document23

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

Subject/Unit Name/Module Name http://10.20.3.1:8080/engineering/I_YEAR/EM/M23...

1 of 19 Friday 19 August 2011 10:01 AM

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.



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



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.



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)



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.



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:



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.



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



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.



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



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



(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.



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



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.



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



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



-

- 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



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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