longer gage length, a larger
deformation is observed, then when the gage length is small. Therefore, it is
more fundamental to refer to the observed deformation per unit of length of the
gage, i.e., to the intensity of deformation.
If Lo is the initial gage length and L is the observed length under a given load, the gage elongation ΔL = L -
Lo. The elongation and ε per unit of initial gage lengthis then given as
(1)
This expression defines the extensional strain. Since this strain is
associated with the normal stress, it is usually called the normal strain. It is a
dimensionless quantity, but it is customary to refer to it as having the
dimensions of in/in, rn/rn, or µm/m (microstrain). Sometimes it is given as a
percentage. The quantity generally is very small. In most engineering
applications of the type considered in this text, it is of the order of magnitude of
0.1 percent.
For small strains, this definition essentially coincides with the
conventional strain ε. If under the integral, the length L is set equal to Lo, the
strain definition given by Eq. 1 is obtained.
Natural strains are useful in theories of viscosity and viscoplasticity for
expressing an instantaneous rate of deformation. Natural strains are not
discussed elsewhere in this text.
Since the strains generally encountered are very small, it is possible to employ a highly versatile means for
measuring them, using expendable electric strain gages. These are made of very fine wire or foil that is glued to
the member being investigated. As the forces are applied to the member, elongation or contraction of the wires or
foil takes place concurrently with similar changes in the material. These changes in length alter the electrical
resistance of the gage, which can be measured and calibrated to indicate the strain taking place. Such gages,
suitable for different environmental conditions, are available in a range of lengths, varying from 4 to 150 mm (0.15
to 6 in). A schematic diagram of a wire gage is shown in Fig. 3 , and a photograph of a typical small foil gage is
shown in Fig. 4.
2. Stress-Strain Relationships
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In solid mechanics, the mechanical behavior of real materials under
load is of primary importance. Experiments, mainly tension or compression
tests, provide basic information on this behavior. In these experiments,
macroscopic (overall) response of specimens to the applied loads is observed
in order to determine empirical force-deformation relationships. Researchers in
material science attempt to provide reasons for the observed behavior.
It should be apparent from the previous discussion that for general purposes, it is more fundamental to
report the strain of a member in tension or compression than to report the elongation of its gage. Similarly, stress is
a more significant parameter than force since the effect on a material of an applied force P depends primarily on
the cross sectional area of the member. As a consequence, in the experimental study of the mechanical properties
of materials, it is customary to plot diagrams of the relationship between stress and strain in a particular test. Such
diagrams, for most practical purposes, are assumed to be independent of the size of the specimen and of its gage
length.
There are two ways in which these diagrams can be described. Both of them are discussed in this section.
Engineering Stress-Strain Diagrams Assuming that the stress is constant over the cross section of the central
portion of the specimen and along the gage length, the nominal or engineering stress, σ, can be determined. Thus,
dividing the applied force P by the specimen's original cross-sectional area A0.
(2)
Likewise, the nominal or engineering strain, ε, is found directly from the strain gage reading or by dividing
the change in the gage length ΔL by the specimen's original gage length L0 and applying Eq.1. Here the strain is
assumed to be constant throughout the gage length.
If the computed values of σ and corresponding ε are plotted on a graph, for which the ordinate is the stress
and abscissa is the strain, the resulting curve is called the engineering stress-strain diagram. This diagram is very
important in engineering since it provides the means for obtaining various mechanical properties of a material
without regard to its physical size or shape. As an example, the characteristics of the engineering stress-strain
diagram for ductile steel, a commonly used material for making structural members and mechanical elements, will
be discussed.
The general shape of the stress-strain diagram for a ductile steel specimen loaded in tension to failure for a
monotonically increasing load is well known from numerous tests. A plot of the normal stress σ versus engineering
strain ε, shown in Fig. 5, can be subdivided into four well-defined regions:
The linear elastic region1.
The yield plateau2.
The strain-hardening region3.
The postultimate stress or strain-softening region.4.
The linear elastic region 0 ≤ εs
≥ εy
of the stress-strain curve, Where εy
is the yield strain, is a straight
line (see Fig. 5).
In the yield plateau region εy < ε
s < ε
sh, where ε
sh is the strain at initiation of hardening strain, which
begins at the point A(εy ,
σy), the steel behaves plastically. This specific region of the stress-strain curve is
shown in the inset of Fig.5 and is assumed to be horizontal. The yield stress, σy
, corresponding to the idealized
yield plateau must therefore be taken as an arbitary average value within the range of this plateau.
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The point at which the yield plateau ends and strain hardening begins is not obvious. Before strainhardening initiates, a dip generadlly occurs in the yield plateau, followed by a steep increase that suddenlychanges slope into the relatively smooth strain-hardening region. The strain-hardening region (see Fig.5) ranges
from the idealized point B(εsh ,
σy), at which starin hardening begins, to the ultimate point C(ε
su , σ
su), that
corresponds to the moment at which the maximum tensile stress is resisted and the process of necking begins.Necking is displayed by contraction of the specimen, as shown in Fig.6.
In the postultimate region εs
≥ ε
su, the shape of stress-strain curve is related to the location and gage
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length over which experimental data are collected. Therefore, it is assumed that the ultimate point C(εsu ,
σsu
)
marks the end of useful region of the stress-strain curve.
In the past it was generally assumed that the monotonic stress-strain curve of ductile steel subjected tocompression is equal and opposite to the tension curve. However, the experimental data from monotonic testsshow that the tension and compression engineering stress-strain curves are practically coincident only when thestrain is small. The differences between the two diagrams, shown exaggerated in Fig.7, begin to appear in thestrain-hardening region, where the extent of the strain becomes more pronounced, when necking/barrelingdevelops in the tensile/compression test.
True Stress-Strain Diagrams In some engineering applications (for example, in metal forming), the strains maybe large. For such purposes teh total strain is defined as the sum of incremental strains Δ ; thus
(3)
where L is the current gage length of the specimen when the increment of elongation (contraction) ΔL occurs. If L0
is the initial gage length of the specimen, then in the limit as ΔL → 0 the strain corresponding to the gage lengthL
f can be defined by the following integral:
(4)
This strain, obtained by adding up the increments of strains, which are based on the current dimensions of aspecimen, is called a natural or true strain. Sometimes the true strain is called logarithmic strain because of theform of Eq.4.
For small strains, the true strain defined by Eq.4 essentially coincides with the engineering strain ε. If,under the integral, the length L is set equal to L
0, the strain definition given by Eq.4 is obtained.
During plastic strain of a uniform specimen subjected to axial tension (compression), the cross-sectionalarea gets smaller (larger) as the specimen elongates (shortens). A more accurate description of the actual stress
experienced by the specimen can be given by the true stress concept. The true stress, is related to theinstantaneous cross-sectional area, A, and the applied force F as
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(5)
Since plastic strain involves no volume change--that is, A0 L0 = AL and L = L
0(1+ ε )
(6)
which, using Eq.2 and noting that F = P, allows to relate the true stress and engineering stress as follows:
(7)
If the value of the is so defined and the corresponding are plotted on a graph, for which the ordinate is
the true stress and the abscissa is the true strain, the resulting curve is called the true stress-strain diagram. The
true stress-strain diagrams for ductile materials (such as a ductile steel), the compression, and the tension true
stress-strain diagrams practically coincide, whereas the two engineering stress-strain diagrams drift apart.
In Fig. 7, in the same quadrant, compression and tension stress-strain diagram are illustrated for a
monotonic test of a ductile steel specimen plotted in true and engineering coordinate systems. As can be seen,
both compression and tension true stress-strain diagrams are similar until the effect of bucking becomes
noticeable at a strain level of approximately 6% in the compression test. Comparision of the engineering and the
true stress-strain diagrams shows that in tension, since the cross-sectional area decreases as the specimen
elongates, the true stress is grated than engineering stress, whereas in compression as the specimen shortens,
the cross-sectional area increases and thus the true stress is less than the corresponding engineering stress.
It is important to recognize that experimentally determined stress-strain diagrams differ widely for different
materials. Even for the same material they differ depending on the temperature at which the test was conducted,
the speed of the test, and a number of other variables. Conventional stress-strain diagrams for a few representative
materials are illustrated in Figs. 8 and 9. These are shown to larger scale in Fig. 9, particularly for strain.
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Since for most engineering applications,
deformations must be limited, the lower range of
strains is particularly important. The large
deformations of materials in the analysis of such
operations as forging, forming, and drawing are not
pursued.
In calculating engineering stress using Eq.2,
the original cross-sectional Ao is generally
designated by A.
An illustration of fractured tension specimens
after static tension tests, i.e., where the loads were
gradually applied, is shown in Fig. 10. Steel and
aluminum alloy specimens exhibit ductile behavior, and a fracture occurs only after a considerable amount of
deformation. This behavior is clearly exemplified in their respective stress-strain diagrams; see Fig. 9. These
failures occur primarily due to slip in shear along the planes forming approximately 45° angles with the axis of the
rod. A typical “cup and cone” fracture may be detected in the photographs of steel and aluminum alloy specimens.
By contrast, the failure of a castiron specimen typically occurs very suddenly, exhibiting a square fracture across
the cross section. Such cleavage or separation fractures are typical of brittle materials.
3. Hooke's Law
For a limited range from the origin, the experimental values ot stress vs. strain lie essentially on a straight
line. This holds true almost without reservations for the entire range for glass at room temperature. It is true for mild
steel up to some point, as A in Fig.5. It holds nearly true up to very close to the failure point for many high-grade
alloy steels. On the other hand, the straight part of the curve hardly exists in concrete, soil, annealed copper,
aluminum, or cast iron. Nevertheless, for all practical purposes, up to some such point, such as A, also in Fig. 11,
the rela tionship between stress and strain may be said to be linear for all materials. This sweeping idealization
and generalization applicable to all materials is known as Hooke's law. is Symbolically, this law can be expressed
by the equation
(8)
which simply means that stress is directly
proportional to strain, where the constant of proportionality
is E. This constant E is called the elastic modulus, modulus
of elasticity, or Young’s modulus. As ε is dimensionless, E
has the units of stress in this relation. In the U.S. customary
system of units, it is usually measured in pounds per square
inch, and in the SI units, it is measured in newtons per
square meter (or pascals).
Graphically, E is interpreted as the slope of a straight
line from the origin to the rather vague point A on a uniaxial
stress-strain diagram. The stress corresponding to the latter
point is termed the proportional or elastic limit of the
material. Physically, the elastic modulus represents the
stiffness of the material to an imposed load. The value of
the elastic modulus is a definite property of a material. From
experiments, it is known that is always a very smnall
quantity; hence, E must be large. Its approximate values are
tabulated for a few materials in Tables 1A and B of the
Appendix. For all steels, E at room temperature is between
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29 and 30 x 106 psi, or 200 and 207 GPa.
It follows from the foregoing discussion that Hooke’s law applies only up to the proportional limit of the
material. This is highly significant as in most of the subsequent treatment, the derived formulas are based on this
law. Clearly, then, such formulas are limited to the material’s behavior in the lower range of stresses.
Some materials, notably single crystals and wood, possess different elastic moduli in different directions.
Such materials, having different physical properties in different directions, are called anisotropic. A consideration of
such materials is excluded from this text. The vast majority of engineering materials consist of a large number of
randomly oriented crystals. Because of this random orientation, properties of materials become essentially alike in
any direction. Such materials are called isotropic. .With some exceptions, such as wood, in this text, complete
homnogeneity (sameness from point to point) and isotropy of materials is generallY assumed.
4. Further Remarks on Stress-strain Relationships
In addition to the proportional limit defined in Section-3, several
other interesting points can be observed on the stress-strain diagrams.
For instance, the highest points (see the ultimate stress point C in Fig. 5)
correspond to the ultimate strength of a material. Stress associated with
the long plateau of the stress-strain curve (see the inset of Fig. 5) is called
the yield strength of a material. As will be brought out later, this
remarkable property of mild steel, in common with other ductile materials,
is significant in stress analysis. For the present, note that at an essentially
constant stress, strains 15 to 20 times those that take place up to the
proportional limit occur during yielding. At the yield stress, a large amount
of deformation takes place at a constant stress. The yielding phenomenon
is absent in most materials.
A study of stress-strain diagrams shows that the yield strength
(stress) is so near the proportional limit that, for most purposes, the two
may be taken to be the same. However, it is much easier to locate the
former. For materials that do not possess a well-defined yield strength, one is sometimes “invented” by the use of
the so-called “offset method.” This is illustrated in Fig. 12, where a line offset an arbitraiy amount of 0.2 percent of
strain is drawn parallel to the straight-line portion of the initial stress-strain diagram. Point C is then taken as the
yield strength of the material at 0.2-percent offset.
That a material is elastic usually implies that stress is directly proportional to strain, as in Hooke’s law. Such
materials are linearly elastic or Hookean. A material responding in a nonlinear manner and yet, when unloaded,
returning back along the loading path to its initial stress-free state of deformation is also an elastic material. Such
materials are called nonlinearly elastic. The difference between the two types of elastic materials is highlighted in
Figs. 13(a) and (b). if in stressing a material its elastic limit is exceeded, on unloading it usually responds
approximately in a linearly elastic manner, as shown in Fig. 13(c), and a permanent deformation, or set, develops
at no external load. The area enclosed by the loop corresponds to dissipated energy released through heat. Ideal
elastic materials are considered not to dissipate any energy under monotonic or cyclic loading.
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For ductile materials, stress-strain diagrams obtained for short compressions blocks are reasonably close to
those found in tension. Brittle materials, such as cast iron and concrete are very weak in tension but not in
compression. For these materials, the diagrams differ considerably, depending on the sense of the applied force.
It is well to note that in some of the subsequent analyses, it will be advantageous to refer to elastic bodies
and systems as springs. Sketches such as shown in Fig. 14 are frequently used in practice for interpreting the
physical behavior of mechanical systems.
1.6. Poisson’s Ratio
In addition to the deformation of materials in the direction of the applied
force, another remarkable property can be observed in all solid materials, namely,
that at right angles to the applied force, a certain amount of lateral (transverse)
expansion or contraction takes place. This phenomenon is illustrated in Fig. 15,
where the deformations are greatly exaggerated. For clarity, this physical fact may
be restated thus: if a solid body is subjected to an axial tension, it contracts
laterally; on the other hand, if it is compressed, the material “squashes out”
sideways. With this in mind, directions of lateral deformations are easily
determined, depending on the sense of the applied force.
For a general theory, it is preferable to refer to these lateral deformations on
the basis of deformations per unit of length of the transverse dimension. Thus, the
lateral deformations on a relative basis can be expressed in in/in or m/m. These
relative unit lateral deformations are termed lateral strains. Moreover, it is known
from experiments that lateral strains bear a constant relationship to the longitudinal
or axial strains caused by an axial force, provided a material remains elastic and is
homogeneous and isotropic. This constant is a definite property of a material, just
like the elastic modulus E, and is called Poisson’s ratio. It will be denoted by (nu)
and is defined as follows:
(9)
where the axial strains are caused by uniaxial stress only, i.e., by simple tension or compression. The
second, alternative form of Eq. (9) is true because the lateral and axial strains are always of opposite sign for
uniaxial stress.
The value of fluctuates for different materials over a relatively narrow range. Generally, it is on the order
of 0.25 to 0.35. In extreme cases, values as low as 0.1 (some concretes) and as high as 0.5 (rubber) occur. The
latter values is the largest possible. It is normally attained by materials during plastic flow and signifies constancy
of volume. In this text, Poisson’s ratio will be used only when materials behave elastically.
In conclusion, note that the Poisson effect exhibited by materials causes no additional stresses other than
those considered earlier unless the transverse deformation is inhibited or prevented.
1-9. Thermal Strain and Deformation
With changes of temperature, solid bodies expand on increase of temperature
and contract on its decrease. The thermal strain εT caused by a change in temperature
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from T0 to T measured in degrees Celsius or Fahrenheit, can be expressed as
εT = α(T – T
o)
(10)
Where α is an experimentally determined coefficient of linear thermal
expansion. For moderately narrow ranges in temperature, α remains reasonably
constant.
Equal thermal strains develop in every direction for unconstrained
homogeneous isotropic materials. For a body of length L subjected to a uniform
temperature, the extensional deformation ΔT due to a change in temperature of δT = T – T
o is
ΔT = α(δT)L (11)
For a decrease in temperature, δT assumes negative values.
An illustration of the thermal effect on deformation of bars due to an increase in temperature is shown in Fig.
16.
Professor Notes
Mechanical properties of material:
Properties of material are
· Elasticity = mild steel, ss, alloy steels
· Plasticity – Polymers, rubber, steel beyond elastic limit
· Brittleness – Glass, cast iron
· Malleability – Cu, tin, CRCA sheet, silver, gold
· Ductility – Cu, Al, silver, gold
Stress strain diagram of mild steel of medium strength of axially load bar
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- Stress reduce to maintain equilibrium in AB, quickly
- In BC for same stress strain continuous. A to B to C is called yielding without increase in stress strainincreases.
- At C hardening of material takes place requiring increase in stress for strain to increase, but this is a nonlinear length change.
- D is the hight stree reached by the member of given area(original) stress at 'D' is called the ultimatestress. At Ultimate stree point waist necking stars to form at D forms.
- At point E material fails
Properties of engineering materials:
Working stress is the stress the material is subjected to allowable stress or with the load.
Permissible stress: of material is highest stress allowed to junction
Working stress to be < Permissible stress to function safely
Permissible stress may be elastic limit, ultimate stress
Example 1
Yielded at F = 82.5 KN specimen attained a maximum load of
155KN and Ultimately broke at 72.5 KN
Find : (i) Tensile strength at yield point
(ii) Ultimate stress
(iii) Average stress at breaking if dia. of neck
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Original area of specimen section =
Tensile stress (strength ) at yield =
(Used for design Estimates)
Ultimate stree (strength ) =
(Used for factor of safety)
Actually, at breaking point though load is less compared to ultimate
Example 2
A bar subjected to tensile test yielded at 47.25 KN
Find:
(i) Tensile Stress at yield point
Original area =
Tensile Stress=
(ii) Ultimate stress at point breaking
Ultimate stress =
(iii) Average stree at break point y neck dil is 8.05 mm
Average stree at break-point =
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