Chapter 1

Tension, Compression, and

Shear

1
 Learning Objectives
• Define “mechanics of materials”.
• Study normal stress (σ) and normal strain (ε).
• Identify key properties of various materials:
• Modulus of elasticity (E), yield stress (σy) and ultimate stress (σu).
• Plot shear stress (τ) vs. shear strain (γ).
• Identify the shearing modulus of elasticity (G).
• Study Hooke’s law for normal and shear stresses and strains.
• Investigate changes in lateral dimensions and volume of a bar.
• Study normal, shear, and bearing stresses in bolted connections.
• Use factors of safety to establish allowable values of stresses.
• Introduce basic concepts of design.

2
 1.1 Introduction and Statics Review
• “Mechanics of Materials”– a branch of applied mechanics that deals with the
behavior of solid bodies subjected to various types of loading.
• Other names for this field of study are “Strength of Materials” and
“Mechanics of Deformable Bodies”

• Principal Objective– determine the stresses, strains, and displacements in

structures and their components due to loads acting on them.

• Analytical methodologies will be applied to real bodies– that is, bodies of

finite dimensions that deform under loads.

3
 Problem-Solving Approach
• Step One: Conceptualize [Hypothesize, Sketch]
• List all relevant data provided within the problem statement.
• Draw a sketch showing all applied forces, support/ boundary conditions, and
interactions between adjacent bodies.
• Step Two: Categorize [Simplify, Classify]
• Identify all unknowns within the problem.
• Make any necessary assumptions to simplify the problem.
• Step Three: Analyze [Evaluate, Select relevant equations, Carry out
mathematical solution]
• Apply appropriate theory and equations and solve for the unknowns.
• Step Four: Finalize [Conclude, Examine Answer– Sensible? Appropriate Units?]
• Compare answer to similar problem solutions, scrutinize the units.
• Vary key parameters to see how the results change.

4
 Equilibrium Equations
• Vector Form

Resultant Force R Resultant Moment M

R  F  0 M   M   (r  F )  0

• Scalar Form

F x 0 F
y 0 F
z 0 M x 0 M y 0 M z 0

5
 Applied Forces

• Concentrated Forces/Moments– applied to a single point.

• Distributed Forces– applied along a portion of the member’s length.

• Free-Body Diagram– includes all applied forces and moments, reaction

forces and moments, and connection forces between individual
components.

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10
 Reactive Forces and Support Conditions
• Forces at structural supports which usually result from the action of
applied forces to the overall structure.

• Reaction Force– shown as a single arrow with a slash drawn through it.

• Reaction Moment– shown as a double-headed arrow or curve with a slash.

• Table 1-1 lists a wide range of support or connection types along with their
corresponding sketch and reaction forces.

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Reactions and Support Conditions (Table 1-1)
• Roller

12
Reactions and Support Conditions (Table 1-1)
• Pin

13
 Reactions and Support Conditions (Table 1-1)

• Ball-and-Socket

14
Reactions and Support Conditions (Table 1-1)
• Sliding Support

15
Reactions and Support Conditions (Table 1-1)
• Fixed Support

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 Reactions and Support Conditions (Table 1-1)
• Elastic or Spring Support

Translational spring support for

heavy equipment

Rotational spring in a clothespin

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 Reactions and Support Conditions (Table 1-1)
• Wheel on Rail Support

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Reactions and Support Conditions (Table 1-1)
• Bearing

19
 Internal Forces (Stress Resultants)
• The internal axial force, torsional moment, transverse shear, and bending
moment present within the member.

• A sectional cut normal to the axis of each member allows for internal forces
to be represented within the FBD.

• Deformation Sign Convention

• Tension is positive
• Compression is negative

20
 EXAMPLE 1-1
The plane truss shown has four joints and five members. Find support
reactions at A and B and then use the methods of joints and sections to find
all member forces. Let P = 150 kN and c = 3 m.

21
 EXAMPLE 1-1
Solution:
1. Conceptualize [hypothesize, sketch]: First
sketch a free-body diagram of the entire truss model.

2. Categorize [simplify, classify]: Overall

equilibrium requires that the force components in x
and y directions and the moment about the z axis
must sum to zero; this leads to reaction force
components Ax, Ay, and By.

22
 EXAMPLE 1-1
3. Analyze [evaluate; select relevant equations, carry out
mathematical solution]:
Law of sines to find member lengths a and b: Use known angles θA, θB,
and θC and c = 3 m to find lengths a and b:

Support reactions: Using the truss model free-body diagram in Fig.

1-6, sum forces in x and y directions and moments about joint A:

23
 EXAMPLE 1-1
Member forces using method of joints: Begin by drawing free-body
diagrams of the pin at each joint (Fig. 1-7). The forces are concurrent at
each joint, so use force equilibrium at each location to find the unknown
member forces.

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 EXAMPLE 1-1
First sum forces in the y direction at joint A to find member force AC, and
then sum forces in the x direction to get member force AD:

Summing forces at joint B gives member forces BC and BD as

The minus sign means that member BC is in compression, not in tension as

assumed.
Finally, observe that CD is a zero-force member because forces in the y
direction must sum to zero at joint D.

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 EXAMPLE 1-1
Finally, summing moments about A in Fig. 1-8 confirms member force BC:

4. Finalize [conclude; examine answer—Does it make sense? Are

units correct? How does it compare to similar problem solutions?]:
There are 2 j = 8 equilibrium equations for the simple plane truss considered,
and using the method of joints, these are obtained by applying ΣFx = 0 and
ΣFy = 0 at each joint in succession.

26
 EXAMPLE 1-2
A simple beam with an overhang is supported at points A and B (Fig. 1-9). A
linearly varying distributed load of peak intensity q0 = 160 N/m acts on span
AB. Concentrated moment M0 = 380 N·m is applied at A, and an inclined
concentrated load P = 200 N acts at C. The length of segment AB is L = 4 m,
and the length of the overhang BC is 2 m.
Find support reactions at A and B and then calculate the axial force N, shear
force V, and bending moment M at midspan of AB.

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 EXAMPLE 1-2
Solution:
1. Conceptualize: Find the reaction forces Ay, Bx, and By using the FBD of
The overall structure shown in Fig. 1-10.

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 EXAMPLE 1-2
2. Categorize: Start by finding reaction forces Ay, Bx, and By then use
either the left-hand or right-hand free-body diagram in Fig. 1-11 to find N, V,
and M.

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 EXAMPLE 1-2
3. Analyze:
Solution for external reactions:

Reaction Ay is negative, so in accordance with a statics sign convention, it

acts downward. Reaction components Bx and By are positive, so they act in
the directions shown in Figs. 1-10 and 1-11. The resultant reaction force at
B is

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 EXAMPLE 1-2
Solution for internal axial force N, shear force V, and moment M at
midspan of AB: Using the left-hand free-body diagram in Fig. 1-11,

Right-hand: Note that the trapezoidal distributed load segment is treated as

two triangular loads when moments are summed to find internal moment M:

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 EXAMPLE 1-2
4. Finalize: The computed internal forces (N and V) and internal moment
(M) can be determined using either the left- or right-hand free-body
diagram. This applies for any section taken through the beam at any point
along its length.

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 EXAMPLE 1-3
A stepped circular shaft is fixed at A and has three gears that transmit the
torques shown in Fig. 1-12. Find the reaction torque MAx at A and then find
the internal torsional moments in segments AB, BC, and CD. Use properly
drawn free-body diagrams in your solution.

33
 EXAMPLE 1-3
Solution:
1. Conceptualize: The solution for the reaction moment at A(MAx) must
begin with a proper drawing of the FBD of the overall structure (Fig. 1-13).
Separate FBDs showing internal torques T in each segment are obtained by
cutting the shaft in regions AB, BC, and CD in succession and are given in
Fig. 1-14(a–c). Each cut produces a left-hand and a right-hand free-body
diagram.

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35
 EXAMPLE 1-3
2. Categorize:
There is no distributed torque acting on this shaft, so the internal torsional
moment T is constant within each segment.
3. Analyze:
Solution for external reaction moment MAx:

36
 EXAMPLE 1-3
Solution for internal torsional moments T in each shaft segment:
Start with segment AB and use either FBD in Fig. 1-14a to find:

Next consider segment BC. Summing moments about the x axis in Fig. 1-
14b gives

37
 EXAMPLE 1-3
Finally, moment equilibrium about the x axis leads to a solution for the
internal torsional moment in segment CD:

In each segment, the internal torsional moments computed using either the
left or right FBDs are the same.

4. Finalize: Segment BC has the maximum positive internal torsional

moment, and segment AB has the maximum negative torsional moment.
This is important information for the designer of the shaft. Properly drawn
free-body diagrams are essential to a correct solution. Either the left or right
free-body diagram can be used to find the internal torque at any section.

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 1.2 Normal Stress and Strain
Normal Stress
•Symbol: σ (sigma)
•Equation:
P

A

•Units: force per unit of area

• IP:
• Pounds per square inch (psi)

• Kips per square inch (ksi)

• SI:
• Newtons per square meter (N/m2)

• Pascals (Pa)

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 1.2 Normal Stress and Strain
Normal Strain (elongation per unit length)
• Symbol: ε (epsilon)
• Equation:


L
• Units: none (dimensionless)

Requirements for Normal Stress and Strain

• The bar is prismatic.

• The loads act through the centroids of the cross section.

• The material is homogeneous.

40
The loads P are transmitted to the bar by pins that pass
through the holes (or eyes) at the ends of the bar. The
stress distribution around the holes is quite complex.

As a practical rule, the formula σ=P/A may be used with good

accuracy at any point within a prismatic bar that is at least as far
away from the stress concentration as the largest lateral
dimension of the bar.

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Line of Action of the Axial Forces for a Uniform
Stress Distribution

42
 Location of the force

the coordinates of the (1-9a,b)

centroid of an area

In order to have uniform tension or compression in a prismatic

bar, the axial force must act through the centroid of the cross-
sectional area.

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 EXAMPLE 1-4
An elastic spring rests on a base plate that is on top of rigid tube B (see Fig.
1-18). The spring is enclosed by rigid tube A but is longer than tube A by
an amount s. Force P is then applied to a cap plate to compress the spring.
Both tubes have outer diameter dO, but the inner diameters are dA and dB
for tubes A and B, respectively. Assume that spring stiffness k = 4200
kN/m, dO = 75 mm, dA =62 mm, dB = 57 mm, and s = 3.2 mm.

(a) If applied load P = 11 kN, what are the axial normal stresses in tubes A
and B?
(b) Repeat part (a) if P = 22 kN.
(c) What is P if the normal stress in tube A is 5.5 MPa? What is the
associated stress in tube B?

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45
46
 EXAMPLE 1-4
Solution:
Use the following four-step problem-solving approach.

1. Conceptualize: The two possible states of the assemblage are shown in

the freebody diagrams in Fig. 1-19. In Fig. 1-19a, an upper-section cut
through both the spring and tube A creates an upper free-body diagram that
reveals a spring force (k)(x) for the case of downward cap displacement x
that is less than gap width s. In Fig. 1-19b, cap displacement x is equal to
gap width s, so the spring force now equals (k)(s). Figure 1-19 also shows
lower free-body diagrams for both cases in which a section cut through tube
B shows that the internal compressive force in tube B is equal to applied
load P. (Internal forces in tubes A and B are shown as two arrows, one at
each tube wall, indicating that NA and NB are actually uniformly distributed
forces acting on the circular cross section of each tube.)

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 EXAMPLE 1-4
2. Categorize: The force P required to close gap s is (k)(s). This is also the
maximum force that can be developed in the spring. If applied force P is too
small to close the gap s, force P will be transferred to the base plate and into
rigid tube B; tube A will be unaffected by the load. However, if force P is
large enough to compress the spring to close the gap s, the spring and tube
A will share the load P applied to the cap plate and together will transfer it to
tube B through the base plate. In summary, the free-body diagrams in Fig.
1-19 show that, if the spring is compressed by load P an amount x, the
compressive internal forces in the spring and the two tubes are

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 EXAMPLE 1-4
3. Analyze:
Force P required to close gap s: The gap closes when force P is equal to
ks

Tube stresses for applied load P = 11 kN: The cap will displace
downward a distance x = P/k = 2.619 mm (< s), so tube internal forces are
NA = 0 and NB = P. Tube cross-sectional areas are

The resulting axial normal compressive stresses in tubes A and B are

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 EXAMPLE 1-4
Tube stresses for applied load P = 22 kN: Cap downward displacement
is now x = P/k = 5.24 mm (> s), so tube internal forces are NA = P - (k)(s)
= (22 - 13.44) kN = 8.56 kN and NB = P. The normal stresses in tubes A
and B are now:

Applied load P if stress in tube A is 5.5 MPa: Force P must exceed (k)(s)
= 13.44 kN for the gap to close, leading to a force in tube A and a normal
stress of σA = 5.5 MPa. The normal compressive force in tube A is NA = (σA)
(AA) =7.69kN. It follows that applied force P is now P = NA + ks = 7.69 kN
+ 13.44 kN = 21.1 kN. Internal force NB = P, so the normal compressive

stress in tube B is now

50
 EXAMPLE 1-4
4. Finalize: If tube A is elastic instead of rigid as assumed here, tube A can
be modeled as another spring that is parallel to the spring it encloses. Now a
more advanced analysis procedure will be needed to find tube force NA for
the case of P > (k)(s). Force NA is no longer equal to P - (k)(s), and
downward displacement x can be larger than s.

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 EXAMPLE 1-5
An antenna and receiver are suspended on a steel wire from a helicopter to
measure the effects of wind turbines on a local radar installation (see Fig. 1-
20). Obtain a formula for the maximum stress in the wire, taking into
account the weight of the wire itself. Calculate the maximum stress in the
wire in MPa using the following numerical properties: L1 = 6 m, L2 = 5 m, d
= 9.5 mm; antenna weight is W1 = 380 N; receiver weight is W2 = 700 N.
Note that the weight density γ of steel is 77.0 kN/m3.

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53
 EXAMPLE 1-5
Solution:
Use the following four-step problem-solving approach.

1. Conceptualize: A free-body diagram of the suspended instrument

package is shown in Fig. 1-21. The antenna (W1) and receiver (W2) weights
are concentrated forces at specified locations along the wire; the weight of
the wire is a uniformly distributed axial force expressed as w(x) = γA ,
where A is the cross-sectional area of the wire. Cutting the wire at some
point x leads to upper and lower free-body diagrams (Fig. 1-22); either can
be used to find the internal axial force N(x) at the location of the cut section.
The internal axial force in the wire is a maximum at the point at which it is
attached to the helicopter (x = 0).

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55
 EXAMPLE 1-5
2. Categorize: Start by solving for the reaction force R at the top of the
wire and then cut the wire a short distance below the support to find Nmax.
The wire is prismatic, so simply divide Nmax by cross-sectional area A to find
the maximum axial normal stress σmax.

3. Analyze:
Reaction force R: Use the free-body diagram in Fig. 1-21 to obtain

The minus sign indicates that reaction force R acts in the (-x) direction, or
upward in Figs. 1-21 and 1-22.

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 EXAMPLE 1-5
Internal axial forces N(x) in hanging wire: The internal axial force in
the wire varies over the length of the wire. Cutting through the wire in upper
and Lower segments (the lower segment is cut in Fig. 1-22) gives

Internal force N(x) is shown as a pair of forces acting away from the cut
section in accordance with a deformation sign convention in which the wire
is initially assumed to be in tension and that tension is positive. The
maximum force in the wire is at x = 0: Nmax = N(0) = W1 +W2 + w(L1 +L2).

57
 EXAMPLE 1-5
Formula for maximum stress in the wire: The cross-sectional area A of
the wire is constant, so dividing Nmax by A leads to a formula for maximum
stress in the wire:

Numerical calculations: The cross-sectional area of the wire is

Solving for maximum normal stress gives

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 EXAMPLE 1-5
4. Finalize: If the weight of the wire is ignored, the maximum normal
stress is reduced to 15.24 MPa, which is a decrease of more than 5%.
Although the stresses are low here, eliminating the self-weight of the wire
from the stress calculation is not recommended.

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1.3 Mechanical Properties of Materials

60
Figure 1-25

Rock sample being tested

in compression to obtain
compressive strength, elastic
modulus and Poisson’s ratio
(Courtesy of MTS Systems
Corporation)

61
 Stress-Strain Diagrams
 Tension or compression
tests are performed on a
specimen with the stress
and strain being
determined at various
magnitudes of loading.

 The stress is plotted

against strain to achieve a
stress-strain diagram.

Figure 1-26
Stress-strain diagram for a typical structure steel in
tension (not to scale)

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 Proportional Limit– The point in which the stress and strain
are no longer proportional to one another.

 Modulus of Elasticity (E)– The slope of the straight line the

plotting of stress and strain produces within the linear region
of the diagram (before proportional limit is reached).

 Yield Stress (σy)– The corresponding magnitude of stress at

which point the material “yields” under the loading and
allows for rapid increase of strain.

 Ultimate Stress (σu)– The corresponding magnitude of stress

at which point the material begins to “neck,” thereby allowing
a decrease in cross-sectional area.
 Strength – The yield stress and ultimate stress of a material
are also called the yield strength and ultimate strength,
respectively. Strength is a general term that refers to the
capacity of a structure to resist loads.
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64
 Ductility
FIGURE 1-28
Stress-strain diagram for • Metals such as structural steel that
mild steel drawn to scale
undergo large permanent strains before
failure are classified as ductile. Ductility
is the property that enables a bar of
steel to be bent into a circular arc or
drawn into a wire without breaking. A
desirable feature of ductile materials is
that visible distortions occur if the loads
become too large, thus providing an
opportunity to take remedial action
before an actual fracture occurs.
• Also, materials exhibiting ductile
behavior are capable of absorbing large
amounts of strain energy prior to
fracture.
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 Aluminum Alloys

• Although they have considerable ductility,

aluminum alloys typically do not have a
clearly definable yield point. However,
they do have an initial linear region with
a recognizable proportional limit.
• When a material does not have an
obvious yield point yet undergoes large
strains after the proportional limit is
exceeded, an arbitrary yield stress may
be determined by the offset method.

66
 Offset Method
 Used to determine an arbitrary yield
stress.

 A straight line is drawn on the

stress-strain diagram parallel to the
initial linear part of the curve at an
offset of 0.002 (or 0.2%) strain.

 The intersection of the offset line

and the stress-strain curve (point A)
defines the “offset yield stress”.

67
 Rubber
• Rubber maintains a linear relationship
between stress and strain up to relatively
large strains (as compared to metals). The
strain at the proportional limit may be as
high as 0.1 or 0.2 (10 or 20%).
• Beyond the proportional limit, the behavior
depends upon the type of rubber. Some
kinds of soft rubber stretch enormously
without failure, reaching lengths several
times their original lengths. The material
eventually offers increasing resistance to
the load, and the stress-strain curve turns
markedly upward.
• Although rubber exhibits very large strains,
it is not a ductile material because the
strains are not permanent. It is an elastic
material.
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 Ductility and Elongation
 A measure of the amount of permanent strain the material
undergoes before failure occurs.

 Can be quantified by “percent elongation” and “percent

reduction in area”.

L1  Lo
 Percent Elongation: Percent elongation  (100)
Lo
For structural steel, values of 20 or 30% are common.

Ao  A1
 Percent Reduction in Area: Percet reduction  (100)
Ao

For ductile steels, the reduction is about 50%.

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Brittle Materials

Materials that fail in tension at

relatively low values of strain are
classified as brittle. Examples are
concrete, stone, cast iron, glass,
ceramics, and a variety of metallic
alloys.

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Compression

• Stress-strain curves for materials in

compression differ from those in
tension.
• Ductile metals such as steel, aluminum,
and copper have proportional limits in
compression very close to those in
tension, and the initial regions of their
compressive and tensile stress-strain
diagrams are about the same. However,
after yielding begins, the behavior is
quite different.

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 1.4 Elasticity and Plasticity
 Elasticity– The property of a material by
which it returns to its original
dimensions during unloading.

 Partially Elastic– During unloading, the

bar only returns partially to its original
shape, thereby undergoing some “elastic
recovery” but with “residual strain”
remaining.

 Plasticity– The characteristic of a

material that undergoes inelastic strains
beyond the stress (E) at the elastic limit.

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 1.4 Elasticity and Plasticity
Reloading
 In elastic range – Can be loaded,
unloaded, and reloaded without much
change in behavior.

 In plastic range – When initially loaded

and unloaded, residual strain is
introduced and the properties of the
material are changed (the linearly elastic
region, proportional limit, and elastic
limit are all increased).

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 EXAMPLE 1-6
A machine component slides along a horizontal bar at A and moves in a
vertical slot B. The component is represented as a rigid bar AB (length L =
1.5 m, weight W = 4.5 kN) with roller supports at A and B (neglect friction).
When not in use, the machine component is supported by a single wire
(diameter d = 3.5 mm) with one end attached at A and the other end
supported at C (see Fig. 1-36).

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 EXAMPLE 1-6
The wire is made of a copper alloy; the stress-strain relationship for the wire is

(a)Plot a stress-strain diagram for the material; What is the modulus of elasticity E
(GPa)? What is the 0.2% offset yield stress (MPa)?
(b) Find the tensile force T (kN) in the wire.
(c) Find the normal axial strain ε and elongation δ (mm) of the wire.
(d) Find the permanent set of the wire if all forces are removed.

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 EXAMPLE 1-6
Solution:

1. Conceptualize: The copper alloy has considerable ductility but will have
a stress-strain curve without a well-defined yield point. Define the yield
point using an offset method as illustrated in Fig. 1-30. Find the residual
strain and then the permanent set of the wire, as shown in Fig. 1-34.

2. Categorize: The given analytical expression for the stress-strain curve

is based on measured laboratory data for the copper alloy used to
manufacture this wire. Hence, the analytical expression is an approximation
of the actual behavior of this material and was formulated based on test
data. Analytical representations of actual stress-strain curves are often used
in computer programs to model and analyze structures of different materials
under applied loads of various kinds.

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 EXAMPLE 1-6
3. Analyze:

Part (a): Plot a stress-strain diagram for the material; What is the
modulus of elasticity E (GPa)? What is the 0.2% offset yield stress
(MPa)?

Plot the function σ(ε) for strain values between 0 and 0.03 (Fig. 1-37). The
stress at strain ε = 0.03 is 454 MPa.

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EXAMPLE 1-6

78
 EXAMPLE 1-6
The slope of the tangent to the stress-strain curve at strain ε = 0 is the
modulus of elasticity E (see Fig. 1-38). Take the derivative of σ(ε) to get the
slope of the tangent to the σ(ε) curve, and evaluate the derivative at strain ε
= 0 to find E:

Next, find an expression for the yield strain εy, the point at which the 0.2%
offset line crosses the stress-strain curve (see Fig. 1-38). Substitute the
expression εy into the σ(ε) expression and then solve for yield (εy) = σy：

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 EXAMPLE 1-6
Rearranging the equation in terms of σy gives

Solving this quadratic equation for the 0.2% offset yield, stress σy gives σy =
255 MPa.

The yield strain is computed as

80
EXAMPLE 1-6

81
 EXAMPLE 1-6
Part (b): Find the tensile force T (kN) in the wire. Recall that bar
weight W = 4.5 kN.

Find the angle between the x-axis and cable attachment position at C:

Sum the moments about A to obtain one equation and one unknown. The
reaction Bx acts to the left:

Next, sum the forces in the x direction to find the cable force TC:

82
 EXAMPLE 1-6
Part (c): Find the normal axial strain ε and elongation δ (mm) of the
wire.
Compute the normal stress then find the associated strain from stress-strain
plot (or from the σ(ε) equation). The wire elongation is strain times wire
length.

The wire diameter, cross-sectional area, and length are

Now compute the stress and strain in the wire and the elongation of the wire
as

83
 EXAMPLE 1-6
Note that the stress in the wire exceeds the 0.2% offset yield stress of 255
MPa.
The corresponding normal strain is found from the σ(ε) plot or by
rearranging the σ(ε) equation to give

Then,

Finally, the wire elongation is

84
 EXAMPLE 1-6
Part (d): Find the permanent set of the wire if all forces are removed.
If the load is removed from the wire, the stress in the wire will return to zero
following the unloading line in Fig. 1-39 (see also Fig. 1-34b). The elastic
recovery strain is

The residual strain is the difference between the total strain (εC) and the
elastic recovery strain (εer)

Finally, the permanent set of the wire is the product of the residual strain
and the length of the wire:

85
 EXAMPLE 1-6
4. Finalize: This example presents an analytical model of the stress-strain
relationship for a copper alloy. The computed values of modulus of elasticity
E and yield stress σy are consistent with values listed in Appendix F. The
tensile force, normal strain and elongation, and permanent set are computed
for the wire when stressed beyond the apparent yield point of the material.

86
EXAMPLE 1-6

87
 1.5 Linear Elasticity and Hooke’s Law
 Linear Elastic– When a material exhibits a linear relationship
between stress and strain. Typically engineers design
structures and machines to function within this region to
avoid permanent deformations due to yielding.

Hooke’s Law
 Expression for the linear relationship between stress and
strain.
 AKA Young’s Modulus (Modulus of elasticity)

 Equation:   E
 Limitations– Relates only to the longitudinal stresses and
strains in simple tension or compression of a bar (i.e. uniaxial
stresses).
88
 Poisson’s Ratio
 Lateral contraction accompanies axial elongation and
Poisson’s Ratio relates these two strains to one another.
 Symbol:  (nu)
 Equation:

 Units: none (dimensionless)

 Limitations:
 The material must be homogeneous.
 Elastic properties must be the same in all directions
perpendicular to the longitudinal axis

89
Known ν, you can obtain the lateral strain from the axial strain:

When using Eqs. (1-13) and (1-14), always keep in mind that they apply
only to a bar in uniaxial stress, that is, a bar for which the only stress is the
normal stress in the axial direction.

For isotropic materials, by a molecular theory of materials, v = 1/4;

based upon better models of atomic structure, v = 1/3. The actual
measured values, are in the range 0.25 to 0.35 for most metals and
many other materials.

When the strains in a material become large, Poisson’s ratio changes.

For instance, in the case of structural steel, the ratio becomes almost 0.5
when plastic yielding occurs. Thus, Poisson’s ratio remains constant
only in the linearly elastic range. When the material behavior is
nonlinear, the ratio of lateral strain to axial strain is often called the
contraction ratio.

90
 Limitations

For constant Poisson’s ratio, 3 conditions required:

• The material is linearly elastic.

• The material must be homogeneous, that is, it must have the same
composition (and hence the same elastic properties) at every point.

• The material must be isotropic, that is, it must have the same properties in
all directions (whether axial, lateral, or any other direction). If the
properties differ in various directions, the material is anisotropic.

91
 EXAMPLE 1-7
A hollow plastic circular pipe (length Lp, inner and outer diameters d1 and d2,
respectively; see Fig. 1-41) is inserted as a liner inside a cast iron pipe
(length Lc, inner and outer diameters d3 and d4, respectively).

92
 EXAMPLE 1-7
(a) Derive a formula for the required initial
length Lp of the plastic pipe so that, when it
is compressed by some force P, the final
length of both pipes is the same and at the
same time the final outer diameter of the
plastic pipe is equal to the inner diameter
of the cast iron pipe.
(b) Using the numerical data given, find the
initial length Lp (m) and final thickness tp
(mm) for the plastic pipe.
(c) What is the required compressive force P
(N)? What are the final normal stresses
(MPa) in both pipes?
(d) Compare the initial and final volumes (mm3)
for the plastic pipe.

93
 EXAMPLE 1-7
Numerical data and pipe cross-section properties are

94
 EXAMPLE 1-7
Solution:

1. Conceptualize: Application of a compressive force P results in

compressive normal strains and extensional lateral strains in the plastic pipe,
while the cast iron pipe is stress-free. The initial length of the plastic pipe (Lp)
is greater than that of the cast iron pipe (Lc). With full application of force P,
the lengths are made equal.

The initial cross-sectional areas of the plastic and cast iron pipes are

95
 EXAMPLE 1-7
2. Categorize: The two requirements are (a) compression of the plastic
pipe must close the gap (d3 - d2) between the plastic pipe and the inner
surface of the cast iron pipe and (b) the final lengths of the two pipes are
the same. The first requirement depends on lateral strain and the second on
normal strain. Each requirement leads to an expression for shortening of the
plastic pipe. Equating the two expressions (i.e., enforcing compatibility of
displacements) leads to a solution for the required length of the plastic pipe.

3. Analyze:
Part (a): Derive a formula for the required initial length Lp of the
plastic pipe.
The lateral strain resulting from compression of the plastic pipe must close
the gap (d3 - d2) between the plastic pipe and the inner surface of the cast
iron pipe. The required extensional lateral strain is positive (here, εlat = ε'.):

96
 EXAMPLE 1-7
The accompanying compressive normal strain in the plastic pipe is obtained
using Eq. 1-14, which requires Poisson’s ratio for the plastic pipe and also
the required lateral strain:

Use the compressive normal strain εp to compute the shortening δp1 of the
plastic pipe as

The required shortening of the plastic pipe (so that it has the same final
length as that of the cast iron pipe) is

97
 EXAMPLE 1-7
Equating δp1 and δp2 leads to a formula for the required initial length Lp of the
plastic pipe:

Part (b): Now substitute the numerical data to find the initial length
Lp, change in thickness Δtp, and final thickness tpf for the plastic pipe.
As expected, Lp is greater than the length of the cast iron pipe, Lc = 0.25 m,
and the thickness of the compressed plastic pipe increases by Δtp:

98
 EXAMPLE 1-7
Part (c): Next find the required compressive force P and the final
normal Stresses in both pipes.

A check on the normal compressive stress in the plastic pipe, computed

using Hooke’s law (Eq. 1-12), shows that it is well below the ultimate stress
for selected plastics (see Appendix F); this is also the final normal stress in
the plastic pipe:

The required downward force to compress the plastic pipe is

Both the initial and final stresses in the cast iron pipe are zero because no
force is applied to the cast iron pipe.

99
 EXAMPLE 1-7
Part (d): Lastly, compare the initial and final volumes of the plastic
pipe.

The initial cross-sectional area of the plastic pipe is

The final cross-sectional area of the plastic pipe is

The initial volume of the plastic pipe is

100
 EXAMPLE 1-7
and the final volume of the plastic pipe is

4. Finalize: The ratio of final to initial volume reveals little change in the
volume of the plastic pipe:

The numerical results obtained in this example illustrate that the

dimensional changes in structural materials under normal loading conditions
are extremely small. In spite of their smallness, changes in dimensions can
be important in certain kinds of analysis (such as the analysis of statically
indeterminate structures) and in the experimental determination of stresses
and strains.

101
 1.6 Shear Stress and Strain
Bearing Stress
 Contact stresses which develop under tensile loads between
connections.
The bearing area Ab is defined as the
 Equation: 1-15 projected area of the curved bearing surface

 Note: This calculates the “average” bearing stress.

Double shear
102
 Shear Stress
 Stress that acts tangential to
the surface of the material.

 Symbol:  (Tau)

 Equation: Single shear

 Units: Same as normal stress

(Pa, psi, etc.)

103
• The loading arrangements shown in Figs. 1-42 and 1-43 are
examples of direct shear (or simple shear) in which the shear
stresses are created by the direct action of the forces trying to
cut through the material.
• Direct shear arises in the design of bolts, pins, rivets, keys,
welds, and glued joints.

104
 Equality of Shear Stresses on
Perpendicular Planes

 Shear stresses on opposite (and

parallel) faces of an element are
equal in magnitude and opposite
in direction.

 Shear stresses on adjacent (and

perpendicular) faces of an
element are equal in magnitude
and have directions such that Pure shear: no normal stress
both stresses point toward, or
both point away from, the line of
intersection of the faces.

105
 Shear Strain
 A measure of the “distortion” (or, the
change in shape) of the element.

 Symbol:  (gamma)

 Units: Degrees or Radians (angle)

 Note: The lengths of the sides of the

element do not change under shear
stress. Instead, the element changes
in shape only. Shear strain is an
angle.

106
 Sign Conventions
Shear Stress
 Positive Face:
 Positive if shear stress acts in the positive direction of a coordinate
axis.
 Negative if shear stress acts in the negative direction of a
coordinate axis.
 Negative Face:
 Positive if shear stress acts in a negative direction of a coordinate
axis.
 Negative if shear stress acts in a positive direction of a coordinate
axis.
Shear Strain
 Positive when the angle between 2 positive (or 2 negative) faces is
reduced.
 Negative when the angle between 2 positive (or 2 negative) faces
is increased.
107
 Shear Stress and Strain
Hooke’s Law in Shear:

Shear Modulus of Elasticity (G):

 AKA Modulus of Rigidity

 Units are same as the elastic modulus (i.e. Pa, psi, etc.)
 The yield stress for structural steel in shear is 0.5 to 0.6
times the yield stress in tension.
 Because the value of Poisson’s ratio for ordinary materials is
between zero and one-half, G must be from one-third to one-
half of E.
108
 EXAMPLE 1-8
A punch for making holes in steel plates is shown in Fig. 1-47a. Assume that
a punch having diameter d = 20 mm is used to punch a hole in an 8 mm
plate, as shown in the cross-sectional view (Fig. 1-47b).

If a force P = 110 kN is required to create the hole, what is the average

shear Stress in the plate and the average compressive stress in the punch?

109
 EXAMPLE 1-8
Solution:

1. Conceptualize: Assume that the shaft of the punch is in compression

over its entire length due to applied load P. Force P acts downward on the
plate and is applied as a uniformly distributed force along a circle of
diameter d as the punch passes through the plate.

2. Categorize: The average shear stress in the plate is obtained by dividing

the force P by the shear area of the plate. The shear area is the cylindrical
area of the plate that is exposed when the punch passes through the plate.
The compressive stress of interest is the one acting on a circular cross
section through the lower segment of the punch (Fig. 1-47).

110
 EXAMPLE 1-8
3. Analyze: The shear area As is equal to the circumference of the hole
times the thickness of the plate, or

in which d is the diameter of the punch and t is the thickness of the plate.
Therefore, the average shear stress in the plate is

The average compressive stress in the punch is

111
 EXAMPLE 1-8
4. Finalize: The normal and shear stress distributions are not uniform due
to stress concentration effects; hence, the calculations result in “average”
stresses. In addition, this analysis is highly idealized because impact effects
that occur when a punch is rammed through a plate are not part of this
analysis.

112
 EXAMPLE 1-9
A bearing pad of the kind used to support machines and bridge girders
consists (see photos) of a linearly elastic material (usually an elastomer,
such as rubber) capped by a steel plate (Fig. 1-48a). Assume that the
thickness of the elastomer is h, the dimensions of the plate are a × b, and
the pad is subjected to a horizontal shear force V.
Obtain formulas for the average shear stress aver in the elastomer and
the horizontal displacement d of the plate (Fig. 1-48b).

113
 EXAMPLE 1-9
Solution:

1. Conceptualize: The distortion of the bearing pad under shear force V is

assumed to be linear through the thickness h, as shown in Fig. 1-48b.

2. Categorize: Assume that the shear stresses in the elastomer are

uniformly distributed throughout its entire volume and that the shear strain
γ is small.

3. Analyze: The shear stress on any horizontal plane through the elastomer
equals the shear force V divided by the area ab of the plane (Fig. 1-48a):

114
 EXAMPLE 1-9
The corresponding shear strain [from Hooke’s law in shear; Eq. (1-18)] is

in which Ge is the shear modulus of the elastomeric material. Finally, the

horizontal displacement d is equal to h tan γ (from Fig. 1-48b):

In most practical situations, the shear strain γ is a small angle, and in such
cases, replace tang with γ and obtain

115
 EXAMPLE 1-9
For example, if V = 0.8 kN, a = 75 mm, b = 60 mm, h = 20 mm, and Ge =
1.25 MPa, Eq. (1-22) results in d = 2.86 mm, while Eq. (1-23) gives d =
2.84 mm.

4. Finalize: Equations (1-22) and (1-23) give approximate results for the
horizontal displacement of the plate because they are based upon the
assumption that the shear stress and strain are constant throughout the
volume of the elastomeric material. In reality, the shear stress is zero at the
edges of the material (because there are no shear stresses on the free
vertical faces), and therefore, the deformation of the material is more
complex than pictured in Fig. 1-48b. However, if the length a of the plate is
large compared with the thickness h of the elastomer, the preceding results
are satisfactory for design purposes.

116
 1.7 Allowable Stresses and
Allowable Loads
 Strength– The capacity of the object to support or transmit
loads.

Factor of Safety
 For successful structures, the actual strength must exceed
the required strength.
 Equation:

 To avoid failure, Factor of Safety > 1

 Margin of Safety:

117
 Allowable Stresses
 AKA working stress

 Equation: Tension Shear

 Yielding begins when the yield stress is reached at any point

within the structure.
 For brittle materials, the ultimate stress can be used instead
of the yield stress.

118
 Allowable Loads
 Equation:

 Tension/Compression: [Area: Net Area]

 Direct Shear: [Area: Cross-sectional Area]

 Bearing: [Area: Projected Area]

119
 EXAMPLE 1-10
A steel bar serving as a vertical hanger to support
heavy machinery in a factory is attached to a
support by the bolted connection shown in Fig. 1-49.
Two clip angles (thickness tc = 9.5 mm) are
fastened to an upper support by bolts 1 and 2 each
with a diameter of 12 mm; each bolt has a washer
with a diameter of dw = 28 mm. The main part of
the hanger is attached to the clip angles by a single
bolt (bolt 3 in Fig. 1-49a) with a diameter of d = 25
mm. The hanger has a rectangular cross section
with a width of b1 = 38 mm and thickness of t = 13
mm, but at the bolted connection, the hanger is
enlarged to a width of b2 = 75 mm. Determine the
allowable value of the tensile load P in the hanger
based upon the following considerations.

120
121
 EXAMPLE 1-10
(a) The allowable tensile stress in the main part of the hanger is 110 MPa.
(b) The allowable tensile stress in the hanger at its cross section through the
bolt 3 hole is 75 MPa. (The permissible stress at this section is lower
because of the stress concentrations around the hole.)
(c) The allowable bearing stress between the hanger and the shank of bolt 3
is 180 MPa.
(d) The allowable shear stress in bolt 3 is 45 MPa.
(e) The allowable normal stress in bolts 1 and 2 is 160 MPa.
(f) The allowable bearing stress between the washer and the clip angle at
either bolt 1 or 2 is 65 MPa.
(g) The allowable shear stress through the clip angle at bolts 1 and 2 is 35
MPa.

122
123
 EXAMPLE 1-10
Solution:

1. Conceptualize: Start by sketching a series of free-body diagrams to find

the forces acting on each connection component. Express the force on each
component in terms of an allowable stress times the associated area upon
which it acts. This force is the allowable value of applied load P for that
stress condition. Each of the seven stress states [(a)–(g) in the problem
statement] and the associated applied load are illustrated in this example’s
figures; each is adjacent to the corresponding calculations in Step (3).

124
 EXAMPLE 1-10
2. Categorize: Compute seven different values of the allowable load P,
each based on an allowable stress and a corresponding area. The minimum
value of load P will control.
Numerical data for the hanger connection design shown in Fig. 1-49 are
as follows.
(a) Connection component dimensions:

(b) Allowable stresses:

125
 EXAMPLE 1-10
3. Analyze:
Part (a): Find the allowable load based upon the stress in the
main part of the hanger (Fig. 1-49c). This is equal to the
allowable stress in tension (110 MPa) times the cross-sectional area
of the hanger (Eq. 1-30):

A load greater than this value will overstress the main part of the
hanger (that is, the actual stress will exceed the allowable stress),
thereby reducing the factor of safety.

126
 EXAMPLE 1-10
Part (b): Find the allowable load based upon the allowable
tensile stress (75 MPa) in the hanger at its cross section
through the bolt 3 hole.

At the cross section of the hanger through the bolt hole (Fig. 1-
49d), make a similar calculation but with a different allowable
stress and a different area. The net cross-sectional area (that is,
the area that remains after the hole is drilled through the bar) is
equal to the net width times the thickness. The net width is equal
to the gross width b2 minus the diameter d of the hole. Thus, the
equation for the allowable load Pb at this section is

127
 EXAMPLE 1-10
Part (c): Now find the allowable load based upon the
allowable bearing stress (180 MPa) between the hanger and
the shank of bolt 3.
The allowable load based upon bearing between the hanger and the
bolt (Fig.1-49e) is equal to the allowable bearing stress times the
bearing area. The bearing area is the projection of the actual contact
area, which is equal to the bolt diameter times the thickness of the
hanger. Therefore, the allowable load (Eq. 1-32) is

128
 EXAMPLE 1-10
Part (d): Determine the allowable load based upon the allowable
shear stress (45 MPa) in bolt 3.
The allowable load Pd based upon shear in the bolt (Fig. 1-49f) is equal to
the allowable shear stress times the shear area (Eq. 1-31). The shear
area is twice the area of the bolt because the bolt is in double shear; thus,

129
 EXAMPLE 1-10
Part (e): Find the allowable load based upon the allowable normal
stress (160 MPa) in bolts 1 and 2.

The allowable normal stress in bolts 1 and 2 is 160 MPa. Each bolt carries
one half of the applied load P (see Fig. 1-49g). The allowable total load Pe is
the product of the allowable normal stress in the bolt and the sum of the
crosssectional areas of bolts 1 and 2:

130
 EXAMPLE 1-10
Part (f): Now find the allowable load based upon the allowable
bearing stress (65 MPa) between the washer and the clip angle at
either bolt 1 or 2.

The allowable bearing stress between the washer and the clip angle at either
bolt 1 or 2 is 65 MPa. Each bolt (1 or 2) carries one half of the applied load P
(see Fig. 1-49h). The bearing area here is the ring-shaped circular area of
the washer (the washer is assumed to fit snugly against the bolt). The
allowable total load Pf is the allowable bearing stress on the washer times
twice the area of the washer:

131
 EXAMPLE 1-10
Part (g): Finally, determine the allowable load based upon the
allowable Shear stress (35 MPa) through the clip angle at bolts 1
and 2.

The allowable shear stress through the clip angle at bolts 1 and 2 is 35 MPa.
Each bolt (1 or 2) carries one half of the applied load P (see Fig. 1-49i). The
shear area at each bolt is equal to the circumference of the hole (ϖ × dw)
times the thickness of the clip angle (tc).

The allowable total load Pg is the allowable shear stress times twice the
shear area:

132
 EXAMPLE 1-10
4. Finalize: All seven conditions were used to find the allowable tensile
loads in the hanger. Comparing the seven preceding results shows that the
smallest value of the load is Pallow = 36.2 kN. This load is based upon normal
stress in bolts 1 and 2 [see part (e)] and is the allowable tensile load for the
hanger.

133
 1.8 Design For Axial Loads and
Direct Shear
 Design– The determination of structural properties in order
for the structure to support given loads and perform its
intended function.

 Design of Areas for Members in Simple Tension or

Compression:

134
 EXAMPLE 1-11
Continuous cable ADB runs over a small frictionless pulley at D to support
beam OABC, which is part of an entrance canopy for a building (see Fig. 1-
50). Load P = 4.5 kN is applied at the end of the canopy at C. Assume that
the canopy segment has weight W = 7.5 kN.

135
 EXAMPLE 1-11
(a) Find cable force T and pin support reactions at O and D.
(b) Find the required cross-sectional area of cable ADB if the allowable
normal stress is 125 MPa.
(c) Determine the required diameter of the pins at O, A, B, and D if the
allowable stress in shear is 80 MPa.
(Note: The pins at O, A, B, and D are in double shear. Also, consider only
load P and the weight W of the canopy; disregard the weight of cable ADB.)

136
 EXAMPLE 1-11
Solution:

1. Conceptualize: Begin with a free-body diagram of beam OABC (Fig. 1-

51a). Also sketch free-body diagrams of the entire structure (Fig. 1-51b)
and of joint D alone (Fig. 1-51c). Show cable force T and all applied and
reaction force components.

2. Categorize: First, use the free-body diagram of beam OABC (Fig. 1-51a)
to find cable force T and reaction force components at O. Then use Fig. 1-
51b or Fig. 1-51c to find reaction forces at D. Use cable force T and the
allowable normal stress to find the required cross-sectional area of the cable.
Also use force T and the allowable shear stress to find the required diameter
of the pins at A and B. Use the resultant reaction forces at O and D to find
pin diameters at these locations.

137
138
 EXAMPLE 1-11
3. Analyze:
Cable force T: First, find required distances and angles in Fig. 1-51a:

Now sum moments about O in Fig. 1-51a to find tension T in continuous

cable ADB:

where

139
 EXAMPLE 1-11
Reaction force at O: Sum forces in Fig. 1-51a to find reaction force
components at O:

The resultant reaction force at O is

Reaction force at D: Sum forces in Fig. 1-51c to find reaction force

components at D:

The resultant reaction force at D is

140
 EXAMPLE 1-11
Cross-sectional area of cable ADB: Use the allowable normal stress of
125 MPa and cable force T = 9.527 kN [Eq. (a)] to find the required cross-
sectional Area of the cable:

Required diameter of the pins at O, A, B, and D: All pins are in double

shear. The allowable shear stress is τallow = 80MPa. Required diameters of
each pin are computed as

141
 EXAMPLE 1-11
4. Finalize: In practice, other loads besides the weight of the canopy would
have to be considered before making a final decision about the sizes of the
cables and pins. Loads that could be important include wind loads,
earthquake loads, and the weights of objects that might have to be
supported temporarily by the structure. In addition, if cables AD and BD are
separate cables (instead of one continuous cable ADB), the forces in the two
cables are not equal in magnitude. The structure is now statically
indeterminate, and the cable forces and the reactions at O and D cannot be
determined using the equations of static equilibrium alone. Problems of this
type are discussed in Chapter 2, Section 2.4.

142
 Summary
 Principal Objective– to determine stresses, strains, and
displacements.

 Normal Stress: Normal Strain:

 Stress-Strain Diagram expresses the mechanical behavior

of various materials.

 Hooke’s Law: [Applicable only in the linearly elastic region]

 Poisson’s Ratio:

 Bearing Stress: Shear Stress:

143
 Summary (cont.)
 Hooke’s Law in Shear: [shear strain (τ) measures
distortion]
 Shear Modulus of Elasticity (G):

Actual Strength
 Factor of Safety: Factor of Safety n 
Required Strength

 Yield Strength is used for ductile materials; Ultimate

Strength if brittle.
 Factor of Safety helps determine Allowable Stresses and
Allowable Loads.
 Required areas of members can be determined from these
allowable factors.

144