Normal stresses in straight bars and bars with slight curvature
Definition of internal forces
Let a straight bar be loaded by forces of arbitrary direction, with the only assumption that they all lie in one plane (the plane of forces) passing through the axis of the bar (the line connecting the centroids of the cross-sections).
Fig. 1
In order to calculate the stresses acting in a certain cross-section n – n (Fig. 1), it is necessary to know the force that is transmitted through this cross-section from one part of the bar to the other. In magnitude and position, it is obviously equal to the resultant of all forces acting on one part of the bar. This resultant is resolved into components in a coordinate system whose x-axis coincides with the axis of the bar, and whose z-axis is the line along which the plane of the cross-section intersects the force plane. The component in the direction of the z axis is called the transverse force Q or Qz and, in a simple beam, is considered positive if it is directed upward to the left of the section and downward to the right.
When the bar is curved, the coordinate axes x and z have a different direction in every cross-section, always such that the x-axis is tangent to the axis of the bar. If the plane in which the axis of the bar lies does not coincide with the plane of forces, or if the axis of the bar is a spatial curve, the resultant of the forces to the left of the section also has a third component, the transverse force Qy. With N, Qz and Qy, the magnitude and direction of the force transmitted through the section are determined, but not its position. For this, its moments with respect to the coordinate axes must also be known. In the case of plane bending, it is sufficient to know the bending moment M or My. In the most general case, another bending moment Mz and the moment about the x-axis, Mx, appear.
The six forces and moments defined in the preceding discussion are collectively referred to as “section forces.” Their determination is one of the fundamental tasks of the statics of building structures.
In the case of plane bending, there is a simple relationship between the continuously distributed load p (dimension: force per unit length), the transverse force Q, and the bending moment M. Namely, from the equilibrium conditions of a bar element (fig. 2) we obtain
dQ + p dx = 0,
dM = Q dx;
therefore
Q = dM/dx,
p = - dQ/dx = - d2M/dx2. (1a, b)
Fig. 2
Distribution of normal stresses
If a bar is loaded only by an axial force, only a uniformly distributed normal stress σ appears in the cross-section (a tensioned, respectively compressed bar). If a bending moment is also present, these stresses must in some way be unevenly distributed, so that their resultant no longer passes through the centroid of the cross-section. To determine this distribution, technical bending theory does not start from a strict solution of the basic equations, but replaces part of these equations with a plausible assumption that in one particularly simple case is also strictly fulfilled. This special case is the bending of a straight prismatic bar without the action of transverse forces (fig. 3). All parts of the bar sufficiently far from its ends undergo the same deformation, from which it directly follows that the axis of the bar becomes a circular arc and that all planes of cross-sections that were originally perpendicular to the axis of the bar remain plane during deformation, because each of them is a plane of symmetry for adjacent parts of the deformed bar. It is natural to assume that cross-sections remain approximately plane even when the bending moment and the cross-section of the bar are not constant, but change gradually along the bar. This assumption, called the Navier hypothesis, underlies bending theory.
Fig. 3
Magnitude of normal stresses
When a straight beam element is bent, one part of its “fibres”, i.e. material lines parallel to the axis of the bar, is stretched, while the other part is shortened (fig. 4). Between these two parts lie the neutral fibres z=0, whose length does not change during bending. Let the radius of curvature of these fibres in the bent state be ϱ and let it lie in the x-z plane. From fig. 4 m one can read
Δ dx/z = _dx/_ϱ
and from that the strain of the fibers at a distance z from the neutral axis, which is
ε = Δ dx/dx = _z/_ϱ
so, on the basis of Hooke’s law, the stress
σ _= Ez/_ϱ (2)
Fig. 4
This stress is proportional to the distance from the neutral fibers. Such a result is based on the assumption that before deformation all fibers of the bar element were of equal length, so that the bar was originally straight, or at least curved so slightly that the effect of curvature can be neglected. The line along which the neutral layer and the plane of the cross-section intersect is called the neutral axis. We take it as the y-axis (fig. 5). The position of the neutral axis in the cross-section is obtained from the condition of equilibrium of forces in the direction of the x-axis: if the normal force N=0 (bending without normal force), then the resultant of the internal forces expressed through the normal stresses in the cross-section must also be equal to zero:
The integral on the right-hand side is the static moment of the area with respect to the y-axis. For it to be equal to zero, this axis must be one of the centroidal axes.
The moment M that causes deformation has, with respect to the y and z axes in the plane of the cross-section (fig. 5), the components:
Eq. (3)
where Iy, Iyz are the equatorial and centrifugal moments of inertia of the cross-section with respect to the indicated axes.
The most important special case is when Mz=0. This case occurs when Iyz=0, i.e. when y and z are the principal inertia axes of the cross-section. This means that the bending plane adopted here, together with the x-z-plane, coincides with the plane in which the bending moment M=My acts, i.e. with the plane of forces.
Fig. 5
If _E/_ϱ is eliminated from equations (2) and (3), the formula for determining the stress under pure bending (Iy=I) is obtained:
σ = M/I * z (4)
for edge points 1 and 2 in particular is:
σ1 = M/I * h1 = M/W1, σ2 = - M/I * h2 = - M/W2. (5)
The quantities W1,2 = I/h1,2 (dimension: length cubed) are called section moduli. If the horizontal centroidal axis is also an axis of symmetry of the cross-section, then W1 = W2, so the edge stresses are equal in absolute value.
Elastic line
Fig. 6
The significance of the above formulas is tied to Hooke’s law. This requires that σ << E, thus according to (2) h << ϱ. The bending may nevertheless be such that ϱ is of the same order of magnitude as the length of the rod l. In the case of bent rods encountered in construction, this situation does not arise; ϱ is always considerably greater than l, and therefore the deflection w << l (fig. 6). Then approximately the following applies:
1/ ϱ = - d2w/dx2 = -w’’
and on the basis of (3):
w’’ = - M/EI.
This equation and the equilibrium condition (1b) form a system of differential equations for a bent bar. By elimination, they can be reduced to a single fourth-order equation:
wIV = + p/EI
or, in the case that the moment of inertia is variable:
(EIw’’)’’ = +p.
By integrating these, the form of the elastic line can be found when the load p=p(x) is given and when two boundary conditions are known for each case of the bar. Of these, at least two must relate to deformation. For details see the article Statics of Building Structures.
Core of the cross-section
If the neutral axis cuts the cross-section, it divides it into a tension zone and a compression zone. For materials with low tensile strength, the question arises where the compressive force should act, if there must be no tension anywhere in the cross-section. The position of the force application point is then limited to the area called the core of the section. The boundary of the core is evidently the locus of points whose neutral axes are tangent to the cross-section along its perimeter. This also defines its construction.
The distance from the kernel boundary to the centroid, measured in any direction, will be designated by k with the index corresponding to that direction. For the principal axes of inertia, according to the equation
σ = N/F + Mz/Iz y + My/Iy z (6)
the terms are as follows:
ky = i2z / ey, kz = i2y / ez. (7)
For a circular cross-section of radius r, the core distance is k = r/4, and for a circular ring (radii: outer R, inner r) k = (R2 + r2) / 4R.
For a rectangle (fig. 7), ky = 1/6 b, kz = 1/6 h. Thus, when a rectangular cross-section is loaded in one of the planes of symmetry (but only then), it is sufficient for the resultant to act within the middle third of the height or width of the cross-section for the stresses in the section to have the same sign.
Fig. 7
Bending in the plastic range
When the yield point has been reached in the edge fibers of a bent bar, its load-bearing capacity is still not exhausted, because while the outer fibers deform plastically at constant stress, in the parts closer to the neutral axis, where the yield point has not yet been reached, the stress increases, and thus the bending moment that the cross-section can carry also increases.
If the exact course of the stress-strain diagram is known, the process for doubly symmetrical sections (height h) can be easily followed computationally by determining, from the stress-strain diagram and under the assumption that the sections remain plane, the bending stress distribution σ corresponding to some adopted strain of the extreme fibers ε__r and then obtaining the bending moment by integration. If this is carried out for different ε__r and, on the other hand, the corresponding curvature of the deflection line is calculated w’’ = 2 ε__r / h, a series of points is obtained through which a curve can be drawn that gives the relationship between bending moment and curvature M = M(w’’); this relationship is no longer linear, and once it is known, the deflection line corresponding to a given moment distribution can be determined by graphical integration. If the section is unsymmetrical, the calculation becomes more complex, in that for each ε__r the position of the neutral axis must also be determined.
Fig. 8
The calculation is somewhat simplified if, in addition to cross-section symmetry, a simplified relation between stresses and strains is assumed according to fig. 8. Then the stresses due to bending, after the yield limit has been exceeded, have the distribution shown in fig. 9 and the bending moment is:
and the curvature of the deflection line is obtained when equation (2) is applied to the part of the cross-section whose stresses in the elastic region are w’’ = σf / Ez’. If the shape of the cross-section is known for each assumed z’, the integrals can be calculated. For example, for the simplest case of a rectangular cross-section of width b and height h, it is
If z’ is expressed here by means of w’’, one obtains
In I-sections and similar sections, the course of the M-w’’ curve in the plastic range is even flatter, and the moment at which yielding begins is then practically equal to the maximum bending moment the section can carry at all. For this reason, the importance of plastic bending in construction does not lie in increasing the bending moment a section can resist, but in the fact that the large local deformation occurring at the yield limit can lead to a favorable redistribution of forces in statically indeterminate systems.
Fig. 9
This can be explained by the example of a continuous beam over two spans. Let the beam (Fig. 10) have spans of equal length, a constant cross-section, and be loaded with a uniformly distributed load. Then, before the yield limit is reached, the moment over the support MB = _pl_2/8. If the load p is increased so much that the moment in the support section reaches the value Mf at which yielding occurs, then with further increase of the load the curvature at that point will also increase without any appreciable increase in moment. A so-called plastic hinge is formed, i.e. under the additional load the beam behaves as if there were a hinge over the intermediate support. The load can be increased further until yielding also begins at the point of maximum moment in the span. Further increase of the bending moment is no longer possible and the bearing capacity of the beam is exhausted. The load at which this state occurs is called the limit load.
Fig. 10
In general, it can be expected that in an n-times statically indeterminate system, (n+1) plastic hinges must form before the system becomes movable and thus reaches its load-bearing limit. This rule, however, has exceptions in both directions. As the example given shows, at the load-bearing limit two new plastic hinges can form simultaneously (in both spans), so that a system that has hitherto still been immovable suddenly turns into a system with two degrees of freedom. On the other hand, it is possible for load capacity to be locally exhausted even with fewer than (n+1) plastic hinges, for example when only one span of a continuous beam fails. Fig. 11 shows how this can happen in a statically twice indeterminate system with n=2 plastic hinges. If a fourth span is added on the right side of the beam, the beam becomes three times statically indeterminate, so (n-1) plastic hinges are sufficient to bring only the left span to the load-bearing limit.
Fig. 11
Shear stresses in bending
Elementary calculation
Fig. 12
Normal stresses σ, i.e. the stresses that actually arise due to bending, are not the only stresses in a bent bar; shear stresses also occur due to the action of the transverse force. Since the basic characteristic of the engineering theory of bending is that deformations due to shear stresses are neglected, these can be determined only from the conditions of equilibrium. To set this up, the beam element dx is cut, as indicated in fig. 12, by a horizontal plane into two parts, and the equilibrium condition of the horizontal forces acting on the lower part is established. On the left side these are the internal forces expressed by the stresses σ due to bending, which should be integrated from z’ = z to z’ = h2
On the right side the same stresses act, but increased by the differential in accordance with the corresponding bending moment M + dM. The difference between the forces on both sides of the element must be in equilibrium with the force due to the shear stresses τ acting in the horizontal cross-sectional area t dx:
Here the integral represents the static moment of the part of the cross-section cut off by a horizontal transverse plane, relative to the centroidal axis of the section. It is denoted by S. From dM/dx = Q, the following is obtained from the equilibrium condition:
τ = QS/It. (8)
According to the principle of conjugation of shear stresses, this is at the same time the magnitude of the vertical shear stress acting on the cross-sectional area and which, together with S and t, is a function of z.
Fig. 13
Fig. 13.1
Equation (8) provides reliable data for all thin-walled sections if the section plane is not set parallel to the neutral axis, but in the direction of the wall thickness. Accordingly, the I-section is cut transversely through the web and transversely through the flange, which gives the stress distribution shown in fig. 13.1. In the ring section (fig. 13) the shear stresses are distributed symmetrically with respect to the vertical diameter and reach a maximum to the left and right of the neutral line
τ = Q / πrt.
If the cross-section is cut at the highest or lowest point, the distribution of shear stresses does not change; however, if it is cut at the right end of the horizontal diameter (fig. 14), the shear stress at that point is zero, while its value doubles at the left end of the same diameter.
Fig. 14
Solid sections
For solid sections of arbitrary shape, the problem of shear stresses is closely related to torsion. The elementary equation (8) is no longer reliable in that case. For a circular section of radius r, equation (8) gives
τ = 4Q / 3__πr2
as the mean value of the shear stress along the neutral axis, while the exact value is only a few percent higher.
Shear center
If the shear stresses for the U-section are calculated in the manner described above, the stress distribution shown in fig. 15 is obtained. The resultant of the corresponding internal forces is the transverse force Q. From the figure its position relative to the section can be seen, namely that it neither passes through the centroid of the section nor lies in the plane of the web. If the transverse force calculated on the basis of external forces does not have the position shown in fig. 15, the beam is loaded not only in bending, but also in torsion.
If the resultant corresponding to the shear stresses for a given horizontal load is determined for the same cross-section, the line of action of the horizontal transverse force is obtained. The intersection of these two lines of action gives the characteristic point for the given cross-section, called the shear center M. In a member, stresses due to torsion do not occur when the shear centers of all cross-sections lie on one straight line and all loads perpendicular to the axis of the rod pass through that line. Otherwise, the moment of each load with respect to the line connecting the shear centers represents torsional shear.
Fig. 15
Effect of transverse force on bending
The technical theory of bending is based on disregarding deformations caused by shear stresses. Accordingly, it can actually be applied to slender members, for which this neglect is justified and where the transverse force has no significant effect on bending.
Experience, however, has shown that the technical theory of bending can, with some success, also be applied to short, deep members, for which it was in fact not intended at all. This makes it necessary to take the influence of deformation due to shear stresses, which is neglected in this theory, into account afterward, at least approximately.
Just as the shear stresses τ due to transverse forces and the corresponding slips γ_’ =_ τ_/G_ are distributed unevenly over the cross-section, so too are they distributed unevenly over the cross-section. For this reason, their mean value γ is introduced, determined so that the transverse force, at this mean value of deformation, performs on the beam element dx a deformation work equal to that which would be obtained by integrating the work of the shear stresses over the individual volume elements making up the observed beam element:
from which it follows
The integral can be calculated when the distribution of shear stresses over the cross-section is known. Since all shear stresses are proportional to Q/F, its value is proportional to Q2/F2 * F, so if we set it equal to κ__Q2/F, we obtain
γ = κ__Q / GF.
The constant κ is called the shear stress distribution coefficient. For rectangular cross-sections, κ=1,2.
Fig. 16
When the coefficient κ is known, and thus the average slip γ, the additional deflection wQ, caused by the action of transverse forces, can be calculated and added to the deflections w: because according to fig. 16 it is
and therefore because of Q = dM/dx:
where the integration constant is obtained from a boundary condition.