171 CHAPTER VII MECHANICS AN understanding of stresses first became possible when the laws of elasticity began to get known, but since then the subject has had such fascination for mathematicians, that it is now difficult for men with little leisure to follow their investigations, even although they have been most admirably collected and edited by Mr. K. Pearson (J. Todhunter and K. Pearson). Unfortunately, too, these volumes do not give the results of researches in such a form as to make them of value to practical men. The matter is unquestionably a complicated one, as can be seen by the few formulæ in this chapter, most of which it has been found impracticable to work out in detail, although the necessary hints or references are given. One object of going into these matters is to place in the hands of engineers the means of analysing and criticising the various restrictive rules which from time to time are evolved; the other and more important object is to place in the hands of engineers the means of analysing experiences and experiments notably for ascertaining the stresses when the strains or deformations have been measured. Attention has therefore been paid to the relation between stresses and strains, and as a consequence Poisson's ratio has often to be mentioned. Most text-books content themselves with dealing with elastic deformations, but in boilers, where plates sometimes grow red hot and plastic, and also in all experiments where the metal ruptures, it is of importance to bear in mind that we are not dealing with a purely elastic material, and that factors of safety which are based on ultimate strength are not all of the same standard. For this reason some remarks will be made about plasticity and ultimate strengths. 1 Elasticity. When a prismatic bar is placed in a testing machine and pulled at or compressed, it elongates or contracts, and the amount of deformation per unit of length, compared with the stress per unit of area, is the elasticity of the material. For steel it is about 30,000,0007 using the inch and pound; i.e. a bar 1 in. long would elongate 1: 30,000,000 in. if subjected to a stress of one pound per square inch. If, therefore, a wire were to be placed round a boiler, as explained in connection with fig. 428, and if it were found that during the test a displacement of the vernier took place of in., the length of the wire being say 300 ins., then the elongation per unit of length would be one in 3,000, which corresponds to a stress of 10,000 lbs. per square inch. The reciprocal value of elasticity is the modulus of elasticity, viz. E=30,000,000 lbs. per square inch, for iron or steel; this is 13,363 tons per square inch or 21,000 kilograms per square millimeter. If instead of a cylindrical boiler we were to experiment on a spherical shell, then we would have to remember that there are circumferential stresses in all directions, and further that as a test piece contracts cross-wise when pulled longitudinally, so also does the circumference of the sphere contract, partially due to the cross stresses, and this amount deducts itself from the circumferential elongation. The ratio of this cross contraction varies for different materials; it 1 μ is about 03 for mild steel (C. E. Stromeyer, Proceedings,' 1894, vol. lv. p. 377). We must therefore divide the stresses as estimated from the strain by 07, so that with the same dimensions and displacements as above, the stresses in a shell of a sphere would be not 10,000 lbs., but 14,286 lbs. A wrong view would therefore have led to an underestimation by 30 per cent. In structures which are of less simple form one should measure the strains in at least three directions, so as to find the angles of the principal strains, and then the cross contraction due to each stress has to be taken as a correction for the other. Thus when testing cylindrical furnaces there is, besides the circumferential compression stress, also a very appreciable longitudinal compression stress which should not be neglected. Shearing Elasticity. It has been explained in connection with fig. 125, that a shearing stress is a compound of tension stress and compression stress. The shearing elasticity is the angular displacement of the two diagonals as compared with the shearing stress producing it. As the cross contraction due to the tension stress has to be added to the contraction stress and vice versá, it is evident that under this compound stress the material is more elastic and the modulus of shearing elasticity is smaller than E, the correct formula 30,000,000 being σ = E. = €11,600,000 lbs. per square 2.6 inch. 1 2 (1+1) = Flexibility, or bending elasticity, is best reduced to E. As will shortly be explained, here also the co-efficient of cross contraction has an important influence. Plasticity. The elastic limit and breakdown point have already been explained (p. 158). Resolution of Stresses. Investigations in statics show that several forces acting through one point can always be replaced by a single resultant, and similarly several stresses can be replaced by resultants; but there will be three instead of one, unless they are all acting parallel to one plane, and in that case there will be two resultant stresses as against one resultant force. To resolve stresses which are irregularly distributed in space is a problem which need not be discussed in this chapter, particularly as it leads to rather complicated formulæ (Rankine, R. Soc. Edinburgh,' vol. xxvi. p. 715). The Resolution of Stresses parallel to one plane is comparatively simple. Let there be four stresses, a, b, c, d (fig. 134), acting in the direc 1 tions a1, a2, 3, 4. Double each of these angles and construct the polygon A, a, b, c, d, B (fig. 135). The points A, B may then be looked upon as the two foci of an ellipse, the sum of the radii vectores A, a,, B being equal to the sum of the stresses. By halving the angles 23, and 232, and drawing the stresses a, and b, from a point (fig. 136), a system of two resultants is obtained which would produce exactly the same strain as the four original stresses; but they are not necessarily at right angles to each other. In order to fulfil this condition, radii vectores ACB (fig. 135) will have to be drawn parallel to the major axis, and as 272 27 +180° the resultants will be at right angles to each other, as shown in fig. 137. Their intensities are represented by the lengths AC = x and CB = y. = The following is the algebraical solution of this question. As regards the angles we have AB. sin 271 =a sin 2a, + b sin 2a2+ &c. AB. cos 2y=a cos 2a, + b cos 2a2+ &c. +b sin 2a + &c. a cos 2a + b cos 2a2 + &c. Divide 27, by 2; this determines y1; and add 90°, which is the angle 72. As regards the stresses we have x= y= (a + b + &c.) AB_Σ (S) + √Σ (S . sin 2a)2 + Σ (S. cos 2a)2 2 + = 2 (a + b + &c.) _ AB_Σ (S) — √ ≥ (S. sin 2a)2 + Σ (S. cos 2a)2. 2 If some of the stresses are positive and the others negative, i.e. tension or compression, they should be placed in proper order in the polygon (fig. 135). But then instead of an ellipse, a hyperbola will be the boundary line. It can also be proved that there will only be one resultant to several shearing stresses. Thus, if in fig. 134 they are represented by a, b, c, d, the intensity of the resultant stress would be AB (fig. 135) and its angle 71. The Elastic Beam.-When loaded at one end (fig. 138) the bending moment of a beam, at the distance a from its attachment, is m = Q. (lx). If the longitudinal stresses at this point vary from S at the bottom fibres to S at the top fibres, they combine to form a resisting moment which is equal to m, which for a rectangular section is: Here b is the breadth of the beam, and t its thickness. For a round bar of the diameter d we have In flat plates b can be taken as unity, provided the pressures and forces are measured per unit of width. The Radius of Curvature p can be determined from the following formula: 1 = d2y m 2.S = = Ρ dx2 E.I t.E Here x and y are the co-ordinates of any point of the elastic line, E is the modulus of elasticity, and I is the moment of inertia of the section t3 12 2.h 12, for a flat beam 1 in. wide). When is constant we have d2y dx2 Cross Curvature.-Besides bending lengthways, the edges of the beam tend to curl up (fig. 139); this is due to the cross contraction of the upper side caused by longitudinal tension, and the cross extension of the lower side caused by longitudinal compression. If the radius of this cross curvature be called p1, then traction. In wide plates, which cannot curl crossways when bent, or in such bent plates as are constrained to remain flat crossways, it is evident that FIG. 139 there are internal or external forces at work reducing the cross curvature to a straight line, viz. p1 = ∞, which again means that there are cross stresses C due to bending: Also that the longitudinal bending has been reduced, and p has been increased to p2. 1 1 1 P2 ρ μ.ρι The formulæ for a wide beam which remains straight axially are: Compound Curvatures. If the bent beam is not straight axially, then the stresses, as estimated from the curvatures, have to be combined as follows. Let the radii of the two curvatures be Ρι and P2, then, always assuming 0.3 we have: 1 |