Structural Elements for Architects and Builders: Design of Columns, Beams, and Tension Elements in Wood, Steel, and Reinforced Concrete

By Jonathan Ochshorn

Concise but comprehensive, Structural Elements for Architects and Builders is primarily focused on the design and analysis of structural elements: columns, beams, tension members and their connections. The material is organized into a single, self-sufficient volume, including all necessary data for the preliminary design and analysis of these structural elements in wood, steel, and reinforced concrete.

Every chapter contains insights developed by the author and generally not found elsewhere. Additionally, the Appendices included at the end of the text contain numerous tables and graphs, based on material contained in industry publications, but reorganized and formatted especially for this text to improve clarity and simplicity, without sacrificing comprehensiveness.

• Based on the standards and codes from For timber: The American Institute of Timber Construction's (AITC), The American Institute of Steel Construction's (AISC) The American Concrete Institute's (ACI) Building Code Requirements for Structural Concrete and Commentary (ACI 318 and ACI 318R)and the American Society of Civil Engineers
• Contains graphs, charts, and tables to solve basic structural design problems
• Step-by-step illustrative examples
• Cover common connectors such as nails, bolts, and welds

• Language: English
• Publisher: Elsevier Science
• Release date: Oct 12, 2009
• ISBN: 9780080962160

Book preview

Statics

The study of structural behavior and structural design begins with the concept of load. We represent loads with arrows indicating direction and magnitude. The magnitude is expressed in pounds (lb), kips (1 kip = 1000 lb), or appropriate SI units of force; the direction is usually vertical (gravity) or horizontal (wind, earthquake), although wind loads on pitched roofs can be modeled as acting perpendicular to the roof surface (Figure 1.1).

Figure 1.1 Direction of loads can be (a) vertical, (b) horizontal, or (c) inclined

Where loads are distributed over a surface, we say, for example, 100 pounds per square foot, or 100 psf. Where loads are distributed over a linear element, like a beam, we say, for example, 2 kips per linear foot, or 2 kips per foot, or 2 kips/ft (Figure 1.2). Where loads are concentrated at a point, such as the vertical load transferred to a column, we say, for example, 10 kips or 10 k.

Figure 1.2 Distributed loads on a beam

Finding out what the loads are that act on a structure and how these loads are supported is the prerequisite to all structural design. There are two main reasons for this. First, the fact that a structural element is supported at all means that the supporting element is being stressed in some way. To find the magnitude of the reactions of an element is thus to simultaneously find the magnitude of the loads acting on the supporting element. Each action, or load, has an equal reaction; or, as Newton said in defense of this third law: If you press a stone with your finger, the finger is also pressed by the stone.

The second reason for finding reactions of the structural element is that doing so facilitates the further analysis or design of the element itself. That is, determining reactions is the prerequisite to the calculation of internal loads and internal stresses, values of which are central to the most fundamental questions of structural engineering: Is it strong enough? Is it safe?

Tributary Areas

When loads are evenly distributed over a surface, it is often possible to assign portions of the load to the various structural elements supporting that surface by subdividing the total area into tributary areas corresponding to each member. In Figure 1.3, half the load of the table goes to each lifter.

Figure 1.3 Tributary areas divide the load among the various supports

In Figure 1.4, half the 20-psf snow load on the cantilevered roof goes to each column; the tributary area for each column is 10 ft × 10 ft, so the load on each column is 20(10 × 10) = 2000 lb = 2 kips.

Figure 1.4 Distributed load on a floor carried by two columns

Figure 1.5 shows a framing plan for a steel building. If the total floor load is 100 psf, the load acting on each of the structural elements comprising the floor system can be found using appropriate tributary areas. Beam A supports a total load of 100(20 × 10) = 20,000 lb = 20 kips; but it is more useful to calculate the distributed load acting on any linear foot of the beam—this is shown by the shaded tributary area in Figure 1.6a and is 100(1 × 10) = 1000 lb = 1 kip. Since 1000 lb is acting on a 1-ft length of beam, we write 1000 lb/ft or 1.0 kip/ft, as shown in Figure 1.6b.

Figure 1.5 Framing plan showing tributary areas for beams, girders, and columns

Figure 1.6 Distributed load on a steel beam, with (a) one linear foot of its tributary area shown and (b) load diagram showing distributed load in kips per foot

As shown in Figure 1.7a, Beam B (or Girder B) supports a total tributary area of 17.5 × 20 = 350 ft². The load at point a is not included in the beam’s tributary area. Rather, it is assigned to the edge, or spandrel, girder where it goes directly into a column, having no effect on Beam B. Unlike Beam A, the floor loads are transferred to Beam B at two points: each concentrated load corresponds to a tributary area of 17.5 × 10 = 175 ft²; therefore, the two loads each have a magnitude of 100 × 175 = 17,500 lb = 17.5 kips. The load diagram for Beam B is shown in Figure 1.7b.

Figure 1.7 Concentrated loads on a girder (a) derived from tributary areas on framing plan and (b) shown on load diagram

Spandrel girders

Beam C (or Spandrel Girder C), shown in Figure 1.5, is similar to Beam B except that the tributary area for each concentrated load is smaller, 7.5 × 10 = 75 ft², as shown in Figure 1.8a. The two concentrated loads, therefore, have a magnitude of 100 × 75 = 7500 lb = 7.5 kips, and the load diagram is as shown in Figure 1.8b.

Figure 1.8 Concentrated loads on a spandrel girder (a) derived from tributary areas on framing plan and (b) shown on load diagram

There are three reasons spandrel girders are often larger than otherwise similar girders located in the interior of the building, even though the tributary areas they support are smaller. First, spandrel girders often support cladding of various kinds, in addition to the floor loads included in this example. Second, aside from the added weight to be supported, spandrels are often made bigger so that their deflection, or vertical movement, is reduced. This can be an important consideration where nonstructural cladding is sensitive to movement of the structural frame. Third, when the girders are designed to be part of a moment-resisting frame, their size might need to be increased to account for the stresses introduced by lateral forces such as wind and earthquake.

Columns

One way or another, all of the load acting on the floor must be carried by columns under that floor. For most structures, it is appropriate to subdivide the floor into tributary areas defined by the centerlines between columns so that every piece of the floor is assigned to a column.

It can be seen from Figure 1.9 that typical interior columns carry twice the load of typical exterior columns, and four times the load of corner columns. However, two of the conditions described earlier with respect to the enlargement of spandrel girders can also increase the size of exterior and corner columns: the need to support additional weight of cladding and the possibility of resisting wind and earthquake forces through rigid connections to the spandrel girders.

Figure 1.9 Framing plan showing tributary areas for columns (one floor only)

Column D supports a tributary area of 35 × 20 = 700 ft² so that the load transferred to Column D from the floor above is 100 × 700 = 70,000 lb = 70 kips, assuming that the floor above has the same shape and loads as the floor shown. But every floor and roof above also transfers a load to Column D. Obviously, columns at the bottom of buildings support more weight than columns at the top of buildings, since all the tributary areas of the floors and roof above are assigned to them. As an example, if there are nine floors and one roof above Column D, all with the same distributed load and tributary area, then the total load on Column D would be, not 70 k, but (9 + 1) × 70 = 700 kips.

In practice, the entire load as previously calculated is not assigned to columns or to other structural elements with large total tributary areas. This is because it is unlikely that a large tributary area will be fully loaded at any given time. For example, if the live load caused by people and other movable objects is set at 60 psf, and one person weighed 180 lb, then a tributary area of 7000 ft² (as in the example of Column D) would have to be populated by more than 2000 people, each occupying 3 ft², in order to achieve the specified load. That many people crowded into that large a space is an unlikely occurrence in most occupancies, and a live load reduction is often allowed by building codes. As the tributary area gets smaller, however, the probability of the full live load being present increases, and no such reduction is permitted. Permanent and immovable components of the building, or dead loads, have the same probability of being present over large tributary areas as small tributary areas, so they are never included in this type of probability-based load reduction. Calculations for live load reduction are explained in the next chapter.

The path taken by a load depends on the ability of the structural elements to transfer loads in various directions. Given the choice of two competing load paths such as (1) and (2) in Figure 1.10, the load is divided between the two paths in proportion to the relative stiffness of each path. Since the corrugated steel deck shown in Figure 1.10 is much stiffer in the direction of load path (1), and, in fact, is designed to carry the entire load in that direction, we neglect the possibility of the load moving along path (2).

Figure 1.10 Competing load paths on a corrugated steel deck

For two-way systems, generally only used in reinforced concrete (Figure 1.11), or for indeterminate systems in general, the assignment of loads to beams and columns also becomes a function of the relative stiffness of the various components of the system. Stiffer elements attract more load to them, and the simplistic division into tributary areas becomes inappropriate, except in certain symmetrical conditions.

Figure 1.11 Competing load paths on a two-way slab

Equilibrium

Where loads or structural geometries are not symmetrical, using tributary areas may not accurately predict the effects of loads placed on structures, and other methods must be used. We can determine the effects of loads placed on statically determinate structures by assuming that such structures remain at rest, in a state of equilibrium. The implication of this condition, derived from Newton’s second law, is that the summation of all forces (or moments) acting on the structure along any given coordinate axis equals zero. For a plane structure—that is, one whose shape and deflection under loads occurs on a planar surface—three equations uniquely define this condition of equilibrium: two for loads (forces) acting along either of the perpendicular axes of the plane’s coordinate system and one for moments acting about the axis perpendicular to the structure’s plane. Some examples of plane structures are shown in Figure 1.12.

Figure 1.12 Examples of plane structures: simply supported beam, three-hinged arch, and rigid (moment-resisting) frame

In words, the equations of equilibrium state that the sum of all horizontal forces is zero; the sum of all vertical forces is zero; and—take a deep breath here—the sum of all moments about any point, including those resulting from any force multiplied by its distance (measured perpendicular to the line of action of the force) to the point about which moments are being taken, is zero.

Horizontal and vertical can be taken as any perpendicular set of coordinate axes. Where x is used for the horizontal axis and y for the vertical, moments in the plane of the structure are acting about the z-axis. This conventional way of representing coordinate systems for the consideration of equilibrium is inconsistent with the labeling typically used to distinguish between axes of bending. Compare the typical axes of bending shown in Figure 1.13 with the equilibrium coordinate axes in Figure 1.12. Written symbolically, the equations are:

(1.1)

Figure 1.13 Coordinate axes for a steel W-shape

For any plane, rigid-body structure (just structure or structural element from now on) subjected to various loads, the three equations of equilibrium provide the mathematical basis for determining values for up to three unknown forces and moments—the reactions of the structure to the loads. Structural elements of this type are statically determinate because the magnitudes of the unknown reactions can be determined using only the equations of static equilibrium.

Free-body diagrams

Any structure (or part of a structure) so defined can be represented as a free-body diagram (FBD). All external loads acting on the FBD, all unknown external moments or forces at the points where the FBD is connected to other structural elements (i.e., all reactions), and all unknown internal moments or forces at points where a FBD is cut must be shown on the diagram.

Single or multiple reactions occurring at a given point are often represented by standard symbols. These pictures graphically indicate the types of forces and moments that can be developed (Figure 1.14). Other combinations of forces and moments can be represented graphically; the three symbols shown, however, cover most commonly encountered conditions.

Figure 1.14 Abstract symbols for reactions, including (a) hinge or pin-end, (b) roller, (c) fixed, and (d) free end

Where an FBD is cut at a point other than at the reactions of the structural element, an internal moment as well as two perpendicular internal forces are typically present, unless an internal constraint, such as a hinge, prevents one or more of those forces (or moments) from developing.

Where there are more reactions than equations of equilibrium, the structure is said to be statically indeterminate, and equilibrium alone is insufficient to determine the values of the reactions; other techniques have been developed to find the reactions of indeterminate structures, but these are beyond the scope of this text.

Reactions

The following examples show how the equations of equilibrium can be used to find reactions of various common determinate structures. The procedures have been developed so that the equations need not be solved simultaneously. Alternatively, where determinate structures are symmetrical in their own geometry as well as in their loading (assumed to be vertical), reactions can be found by assigning half of the total external loads to each vertical reaction.

Example 1.1

Find reactions for simply supported beam

Problem definition

Find the three reactions for a simply supported beam supporting a distributed load of 100 kips/ft over a span of 20 ft. Simply supported means that the beam is supported by a hinge and a roller, and is therefore determinate.

Solution overview

Draw load diagram with unknown forces and/or moments replacing the reaction (constraint) symbols; use the three equations of equilibrium to find these unknown reactions.

Problem solution

1. Redraw load diagram (Figure 1.15a) by replacing constraint symbols with unknown forces, HA, RA, and RB, and by showing a resultant for all distributed loads (Figure 1.15b).

Figure 1.15 Load diagram for simply supported beam for Example 1.1 showing (a) constraint symbols and (b) unknown forces replacing constraint symbols, and resultant corresponding to distributed load

2. The solution to the horizontal reaction at point A is trivial, since no horizontal loads are present: ΣFx = HA = 0. In this equation, we use a sign convention, where positive corresponds to forces pointing to the right and negative to forces pointing to the left.

3. The order in which the remaining equations are solved is important: moment equilibrium is considered before vertical equilibrium in order to reduce the number of unknown variables in the vertical equilibrium equation. Moments can be taken about any point in the plane; however, unless you wish to solve the two remaining equations simultaneously, it is suggested that the point be chosen strategically to eliminate all but one of the unknown variables. Each moment is the product of a force times a distance called the moment arm; this moment arm is measured from the point about which moments are taken to the line of action of the force and is measured perpendicular to the line of action of the