mechanics of materials ce 211 chapter -4- part -1- … · mechanics of material 9/27/2017 2 3 -r -s...

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;


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mechanics of materials ce 211 chapter -4- part -1- … · mechanics of material 9/27/2017 2 3 -r -s...

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