Uplift Shear Connection

In the preceding examplem the stress normal to the interface AOB (Fig.2.2) was everywhere compressive and equal to w/2b except at the ends of the beam .The stress would have been tensile if the load w had been applied to the lower member. Such loading is unlikely, except when traveling cranes are suspended from the steelwork of a composite floor above: but there are other situations in which stresses tending to cause uplift can occur at the interface. These arise from complex effects such as the torsional stiffness of reinforced concrete slabs forming flanges of composite beams, the triaxial stresses in the vicinity of shear connectors and, in box-girder bridges, the torsional stiffness of the steel box.
Tension across the interface can also occur in beams of non-uniform, Section or with partially completed flanges. Two members without shear connection, as shown in Fig. 2.5,provide a simple example. AB. Is sup-ported on CD and carries distributed loading. It can easily be shown by elastic theory that if the flexural rigidity of AB exceeds about one-tenth of that of CD, then the whole of the load on AB is transferred to CD at points A and B, with separation of the beams between these points. If AB was connected to CD, there would be uplift forces at midspan.

Almost all connectors used in practice are therefore so shaped that they provide resistance to uplift as well as to slip. Uplift forces are so much less than shear forces that it is not normally necessary to calculate or estimate them for design purposes, provided that connectors with some uplift resistance are used.

No shear connection

We assume first that there is no shear connection or friction on the interface AB. The upper beam
cannot deflect more than the lower one, so each carries load w/2 per unit length as if it were an
isolated beam of second moment of area bh3/12, and the vertical compressive stress across the
interface is w/2b. The midspan bending moment in each beam is wL2/16. By elementary beam
theory, the stress distribution at midspan is as in Fig. 2.2.(c), and the maximum bending stress in
each component, σ , is given by

There is an equal and opposite strain in the top fibre of the lower beam, so that the difference between the strains in these adjacent fibres, known as the slip strain, is x 2ε ,

It is easy to show by experiment with two or more flexible wooden laths or rulers that under load, the end faces of the two-component beam have the shape shown in Fig,2.3(a).The slip at the interface, s, is zero at x = 0 is the only one where plane sections remain plane. The slip strain, defined above, is not the same as slip. In the same way that strain is rate of change of displacement,
slip strain is the rate of change of slip along the beam. Thus from(2.4),

The constant of integration is zero, since s = 0 when x = 0, so that (2.6) gives the distribution of slip along the beam.
Results(2.5)and(2.6)for the beam studied in Section 2.7 are plotted in fig.2.3.This shows that at midspan, slip strain is a maximum and slip is zero, and at the ends of the beam, slip is a maximum and slip strain is zero. From (2.6), the maximum slip (when x = L / 2 )is wL3 / 4Ebh2 .Some idea of the magnitude of this slip is given by relating it to the maximum deflection of the two beams. From (2.3), the ratio of slip to deflection is3.2h / L , The ratio L / 2h for a beam is typically about 20,so that the end slip is less than a tenth of the deflection. We conclude that shear
connection must be very stiff if it is to be effective.

Design Provisions for Flexural Members

Design of flexural members requires consideration primarily of bending and shear strength, deflection, and end bearing.

Strength of Flexural Members

The stress induced in a beam (or other flexural member) when subjected to design loads should not exceed the strength of the member. The maximum bending stress ƒb at any section of a beam is given by the flexural formula

ƒb =  M/S

where M is the bending moment and S the section modulus. For a rectangular beam, the section modulus is bd2 /6 and Eq. (10.14) transforms into

ƒ = 6M/bd^2

where b is the beam width and d the depth. At every section of the beam, ƒb should be equal to or less than the design value for bending Fb adjusted for all end-use modification factors (Art. 10.5).
Shear stress induced by design loads in a member should not exceed the allowable design value for shear FV. For wood beams, the shear parallel to the grain, that is, the horizontal shear, controls the design for shear. Checking the shear stress perpendicular to the grain is not necessary inasmuch as the vertical shear will never be a primary failure mode.
The maximum horizontal shear stress ƒV in a rectangular wood beam is given by

ƒ = 3V/2bd

where V is the vertical shear. In calculation of V for a beam, all loads occurring within a distance d from the supports may be ignored. This is based on the assumption that loads causing the shear will be transmitted at a 45 angle through the beam to the supports.
(K. F. Faherty and T. G. Williamson, ‘‘Wood Engineering and Construction Handbook.’’ McGraw-Hill Publishing Company, New York.)

Beam Stability

Beams may require lateral support to prevent lateral buckling. Need for such bracing depends on the unsupported length and cross-sectional dimensions of the members.
When buckling occurs, a member deflects in the direction of its least dimension b.
In a beam, b usually is taken as the width. If bracing precludes buckling in that direction, deflection can still occur, but in the direction of the strong dimension.

Thus, the unsupported length L, width b, and depth d are key variables in formulas for lateral support and for reduction of design values for buckling.
Design for lateral stability of flexural members is based on a function of Ld/b2.
The beam stability factor, CL, for lumber beams of rectangular cross section having maximum depth-width ratios based on nominal dimensions, as summarized in Table 10.13, can be taken as unity.
No lateral support is required when the depth does not exceed the width of a beam. In that case also, the design value for bending does not have to be adjusted for lateral stability. Similarly, if continuous support prevents lateral movement of the compression flange, and the ends at points of bearing are braced to prevent rotation, then lateral buckling cannot occur and the design of value Fb need not be reduced.
When the beam depth exceeds the width, lateral support should be provided at end bearings. This support should be so placed as to prevent rotation of the beam about the longitudinal axis. Unless the compression flange is braced at sufficiently close intervals between supports, the design value should be adjusted for lateral buckling.
The slenderness ratio RB for beams is defined by

The effective length Le for Eq. (10.17) is given in Table 10.14 in terms of the unsupported length of beam. Unsupported length is the distance between supports or the length of a cantilever when the beam is laterally braced at the supports to prevent rotation and adequate bracing is not installed elsewhere in the span. When both rotational and lateral displacement are also prevented at intermediate points, the unsupported length may be taken as the distance between points of lateral support. If the compression edge is supported throughout the length of the beam and adequate bracing is installed at the supports, the unsupported length is zero.

Deflection of Wood Beams

The design of many structural systems, particularly those with long span, may be governed by deflection. Verifying structural adequacy based on allowable stresses alone may not be sufficient to prevent excessive deflection. Limitations on deflection may increase member stiffness.

Deflection of wood beams is calculated by conventional elastic theory. For example, for a uniformly loaded, simple-span beam, the maximum deflection is computed from

Deflection should not exceed limitations specified in the local building code nor industry-recommended limitations. (See, for example, K. F. Faherty and T. G. Williamson, ‘‘Wood Engineering and Construction Handbook,’’ McGraw-Hill Publishing Company, New York.) Deflections also should be evaluated with respect to other considerations, such as possibility of binding of doors or cracking of partitions or glass.
Table 10.15 gives recommended deflection limits, as a fraction of the beam span, for timber beams. The limitation applies to live load or total load, whichever governs.
Glulam beams may be cambered to offset the effects of deflections due to design loads. These members are cambered during fabrication by creation of curvature  opposite in direction to that of deflections under load. Camber, however, does not increase stiffness. Table 10.16 lists recommended minimum cambers for glulam  beams.
Minimum Roof Slopes. Flat roofs have collapsed during rainstorms even though they were adequately designed for allowable stresses and definite deflection limitations.
The failures were caused by ponding of water as increasing deflections permitted more and more water to collect.
Roof beams should have a continuous upward slope equivalent to 1⁄4 in / ft between  a drain and the high point of a roof, in addition to minimum recommended camber (Table 10.16), to avoid ponding. As a general guideline, when flat roofs have insufficient slope for drainage (less than 1⁄4 in / ft), the stiffness of supporting members should be such that a 5-lb / ft2 load will cause no more than 1⁄2-in deflection.

Because of ponding, snow loads or water trapped by gravel stops, parapet walls, or ice dams magnify stresses and deflections from existing roof loads by

where Cp  factor for multiplying stresses and deflections under existing loads to determine stresses and deflections under existing loads plus ponding
W  weight of 1 in of water on roof area supported by beam, lb
L  span of beam, in
E  modulus of elasticity of beam material, psi
I  moment of inertia of beam, in4
(Kuenzi and Bohannan, ‘‘Increases in Deflection and Stresses Caused by Ponding of Water on Roofs,’’ Forest Products Laboratory, Madison, Wis.)

Bearing Stresses in Beams

Bearing stresses, or compression stresses perpendicular to the grain, in a beam occur at the supports or at places where other framing members are supported on the beam. The compressive stress in the beam ƒc is given by

where P = load transmitted to or from the beam and A  bearing area. This stress should be less than the design value for compression perpendicular to the grain Fc  multiplied by applicable adjustment factors (Art. 10.5). (The duration-of-load factor does not apply to Fc for either solid sawn lumber of glulam timber.)

Limitations on compressive stress perpendicular to the grain are set to keep deformations within an acceptable range. An expected failure mode is excessive localized deformation rather than a catastrophic type of failure.
Design values for Fc are averages based on a maximum deformation of 0.04 in in tests conforming with ASTM D143. Design values Fc for glulam beams are generally lower than for solid sawn lumber with the same deformation limit. This is due partly to use of larger-size sections for glulam beams, length of bearing and partly to the method used to derive the design values.
Where deformations are critical, the deformation limit may be decreased, with resulting reduction in Fc. For example, for a deformation maximum of 0.02 in.
the ‘‘National Design Specification for Wood Construction,’’ American Forest & Paper Association, recommends that Fc, psi, be reduced to 0.73 Fc.

Cantilevered-Span Construction

Cantilever systems may be composed of any of the various types and combinations of beam illustrated in Fig. 10.1. Cantilever systems permit longer spans or larger loads for a given size member than do simple-span systems, if member size is not controlled by compression perpendicular to grain at the supports or by horizontal shear. Substantial design economies can be effected by decreasing the depths of the members in the suspended portions of a cantilever system.
For economy, the negative bending moment at the supports of a cantilevered beam should be equal in magnitude to the positive moment.
Consideration must be given to deflection and camber in cantilevered multiple spans. When possible, roofs should be sloped the equivalent of 1⁄4 in per foot of horizontal distance between the level of drains and the high point of the roof to eliminate water pockets, or provisions should be made to ensure that accumulation of water does not produce greater deflection and live loads than anticipated. Unbalanced loading conditions should be investigated for maximum bending moment, deflection, and stability.
(For further information on the design of cantilevered beam systems, see K. F. Faherty and T. G. Williamson, ‘‘Wood Engineering and Construction Handbook,’’ 2d ed., McGraw-Hill Publishing Company, New York; D. E. Breyer, ‘‘Design of Wood Structures,’’ 3d ed., McGraw-Hill Publishing Company, New York; ‘‘Wood Structural Design Data,’’ American Forest and Paper Association, Washington, D.C.)

Composite Construction

In composite construction, rolled or built-up steel shapes are combined with reinforced concrete to form a structural member. Examples of this type of construction include: (a) concrete-encased steel beams (Fig. 7.37c), (b) concrete decks interactive with steel beams (Fig. 7.37a and b), (c) concrete encased steel columns, and (d) concrete filled steel columns. The most common use of this type of construction is for composite beams, where the steel beam supports and works with the concrete slab to form an economical building element.
Design procedures require that a decision be made regarding the use of shoring for the deck pour. (Procedures for ASD and LRFD differ in this regard.) If shoring is not used, the steel beam must carry all dead loads applied until the concrete hardens, even if full plastic capacity is permitted for the composite section afterward.
The assumed composite cross section is the same for ASD and LRFD procedures.
The effective width of the slab is governed by beam span and beam spacing or edge distance (Fig. 7.37a and b).
Slab compressive stresses are seldom critical for interior beams but should be investigated, especially for edge beams. Thickening the slab key and minimum requirements for strength of concrete can be economical.

Connector Details. In composite construction, shear connectors welded to the top flange of the steel beam are typically used to ensure composite action by transferring shear between the concrete deck and steel beam. Location, spacing, and size limitations for shear connectors are the same for ASD and LRFD procedures. Connectors, except those installed in ribs of formed steel decks, should have a minimum lateral concrete cover of 1 in. The diameter of a stud connector, unless located directly over the beam web, is limited to 2.5 times the thickness of the beam flange to which it is welded. Minimum center-to-center stud spacing is 6 diameters along the longitudinal axis, 4 diameters transversely. Studs may be spaced uniformly, rather than in proportion to horizontal shear, inasmuch as tests show a redistribution of shear under high loads similar to the stress redistribution in large bolted joints.
Maximum spacing is 8 times the slab thickness.
Formed Steel Decking. Concrete slabs are frequently cast on permanent steel decking with a ribbed, corrugated, cellular, or blended cellular cross section (see Sec. 8). Two distinct composite-design configurations are inherent: ribs parallel or ribs perpendicular to the supporting beams or girders (Fig. 7.38) The design procedures, for both ASD and LRFD, prescribed for composite concrete-slab and steelbeam construction are also applicable for systems utilizing formed steel decking, subject to additional requirements of the AISC ‘‘Specification for Structural Steel for Buildings’’ and as illustrated in Fig. 7.38.

Shear and Deflection of Composite Beams. In ASD and LRFD, shear forces are assumed to be resisted by the steel beam. Deflections are calculated based on composite section properties. It should be noted that, because of creep of the concrete, the actual deflections of composite beams under long-term loads, such as dead load, will be greater than those computed.

ASD of Encased Beams

Two design methods are allowed for encased beams. In one method, stresses are computed on the assumption that the steel beam alone supports all the dead load applied prior to concrete hardening (unless the beam is temporarily shored), and the composite beam supports the remaining dead and live loads. Then, for positive bending moments, the total stress, ksi, on the steel-beam bottom flange is

ASD of Beams with Shear Connectors

For composite construction where shear connectors transfer shear between slab and beam, the design is based on behavior at ultimate load. It assumes that all loads are resisted by the composite section, even if shores are not used during construction to support the steel beam until the concrete gains strength. For this case, the computed stress in the bottom flange for positive bending moment is

In continuous composite beams, where shear connectors are installed in negativemoment regions, the longitudinal reinforcing steel in the concrete slab may be considered to act compositely with the steel beam in those regions. In such cases, the total horizontal shear to be resisted by the shear connectors between an interior support and each adjacent infection point is

where Asr  total area, in2, of longitudinal reinforcing steel within the effective width of the concrete slab at the interior support Fyr  specified yield stress of the reinforcing steel, ksi
These formulas represent the horizontal shear at ultimate load divided by 2 to approximate conditions at working load.
Number of Connectors. The minimum number of connectors N1, spaced uniformly between the point of maximum moment and adjacent points of zero moment, is Vh /q, where q is the allowable shear load on a single connector, as given in Table 7.22. Values in this table, however, are applicable only to concrete made with aggregates conforming to ASTM C33. For concrete made with rotary-kiln-produced aggregates conforming to ASTM C330 and with concrete weight of 90 pcf or more, the allowable shear load for one connector is obtained by multiplying the values in Table 7.22 by the factors in Table 7.23.

The allowable shear loads for connectors incorporate a safety factor of about 2.5 applied to ultimate load for the commonly used concrete strengths. Not to be confused with shear values for fasteners given in Art. 7.30, the allowable shear loads for connectors are applicable only with Eqs. (7.67) to (7.69).
The allowable horizontal shear loads given in Tables 7.22 and 7.23 may have to be adjusted for use with formed steel decking. For decking with ribs parallel to supports (Fig. 7.38a), the allowable loads should be reduced when w/h is less than 1.5 by multiplying the tabulated values by

LRFD of Encased Beams

Two methods of design are allowed, the difference being whether or not shoring is used. In both cases, the design strength is bMn, where b  0.90. Mn is calculated for the elastic stress distribution on the composite section if shoring is used or the plastic stress distribution on the steel section alone if shoring is not used.

LRFD of Composite Beams

As with ASD, the use of shoring to carry deal loads prior to the time the concrete has hardened determines which design procedures are used. For composite construction where the steel beams are exposed, the design flexural strength for positive moment (compression in the concrete) is bMn. It is dependent on the depththickness ratio hc / tw of the steel beam, where tw is the web thickness and, for webs of rolled or formed sections, hc is twice the distance from the neutral axis to the toe of the fillet at the compression flange, and for webs of built-up sections, hc is twice the distance from the neutral axis to the nearest line of fasteners at the compression flange or the inside face of a welded compression flange.

Continuous Beams and Frames

Fixed-end beams, continuous beams, continuous trusses, and rigid frames are statically indeterminate. The equations of equilibrium are not sufficient for the deter mination of all the unknown forces and moments. Additional equations based on a knowledge of the deformation of the member are required.

Hence, while the bending moments in a simply supported beam are determined only by the loads and the span, bending moments in a statically indeterminate member are also a function of the geometry, cross-sectional dimensions, and modulus of elasticity.

Sign Convention

For computation of end moments in continuous beams and frames, the following sign convention is most convenient: A moment acting at an end of a member or at a joint is positive if it tends to rotate the joint clockwise, negative if it tends to rotate the joint counterclockwise.
Similarly, the angular rotation at the end of a member is positive if in a clockwise direction, negative if counterclockwise. Thus, a positive end moment produces a positive end rotation in a simple beam.
For ease in visualizing the shape of the elastic curve under the action of loads and end moments, bending-moment diagrams should be plotted on the tension side of each member. Hence, if an end moment is represented by a curved arrow, the arrow will point in the direction in which the moment is to be plotted.

Carry-Over Moments

When a member of a continuous beam or frame is loaded, bending moments are induced at the ends of the member as well as between the ends. The magnitude of the end moments depends on the magnitude and location of the loads, the geometry of the member, and the amount of restraint offered to end rotation of the member by other members connected to it. Because of the restraint, end moments are induced in the connecting members, in addition to end moments that may be induced by loads on those spans.
If the far end of a connecting member is restrained by support conditions against rotation, a resisting moment is induced at that end. That moment is called a carryover moment. The ratio of the carry-over moment to the other end moment is called carry-over factor. It is constant for the member, independent of the magnitude and direction of the moments to be carried over. Every beam has two carry-over factors, one directed toward each end.
As pointed out in Art. 5.10.6, analysis of a continuous span can be simplified by treating it as a simple beam subjected to applied end moments. Thus, it is convenient to express the equations for carry-over factors in terms of the end rotations of simple beams: Convert a continuous member LR to a simple beam with the same span L. Apply a unit moment to one end (Fig. 5.60). The end rotation at the support where the moment is applied is , and at the far end, the rotation is B. By the dummy-load method (Art. 5.10.4), if x is measured from the B end,

Carry-Over Factors. The preceding equations can be used to determine carryover factors for any magnitude of end restraint. The carry-over factors toward fixed ends, however, are of special importance.
The bending-moment diagram for a continuous span LR that is not loaded except for a moment M applied at end L is shown in Fig. 5.61a. For determination of the carry-over factor CR toward R, that end is assumed fixed (no rotation can occur there). The carry-over moment to R then is CRM. The moment diagram in Fig. 5.61a can be resolved into two components: a simple beam with M applied at L (Fig. 5.61b) and a simple beam with CRM applied at R (Fig. 5.61c). As indicated in Fig. 5.61d, M causes an angle change at R of . As shown in Fig. 5.61e, CR M induces an angle change at R of CRMR.

With the use of Eqs. (5.107) and (5.111), the stiffness of a beam with constant moment of inertia is given by

This equation indicates that a prismatic beam hinged at only one end has threefourths the stiffness, or resistance to end rotation, of a beam fixed at both ends.

Fixed-End Moments

A beam so restrained at its ends that no rotation is produced there by the loads is called a fixed-end beam, and the end moments are called fixed-end moments. Fixedend moments may be expressed as the product of a coefficient and WL, where W is the total load on the span L. The coefficient is independent of the properties of other members of the structure. Thus, any member can be isolated from the rest of the structure and its fixed-end moments computed.
Assume, for example, that the fixed-end moments for the loaded beam in Fig. 5.63a are to be determined. Let MF be the moment at the left end L and MF the L R
moment at the right end R of the beam. Based on the condition that no rotation is permitted at either end and that the reactions at the supports are in equilibrium with the applied loads, two equations can be written for the end moments in terms of the simple-beam end rotations, L at L and R, at R for the specific loading.
Let KL be the fixed-end stiffness at L and KR the fixed-end stiffness at R, as given by Eqs. (5.112) and (5.113). Then, by resolution of the moment diagram into simple-beam components, as indicated in Fig. 5.63ƒ to h, and application of the superposition principle (Art. 5.10.6), the fixed-end moments are found to be

Deflection of Supports. Fixed-end moments for loaded beams when one support is displaced vertically with respect to the other support may be computed with the use of Eqs. (5.116) to (5.121) and the principle of superposition: Compute the fixedend moments induced by the deflection of the beam when not loaded and add them to the fixed-end moments for the loaded condition with immovable supports.
The fixed-end moments for the unloaded condition can be determined directly from Eqs. (5.116) and (5.117). Consider beam LR in Fig. 5.64, with span L and support R deflected a distance d vertically below its original position. If the beam were simply supported, the angle change caused by the displacement of R would be very nearly d/L. Hence, to obtain the fixed-end moments for the deflected conditions, set @L =@R = d/L and substitute these simple-beam end rotations in Eqs. (5.116) and (5.117):

where MF is the fixed-end moment at the left support and MF at the right support. L R
As an example of the use of the curves, find the fixed-end moments in a prismatic beam of 20-ft span carrying a triangular loading of 100 kips, similar to the loading shown in Case 4, Fig. 5.70, distributed over the entire span, with the maximum intensity at the right support.

Slope-Deflection Equations

In Arts. 5.11.2 and 5.11.4, moments and displacements in a member of a continuous beam or frame are obtained by addition of their simple-beam components. Similarly, moments and displacements can be determined by superposition of fixed-end-beam components. This method, for example, can be used to derive relationships between end moments and end rotations of a beam known as slope-deflection equations.
These equations can be used to compute end moments in continuous beams.
Consider a member LR of a continuous beam or frame (Fig. 5.72). LR may have a moment of inertia that varies along its length. The support R is displaced vertically

The slope-deflection equations can be used to determine end moments and rotations of the spans of continuous beams by writing compatibility and equilibrium equations for the conditions at each support. For example, the sum of the moments at each support must be zero. Also, because of continuity, the member must rotate through the same angle on both sides of every support. Hence, ML for one span, given by Eq. (5.133) or (5.135), must be equal to MR for the adjoining span, given by Eq. (5.134) or (5.136), and the end rotation  at that support must be the same on both sides of the equation. One such equation with the end rotations at the supports as the unknowns can be written for each support. With the end rotations determined by simultaneous solution of the equations, the end moments can be computed from the slope-deflection equations and the continuous beam can now be treated as statically determinate.
(C. H. Norris et al., ‘‘Elementary Structural Analysis,’’ 4th ed., McGraw-Hill Book Company, New York.)

Moment Distribution

The frame in Fig. 5.74 consists of four prismatic members rigidly connected together
at O at fixed at ends A, B, C, and D. If an external moment U is applied at  O, the sum of the end moments in each member at O must be equal to U. Furthermore, all members must rotate at O through the same angle , since they are assumed to be rigidly connected there. Hence, by the definition of fixed-end stiffness, the proportion of U induced in the end of each member at O is equal to the ratio of the stiffness of that member to the sum of the stiffnesses of all the members at the joint (Art. 5.11.3).

can be taken when a member has a hinged end to reduce the work of distributing moments. This is done by using the true stiffness of a member instead of the fixedend stiffness. (For a prismatic beam with one end hinged, the stiffness is threefourth the fixed-end stiffness; for a beam with variable I, it is equal to the fixedend stiffness times 1 - CLCR, where CL and CR are the carry-over factors for the beam.) Naturally, the carry-over factor toward the hinge is zero.
When a joint is neither fixed nor pinned but is restrained by elastic members connected there, moments can be distributed by a series of converging approximations.
All joints are locked against rotation. As a result, the loads will create fixed-end moments at the ends of every member. At each joint, a moment equal to the algebraic sum of the fixed-end moments there is required to hold it fixed. Then, one joint is unlocked at a time by applying a moment equal but opposite in sign to the moment that was needed to prevent rotation. The unlocking moment must be distributed to the members at the joint in proportion to their fixed-end stiffnesses and the distributed moments carried over to the far ends.
After all joints have been released at least once, it generally will be necessary to repeat the process—sometimes several times—before the corrections to the fixed  end moments become negligible. To reduce the number of cycles, the unlocking of joints should start with those having the greatest unbalanced moments.

Suppose the end moments are to be found for the prismatic continuous beam ABCD in Fig. 5.75. The I /L values for all spans are equal; therefore, the relative fixed-end stiffness for all members is unity. However, since A is a hinged end, the computation can be shortened by using the actual relative stiffness, which is 3⁄4.
Relative stiffnesses for all members are shown in the circle on each member. The distribution factors are shown in boxes at each joint.
The computation starts with determination of fixed-end moments for each member (Art. 5.11.4). These are assumed to have been found and are given on the first line in Fig. 5.75. The greatest unbalanced moment is found from inspection to be at hinged end A; so this joint is unlocked first. Since there are no other members at the joint, the full unlocking moment of +400 is distributed to AB at A and onehalf of this is carried over to B. The unbalance at B now is +400 - 480 plus the carry-over of +200 from A, or a total of -120. Hence, a moment of 120 must be applied and distributed to the members at B by multiplying by the distribution factors in the corresponding boxes.
The net moment at B could be found now by adding the entries for each member at the joint. However, it generally is more convenient to delay the summation until the last cycle of distribution has been completed.
The moment distributed to BA need not be carried over to A, because the carryover factor toward the hinged end is zero. However, half the moment distributed to BC is carried over to C.
Similarly, joint C is unlocked and half the distributed moments carried over to B and D, respectively. Joint D should not be unlocked, since it actually is a fixed end. Thus, the first cycle of moment distribution has been completed.
The second cycle is carried out in the same manner. Joint B is released, and the distributed moment in BC is carried over to C. Finally, C is unlocked, to complete the cycle. Adding the entries for the end of each member yields the final moments.

Maximum Moments in Continuous Frames

In design of continuous frames, one objective is to find the maximum end moments and interior moments produced by the worst combination of loading. For maximum moment at the end of a beam, live load should be placed on that beam and on the  beam adjoining the end for which the moment is to be computed. Spans adjoining these two should be assumed to be carrying only dead load.

For maximum midspan moments, the beam under consideration should be fully loaded, but adjoining spans should be assumed to be carrying only dead load.

For maximum midspan moments, the beam under consideration should be fully loaded, but adjoining spans should be assumed to be carrying only dead load.
The work involved in distributing moments due to dead and live loads in continuous frames in buildings can be greatly simplified by isolating each floor. The tops of the upper columns and the bottoms of the lower columns can be assumed fixed. Furthermore, the computations can be condensed considerably by following the procedure recommended in ‘‘Continuity in Concrete Building Frames.’’
EB033D, Portland Cement Association, Skokie, IL 60077, and indicated in Fig. 5.74.
Figure 5.74 presents the complete calculation for maximum end and midspan moments in four floor beams AB, BC, CD, and DE. Building columns are assumed to be fixed at the story above and below. None of the beam or column sections is known to begin with; so as a start, all members will be assumed to have a fixedend stiffness of unity, as indicated on the first line of the calculation.
On the second line, the distribution factors for each end of the beams are shown, calculated from the stiffnesses (Arts. 5.11.3 and 5.11.4). Column stiffnesses are not shown, because column moments will not be computed until moment distribution to the beams has been completed. Then the sum of the column moments at each joint may be easily computed, since they are the moments needed to make the sum of the end moments at the joint equal to zero. The sum of the column moments at each joint can then be distributed to each column there in proportion to its stiffness.
In this example, each column will get one-half the sum of the column moments.
Fixed-end moments at each beam end for dead load are shown on the third line, just above the heavy line, and fixed-end moments for live plus dead load on the fourth line. Corresponding midspan moments for the fixed-end condition also are shown on the fourth line and, like the end moments, will be corrected to yield actual midspan moments.

Moment-Influence Factors

In certain types of framing, particularly those in which different types of loading conditions must be investigated, it may be convenient to find maximum end moments from a table of moment-influence factors. This table is made up by listing for the end of each member in the structure the moment induced in that end when a moment (for convenience, +1000) is applied to every joint successively. Once this table has been prepared, no additional moment distribution is necessary for computing the end moments due to any loading condition.
For a specific loading pattern, the moment at any beam end MAB may be obtained from the moment-influence table by multiplying the entries under AB for the various  joints by the actual unbalanced moments at those joints divided by 1000, and summing (see also Art. 5.11.9 and Table 5.6).

Procedure for Sidesway

Computations of moments due to sidesway, or drift, in rigid frames is conveniently executed by the following method:
1. Apply forces to the structure to prevent sidesway while the fixed-end moments due to loads are distributed.
2. Compute the moments due to these forces.
3. Combine the moments obtained in Steps 1 and 2 to eliminate the effect of the forces that prevented sidesway.

Suppose the rigid frame in Fig. 5.77 is subjected to a 2000-lb horizontal load acting to the right at the level of beam BC. The first step is to compute the moment-
influence factors (Table 5.6) by applying moments of 1000 at joints B and C,  assuming sidesway prevented.

Since there are no intermediate loads on the beams and columns, the only fixed-end moments that need be considered
are those in the columns resulting from lateral deflection of the frame caused by the horizontal load. This deflection, however is not known initially.
So assume an arbitrary deflection, which produces a fixed-end moment of -1000M at the top of column CD. M is an unknown constant to be determined fr  the columns and hence are equal in AB to -1000M x 6⁄2 = -3000M. The columnom the fact that the sum of the shears in the deflected columns must be equal to the 2000-lb load. The same deflection also produces a moment of -1000M at the bottom of CD [see Eq. (5.126)].

From the geometry of the structure, furthermore, note that the deflection of B  relative to A is equal to the deflection of C relative to D. Then, according to Eq. (5.126) the fixed-end moments in the columns are proportional to the stiffnesses of  fixed-end moments are entered in the first line of Table 5.7, which is called a moment-collection table.

from which M = 2.30. This value is substituted in the sidesway total in Table 5.7 to yield the sidesway moments for the 4000-lb load. The addition of these moments to the totals for no sidesway yields the final moments.
This procedure enables one-story bents with straight beams to be analyzed with the necessity of solving only one equation with one unknown regardless of the number of bays. If the frame is several stories high, the procedure can be applied to each story. Since an arbitrary horizontal deflection is introduced at each floor or roof level, there are as many unknowns and equations as there are stories.
The procedure is more difficult to apply to bents with curved or polygonal members between the columns. The effect of the change in the horizontal projection of the curved or polygonal portion of the bent must be included in the calculations.
In many cases, it may be easier to analyze the bent as a curved beam (arch).
(A. Kleinlogel, ‘‘Rigid Frame Formulas,’’ Frederick Ungar Publishing Co., New York.)

Rapid Approximate Analysis of Multistory Frames

Exact analysis of multistory rigid frames subjected to lateral forces, such as those from wind or earthquakes, involves lengthy calculations, and they are timeconsuming and expensive, even when performed with computers. Hence, approximate methods of analysis are an alternative, at least for preliminary designs and, for some structures, for final designs.
It is noteworthy that for some buildings even the ‘‘exact’’ methods, such as those described in Arts. 5.11.8 and 5.11.9, are not exact. Usually, static horizontal loads are assumed for design purposes, but actually the forces exerted by wind and earthquakes are dynamic. In addition, these forces generally are uncertain in intensity, direction, and duration. Earthquake forces, usually assumed as a percentage of the mass of the building above each level, act at the base of the structure, not at each floor level as is assumed in design, and accelerations at each level vary nearly linearly with distance above the base. Also, at the beginning of a design, the sizes of the members are not known. Consequently, the exact resistance to lateral deformation cannot be calculated. Furthermore, floors, walls, and partitions help resist the lateral forces in a very uncertain way. See Art. 5.12 for a method of calculating the distribution of loads to rigid-frame bents.
Portal Method. Since an exact analysis is impossible, most designers prefer a wind-analysis method based on reasonable assumptions and requiring a minimum of calculations. One such method is the so-called ‘‘portal method.’’
It is based on the assumptions that points of inflection (zero bending moment) occur at the midpoints of all members and that exterior columns take half as much shear as do interior columns. These assumptions enable all moments and shears throughout the building frame to be computed by the laws of equilibrium.
Consider, for example, the roof level (Fig. 5.78a) of a tall building. A wind load of 600 lb is assumed to act along the top line of girders. To apply the portal method, we cut the building along a section through the inflection points of the top-story columns, which are assumed to be at the column midpoints, 6 ft down from the top of the building. We need now consider only the portion of the structure above this section.
Since the exterior columns take only half as much shear as do the interior columns, they each receive 100 lb, and the two interior columns, 200 lb. The moments at the tops of the columns equal these shears times the distance to the inflection point. The wall end of the end girder carries a moment equal to the moment in the column. (At the floor level below, as indicated in Fig. 5.78b, that end of the end girder carries a moment equal to the sum of the column moments.) Since the inflection point is at the midpoint of the girder, the moment at the inner end of the girder must the same as at the outer end. The moment in the adjoining girder can be found by subtracting this moment from the column moment, because the sum of the moments at the joint must be zero. (At the floor level below, as shown in Fig. 5.78b, the moment in the interior girder is found by subtracting the moment in the exterior girder from the sum of the column moments.)
Girder shears then can be computed by dividing girder moments by the half span. When these shears have been found, column loads can be easily computed from the fact that the sum of the vertical loads must be zero, by taking a section around each joint through column and girder inflection points. As a check, it should be noted that the column loads produce a moment that must be equal to the moments of the wind loads above the section for which the column loads were computed.
For the roof level (Fig. 5.78a), for example, -50 x 24 + 100  x 48 = 600 x 6.

Cantilever Method. Another wind-analysis procedure that is sometimes employed is the cantilever method. Basic assumptions here are that inflection points are at the midpoints of all members and that direct stresses in the columns vary as the distances of the columns from the center of gravity of the bent. The assumptions are sufficient to enable shears and moments in the frame to be determined from the laws of equilibrium.
For multistory buildings with height-to-width ratio of 4 or more, the Spurr modification
is recommended (‘‘Welded Tier Buildings,’’ U.S. Steel Corp.). In this
method, the moments of inertia of the girders at each level are made proportional to the girder shears.
The results obtained from the cantilever method generally will be different from those obtained by the portal method. In general, neither solution is correct, but the answers provide a reasonable estimate of the resistance to be provided against lateral deformation. (See also Transactions of the ASCE, Vol. 105, pp. 1713–1739, 1940.)

Beams Stressed into the Plastic Range

When an elastic material, such as structural steel, is loaded in tension with a gradually increasing load, stresses are proportional to strains up to the proportional limit (near the yield point). If the material, like steel, also is ductile, then it continues to carry load beyond the yield point, though strains increase rapidly with little increase in load (Fig. 5.79a).

Similarly, a beam made of a ductile material continues to carry more load after the stresses in the outer surfaces reach the yield point. However, the stresses will no longer vary with distance from the neutral axis, so the flexure formula [Eq. (5.54)] no longer holds. However, if simplifying assumptions are made, approximating the stress-strain relationship beyond the elastic limit, the load-carrying capacity of the beam can be computed with satisfactory accuracy.

Modulus of rupture is defined as the stress computed from the flexure formula for the maximum bending moment a beam sustains at failure. This is not a true stress but it is sometimes used to compare the strength of beams.
For a ductile material, the idealized stress-strain relationship in Fig. 5.79b may be assumed. Stress is proportional to strain until the yield-point stress ƒy is reached, after which strain increases at
a constant stress.
For a beam of this material, the following assumptions will also be made:
1. Plane sections remain plane, strains thus being proportional to distance from the neutral axis.
2. Properties of the material in tension are the same as those in compression.
3. Its fibers behave the same in flexure as in tension.
4. Deformations remain small.

Strain distribution across the cross section of a rectangular beam, based on these assumptions, is shown in Fig. 5.80a. At the yield point, the unit strain is y and the curvature y, as indicated in (1). In (2), the strain has increased several times, but the section still remains plane. Finally, at failure, (3), the strains are very large and nearly constant across upper and lower halves of the section.
Corresponding stress distributions are shown in Fig. 5.80b. At the yield point, (1), stresses vary linearly and the maximum if ƒy . With increase in load, more and more fibers reach the yield point, and the stress distribution becomes nearly constant, as indicated in (2). Finally, at failure, (3), the stresses are constant across the top and bottom parts of the section and equal to the yield-point stress.
The resisting moment at failure for a rectangular beam can be computed from the stress diagram for stage 3. If b is the width of the member and d its depth, then the ultimate moment for a rectangular beam is

Since the resisting moment at stage 1 is My  ƒybd2 / 6, the beam carries 50% more moment before failure than when the yield-point stress is first reached at the outer surfaces.

Curved Beams

Structural members, such as arches, crane hooks, chain links, and frames of some machines, that have considerable initial curvature in the plane of loading are called curved beams. The flexure formula of Art. 5.5.10, ƒ = Mc/ I, cannot be applied to them with any reasonable degree of accuracy unless the depth of the beam is small compared with the radius of curvature.

Unlike the condition in straight beams, unit strains in curved beams are not proportional to the distance from the neutral surface, and the centroidal axis does not coincide with the neutral axis. Hence the stress distribution on a section is not linear but more like the distribution shown in Fig. 5.42c.

Stresses in Curved Beams

Just as for straight beams, the assumption that plane sections before bending remain plane after bending generally holds for curved beams. So the total strains are proportional
to the distance from the neutral axis. But since the fibers are initially of unequal length, the unit strains are a more complex function of this distance. In Fig. 5.42a, for example, the bending couples have rotated section AB of the curved beam into section A B through an angle d. If o is the unit strain at the centroidal axis and  is the angular unit strain d /d, then the unit strain at a distance y from the centroidal axis (measured positive in the direction of the center of curvature) is

where R  radius of curvature of centroidal axis.
Equation (5.75) can be expressed in terms of the bending moment if we take advantage of the fact that the sum of the tensile and compressive forces on the section must be zero and the moment of these forces must be equal to the bending moment M. These two equations yield

Curved Beams with Various Cross Sections

Equation (5.78) for bending stresses in curved beams subjected to end moments in the plane of curvature can be expressed for the inside and outside beam faces in the form:

where c  distance from the centroidal axis to the inner or outer surface. Table 5.4 gives values of K calculated from Eq. (5.78) for circular, elliptical, and rectangular cross sections.
If Eq. (5.78) is applied to 1 or T beams or tubular members, it may indicate circumferential flange stresses that are much lower than will actually occur. The error is due to the fact that the outer edges of the flanges deflect radially. The effect is equivalent to having only part of the flanges active in resisting bending stresses.
Also, accompanying the flange deflections, there are transverse bending stresses in the flanges. At the junction with the web, these reach a maximum, which may be greater than the maximum circumferential stress. Furthermore, there are radial stresses (normal stresses acting in the direction of the radius of curvature) in the web that also may have maximum values greater than the maximum circumferential stress.
A good approximation to the stresses in I or T beams is as follows: for circumferential stresses, Eq. (5.78) may be used with a modified cross section, which is obtained by using a reduced flange width. The reduction is calculated from b  b, where b is the length of the portion of the flange projecting on either side from the web, b is the corrected length, and  is a correction factor determined from equations developed by H. Bleich,  is a function of b2 / rt, where t is the flange thickness and r the radius of the center of the flange:

where I  moment of inertia of modified section about its centroidal axis R  radius of curvature of centroidal axis c  distance from centroidal axis to center of the more sharply curved flange Because of the high stress factor, it is advisable to stiffen or brace curved I-beam flanges.
The maximum radial stress will occur at the junction of web and flange of I beams. If the moment is negative, that is, if the loads tend to flatten out the beam, the radial stress is tensile, and there is a tendency for the more sharply curved flange to pull away from the web. An approximate value of this maximum stress is

where K is a correction factor for the curvature [see Eq. (5.80)]. The sign of M is taken positive in this equation when it increases the curvature, and P is positive when it is a tensile force, negative when compressive.

Slope and Deflection of Curved Beams

If we consider two sections of a curved beam separated by a differential distance ds (Fig. 5.42), the change in angle d between the sections caused by a bending moment M and an axial load P may be obtained from Eq. (5.76), noting that d  ds/R.

be taken into account; but unless the curvature is sharp, its effect on deformations may be neglected. So only Eq. (5.86) and the first term in Eq. (5.85) need be used.
(S. Timoshenko and D. H. Young, ‘‘Theory of Structures,’’ McGraw-Hill Publishing Company, New York.) See also Arts. 5.14.1 to 5.14.3.

Straight Beams

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.

Reactions

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):

Internal Forces

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

Shear Diagrams

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.

Bending-Moment Diagrams

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.

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.

Shear-Moment Relationship

The slope of the bending-moment curve for any point on a beam is equal to the shear at that point; i.e.,

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 1⁄2 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 3⁄4 and 40⁄45 x 3⁄4. Hence, the maximum shear is 20 x 3⁄4 + 10 x 40⁄45 x 3⁄4 = 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, ƒ 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 ƒ 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.

Section Modulus

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 ƒ on an inclined plane caused by a combination of shear and bending stress—v and ƒ, respectively. This stress ƒ , which may be either tension or compression, is greater than the normal stress. Its value may be obtained by application of Mohr’s circle (Art. 5.3.6), as indicated in Fig. 5.10, but with ƒy = 0, and is

Beam Deflections

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.

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.

Unsymmetrical Bending

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 ƒ =  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 beam’s 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.)

Allowable-Stress Design of Bridge with Continuous, Composite Stringers

The structure is a two-lane highway bridge with overall length of 298 ft. Site conditions require a central span of 125 ft. End spans, therefore, are each 86.5 ft (Fig. 12.66a). The typical cross section in Fig. 12.66b shows a 30-ft roadway, flanked on one side by a 21-inwide barrier curb and on the other by a 6-ft-wide sidewalk. The deck is supported by six rolled-beam, continuous stringers of Grade 36 steel. Concrete to be used for the deck is Class A, with 28-day strength f’c = 4,000 psi and allowable compressive stress ƒc = 1,600 ƒc psi. Loading is HS20-44. Appropriate design criteria given in Sec. 11 will be used for this structure.
Concrete Slab. The slab is designed to span transversely between stringers..
A 9-in-thick, two-course slab will be used. No provision will be made for a future 2-in wearing course.
Stringer Loads. Assume that the stringers will not be shored during casting of the concrete slab. Then, the dead load on each stringer includes the weight of a strip of concrete slab plus the weights of steel shape, cover plates, and framing details. This dead load will be referred to as DL and is summarized in Table 12.81.

Sidewalks, parapet, and barrier curbs will be placed after the concrete slab has cured.
Their weights may be equally distributed to all stringers. Some designers, however, prefer to calculate the heavier load imposed on outer stringers by the cantilevers by taking moments of the cantilever loads about the edge of curb, as shown in Table 12.82. In addition, the six composite beams must carry the weight, 0.016 ksf, of the 30-ft-wide latex-modified concrete wearing course. The total superimposed dead load will be designated SDL.
The HS20-44 live load imposed may be a truck load or lane load. For these spans, truck loading governs. With stringer spacing S  6.5 ft, the live load taken by outer stringers S1 and S3 is

Stringer Moments. The steel stringers will each consist of a single rolled beam of Grade 36 steel, composite with the concrete slab only in regions of positive moment. To resist negative moments, top and bottom cover plates will be attached in the region of the interior supports. To resist maximum positive moments in the center span, a cover plate will be added to the bottom flange of the composite section. In the end spans, the composite section  with the rolled beam alone must carry the positive moments.
For a precise determination of bending moments and shears, these variations in moments of inertia of the stringer cross sections should be taken into account. But this requires that the cross sections be known in advance or assumed, and the analysis without a computer is tedious. Instead, for a preliminary analysis, to determine the cross sections at critical points the moment of inertia may be assumed constant and the same in each span. This assumption considerably simplifies the analysis and permits use of tables of influence coefficients. (See, for example, ‘‘Moments, Shears, and Reactions for Continuous Highway Bridges,’’American Institute of Steel Construction.) The resulting design also often is sufficiently accurate to serve as the final design. In this example, dead-load negative moment at the supports. computed  for constant moment of inertia, will be increased 10% to compensate for the variations in moment of inertia.
Curves of maximum moment (moment envelopes) are plotted in Figs. 12.67 and 12.68 for S1 and S2 , respectively. Because total maximum moments at critical points are nearly equal for S1 , S2 , and S3 , the design selected for S1 will be used for all stringers. (In some cases, there may be some cost savings in using shorter cover plates for the stringers with smaller moments.)
Properties of Negative-Moment Section. The largest bending moment occurs at the interior supports, where the section consists of a rolled beam and top and bottom cover plates. With the dead load at the supports as indicated in Fig. 12.67 increased 10% to compensate for the variable moment of inertia, the moments in stringer S1 at the supports are as follows:

Allowable Compressive Stress near Supports. Because the bottom flange of the beam is in compression near the supports and is unbraced, the allowable compressive stress may have to be reduced to preclude buckling failure. AASHTO specifications, however, permit a 20% increase in the reduced stress for negative moments near interior supports. The unbraced length should be taken as the distance between diaphragms or the distance from interior support to the dead-load inflection point, whichever is smaller. In this example, if distance between diaphragms is assumed not to exceed about 22 ft, the allowable bending stress for a flange width of 16.6 in is computed as follows:

distance from an interior support, the top and bottom cover plates can be terminated where the rolled beam alone has sufficient capacity to carry the bending moment. The actual cutoff points, however, may be determined by allowable fatigue stresses for the base metal adjacent to the fillet welds between flanges and ends of the cover plates. The number of cycles of load to be resisted for HS20-44 loading is 500,000 for a major highway. For Grade 36 steel and these conditions, the allowable fatigue stress range for this redundant-load-path structure

In the center span, the resisting moment of the W36 equals the bending moment about 8 ft 4 in from the interior support. With allowance for the terminal distance, the plates may be cut off 10 ft 6 in from the support. Fatigue does not govern there.
Properties of End-Span Composite Section. The 9-in-thick roadway slab includes an allowance
of 0.5 in for wear. Hence, the effective thickness of the concrete slab for composite
action is 8.5 in.
The effective width of the slab as part of the top flange of the T beam is the smaller of the following:

Hence the effective width is 76.5 in (Fig. 12.69).
To resist maximum positive moments in the end span, the W36 x 280 will be made composite with the concrete slab. As in Art. 12.2, the properties of the end-span composite
section are computed with the concrete slab, ignoring the haunch area, transformed into an equivalent steel area. The computations for neutral-axis locations and section moduli for the
composite section are tabulated in Table 12.83. To locate the neutral axes for n = 24 and n  8 moments are taken about the neutral axis of the rolled beam.
Stresses in End-Span Composite Section. Since the stringers will not be shored when the concrete is cast and cured, the stresses in the steel section for load DL are determined with the section moduli of the steel section alone. Stresses for load SDL are computed with section moduli of the composite section when n = 24. And stresses in the steel for live loads and impact are calculated with section moduli of the composite section when n  8. See Table 12.68. Maximum positive bending moments in the end span are estimated from Fig. 12.67:

Continuous-Beam Bridges

Articles 12.1 and 12.3 recommended use of continuity for multispan bridges. Advantages over simply supported spans include less weight, greater stiffness, smaller deflections, and fewer bearings and expansion joints. Disadvantages include more complex fabrication and erection and often the costs of additional field splices.
Continuous structures also offer greater overload capacity. Failure does not necessarily occur if overloads cause yielding at one point in a span or at supports. Bending moments are redistributed to parts of the span that are not overstressed. This usually can take place in bridges because maximum positive moments and maximum negative moments occur with loads in different positions on the spans. Also, because of moment redistribution due to yielding, small settlements of supports have no significant effects on the ultimate strength of continuous spans. If, however, foundation conditions are such that large settlements could occur, simple-span construction is advisable.
While analysis of continuous structures is more complicated than that for simple spans, design differs in only a few respects. In simple spans, maximum dead-load moment occurs at midspan and is positive. In continuous spans, however, maximum dead-load moment occurs at the supports and is negative. Decreasing rapidly with distance from the support, the negative moment becomes zero at an inflection point near a quarter point of the span.

As for simple spans, live loads are placed on continuous spans to create maximum stresses at each section. Whereas in simple spans maximum moments at each section are always positive, maximum live-load moments at a section in continuous spans may be positive or negative. Because of the stress reversal, fatigue stresses should be investigated, especially in the region of dead-load inflection points. At interior supports, however, design usually is governed by the maximum negative moment, and in the midspan region, by maximum positive moment. The sum of the dead-load and live-load moments usually is greater at supports than at midspan. Usually also, this maximum is considerably less than the maximum moment in a simple beam with the same span. Furthermore, the maximum negative moment decreases rapidly with distance from the support.
The impact fraction for continuous spans depends on the length L, ft, of the portion of the span loaded to produce maximum stress. For positive moment, use the actual loaded length. For negative moment, use the average of two adjacent loaded spans.
Ends of continuous beams usually are simply supported. Consequently, moments in threespan and four-span continuous beams are significantly affected by the relative lengths of interior and exterior spans. Selection of a suitable span ratio can nearly equalize maximum positive moments in those spans and thus permit duplication of sections. The most advantageous ratio, however, depends on the ratio of dead load to live load, which, in turn, is a function of span length. Approximately, the most advantageous ratio for length of interior to exterior span is 1.33 for interior spans less than about 60 ft, 1.30 for interior spans between about 60 to 110 ft, and about 1.25 for longer spans.
When composite construction is advantageous (see Art. 12.1), it may be used either in the positive-moment regions or throughout a continuous span. Design of a section in the positive-moment region in either case is similar to that for a simple beam. Design of a section in the negative-moment regions differs in that the concrete slab, as part of the top flange, cannot resist tension. Consequently, steel reinforcement must be added to the slab to resist the tensile stresses imposed by composite action.
Additionally, for continuous spans with a cast-in-place concrete deck, the sequence of  concrete pavement is an important design consideration. Bending moments, bracing requirements, and uplift forces must be carefully evaluated.

Characteristics of Beam Bridges

Rolled wide-flange shapes generally are the most economical type of construction for shortspan bridges. The beams usually are used as stringers, set, at regular intervals, parallel to the direction of traffic, between piers or abutments (Fig. 12.1). A concrete deck, cast on the top flange, provides lateral support against buckling. Diaphragms between the beams offer additional bracing and also distribute loads laterally to the beams before the concrete deck has cured.

Spacing. For railroad bridges, two stringers generally carry each track. They may, however,
be more widely spaced than the rails, for stability reasons. If a bridge contains only two
stringers, the distance between their centers should be at least 6 ft 6 in. When more stringers
are used, they should be placed to distribute the track load uniformly to all beams.
For highway bridges, one factor to be considered in selection of stringer spacing is the
minimum thickness of concrete deck permitted. For the deck to serve at maximum efficiency, its span between stringers should be at least that requiring the minimum thickness. But when stringer spacing requires greater than minimum thickness, the dead load is increased, cutting into the savings from use of fewer stringers. For example, if the minimum thickness of concrete slab is about 8 in, the stringer spacing requiring this thickness is about 8 ft for 4,000-psi concrete. Thus, a 29-ft 6-in-wide bridge, with 26-ft roadway, could be carried on four girders with this spacing. The outer stringers then would be located 1 ft from the curb into the roadway, and the outer portion of the deck, with parapet, would cantilever 2 ft 9 in beyond the stringers.

If an outer stringer is placed under the roadway, the distance from the center of the stringer to the curb preferably should not exceed about 1 ft.
Stringer spacing usually lies in the range 6 to 15 ft. The smaller spacing generally is desirable near the upper limits of rolled-beam spans.
The larger spacing is economical for the longer spans where deep, fabricated, plate girders are utilized. Wider spacing of girders has resulted in development of long-span stay-in-place forms. This improvement in concrete-deck forming has made steel girders with a concrete deck more competitive.
Regarding deck construction, while conventional cast-in-place concrete decks are commonplace, precast-concrete deck slab bridges are often used and may prove practical and economical if stage construction and maintenance of traffic are required. Additionally, use of lightweight concrete, a durable and economical product, may be considered if dead weight is a problem.
Other types of deck are available such as steel orthotropic plates (Arts. 12.14 and 12.15).
Also, steel grating decks may be utilized, whether unfilled, half-filled, or fully filled with concrete. The latter two deck-grating construction methods make it possible to provide composite action with the steel girder.
Short-Span Stringers. For spans up to about 40 ft, noncomposite construction, where beams act independently of the concrete slab, and stringers of AASHTO M270 (ASTM A709), Grade 36 steel often are economical. If a bridge contains more than two such spans in succession, making the stringers continuous could improve the economy of the structure.
Savings result primarily from reduction in number of bearings and expansion joints, as well as associated future maintenance costs. A three-span continuous beam, for example, requires four bearings, whereas three simple spans need six bearings.
For such short spans, with relatively low weight of structural steel, fabrication should be kept to a minimum. Each fabrication item becomes a relatively large percentage of material cost. Thus, cover plates should be avoided. Also, diaphragms and their connections to the stringers should be kept simple. For example. they may be light channels field bolted or welded to plates welded to the beam webs (Fig. 12.2).

For spans 40 ft and less, each beam reaction should be transferred to a bearing plate through a thin sole plate welded to the beam flange. The bearing may be a flat steel plate or an elastomeric pad. At interior supports of continuous beams, sole plates should be wider than the flange. Then, holes needed for anchor bolts can be placed in the parts of the plates extending beyond the flange. This not only reduces fabrication costs by avoiding holes in the stringers but also permits use of lighter stringers, because the full cross section is available for moment resistance.
At each expansion joint, the concrete slab should be thickened to form a transverse beam, to protect the end of the deck. Continuous reinforcement is required for this beam. For the purpose, slotted holes should be provided in the ends of the steel beams to permit the reinforcement to pass through.
Live Loads. Although AASHTO ‘‘Standard Specifications for Highway Bridges’’ specify for design H15-44, HS15-44, H20-44, and HS20-44 truck and lane loadings (Art. 11.4), many state departments of transportation are utilizing larger live loadings. The most common is HS20-44 plus 25% (HS25). An alternative military loading of two axles 4 ft apart, each axle weighing 24 kips, is usually also required and should be used if it causes higher stresses.
Some states prefer 30 kip axles instead of 24 kips.
Bridges with a two-course deck slab generally do not include this additional dead load. The assumption is that during repaving of the adjoining roadway, the 11⁄4-in wearing course (possibly latex modified concrete) will be removed and replaced only if necessary.
If metal stay-in-place forms are permitted for deck construction, consideration should be given to providing for an additional 8 to 12 psf to be included for the weight of the permanent  steel form plus approximately 5 psf for the additional thickness of deck concrete required.