1 Introduction to Stress and Strain

DMV 4343JAN ~ JUN ‘07

DEPARTMENT MANUFACTURING / PRODUCT DESIGN / MOULD / TOOL AND DIE

SEMESTER 4 / 6

COURSE MECHANICS OF MATERIALS DURATION 8 hrsCOURSE CODE DMV 4343 / DMV 5343 REF. NO.VTO’S NAME MISS AFZAN BINTI ROZALI PAGE 29

TOPICINTRODUCTION TO STRESS AND STRAIN

SUB TOPIC1.1 Normal Stress and Strain 1.2 Stress – Strain Relationship and Diagram 1.3 Elasticity and Plasticity1.4 Linear Elasticity and Hooke’s Law1.5 Allowable Stress and Allowable Load1.6 Deformation of Axially Loaded Member

Chapter 1 INTRODUCTION TO STRESS AND STRAIN p0

INFORMATION SHEET

REF NO. :PAGE : 29

DMV 4343JAN ~ JUN `07

INTRODUCTION TO STRESS AND STRAIN

Figure 1.0a shows an L-shaped bracket loaded as a two-force member, and Figure 1.0b

shows the internal resultants, F(x), V(x), and M(x), that are required to maintain the

equilibrium of the two sectioned parts of the bracket. Although we could compute the

internal resultants shown on Figure 1.0b by using static equilibrium procedures (free-body

diagrams and equations of equilibrium), those procedures are clearly insufficient for

determining the complex internal force distribution making up those resultants. The concept

of stress is introduced in this chapter to enable us to quantify internal force distributions.

The shape of the bracket also changes due to the applied loads; that is, the member

deforms. The concept of strain is introduced to permit us to give a detailed analytical

description of such deformation. Finally, stress and strain are related to each other. This

relationship, which depends on the material(s) used in the fabrication of the member, must

be determined by performing certain stress-strain tests, which are described in this chapter.

(a) A two-force member (b) Internal resultants: F, V, and M

FIGURE 1.0 An illustration of internal resultants.

1.1 Normal Stress and strain

Equal and opposite forces of magnitude P acting on a straight bar cause it to elongate, and

also to get narrower, as can be seen by comparing Figure 1.1a and 1.1b. The bar is said to

be in tension. If the external forces had been applied in the opposite sense, that is, pointing

toward each other, the bar would have shortened and would then be said to be in

compression.



FIGURE 1.1

A straight bar

undergoing axial

loading, (a) The

undeformed bar, with

vertical lines indicating

cross sections. (b) The

deformed bar. (c) The

distribution of internal

force at section A. (d)

The distribution of

internal force at section

B.

Definition of Normal Stress. The thin red arrows in Figure 1.1c and 1.1d represent the

distribution of force on cross sections at A and B, respectively. (A cross section is a plane

that is perpendicular to the axis of the bar.) Near the ends of the bar, for example at section

A, the resultant normal force, FA, is not uniformly distributed over the cross section; but at

section B, farther from the point of application of force P, the force distribution is uniform. In

mechanics, the term stress is used to describe the distribution of a force over the area

on which it acts and is expressed as force intensity, that is, as force per unit area.

Stress =Force

Area

The units of stress are units of force divided by units of area. In the U.S. Customary

System of units (USCS), stress is normally expressed in pounds per square inch (psi) or in

kips per square inch, that is, kilopounds per square inch (ksi). In the International System of

units (SI), stress is specified using the basic units of force (newton) and length (meter) as

newtons per meter squared (N/m2). This unit, called the pascal (1 Pa = 1 N/m2), is quite

small, so in engineering work stress is normally expressed in kilopascals (1 kPa = 103 N/

m2), megapascals (1 MPa = 106 N/ m2), or gigapascals (1 GPa = 109 N/ m2). For example, 1

psi = 6895 Pa = 6.895 kPa.

There are two types of stress, called normal stress and shear stress. In this section we will

consider only normal stress; shear stress is introduced in Chapter 2. In words, normal stress

is defined by

Normal stress = Force normal (i.e., perpendicular) to an area



Area on which the force acts

The symbol used for normal stress is the lowercase Greek letter sigma (σ). The

normal stress at a point is defined by the equation

σ(x, y, z) =lim

ΔA →0( ΔF

ΔA) Normal

Stress1.1

where, as shown in Figure 1.2a, ΔF is the normal force (assumed positive in tension) acting

on an elemental area ΔA containing the point (x, y, z) where the stress is to be determined.

The sign convention for normal stress is as follows:

A positive value for σ indicates tensile stress, that is, the stress due to a force

ΔF that pulls on the area on which it acts.

A negative value for σ indicates compressive stress.

(a) Distributed normal

stress on a cross section

(b) Resultant of distributed

normal stress in (a)

FIGURE 1.2 Normal force on a cross section

Thus, the equation σ = 6.50 MPa signifies that σ is a tensile stress of magnitude 6.50 MPa,

or 6.50 MN/m2, and the equation σ = -32.6 ksi indicates a compressive stress of magnitude

32.6 kips/in2.

Average Normal Stress. Even when the normal stress varies over a cross section, as it

does in Figure 1.1c, we can compute the average normal stress on the cross section by

letting

σavg =F

A

Average

Normal

Stress

1.2

Thus, for Figures 1.1c and 1.1d we get

(σavg)A =F A

A=

P

A, (σavg)B =

F B

A=

P

A

Much of the rest of this textbook is devoted to determining how stress is distributed on cross

sections of structural members under various loading conditions. However, in many



situations the normal stress on a cross section is either constant or very nearly constant, as

in the next two examples.



EXAMPLE 1.1

In the evening, a contractor attaches a steel wire to an eyebolt

at point A on his air compressor (Figure 1), and, with the boom of

his construction crane, he raises the compressor a safe distance

above the ground to prevent mischief from being done to the

compressor overnight. While the compressor is above the

ground, the wire will be in tension, and the normal stress on any

cross section of the wire, except near ends A and B, can be

assumed to be constant. If the compressor weighs 600 N and

the diameter of the wire is d = 10 cm, what is the average

tensile stress in the wire?

FIGURE 1

Solution From the free-body diagram in Figure 2, the tensile

force in the wire is equal to the weight of the compressor,

so

∑ F = 0 F = 600 N

and from Eq. 1.2,

σavg = F/A = F/(πd2/4) = 4(600N)/π(0.1m)2 = 76,394 N/m2

Rounded to three significant figures, the average normal stress

on typical cross section of the wire is

σavg = 76.4 kPa (T)

FIGURE 2 Free body

diagram

where the (T) stands for tension. Actually, this is quite a high

stress, as you will discover in Section 1.4. Figure 3 illustrates

how this average normal stress would be distributed over a

cross section of the wire.

FIGURE 3 Tensile stress

on a typical cross section



EXAMPLE 1.2

The Washington Monument (Figure 1a) stands 555 m

high and weighs 181,700 kN. The monument was made

from over 36,000 blocks of marble and granite. As shown

in Figure 1b, the base of the monument is a square that

is 665.5 m long on each side, and the stone walls at the

base are 180 m thick.

Determine the compressive stress that the foundation

exerts over the cross section at the base of the monument,

assuming that this normal stress is uniform.FIGURE 1

Solution From the free-body diagram in Figure 2, the

total normal force on the base of the monument is equal

to negative of the weight of the monument,

so

∑ F = 0 F = -181,700 kN

(Note: In accordance with the sign convention for normal

stress, the normal force F is taken positive in tension. The

negative value for F indicates that it is a compressive

force, as is clearly evident in this case.) The cross-

sectional area of the base is

A =(665.5 m)2 - (665.5 m - 360

m)2

= 349,600 m2

Therefore, from Eq. 1.2,

σavg =F

A=

-181,700kN

349,600 m2= - 519.8 Pa

Rounded to three significant figures, the average normal

stress on the cross section of the monument at its base is

σavg = 520 Pa (C)

FIGURE 2 Free body diagram

where the (C) stands for compression. This is a very low

stress, even for stone. Figure 3 illustrates how this uniform

compressive stress would be distributed over the

foundation at the base of the monument.

FIGURE 3 Compressive stress

at the base



The compressive stress that results when one object bears on another, like the stress that

the monument exerts on the foundation in the above example problem, is frequently called

bearing stress. Bearing stress is just a special case of compressive normal stress.

Stress Resultant.

Given the distribution of normal stress on a cross section, σ = σ (x, y, z), we can integrate

over the cross section to determine the magnitude and point of application of the resultant

normal force:

The two moment equations are used to locate the line of action of the force F(x). Note that

the sign convention for σ implies that the force F in Eq. 1.3 is to be taken positive in tension.

This is the reason that we will consistently, take normal force resultants to be positive in

tension.

Resultant of Constant Normal Stress on a Cross Section: Let us determine the resultant of

normal stress on the cross section at x (Figure 1.2a) if the normal stress is constant over the

cross section. We will prove that normal stress that is constant on a cross section

corresponds to an axial force F(x) = A σ (x) acting through the centroid of the cross section

at x.

Let the resultant be assumed to be a force F(x) acting parallel to the x axis and passing

through point (yR, ZR), as in Figure 1.2b. We must show that

F(x

)

= A(x) σ (x)

,

yR =

y,

ZR =

Z

For this we can use Eqs. 1.3. Substituting the condition σ (x, y, z) = σ (x) into Eqs. 1.3, we get


∑ Fx : F(x) = ∫Aσ dA

∑ My : zR F(x) = ∫Azσ dA

1.3

∑ Mz : -yR F(x) = -∫Ayσ dA

∑ Fx : F(x) = σ (x) ∫A dA σ (x)dA

∑ My : zR F(x) = σ (x) ∫A zdA σ (x)z A

∑ Mz : -yR F(x) = - σ (x) ∫A ydA -σ(x) yA


For generality, the normal force has been permitted to be a function of x in Figure 1.2b and in

Eqs. 1.3. Of course, F(x) = P = const in the axial-loading case illustrated in Figure 1.1.

Therefore, if the normal stress is uniform over a cross section, the normal stress on the

cross section, also called the axial stress, is given by

σ

(x)=

F(x)

A(x)

Axial-

Stress

Equations

1.4

and corresponds to a force F(x) (tension positive) acting at the centroid of the cross section,

that is, at ZR = z, yR = y.

We would certainly expect uniform stress on a circular rod to correspond to a force acting

along the axis of the rod, and similarly for a square or rectangular bar. Hence, it is

"reasonable" that a uniform normal stress distribution acting over a cross section of general

shape produces a resultant force acting through the centroid of the cross section. In most

cases, the cross-sectional area is constant throughout the length of the member, but Eq. 1.4

may also be used if the cross-sectional area varies slowly with x.

Uniform Normal Stress in an Axially Loaded Bar: Under certain assumptions, an axially

loaded bar will have the same uniform normal stress on every cross section; that is, σ (x, y,

z) = σ = constant. These assumptions are:

The bar is prismatic; that is, the bar is straight and it has the same cross section

throughout its length.

The bar is homogeneous; that is, the bar is made of the same material

throughout.

The load is applied as equal and opposite uniform stress distributions over

the two end cross sections of the bar.

So long as the resultant force at each end of the bar is applied at the centroid of the end

cross section, the last assumption—that the loads are applied as uniform normal stress

distributions on the end cross sections—can be relaxed. As illustrated in Figure 1.1 a-d, the

stress is uniform on every cross section, except on cross sections that are very near the

points of application of load. This is an application of Saint-Venant's Principle.



The uniform, prismatic bar in Figure 1.3a is labeled as member "i" and is subjected to equal

and opposite axial forces Fi, acting through the centroids at its ends. Its cross-sectional area

is Ai.

The normal stress on cross sections of an axially loaded member, like the one in Figure 1.3,

is called the axial stress. Since, from the free-body diagram in Figure 1.3b, the resultant

force, F(x), on every cross section of the bar is equal to the applied load Fi, and since the

cross-sectional area is constant, from Eq. 1.4 we get the following formula for the uniform

axial stress:

σi =Fi

Ai

= constant

Axial-

Stress

Equations

1.5

FIGURE 1.3 Uniform stress in an axially loaded prismatic bar

Example 1.3 shows one application of the axial-stress equation.



EXAMPLE 1.3

Two solid circular rods are welded to a plate at B

to form a single rod, as shown in Figure 1.

Consider the 30-kN force at B to be uniformly

distributed around the circumference of the collar

at B and the 10 kN load at C to be applied at the

centroid of the end cross section. Determine the

axial stress in each portion of the rod. FIGURE 1

Plan the Solution

Since each segment of the rod satisfies the

conditions for uniform axial stress, we can use

Eq. 1.5 to calculate the two required axial

stresses. First, however, we need to compute the

force in each rod by using an appropriate free-

body diagram and equation of equilibrium. FIGURE 2

Solution

Free-body Diagrams: First we draw free-body diagrams that expose the rod forces F1 (or

FAB) and F2 (or FBC). We show F1 and F2 positive in tension.

Equations of Equilibrium: From free-body diagram 1 (Figure 2a),

And, from free-body diagram 2 (Figure 2b),

Axial Stresses: Using Eq. 1.4, we obtain the axial stresses

σ1 =F1

A1

=- 20 kN

341.2 mm2= - 63.7 MN / m2 @ 63.7 MPa (C)

σ2 =F2

A2

=10 kN

176.7 mm2= 56.6 MN / m2 @ 56.6 MPa (T)


+

→∑ (F)1 = 0: - F1 - 30kN + 10kN = 0, F1 = - 20 kN

+

→∑ (F)2 = 0: - F2 + 10kN = 0, F2 = 10 kN

A1 =Π d 1

2

4=

Π (20 mm) 2

4= 314.2 mm2

A2 =Π d 2

2

4=

Π (15 mm) 2

4= 176.7 mm2


The free-body diagrams in Figure 2 of Example 1.3 illustrate the application of the method of

sections to structures that have axially loaded segments that are collinear. For example, the

free-body diagram in Figure 2a makes it possible with a single free-body diagram to relate

the internal force F1 to the external forces applied at the two nodes (joints) B and C; this is

the most efficient way to determine F1.

STRAIN

Figure 1.4a illustrates an underformed tension specimen with two points on the specimen

marking the original gage length, L0. The notation L0 is used here to emphasize that this is

the original gage length, not the total length of the specimen. An axial load P causes the

portion of the specimen between the gage marks to elongate, as indicated in Figure 1.4b.

(a) Undeformed

specimen.

(b) Deformed

specimen.

FIGURE 1.4 Tension and compression test specimens.

As the specimen is pulled, the load P is measured by the testing machine and recorded. The

extensometer provides a simultaneous measurement of the corresponding length. L* =

L*(P), of the test section, or else it directly measures the elongation

ΔL = L* - L0 1.6

In a static tension test the length of the specimen is increased very slowly, in which case the

loading rate need not be measured. In some situations, however, a dynamic test must be

performed. Then, the rate of loading must be measured and recorded, since material

properties are affected by high rates of loading.

Є =L* - L 0

L0

Normal

Strain



1.2 Stress – Strain Relationship and Diagram

In order to relate the loads on engineering structures to the deformation produced by the

loads, experiments must be performed to determine the load-deformation behavior of the

materials (e.g., aluminum, steel, and concrete) used in fabricating the structures. Many

useful mechanical properties of materials are obtained from tension tests or from

compression tests, and these properties are listed in tables like those in Appendix F. This

section describes how a tension test is performed and discusses the material properties that

are obtained from this type of test.

Tension Tests and Compression Tests.

Figure 1.5 shows a computer-controlled, hydraulically actuated testing machine that may be

used to apply a tensile load or a compressive load to a test specimen, like the steel tension

specimen in Figure 1.6a or the concrete compression specimen in Figure 1.6c. Figure 1.6b

shows a close-up view of a ceramic tension specimen mounted in special testing-machine

grips. Electromechanical extensometers are mounted on the specimens in Figure 1.6a and

1.6b to measure the extension (i.e., the elongation) that occurs over the gage length of the

test section.

FIGURE 1.5 A computer-controlled hydraulically actuated testing machine, (MTS Systems Corp. photo.)



(a) A metal tension

specimen with

extensometer attached

(b) A ceramic tension

specimen with

extensometer attached.

(MTS Systems Corp.

photo.)

(c) A concrete cylinder

before and after

compression testing.

FIGURE 1.6 Tension and compression test specimens.

Stress-Strain Diagrams.

A plot of stress versus strain is called a stress-strain diagram, and from such stress-strain

diagrams we can deduce a number of significant mechanical properties of materials.1" The

values of normal stress and extensional strain that are used in plotting a conventional stress-

strain diagram are the engineering stress (load divided by original cross-sectional area of the

test section) and engineering strain (elongation divided by original gage length), that is,

σ =P

A0

, Є =L* - L 0

L0

1.7

Mechanical Properties of Materials.

Figures 1.7a and 1.7b are stress-strain diagrams for structural steel (also called mild steel,

or low-carbon steel), which is the metal commonly used in fabricating bridges, buildings,

automotive and construction vehicles, and many other machines and structures. A number

of important mechanical properties of materials that can be deduced from stress-strain

diagrams are illustrated in Figure 1.7. In Figure 1.7a the stress is plotted accurately, but the

strain is plotted to a variable scale so that all important features can be shown and

discussed. In Figure 1.7b, which gives typical numerical values of stress and strain for

structural steel, one stress-strain curve, the lower one is plotted against a strain scale that



emphasizes the low-strain region; the upper curve is plotted against a strain scale that

emphasizes the high-strain region and puts the entire stress-strain history into perspective.

Starting at the origin A in Figure 1.7a and continuing to point B, there is a linear relationship

between stress and strain. The stress at point B is called the proportional limit, σPL.

(a) Strain not plotted to scale

(b) Strain plotted to two different scales

FIGURE 1.7 Stress-strain diagrams for structural steel in tension.

The ratio of stress to strain in this linear region of the stress-strain diagram is called Young's

modulus, or the modulus of elasticity, and is given by

E =Δσ

Δ Є , σ < σPL

Young’s

Modulus 1.12



Typical units for E are ksi or GPa.

At B the specimen begins yielding, that is, smaller and smaller increments at load are

required to produce a given increment of elongation. The stress at C is called the upper

yield point, (σYP)u, while the stress at D is called the lower yield point, (σYP)l. The upper

yield point has little practical importance, so the lower yield point is usually referred to simply

as the yield point, σYP. From D to E the specimen continues to elongate without any increase in

stress. The region DE is referred to as the perfectly plastic zone. The stress begins to

increase at E, and the region from E to F is referred to as the zone of strain hardening. The

stress at F is called the ultimate stress, or ultimate strength, σF. At F the load begins to drop,

and the specimen begins to "neck down." This neck-down behavior continues until, at G, fracture

occurs at the fracture stress, σF. Figure 1.8a, shows a hot-rolled steel specimen at three stages of

tensile testing: (1) before testing, (2) as removed from the testing machine at a point between F and

G with pronounced reduction in area referred to as necking or neck-down, and (3) after fracture.

Figure 1.8 b shows the typical cup and cone fracture of a hot-rolled steel tensile specimen.

(a) (b)

FIGURE 1.8 A hot-rolled steel tensile specimen

The true stress, σtrue, is the load at some instant during the test divided by the actual

minimum cross-sectional area of the specimen at that instant. Thus, when a specimen

starts to neck down, the true stress is taken as the load divided by the minimum cross-

sectional area in the neck-down region. The true strain, Єtrue, is the instantaneous change

in length of a test section divided by the instantaneous length of that test section .

True stress and true strain are given by the formulas

σtrue =P

Amin

, Єtrue = ln (1 + Є) 1.13



True strain can also be expressed in terms of area change

Єtrue = ln ( A0

Amin

)

The solid-line curve in Figure 1.7a is a conventional stress-strain diagram of engineering

stress versus engineering strain, while the dashed curve is a sketch of true stress versus

true strain. The curves differ only when strain is large and when the cross-sectional area is

decreasing significantly.

Design Properties.

Now that you have some idea of the stress-strain behavior of several common metallic

materials, let us note the material properties that are of primary interest to an engineer

designing some structure or machine. From the design standpoint the most significant

stress-strain properties can be categorized under the three headings—strength, stiffness,

and ductility.

Strength—There are three strength values of interest,

1. The yield strength, σY, is the highest stress that the material can withstand

without undergoing significant yielding. The yield-point stress or the offset

yield stress, whichever is appropriate for the particular material, is taken as

the yield strength (i.e., σY = σYP or σY = σYS as appropriate).

2. The ultimate strength, σU is the maximum value of stress (i.e., the maximum

value of engineering stress) that the material can withstand. Finally,

3. The fracture stress, σF, if different from the ultimate stress, may be of

interest. It is the value of the stress at fracture.

Stiffness—The stiffness of a material is basically the ratio of stress to strain.

Stiffness is of interest primarily in the linearly elastic region; therefore, Young's

modulus, E, is the value used to represent the stiffness of a material.

Ductility—Materials that can undergo large strain before fracture are classified as

ductile material; those that fail at small values of strain are classified as brittle

materials. Strictly speaking, the terms ductile and brittle refer to modes of fracture,

and a material like structural steel, which behaves in a ductile manner at room

temperature, may exhibit brittle behavior at very low temperatures. Therefore, when

we speak of a "brittle material" or a "ductile material," we are referring to the normal

(room temperature) behavior of the material.

Ductility. The two commonly used measures of ductility are the percent elongation (the final

elongation expressed as a percentage of the original gage length) and the percent reduction



in area at the section where fracture occurs (the area reduction expressed as a percentage

of the original area). The percent elongation is given by the formula

Percent elongation =LF - L 0

L0

(100%)

where L0 is the original gage length and LF is the length of the gage section at fracture.

When percent elongation is stated, the gage length should also be stated, since the

elongation at fracture is not uniform over the entire gage length but is concentrated in the

necked-down region. The percent reduction in area is given by the formula

Percent reduction in area =A0 – A F

A0

(100%)

FIGURE 1.9 A superplastically deformed steel-alloy specimen tested over 1100%

elongation. (Courtesy of Prof. Oleg D. Sherby, Standford University).



1.3 Elasticity and Plasticity

In the previous section we considered the stress-strain behavior of tension specimens under

loading. Specifically, the tensile tests were assumed to be performed in a relatively short

time with monotonically increasing tensile strain. We consider now what happens when the

loading is reversed, that is, when the strain is allowed to decrease.

Elastic Behavior and Plastic Behavior.

Consider the loading and unloading behavior of a material, as illustrated in Figure 1.10. The

upward-pointing arrows in Figure 1.10a and Figure 1.10b indicate the loading curves, that is,

the curves that would be followed for monotonically increasing initial loading. The stress-

strain behavior of a material is said to be elastic if the unloading path retraces the

loading path. The stress at C is called the elastic limit, σEL. Since the stress-strain curve

from the origin A in Figure 1.10a to the elastic limit at C is not a straight line, material

behavior in this range is termed nonlinear elastic behavior. Unloading from a point below

point C (e.g., point B) retraces the loading path indicated in Figure 1.10a, but unloading from

a point beyond C (e.g., point D) follows a path DE that is different than the loading path.

Unloading from a point on the stress-strain curve beyond the elastic limit typically follows a

straight-line path whose slope is parallel to the tangent of the stress-strain curve at the

origin. The strain that remains at point E when the stress returns to zero is called the

permanent set, or residual strain, as illustrated in Figure 1.10b.

FIGURE 1.10 Illustrations of elastic and plastic stress-strain behaviour.



1.4 Linear Elasticity and Hooke’s Law

For structural steel the elastic limit is very close to the proportional limit and also to the yield

point. Therefore, for structural steel the proportional limit stress and the elastic limit stress

are usually assumed to be equal to the yield-point stress. It is sometimes convenient to

approximate the stress-strain behavior of mild steel and similar materials by the linearly

elastic, perfectly plastic representation of Figure 1.11.

FIGURE 1.11 Linearly elastic, perfectly plastic material model.

In order to keep deformations small and stresses at safe levels, most structures and

machine parts are designed so that stresses remain well below the yield stress. Fortunately,

most engineering materials exhibit a linear stress-strain behavior at these lower stress

levels.

Hooke's Law.

Let us, for the present, restrict our discussion to the case of uniaxial stress applied to a

homogeneous (same properties throughout), isotropic (same properties in every direction)

member oriented along the x axis (Figure 1.12). The linear relationship between stress and

strain, given by Eq. 2.12, applies for 0 ≤ σ ≤ σY. Therefore,

σx = EЄx Hooke’s Law 1.11

This equation is called Hooke's Law." The subscripts on σ and Є identify the axis of the

particular stress and strain.



FIGURE 1.12 The deformation (much exaggerated)

of a homogeneous, isotropic specimen under uniaxial

stress.

Hooke's Law is valid for uniaxial tension or compression within the linear portion of the

stress-strain diagram. As noted earlier, E is called the modulus of elasticity, or Young's

modulus. It has the units of stress, typically ksi (or psi) in U.S. Customary units, and GPa (or

MPa) in SI units. Representative approximate values of E are: 30 X 103 ksi (200 GPa) for

steel, 10 X 103 ksi (70 GPa) for aluminum, and 300 ksi (2 GPa) for nylon. Values of the

modulus of elasticity for selected materials are listed in Table F.2.

Poisson's Ratio.

Associated with the elongation of a member in axial tension, there is a transverse

contraction, which is illustrated in Figure 1.12. The transverse contraction during a tensile test

is related to the longitudinal elongation by

Єtranvs = - vЄlongit Poisson’s Ratio 1.12

where v (Greek symbol nu) is Poisson's ratio. This expression also holds when the

longitudinal strain is compressive; then, the lateral strain results in an expansion of the

transverse dimensions. Poisson's ratio is dimensionless, with typical values in the 0.25-0.35

range. For the orientation of axes in Figure 1.12 the transverse strains are related to the

longitudinal strain by

Єy = Єz = - vЄx 1.13

Equations 1.11 and 1.13 apply to the simple case of uniaxial stress.

All the examples considered so far in this chapter have dealt with slender members subjected

to axial loads, i.e., to forces directed along a single axis. Choosing this axis as the X- axis, and

denoting by P the internal force at a given location, the corresponding stress components

were found to be σx = P/A, σy = 0, and σz = 0.

Let us now consider structural elements subjected to loads acting in the directions of the

three coordinate axes and producing normal stresses σx, σy, and σz which are all different from

zero (Figure 1.13). This condition is referred to as a multiaxial loading. Note that there is no

shearing stresses are included among the stresses shown in Figure 1.13.



FIGURE 1.13 A body subjected to a multiaxial loadings

The approach we propose here will be used repeatedly in this text, and is based on the

principle of superposition. This principle states that the effect of a given combined loading on a

structure can be obtained by determining separately the effects of the various loads and

combining the results obtained, provided that the following conditions are satisfied:

1. Each effect is linearly related to the load that produces it.

2. The deformation resulting from any given load is small and does not affect the

conditions of application of the other loads.

In the case of a multiaxial loading, the first condition will be satisfied if the stresses do not

exceed the proportional limit of the material, and the second condition will also be satisfied if the

stress on any given face does not cause deformations of the other faces that are large enough

to affect the computation of the stresses on those faces.

Considering first the effect of the stress component σx, recall that Єx = σx/E in the x direction,

and strains equal to Єx = -νσx/E in each of the y and z directions. Similarly, the stress

component σy if applied separately, will cause a strain Єy = σy/E in the y direction and strains Єy

= -νσy/E in the other two directions. Finally, the stress component σz causes a strain σz /E in the

z direction and strains Єy = -νσy/E in the x and y directions. Combining the results obtained, we

conclude that the components of strain corresponding to the given multiaxial loading are

Єx = +σx

-νσy

-νσz

Generalized Hooke’s

Law

E E E

Єy = -νσx

+σy

-νσz

E E E

Єz = -νσx

-νσy

+σz

E E E

The relations (2.28) are referred to as the generalized Hooke's law for the multiaxial loading of

a homogeneous isotropic material. As we indicated earlier, the results obtained are valid only

as long as the stresses do not exceed the proportional limit, and as long as the deformations

involved remain small. We also recall that a positive value for stress component signifies

tension, and a negative value compression. Similarly, a positive value for a strain component

indicates expansion in the corresponding direction, and a negative value contraction.



1.5 Allowable Stress and Allowable Load

An engineer in charge of the design of a structural member or mechanical clement must

restrict the stress in the material to a level that will be safe. Furthermore, a structure or

machine that is currently in use may, on occasion, have to be analyzed to see what additional

loadings its members or parts can support. So again it becomes necessary to perform the

calculations using a safe or allowable stress.

To ensure safety, it is necessary to choose an allowable stress that restricts the applied load to

one that is less than the load the member can fully support. There are several reasons for

this. For example, the load for which the member is designed may be different from actual

loadings placed on it. The intended measurements of a structure or machine may not be

exact due to errors in fabrication or in the assembly of its component parts. Unknown

vibrations, impact, or accidental loadings can occur that may not be accounted for in the design.

Atmospheric corrosion, decay, or weathering tends to cause materials to deteriorate during

service. And lastly, some materials, such as wood, concrete, or fiber-reinforced composites, can

show high variability in mechanical properties.

One method of specifying the allowable load for the design or analysis of a member is to

use a number called the factor of safety. The factor of safety (F.S.) is a ratio of the failure

load Ffail divided by the allowable load, Fallow. Here Ffail is found from experimental testing of the

material, and the factor of safety is selected based on experience so that the above mentioned

uncertainties are accounted for when the member is used under similar conditions of loading

and geometry. Stated mathematically,

If the load applied to the member is linearly related to the stress developed within the

member, as in the case of using σ = P/ A then, we can express the factor of safety as a ratio

of the failure stress σ fail to the allowable stress σ allow that is,

F.S =σ fail

σ allow

Where F.S must be greater than 1. Normally: 1< F.S < 3.


F.S =Ffail

Fallow


In any of these equations, the factor of safety is chosen to be greater than 1 in order to avoid

the potential for failure. Specific values depend on the types of materials to be used and the

intended purpose of the structure or machine. For example, the F.S. used in the design of

aircraft or space-vehicle components may be close to 1 in order to reduce the weight of the

vehicle. On the other hand, in the case of a nuclear power plant, the factor of safety for some

of its components may be as high as 3 since there may be uncertainties in loading or

material behavior. In general, however, factors of safety and therefore the allowable loads or

stresses for both structural and mechanical elements have become well standardized, since

their design uncertainties have been reasonably evaluated. Their values, which can be found

in design codes and engineering handbooks, are intended to form a balance of insuring

public and environmental safety and providing a reasonable economic solution to design.

DESIGN OF SIMPLE CONNECTIONS

By making simplifying assumptions regarding the behavior of the material, the equations σ =

P/ A can often be used to analyze or design a simple connection or a mechanical element. In

particular, if a member is subjected to a normal force at a section, its required area at the

section is determined from

A =P

σ allow

Cross-Sectional Area of a Tension Member.

The cross-sectional area of a prismatic member subjected to a tension force can be

determined provided the force has a line of action that passes through the centroid of the

cross section. For example, consider the "eye bar" shown in Fig. 1-27a. At the intermediate

section a-a, the stress distribution is uniform over the cross section and the required area A

is determined, as shown in Fig. 1-27b

Required Area to Resist Bearing.

A normal stress that is produced by the compression of one surface against another is called

a bearing stress. If this stress becomes large enough, it may crush or locally deform one or

both of the surfaces. Hence, in order to prevent failure it is necessary to determine the

proper bearing area for the material using an allowable bearing stress. For example, the

area A of the column base plate B shown in Fig. 1-29 is determined from the allowable

bearing stress of the concrete using A = P/(σb)allow. This assumes, of course, that the

allowable bearing stress for the concrete is smaller than that of the base plate material, and



furthermore the bearing stress is uniformly distributed between the plate and the concrete as

shown in the figure.


1.6 Deformation of Axially Loaded Member

Axial deformation is characterized by two fundamental kinematic assumptions:

1. The axis of the member remains straight.

2. Cross sections, which are plane and are perpendicular to the axis before

deformation, remain plane and remain perpendicular to the axis after deformation.

And, the cross sections do not rotate about the axis.

These assumptions are illustrated in Figure 1.13, where A and B designate cross sections at

x and (x + Δx) prior to deformation, and where A* and B* designate these same cross

sections after deformation.

The distance that a cross section moves in the axial direction is called its axial displacement.

The displacement of cross section A is labeled u(x), while the neighboring section B

displaces an amount u(x + Δx). The displacement u(x) is taken to be positive in the +x

direction. We can derive a strain-displacement expression that relates the axial strain Є to

this axial displacement u by considering the fundamental definition of extensional strain:

Є =Final length – Initial length

Initial length

FIGURE 1.13 The geometry of axial deformation.

The axial strain of any fiber1 of infinitesimal length Δx that is parallel to the x axis and

extends from section A to section B of the undeformed member may be determined from the

fundamental definition of extensional strain at a point. By letting the initial length of a typical

fiber be Δx, and then letting Δx approach zero, we can write the following expression for the

axial strain:

Єx(x) =lim

Δx→0(

Δx* - Δx) =

lim

Δx→0[

u(x + Δx)- ux] =

du

Δx Δx dx

DMV 4343 / DMV 5343JAN~JUN 2007

Therefore, the axial strain at section x is the derivative (with respect to x) of the axial

displacement, or

Є(x

)=

du(x)

Strain

Displacemen

t

Equation

dx

This equation relating axial strain to axial displacement is called the strain-displacement

equation for axial deformation. The two fundamental kinematic assumptions stated above

imply that the axial strain Є may be a function of x, but that it is not a function of position in

the cross section, that is, of y or z. To emphasize this point, a plot of the strain distribution at

an arbitrary cross section at x is shown in Figure 1.14 superimposed on a sketch of a portion

of the member. To reiterate, axial deformation is characterized by extensional strain that is

not a function of position in the cross section.

Figure 1.14 Extensional stress

distribution for a member undergoing

axial deformation.

Figure 1.15 Definition

of the total elongation

e.

As indicated in Figure 1.15, the total elongation of the member is the difference between the

displacements of its two ends, that is,

e = u (L) - u (0)

By summing up the changes in length of increments dx over the entire length of the

member, we get the following equation for the elongation of an axial-deformation member of

initial length L:


DMV 4343 / DMV 5343JAN~JUN 2007

e = ∫L

Є(x) dxElongation

Formula 0

How the strain Є varies with x depends on the loads that are applied to the member,

whether is has a constant cross section or the cross section varies with x, and the behavior

of the material (or materials) from which the member is constructed. Next, we will consider

this material behavior.

Material Behavior.

Having the strain distribution given by Eq. 1.14 and illustrated in Figure 1.14, we can

now employ the uniaxial stress-strain behavior of materials to determine the stress

distribution in a member undergoing axial deformation. Let us first take the simplest

case—linearly elastic behavior—and let us assume that the temperature remains

constant (i.e., ΔT = 0) and that σY = σZ = 0. Then, from Eq. 1.14 we get the uniaxial

stress-strain equation

σ = σx = EЄx Hooke’s Law 1.14

Equation 1.14 gives the distribution of the axial stress σ on the cross section at x. As

indicated in Eq 1.14, Є may only vary with x, but not with position in the cross section. Most

axial-deformation members are homogeneous, so that Young's modulus, E, is constant

throughout the member.

Stress Resultant. Deformable-body mechanics problems are simplified by making

assumptions that reduce a basically three-dimensional problem to a one-dimensional

problem, like the axial-deformation kinematic assumptions discussed earlier in this section.

Given the distribution of axial stress σ (Eq. 1.14), we can replace the distributed stress by a

single resultant axial force and relate that force to the axial stress on the cross section.

Figure 1.15 shows the three stress resultants that are related to the axial stress σ, and

defines the sign convention for these stress resultants. The resultants in Figure 1.15 are

shown for the +x face: equal and opposite resultants act on the -x face at the cross section.

The axial force, F, on the cross section will always be taken to be positive in tension. The

bending moments My and Mz are taken positive according to the right-hand rule. By

summing up the contributions to these resultants of the forces dF on infinitesimal areas, dA,

we get


DMV 4343 / DMV 5343JAN~JUN 2007

Equations 1.15 define the three stress resultants associated with the axial stress σ.

However, when the distribution of stress on the cross section is known, we can combine

these to form a single resultant force on the cross section. Consider the most common case,

a homogeneous linearly elastic member. Then, Young's modulus is constant throughout

the entire member, so Eq. 1.14 takes the form

σ(x) = EЄ(x)

Stress

Distribution

(E = const)

(a) Stress resultants on the cross section at x.

(b) Axial stress at a point in the cross section at x.

FIGURE 1.15

In previous section we have cover that strain, Є is

Є =L* - L 0

L0

Normal

Strain

Where L* - L0 is the change in length in the axis of applied load. If we let L* - L0 = δ and L0 =

L, then

Є =δ

L

Stress,

σ = P / A


F(x) = ∫Aσ dA

Stress

ResultantsMy(x) = ∫A

zσ dA

Mz(x) = -∫Ayσ dA

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From Hooke’s Law

E =σx

Єx

Substituting in Hooke’s Law we get,

E =P / A

=P L

δ L A δ

Since E = constant, we can manipulate this equation

δ =P L

A E

EXAMPLE 1.4

The cylindrical rod in Figure 1 is made of steel with E = 30 MPa, v = 0.3, and σY = 50 MPa. If

the initial length of the rod is L = 4 m and its original diameter is d = 10 cm, what is the

change in length, ΔL, and what is the change in diameter, Δd, due to the application of an

axial load P = 10 kN?

FIGURE 1

Plan the Solution

From the load P and the cross-sectional area A we can determine the stress σ. Then, if σ ≤

σY, we can use Hooke's Law to relate the stress σ to the strain Є. Then we can relate the

uniform strain, Є, to the elongation ΔL. The change in diameter is due to the Poisson’s ratio

effect.

Solution


DMV 4343 / DMV 5343JAN~JUN 2007

FIGURE 2 Free-body diagram

Equilibrium: From the free-body diagram in Figure 2,

∑ F x = 0 F(x) = P = const

From Eq. 1.5,

σx = P/A = 10 kN / π (0.05 m)2 = 1.273 MPa

Therefore, linearly elastic behaviour occurs when the load P = 10 kN is applied to the bar.

Material Behavior: From Eq 1.14

Єx =σx

=P L

E AE

And, from Eq 1.7

Єradial = -v Єx =-v P

AE

Strain-Displacement: From Eq. 2.7,

ΔL = L Єx =P L

AE

And

Δd = dЄradial =- v Pd

AE

Substituting numerical values into Eqs (3), we get

ΔL = L Єx =P L

=(10 kN)(4 m)

Π (0.05 m)2 (30 MPa)= 0.17 m.

AE

And

Δd = dЄradial =- v Pd

AE=

- 0.3(10 kN)(0.1 m)

Π (0.05 m)2 (30 MPa)= -1.273 mm.

Review the Solution

The changes in length and diameter are quite small in comparison with the original length

and the original diameter, respectively, as they should be. Δd should definitely be smaller


DMV 4343 / DMV 5343JAN~JUN 2007

than ΔL, and the signs should be different, which is the case. The extensometer in Figure

1.5a is capable of measuring both Δd and ΔL.


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1 Introduction to Stress and Strain

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