4 load tracing
TRANSCRIPT
195
4 Load Tracing
4.1 LOAD TRACINGEarly in the structural design phase of a project, an initialassumption is made by the designer about the path acrosswhich forces must travel as they move throughout thestructure to the foundation (ground). Loads (forces) travelalong load paths, and the analysis method is often referredto as load tracing (Figure 4.1).
Figure 4.1 Load tracing can be visualized in the exposed light-frame construction of an apartmentcomplex through the roof trusses, stud walls (bearing and partition), floor joists, and foundation.Photo courtesy of the Southern Forest Products Association.
196 Chapter 4
Figure 4.3 Loads and load paths.
Figure 4.2 Load paths through a simplebuilding.
Engineers often view structures as interdependent mecha-nisms by which loads are distributed to their individualmembers, such as roof sheathing, floor slabs, rafters, joists(a regular, relatively closely spaced series of secondarybeams), beams, and columns (Figure 4.2). The structuraldesigner makes judgments regarding the amount of loadassigned to each member and the manner in which loadstravel throughout the structure (load path).
Load tracing involves the systematic process of deter-mining loads and support reactions of individual struc-tural members as they in turn affect the loading of otherstructural elements (Figure 4.3). Simple determinatestructures can be thoroughly analyzed using FBDs inconjunction with the basic equations of equilibrium stud-ied previously.
Usually, the process begins with the very uppermostmember or level, tracing loads downward, layer by layer,until the last affected member under investigation issolved. In other words, start from the uppermost roof ele-ment, and work your way down through the structure un-til you reach the foundation.
Figure 4.5(d) FBD of the column.
Figure 4.5(c) FBD of the girder.
Figure 4.5(b) FBD of the beam.
Figure 4.5(a) FBD of the decking.
Figure 4.4 Decking-beam-girder-columnload path.
Load Trac ing 197
Load PathsIn general, the shorter the load path to its foundation andthe fewer elements involved in doing so, the greater theeconomy and efficiency of the structure. The most efficientload paths also involve the unique and inherent strengthsof the structural materials used—tension in steel, compres-sion in concrete, and so on. Bending, however, is a relativelyinefficient way to resist loads, and as a result, beamsbecome relatively large as loads and spans increase.
Sketches of structural members in the form of FBDs areused extensively to clarify the force conditions of individ-ual elements as well as other interconnected members.Simple determinate structures can be thoroughly analyzedusing FBDs in conjunction with the basic equations ofequilibrium studied previously. As long as each element isdeterminate, the equations of equilibrium are sufficient todetermine all supporting reactions. Load tracing requiresan initial assessment of the general structural frameworkto determine where the analysis should begin (Figure 4.4).
Each time the load path is redirected, a support conditionis created, and the loads and reactions at each transfermust be analyzed (using FBDs) and solved (using equa-tions) (Figure 4.5).
198 Chapter 4
Tributary AreaLoads uniformly distributed over an area of roof or floorare assigned to individual members (rafters, joists, beams,girders) based on the concept of distributive area, tributaryarea, or contributory area. This concept typically considersthe area that a member must support as being halfwaybetween the adjacent similar members.
A section of a wood floor framing system (Figure 4.6) willbe used to illustrate this concept. Assume that the generalload over the entire deck area is a uniform
■ The tributary width contributing to the load onbeam B is the distance (plank span) between
A and B plus the distance between B and C.12
12
50 #>ft.2.
Figure 4.6 Wood floor system with deckingand beams.
■ The tributary width of load on beam The same is true for beam C.
■ Similarly, the tributary width for edge beams Aand
Beam loads resulting from a uniformly applied load con-dition are determined by multiplying the load in poundsper square foot by the tributary width of load:
The load on each beam may be expressed as
ωA,D = 150 #>ft.2 * 2¿2 = 100 #>ft.
ω = 1#>ft.22 * 1tributary width21#>ft.22
D = 2¿.
2 = 4¿.B = 2 +
FBD of beams A and D.
ωB,C = 150 #>ft.2 * 4¿2 = 200 #>ft.
FBD of beams B and C.
Load tracing involves the systematic process of determin-ing loads and support reactions of individual structuralmembers as they in turn affect the loading of other struc-tural elements.
Figure 4.7 Various framing options.
Load Trac ing 199
Framing Design Criteria: Direction of SpanArchitectural characterThe structural framing, if exposed, can contribute signifi-cantly to the architectural expression of buildings.
Short joists loading relatively long beams yield shallowjoists and deep beams. The individual structural bays aremore clearly expressed in Figure 4.7.
Structural efficiency and economyConsiderations should include the materials selected forthe structural system, the span capability, and the avail-ability of material and skilled labor. Standard sections andrepetitive spacing of uniform members are generally moreeconomical.
Mechanical and electrical system requirementsThe location and direction of mechanical systems should becoordinated with the intended structural system. Layeringthe structural system provides space for ducts and pipes tocross structural members, eliminating the need to cut open-ings in the beams. Flush or butt framing saves space insituations where floor-to-floor dimensions are limited byheight restrictions.
Openings for stairs and vertical penetrationsMost framing systems accommodate openings, but gener-ally, it is more economical to make openings parallel to thedominant spanning direction. Additional headers andconnections create point loads on members that are typi-cally designed for light, uniform loads, thus increasingtheir size.
(a) Long, lightly loaded joists bearing onshorter beams create a more uniformstructural depth. Space can be conserved if thejoists and beams are flush framed.
(b) Short joists loading relatively long beamsyield shallow joists and deep beams. Theindividual structural bays are more clearlyexpressed.
(c) Loads can be reduced on selected beams byintroducing intermediate beams.
(d) The span capability of the decking materialcontrols the spacing of the joists, whereasbeam spacing is controlled by the allowablejoist span.
(e) Three-level framing system. Photo by author.
200 Chapter 4
Load Paths: Pitched Roof SystemsSingle-level framingRafters and ceiling joists combine to form a simple trussspanning from wall to wall. In addition to the truss action(rafters pick up compression forces, and ceiling joists de-velop tension to resist the horizontal thrust), rafters expe-rience bending due to the uniform load along their length,as shown in Figures 4.8(a) and 4.8(b).
Figure 4.8 Framing for pitched roofs.
(a)
(c)
(e)
(b)
(d)
(f)
Double-level framingRoof joists or beams are supported by a ridge beam at oneend and a bearing wall or header beam at the other. Noceiling ties are used, because this arrangement does notdevelop a horizontal thrust (as in the previous example).Notice that each level of structural framing spans in a di-rection perpendicular to the next layer, as shown inFigures 4.8(c) and 4.8(d).
Three-level framingThe load path sequence in this arrangement starts with theloads transferred from the sheathing (decking) onto thepurlins, which distribute concentrated loads onto the roofbeams, which in turn transmit load to the ridge beam atone end and a bearing wall or wall beam at the other.Columns or wall framing support the ridge beam at eitherend, as shown in Figures 4.8(e) and 4.8(f).
Load Trac ing 201
Construction: Pitched Roof SystemsSingle-level framingA common roof system for residential structures is arafter/ceiling joist arrangement. Loads onto the roof areinitially supported by the sheathing (plywood or otherstructural panels or skip sheathing, usually boards spaced some distance apart), which in turn loadsthe rafters (Figure 4.9).
1– * 4–
Figure 4.9 Single-level framing.
(b) Typical light-frame structure. Photocourtesy of the Southern Forest ProductsAssociation.(a) FBD—rafters.
Double-level framingAnother common roof framing arrangement involves roofjoists or beams that are supported by a ridge beam at oneend and a bearing wall or header beam at the other. Theridge beam must be supported at each end by a column orbearing wall (Figure 4.10).
(a) FBD—rafters.
Figure 4.10 Double-level framing.
(c) Post-and-beam construction. Photo byauthor.(b) FBD—ridge beam.
Three-level framingA method used to achieve a heavier beam appearance isspacing the roof beams (rather than the rafters) fartherapart, typically 4 to 12 feet o.c. Perpendicular to the roofbeams are purlins, spaced from one foot, six inches to fourfeet o.c., supporting sheathing, decking, or a metal roof. Inboth the two- and three-level framing systems, the ceilingplane can follow the roof slope (Figure 4.11).
(a) FBD—purlins.
Figure 4.11 Three-level framing.
(b) FBD—roof beams. (c) FBD—ridge beam.
202 Chapter 4
Load Paths: Wall SystemsA bearing wall is a vertical support system that transmitscompressive forces through the wall plane and to thefoundation. Uniform compressive forces along the lengthof the wall result in a relatively uniform distribution offorce. Concentrated loads or disruptions in the structuralcontinuity of the wall, such as large window or dooropenings, will result in a nonuniform distribution of com-pressive forces on the footing. Bearing wall systems canbe constructed with masonry, cast-in-place concrete, site-cast tilt-up concrete, or studs (wood or light-gauge metalframing).
Uniform slab loads are distributed along the top of thebearing wall as A masonry or concrete wall footing willbe required to support plus the additional wall weight.The load and remains a uniformload (Figure 4.12).
Uniform distributionRoof or floor joists (in typical light-wood framing) arespaced 16 or 24 inches on center. This regular, close spac-ing is assumed as a uniform load along the top of the wall.If no openings disrupt the load path from the top of thewall, a uniform load will result on top of the footing(Figure 4.13).
Nonuniform distributionConcentrated loads develop at the top of a wall whenbeams are spaced at wide intervals. Depending on thewall material, the concentrated load distributes along anangle of to as it moves down the wall. The result-ing footing load will be nonuniform, with the largestforces directly under the applied load (Figure 4.14).
60°45°
ω2 = 1ω1 + wall weight2ωω.
Figure 4.14 Concentrated loads from widelyspaced beams.
Figure 4.13 Uniform wall load from raftersand joists.
Figure 4.12 Uniform wall load from a slab.
Load Trac ing 203
“Arching action” over openingOpenings in walls also redirect the loads to either side ofthe opening. The natural stiffness of a concrete wall undercompression produces an “arching action” that contributesto the lateral distribution of the loads (Figure 4.15).
Opening in a stud wallStud walls (wood and metal) are generally idealized asmonolithic walls (except for openings) when loaded uni-formly from above. Openings require the use of headers(beams) that redirect the loads to either side. Concentratedloads from the header reactions must be supported by abuildup of studs resembling a column (Figure 4.16).
Concentrated loads—pilastersIn special cases where the concentrated loads are verylarge, walls may need to be reinforced with pilasters di-rectly under the beam. Pilasters are essentially columnsand carry the large concentrated loads directly to the foot-ing. The walls between the pilasters are now considered asnonbearing walls except for carrying their own weight(Figure 4.17).
Figure 4.15 Arching over wall openings.
Figure 4.17 Pilasters supportingconcentrated beam loads.
Figure 4.16 Stud wall with a windowopening.
Figure 4.18 One-level framing (roofand floor).
204 Chapter 4
Load Paths: Roof and Floor SystemsOne-level framingAlthough it is not a common framing system, relativelylong-spanning decking materials may transmit roof orfloor loads directly to bearing walls (Figure 4.18).
Framing plan.
Two-level framingThis is a very common floor system that uses joists to sup-port a deck. The decking is laid perpendicular to the joistframing. Span distances between bearing walls and beamsaffect the size and spacing of the joists (Figure 4.19).
Figure 4.20 Three-level framing.
Figure 4.19 Two-level framing. Framing plan.
Three-level framingWhen bearing walls are replaced by beams (girders ortrusses) spanning between columns, the framing involvesthree levels. Joist loads are supported by major beams,which transmit their reactions to girders, trusses, orcolumns. Each level of framing is arranged perpendicularto the level directly above it (Figure 4.20).
Framing plan.
Load Trac ing 205
Load Paths: Roof and Floor SystemsOne-level framingPrecast hollow-core concrete planks or heavy-timber-plank decking can be used to span between closely spacedbearing walls or beams. Spacing of the supports (the dis-tance between bearing walls) is based on the span capabil-ity of the concrete planks or timber decking (Figure 4.21).
(a) FBD—decking.
Figure 4.22 Two-level framing.
FBD—plank.
Figure 4.21 One-level framing.
Two-level framingEfficient structural sections in wood and steel joists allowrelatively long spans between bearing walls. Lighter deckmaterials, such as plywood panels, can be used to span be-tween the closely spaced joists (Figure 4.22).
(b) FBD—joists.(c) Light-framed joist-beam assembly. Photoby Chris Brown.
Figure 4.23 Three-level framing.
Three-level framingBuildings requiring large open floor areas, free of bearingwalls and with a minimum number of columns, typicallyrely on the long span capability of joists supported bytrusses or girders. The spacing of the primary structureand the layering of the secondary structural members es-tablish regular bays that subdivide the space (Figure 4.23).
(e) Joist-beam truss; three level-framing.Photo by Chris Brown.
(a) FBD—decking. (b) FBD—joists.
(c) FBD—beams. (d) FBD—girder.
206 Chapter 4
Load Paths: Foundation SystemsThe foundation system for a particular structure or build-ing depends on the size of the building, the use of thestructure, the subsurface conditions at the site, and thecost of the foundation system to be used.
A large building with heavy loads can often be supportedon relatively inexpensive shallow footings if the subsur-face soils are dense and stable. However, the same build-ing constructed at a site containing soft soils or expansiveclay may require pile or caisson foundations. Foundationsare generally subdivided into two major categories: (a)shallow foundations and (b) deep foundations.
Shallow foundationsShallow foundations essentially obtain their support on soilor rock just below the bottom of the structure in direct bear-ing. Vertical loads are transmitted from walls or columns toa footing that distributes the load over an area large enoughthat the allowable load-carrying capacity of the soil is not ex-ceeded and/or settlement is minimized. Shallow founda-tions are of three basic types: (a) spread footings—individualcolumn footing, (b) continuous strip footings—supportinga bearing wall, and (c) mat foundations—covering the entireplan area of the building.
Spread footing. This footing type is usually square orsometimes circular in plan and is generally simple andeconomical for moderate to high soil-bearing capacities.The purpose of this type of footing is to distribute the loadover a large area of soil. Pedestal and footing are rein-forced with steel (Figure 4.24).
Wall footing. Wall footings are one of the most commonfooting types and support relatively uniform bearing wallloads through a continuous foundation wall. The wallfooting width remains constant throughout its length if nolarge concentrated loads occur (Figure 4.25).
Mat or raft foundations. Mat foundations are usedwhen soil bearing is relatively low or the loads are heavyin relation to soil-bearing capacities. This foundation typeis essentially one large footing under the entire building,thus distributing the load over the entire mat. A mat iscalled a raft foundation when it is placed deep enough inthe soil that the soil removed during excavation equalsmost or all of the building’s weight (Figure 4.26).
Figure 4.25 Wall footing.
Figure 4.24 Spread footing.
Figure 4.26 Mat or raft foundation.
Load Trac ing 207
Deep foundationsThe function of a deep foundation is to carry buildingloads beneath a layer of unsatisfactory soil to a satisfac-tory bearing stratum. Deep foundations are generallypiles, piers, or caissons installed in a variety of ways.Normally, there is no difference between a drilled caissonand a drilled pier, and most often, only a modest differ-ence in diameter exists between them. Piles, the most com-mon deep-foundation system, are driven into the earthusing pile-driving hammers powered by drop hammers,compressed air, or diesel engines. Building loads are dis-tributed to the soil in contact with the surface area of thepile through skin friction (friction piles), in direct bearing(bearing piles) at the bottom of the pile on a sound stra-tum of earth or rock, or a combination of skin friction anddirect bearing.
Pile foundations. Timber piles are normally used as fric-tion piles, whereas concrete and steel piles are generallyused as bearing piles. When bearing piles must be drivento great depths to reach suitable bearing, a combination ofsteel and concrete is used. Hollow steel shells are driveninto the ground to a predetermined bearing point, andthen the casings are filled with concrete (Figure 4.27).
Pile caps. Individual building columns are generallysupported by a group (cluster) of piles. A thick reinforcedcap is poured on top of the pile group and distributes thecolumn load to all the piles in the cluster (Figure 4.28).
Grade beams. Piles or piers supporting bearing wallsare generally spaced at regular intervals and connectedwith a continuous reinforced concrete grade beam. Thegrade beam is intended to transfer the loads from thebuilding wall to the piles (Figure 4.29).
Figure 4.29 Grade beam supporting abearing wall.
Figure 4.28 Pile cap on one pile group.
Figure 4.27 Pile foundations.
208 Chapter 4
Example Problems: Load Tracing
The example problems that follow will illustrate the load-tracing methodology as it applies to a variety of structuralframeworks and arrangements. Note that most of the exam-ples illustrated are wood-frame structures, such as thoseshown in Figures 4.1 and 4.30. Wood framing is the onestructural material type that generally results in a determi-nate framing system, whereas steel and, particularly, cast-in-place concrete are often designed to capitalize on theadvantages of indeterminacy through the use of rigidconnections and/or continuity.
Figure 4.30 Light-frame construction—a drugstore in Quincy, Washington. Photo by Phil Lust.
Load Trac ing 209
4.1 In the single-bay, post-and-beam deck illustrated,planks typically are available in nominal widths of four orsix inches, but for the purposes of analysis, it is permissi-ble to assume a unit width equal to one foot. Determinethe plank, beam, and column reactions.
Solution:
PLANK REACTION Looking at the deck in elevation, theload is determined by multiplying the pounds persquare foot load by the tributary width of the plank.Therefore:
ω = 68 #>ft.211¿2 = 68 #>ft.
ω
Total load 1LL + DL2 = 68 psf
Deck weight 1dead load, or DL2 = 8 psf Load on the deck 1live load, or LL2 = 60 psf
BEAM REACTION The planks load the beams with a loadof 68# per foot of the plank span. Half of the plank load istransferred to each beam. The beams are loaded by theplanks with a load of 272# per foot of the beam span.
In addition, the beam has a self-weight equivalent to 10 #/ft.
R =ωL2
=68 #>ft.218¿2
2= 272 # 1Beam reaction2
COLUMN REACTION Half of each beam load is trans-ferred to the column at each corner of the deck. Thecolumns are loaded by the beams with loads of 1,692#at each column. Assume each column has a self-weightof 100#.
GROUND REACTION The load at each column is resistedby an equivalent ground reaction of 1,792#.
= 1,692 # 1Column reaction2R =
ωL2
=1272 + 102 #>ft.112¿2
2
210 Chapter 4
4.2 This problem represents an expansion of ExampleProblem 4.1, where the decking has an additional six feetto span and the beams are extended another bay. Loads onthe structural system remain the same.
Determine the loads developed in each column support.Assume that columns are located at grids 1-A, 2-A, 3-A, 1-B,2-B, 3-B, 1-C, 2-C, and 3-C.
Solution:
PLANK REACTION
ω = 68 #>ft.211¿2 = 68 #>ft.
Column self-weight = 100#Beam self-weight = 10 #>ft.Total load = 68 psfDeck LL = 60 psfDeck DL = 8 psf
BEAM REACTION First, analyze the planks that span sixfeet between grid lines A and B.
R =ωL2
=68 #>ft.216¿2
2= 204# 1Beam reaction2
Next, analyze the plank loads and beam support for theeight foot span between grids B and C.
R =ωL2
=68 #>ft.218¿2
2= 272# 1Beam reaction2
Load Trac ing 211
COLUMN REACTION All of the beam cases below repre-sent uniformly loaded conditions with simply supportedends, which result in reactions that are Theresulting reactions of the beams represent the loads pre-sent in each column.
First, analyze the beams along grid line A:
R = ωL>2.
Next, analyze the beams along grid line B:
Then, analyze the beams along grid line C:
212 Chapter 4
COLUMN LOADS AND REACTIONS The perimetercolumns along grid lines 1 and 3 receive half of the load ofeach beam. The interior columns along grid line 2 receiveloads from two beams, which are added together to calcu-late the column loads.
Column B-2
FBD of Column B-2
Load Trac ing 213
Example Problem 4.2—Alternate Method
Another technique may be employed in determining thebeam reactions without going through an analysis of theplanks. This may be accomplished by evaluating the tribu-tary widths of load for each beam and directly calculatingthe for each beam. For example, in the following figure,the tributary width of load assigned to the beams along gridline A is three feet. Therefore:
This value corresponds to the result obtained in the previ-ous method.
And along grid line C, with tributary
ω = 68 #>ft.214¿2 + 10 #>ft. = 282 #>ft.Q CHECKS
width = 4¿,
ω
ω = 68 #>ft.213¿2 + 10 #>ft. = 214 #>ft.
ω
Similarly, for beams along gridline B:
‹ ω = 68 #>ft.217¿2 + 10 #>ft. = 486 #>ft.Q CHECKS
Tributary width = 3¿ + 4¿ = 7¿
214 Chapter 4
4.3 A steel-framed floor for an office building was de-signed to support a load condition as follows:
Loads:
Dead Loads:
Using appropriate FBDs, determine the reaction forces forbeams B-1, B-2, and B-3 and for girder G-1.
Steel girders = 35 #>ft.Steel beams = 25 #>ft.Suspended ceiling = 5 psfMechanical equipment = 10 psfSteel decking = 5 psfConcrete = 150 #>ft.3
Live Load = 50 psf
Solution:
Loads:
DL + LL = 70 psf + 50 psf = 120 psf Total DL = 70 psf
+ 5 psf 1ceiling2 + 10 psf 1mechanical eq.2 + 5 psf 1decking2 = 50 psf 1slab2
Dead Loads (DL) = a 4–12 in.>ft. b A150 #>ft.3 B
Load Trac ing 215
Beam B-1: (Tributary width of load is )
ω1 = 120 #>ft.216¿2 + 25 #>ft. = 745 #>ft.6¿
Beam B-2: (Tributary width of load is )
ω2 = 120 #>ft.2112¿2 + 25 #>ft. = 1,465 #>ft.6¿ + 6¿ = 12¿
Beam B-3: This beam has two different load conditions dueto the changing tributary width created by the opening.
For 12 feet of span,
For six feet of span,
ω4 = 120 #>ft.216¿2 + 25 #>ft. = 745 #>ft.
ω3 = 120 #>ft.2112¿2 + 25 #>ft. = 1,465 #>ft.
216 Chapter 4
‹ Ay = 10,042#3©Fy = 04 - 1745 #>ft.218¿2 - 11465 #>ft.2112¿2 + 13,498# + Ay = 0
‹ By = 13,498#3©Ma = 04 - 1745 #>ft.218¿214¿2 - 11,465 #>ft.2112¿2114¿2 + By120¿2 = 0
Girder G-1: Girder G-1 supports reactions from beams B-2and B-3. Beam B-1 sends its reaction directly to the columnand causes no load to appear in girder G-1.
‹ Ay = 14,896#3©Fy = 04 - 14,650# - 13,498# + 14,512# + Ay = 0
‹ By = 14,512#- 135 #>ft. * 36¿2118¿2 + By136¿2 = 0
3©Ma = 04 - 14,650#112¿2 - 13,498#124¿2
Load Trac ing 217
4.4 In this example, the load trace will involve the fram-ing for a small deck addition to a residence. Once post re-actions have been determined, a preliminary footing sizewill be designed assuming the soil capacity of 3,000 psf isknown from a geotechnical investigation.
Loads:
Dead Loads:
γconcrete = 150 #>ft.3 (density) Girder = 10 #>ft. Beams = 5 #>ft.Decking = 5 #>ft.2
Live Load = 60 #>ft.2
For this load-trace problem, we will investigate thefollowing:
1. Draw an FBD of the typical beam with its loadcondition shown.
2. Draw an FBD of the girder with its loadconditions shown.
3. Determine the load in each post.4. Determine the size x of the critical pier footing
(account for the weight of the concrete).
218 Chapter 4
Solution:
1. Beam (typical interior):
Dead Load:
Live Load:
ωDL+LL = 265 #>ft.60 psf 14¿2 = 240 #>ft.
ωDL = 25 #>ft.Beam wt. = 5 #>ft.5 psf 14¿2 = 20 #>ft.
2. Beam (typical exterior):
Dead Load:
Live Load:
ωDL+LL = 135 #>ft.60 psf 12¿2 = 120 #>ft.
ωDL = 15 #>ft.Beam wt. = 5 #>ft.5 psf 12¿2 = 10 #>ft.
3/4. Girder and post:
‹ Ay = 7,032#- 110 #>ft.2112¿216¿2 + Ay18¿2 = 0
3©MB = 04 - 3,066#14¿2 - 3,066#18¿2 - 1,562#112¿2
Load Trac ing 219
The total reaction at post B is the sum of the reactions:
5. Critical footing:
The soil is capable of resisting a total bearing pressure of
By setting (allowable capacity of the soil),we need to deduct the weight of the footing itself to deter-mine the footing’s capacity to resist applied load fromabove. Therefore,
Footing weight can be solved by converting the density ofconcrete into equivalent poundsper square foot units by multiplying:
The remaining soil capacity to resist point loads is ex-pressed as
Practical size may be x = 2¿0–
‹ x = 1.57¿ = 1¿7– square footing 1theoretical size2‹ x2 =
Pqnet
=7,032#
2,875 #>ft.2 = 2.45 ft.2
qnet =PA
=P
x2
Because pressure =loadarea
qnet = 3,000 #>ft.2 - 125 #>ft.2 = 2,875 #>ft.2
= 125 #>ft.2‹ footing wt. = 1150 #>ft.32110–>12 in.>ft.2
in feet2footing wt. 1psf2 = 1γconcrete21thickness of concrete
1γconcrete = 150 #>ft.32
in psf2qnet = q - footing weight 1as pressure measured
q = 3,000 #>ft.2Note: pressure =
loadarea
; q =PA
3,000 #>ft.2
ByR + ByL = 4,630#
‹ ByR = 2,315#- 110 #>ft.2112¿2 + By1right side2 + 7,032# = 0
3©Fy = 04 -3,066#
2- 3,066# - 3,066# - 1,562#
220 Chapter 4
4.5 Calculate the load trace for a sloped roof structure.
Roofing = 5 psfRoof Sheathing = 3 psf }Along rafter
Rafters = 4 #/ft.
Snow load (SL): 40 psf on horizontal projection ofrafter*
*Snow loads are normally given as a load on the horizontal projec-tion of a roof.
Solution:
1. Rafter analysis:
1Along rafter length220 #>ft.
Rafter: + 4 #>ft.18 psf2124–>12 in.>ft.2 = 16 #>ft.
Sheathing: 3 psfRoofing: 5 psf
Beam: 16 #>ft.
Adjust to horizontal projection:
Note: Rafter horizontally projected for ease of computation.
ω = 21.7 #>ft. + 80 #>ft. = 101.7 #>ft.SL = 40 #>ft.2 *
24–12 in.>ft. = 80 #>ft.
1312120 #>ft.2 = 21.7 #>ft.
Load Trac ing 221
Using FBD sketches and computations, determine theload condition on (a) beam, (b) column, and (c) wall (typi-cal stud).
1. Beam analysis:
The reaction from a typical roof joist on top of the beamand stud wall is 813.6#. However, because the roof joistsoccur every two feet, the equivalent load is equal to
= 422.8 #>ft.ω =
813.6#2¿
= 406.8 #>ft. + 16 #>ft.1beam weight2ω
2. Column analysis:
The column load is computed as
Note: This equation simply divides the total load on the rafterin half, because the rafter is symmetrically loaded.
3. Stud wall:
Tributary wall length per stud is
16– =16–
12 in.>ft. = 1.33¿
P =ωL2
=422.8 #>ft.124¿2
2= 5,074#
222 Chapter 4
4.6 A simple light-framed wood building is subjected tothe load conditions as specified. Using FBDs and equa-tions of equilibrium, trace the loads through the buildingfor the following elements:
1. Determine the equivalent (horizontallyprojected) load on the rafters.
2. Determine the load per foot on the bearing wall.3. Determine the load on the ridge beam.4. Determine the column loads.5. Determine the minimum width of the continuous
foundation.6. Determine the size of the interior footings.
Load Conditions:
Along Rafter Length:
Ridge beam spans 16 feet from column support to columnsupport.
Ceiling = 2 psfRafters = 2 psfSheathing = 3 psfRoofing = 5 psf
Walls = 7 psfSL = 25 psfLL 1occupancy2 = 40 psfJoists = 4 psfSubfloor = 5 psfFlooring = 2 psfSoil Bearing Pressure = 2,000 psf
Load Trac ing 223
Solution:
1. Rafters:
1Equivalent load horizontally projected2ω¿DL = a 13
12b 116 #>ft.2 = 17.3 #>ft.
1Along rafter length2ωDL = 112 #>ft.22 a 16
12b¿ = 16 #>ft.
Roof DL = 12 psf1Horizontal projection2ωSL = 125 #>ft.22 a 16
12b ¿
= 33.3 #>ft.SL = 25 psf: rafters spaced 16– on center.
ωtotal = 33.3 #>ft. + 17.3 #>ft. = 50.6 #>ft.ωtotal = ωSL + ω¿DL
The reaction at each rafter support may be determined us-ing equilibrium equations. When uniform loads on a sim-ply supported member are present, a simple formula maybe used where
R =ωL2
=150.6 #>ft.2114¿2
2= 354#
224 Chapter 4
2. Bearing wall:
The reaction of the rafter onto the bearing wall is 354#every 16 inches. A conversion should be done to expressthe load at the top of the wall in pounds per lineal foot.
A strip of wall foot wide and seven feet, six inches tallweighs
3. Ridge beam:
Rafter reactions are equal to 354# per 16 inches, or 266#/ft. Because the ridge beam is required to support raftersfrom both sides,
ω = 21266 #>ft.2 = 532 #>ft.
ω = 266 #>ft. + 52.5 #>ft. = 318.5 #>ft.ωwall = 7 #>ft.217.5¿2 = 52.5 #>ft.
1
ω = 1354 #>16–2 a 1216b ¿
= 266 #>ft.
Note: The ridge beams are treated as two simple span beams,each 16 feet in length.
Exterior columns supporting the ridge beam carry
Note: The eight feet represents the tributary beam length that issupported by the exterior columns.
Interior columns support a tributary beam length of 16feet; therefore,
Pint = 1532 #>ft.2116¿2 = 8,512#
Pext = 1532 #>ft.218¿2 = 4,256#
Load Trac ing 225
4. Floor joists:
Joists are spaced 16 inches on center, which also representsthe tributary width of load assigned to each joist.
ωD + L = 51 #>ft.2 a 1612b¿ = 68 #>ft.
1Floor Joist Loads and Reactions2Loads: DL + LL = 11 psf + 40 psf = 51 psf
Foundation reactions may be obtained by
The central floor beam supports a floor joist reaction equalto
Conversion of the floor joist reactions into load per footresults in
5. Continuous foundation:
The stem wall measures eight inches thick and two feettall. The footing base is eight inches thick and x wide.
Loads from the roof, wall, and floor are combined as atotal load on top of the foundation stem:
The foundation wall stem adds additional load on thefooting equal to
Because the footing width x is unknown, the weight of thefooting base must be computed in terms of pounds persquare foot.
Footing weight = a 812b¿ 1150 #>ft.32 = 100 #>ft.2
Stem weight = a 812b¿ 12¿21150 #>ft.32 = 200 #>ft.
ωtotal = 318.5 #>ft. + 357 #>ft. = 675.5 #>ft.
Beam: ω = 1952 #>16 in.2 a 1216b¿ = 714#
Foundation: ω = 1476 #>16 in.2 a 1216b¿ = 357 #
= 952 #>16–Beam = 168 #>ft.2114¿ 5tributary length62
= 476 #>16–Foundation = 168 #>ft.217¿ 5tributary length62
226 Chapter 4
In determining the footing width, examine a unit lengthof foundation as a representation of the entire length.
This value of represents the resistance of the soil avail-able to safely support the loads of roof, walls, floor, andfoundation stem.
The minimum required resistance area of the footing perunit length is
Note that a footing base of six inches would be less thanthe thickness of the stem wall. A minimum footing widthfor a one-story, light-framed assembly should be 12inches. If a 12 inch base width is provided, the actual pres-sure on the soil will be
6. Interior spread footings:
A determination of individual post loads is necessarybefore footing sizes can be computed.
Assume the spread footings have a thickness of eightinches and psf:
Critical center column:
Other footing:
‹ x = 1.73¿ L 1¿9– square
x2 =5,712#
1,900 #>ft.2 = 3.00 ft.2
‹ x = 2.74¿ L 2¿9– square
A = x2 =P
qnet=
14,224#
1,900 #>ft.2 = 7.5 ft.2
= 1,900 psfqnet = q - footing wt. = 2,000 psf - 100 psf
ωfooting = a 812b¿ 1150 #>ft.32 = 100 #>ft.2
q = 2,000
‹ OK.= 875.5 #>ft.2 6 qnet
Actual pressure =ω
x = 1¿=
875.5 #>ft.1¿
x =875.5 #>ft.1,900 #>ft.2 = 0.46¿ L 6–
qnet =ωx
; x =ω
qnet
A = 11¿21x2
ωtotal = 675.5 #>ft. + 200 #>ft. = 875.5 #>ft.
qnet
= 1,900 psfqnet = q - footing weight = 2,000 psf - 100 psfq = allowable soil bearing pressure = 2,000 psf
11¿2
Load Trac ing 227
Problems
In each of the load-tracing problems below, construct a se-ries of FBDs and show the propagation of loads through thevarious structural elements.
4.1 Determine the column loads assuming:
Beam B-2 spans between girder G-3 and column B-2 andbeam B-3 spans between girder G-3 and column C-2.
4.2 Loads:
1. Determine the equivalent (horizontallyprojected) load on the rafters spaced at two feet,zero inches 2’0” on center.
2. Determine the load per foot on the bearing walls.3. Determine the loading and beam reactions for
each of the steel wide-flange beams.
LL = 40 psf 12nd and 3rd floors2Floors: DL = 20 psf 12nd and 3rd floors2Bearing walls: DL = 10 psf 12nd and 3rd floors2
LL = 10 psfCeiling: DL = 5 psf
1snow horizontally projected2 LL = 25 psf
Roof: DL = 10 psf
Total 50 psfLL 1occupancy2 = 40 psfDL 1decking, flooring, etc.2 = 10 psf
228 Chapter 4
4.3 Trace the loads through the following elements inthis structure. Occupancy live load is 40 psf, with a floordead load of 5 psf.
1. Rafters.2. Stud walls.3. Roof beam.4. Columns (interior and exterior).5. Floor joist.6. Floor beam.7. Load on top of continuous footings.8. Critical interior footing load.
4.4 Draw FBDs and show load conditions for B-1, G-1,interior column, B-2, and G-2.
Loads:
Girders G-1 and G-2 = 50 #>ft.Beams B-1 and B-2 = 15 #>ft.Insul., mech., elec. = 5 psfTruss joist = 3 psf
1deck2Roofing and joists = 10 psfSL = 25 psf
Load Trac ing 229
4.5 Roof Loads:
1. Sketch the load and its magnitude acting on the34-foot-long ridge beam.
2. What is the force in columns A and B, which sup-port the ridge beam?
4.6 Show graphically (FBDs for each element) the loadtrace (load condition) for:
1. Rafter(s).2. Roof beam.3. Exterior stud wall(s).4. Interior columns.5. Floor joist(s).6. Floor beam.7. Floor post.8. Exterior foundation width(s) (adequacy?).9. Size of critical pier footing.
Load Conditions:
Along Rafter Length:
Soil bearing pressure = 2,000 psfInsulation = 2 psfCeiling = 3 psfRafters = 3 psfSheathing = 2 psfRoofing = 8 psf
1density of concrete2γconcrete = 150 #>ft.3Bearing walls = 10 psfOccupancy 1LL2 = 40 psfInsulation = 2 psfJoists = 3 psfFinish floor>subfloor = 5 psfSL 1horiz. proj.2 = 30 psf
Ridge beam = 40 #>ft.Joists = 4 #>ft.Insulation = 5 psfPlywood = 2 psfShakes = 5 psfSL = 20 psf
230 Chapter 4
4.7 For the illustrated hipped roof, evaluate the load con-ditions on:
1. Typical jack rafter.2. Hip rafter.3. Ceiling joist.4. Beams B-1, B-2, and B-3.5. Interior column.
Roof Live Loads:
Roof Dead Loads:
Ceiling Loads:
LL = 20 psfDL = 7 psf
Joist framing = 4 #>ft.Plywood roof sheathing = 1.5 psfRoofing = 6 psf
SL = 25 psf
Figure 4.31 Examples of lateral instability.The top photo is of an old carport with kneebraces added a bit too late. The other twophotographs are of wooden structuresdamaged in the Kobe, Japan, earthquake of1995. Photos by authors.
Load Tr ac ing 231
4.2 LATERAL STABILITY LOAD TRACING
The structure, be it large or small, must be stable and last-ing, must satisfy the needs for which it was built, andmust achieve the maximum result with minimum means.
These conditions: stability, durability, function, and maxi-mum results with minimum means—or in current terms,economic efficiency—can be found to an extent in all con-struction from the mud hut to the most magnificent build-ing. They can be summed up in the phrase “buildingcorrectly,” which seems to me more suitable than the morespecific: “good technical construction.” It is easy to seethat each of these characteristics, which at first seem onlytechnical and objective, has a subjective—and I would addpsychological—component which relates it to the aestheticand expressive appearance of the completed work.
Stable resistance to loads and external forces can be achievedeither by means of structures that the beholder can immedi-ately or easily perceive or by means of technical artifices andunseen structures. It is evident that each approach causes adifferent psychological reaction which influences the expres-sion of a building. No one could feel a sense of tranquil aes-thetic enjoyment in a space whose walls or whose roof gavethe visual sensation of being on the verge of collapse, even ifin reality, because of unseen structural elements, they wereperfectly safe. Similarly, an apparent instability might,under certain circumstances, create a feeling of particularaesthetic—though anti-architectural—expression.
Thus one can see that even the most technical and basicquality of construction, that of stability, can, through thedifferent building methods employed to assure it, con-tribute greatly to the achievement of a determined and de-sired architectural expression.
—Pier Luigi Nervi, Aesthetics and Technology inBuilding, Harvard University Press, Cambridge,
Mass., 1966. Pages 2 and 3.
Section 4.1 on load tracing followed the path of loadsthrough a structural framework to the foundation. Deadand live loads on the structure were gravity-induced andassumed to be acting in a vertical, downward direction.Each joist, beam, girder, column, and the like could be
Figure 4.34 Wind acting perpendicular tothe shearwalls.
232 Chapter 4
analyzed using appropriate FBDs along with equations ofequilibrium (in the case of statically determinate systems).Although the conditions of equilibrium need to be satis-fied for each element or member in the structural frame-work, it is not a sufficient condition to ensure thegeometric stability of the whole structure (see Figure 4.31).
Stability can be problematic for a single structural ele-ment, such as an overloaded beam (Figure 4.32[a]) or abuckled column (Figure 4.32[b]), but sometimes an entirestructural assembly may become unstable under certainload conditions. Geometric stability refers to a configura-tional property that preserves the geometry of a structurethrough its elements being strategically arranged and in-teracting together to resist loads.
Figure 4.33 Wind acting parallel to theshearwalls.
Figure 4.32(a) Excessive beam deflection.
Figure 4.32(b) Column buckling.
All building structures require a certain set(s) of elements,referred to as a bracing system (Figure 4.33), which pro-vides the requisite stability for the entire structural geom-etry. Decisions about the type and location of the bracingsystem to be used directly affect the organizational plan ofthe building and, ultimately, its final appearance.
A primary concern in the design of any structure is to pro-vide sufficient stability to resist collapse and also preventexcessive lateral deformation (racking; see Figure 4.34),which may result in the cracking of brittle surfaces andglass. Every building should be adequately stiffened againsthorizontal forces coming from two perpendicular directions.
Wind and seismic forces on buildings are assumed to acthorizontally (laterally) and must be resisted in combina-tion with gravity loads. For example, when wind forcespush laterally on the side of a one-story, wood-framedbuilding, these horizontal forces are transmitted by thesheathing or cladding to vertical framing elements (wallstuds), which in turn transmit the loads to the roof andfloor. The horizontal planes (roof and floor) must be sup-ported against lateral movement. Forces absorbed by thefloor plane are sent directly into the supporting founda-tion system, whereas the roof plane (referred to as the roofdiaphragm) must be supported by the walls aligned paral-lel to the wind direction. In typical light-wood-framedstructures, these lateral force-resisting walls are calledshearwalls. The use of vertical wall framing and horizontaldiaphragms is the most common system in wood-frame
Figure 4.36 Simple frame with four hinges.
(a) Arrangement of lateral elements. (b) Idealized frame.
Figure 4.35 Structure with two parallel shearwalls.
Load Tr ac ing 233
If we assume the construction to be of wood, the beamand post connections are appropriately assumed as simplepins, and the base supports function as hinges. This sim-ple rectilinear geometry with four hinges (Figure 4.36) is
buildings, because the roof sheathing can be designed eco-nomically to function as both a vertical load-carrying andlateral load-carrying element.
Roof and floor diaphragms must be capable of transmittingthe applied lateral forces to the shearwalls through their pla-nar strength; alternately, bracing must be provided in thehorizontal plane. Loads transferred from the roof diaphragmto the wall plane are then channeled to the foundation.
Let’s return to the earlier diagram of the simple roof struc-ture shown in Figure 4.35(a), which is supported by thetwo parallel (N-S) walls and which has two nonstructuralwalls (glass) on the other two parallel planes. A tributaryslice (Figure 4.35[b]) through the framework reveals a rec-tilinear arrangement that is simplified, for analysis pur-poses, as a beam supported by two posts.
Figure 4.39 Knee-bracing.
Figure 4.38 Diagonal tension counters.
234 Chapter 4
inherently unstable and requires the addition of otherstructural elements to prevent lateral collapse from hori-zontally applied loads or uneven vertical loading.
There are several ways of achieving stability and counter-acting the racking of the frame under vertical and/or hor-izontal loading. Note that each solution has obviousarchitectural implications, and selection of the bracingsystem must be made for reasons beyond simply being themost “efficient structurally.”
Diagonal Truss MemberA simple way of providing lateral stability is to introducea simple diagonal member connecting two diagonally op-posite corners. In effect, a vertical truss is created, and sta-bility is achieved through triangulation. If a singlediagonal member is used, it must be capable of resistingboth tension and compression forces, because lateral loadsare assumed to occur in either direction. Members sub-jected to compression have a tendency to “buckle” (sud-den loss of member stability) when too slender (smallcross-sectional dimension with respect to length); there-fore, the members need to be proportioned similarly totruss members in compression (Figure 4.37).
X-Bracing MembersAnother strategy involves the use of two smaller, cross-sectionally dimensioned X-bracing members. These X-bracesare also known as diagonal tension counters (discussed inSection 3.3), where only one counter is effective in resistinga directional lateral load (Figure 4.38).
Knee-BracingA commonly used arrangement in carports and elevatedwood decks is knee-bracing. This stiffening method trian-gulates the beam-column connection to provide a degreeof rigidity at the joint. The larger the knee-braces are, themore effective their ability to control racking. Bracing isusually placed as close to as possible but will some-times range between and Knee-braces developtension and compression forces (like truss members) de-pending on the lateral force direction (Figure 4.39).
Gusset PlatesLarge gusset plates at each beam-column connection canalso provide the required rigidity to stabilize the frame.However, in both the knee-brace and gusset-platearrangements, some movement will still occur because ofthe pin connections at the base of the columns. Modifyingthe base into a more rigid connection can certainly add tothe overall rigidity of the frame. Rigid connections inducebending moments in the beams and columns (Figure 4.40).
60°.30°45°
Figure 4.37 Diagonal truss member.
Figure 4.40 Rigid joint connection.
Figure 4.41 Pole structure—columns withrigid bases.
Load Trac ing 235
Rigid Base ConditionColumns placed at some depth into the ground and set inconcrete can provide a rigid base condition. Resistance tolateral loads comes through the columns acting as largevertical cantilevers and the horizontal beam transferringloads between the columns (Figure 4.41).
Combination Knee-Brace and RigidColumn BaseWhen knee-braces are used in conjunction with rigid col-umn bases, all connections of the frame are now rigid, andlateral loads are resisted through the bending resistanceoffered by the beam and columns. The lateral displace-ments would be less than in the three previous examples(Figure 4.42).
Rigid Beam/Column JointsMoment-resisting frames consist of floor or roof membersin plane with, and connected to, column members withrigid or semirigid joints. The strength and stiffness of aframe is proportional to the beam and column sizes and isinversely proportional to the column’s unsupportedheight and spacing. A moment-resisting frame may be in-ternally located within the building, or it may be in theplane of the exterior walls. Moment frames require consid-erably larger beams and columns, especially at the lowerlevels of tall structures.
All elements in a moment frame are actually beam-columnssubjected to combined stresses (bending and tension orcompression). Structural steel beams and columns may beconnected together to develop moment frame action bymeans of welding, high-strength bolting, or a combinationof the two (Figures 4.43 and 4.44).
Figure 4.43 Steel or concrete frame (rigid joints).
Figure 4.42 Rigid base and knee-bracing.
Figure 4.46 Knee-braced structure with roof trusses.
Figure 4.45 Geometrically unstable truss roof structure.
(b) Concrete moment connection.
Figure 4.44 Rigid connections in steel andconcrete.
236 Chapter 4
Cast-in-place concrete, or precast concrete with cast-in-place joints, provides the rigid or semirigid monolithicjoints required. Frames may consist of beams andcolumns, flat slabs and columns, and slabs with bearingwalls. The inherent continuity that occurs in the mono-lithic casting of concrete provides a naturally occurringmoment-resistant connection and, thus, enables membersto have cantilevers with very simple detailing of the rein-forcing steel.
Joint connections become quite complex and labor inten-sive for a three-dimensional rigid frame in two directions.This translates into higher cost, so alternative lateral resis-tance systems, such as braced frames or shear walls in oneof the directions, are preferred.
If we examine the idea of a beam on two columns andimagine a roof truss spanning between two columns,again the issue of lateral stability must be resolved.Although trusses are stable configurations as a result oftriangulation, a truss supported on two pin-connectedcolumns is unstable (Figure 4.45).
(a) Steel moment-resisting joint.
Because knee-braces help to develop rigidity at the cornerconnections for the beam and columns, a similar arrange-ment can be provided with the truss to develop resistanceto racking. Knee-braces attach to continuous columns,thereby developing lateral resistance through thecolumns’ bending capacity (Figure 4.46).
(a) Pratt truss. (b) Fink truss.
Figure 4.47 Modified Fink truss with side shedsand monitor.
Load Trac ing 237
The continuous column from the ground through the en-tire depth of the truss provides a very stiff bend (rigid con-nection) to resist lateral loads (Figure 4.47). Columns mustbe designed to resist the potentially large bending mo-ments that develop.
Many residential and small to midscale commercial build-ings depend on the walls (bearing and nonbearing) of thestructure to develop the necessary resistance to lateralforces (Figure 4.48). This type of lateral restraint, referredto earlier as a shearwall, depends on the vertical cantilevercapacity of the wall. The span of the cantilever is equal tothe height of the wall.
Figure 4.48 Exploded view of a light-framed wood building showing the various lateralresisting components.
(b) Narrow shearwall.
Figure 4.49 Shearwall proportions.
238 Chapter 4
ShearwallsA shearwall acts as a vertical cantilever beam to providelateral support (see Figures 4.48, 4.49, 4.52, and 4.54through 4.56) and experiences both shear and bendingstresses and deformations. The lateral resistance offeredby the shearwall depends on the relative rigidities of thewall and the horizontal diaphragms. Shearwalls for steel-framed buildings are relatively solid masses usually madeof reinforced concrete for a variety of building scales andconfigurations.
When concrete walls are used as fire-resisting compart-ment walls, stair and elevator cores, and vertical serviceshafts, it is reasonable to utilize them for stiffening thebuilding against lateral loads. A common structuralstrategy is to use a braced or rigid frame in combinationwith concrete walls. The beams of the steel frame mustbe connected to the concrete cores or shearwalls to trans-mit the vertical and horizontal forces to them. Carefuldetailing of the framing elements to the shearwall isessential.
In Figure 4.49, the width of the shearwall d is relativelylarge compared with the height h; therefore, shear deforma-tion replaces bending as the significant issue. Commonlyused materials for shearwalls are concrete, concreteblocks, bricks, and wood sheathing products, such as ply-wood, oriented strand board (OSB), and wafer boards.
Multiple BaysThus far, the discussion of frame stability from lateralloads has been limited to single-bay (panel) frames; how-ever, most buildings contain multiple bays in the hori-zontal and vertical directions. The principles that applyto single-bay frames also hold true for multiple-framestructures.
Remember, single diagonals must be capable of tension orcompression. The length of diagonals can become criticalwhen subjected to compression due to buckling. Bracingdiagonals should be kept as short as possible.
In the examples shown in Figure 4.50, it is quite possiblethat only one panel needs to be braced for the entire frameto be stabilized. It is rarely necessary for every panel to bebraced to achieve stability.
(a) Wide shearwall.
Figure 4.50 Bracing systems in multiple-baystructures.
(a) Diagonal tension counters.
(b) Diagonal truss brace.
(c) Shearwall.
Load Trac ing 239
Multistory, Multibay StructuresMultistory, multibay structures also use the same bracingprinciples. However, as the structures become much taller(height greater than three times the least building dimen-sion), only certain types of bracing systems and materialsof construction remain practical from a structural and/oreconomic standpoint. Knee-braces, although appropriatefor smaller, one- or two-story structures, are not nearly aseffective for larger structures. The horizontal force com-ponent of the knee-brace onto the column produces sig-nificant bending moments, which require larger columnsizes. Larger diagonal braces that go across an entirepanel from opposite diagonal points are much more effec-tive structurally.
Diagonals, X-bracing, and K-trussing on multistoryframes essentially form vertical cantilever trusses thattransmit lateral loads to the foundation (Figure 4.51).These bracing techniques are generally limited to the exte-rior wall planes of the building to permit more flexibilityfor the interior spaces. Reinforced concrete (or masonry)and braced steel framing used for stairwells and elevatorsare often utilized as part of the lateral force strategy.
Combinations of bracing, shearwall, and/or rigid framesare used in many buildings (Figure 4.52). Larger, multi-story buildings contain utility/service cores, which in-clude elevators, stairs, ducts, and plumbing chases, thatare strategically placed to meet functional and structuralcriteria. Because these cores are generally solid to meetfireproofing requirements, they can function as excellentlateral resisting elements, either in isolation or as part of alarger overall strategy.
Figure 4.52 Combination of lateral resistingsystems—steel frame with a central shearwallcore.
(e) Rigid frame.
Figure 4.51 Types of multistory bracingsystems.
(a) X-bracing.
(b) Eccentric braced frame.
(c) K-trussing.
(d) Shearwall.
Roof or floor diaphragmsto collect and distributelateral loads to the vertical resisting planes
Horizontal roof/floor bracing
Shear wall
Diagonal or X-bracingRigid frame
Figure 4.53 Bracing for a three-dimensionalframework.
(X-bracing, truss diagonals, knee-braces,shearwalls, and rigid beam-columnconnections could be used to stabilize any ofthese planes.)
240 Chapter 4
Three-Dimensional FramesFigure 4.53 reminds us that buildings are, indeed, three-dimensional frameworks and not planar, two-dimensionalframes. All of the frame examples illustrated previouslyassume that to understand the whole, only a represen-tative portion of an entire structure needs to be exam-ined. Each planar frame represents just one of several(or many) frames that constitute the structure. It is im-portant to note, however, that a fundamental require-ment of geometric stability for a three-dimensionalstructure is its ability to resist loads from three orthogo-nal directions.
A three-dimensional frame can be stabilized by use ofbracing elements or shearwalls in a limited number ofpanels in the vertical and horizontal planes. In multistorystructures, these bracing systems must be provided ateach and every story level.
The transverse exterior walls of a building transmit thewind forces to the roof and floors, which in turn directthem to the utility/service cores, shearwalls, or bracedframes. In most cases, the roof and floor systems form hor-izontal diaphragms, which can perform this function.
In wood-framed buildings or buildings with wood roofand floor systems, the roof or floor sheathing is designedand connected to the supporting framing members tofunction as a horizontal diaphragm capable of transferringlateral load to the shearwalls. In buildings with concreteroof and floor slabs, the slabs are also designed to functionas diaphragms.
It is unlikely that the horizontal system used in one direc-tion of loading will be different from the horizontal sys-tem used in the other direction. If the wood sheathing orreinforced concrete slab is designed to function as a hori-zontal diaphragm for lateral forces in one direction, itprobably can be designed to function as a diaphragm forforces applied in the other direction.
Occasionally, when the roof or floor sheathing is too lightor flexible and, thus, is unable to sustain diaphragmforces, the horizontal framework must be designed to in-corporate bracing similar to the braced walls or shear-walls.
Horizontal bracing may consist of tension counters, trusses,or stiff panels in strategic locations (see Figure 4.53).
Figure 4.54 Various shearwall arrangements—some stable, others unstable.
Load Trac ing 241
Bracing ConfigurationsOnce the roof plane (or floors) has been configured tofunction as a diaphragm, a minimum requirement forstabilizing the roof is three braced (or shear) walls thatare neither all parallel nor concurrent at a common point.The arrangement of the walls in relationship to one an-other is crucial in resisting loads from multiple directions(Figure 4.54). More than three braced (shear) walls areusually provided, thus increasing the structural stiffnessof the framework in resisting lateral displacements(shear deformation).
Braced (shear) walls should be located strategicallythroughout a structure to minimize the potential of tor-sional displacements and moments. A common solution isto have two shearwalls parallel to one another (a reason-able distance apart) and a third wall (or perhaps more)perpendicular to the other two.
(a) (b) Plan diagram.
Figure 4.57 Rigid frames at end bays (can also comprise the entire skeleton).
242 Chapter 4
Multistory StructuresIn multistory structures, lateral loads (from wind or earth-quake forces) are distributed to each of the floor (di-aphragm) levels. At any given floor level, there must be arequisite number of braced (shear) walls to transfer the cu-mulative lateral forces from the diaphragms above. Eachstory level is similar to the simple structures examinedpreviously, in which the diaphragm load was transferredfrom the upper level (roof) to the lower level (ground).
Multistory structures are generally braced with a mini-mum of four braced planes per story, with each wall beingpositioned to minimize torsional moments and displace-ment (Figures 4.55 through 4.57). Although it is often de-sirable to position the braced walls in the same position ateach floor level, it is not always necessary. The transfer ofshear through any one level may be examined as an iso-lated problem.
(a) (b) Plan diagram.
Figure 4.55 Shearwalls at the central circulation core.
(a) (b) Plan diagram.
Figure 4.56 Shearwalls at the exterior corners.
Solution:
Recall from Section 3.3 that diagonal tension countersare always in pairs, because one counter will buckle asa result of compression loading. The method of sec-tions was used earlier in determining the effectivecounter.
In this case, counters AF and CH are effective in tension,whereas members BE and DG are assumed as zero-forcemembers.
Utilizing the method of sections, draw an FBD of theframe above section cut a-a.
Only the x components of members AF and CH are capableof resisting the 1,500# lateral force. Assuming then
Completing the analysis using the method of joints fortrusses:
AF = CH =75012
= 1,061# 1T2 AFx = CHx = 750# 3©Fx = 04 AFx + CHx = 1,500#
AFx = CHx,
CHx = CHy =CH12
AFx = AFy =AF12
V =ωL2
=1100 #>ft.2130 ft.2
2= 1,500#
ω = 20 psf * 5¿ = 100 #>ft.
Load Trac ing 243
Example Problems: Lateral Stability/Diaphragms and Shearwalls
4.7 An industrial building with a plan dimension ofand a height of 10 feet is subjected to a
wind load of 20 psf. Two braced exterior walls parallel tothe wind direction are used to resist the horizontal di-aphragm force in the roof. Assuming diagonal tensioncounters in two of the three bays, determine the magni-tude of force developed in each diagonal.
30 feet * 30 feet
244 Chapter 4
4.8 A simple carport is framed using knee-braced framesspaced at five feet on center. Assuming wind pressure at20 psf, analyze a typical interior frame.
Solution:
Each interior knee-braced frame is required to resist a loadapplied to a tributary wall area of 20 square feet at 20 psf.
Notice that in the FBD of the frame, a total of four supportreactions develop at A and B. Because only three equa-tions of equilibrium are permitted, all support reactionscannot be solved unless an assumption(s) is made aboutthe frame or its load distribution characteristics. In thiscase, only one assumption is necessary, because the exter-nal support condition is indeterminate to the first degree.
Assume
Then,
Writing the other equations of equilibrium, we get
Once the support reactions have been determined, pass avertical section a-a through one knee-brace. Isolate the leftcolumn, and draw an FBD.
3©Fx = 04 + 400# - Ax + CEx - DEx = 0CEy = 800#CE = 80012#‹ CEx = 800#3©MD = 04 + CEx12¿2 - 200#18¿2 = 0
Ay = 320#3©Fy = 04 - Ay + By = 0By = 320#3©MA = 04 - 400#18¿2 + By110¿2 = 0
‹ Ax = Bx = 200#3©Fx = 04 Ax + Bx = 400#
Ax = Bx.
F = p * A = 20 #>ft.2 * 20 ft.2 = 400#
Load Trac ing 245
Substituting for and
In a similar manner, isolate the right column using sectioncut b-b.
FGx = 800# - 200# = 600# 3©Fx = 04 - FGx + FHx - Bx = 0
FGy = 800# - 320# = 480# 3©Fy = 04 + FGy - FHy + By = 0
FHy = 800# FH = 80012# FHx = +800# 3©MG = 04 - 200#18¿2 + FHx12¿2 = 0
DEy = +480# DEy = 800# - 320# 3©Fy = 04 - DEy + CEy - Ay = 0 DEx = +1,000# DEx = 400# - 200# + 800#
CEx,Ax
4.9 In this section, an analysis will be performed on asimple rectangular, one-story wood building, illustratingthe load propagation and transfer that occurs in a roof di-aphragm and shearwall resistive system.
Wood-framed roofs and walls are stiffened considerablyby the use of plywood or OSB/waferboard sheathing act-ing as diaphragms and shearwalls.
Wind force (assumed as a uniform pressure on the wind-ward front wall) is initially distributed to the roof andfloor diaphragm (or foundation). Conventional wood studframing in the walls functions as vertical beams and dis-tributes half of the wind load to the roof and the other halfto the floor construction.
Pressures on the windward wall and suction on the lee-ward wall convert into uniformly distributed loads along the windward and leeward boundary edges of theroof diaphragm. Often, the loads on the windward andleeward edges are combined as one load distributionalong the windward edge.
ω
The wall height dimension represents the tributary walldimension that loads the roof diaphragm. Uniform load carries the units of pounds per lineal foot, the same as auniform load on a beam.
In fact, roof diaphragms are essentially treated as flat,deep beams spanning from wall support to wall support.
ω = wind pressure p * 12 the wall height
ω12
246 Chapter 4
Shearwalls represent supports for the roof diaphragmwith resulting reactions V, where
The intensity of the shear reaction is expressed as lower-case v, where
Shear load V is applied to the top edge of the shearwall.Wall equilibrium is established by developing an equaland opposite shear reaction at the foundation, accom-panied by a tension (T) and compression (C) couple atthe wall edges to counteract the overturning momentcreated by V. The tension T is normally referred to as thetie-down force.
Because wind cannot be assumed to act in a prescribeddirection, another analysis is required with the wind pres-sure applied to the end walls, perpendicular in directionto the earlier analysis. Each wall is designed with the req-uisite wall sheathing thickness, nail size, and spacing toreflect the results of the wind analysis. Details of thisdesign procedure are covered more thoroughly in subse-quent courses on timber design. An excellent discussion,with examples of shearwalls in actual structures, can befound on the Internet at http://www.mcvicker.com/vwall/dti.htm.
V¿
v =Vd1pounds>ft.2
V =ωL21pounds2
Load Trac ing 247
4.10 A beach cabin on the Washington coast (100 mphwind velocity) is required to resist a wind pressure of 35psf. Assuming wood-frame construction, the cabin utilizesa roof diaphragm and four exterior shearwalls for its lat-eral resisting strategy.
Draw an exploded view of the building, and perform a lat-eral load trace in the N-S direction. Show the magnitudeof shear V and intensity of shear v for the roof and criticalshearwall. Also, determine the theoretical tie-down forcenecessary to establish equilibrium of the shearwall. Notethat the dead weight of the wall can be used to aid in thestabilizing of the wall.
248 Chapter 4
Solution:
Examining the roof diaphragm as a deep beam spanning42 feet between shearwalls:
An FBD of the shearwall shows a shear developing atthe base (foundation) to equilibrate the shear V at the topof the wall. In addition to equilibrium in the horizontal di-rection, rotational equilibrium must be maintained by thedevelopment of a force couple T and C at the edges of thesolid portion of wall.
Tie-down force T is determined by writing a moment equa-tion of equilibrium. Summing moments about point A,
T = 4,390#
T =182,695 #ft.2 - 116,875 #-ft.2
15
15T = 5,513#115¿2 - 2,250#17.5¿23©MA = 04 - V115¿2 + W115¿>22 + T115¿2 = 0
W = 10 psf * 15¿ * 15¿ = 2,250# W = dead load of the wall v = V>shearwall length = 5,513#>15¿ = 368#>ft.
V¿
V =ωL2
=262.5 #>ft.142¿2
2= 5,513#
ω = 35 psf * 7.5¿ = 262.5 #>ft.
Load Trac ing 249
Problems
4.8 Determine the forces in each of the members, includ-ing the effective tension counters. Assume the lateral forceto be resisted equally by each tension counter.
4.9 Determine the reaction forces A and B and all othermember forces. Unlike the previous problem, knee-braceelements are capable of carrying both tension and com-pression forces. Assume
4.10 A two-story warehouse is subjected to lateral forcesas shown. Determine the effective tension counters andforces in all other members. Assume the effective diagonaltension counters at the lower level share equally in resist-ing the horizontal forces.
4.11 A barn structure is subjected to wind velocities withthe equivalent of 20 psf pressure on the vertical projection(including the roof).
Analyze the diaphragm and shearwall forces assuming thewind hitting the long dimension of the barn. Use exploded-view FBDs in tracing the loads. Determine the shear reac-tion and hold-down forces at the base of the walls.
Note: The dead load of the walls may be used in helping to sta-bilize against rotation due to overturning moments.
Ax = Bx.
250 Chapter 4
4.12 A small garage utilizes premanufactured trussesspaced at two feet, zero inches 2’ 0” on center. One wall ofthe garage has a 22 foot opening, framed with a large glue-laminated girder. Trace the loads from the roof, and deter-mine the load at each end of the header. Also, size theconcrete footing that supports the header post (assume thefooting is square, thick).10–
Summary
■ Load tracing involves the systematic process of deter-mining load and support reactions of individual struc-tural members as they in turn affect the loading of otherstructural members. Simple determinate structures canbe thoroughly analyzed using FBDs in conjunction withthe basic equations of equilibrium for rigid bodies.
■ Loads uniformly distributed over an area of roof orfloor are assigned to individual members based on theconcept of tributary (contributory) area. Member loadareas are assumed to be halfway between the adjacentsimilar members.
■ All building structures require a certain set(s) of ele-ments (bracing system) that provides the necessary lat-eral stability for the entire structural geometry. Windand seismic forces on buildings are assumed to act hori-zontally (laterally) and must be resisted in combinationwith gravity loads.