Constants, Variables, and Parameters

Normal Stresses in Bending

In Art. 30, it has been stated that in a beam subjected to pure bending (that is, acted upon by end couples only) the internal forces are normal to the cross section. If transverse forces are present, then there usually is set up on the cross section a system of shearing forces as well as a system of normal forces. It is the concern of this chapter to study the distribution of these internal forces and to calculate their intensities. The consideration of these two kinds of stress will be treated entirely separately: the question of the normal stresses (variously called bending, flexure, or fiber stresses) will be dealt with first.

Imagine, then, a beam subjected to pure bending. To find the distribution of these internal forces over the cross section, the deformation of the bar must be considered. For the simple case of a bar having a longitudinal plane of symmetry with the external bending couples acting in this plane, bending will take place in this same plane. If the bar is of rectangular cross section and two adjacent vertical lines m m and p p are drawn on its sides, direct experiment shows that these lines remain straight during bending and rotate so as to remain perpendicular to the longitudinal fibers¹ of the bar (Fig. 130). The following theory of bending is based on the assumption that not only such lines as \mathrm{mm} remain straight but that:

1. The entire transverse section of the bar, originally plane, remains plane and normal to the longitudinal fibers of the bar after bending.
2. The material is homogeneous and obeys Hooke’s law.
3. Every longitudinal fiber acts as if separate from every other fiber, i.e., there are no lateral pressures nor shearing stresses between the fibers.
4. The beam is straight and of uniform cross section.
5. The moduli of elasticity in tension and compression are equal.

Experiment shows that the theory based on these assumptions gives very accurate results for the deflection of bars and the strain of longitudinal fibers. From the first assumption, it follows that during bending the cross sections m m and p p rotate with respect to each other about axes perpendicular to the plane of bending, so that longitudinal fibers on the convex side undergo extension and those on the concave side compression. The line n q is the trace of the surface in which the fibers do not undergo strain during bending. This surface is called the neutral surface and its intersection with any cross section is called the neutral axis. The elongation st of any fiber, at distance y below the neutral surface, is obtained by drawing the line q s parallel to mm (Fig. 130a). Denoting by \rho the radius of curvature of the deflected axis² of the bar and using the similarity of the triangles n O q and t q s, the unit elongation of the fiber r s is:

\epsilon_{x}=\frac{s t}{n q}=\frac{q s}{n O}=-\frac{y}{\rho}

The minus sign in eq. (a) is needed so that a negative value of y ( q t in Fig. 130a) will make \epsilon_{x} positive when it is a tensile strain corresponding, for instance, to s t in the figure. It can be seen that the strains of the longitudinal fibers are proportional to the distance y from the neutral surface and inversely proportional to the radius of curvature. { }^{3}

[Insert Figure 130 here]

From the strains of the longitudinal fibers, the corresponding stresses follow from Hooke’s law, \sigma_{x}=E \epsilon_{x}, and:

\sigma_{x}=-\frac{E y}{\rho} .

The distribution of these stresses is shown in Fig. 131. The stress in any fiber is proportional to its distance from the neutral axis n n. The position of the neutral axis and the radius of curvature \rho, the two unknowns in eq. (51), can now be determined from the condition that the forces distributed over any cross section of the bar must give rise to a resisting couple which balances the external couple M (see Art. 30).

[Insert Figure 131 here]

The moment of the force on the above element with respect to the neutral axis is³:

d M=-y d F=-y\left(-\frac{E y}{\rho} d A\right) Summarizing such moments over the cross section and putting the resultant equal to the moment M of the external forces, the following equation for determining the radius of curvature \rho is obtained: M=\int \frac{E}{\rho} y^{2} d A=\frac{E I_{z}}{\rho} \text{ or } \frac{1}{\rho}=\frac{M}{E I_{z}}, where I_{z}=\int y^{2} d A is the moment of inertia of the cross section with respect to the neutral axis Z. From eq. (60) it is seen that the curvature 1 / \rho varies directly as the bending moment and inversely as the quantity E I_{z}, which is called the flexural rigidity of the bar. Elimination of \rho from eqs. (59) and (60) gives the following equation for the stresses:

\sigma_{x}=-\frac{M y}{I_{z}} In eq. (61) M is positive when it produces a deflection of the bar convex down, as in Fig. 130; y is positive upwards.

The maximum tensile and compressive stresses occur in the outermost fibers and these maximum stresses are given by the formula obtained from eq. (61):

\left(\sigma_{x}\right)_{\max }=-\frac{M c}{I_{z}}

where c is the distance in inches to the outside fiber being investigated. When the neutral axis is an axis of symmetry, c is the same for both tensile and compressive fibers.

In much of the work in this book, no confusion will result if the subscripts be dropped from \sigma_{x} and I_{x} and the minus sign omitted. This will result in simplification of formula (c) and will bring it into accord with that used by practicing engineers in this country, namely,

\sigma=\frac{M c}{I}.

It is thus understood that s is the normal stress parallel to the longitudinal axis of the beam, M is the bending moment in inch lbs., c is the distance in inches to the extreme fiber and I is the moment of inertia of the cross sectional area about the neutral axis expressed in in. 4 units. If necessary to determine whether a stress be tension or compression, it may readily be found by picturing whether the fiber is being extended or contracted by the bending action.

The quantity I / c is called the section modulus and is frequently denoted by S or Z which is in in. {}^3 units. Formula (62) may be expressed as:

\sigma=\frac{\text{bending moment}}{\text{section modulus}}

In the case of a rectangular cross section (Fig. 130b) we have:

I=\frac{b h^{3}}{12} ; \frac{I}{c}=\frac{b h^{2}}{6}

For a circular cross section of diameter d:

I=\frac{\pi d^{4}}{64} ; \frac{I}{c}=\frac{\pi d^{3}}{32}

For the profile sections used in structural design, the magnitudes of I and I / c are tabulated in handbooks.

The preceding derivation of eq. (61) was for the case of a rectangular cross section. It is readily seen that the same results are also valid for any cross section which has a longitudinal plane of symmetry. If the external bending couples act in such a plane, the internal couple must also act in this plane. The moments of the internal forces (such as shown in Fig. 131) about the horizontal axis are balanced by the external couple. The moments of these forces about the vertical axis must and do cancel each other, because the moments of the forces on one side of this axis are just balanced by the moments of the corresponding forces on the other side.

These results can also be used when no such plane of symmetry exists, provided the bending couples act in an axial plane which contains one of the two principal axes of the cross section. These planes are called the principal planes of bending.

When there is no plane of symmetry, but the bending couples act in an axial plane through one principal axis of the cross section, plane X Y in Fig. 132, a distribution of the stresses following eq. (61) will still satisfy all the conditions of equilibrium. This distribution gives a couple about the horizontal axis which balances the external couple. About the vertical y-axis, it gives a resultant moment:

M_{y}=\int z \frac{E y}{\rho} d A=\frac{E}{\rho} \int y z d A .

[Insert Figure 132 here]

This integral is the product of inertia of the cross section and it is zero if Y and Z are the principal axes of the section. Therefore this couple is zero and, since the component about the y-axis of the applied couple is zero, the conditions of equilibrium are satisfied.

The formulas of this article have been derived on the basis that the beam is subjected to pure bending only. A more accurate analysis, however, shows that the normal stress distribution on a cross section is not affected by the presence of shearing forces caused by transverse loads. The above formulas for normal stresses will therefore be valid for the more usual case of bending occurring under the action of transverse loads which produce shearing components as well as normal components of stress.

Various Shapes of Cross Sections of Beams

From the discussion in the previous paragraph, it follows that the maximum tensile or compressive stresses in a bar in pure bending are proportional to the distances of the most remote fibers from the neutral axis of the cross section. Hence if the material has the same strength in tension and compression, it will be logical to choose those shapes of cross section in which the centroid is at the middle of the depth of the beam. In this manner, the same factor of safety for fibers in tension and fibers in compression will be obtained. This is the underlying idea in the choice of sections symmetrical with respect to the neutral axis for materials such as structural steel, which have about the same yield point in tension and compression. If the section is not symmetrical with respect to the above axis, for example a rail section, the material is frequently distributed between the head and the base so as to have the centroid near the middle of its height.

For a material of low strength in tension and high strength in compression, for example, cast iron or concrete, the advisable cross section for a beam is not symmetrical with respect to the neutral axis but is such that the distances c_{1} and

c_{2} from the neutral axis to the most remote fibers in tension and compression are in the same proportion as the strengths of the material in tension and in compression. In this manner, equal strength in tension and compression is obtained. For example, with a T-section, the centroid of the section may be put in any prescribed position along the height of the section by properly proportioning its flange and web.

For a given bending moment, the maximum stress depends upon the section modulus and it is interesting to note that there are cases in which an increase in area does not give a decrease in this stress. As an example, a bar of square cross section bent by couples acting in the vertical plane through a diagonal of the cross section will have a lower maximum stress if the corners at top and bottom are cut off.

In designing a beam to undergo pure bending, not only the conditions of strength should be satisfied but also the condition of economy in the weight of the beam. Of two cross sections having the same section modulus, that is, satisfying the condition of strength with the same factor of safety, that with the smaller area is more economical.

In comparing various shapes of cross sections, consider first the rectangular section of depth h and width b. The section modulus is:

\frac{I}{c}=\frac{b h^{2}}{6}=\frac{1}{6} A h,

where A denotes the cross-sectional area. It is seen that the rectangular cross section becomes more and more economical with an increase in its depth h. However, there is a certain limit to this increase and the question of the stability of the beam arises as the section becomes narrower. The collapse of a beam of very narrow rectangular section may be due not to overcoming the strength of the material but to sidewise buckling.

In the case of a circular cross section:

\frac{I}{c}=\frac{\pi d^{2}}{32}=\frac{1}{8} A d .

Comparing circular and square cross sections of the same area, the side h of the square will be h=\frac{d \sqrt{\pi}}{2}, for which eq. (a) gives:

I / c=.147 Ad

Comparison of this with (b) shows a square cross section to be more economical than a circular one.

Consideration of the stress distribution along the depth of the cross section (Fig. 131) leads to the conclusion that for economical design most of the material of the beam should be put as far as possible from the neutral axis. This is accomplished in I-beams by putting most of the material in the flanges. At the same time, due to its wide flanges, an I-beam will always be more stable with respect to sidewise buckling (to be discussed in the next paragraph) than a beam of rectangular cross section of the same depth and section modulus.