38.Methods of determining the Stress in a Body subjected to given Forces.—To determine the state of stress, or the state of strain, in an isotropic solid body strained within its limits of elasticity by given forces, we have to use (i.) the equations of equilibrium, (ii.) the conditions which hold at the bounding surface, (iii.) the relations between stress-components and strain-components, (iv.) the relations between strain-components and displacement. The equations of equilibrium are (with notation already used) three partial differential equations of the type∂Xx+∂Xy+∂Zz+ ρX = 0.∂x∂y∂z(1)The conditions which hold at the bounding surface are three equations of the typeXxcos (x, ν) + Xycos (y, ν) + Zxcos (z, ν) =Xν,(2)where ν denotes the direction of the outward-drawn normal to the bounding surface, andXνdenotes the x-component of the applied surface traction. The relations between stress-components and strain-components are expressed by either of the sets of equations (1) or (3) of § 26. The relations between strain-components and displacement are the equations (1) of § 11, or the equivalent conditions of compatibility expressed in equations (1) and (2) of § 16.39. We may proceed by either of two methods. In one method we eliminate the stress-components and the strain-components and retain only the components of displacement. This method leads (with notation already used) to three partial differential equations of the type(λ + μ)∂(∂u+∂v+∂w)+ μ(∂²u+∂²u+∂²u)+ ρX = 0,∂x∂x∂y∂z∂x²∂y²∂z²(3)and three boundary conditions of the typeλ cos (x, ν)(∂u+∂v+∂w)+ μ{2 cos (x, ν)∂u+ cos (y, ν)(∂v+∂u)∂x∂y∂z∂x∂x∂y+ cos (z, ν)(∂u+∂w) }=Xν.∂z∂x(4)In the alternative method we eliminate the strain-components and the displacements. This method leads to a system of partial differential equations to be satisfied by the stress-components. In this system there are three equations of the type∂Xx+∂Xy+∂Xz+ ρX = 0,∂x∂y∂z(1bis)three of the type∂²Xx+∂²Xx+∂²Xx+1∂²(Xx+ Yy+ Zz) =∂x²∂y²∂z²1 + σ∂x²−σρ(∂X+∂Y+∂Z)− 2ρ∂X,1 − σ∂x∂y∂z∂x(5)and three of the type∂²Yz+∂²Yz+∂²Yz+1∂²(Xx+ Yy+ Zz) = − ρ(∂Z+∂Y),∂x²∂y²∂z²1 + σ∂y∂z∂y∂z(6)the equations of the two latter types being necessitated by the conditions of compatibility of strain-components. The solutions of these equations have to be adjusted so that the boundary conditions of the type (2) may be satisfied.40. It is evident that whichever method is adopted the mathematical problem is in general very complicated. It is also evident that, if we attempt to proceed by help of some intuition as to the nature of the stress or strain, our intuition ought to satisfy the tests provided by the above systems of equations. Neglect of this precaution has led to many errors. Another source of frequent error lies in the neglect of the conditions in which the above systems of equations are correct. They are obtained by help of the supposition that the relative displacements of the parts of the strained body are small. The solutions of them must therefore satisfy the test of smallness of the relative displacements.
38.Methods of determining the Stress in a Body subjected to given Forces.—To determine the state of stress, or the state of strain, in an isotropic solid body strained within its limits of elasticity by given forces, we have to use (i.) the equations of equilibrium, (ii.) the conditions which hold at the bounding surface, (iii.) the relations between stress-components and strain-components, (iv.) the relations between strain-components and displacement. The equations of equilibrium are (with notation already used) three partial differential equations of the type
(1)
The conditions which hold at the bounding surface are three equations of the type
Xxcos (x, ν) + Xycos (y, ν) + Zxcos (z, ν) =Xν,
(2)
where ν denotes the direction of the outward-drawn normal to the bounding surface, andXνdenotes the x-component of the applied surface traction. The relations between stress-components and strain-components are expressed by either of the sets of equations (1) or (3) of § 26. The relations between strain-components and displacement are the equations (1) of § 11, or the equivalent conditions of compatibility expressed in equations (1) and (2) of § 16.
39. We may proceed by either of two methods. In one method we eliminate the stress-components and the strain-components and retain only the components of displacement. This method leads (with notation already used) to three partial differential equations of the type
(3)
and three boundary conditions of the type
(4)
In the alternative method we eliminate the strain-components and the displacements. This method leads to a system of partial differential equations to be satisfied by the stress-components. In this system there are three equations of the type
(1bis)
three of the type
(5)
and three of the type
(6)
the equations of the two latter types being necessitated by the conditions of compatibility of strain-components. The solutions of these equations have to be adjusted so that the boundary conditions of the type (2) may be satisfied.
40. It is evident that whichever method is adopted the mathematical problem is in general very complicated. It is also evident that, if we attempt to proceed by help of some intuition as to the nature of the stress or strain, our intuition ought to satisfy the tests provided by the above systems of equations. Neglect of this precaution has led to many errors. Another source of frequent error lies in the neglect of the conditions in which the above systems of equations are correct. They are obtained by help of the supposition that the relative displacements of the parts of the strained body are small. The solutions of them must therefore satisfy the test of smallness of the relative displacements.
41. Torsion.—As a first example of the application of the theory we take the problem of the torsion of prisms. This problem, considered first by C.A. Coulomb in 1784, was finally solved by B. de Saint-Venant in 1855. The problem is this:—A cylindrical or prismatic bar is held twisted by terminal couples; it is required to determine the state of stress and strain in the interior. When the bar is a circular cylinder the problem is easy. Any section is displaced by rotation about the central-line through a small angle, which is proportional to the distance z of the section from a fixed plane at right angles to this line. This plane is a terminal section if one of the two terminal sections is not displaced. The angle through which the section z rotates is τz, where τ is a constant, called the amount of the twist; and this constant τ is equal to G/μI, where G is the twisting couple, and I is the moment of inertia of the cross-section about the central-line. This result is often called “Coulomb’s law.” The stress within the bar is shearing stress, consisting, as it must, of two sets of equal tangential tractions on two sets of planes which are at right angles to each other. These planes are the cross-sections and the axial planes of the bar. The tangential traction at any point of the cross-section is directed at right angles to the axial plane through the point, and the tangential traction on the axial plane is directed parallel to the length of the bar. The amount of either at a distance r from the axis is μτr or Gr/I. The result that G = μτI can be used to determine μ experimentally, for τ may be measured and G and I are known.
42. When the cross-section of the bar is not circular it is clear that this solution fails; for the existence of tangential traction, near the prismatic bounding surface, on any plane which does not cut this surface at right angles, implies the existence of traction applied to this surface. We may attempt to modify the theory by retaining the supposition that the stress consists of shearing stress, involving tangential traction distributed in some way over the cross-sections. Such traction is obviously a necessary constituent of any stress-system which could be produced by terminal couples around the axis.We should then know that there must be equal tangential traction directed along the length of the bar, and exerted across some planes or other which are parallel to this direction. We should also know that, at the bounding surface, these planes must cut this surface at right angles. The corresponding strain would be shearing strain which could involve (i.) a sliding of elements of one cross-section relative to another, (ii.) a relative sliding of elements of the above mentioned planes in the direction of the length of the bar. We could conclude that there may be a longitudinal displacement of the elements of the cross-sections. We should then attempt to satisfy the conditions of the problem by supposing that this is the character of the strain, and that the corresponding displacement consists of (i.) a rotation of the cross-sections in their planes such as we found in the case of the circle, (ii.) a distortion of the cross-sections into curved surfaces by a displacement (w) which is directed normally to their planes and varies in some manner from point to point of these planes. We could show that all the conditions of the problem are satisfied by this assumption, provided that the longitudinal displacement (w), considered as a function of the position of a point (x, y) in the cross-section, satisfies the equation
(1)
and the boundary condition
(2)
where τ denotes the amount of the twist, and ν the direction of the normal to the boundary. The solution is known for a great many forms of section. (In the particular case of a circular section w vanishes.) The tangential traction at any point of the cross-section is directed along the tangent to that curve of the family ψ = const. which passes through the point, ψ being the function determined by the equations
The amount of the twist τ produced by terminal couples of magnitude G is G/C, where C is a constant, called the “torsional rigidity” of the prism, and expressed by the formula
the integration being taken over the cross-section. When the coefficient of μ in the expression for C is known for any section, μ can be determined by experiment with a bar of that form of section.
43. The distortion of the cross-sections into curved surfaces is shown graphically by drawing the contour lines (w = const.). In general the section is divided into a number of compartments, and the portions that lie within two adjacent compartments are respectively concave and convex. This result is illustrated in the accompanying figures (fig. 4 for the ellipse, given by x²/b² + y²/c² = 1; fig. 5 for the equilateral triangle, given by (x +1⁄3a) (x² − 3y² −4⁄3ax +4⁄9a²) = 0; fig. 6 for the square).
44. The distribution of the shearing stress over the cross-section is determined by the function ψ, already introduced. If we draw the curves ψ = const., corresponding to any form of section, for equidifferent values of the constant, the tangential traction at any point on the cross-section is directed along the tangent to that curve of the family which passes through the point, and the magnitude of it is inversely proportional to the distance between consecutive curves of the family. Fig. 7 illustrates the result in the case of theequilateraltriangle. The boundary is, of course, one of the lines. The “lines of shearing stress” which can thus be drawn are in every case identical with the lines of flow of frictionless liquid filling a cylindrical vessel of the same cross-section as the bar, when the liquid circulates in the plane of the section with uniform spin. They are also the same as the contour lines of a flexible and slightly extensible membrane, of which the edge has the same form as the bounding curve of the cross-section of the bar, when the membrane is fixed at the edge and slightly deformed by uniform pressure.
45. Saint-Venant’s theory shows that the true torsional rigidity is in general less than that which would be obtained by extending Coulomb’s law (G = μτI) to sections which are not circular. For an elliptic cylinder of sectional area ω and moment of inertia I about its central-line the torsional rigidity is μω4/ 4π²I, and this formula is not far from being correct for a very large number of sections. For a bar of square section of side a centimetres, the torsional rigidity in C.G.S. units is (0.1406) μa4approximately, μ being expressed in dynes per square centimetre. How great the defect of the true value from that given by extending Coulomb’s law may be in the case of sections with projecting corners is shown by the diagrams (fig. 8 especially no. 4). In these diagrams the upper of the two numbers under each figure indicates the fraction which the true torsional rigidity corresponding to the section is of that value which would be obtained by extending Coulomb’s law; and the lower of the two numbers indicates the ratio which the torsional rigidity for a bar of the corresponding section bears to that of a bar of circular section of the same material and of equal sectional area. These results have an important practical application, inasmuch as they show that strengthening ribs and projections, such as are introduced in engineering to give stiffness to beams, have the reverse of a good effect when torsional stiffness is an object, although they are of great value in increasing the resistance to bending. The theory shows further that the resistance to torsion is very seriously diminished when there is in the surface any dent approaching to a re-entrant angle. At such a place the shearing strain tends to become infinite, and somepermanent set is produced by torsion. In the case of a section of any form, the strain and stress are greatest at points on the contour, and these points are in many cases the points of the contour which are nearest to the centroid of the section. The theory has also been applied to show that a longitudinal flaw near the axis of a shaft transmitting a torsional couple has little influence on the strength of the shaft, but that in the neighbourhood of a similar flaw which is much nearer to the surface than to the axis the shearing strain may be nearly doubled, and thus the possibility of such flaws is a source of weakness against which special provision ought to be made.
46.Bending of Beams.—As a second example of the application of the general theory we take the problem of the flexure of a beam. In this case also we begin by forming a simple intuition as to the nature of the strain and the stress. On the side of the beam towards the centre of curvature the longitudinal filaments must be contracted, and on the other side they must be extended. If we assume that the cross-sections remain plane, and that the central-line is unaltered in length, we see (at once from fig. 9) that the extensions (or contractions) are given by the formula y/R, where y denotes the distance of a longitudinal filament from the plane drawn through the unstrained central-line at right-angles to the plane of bending, and R is the radius of curvature of the curve into which this line is bent (shown by the dotted line in the figure). Corresponding to this strain there must be traction acting across the cross-sections. If we assume that there is no other stress, then the magnitude of the traction in question is Ey/R, where E is Young’s modulus, and it is tension on the side where the filaments are extended and pressure on the side where they are contracted. If the plane of bending contains a set of principal axes of the cross-sections at their centroids, these tractions for the whole cross-section are equivalent to a couple of moment EI/R, where I now denotes the moment of inertia of the cross-section about an axis through its centroid at right angles to the plane of bending, and the plane of the couple is the plane of bending. Thus a beam of any form of section can be held bent in a “principal plane” by terminal couples of moment M, that is to say by a “bending moment” M; the central-line will take a curvature M/EI, so that it becomes an arc of a circle of radius EI/M; and the stress at any point will be tension of amount My/I, where y denotes distance (reckoned positive towards the side remote from the centre of curvature) from that plane which initially contains the central-line and is at right angles to the plane of the couple. This plane is called the “neutral plane.” The restriction that the beam is bent in a principal plane means that the plane of bending contains one set of principal axes of the cross-sections at their centroids; in the case of a beam of rectangular section the plane would bisect two opposite edges at right angles. In order that the theory may hold good the radius of curvature must be very large.
47. In this problem of the bending of a beam by terminal couples the stress is tension, determined as above, and the corresponding strain consists therefore of longitudinal extension of amount My/EI or y/R (contraction if y is negative), accompanied by lateral contraction of amount σMy/EI or σy/R (extension if y is negative), σ being Poisson’s ratio for the material. Our intuition of the nature of the strain was imperfect, inasmuch as it took no account of these lateral strains. The necessity for introducing them was pointed out by Saint-Venant. The effect of them is a change of shape of the cross-sections in their own planes. This is shown in an exaggerated way in fig. 10, where the rectangle ABCD represents the cross-section of the unstrained beam, or a rectangular portion of this cross-section, and the curvilinear figure A′B′C′D′ represents in an exaggerated fashion the cross-section (or the corresponding portion of the cross-section) of the same beam, when bent so that the centre of curvature of the central-line (which is at right angles to the plane of the figure) is on the line EF produced beyond F. The lines A′B′ and C′D′ are approximately circles of radii R/σ, when the central-line is a circle of radius R, and their centres are on the line FE produced beyond E. Thus the neutral plane, and each of the faces that is parallel to it, becomes strained into ananticlastic surface, whose principal curvatures are in the ratio σ : 1. The general appearance of the bent beam is shown in an exaggerated fashion in fig. 11, where the traces of the surface into which the neutral plane is bent are dotted. The result that the ratio of the principal curvatures of the anticlastic surfaces, into which the top and bottom planes of the beam (of rectangular section) are bent, is Poisson’s ratio σ, has been used for the experimental determination of σ. The result that the radius of curvature of the bent central-line is EI/M is used in the experimental determination of E. The quantity EI is often called the “flexural rigidity” of the beam. There are two principal flexural rigidities corresponding to bending in the two principal planes (cf. § 62 below).
48. That this theory requires modification, when the load does not consist simply of terminal couples, can be seen most easily by considering the problem of a beam loaded at one end with a weight W, and supported in a horizontal position at its other end. The forces that are exerted at any section p, to balance the weight W, must reduce statically to a vertical force W and a couple, and these forces arise from the action of the part Ap on the part Bp (see fig. 12),i.e.from the stresses across the section at p. The couple is equal to the moment of the applied load W about an axis drawn through the centroid of the section p at right angles to the plane of bending. This moment is called the “bending moment” at the section, it is the product of the load W and the distance of the section from the loaded end, so that it varies uniformly along the length of the beam. The stress that suffices in the simpler problem gives rise to no vertical force, and it is clear that in addition to longitudinal tensions and pressures there must be tangential tractions on the cross-sections. The resultant of these tangential tractions must be a force equal to W, and directed vertically;but the direction of the traction at a point of the cross-section need not in general be vertical. The existence of tangential traction on the cross-sections implies the existence of equal tangential traction, directed parallel to the central-line, on some planes or other which are parallel to this line, the two sets of tractions forming a shearing stress. We conclude that such shearing stress is a necessary constituent of the stress-system in the beam bent by terminal transverse load. We can develop a theory of this stress-system from the assumptions (i.) that the tension at any point of the cross-section is related to the bending moment at the section by the same law as in the case of uniform bending by terminal couples; (ii.) that, in addition to this tension, there is at any point shearing stress, involving tangential tractions acting in appropriate directions upon the elements of the cross-sections. When these assumptions are made it appears that there is one and only one distribution of shearing stress by which the conditions of the problem can be satisfied. The determination of the amount and direction of this shearing stress, and of the corresponding strains and displacements, was effected by Saint-Venant and R.F.A. Clebsch for a number of forms of section by means of an analysis of the same kind as that employed in the solution of the torsion problem.
49. Let l be the length of the beam, x the distance of the section p from the fixed end A, y the distance of any point below the horizontal plane through the centroid of the section at A, then the bending moment at p is W (l − x), and the longitudinal tension P or Xxat any point on the cross-section is −W (l − x)y/I, and this is related to the bending moment exactly as in the simpler problem.50. The expressions for the shearing stresses depend on the shape of the cross-section. Taking the beam to be of isotropic material and the cross-section to be an ellipse of semiaxes a and b (fig. 13), the a axis being vertical in the unstrained state, and drawing the axis z at right angles to the plane of flexure, we find that the vertical shearing stress U or Xyat any point (y, z) on any cross-section is2W [(a² − y²) {2a² (1 + σ) + b²} − z²a² (1 − 2σ)].πa³b (1 + σ) (3a² + b²)The resultant of these stresses is W, but the amount at the centroid, which is the maximum amount, exceeds the average amount, W/πab, in the ratio{4a² (1 + σ) + 2b²} / (3a² + b²) (1 + σ).If σ = ¼, this ratio is7⁄5for a circle, nearly4⁄3for a flat elliptic bar with the longest diameter vertical, nearly8⁄5for a flat elliptic bar with the longest diameter horizontal.In the same problem the horizontal shearing stress T or Zxat any point on any cross-section is of amount−4Wyz {a² (1 + σ) + b²σ}.πa³b (1 + σ) (3a² + b²)The resultant of these stresses vanishes; but, taking as before σ = ¼, and putting for the three cases above a = b, a = 10b, b = 10a, we find that the ratio of the maximum of this stress to the average vertical shearing stress has the values3⁄5, nearly1⁄15, and nearly 4. Thus the stress T is of considerable importance when the beam is a plank.As another example we may consider a circular tube of external radius r0and internal radius r1. Writing P, U, T for Xx, Xy, Zx, we findP = −4W(l − x)y,π (r04− r14)U =W[(3 + 2σ){r0² + r1² − y² −r0² r1²(y² − z²)}− (1 − 2σ) z²]2(1 + σ) π (r04− r14)(y² + z²)²T = −W{1 + 2σ + (3 + 2σ)r0² r1²}yz;(1 + σ) π (r04− r14)(y² + z²)²and for a tube of radius r and small thickness t the value of P and the maximum values of U and T reduce approximately toP = − W (l − x)y / πr³tUmax.= W / πrt, Tmax.= W / 2πrt.The greatest value of U is in this case approximately twice its average value, but it is possible that these results for the bending of very thin tubes may be seriously at fault if the tube is not plugged, and if the load is not applied in the manner contemplated in the theory (cf. § 55). In such cases the extensions and contractions of the longitudinal filaments may be practically confined to a small part of the material near the ends of the tube, while the rest of the tube is deformed without stretching.
49. Let l be the length of the beam, x the distance of the section p from the fixed end A, y the distance of any point below the horizontal plane through the centroid of the section at A, then the bending moment at p is W (l − x), and the longitudinal tension P or Xxat any point on the cross-section is −W (l − x)y/I, and this is related to the bending moment exactly as in the simpler problem.
50. The expressions for the shearing stresses depend on the shape of the cross-section. Taking the beam to be of isotropic material and the cross-section to be an ellipse of semiaxes a and b (fig. 13), the a axis being vertical in the unstrained state, and drawing the axis z at right angles to the plane of flexure, we find that the vertical shearing stress U or Xyat any point (y, z) on any cross-section is
The resultant of these stresses is W, but the amount at the centroid, which is the maximum amount, exceeds the average amount, W/πab, in the ratio
{4a² (1 + σ) + 2b²} / (3a² + b²) (1 + σ).
If σ = ¼, this ratio is7⁄5for a circle, nearly4⁄3for a flat elliptic bar with the longest diameter vertical, nearly8⁄5for a flat elliptic bar with the longest diameter horizontal.
In the same problem the horizontal shearing stress T or Zxat any point on any cross-section is of amount
The resultant of these stresses vanishes; but, taking as before σ = ¼, and putting for the three cases above a = b, a = 10b, b = 10a, we find that the ratio of the maximum of this stress to the average vertical shearing stress has the values3⁄5, nearly1⁄15, and nearly 4. Thus the stress T is of considerable importance when the beam is a plank.
As another example we may consider a circular tube of external radius r0and internal radius r1. Writing P, U, T for Xx, Xy, Zx, we find
and for a tube of radius r and small thickness t the value of P and the maximum values of U and T reduce approximately to
P = − W (l − x)y / πr³t
Umax.= W / πrt, Tmax.= W / 2πrt.
The greatest value of U is in this case approximately twice its average value, but it is possible that these results for the bending of very thin tubes may be seriously at fault if the tube is not plugged, and if the load is not applied in the manner contemplated in the theory (cf. § 55). In such cases the extensions and contractions of the longitudinal filaments may be practically confined to a small part of the material near the ends of the tube, while the rest of the tube is deformed without stretching.
51. The tangential tractions U, T on the cross-sections are necessarily accompanied by tangential tractions on the longitudinal sections, and on each such section the tangential traction is parallel to the central line; on a vertical section z = const. its amount at any point is T, and on a horizontal section y = const. its amount at any point is U.
The internal stress at any point is completely determined by the components P, U, T, but these are not principal stresses (§ 7). Clebsch has given an elegant geometrical construction for determining the principal stresses at any point when the values of P, U, T are known.
From the point O (fig. 14) draw lines OP, OU, OT, to represent the stresses P, U, T at O, on the cross-section through O, in magnitude, direction and sense, and compound U and T into a resultant represented by OE; the plane EOP is a principal plane of stress at O, and the principal stress at right angles to this plane vanishes. Take M the middle point of OP, and with centre M and radius ME describe a circle cutting the line OP in A and B; then OA and OB represent the magnitudes of the two remaining principal stresses. On AB describe a rectangle ABDC so that DC passes through E; then OC is the direction of the principal stress represented in magnitude by OA, and OD is the direction of the principal stress represented in magnitude by OB.
From the point O (fig. 14) draw lines OP, OU, OT, to represent the stresses P, U, T at O, on the cross-section through O, in magnitude, direction and sense, and compound U and T into a resultant represented by OE; the plane EOP is a principal plane of stress at O, and the principal stress at right angles to this plane vanishes. Take M the middle point of OP, and with centre M and radius ME describe a circle cutting the line OP in A and B; then OA and OB represent the magnitudes of the two remaining principal stresses. On AB describe a rectangle ABDC so that DC passes through E; then OC is the direction of the principal stress represented in magnitude by OA, and OD is the direction of the principal stress represented in magnitude by OB.
52. As regards the strain in the beam, the longitudinal and lateral extensions and contractions depend on the bending moment in the same way as in the simpler problem; but, the bending moment being variable, the anticlastic curvature produced is also variable. In addition to these extensions and contractions there are shearing strains corresponding to the shearing stresses T, U. The shearing strain corresponding to T consists of a relative sliding parallel to the central-line of different longitudinal linear elements combined with a relative sliding in a transverse horizontal direction of elements of different cross-sections; the latter of these is concerned in the production of those displacements by which the variable anticlastic curvature is brought about; to see the effect of the former we may most suitably consider, for the case of an elliptic cross-section, the distortion of the shape of a rectangular portion of a plane of the material which in the natural state was horizontal; all the boundaries of such a portion become parabolas of small curvature, which is variable along the length of the beam, and the particular effect under consideration is the change of the transverse horizontal linear elements from straight lines such as HK to parabolas such as H’K’ (fig. 15); the lines HL and KM are parallel to the central-line, and the figure is drawn for a plane above the neutral plane. When the cross-section is not an ellipse the character of the strain is the same, but the curves are only approximately parabolic.
The shearing strain corresponding to U is a distortion which has the effect that the straight vertical filaments become curved lines which cut the longitudinal filaments obliquely, and thus the cross-sections do not remain plane, but become curved surfaces, and the tangent plane to any one of these surfaces at the centroid cuts the central line obliquely (fig. 16). The angle between these tangent planes and the central-line is the same at all points of the line; and, if it is denoted by ½π + s0, the value of s0is expressible as
and it thus depends on the shape of the cross-section; for the elliptic section of § 50 its value is
for a circle (with σ = ¼) this becomes 7W / 2Eπa². The vertical filament through the centroid of any cross-section becomes a cubical parabola, as shown in fig. 16, and the contour lines of the curved surface into which any cross-section is distorted are shown in fig. 17 for a circular section.
53. The deflection of the beam is determined from the equation
curvature of central line = bending moment ÷ flexural rigidity,
and the special conditions at the supported end; there is no alteration of this statement on account of the shears. As regards the special condition at an end which isencastrée, or built in, Saint-Venant proposed to assume that the central tangent plane of the cross-section at the end is vertical; with this assumption the tangent to the central line at the end is inclined downwards and makes an angle s0with the horizontal (see fig. 18); it is, however, improbable that this condition is exactly realized in practice. In the application of the theory to the experimental determination of Young’s modulus, the small angle which the central-line at the support makes with the horizontal is an unknown quantity, to be eliminated by observation of the deflection at two or more points.
54. We may suppose the displacement in a bent beam to be produced by the following operations: (1) the central-line is deflected into its curved form, (2) the cross-sections are rotated about axes through their centroids at right angles to the plane of flexure so as to make angles equal to ½π + s0with the central-line, (3) each cross-section is distorted in its own plane in such a way that the appropriate variable anticlastic curvature is produced, (4) the cross-sections are further distorted into curved surfaces. The contour lines of fig. 17 show the disturbance from the central tangent plane, not from the original vertical plane.
55.Practical Application of Saint-Venant’s Theory.—The theory above described is exact provided the forces applied to the loaded end, which have W for resultant, are distributed over the terminal section in a particular way, not likely to be realized in practice; and the application to practical problems depends on a principle due to Saint-Venant, to the effect that, except for comparatively small portions of the beam near to the loaded and fixed ends, the resultant only is effective, and its mode of distribution does not seriously affect the internal strain and stress. In fact, the actual stress is that due to forces with the required resultant distributed in the manner contemplated in the theory, superposed upon that due to a certain distribution of forces on each terminal section which, if applied to a rigid body, would keep it in equilibrium; according to Saint-Venant’s principle, the stresses and strains due to such distributions of force are unimportant except near the ends. For this principle to be exactly applicable it is necessary that the length of the beam should be very great compared with any linear dimension of its cross-section; for the practical application it is sufficient that the length should be about ten times the greatest diameter.
56. In recent years the problem of the bending of a beam by loads distributed along its length has been much advanced. It is now practically solved for the case of a load distributed uniformly, or according to any rational algebraic law, and it is also solved for the case where the thickness is small compared with the length and depth, as in a plate girder, and the load is distributed in any way. These solutions are rather complicated and difficult to interpret. The case which has been worked out most fully is that of a transverse load distributed uniformly along the length of the beam. In this case two noteworthy results have been obtained. The first of these is that the central-line in general suffers extension. This result had been found experimentally many years before. In the case of the plate girder loaded uniformly along the top, this extension is just half as great as the extension of the central-line of the same girder when free at the ends, supported along the base, and carrying the same load along the top. The second noteworthy result is that the curvature of the strained central-line is not proportional to the bending moment. Over and above the curvature which would be found from the ordinary relation—
curvature of central-line = bending moment ÷ flexural rigidity,
there is an additional curvature which is the same at all the cross-sections. In ordinary cases, provided the length is large compared with any linear dimension of the cross-section, this additional curvature is small compared with that calculated from the ordinary formula, but it may become important in cases like that of suspension bridges, where a load carried along the middle of the roadway is supported by tensions in rods attached at the sides.
57. When the ordinary relation between the curvature and the bending moment is applied to the calculation of the deflection ofcontinuous beamsit must not be forgotten that a correction of the kind just mentioned may possibly be requisite. In the usual method of treating the problem such corrections are not considered, and the ordinary relation is made the basis of the theory. In order to apply this relation to the calculation of the deflection, it is necessary to know the bending moment at every point; and, since the pressures of the supports are not among the data of the problem, we require a method of determining the bending moments at the supports either by calculation or in some other way. The calculation of the bending moment can be replaced by a method of graphical construction, due to Mohr, and depending on the two following theorems:—
(i.) The curve of the central-line of each span of a beam, when the bending moment M is given,1is identical with the catenary or funicular curve passing through the ends of the span under a (fictitious) load per unit length of the span equal to M/EI, the horizontal tension in the funicular being unity.
(ii.) The directions of the tangents to this funicular curve at the ends of the span are the same for all statically equivalent systems of (fictitious) load.
When M is known, the magnitude of the resultant shearing stress at any section is dM/dx, where x is measured along the beam.
58. Let l be the length of a span of a loaded beam (fig. 19), M1and M2the bending moments at the ends, M the bending moment at a section distant x from the end (M1), M′ the bending moment at the same section when the same span with the same load is simply supported; then M is given by the formulaM = M′ + M1l − x+ M2x,llFig. 19.and thus a fictitious load statically equivalent to M/EI can be easily found when M′ has been found. If we draw a curve (fig. 20) to pass through the ends of the span, so that its ordinate represents the value of M′/EI, the corresponding fictitious loads are statically equivalent to a single load, of amount represented by the area of the curve, placed at the point of the span vertically above the centre of gravity of this area. If PN is the ordinate of this curve, and if at the ends of the span we erect ordinates in the proper sense to represent M1/EI and M2/EI, the bending moment at any point is represented by the length PQ.2For a uniformly distributed load the curve of M’ is a parabola M′ = ½wx (l − x), where w is the load per unit of length; and the statically equivalent fictitious load is1⁄12wl³ / EI placed at the middle point G of the span; also the loads statically equivalent to the fictitious loads M1(l − x) / lEI and M2x / lEI are ½M1l / EI and ½M2l / EI placed at the points g, g′ of trisection of the span. The funicular polygon for the fictitious loads can thus be drawn, and the direction of the central-line at the supports is determined when the bending moments at the supports are known.Fig. 21.Fig. 22.Fig. 23.59. When there is more than one span the funiculars in question may be drawn for each of the spans, and, if the bending moments at the ends of the extreme spans are known, the intermediate ones can be determined. This determination depends on two considerations: (1) the fictitious loads corresponding to the bending moment at any support are proportional to the lengths of the spans which abut on that support; (2) the sides of two funiculars that end at any support coincide in direction. Fig. 21 illustrates the method for the case of a uniform beam on three supports A, B, C, the ends A and C being freely supported. There will be an unknown bending moment M0at B, and the system3of fictitious loads is1⁄12wAB³/EI at G the middle point of AB,1⁄12wBC³ / EI at G′ the middle point of BC, −½M0AB / EI at g and −½M0BC / EI at g′, where g and g′ are the points of trisection nearer to B of the spans AB, BC. The centre of gravity of the two latter is a fixed point independent of M0, and the line VK of the figure is the vertical through this point. We draw AD and CE to represent the loads at G and G’ in magnitude; then D and E are fixed points. We construct any triangle UVW whose sides UV, UW pass through D, B, and whose vertices lie on the verticals gU, VK, g′W; the point F where VW meets DB is a fixed point, and the lines EF, DK are the two sides (2, 4) of the required funiculars which do not pass through A, B or C. The remaining sides (1, 3, 5) can then be drawn, and the side 3 necessarily passes through B; for the triangle UVW and the triangle whose sides are 2, 3, 4 are in perspective.The bending moment M0is represented in the figure by the vertical line BH where H is on the continuation of the side 4, the scale being given byBH=½M0BC;CE1⁄12wBC³this appears from the diagrams of forces, fig. 22, in which the oblique lines are marked to correspond to the sides of the funiculars to which they are parallel.In the application of the method to more complicated cases there are two systems of fixed points corresponding to F, by means of which the sides of the funiculars are drawn.
58. Let l be the length of a span of a loaded beam (fig. 19), M1and M2the bending moments at the ends, M the bending moment at a section distant x from the end (M1), M′ the bending moment at the same section when the same span with the same load is simply supported; then M is given by the formula
and thus a fictitious load statically equivalent to M/EI can be easily found when M′ has been found. If we draw a curve (fig. 20) to pass through the ends of the span, so that its ordinate represents the value of M′/EI, the corresponding fictitious loads are statically equivalent to a single load, of amount represented by the area of the curve, placed at the point of the span vertically above the centre of gravity of this area. If PN is the ordinate of this curve, and if at the ends of the span we erect ordinates in the proper sense to represent M1/EI and M2/EI, the bending moment at any point is represented by the length PQ.2For a uniformly distributed load the curve of M’ is a parabola M′ = ½wx (l − x), where w is the load per unit of length; and the statically equivalent fictitious load is1⁄12wl³ / EI placed at the middle point G of the span; also the loads statically equivalent to the fictitious loads M1(l − x) / lEI and M2x / lEI are ½M1l / EI and ½M2l / EI placed at the points g, g′ of trisection of the span. The funicular polygon for the fictitious loads can thus be drawn, and the direction of the central-line at the supports is determined when the bending moments at the supports are known.
59. When there is more than one span the funiculars in question may be drawn for each of the spans, and, if the bending moments at the ends of the extreme spans are known, the intermediate ones can be determined. This determination depends on two considerations: (1) the fictitious loads corresponding to the bending moment at any support are proportional to the lengths of the spans which abut on that support; (2) the sides of two funiculars that end at any support coincide in direction. Fig. 21 illustrates the method for the case of a uniform beam on three supports A, B, C, the ends A and C being freely supported. There will be an unknown bending moment M0at B, and the system3of fictitious loads is1⁄12wAB³/EI at G the middle point of AB,1⁄12wBC³ / EI at G′ the middle point of BC, −½M0AB / EI at g and −½M0BC / EI at g′, where g and g′ are the points of trisection nearer to B of the spans AB, BC. The centre of gravity of the two latter is a fixed point independent of M0, and the line VK of the figure is the vertical through this point. We draw AD and CE to represent the loads at G and G’ in magnitude; then D and E are fixed points. We construct any triangle UVW whose sides UV, UW pass through D, B, and whose vertices lie on the verticals gU, VK, g′W; the point F where VW meets DB is a fixed point, and the lines EF, DK are the two sides (2, 4) of the required funiculars which do not pass through A, B or C. The remaining sides (1, 3, 5) can then be drawn, and the side 3 necessarily passes through B; for the triangle UVW and the triangle whose sides are 2, 3, 4 are in perspective.
The bending moment M0is represented in the figure by the vertical line BH where H is on the continuation of the side 4, the scale being given by
this appears from the diagrams of forces, fig. 22, in which the oblique lines are marked to correspond to the sides of the funiculars to which they are parallel.
In the application of the method to more complicated cases there are two systems of fixed points corresponding to F, by means of which the sides of the funiculars are drawn.
60.Finite Bending of Thin Rod.—The equation
curvature = bending moment ÷ flexural rigidity
may also be applied to the problem of the flexure in a principal plane of a very thin rod or wire, for which the curvature need not be small. When the forces that produce the flexure are applied at the ends only, the curve into which the central-line is bent is one of a definite family of curves, to which the nameelasticahas been given, and there is a division of the family into two species according as the external forces are applied directly to the ends or are applied to rigid arms attached to the ends; the curves of the former species are characterized by the presence of inflections at all the points at which they cut the line of action of the applied forces.