V. DEMENKO MECHANICS OF MATERIALS 2015 1 LECTURE 4 Stresses in Elastic Solid 1 Concept of Stress In order to characterize the law of internal forces distribution over the section a measure of their intensity should be introduced. The measure used is stress. Fig. 1 Consider an arbitrary cross-section A of elastic body (Fig. 1). We isolate an element of area average stress on the area We reduce the area A enclosing a point the internal force P[N] is acting on. The 2 A[m ] is taken to be ur p av ur P =. (1) A A by contracting it to the point K. Since the medium is continuous, a limiting process is applied as A 0. In the limit we obtain The vector quantity of true stress p r distribution at the point K within section A. r r r P dp p = lim =. (2) A 0 A da represents the intensity of internal forces Stress has the dimension of force divided by area. The SI unit of stress is Pascal: N Pa = or megapascal: 2 m 6 MPa = 10 Pa.
2 V. DEMENKO MECHANICS OF MATERIALS 2015 Fig. 2 The stress in given point of a section considered as a vector quantity, i.e. it is characterized by magnitude and direction. Let us resolve the total stress vector into two components: one, directed along the normal to the section and the other one, lying in the section plane (see Fig. 2). Stress component directed along the normal to the section is called normal stress and designated with greek letter σ. The component lying in the section plane will be called the tangential (shearing) stress and designated with greek letter τ. 2 Relations Between Stresses and Internal Forces Fig. 3 In majority of cases, it is convenient to resolve the total stress into three components directed parallel to coordinate aes. For the point in the cross section of a bar it is shown in Fig. 3. For these components, the following rule of subscripts has been adopted: the first subscript corresponds the coordinate ais perpendicular to the plane; the second subscript gives the coordinate ais to which the given stress is parallel. According to this rule, normal stresses should be written with two subscripts, i.e. σ, but usually one of two subscripts is omitted.
V. DEMENKO MECHANICS OF MATERIALS 2015 3 Let us establish how the stresses and internal force factors in a bar cross section are interrelated. Multiplying the stresses elementary internal forces: σ, τ z and τ y by the area da, we obtain dn dq dq y z = σ = τ = τ y z da, da, da. (3) Summing these elementary forces over the area of the section we find epressions for main vector components of internal forces: N Q Q y z = = = A A A σ τ τ y z da, da, da. (4) Multiplying each of the elementary forces by the distance to the corresponding ais, we obtain the elementary moments of internal forces: ( τ ) ( τ ) dm = zda y+ yda z, dmy = ( σda) z, dmz = ( σda) y. Summing the elementary moments over the entire area of the section, we get the epressions for main moment components of internal forces: (5) A ( τ τ ), M = yz z y da My Mz = σzda, A = σyda. A (6) These formulas will be used in the further discussion for solving one of the principal problems in the theory of strength of materials: the determination of stresses with known internal force factors.
4 V. DEMENKO MECHANICS OF MATERIALS 2015 3 Saint-Venant's Principle The features of eternal forces application are manifested, as a rule, at distances not eceeding the characteristic dimensions of bar cross section (see Fig. 4). Fig. 4 Therefore one should neglect the portion of the bar located in the zone of eternal forces application if you will estimate internal stresses in the bar using the methods of mechanics of materials. Eample of Saint-Venant s principle is illustrated by Fig. 5, where vicinity of P-force application is the Saint-Venant s zone with nonuniform stress distribution. Fig. 5
V. DEMENKO MECHANICS OF MATERIALS 2015 5 4 Principle of Superposition Since deformations the mechanics of materials deals with, are relatively small (elastic), eternal forces may be assumed to act independently from one another, i.e. the deformations and internal forces appearing in elastic bodies do not depend on the order the eternal forces are applied in. Besides, the total effect of the whole system of forces acting on a body is assumed to be the sum of effects produced by each force separately. This is the principle of superposition. 5 Construction of Internal Force Factors Diagrams A graph (plot) of internal force factor distribution along the length of bar is called diagram. There are some rules of significant importance for diagrams constructing: 1. Diagrams of internal force factors are always constructed on aial line which is parallel to aial line of a bar. 2. The force factor magnitude is laid off along a normal to the ais. 3. The diagrams include denominations, dimensions, signs, numerical information. 6 Diagrams of Normal Forces In case only normal force occurs at cross sections of a rod we have tension or compression deformation. Bar in tension-compression will be named as a rod. Eamples of the rods and rod systems are shown in Fig. 6.
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V. DEMENKO MECHANICS OF MATERIALS 2015 7 Fig. 6 Let us consider a bar of constant cross-sectional area (prismatic bar) with two oppositely directed forces applied to its ends: Fig. 7
8 V. DEMENKO MECHANICS OF MATERIALS 2015 Both in tension and compression, the bar is in equilibrium, i.e. the sum of all forces projections onto ais is equal to zero. The normal force N can be found by the method of sections: let the bar be virtually cut in the point where the force N is to be determined and the effect of rejected portion of a bar onto the remaining portion to be replaced by internal force N. In tension, the force Fig. 8 N is directed away from the section, having plus sign. In compression, it is directed to the section, having minus sign. the force Rule If the force N is directed away from the section then it has plus sign; if N is directed to the section, then it has minus sign. Generally, the value of normal force in any cross-section numerically equals to the algebraic sum of eternal forces applied to the part of a rod under consideration. Eample 1 Calculation of normal force in tensile rod Given: The bar of the total length l under the action of the force F. R.D: the normal force along the length of the bar. Consider the equilibrium of remaining portion: In our case, F = 0, F N = 0, N =+ F. N = F = const and graph is a straight line with abrupt changes in the points of forces F, R B Fig. 9 application, i.e. at the ends of the bar, by the magnitude of forces F = R. B
V. DEMENKO MECHANICS OF MATERIALS 2015 9 Eample 2 Calculation of normal force in compressed rod plane A B I F In this case, N = F = const and the graph is a straight line, but N < 0. N N F + F Eample 3 Calculation of normal forces in stepped prismatic bar Given: Bar of a stepped profile: F 1 = 60kN, F 2 = 40kN, F 3 = 70kN. R.D.: N ( ) Fig. 10 Fig. 11 The construction of normal forces diagram is to be started by division of bar into four portions I, II, III and IV according to the forces applied.
10 V. DEMENKO MECHANICS OF MATERIALS 2015 Herewith, we begin the normal forces diagram construction from free end of the bar. force If a number of forces are applied to a bar, the internal force factor, i.e. the normal N, can be found with the method of sections, using the conditions of equilibrium. In the section under consideration, the force algebraic sum of all forces acting on a bar up to the considered section: N is determined as the N k = F, i= 1 i where F i is an i-th force acting on the bar. If it turns out that N > 0, it will be directed away from section and be a tensile force; otherwise, it is directed to the section and is a compressive force: I ( ) = 1=+ 60kN, N II ( ) F1 60 N F IV III ( ) 1 2 60 40 20 = = kn, N = F F = =+ kn, ( ) 1 2 3 60 40 70 50 N = F F F = = kn. One should know the important rule in diagrams of normal forces constructing, since similar rules will be often used in further discussion. In points where an eternal force is applied, including support reactions, a diagram of normal forces shows a jump (abrupt) equal in magnitude to applied eternal force. In our eample, there are four forces (including the support reaction) and the diagram of normal forces has four jumps equal in magnitude to these forces, respectively. 7 Diagrams of Torsional Moments A bar subjected to torsion is called shaft. In shaft torsion, there appears a single internal force factor, torsional moment, or torque, which acts in the plane of shaft cross section. The eamples of torsion deformation are shown in Fig. 12
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12 V. DEMENKO MECHANICS OF MATERIALS 2015 The internal torque appearing in section can be found from the condition of equilibrium for the right-hand or left-hand portion of the loaded shaft. Internal torsional moment in the cross-section, as earlier, will be equal to the algebraical sum of eternal moments applied to the part of the shaft under consideration (left or right). The rule of signs in torque diagram design: if, watching from the end of a shaft, eternal torsional moment rotates counterclockwise, corresponding component of the torque in cross-section of the shaft will be assumed to be negative and vice versa. A graph of internal torsional moments distribution along the length of a shaft is called the torque diagram. Eample 4 Calculation of internal torque moment of the shaft loaded by concentrated eternal moment M T M in the cross-section The condition of equilibrium for the righthand portion of the twisted shaft is M T = M = 0, whence M ( ) = M. T Fig. 13
V. DEMENKO MECHANICS OF MATERIALS 2015 13 Eample 5 Construction of torque moment diagram for the shaft under an arbitrary loading Given: M 1 = 20kNm, M 2 = 50kNm, M 3 = 40kNm. R.D.: M( ) =? The internal torsional moment in a particular section of a shaft is equal to the algebraic sum of all eternal torsional moments which act on the shaft up to the Fig. 14 section considered: k = ( ), M I ( ) M1 20 M M F II i i= 1 = = knm, ( ) 1 2 20 50kNm 30 M = M M = = knm, III ( ) 1 2 3 20 50 40 10 M = M M + M = + =+ knm. Let us now construct the diagram of torsional moments. For each portion of the shaft, we lay off M on a chosen scale in the same way as has been done for constructing the diagram of normal forces N in tension. A diagram of torsional moments has jumped in points of application of eternal torsional moments, which are equal to the magnitude of applied torsional moment. In the considered case, when moving leftwards along the shaft, we observe the first jump equal to M 1, the second jump equal to M 2, the third jump M 3 and the fourth jump is equal to the reaction moment in the built-in end of the shaft.
14 V. DEMENKO MECHANICS OF MATERIALS 2015 Glossary Normal stress (рус. нормальное напряжение, укр. нормальне напруження) the stress component at a point in a structure which is perpendicular to the reference plane. Tangential stress (рус. касательное напряжение, укр. дотичне напруження) A stress in which the material on one side of a surface pushes on the material on the other side of the surface with a force which is parallel to the surface. Also known as shear stress; tangential stress. Compression (рус. сжатие, укр. стиснення) Reduction in the volume of a substance due to pressure; for eample in building, the type of stress which causes shortening of the fibers of a wooden member.