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Module

Material Properties, Thin-walled

Pressure Vessel and Stress
Concentrations
Intended Learning Outcomes

After studying this chapter, you should be able to do the following:

1. Determine the different properties of materials using material testing.
2. Determine how to calculate the stresses in thin-walled pressure vessel.
3. Calculate stresses with considerations of the stress concentrations.

This module presents the material properties, thin- walled pressure vessel and stress
concentrations in a material. In material properties, different properties of the material will be
determined through material testing and plotting the stress-strain diagram. In thin-walled pressure
vessel, the formula to be used in computing the stresses will also be discussed. A brief discussion
about the stress concentrations will be tackled at the end part of this module.

Mechanical Properties of Material

The mechanical behavior of a material reflects its response or deformation in relation to
an applied load or force. The mechanical properties of materials can be determined by performing
carefully laboratory experiments such as tensile test, compression test, shear test, torsion test,
impact test, fatigue test and hardness test. Factors to be considered in the experiment include
the nature of the applied load and its duration, as well as the environmental conditions. The
applied load can be tensile, compressive, or shear, and its magnitude may be constant with time,
or it may fluctuate. Mechanical Properties of Metals and the publication of these standards are
often coordinated by professional societies such American Society for Testing and Materials
(ASTM), the most active organization in the United States.

Tension and Compression Test

The strength of a material depends on its ability to sustain a load without undue
deformation or failure. This property is inherent in the material itself and must be determined by
experiment. One of the most important tests to perform in this regard is the tension or compression
test. Although several important mechanical properties of a material can be determined from this

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test, it is used primarily to determine the relationship between the average normal stress and
average normal strain in many engineering materials such as metals, ceramics, polymers, and
composites.

One of the instruments used for conducting this test is the Universal Testing Machine. The
Universal testing machine can be shown in figure 1 and in figure 2 is the sample specimen for the
test.

Figure 1. Universal Testing Machine

Source: Callister,2014

Figure 2. Sample Specimen

Source: Callister,2014

Stress- Strain Diagram

From the data of a tension and compression test, it is possible to compute various values
of the stress and corresponding strain in the specimen and then plot the results. A plot of the
results produces a curve called the stress–strain diagram. There are two ways in which it is
normally described.

1. Conventional Stress-Strain
We can determine the nominal or engineering stress by dividing the applied load P by the
specimen’s original cross-sectional area This calculation assumes that the stress is constant over
the cross section and throughout the gauge length.
𝐹
𝜎=
𝐴𝑜

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Likewise, the nominal or engineering strain is found directly from the strain gauge reading,
or by dividing the change in the specimen’s gauge length, by the specimen’s original gauge length
Here the strain is assumed to be constant throughout the region between the gauge points.

𝛿
∈=
𝐿𝑜

Figure 3.4 shows the conventional and true stress-strain diagram for ductile
material. By plotting the stress-strain diagram, different mechanical properties of the material will
be determined.

Figure 3. conventional and true stress-strain diagram for ductile material

Source: Hibbeler, 2011

Elastic Behavior

Elastic behavior of the material occurs when the strains in the specimen are within the
light orange region shown in the figure above. Here the curve is actually a straight line throughout
most of this region, so that the stress is proportional to the strain. The material in this region is
said to be linear elastic. The upper stress limit to this linear relationship is called the proportional
limit, If the stress slightly exceeds the proportional limit, the curve tends to bend and flatten out
as shown. This continues until the stress reaches the elastic limit. Upon reaching this point, if
the load is removed the specimen will still return back to its original shape.

Yielding

A slight increase in stress above the elastic limit will result in a breakdown of the material
and cause it to deform permanently. This behavior is called yielding, and it is indicated by the
rectangular dark orange region of the curve. The stress that causes yielding is called the yield
stress or yield point, and the deformation that occurs is called plastic deformation. Although not

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shown in Fig. 3, for low-carbon steels or those that are hot rolled, the yield point is often
distinguished by two values. The upper yield point occurs first, followed by a sudden decrease in
load-carrying capacity to a lower yield point.

Strain Hardening.

When yielding has ended, an increase in load can be supported by the specimen, resulting
in a curve that rises continuously but becomes flatter until it reaches a maximum stress referred
to as the ultimate stress. The rise in the curve in this manner is called strain hardening, and it is
identified in Fig. 3 as the region in light green.

Necking

Up to the ultimate stress, as the specimen elongates, its cross-sectional area will
decrease. This decrease is fairly uniform over the specimen’s entire gauge length; however, just
after, at the ultimate stress, the cross-sectional area will begin to decrease in a localized region
of the specimen. As a result, a constriction or “neck” tends to form in this region as the specimen
elongates further. This region of the curve due to necking is indicated in dark green in Fig. 3. Here
the stress–strain diagram tends to curve downward until the specimen breaks at the fracture
stress.

2. True Stress–Strain Diagram

Instead of always using the original cross-sectional area and specimen length to calculate
the (engineering) stress and strain, we could have used the actual cross-sectional area and
specimen length at the instant the load is measured. The values of stress and strain found from
these measurements are called true stress and true strain, and a plot of their values is called
the true stress–strain diagram.

Stress–Strain Behavior of Ductile and Brittle Materials

Materials can be classified as either being ductile or brittle, depending on their stress–
strain characteristics.

Ductile Materials
Any material that can be subjected to large strains before it fractures is called a ductile
material. Mild steel, as discussed previously, is a typical example. Engineers often choose ductile
materials for design because these materials are capable of absorbing shock or energy, and if
they become overloaded, they will usually exhibit large deformation before failing.
One way to specify the ductility of a material is to report its percent elongation or percent
reduction in area at the time of fracture. The percent elongation is the specimen’s fracture strain
expressed as a percent. Thus, if the specimen’s original gauge length is and its length at fracture
is then,

𝑙𝑓 − 𝑙𝑜
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 = 𝑥 100%
𝑙𝑜

The percent reduction in area is another way to specify ductility. It is defined within the
region of necking as follows:

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𝐴𝑜 − 𝐴𝑓
% 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝐴𝑟𝑒𝑎 = 𝑥 100%
𝐴𝑜
Brittle Materials

Materials that exhibit little or no yielding before failure are referred to as brittle materials.

Hooke’s Law

The stress–strain diagrams for most engineering materials exhibit a linear relationship
between stress and strain within the elastic region. Consequently, an increase in stress causes a
proportionate increase in strain. This fact was discovered by Robert Hooke in 1676 using springs
and is known as Hooke’s law. It may be expressed mathematically as:

𝜎=𝐸∈

Here E represents the constant of proportionality, which is called the modulus of elasticity
or Young’s modulus, named after Thomas Young, who published an account of it in 1807.

Strain Hardening

If a specimen of ductile material, such as steel, is loaded into the plastic region and then
unloaded, elastic strain is recovered as the material returns to its equilibrium state. The plastic
strain remains, however, and as a result the material is subjected to a permanent set.

Strain Energy

As a material is deformed by an external loading, it tends to store energy internally

throughout its volume. Since this energy is related to the strains in the material, it is referred to as
strain energy. For applications, it is sometimes convenient to specify the strain energy per unit
volume of material. This is called the strain-energy density, and it can be expressed as:

𝛥𝑈 1
𝑈= = 𝜎𝜖
𝛥𝑉 2
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If the material behavior is linear elastic,

1 𝜎2
𝑈= ( )
2 𝜖
Modulus of Resilience

When stress reaches the proportional limit, the strain-energy density is the modulus of
resilience, Ur.

Modulus of Toughness

This quantity represents the entire area under the stress–strain diagram, Fig. 3–16b, and
therefore it indicates the strain-energy density of the material just before it fractures.

Poisson’s Ratio

Poisson’s Ratio states that in the elastic range, the ratio of these strains is a constant
since the deformations are proportional.

The negative sign is included here since longitudinal elongation (positive strain) causes
lateral contraction (negative strain), and vice versa.

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The Shear Stress–Strain Diagram

For pure shear, equilibrium requires equal shear stresses on each face of the element.

When material is homogeneous and isotropic, shear stress will distort the element
uniformly.

For most engineering materials the elastic behaviour is linear, so Hooke’s Law for shear
applies.

Where: G = shear modulus of elasticity or the modulus of rigidity

For isotropic materials, shear and elastic moduli are related to each other and to Poisson’s
ratio according to:

𝐸 = 2𝐺 (1 + 𝑣)

Failure of Materials Due to Creep and Fatigue

Creep

When a material has to support a load for a very long period of time, it may continue to
deform until a sudden fracture occurs or its usefulness is impaired. This time-dependent
permanent deformation is known as creep. Both stress and/or temperature play a significant role
in the rate of creep. y Creep strength will decrease for higher temperatures or higher applied
stresses.

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Fatigue Failure

When a metal is subjected to repeated cycles of stress or strain, it causes its structure to
break down, ultimately leading to fracture. This behavior is called fatigue, and it is usually
responsible for a large percentage of failures in connecting rods and crankshafts of engines;
steam or gas turbine blades; connections or supports for bridges, railroad wheels, and axles; and
other parts subjected to cyclic loading. In all these cases, fracture will occur at a stress that is less
than the material’s yield stress.

Sample Problem No. 1

A tension test for a steel alloy results in the stress–strain diagram shown in Fig. 3–18.
Calculate the modulus of elasticity and the yield strength based on a 0.2% offset. Identify on the
graph the ultimate stress and the fracture stress.

Solution:

Modulus of Elasticity. We must calculate the slope of the initial straight-line portion of the graph.
Using the magnified curve and scale shown in blue, this line extends from point O to an estimated
point A, which has coordinates of approximately (0.0016 in./in., 50 ksi). Therefore,)

Yield Strength. For a 0.2% offset, we begin at a strain of 0.2% or 0.0020in/in and graphically
extend a (dashed) line parallel to OA until it intersects the σ-ϵ curve at A’. The yield strength is
approximately,
𝜎𝑦𝑠 = 68 𝑘𝑠𝑖

Ultimate Stress. This is defined by the peak of the σ-ϵ graph, points B

𝜎𝑢 = 108 𝑘𝑠𝑖

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Fracture Stress. When the specimen is strained to its maximum of ϵf =0.23in/in, it fractures at
points C.
𝜎𝑢 = 90 𝑘𝑠𝑖
Sample Problem No. 2

An aluminum rod shown in Fig. 3–20a has a circular cross section and is subjected to an
axial load of 10 kN. If a portion of the stress–strain diagram is shown in Fig. 3–20b, determine the
approximate elongation of the rod when the load is applied. Take Eal = 70 GPa.

SOLUTION:
In order to find the elongation of the rod, we must first obtain the strain. This is done by
calculating the stress, then using the stress–strain diagram. The normal stress within each
segment is

From the stress–strain diagram, the material in segment AB is strained elastically since
𝜎𝐴𝐵 = 40 𝑀𝑃𝑎. Using Hooke’s law,

The material within segment BC is strained plastically, since 𝜎𝐵𝐶 > 𝜎𝑌 = 40 𝑀𝑃𝑎. From
the graph, for 𝜎𝐵𝐶 = 56.59𝑀𝑃𝑎, 𝜖𝐵𝐶 ≈ 0.045 𝑚𝑚/𝑚𝑚. The approximate elongation of the rod is
therefore

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Thin-Walled Pressure Vessels

Cylindrical or spherical vessels are commonly used in industry to serve as boilers or tanks.
When under pressure, the material of which they are made is subjected to a loading from all
directions. Although this is the case, the vessel can be analyzed in a simple manner provided it
has a thin wall. Thin wall refers to a vessel having an inner-radius to-wall-thickness ratio of 10 or
more ( r/t ≥ 10). For cylindrical vessels under normal loading, there are normal stresses in the
circumferential or hoop direction and in the longitudinal or axial direction.

TANGENTIAL STRESS (Circumferential Stress)

Consider the tank shown being subjected to an internal pressure p. The length of the tank
is L and the wall thickness is t. Isolating the right half of the tank:

𝑭 = 𝒑𝑨 = 𝒑𝑫𝑳
𝑻 = 𝜎𝑡 𝐴𝑤𝑎𝑙𝑙 = 𝜎𝑡 𝒕𝑳
ΣFx = 0
𝑭 = 𝟐𝑻
𝒑𝑫𝑳 = 𝜎𝑡 𝒕𝑳
Then,
𝑝𝐷
𝜎𝑡 =
2𝑡

If there is exist an external pressure po and an internal pressure pi, the formula may be
expressed as:
(𝑝𝑖 − 𝑝𝑜 )𝐷
𝜎𝑡 =
2𝑡

LONGITUDINAL STRESS, 𝜎𝐿
Consider the free body diagram in the transverse section of the tank:

𝑃𝑇 = 𝜎𝐿 𝐴𝑤𝑎𝑙𝑙 = 𝜎𝐿 𝛱Dt
𝛱
𝐹 = 𝑝 𝐷2
4
ΣFx = 0
𝑃𝑇 = 𝐹
Π
𝛔𝐋 ΠDt = 𝐩 D2
4
Then,
𝑝𝐷
𝜎𝑡 =
4𝑡
If there is exist an external pressure p o and an internal pressure pi, the formula may be
expressed as:
(𝑝𝑖 − 𝑝𝑜 )𝐷
𝜎𝑡 =
4𝑡

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Sample Problem No.3

A cylindrical steel pressure vessel 400 mm in diameter with a wall thickness of 20 mm, is
subjected to an internal pressure of 4.5 MN/m2. Calculate the tangential and longitudinal stresses
in the steel.

Given:
Diameter of cylindrical pressure vessel = 400 mm
Wall thickness = 20 mm
Internal pressure = 4.5 MN/m2
Allowable stress = 120 MN/m2

Required: Longitudinal and tangential stress

Solution:
For tangential stress (longitudinal section)

𝑝𝐷
𝜎𝑡 =
2𝑡
𝑀𝑁
4.5 2 (400𝑚𝑚)
𝜎𝑡 = 𝑚
2(20𝑚𝑚)
𝑀𝑁
𝜎𝑡 = 45 2 = 45 𝑀𝑃𝑎
𝑚

For longitudinal Stress (transverse section):

𝑝𝐷
𝜎𝑡 =
4𝑡
𝑀𝑁
4.5 2 (400𝑚𝑚)
𝜎𝑡 = 𝑚
4(20𝑚𝑚)
𝑀𝑁
𝜎𝑡 = 22.5 2 = 22.5 𝑀𝑃𝑎
𝑚

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Sample Problem No. 4

A cylindrical pressure vessel is fabricated from steel plating that has a thickness of 20 mm.
The diameter of the pressure vessel is 450 mm and its length is 2.0 m. Determine the maximum
internal pressure that can be applied if the longitudinal stress is limited to 140 MPa, and the
circumferential stress is limited to 60 MPa.

Given:
Thickness of steel plating = 20 mm
Diameter of pressure vessel = 450 mm
Length of pressure vessel = 2.0 m
Maximum longitudinal stress = 140 MPa
Maximum circumferential stress = 60 MPa

Required: maximum internal pressure that can be applied

Solution:
Based on circumferential stress (tangential)

𝑝𝐷
𝜎𝑡 =
2𝑡
𝑝(450𝑚𝑚)
60 𝑀𝑃𝑎 =
4(20𝑚𝑚)
𝑝 = 5.33 𝑀𝑃𝑎

Based on longitudinal stress

𝑝( 450𝑚𝑚)
140 𝑀𝑃𝑎 =
2(20𝑚𝑚)
𝑝 = 24.89𝑀𝑃𝑎

Therefore, use 𝑝 = 5.33 𝑀𝑃𝑎

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Stress Concentrations
Stress concentrations occur at sections where the cross-sectional area suddenly changes.
The more severe the change, the larger the stress concentration. Stress concentration produced
by discontinuities in structures such as holes, grooves, notches, fillet etc.

For design or analysis, it is only necessary to determine the maximum stress acting on the
smallest cross-sectional area. This is done using a stress concentration factor, K, that has been
determined through experiment and is only a function of the geometry of the specimen. Normally
the stress concentration in a ductile specimen that is subjected to a static loading will not have to
be considered in design; however, if the material is brittle, or subjected to fatigue loadings, then
stress concentrations become important.
Stress concentration factor is the ratio of maximum stress to the average normal stress
acting at the cross section.
𝜎𝑚𝑎𝑥
𝑲=
𝜎𝑎𝑣𝑔
Provided K is known, and the average normal stress has been calculated from 𝜎𝑚𝑎𝑥 =
𝑃/𝐴, where A is the smallest cross-sectional area, then the maximum normal stress at the section
𝑃
is 𝜎𝑚𝑎𝑥 = 𝐾( ). The value of K can be determine using the figure below.
𝐴

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Sample Problem No. 5

The bar in Fig. 4–29a is made of steel that is assumed to be elastic perfectly plastic, with
Determine (a) the maximum value of the applied load P that can be applied without causing the
steel to yield.

Solution:
When the material behaves elastically, we must use a stress-concentration factor
determined from Fig. 4–24 that is unique for the bar’s geometry. Here

From the figure K ≈1.75.The maximum load, without causing yielding, occurs when
𝜎𝑚𝑎𝑥 = 𝜎𝑌 .

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CHAPTER TEST

Directions: Solve the following problems.

Problem 1.0

A tension test was performed on a specimen having an original diameter of 12.5 mm and
a gauge length of 50 mm. The data are listed in the table. Plot the stress–strain diagram, and
determine approximately the modulus of elasticity, the ultimate stress, and the fracture stress.
Use a scale of 20mm=50 MPa and 20mm= 0.05mm/mm.

Problem 2.0

A bar having a length of 5 in. and cross-sectional area of 0.7 is subjected to an axial force
of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The
material has linear-elastic behavior.

Problem 3.0

A spherical gas tank has an inner radius of If it is subjected to an internal pressure of

determine its required thickness if the maximum normal stress is not to exceed 12 MPa.

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Problem 4.0

The member is to be made from a steel plate that is 0.25 in. thick. If a 1-in. hole is drilled
through its center, determine the approximate width w of the plate so that it can support an axial
force of 3350 lb. The allowable stress is sallow = 22 ksi.

Problem 5.0

A short cylindrical block of 6061-T6 aluminum, having an original diameter of 20 mm and

a length of 75 mm, is placed in a compression machine and squeezed until the axial load applied
is 5 kN. Determine (a) the decrease in its length and (b) its new diameter.

References:
1. HIBBELER, R. C., Mechanics of Materials, 8th Ed. (USA: Pearson Prentice Hall, 2011)
2. PYTEL, A. and SINGER, F.L, Strength of Materials 4 th Ed., (New York: Harper Collins
Publisher Inc.)

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