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Dire Dawa Institute of Technology Department of Mechanical And Industrial Engineering. Chapter One &Two SIMPLE STRESS AND STRAIN By: Mesfin Dejene. 1.0 Introduction
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Dire DawaInstitute of TechnologyDepartment of Mechanical And Industrial Engineering Chapter One &Two SIMPLE STRESS AND STRAIN By: MesfinDejene
1.0 Introduction • The fundamental loadings on machine elements are axial loading, direct shear loading, torsion, and bending. Each of these loadings produces stresses in the machine element, as well as deformations, meaning a change in shape. There are only two types of stresses: normal and shear. Axial loading produces a normal stress, direct shear and torsion produce shear stresses, and bending produces both a normal and a shear stress.
1.1 DIRECT OR NORMAL STRESS • When a force is transmitted through a body, the body tends to change its shape or deform. The body is said to be strained. • Direct Stress = Applied Force (F) Perpendicular Cross Sectional Area (A) • Units: Usually N/m2 (Pa), N/mm2, MN/m2, GN/m2 or N/cm2 • Note: 1 N/mm2 = 1 MN/m2 = 1 Mpa • Direct stress may be tensile, or compressive, and result from forces acting perpendicular to the plane of the cross-section.
1.2 Direct or Normal Strain • When loads are applied to a body, some deformation will occur resulting to a change in dimension. • Consider a bar, subjected to axial tensile loading force, F. If the bar extension is dl and its original length (before loading) is L, then tensile strain is: • Alternatively, strain can be expressed as a percentage strain
As strain is a ratio of lengths, it is dimensionless. • Similarly, for compression by amount, dl: Compressive strain = - dl/L • Note: Strain is positive for an increase in dimension and negative for a reduction in dimension. 1.3. Sign convention for direct stress and strain • Tensile stresses and strains are considered POSITIVE in sense producing an increase in length. • Compressive stresses and strains are considered NEGATIVE in sense producing a decrease in length. 1.4. Shear stress • Consider a block or portion of material as shown in Fig. 1.2a subjected to a set of equal and opposite forces Q. There is then a tendency for one layer of the material to slide over another to produce the form of failure shown in Fig. 1.2b. If this failure is restricted, then a shear stress T is set up, defined as follows:
The forces tend to make one part of the material slide over the other part. Shear stress is tangential to the area over which it acts. Fig 1.2 Shear force and resulting shear stress system showing typical form of failure by relative sliding of planes. 1.5. Shear strain • Shear strain is the distortion produced by shear stress on an element or rectangular block as shown in fig 1.3. tan Fig. 1.3 Deformation (shear strain) produced by shear stresses.
1.7. Stress Strain Relation Ship Tensile test • In order to compare the strengths of various materials it is necessary to carry out some standard form of test to establish their relative properties. One such test is the standard tensile test in which a circular bar of uniform cross-section is subjected to a gradually increasing tensile load until failure occurs. • For the first part of the test it will be observed that Hooke’s law is obeyed, i.e. the material behaves elastically and stress is proportional to strain, giving the straight-line graph indicated. Some point A is eventually reached, however when the linear nature of the graph ceases and this point is termed the limit of proportionality.
For a short period beyond this point the material may still be elastic in the sense that deformations are completely recovered when load is removed (i.e. strain returns to zero) but Hooke’s law does not apply. The limiting point B for this condition is termed the elastic limit. • Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will thus be some permanent deformation or permanent set when load is removed. • After the points C, termed the upper yield point, and D, the lower yield point, relatively rapid increases in strain occur without correspondingly high increases in load or stress. • The graph thus becomes much more shallow and covers a much greater portion of the strain axis than does the elastic range of the material. This is applicable for ductile material like steel.
For brittle material like Aluminum, Fig. 1.5
1.6. Elasticity and Hooke’s Law • All solid materials deform when they are stressed, and as stress is increased, deformation also increases. • If a material returns to its original size and shape on removal of load causing deformation, it is said to be elastic. • If the stress is steadily increased, a point is reached when, after the removal of load, not all the induced strain is removed. This is called the elastic limit. • Since loads are proportional to the stresses they produce and deformations are proportional to the strains, this also implies that, whilst materials are elastic, stress is proportional to strain. Hooke’s law, in its simplest form, therefore states that
If a graph of stress and strain is plotted as load is gradually applied, the first portion of the graph will be a straight line. • The slope of this line is the constant of proportionality called modulus of Elasticity, E or Young’s Modulus. It is a measure of the stiffness of a material.
1.8. Factor of Safety • The load which any member of a machine carries is called working load, and stress produced by this load is the working stress. • Obviously, the working stress must be less than the yield stress, tensile strength or the ultimate stress. • This working stress is also called the permissible stress or the allowable stress or the design stress. • Some reasons for factor of safety include the inexactness or inaccuracies in the estimation of stresses and the non-uniformity of some materials. • Or,
1.9. Type of limiting stresses • Ultimate stress • Ultimate stress is used for materials e.g. concrete which do not have a well-defined yield point, or brittle materials which behave in a linear manner up to failure. • Yield stress • Yield stress is used for other materials e.g. steel with well defined yield stress. 1.10. Lateral Strain and Poisson’s Ratio • Under the action of a longitudinal stress, a body will extend in the direction of the stress and contract in the transverse or lateral direction • The reverse occurs under a compressive load. • For most engineering materials the value of v lies between 0.25 and 0.33.
Consider the rectangular bar shown below subjected to a tensile load. Under the action of this load the bar will increase in length by an amount giving a longitudinal strain in the bar of, Fig. 1.6 • The bar will also exhibit, however, a reduction in dimensions laterally, i.e. its breadth and depth will both reduce. The associated lateral strains will both be equal, will be of opposite sense to the longitudinal strain, and will be given by
1.11. Relations between the elastic constants Modulus of rigidity • For materials within the elastic range the shear strain is proportional to the shear stress producing it, • The constant G is termed the modulus of rigidity or shear modulus and is directly comparable to the modulus of elasticity used in the direct stress application. The term modulus thus implies a ratio of stress to strain in each case.
1.12 Deformation in axially loaded members • Consider the rod of uniform cross section under tensile load P along its axis as shown in figure below. • Let that the initial length of the rod be L and the deflection due to load be δ. Using equations (a) and (b), (a) (b) (c) Fig. 1.7 • Equation (c) is obtained under the assumption that the material is homogeneous and has a uniform cross section. • Now, consider another rod of varying cross section with the same axial load P as shown in figure 1.8.
Let us take an infinitesimal element of length dx in the rod that undergoes a deflection due to load P. The strain in the element is • The deflection of total length of the rod can be obtained by integrating above equation, (d) Fig. 1.8 • As the cross sectional area of the rod keeps varying, it is expressed as a function of its length. • If the load is also varying along the length like the weight of the material, it should also be expressed as a function of distance, i.e., P(x) in equation (d).
Also, if the structure consists of several components of different materials, then the deflection of each component is determined and summed up to get the total deflection of the structure. • When the cross section of the components and the axial loads on them are not varying along length, the total deflection of the structure can be determined easily by, (e) 1.13 Statically indeterminate problems • Members for which reaction forces and internal forces can be found out from static equilibrium equations alone are called statically determinate members or structures. • Problems requiring deformation equations in addition to static equilibrium equations to solve for unknown forces are called statically indeterminate problems.
Fig. 1.9 • The reaction force at the support for the bar ABC in figure 1.9 can be determined considering equilibrium equation in the vertical direction. • Now, consider the right side bar MNO in figure 1.9 which is rigidly fixed at both the ends. From static equilibrium, we get only one equation with two unknown reaction forces R1and R2. (f)
Hence, this equilibrium equation should be supplemented with a deflection equation which was discussed in the preceding section to solve for unknowns. • If the bar MNO is separated from its supports and applied the forces, then these forces cause the bar to undergo a deflection R1,R2 and that must be equal to zero. (g) • are the deflections of parts MN and NO respectively in the bar MNO. Individually these deflections are not zero, but their sum must make it to be zero. • Equation (g) is called compatibility equation, which insists that the change in length of the bar must be compatible with the boundary conditions. • Deflection of parts MN and NO due to load P can be obtained by assuming that the • material is within the elastic limit,
Substituting these deflections in equation (g), • Combining equations (f) and (h), one can get, • From these reaction forces, the stresses acting on any section in the bar can be easily determined. 1.14. Temperature stresses • When a material undergoes a change in temperature, it either elongates or contracts depending upon whether heat is added to or removed from the material. • If the elongation or contraction is not restricted, then the material does not experience any stress despite the fact that it undergoes a strain
The strain due to temperature change is called thermal strain and is expressed as (i) • where α is a material property known as coefficient of thermal expansion and ΔT indicates the change in temperature. • Since strain is a dimensionless quantity and ΔT is expressed in K or 0C, α has a unit that is reciprocal of K or 0C. • The free expansion or contraction of materials, when restrained induces stress in the material and it is referred to as thermal stress. • Thermal stress produces the same effect in the material similar to that of mechanical stress and it can be determined as follows. • Consider a rod AB of length L which is fixed at both ends as shown in figure 1.10. • Let the temperature of the rod be raised by ΔT and as the expansion is restricted, the material develops a compressive stress. • In this problem, static equilibrium equations alone are not sufficient to solve for unknowns and hence is called statically indeterminate problem.
To determine the stress due to ΔT, assume that the support at the end B is removed and the material is allowed to expand freely. • Increase in the length of the rod due to free expansion can be found out using equation (a) (j) Fig. 1.10 • Now, apply a compressive load P at the end B to bring it back to its initial position and the deflection due to mechanical load from equation (c), • As the magnitude of and δ are equal and their signs differ, (k)
Minus sign in the equation indicates a compressive stress in the material and with decrease in temperature, the stress developed is tensile stress as ΔT becomes negative. • It is to be noted that the equation (h) was obtained on the assumption that the material is homogeneous and the area of the cross section is uniform. • Thermoplastic analysis assumes significance for structures and components that are experiencing high temperature variations.
1.15. Saint - Venant's Principle: • Consider a slender bar with point loads at its ends as shown in figure above. • The normal stress distribution across sections located at distances b/4 and b from one and of the bar is represented in the figure. • It is found from figure that the stress varies appreciably across the cross section in the immediate vicinity of the application of loads.
The points very near the application of the loads experience a larger stress value whereas, the points far away from it on the same section has lower stress value. • The variation of stress across the cross section is negligible when the section considered is far away, about equal to the width of the bar, from the application of point loads. • Thus, except in the immediate vicinity of the points where the load is applied, the stress distribution may be assumed to be uniform and is independent of the mode of application of loads. This principle is called Saint-Venant's principle.
