Beams are the horizontal members used to support vertically applied loads across an opening. In a more general sense, they are structural members that external loads tend to bend, or curve. Usually, the term beam is applied to members with top continuously connected to bottom throughout their length, and those with top and bottom connected at intervals are called trusses. See also Structural System, Art. 1.7.
Types of Beams
There are many ways in which beams may be supported. Some of the more common methods are shown in Figs. 5.11 to 5.16.
The beam in Fig. 5.11 is called a simply supported, or simple beam. It has
supports near its ends, which restrain it only against vertical movement. The ends of the beam are free to rotate. When the loads have a horizontal component, or when change in length of the beam due to temperature may be important, the supports may also have to prevent horizontal motion. In that case, horizontal restraint at one support is generally sufficient.
The distance between the supports is called the span. The load carried by each support is called a reaction.
The beam in Fig. 5.12 is a cantilever. It has only one support, which restrains it from rotating or moving horizontally or vertically at that end. Such a support is called a fixed end.
If a simple support is placed under the free end of the cantilever, the propped beam in Fig. 5.13 results. It has one end fixed, one end simply supported.
The beam in Fig. 5.14 has both ends fixed. No rotation or vertical movement can occur at either end. In actual practice, a fully fixed end can seldom be obtained.
Some rotation of the beam ends generally is permitted. Most support conditions are intermediate between those for a simple beam and those for a fixed-end beam.
In Fig. 5.15 is shown a beam that overhangs both is simple supports. The overhangs have a free end, like cantilever, but the supports permit rotation.
When a beam extends over several supports, it is called a continuous beam (Fig. 5.16).
Reactions for the beams in Figs. 5.11, 5.12, and 5.15 may be found from the equations of equilibrium. They are classified as statically determinate beams for that reason.
The equations of equilibrium, however, are not sufficient to determine the reactions of the beams in Figs. 5.13, 5.14, and 5.16. For those beams, there are more unknowns than equations. Additional equations must be obtained on the basis of deformations permitted; on the knowledge, for example, that a fixed end permits no rotation. Such beams are classified as statically indeterminate. Methods for finding the stresses in that type of beam are given in Arts. 5.10.4, 5.10.5, 5.11, and 5.13.
As an example of the application of the equations of equilibrium (Art. 5.2.1) to the determination of the reactions of a statically determinate beam, we shall compute the reactions of the 60-ft-long beam with overhangs in Fig. 5.17. This beam carries a uniform load of 200 lb / lin ft over its entire length and several concentrated loads. The supports are 36 ft apart.
To find reaction R1, we take moments about R2 and equate the sum of the moments to zero (clockwise rotation is considered positive, counterclockwise, negative):
Since a beam is in equilibrium under the forces applied to it, it is evident that at every section internal forces are acting to prevent motion. For example, suppose we cut the beam in Fig. 5.17 vertically just to the right of its center. If we total the external forces, including the reaction, to the left of this cut (see Fig. 5.18a), we find there is an unbalanced downward load of 4000 lb. Evidently, at the cut section, an upward-acting internal force of 4000 lb must be present to maintain equilibrium. Again, if we take moments of the external forces about the section, we find an unbalanced moment of 54,000 ft-lb. So there must be an internal moment of 54,000 ft-lb acting to maintain equilibrium.
This internal, or resisting, moment is produced by a couple consisting of a force C acting on the top part of the beam and an equal but opposite force T acting on
the bottom part (Fig. 18b). The top force is the resultant of compressive stresses acting over the upper portion of the beam, and the bottom force is the resultant of tensile stresses acting over the bottom part. The surface at which the stresses change from compression to tension where the stress is zero is called the neutral surface.
s change from compression to tension where the str
The unbalanced external vertical force at a section is called the shear. It is equal to the algebraic sum of the forces that lie on either side of the section. Upward acting forces on the left of the section are considered positive, downward forces negative; signs are reversed for forces on the right.
A diagram in which the shear at every point along the length of a beam is plotted as an ordinate is called a shear diagram. The shear diagram for the beam in Fig. 5.17 is shown in Fig. 5.19b.
The diagram was plotted starting from the left end. The 2000-lb load was plotted downward to a convenient scale.
Then, the shear at the next concentrated load the left support was determined.
This equals 2000 - 200 x 12, or -4400 lb. In passing from must to the left of the support to a point just to the right, however, the shear changes by the magnitude of the reaction. Hence, on
the right-hand side of the left support the shear is -4400 + 14,000, or 9600 lb. At the next concentrated load, the shear is 9600 - 200 x 6, or 8400 lb. In passing the 4000-lb load, however, the shear changes to 8400 - 4000, or 4400 lb. Proceeding in this manner to the right end of the beam, we terminate with a shear of 3000 lb, equal to the load on the free end there.
It should be noted that the shear diagram for a uniform load is a straight line sloping downward to the right (see Fig. 5.21). Therefore, the shear diagram was completed by connecting the plotted points with straight lines.
Shear diagrams for commonly encountered loading conditions are given in Figs. 5.30 to 5.41.
The unbalanced moment of the external forces about a vertical section through a beam is called the bending moment. It is equal to the algebraic sum of the moments about the section of the external forces that lie on one side of the section. Clockwise moments are considered positive, counterclockwise moments negative, when the forces considered lie on the left of the section. Thus, when the bending moment is positive, the bottom of the beam is in tension.
A diagram in which the bending moment at every point along the length of a beam is plotted as an ordinate is called a bending-moment diagram.
Figure 5.20c is the bending-moment diagram for the beam loaded with concentrated loads only in Fig. 5.20a. The bending moment at the supports for this simply supported beam obviously is zero. Between the supports and the first load, the bending moment is proportional to the distance from the support, since it is equal to the reaction times the distance from the support. Hence the bending-moment diagram for this portion of the beam is a sloping straight line.
The bending moment under the 6000-lb load in Fig. 5.20a considering only the force to the left is 7000 x 10, or 70,000 ft-lb. The bending-moment diagram, then, between the left support and the first concentrated load is a straight line rising from??zero at the left end of the beam to 70,000 ft-lb, plotted to a convenient scale, under the 6000-lb load.
The bending moment under the 9000-lb load, considering the forces on the left of it, is 7000 x 20 - 6000 x 10, or 80,000 ft-lb. (It could have been more easily obtained by considering only the force on the right, reversing the sign convention:
8000 x 10 = 80,000 ft-lb.) Since there are no loads between the two concentrated loads, the bending-moment diagram between the two sections is a sloping straight line.
If the bending moment and shear are known at any section of a beam, the bending moment at any other section may be computed, providing there are no unknown forces between the two sections. The rule is:
The bending moment at any section of a beam is equal to the bending moment at any section to the left, plus the shear at that section times the distance between sections, minus the moments of intervening loads. If the section with known moment and share is on the right, the sign convention must be reversed.
For example, the bending moment under the 9000-lb load in Fig. 5.20a could also have been obtained from the moment under the 6000-lb load and the shear to the right of the 6000-lb load given in the shear diagram (Fig. 5.20b). Thus, 80,000 = 70,000 + 1000 x 10. If there had been any other loads between the two concentrated loads, the moment of these loads about the section under the 9000-lb load would have been subtracted.
Bending-moment diagrams for commonly encountered loading conditions are given in Figs. 5.30 to 5.41. These may be combined to obtain bending moments for other loads.
Moments in Uniformly Loaded Beams
When a bean carries a uniform load, the bending-moment diagram does not consist of straight lines. Consider, for example, the beam in Fig. 5.21a, which carries a uniform load over its entire length. As shown in Fig. 5.21c, the bending-moment diagram for this beam is a parabola.
The slope of the bending-moment curve for any point on a beam is equal to the shear at that point; i.e.,
Moving Loads and Influence Lines
One of the most helpful devices for solving problems involving variable or moving loads is an influence line. Whereas shear and moment diagrams evaluate the effect of loads at all sections of a structure, an influence line indicates the effect at a given section of a unit load placed at any point on the structure.
For example, to plot the influence line for bending moment at some point A on a beam, a unit load is applied at some point B. The bending moment is A due to the unit load at B is plotted as an ordinate to a convenient scale at B. The same procedure is followed at every point along the beam and a curve is drawn through the points thus obtained.
Actually, the unit load need not be placed at every point. The equation of the influence line can be determined by placing the load at an arbitrary point and computing the bending moment in general terms. (See also Art. 5.10.5.)
Figure 5.22b shows the influence line for bending moment at the center of a beam. It resembles in appearance the bending-moment diagram for a load at the center of the beam, but its significance is entirely different. Each ordinate gives the moment at midspan for a load at the corresponding location. It indicates that, if a unit load is placed at a distance xL from one end, it produces a bending moment of 1â„2 xL at the center of the span.
Figure 5.22c shows the influence line for shear at the quarter point of a beam.
When the load is to the right of the quarter point, the shear is positive and equal to the left reaction. When the load is to the left, the shear is negative and equal to the right reaction.
The diagram indicates that, to produce maximum shear at the quarter point, loads should be placed only to the right of the quarter point, with the largest load at the quarter point, if possible. For a uniform load, maximum shear results when the load extends from the right end of the beam to the quarter point.
Suppose, for example, that the beam is a crane girder with a span of 60 ft. The wheel loads are 20 and 10 kips, respectively, and are spaced 5 ft apart. For maximum shear at the quarter point, the wheels should be placed with the 20-kip wheel at that point and the 10-kip wheel to the right of it. The corresponding ordinates of the influence line (Fig. 5.22c) are 3â„4 and 40â„45 x 3â„4. Hence, the maximum shear is 20 x 3â„4 + 10 x 40â„45 x 3â„4 = 21.7 kips.
Figure 5.22d shows influence lines for bending moment at several points on a beam. It is noteworthy that the apexes of the diagrams fall on a parabola, as shown by the dashed line. This indicates that the maximum moment produced at any given section by a single concentrated load moving across a beam occurs when the load is at that section. The magnitude of the maximum moment increases when the section is moved toward midspan, in accordance with the equation shown in Fig. 5.22d for the parabola.
Maximum Bending Moment
When there is more than one load on the span, the influence line is useful in developing a criterion for determining the position of the loads for which the bending moment is a maximum at a given section.
Maximum bending moment will occur at a section C of a simple beam as loads move across it when one of the loads is at C. The proper load to place at C is the one for which the expression Wa /a Wb /b (Fig. 5.23) changes sign as that load passes from one side of C to the other.
When several loads move across a simple beam, the maximum bending moment produced in the beam may be near but not necessarily at midspan. To find the maximum moment, first determine the position of the loads for maximum moment at midspan. Then shift the loads until the load P2 that was at the center of the beam is as far from midspan as the resultant of all the loads on the span is on the other side of midspan (Fig. 5.24). Maximum moment will occur under P2.
When other loads move on or off the span during the shift of P2 away from midspan, it may be necessary to investigate the moment under one of the other
loads when it and the resultant are equidistant from midspan.
Bending Stresses in a Beam
To derive the commonly used flexure formula for computing the bending stresses in a beam, we have to make the following assumptions:
1. The unit stress at a point in any plane parallel to the neutral surface of a beam is proportional to the unit strain in the plane at the point.
2. The modulus of elasticity in tension is the same as that in compression.
3. The total and unit axial strain in any plane parallel to the neutral surface are both proportional to the distance of that plane from the neutral surface. (Cross sections that are plane before bending remain plane after bending. This requires that all planes have the same length before bending; thus, that the beam be straight.)
4. The loads act in a plane containing the centroidal axis of the beam and are perpendicular to that axis. Furthermore, the neutral surface is perpendicular to the plane of the loads. Thus, the plane of the loads must contain an axis of symmetry of each cross section of the beam. (The flexure formula does not apply to a beam loaded unsymmetrically. See Arts. 5.5.18 and 5.5.19.)
5. The beam is proportioned to preclude prior failure or serious deformation by torsion, local buckling, shear, or any cause other than bending.
Equating the bending moment to the resisting moment due to the internal stresses at any section of a beam yields M is the bending moment at the section, Æ’ is the normal unit stress in a plane at a distance c from the neutral axis (Fig. 5.25), and I is the moment of inertia of the cross section with respect to the neutral
axis. If Æ’ is given in pounds per square inch (psi), I in in4, and c in inches, then M will be in inch-pounds. For maximum unit stress, c is the distance to the outermost fiber. See also Arts. 5.5.11 and 5.5.12.
Moment of Inertia
The neutral axis in a symmetrical beam is coincidental with the centroidal axis;
i.e., at any section the neutral axis is so located that
Values of I for several common types of cross section are given in Fig. 5.26. Values for structural-steel sections are presented in manuals of the American Institute of Steel Construction, Chicago, Ill. When the moments of inertia of other types of sections are needed, they can be computed directly by application of Eq. (5.56) or by braking the section up into components for which the moment of inertia is known.
If I is the moment of inertia about the neutral axis, A the cross-sectional area, and d the distance between that axis and a parallel axis in the plane of the cross section, then the moment of inertia about the parallel axis is
With this equation, the known moment of inertia of a component of a section about the neutral axis of the component can be transferred to the neutral axis of the complete section. Then, summing up the transferred moments of inertia for all the components yields the moment of inertia of the complete section.
When the moments of inertia of an area with respect to any two perpendicular axes are known, the moment of inertia with respect to any other axis passing through the point of intersection of the two axes may be obtained through the use
The two perpendicular axes through a point about which the moments of inertia are a maximum and a minimum are called the principal axes. The products of inertia are zero for the principal axes.
The ratio S I /c in Eq. (5.54) is called the section modulus. I is the moment of inertia of the cross section about the neutral axis and c the distance from the neutral axis to the outermost fiber. Values of S for common types of sections are given in Fig. 5.26.
Shearing Stresses in a Beam
The vertical shear at any section of a beam is resisted by nonuniformly distributed, vertical unit stresses (Fig. 5.27). At every point in the section, there is also a horizontal unit stress, which is equal in magnitude to the vertical unit shearing stress there [see Eq. (5.34)].
At any distances y from the neutral axis, both the horizontal and vertical shearing unit stresses are equal to
That is, the maximum shear stress is 50% greater than the average shear stress on the section. Similarly, for a circular beam, the maximum is one-third greater than the average. For an I beam, however, the maximum shearing stress in the web is not appreciably greater than the average for the web section alone, if it is assumed that the flanges take no shear.
Combined Shear and Bending Stress
For deep beams on short spans and beams made of low-strength materials, it is sometimes necessary to determine the maximum stress Æ’ on an inclined plane caused by a combination of shear and bending stress v and Æ’, respectively. This stress Æ’, which may be either tension or compression, is greater than the normal stress. Its value may be obtained by application of Mohrs circle (Art. 5.3.6), as indicated in Fig. 5.10, but with Æ’y = 0, and is
The tangential deviation t of a point on the elastic curve is the distance of this point, measured in a direction perpendicular to the original position of the beam, from a tangent drawn at some other point on the elastic curve.
Equation (5.64) indicates that the tangential deviation of any point with respect to a second point on the elastic curve equals the moment about the first point of the M/EI diagram between the two points. The moment-area method for determining the deflection of beams is a technique in which Eqs. (5.63) and (5.64) are utilized.
Suppose, for example, the deflection at midspan is to be computed for a beam of uniform cross section with a concentrated load at the center (Fig. 5.28).
Since the deflection at midspan for this loading is the maximum for the span,
the slope of the elastic curve at the center of the beam is zero; i.e., the tangent is parallel to the undeflected position of the beam. Hence, the deviation of either support from the midspan tangent is equal to the deflection at the center of the beam. Then, by the moment-area theorem [Eq. (5.64)], the deflection yc is given by the moment about either support of the area of the M/EI diagram included between an ordinate at the center of the beam and that support.
This holds, in general, for all simple beams regardless of the type of loading.
The procedure followed in applying Eq. (5.65) to the deflection of the loaded beam in Fig. 5.28 is equivalent to finding the bending moment at D with the M/ EI diagram serving as the load diagram. The technique of applying the M/EI diagram as a load and determining the deflection as a bending moment is known as the conjugate-beam method.
The conjugate beam must have the same length as the given beam; it must be in equilibrium with the M/EI load and the reactions produced by the load; and the bending moment at any section must be equal to the deflection of the given beam at the corresponding section. The last requirement is equivalent to requiring that the shear at any section of the conjugate beam with the M/EI load be equal to the slope of the elastic curve at the corresponding section of the given beam. Figure 5.29 shows the conjugates for various types of beams.
Deflections for several types of loading on simple beams are given in Figs. 5.30 to 5.35 and for overhanging beams and cantilevers in Figs. 5.36 to 5.41.
When a beam carries a number of loads of different types, the most convenient method of computing its deflection generally is to find the deflections separately for the uniform and concentrated loads and add them up.
For several concentrated loads, the easiest solution is to apply the reciprocal theorem (Art. 5.10.5). According to this theorem, if a concentrated load is applied to a beam at a point A, the deflection it produces at point B is equal to the deflection at A for the same load applied at B(dAB dBA).
Suppose, for example, the midspan deflection is to be computed. Then, assume each load in turn applied at the center of the beam and compute the deflection at the point where it originally was applied from the equation of the elastic curve given in Fig. 5.33. The sum of these deflections is the total midspan deflection.
Another method for computing deflections of beams is presented in Art. 5.10.4. This method may also be applied to determining the deflection of a beam due to shear.
Combined Axial and Bending Loads
For stiff beams, subjected to both transverse and axial loading, the stresses are given by the principle of superposition if the deflection due to bending may be neglected without serious error. That is, the total stress is given with sufficient accuracy at any section by the sum of the axial stress and the bending stresses. The maximum stress equals
When the deflection due to bending is large and the axial load produces bending
stresses that cannot be neglected, the maximum stress is given by
An eccentric longitudinal load in the plane of symmetry produces a bending moment Pe where e is the distance of the load from the centroidal axis. The total unit
Figure 5.26 gives values of the radius of gyration for some commonly used cross sections.
For an axial compression load, if there is to be no tension on the cross section, e should not exceed r2 /c. For a rectangular section with width b and depth d, the eccentricity, therefore, should be less than b/6 and d/ 6; i.e., the load should not be applied outside the middle third. For a circular cross section with diameter D, the eccentricity should not exceed D/8.
When the eccentric longitudinal load produces a deflection too large to be neglected in computing the bending stress, account must be taken of the additional bending moment Pd, where d is the deflection. This deflection may be computed by employing Eq. (5.62) or closely approximated by
The principal axes are the two perpendicular axes through the centroid for which the moments of inertia are a maximum or a minimum and for which the products of inertia are zero.
Bending caused by loads that do not lie in a plane containing a principal axis of each cross section of a beam is called unsymmetrical bending. If the bending axis of the beam lies in the plane of the loads, to preclude torsion (see Art. 5.4.1), and if the loads are perpendicular to the bending axis, to preclude axial components, the stress at any point in a cross section is given by
Beams with Unsymmetrical Sections
In the derivation of the flexure formula Æ’ = Mc/ I [Eq. (5.54)], the assumption is made that the beam bends, without twisting, in the plane of the loads and that the neutral surface is perpendicular to the plane of the loads. These assumptions are correct for beams with cross sections symmetrical about two axes when the plane of the loads contains one of these axes. They are not necessarily true for beams that are not doubly symmetrical. The reason is that in beams that are doubly sym
metrical the bending axis coincides with the centroidal axis, whereas in unsymmetrical sections the two axes may be separate. In the latter case, if the plane of the loads contains the centroidal axis but not the bending axis, the beam will be subjected to both bending and torsion.
The bending axis may be defined as the longitudinal line in a beam through which transverse loads must pass to preclude the beams twisting as it bends. The point in each section through which the bending axis passes is called the shear center, or center of twist. The shear center is also the center of rotation of the section in pure torsion (Art. 5.4.1).
Computation of stresses and strains in members subjected to both bending and torsion is complicated, because warping of the cross section and buckling effects should be taken into account. Preferably, twisting should be prevented by use of bracing or avoided by selecting appropriate shapes for the members and by locating and directing loads to pass through the bending axis.
(F. Bleich, Blucking Strength of Metal Structures, McGraw-Hill Publishing Company, New York.)