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Mechanics of Material 9/27/2017
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Ishik University / Sulaimani
Civil Engineering Department
Mechanics of Materials
CE 211
CHAPTER -4- PART -1-
AXIAL LOAD
CHAPTER -4- Axial Load
Outlines of this chapter:
1.Chapter objectives
2.Saint-Venant’s Principle
3.Elastic Deformation of an Axially Loaded Member
4.Principle of Superposition
5.Statically Indeterminate Axially Loaded Member
6.The Force Method of Analysis for Axially Loaded Members
7.Thermal Stress
8.Stress Concentrations
9.Inelastic Axial Deformation
10.Residual Stress 2
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1) Chapter Objectives
In Chapter 1, we developed the method for finding the
normal stress in axially loaded members. In this chapter we
will discuss how to determine the deformation of these
members, and we will also develop a method for finding the
support reactions when these reactions cannot be
determined strictly from the equations of equilibrium. An
analysis of the effects of thermal stress, stress
concentrations, inelastic deformations, and residual stress
will also be discussed.
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Chapter Objectives
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2) Saint-Venant’s Principle
In the previous chapters, we have developed the concept
of stress as a means of measuring the force distribution
within a body and strain as a means of measuring a body’s
deformation. We have also shown that the mathematical
relationship between stress and strain depends on the type
of material from which the body is made. In particular, if
the material behaves in a linear elastic manner, then
Hooke’s law applies, and there is a proportional
relationship between stress and strain.
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Using this idea, consider the manner in which a rectangular bar will deform
elastically when the bar is subjected to a force P applied along its centroidal
axis, Fig. 4–1a. Here the bar is fixed connected at one end, with the force
applied through a hole at its other end. Due to the loading, the bar deforms
as indicated by the once horizontal and vertical grid lines drawn on the bar.
Notice how the localized deformation that occurs at each end tends to even
out and become uniform throughout the midsection of the bar.
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If the material remains elastic then the strains caused by this deformation are
directly related to the stress in the bar. As a result, the stress will be distributed
more uniformly throughout the cross-sectional area when the section is taken
farther and farther from the point where any external load is applied. For
example, consider a profile of the variation of the stress distribution acting at
sections a–a, b–b, and c–c, each of which is shown in Fig. 4–1b. By comparison,
the stress tends to reach a uniform value at section c–c, which is sufficiently
removed from the end since the localized deformation caused by P vanishes. The
minimum distance from the bar’s end where this occurs can be determined
using a mathematical analysis based on the theory of elasticity.
It has been found that this distance should at least be equal to the
largest dimension of the loaded cross section. Hence, section c–c should
be located at a distance at least equal to the width (not the thickness) of the
bar.*
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In the same way, the stress distribution at the support will also even out and become uniform over the cross section located the same distance away from the support.
The fact that stress and deformation behave in this manner is referred to as Saint-Venant’s principle, since it was first noticed by the French scientist Barré de Saint-Venant in 1855. Essentially it states that the stress and strain produced at points in a body sufficiently removed from the region of load application will be the same as the stress and strain produced by any applied loadings that have the same statically equivalent resultant, and are applied to the body within the same region. For example, if two symmetrically applied forces act on the bar, Fig. 4–1c, the stress distribution at section c–c will be uniform and therefore equivalent to as in Fig. 4–1 b. σavg = P/A
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3) Elastic Deformation of an Axially Loaded Member
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**What is the axial load?
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For this general case,
Sign Convention. In order to apply Eq. 4–3, we must develop a sign convention for the internal axial force and the displacement of one end of the bar with respect to the other end. To do so, we will consider both the force and displacement to be positive if they cause tension and elongation, respectively, Fig. 4–4; whereas a negative force and displacement will cause compression and contraction, respectively.
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For example, consider the bar shown in Fig. 4–5a. The internal axial forces “P,” are determined by the method of sections for each segment, Fig. 4–5b. They are PAB = +5 kN, PBC = -3 kN, PCD = -7 kN ,This variation in axial load is shown on the axial or normal force diagram for the bar, Fig. 4–5c. Since we now know how the internal force varies throughout the bar’s length, the displacement of end A relative to end D is determined from;
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Example 4.1
Determine the deformation of the steel rod shown under the given loads.
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Solution;
Divide the rod into components at the load application points.
Apply a free-body analysis
on each component to determine the internal force.
Evaluate the total of the
component deformations.
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• Apply free-body analysis to each component to determine internal forces,
• Evaluate total deformation,
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Example 4.2
The A-36 steel bar shown in Fig. 4–6a is made from two segments having cross-sectional areas of and Determine the vertical displacement of end A and the displacement of B relative to C.
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Internal Force. Due to the application of the external loadings, the internal axial forces in regions AB, BC, and CD will all be different. These forces are obtained by applying the method of sections and the equation of vertical force equilibrium as shown in Fig. 4–6b. This variation is plotted in Fig. 4–6c.
Solution;
Displacement. From the inside back cover Est =
29(103)ksi, Using the sign convention, i.e., internal tensile forces are positive and compressive forces are negative, the vertical displacement of A relative to the fixed support D is
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Solution;
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Example 4.3
Rigid beam AB rests on the two short posts shown in Fig. 4–
8a. AC is made of steel and has a diameter of 20 mm, and
BD is made of aluminum and has a diameter of 40 mm.
Determine the displacement of point F on AB if a vertical
load of 90 kN is applied over this point.
Fig. 4–8
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Solution;
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Solution;
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Solution;
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Solution;
The 20-mm-diameter A-36 steel rod is subjected to the axial forces shown. Determine the displacement of end C with respect to the fixed support at A.
Example 4.4
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Example 4.5
Ans;
Segments AB and CD of the assembly are solid circular rods, and segment BC is a tube. If the assembly is made of 6061-T6 aluminum, determine the displacement of end D with respect to end A.
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4. Principle of Superposition
The principle of superposition is often used to determine the
stress or displacement at a point in a member when the
member is subjected to a complicated loading. By subdividing
the loading into components, the principle of superposition
states that the resultant stress or displacement at the point
can be determined by algebraically summing the stress or
displacement caused by each load component applied
separately to the member.
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5. Statically Indeterminate Axially Loaded Member
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Example 4.6
BC
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Example 4.7
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Solution;
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Example 4.8
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Solution;
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Example 4.9
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Solution;