Chapter 2.pdf

The large 1999 Chi-Chi earthquake (magnitude 7.7) in Taiwan caused the upper floors of the apartmentbuildings shown in the photo to topple over as a unit. Although the columns supporting the building weredesigned for seismic forces, the attachment of rigid concrete and brick partition walls to the columns in theupper floors negated the designer’s intent, and forced the more flexible segments of the lower floorcolumns to fail as the upper mass of the building displaced laterally as a unit.

C H A P T E R

2Design Loads

2.1 Building and Design Code

A code is a set of technical specifications and standards that control majordetails of analysis, design, and construction of buildings, equipment, andbridges. The purpose of codes is to produce safe, economical structuresso that the public will be protected from poor or inadequate design andconstruction.

Two types of codes exist. One type, called a structural code, is writtenby engineers and other specialists who are concerned with the design of aparticular class of structure (e.g., buildings, highway bridges, or nuclearpower plants) or who are interested in the proper use of a specific material(steel, reinforced concrete, aluminum, or wood). Typically, structural codesspecify design loads, allowable stresses for various types of members,design assumptions, and requirements for materials. Examples of structuralcodes frequently used by structural engineers include the following:

1. Standard Specifications for Highway Bridges by the AmericanAssociation of State Highway and Transportation Officials(AASHTO) covers the design and analysis of highway bridges.

2. Manual for Railway Engineering by the American RailwayEngineering and Maintenance of Way Association (AREMA)covers the design and analysis of railroad bridges.

3. Building Code Requirements for Reinforced Concrete (ACI 318) bythe American Concrete Institute (ACI) covers the analysis anddesign of concrete structures.

4. Manual of Steel Construction by the American Institute of SteelConstruction (AISC) covers the analysis and design of steelstructures.

5. National Design Specifications for Wood Construction by theAmerican Forest & Paper Association (AFPA) covers the analysisand design of wood structures.

The second type of code, called a building code, is established to coverconstruction in a given region (often a city or a state). A building code

contains provisions pertaining to architectural, structural, mechanical,and electrical requirements. The objective of a building code is also toprotect the public by accounting for the influence of local conditions onconstruction. Those provisions of particular concern to the structuraldesigner cover such topics as soil conditions (bearing pressures), liveloads, wind pressures, snow and ice loads, and earthquake forces. Todaymany building codes adopt the provisions of the Standard Minimum DesignLoads for Buildings and Other Structures published by the AmericanSociety of Civil Engineers (ASCE) or the more recent InternationalBuilding Code by the International Code Council.

As new systems evolve, as new materials become available, or asrepeated failures of accepted systems occur, the contents of codes arereviewed and updated. In recent years the large volume of research onstructural behavior and materials has resulted in frequent changes to bothtypes of codes. For example, the ACI Code Committee issues a yearlyaddendum and produces a revised edition of the national code every 6 years.

Most codes make provision for the designer to depart from pre-scribed standards if the designer can show by tests or analytical studiesthat such changes produce a safe design.

2.2 Loads

Structures must be proportioned so that they will not fail or deform exces-sively under load. Therefore, an engineer must take great care to anticipatethe probable loads a structure must carry. Although the design loads speci-fied by the codes are generally satisfactory for most buildings, the designermust also decide if these loads apply to the specific structure under consid-eration. For example, if the shape of a building is unusual (and inducesincreased wind speeds), wind forces may deviate significantly from thosespecified by a building code. In such cases, the designer should conductwind tunnel tests on models to establish the appropriate design forces. Thedesigner should also try to foresee if the function of a structure (and conse-quently the loads it must carry) will change in the future. For example, ifthe possibility exists that heavier equipment may be introduced into an areathat is originally designed for a smaller load, the designer may decide toincrease the design loads specified by the code. Designers typically differ-entiate between two types of load: live load and dead load.

2.3 Dead Loads

The load associated with the weight of the structure and its permanentcomponents (floors, ceilings, ducts, and so forth) is called the dead load.

28 Chapter 2 Design Loads

Section 2.3 Dead Loads 29

Since the dead load must be used in the computations to size membersbut is not known precisely until the members are sized, its magnitudemust be estimated initially. After members are sized and architecturaldetails finalized, the dead load can be computed more accurately. If thecomputed value of dead load is approximately equal to (or slightly lessthan) the initial estimate of its value, the analysis is finished. But if a largedifference exists between the estimated and computed values of deadload, the designer should revise the computations, using the improvedvalue of deadweight.

Dead Load Adjustment for Utilities and Partition Walls

In most buildings the space directly under each floor is occupied by avariety of utility lines and supports for fixtures including air ducts, waterand sewage pipes, electrical conduit, and lighting fixtures. Rather thanattempt to account for the actual weight and position of each item, design-ers add an additional 10 to 15 lb/ft2 (0.479 to 0.718 kN/m2) to the weightof the floor system to ensure that the strength of the floor, columns, andother structural members will be adequate.

Normally designers try to position beams directly under heavy masonrywalls to carry their weight directly into supports. If an owner requiresflexibility to move walls or partitions periodically in order to reconfig-ure office or laboratory space, the designer can add an appropriateallowance to the floor dead load. If partitions are light, this may be anadditional dead load of 10 lb/ft2 (0.479 kN/m2) or less. In a factory or alaboratory that houses heavy test equipment, the allowance may be 3 or4 times larger.

Distribution of Dead Load to Framed Floor Systems

Many floor systems consist of a reinforced concrete slab supported on arectangular grid of beams. The supporting beams reduce the span of theslab and permit the designer to reduce the depth and weight of the floorsystem. The distribution of load to a floor beam depends on the geomet-ric configuration of the beams forming the grid. To develop an insightinto how load from a particular region of a slab is transferred to sup-porting beams, we will examine the three cases shown in Figure 2.1. Inthe first case, the edge beams support a uniformly loaded square slab (seeFig. 2.1a). From symmetry we can infer that each of the four beams alongthe outside edges of the slab carries the same triangular load. In fact, ifa slab with the same area of uniformly distributed reinforcement in the xand y directions were loaded to failure by a uniform load, large crackswould open along the main diagonals, confirming that each beam supportsthe load on a triangular area. The area of slab that is supported by a par-ticular beam is termed the beam’s tributary area.

30

y

x

(a)

L

wL2

L

wL2

8

wL16

wL2

8

Ls2

A B

45°

A B

Ls2

L

LB

L2

L1

Ls

1�slab

beam

tributaryarea for

beam B1 B1

B1

CL

CL CL

(e)

assumed

simplified

(d )

wL13

wt =

wL13

wt =

(c)

(b)

w

wLs2

wLs2

L13

L13

B1

B1

RB2

R�B2 R�B2

RB2

B2B3 B3B2

L16

L16

L13

L2

R�B2 R�B2

L2

( f )

(g)

assumed

L1

L13

L13

L16

L16

simplified

L1

RB2 RB2

L13

L13

Figure 2.1: Concept of tributary area.(a) square slab, all edge beams supporta triangular area; (b) two edge beamsdivide load equally; (c) load on a 1 ftwidth of slab in Fig. (b); (d ) tributaryareas for beams B1 and B2 shownshaded, all diagonal lines slope at 45°;(e) top figure shows most likely loadon beam B2 in Fig. (d ); bottom figureshows simplified load distribution onbeam B2; ( f ) most likely load on beamB1 in Fig. (d ); (g) simplified load dis-tribution to beam B1.


In the second case, we consider a rectangular slab supported on oppo-site sides by two parallel beams (Fig. 2.1b). In this case, if we imagine auniformly loaded 1-ft-wide strip of slab that acts as a beam spanning a dis-tance Ls between two edge beams B1 and B2 (Fig. 2.1b), we can see thatthe load on the slab divides equally between the supporting edge beams;that is, each foot of beam carries a uniformly distributed load of wLs/2(Fig. 2.1c), and the tributary area for each beam is a rectangular area thatextends out from the beam a distance Ls/2 to the centerline of the slab.

For the third case, shown in Figure 2.1d, a slab, carrying a uniformlydistributed load w, is supported on a rectangular grid of beams. The trib-utary area for both an interior and an exterior beam is shown shaded inFigure 2.1d. Each interior beam B2 (see Fig. 2.1d ) carries a trapezoidalload. The edge beam B1, which is loaded at the third points by the reac-tions from the two interior beams, also carries smaller amounts of loadfrom three triangular areas of slab (Fig. 2.1f ). If the ratio of the long toshort side of a panel is approximately 2 or more, the actual load distri-butions on beam B2 can be simplified by assuming conservatively thatthe total load per foot, wt � wL1/3, is uniformly distributed over theentire length (see Fig. 2.1e), producing the reaction R�B2. In the case ofbeam B1, we can simplify the analysis by assuming the reaction R�B2

from the uniformly loaded B2 beams is applied as a concentrated load atthe third points (see Fig. 2.1g).

Table 2.1a lists the unit weights of a number of commonly used con-struction materials, and Table 2.1b contains the weights of building com-ponents that are frequently specified in building construction. We willmake use of these tables in examples and problems.

Examples 2.1 and 2.2 introduce computations for dead load.

A three-ply asphalt felt and gravel roof over 2-in-thick insulation boardis supported by 18-in-deep precast reinforced concrete beams with 3-ft-wide flanges (see Fig. 2.2). If the insulation weighs 3 lb/ft2 and theasphalt roofing weighs 5 lb/ft2, determine the total dead load, per footof length, each beam must support.

Solution Weight of beam is as follows:

Flange

Stem10

12 ft �

14

12 ft � 1 ft � 150 lb>ft3 � 145 lb>ft

4

12 ft �

36

12 ft � 1 ft � 150 lb>ft3 � 150 lb>ft

12

E X A M P L E 2 . 1

36� 36�

10�

three-ply feltwith gravel topping

2� rigid insulation

T-beam14�

4�

Figure 2.2: Cross section of reinforced concretebeams.

[continues on next page]

Insulation 3 lb/ft2 � 3 ft � 1 ft � 9 lb/ft

Roofing 5 lb/ft2 � 3 ft � 1 ft � 16.5 lb/ft

Total � 320.5 lb/ft,round to 0.321 kip/ft

Statement of ExampleThe steel framing plan of a small building is shown in Figure 2.3a. Thefloor consists of a 5-in-thick reinforced concrete slab supported on steelbeams (see section 1-1 in Fig. 2.3b). Beams are connected to each otherand to the corner columns by clip angles; see Figure. 2.3c. The clip anglesare assumed to provide the equivalent of a pin support for the beams; thatis, they can transmit vertical load but no moment. An acoustical boardceiling, which weighs 1.5 lb/ft2, is suspended from the concrete slab byclosely spaced supports, and it can be treated as an additional uniformload on the slab. To account for the weight of ducts, piping, conduit, andso forth, located between the slab and ceiling (and supported by hangersfrom the slab), an additional dead load allowance of 20 lb/ft2 is assumed.The designer initially estimates the weight of beams B1 at 30 lb/ft andthe 24-ft girders B2 on column lines 1 and 2 at 50 lb/ft. Establish themagnitude of the dead load distribution on beam B1 and girder B2.

SolutionWe will assume that all load between panel centerlines on either side ofbeam B1 (the tributary area) is supported by beam B1 (see the shadedarea in Fig. 2.3a). In other words, as previously discussed, to computethe dead load applied by the slab to the beam, we treat the slab as a seriesof closely spaced, 1-ft-wide, simply supported beams, spanning betweenthe steel beams on column lines A and B, and between B and C (see thecross-hatched area in Fig. 2.3a). One-half of the load, wL/2, will go toeach supporting beam (Fig. 2.3d ), and the total slab reaction applied perfoot of steel beam equals wL � 8w (see Fig. 2.3e).

Total dead load applied per foot to beam B1:

Weight of slab

Weight of ceiling 1.5 lb/ft2 � 8 ft � 12 lb/ft

Weight of ducts, etc. 20 lb/ft2 � 8 ft � 160 lb/ftEstimated weight of beam � 30 lb/ft

Total � 702 lb/ft,round to0.71 kip/ft

1 ft � 1 ft �5

12 ft � 8 ft � 150 lb>ft3 � 500 lb>ft

12


E X A M P L E 2 . 2

Example 2.1 continues . . .


Sketches of each beam with its applied loads are shown in Figure 2.3e andf. The reactions (8.875 kips) from the B1 beams are applied as concen-trated loads to the third points of girder B2 on column line 2 (Fig. 2.3f ).The uniform load of 0.05 kip/ft is the estimated weight of girder B2.

(a)

wL2

3 @ 8� = 24�

25�

1 1

2

2Cpanel

L Cpanel

L

4� 4�

(c)

(b)

a b

B2

B2

B2

B3 B1 B1 B3

c d

1�

A B D

2

1

C

A

1 2

B

B B C

5� slab

suspended ceiling

Section 2-2

Section 1-1

A

2

B

Duct

L = 8�

(d)

w w

wL2

wL2

wL2

L = 8�

Beam B1

(e)

wD = 0.71 kip/ft

Rf = 8.875 kipsRb = 8.875 kips

A D

2

Beam B2

( f )

Rd = 9.475 kipsRa = 9.475 kips

wD = 0.05 kip/ft

8.875 kips 8.875 kips

24�

8�

25�

8� 8�

Beam B1

e f g h

Figure 2.3: Determination of dead load forbeam and girder.


TABLE 2.1

Typical Design Dead Loads

(a) Material Weights

Substance Weight, lb/ft3 (kN/m3)

Steel 490 (77.0)Aluminum 165 (25.9)Reinforced concrete:

Normal weight 150 (23.6)Light weight 90–120 (14.1–18.9)

Brick 120 (18.9)Wood

Southern pine 37 (5.8)Douglas fir 34 (5.3)

(b) Building Component Weights

Component Weight, lb/ft2 (kN/m2)

CeilingsGypsum plaster on suspended metal lath 10 (0.48)Acoustical fiber tile on rock lath and channel

ceiling 5 (0.24)

FloorsReinforced concrete slab per inch of thickness

Normal weight 12 (0.60)Lightweight 6–10 (0.29–0.48)

RoofsThree-ply felt tar and gravel 5 (0.26)2-in insulation 3 (0.14)

Walls and partitionsGypsum board (1-in thick) 4 (0.19)Brick (per inch of thickness) 10 (0.48)Hollow concrete block (12 in thick)

Heavy aggregate 80 (3.83)Light aggregate 55 (2.63)

Clay tile (6-in thick) 30 (1.44)2 � 4 studs 16 in on center, -in gypsum wall

on both sides 8 (0.38)

12

12

12


Tributary Areas of Columns

To determine the dead load transmitted into a column from a floor slab, thedesigner can either (1) determine the reactions of the beams framing intothe column or (2) multiply the tributary area of the floor surrounding thecolumn by the magnitude of the dead load per unit area acting on the floor.The tributary area of a column is defined as the area surrounding thecolumn that is bounded by the panel centerlines. Use of tributary areasis the more common procedure of the two methods for computing col-umn loads. In Figure 2.4 the tributary areas are shaded for corner columnA1, interior column B2, and exterior column C1. Exterior columns locatedon the perimeter of a building also support the exterior walls as well asfloor loads.

As you can see by comparing tributary areas for the floor system inFigure 2.4, when column spacing is approximately the same length inboth directions, interior columns support approximately 4 times morefloor dead load than corner columns. When we use the tributary areas toestablish column loads, we do not consider the position of floor beams,but we do include an allowance for their weight.

Use of tributary areas is the more common procedure of the two meth-ods for computing columns loads because designers also need the tribu-tary areas to compute live loads given that design codes specify that thepercentage of live load transmitted to a column is an inverse function ofthe tributary areas; that is, as the tributary areas increase, the live loadreduction increases. For columns supporting large areas this reductioncan reach a maximum of 40 to 50 percent. We will cover the ASCE 7-98specification for live load reduction in Section 2.4.1.

C panelL C panelL C panelL

C panelL

C panelL

10�

20�

10� 11� 11� 11�

9�

9�

9�

9�

11�

A

2

2

1

B DC

22� 22�

18�

18�

Figure 2.4: Tributary area of columns A1, B2,and C1 shown shaded.

Using the tributary area method, compute the floor dead loads supportedby columns A1 and B2 in Figure 2.4. The floor system consists of a 6-in-thick reinforced concrete slab weighing 75 lb/ft2. Allow 15 lb/ft2 forthe weight of floor beams, utilities, and a ceiling suspended from thefloor. In addition, allow 10 lb/ft2 for lightweight partitions. The precastexterior wall supported by the perimeter beams weighs 600 lb/ft.

SolutionTotal floor dead load is

D � 75 � 15 � 10 � 100 lb/ft2 � 0.1 kip/ft2

Dead load to column A1 is as follows:

Tributary area At � 9 � 10 � 90 ft2

Floor dead load AtD � 90 � 0.1 kip/ft2 � 9 kips

Weight of exterior wall �weight/ft (length) � (0.6 kip/ft)(10 � 9) � 11.4 kips

Total � 20.4 kips

Dead load to column B2 is as follows:

Tributary area � 18 � 21 � 378 ft2

Total dead load � 378 ft2 � 0.1 kip/ft2 � 37.8 kips

2.4 Live Loads

Buildings Loads

Loads that can be moved on or off a structure are classified as live loads.Live loads include the weight of people, furniture, machinery, and otherequipment. Live loads can vary over time especially if the function of thebuilding changes. The live loads specified by codes for various types ofbuildings represent a conservative estimate of the maximum load likelyto be produced by the intended use of the building. In each region of thecountry, building codes typically specify the design live load. Currently,many state and city building codes base the magnitude of live loads anddesign procedures on the ASCE standard, which has evolved over timeby relating the magnitude of the design load to the successful perfor-mance of actual buildings. When sizing members, designers must alsoconsider short-term construction live loads, particularly if these loads arelarge. In the past a number of building failures have occurred during con-struction when large piles of heavy construction material were concen-


E X A M P L E 2 . 3

Section 2.4 Live Loads 37

trated in a small area of a floor or roof of a partially erected building,when the capacity of members, not fully bolted or braced, is below theirpotential load capacity.

The ASCE standard typically specifies a minimum value of uniformlydistributed live load for various types of buildings (Table 2.2). If certainstructures, such as parking garages, are also subjected to large concen-trated loads, the standard may require that forces in members be investi-gated for both uniform and concentrated loads, and that the design bebased on the loading condition that creates the greatest stresses. For exam-ple, the ASCE standard specifies that, in the case of parking garages, mem-bers be designed to carry either the forces produced by a uniformly dis-tributed live load of 50 lb/ft2 or a concentrated load of 2000 1b actingover an area of 6.25 ft2—whichever is larger.

Live Load Reduction

Recognizing that a member supporting a large tributary area is less likelyto be loaded at all points by the maximum value of live load than a mem-ber supporting a smaller floor area, building codes permit live load reduc-tions for members that have a large tributary area. For this situation, theASCE standard permits a reduction of the design live loads Lo, as listedin Table 2.2, by the following equation when the influence area KLLAT islarger than 400 ft2 (37.2 m2). However, the reduced live load must not beless than 50 percent of Lo for members supporting one floor or a section

TABLE 2.2

Typical Design Live Loads

Occupancy Use Live Load, lb/ft2 (kN/m2)

Assembly areas and theatersFixed seats (fastened to floor) 60 (2.87)Lobbies 100 (4.79)Stage floors 150 (7.18)

LibrariesReading rooms 60 (2.87)Stack rooms 150 (7.18)

Office buildingsLobbies 100 (4.79)Offices 50 (2.40)

ResidentialHabitable attics and sleeping areas 30 (1.44)Uninhabitable attics with storage 20 (0.96)All other areas 40 (1.92)

SchoolsClassrooms 40 (1.92)Corridors above the first floor 80 (3.83)

of a single floor, nor less than 40 percent of Lo for members supportingtwo or more floors:

U.S. customary units (2.1a)

SI units (2.1b)

where Lo � design load listed in Table 2.2L � reduced value of live load

AT � tributary area, ft2 (m2)KLL � live load element factor, equal to 4 for columns and 2 for

beams

For a column or beam supporting more than one floor, the term AT rep-resents the sum of the tributary areas from all floors.

For the three-story building shown in Figure 2.5a and b, calculate thedesign live load supported by (1) floor beam A, (2) girder B, and (3) theinterior column 2B in the first story. Assume a 50 lb/ft2 design live load,Lo, on all floors including the roof.

Solution(1) Floor beam A

Span � 20 ft tributary area AT � 8(20) � 160 ft2 KLL � 2

Determine if live loads can be reduced:

KLLAT � 2AT � 2(160) � 320 ft2 � 400 ft2

therefore, no live load reduction is permitted. Compute the uniform live load per foot to beam:

w � 50(8) � 400 lb/ft � 0.4 kip/ft

See Figure 2.5d for loads and reactions.

(2) Girder B

Girder B is loaded at each third point by the reactions of two floor beams.Its tributary area extends outward 10 ft from its longitudinal axis to themidpoint of the panels on each side of the girder (see shaded area in Fig.2.5a); therefore AT � 20(16) � 320 ft2.

KLLAT � 2(320) � 640 ft2

L � Lo a0.25 �4.572KLLAT

b

L � Lo a0.25 �152KLLAT

b


E X A M P L E 2 . 4

39

wL = 0.4 kip/ft

R = 4 kips R = 4 kips

Elevation(b)

Plan(a)

Tributary area to column C shown shaded(c)

L = 20�

Beam A(d)

A

C B

A

A

B

C

20�

1 2 3

20�

3 @ 8� = 24�3 @ 8� = 24�

8� 16�

12�12�

10�

10�

10�

C

panel CLpanel CL

panel CL

panel CL

AT = 480 ft2

1 2 3

24�

10�

10�

12�

24�

R = 32.3 kips

C

R = 6.736 kips

6.736 kips 6.736 kips

R = 6.736 kips

L = 24�

8� 8�

Beam B(e)

10�

Figure 2.5: Live load reduction.


Since KLLAT � 640 ft2 � 400 ft2, a live load reduction is permitted. UseEquation 2.1a.

Since 42.1 lb/ft2 � 0.5(50) � 25 lb/ft2 (the lower limit), still use w �42.1 lb/ft2.

The resulting design loads are shown in Figure 2.5e.

(3) Column 2B in the first story

The shaded area in Figure. 2.5c shows the tributary area of the interiorcolumn for each floor. Compute the tributary area for each floor:

AT � 20(24) � 480 ft2

Compute the tributary area for three floors:

3AT � 3(480) � 1440 ft2

and KLLAT � 4(1440) � 5760 ft2 � 400 ft2

therefore, reduce live load using Equation 2.1a (but not less than 0.4LO):

Since 22.4 lb/ft2 � 0.4 � 50 lb/ft2 � 20 lb/ft2 (the lower limit), use L� 22.4 lb/ft2.

Load to column � (AT)(22.4 lb/ft2) � (1440)(22.4 lb/ft2) � 32,256 lb� 32.3 kips

Impact

Normally the values of live loads specified by building codes are treatedas static loads because the majority of loads (desks, bookcases, filing cab-inets, and so forth) are stationary. If loads are applied rapidly, they createadditional impact forces. When a moving body (e.g., an elevator comingto a sudden stop) loads a structure, the structure deforms and absorbs thekinetic energy of the moving object. As an alternative to a dynamic analy-sis, moving loads are often treated as static forces and increased empiri-cally by an impact factor. The magnitude of the impact factor I for a num-ber of common structural supports is listed in Table 2.3.

L � Lo a0.25 �152KLLAT

b � 50 lb>ft2 a0.25 �1525760

b � 22.4 lb>ft2

Load at third point � 2 c 42.1

100018 2 110 2 d � 6.736 kips


L�Lo a0.25�152KLLAT

b� 50 a0.25�152640

b� 50 10.843 2�42.1 lb>ft2


Section 2.4 Live Loads 41

Determine the magnitude of the concentrated force for which the beamin Figure 2.6 supporting an elevator must be designed. The elevator, whichweighs 3000 lb, can carry a maximum of six people with an averageweight of 160 lb.

SolutionRead in Table 2.3 that an impact factor I of 100 percent applies to all ele-vator loads. Therefore, the weight of the elevator and its passengers mustbe doubled.

Total load � D � L � 3000 � 6 � 160 � 3960 lb Design load � (D � L)2 � 3960 � 2 � 7920 lb

Bridges

Standards for highway bridge design are governed by AASHTO specifi-cations, which require that the engineer consider either a single HS20truck or the uniformly distributed and concentrated loads shown in Fig-ure 2.7. Typically the HS20 truck governs the design of shorter bridgeswhose spans do not exceed approximately 145 ft. For longer spans thedistributed loading usually controls.

Since moving traffic, particularly when roadway surfaces are uneven,bounces up and down, producing impact forces, truck loads must beincreased by an impact factor I given by

U.S. customary units (2.2a)

SI units (2.2b)

but the impact factor need not be greater than 30 percent, and L � thelength in feet (meters) of the portion of the span that is loaded to producemaximum stress in the member.

I �15.2

L � 38.1

I �50

L � 125

TABLE 2.3

Live Load Impact Factor

Loading Case Impact Factor I, Percent

Supports of elevators and elevator machinery 100Supports of light machinery, shaft or motor-driven 20Supports of reciprocating machinery or power-driven units 50Hangers supporting floors and balconies 33Cab-operated traveling crane support girders and their connections 25

E X A M P L E 2 . 5

cablesupportbeam

elevator

W

Figure 2.6: Beam supporting an elevator.

The position of the span length L in the denominator of Equation 2.2indicates that the additional forces created by impact are an inverse func-tion of span length. In other words, since long spans are more massiveand have a longer natural period than short spans, dynamic loads producemuch larger forces in a short-span bridge than in a long-span bridge.

Railroad bridge design uses the Cooper E80 loading (Fig. 2.8) con-tained in the AREMA Manual for Railway Engineering. This loadingconsists of two locomotives followed by a uniform load representing theweight of freight cars. The AREMA manual also provides an equationfor impact. Since the AASHTO and Cooper loadings require the use ofinfluence lines to establish the position of wheels to maximize forces atvarious positions in a bridge member, design examples illustrating theuse of wheel loads will be deferred to Chapter 9.


(a)

(b)

uniform load640 lb perlinear foot

of lane load

W = Combined weight on the first two axles, which is the same as for the corresponding H truckV = Variable spacing – 14 ft to 30 ft inclusive. Spacing to be used is that which produces maximum stresses.

concentrated load: 18 kips for moment26 kips for shear

6�- 0�2�- 0� 2�- 0�

curb

10�- 0�clearance and

load lane width

8000 lbHS20-44 32,000 lb

0.4 W

32,000 lb

V14�- 0�

0.1 W

0.2

W

0.8

W

0.8

W

0.4 W

0.4 W0.1 W 0.4 W

Figure 2.7: AASHTO HS20-44 design liveloads.

Section 2.5 Wind Loads 43

2.5 Wind Loads

Introduction

As we have all observed from the damage produced by a hurricane or tor-nado, high winds exert large forces. These forces can tear off tree limbs,carry away roofs, and break windows. Since the speed and direction of windare continually changing, the exact pressure or suction applied by windsto structures is difficult to determine. Nevertheless, by recognizing thatwind is like a fluid, it is possible to understand many aspects of its behav-ior and to arrive at reasonable design loads.

The magnitude of wind pressures on a structure depends on the windvelocity, the shape and stiffness of the structure, the roughness and pro-file of the surrounding ground, and the influence of adjacent structures.When wind strikes an object in its path, the kinetic energy of the movingair particles is converted to a pressure qs, which is given by

(2.3)

where m represents the mass density of the air and V equals the windvelocity. Thus the pressure of the wind varies with the density of the air—a function of temperature—and with the square of the wind velocity.

The friction between the ground surface and the wind strongly influ-ences the wind velocity. For example, winds blowing over large, open,paved areas (e.g., runways of an airport) or water surfaces are not slowedas much as winds blowing over rougher, forest-covered areas where thefriction is greater. Also near the ground surface, the friction between theair and ground reduces the velocity, whereas at higher elevations above theground, friction has little influence and wind velocities are much higher.Figure 2.9a shows the approximate variation of wind velocity with height

qs �mV 2

2

8�axle spacing

rail

5�

first locomotive second locomotive

5� 5� 5�9� 6� 5� 5�9� 6�

8 kips/ft

5� 5�5� 5� 5�8� 8�

40E80 loads 80 80 80 5252 52 52 5252 52 5280 80 8040 80

Figure 2.8: AREMA E80 railroad loadings.

above the ground surface. This information is supplied by anemometers—devices that measure wind speeds.

Wind pressure also depends on the shape of the surface that the windstrikes. Pressures are smallest when the body has a streamlined cross sec-tion and greatest for blunt or concave cross sections that do not allow thewind to pass smoothly around (see Fig. 2.10). The influence of shape onwind pressure is accounted for by drag factors that are tabulated in cer-tain building codes.

As an alternative to computing wind pressures from wind velocities,some building codes specify an equivalent horizontal wind pressure.This pressure increases with height above the ground surface (Fig. 2.9b).The force exerted by the wind is assumed to be equal to the product ofthe wind pressure and the surface area of a building or other structure.

When wind passes over a sloping roof (see Fig. 2.11a), it must increaseits velocity to maintain continuity of the flowing air. As the wind veloc-


elevationaboveground

elevationaboveground

increasingwind velocity

wind pressure

0

0

(a)

(b)

Figure 2.9: (a) Variation of wind velocity withdistance above ground surface; (b) variation ofwind pressure specified by typical building codesfor windward side of building.

(a)

(b)

path ofair particle

Figure 2.10: Influence of shape on drag factor:(a) curved profile permits air to pass around bodyeasily (drag factor is small); (b) wind trapped byflanges increases pressure on web of girder (dragfactor is large).

�

(a)

(b)

leewardface

airflowlines

wind

uplift pressure

(2)

(1)

L

A B

A B

B

windwardface

Figure 2.11: (a) Uplift pressure on a sloping roof; the windspeed along path 2 is greater than that along path 1 because ofthe greater length of path. Increased velocity reduces pressureon top of roof, creating a pressure differential between insideand outside of building. The uplift is a function of the roofangle u. (b) Increased velocity creates negative pressure (suc-tion) on sides and leeward face; direct pressure on windwardface AA.


ity increases, the pressure on the roof reduces (Bernoulli’s principle). Thisreduction in pressure exerts an uplift—much like the wind flowing overthe wing of an airplane—that can carry away a poorly anchored roof. Asimilar negative pressure occurs on both sides of a building parallel tothe wind direction and to a smaller extent on the leeward side (see sidesAB and side BB in Fig. 2.11b) as the wind speeds up to pass around thebuilding.

Vortex Shedding. As wind moving at constant velocity passes overobjects in its path, the air particles are retarded by surface friction. Undercertain conditions (critical velocity of wind and shape of surface) smallmasses of restrained air periodically break off and flow away (see Fig.2.12). This process is called vortex shedding. As the air mass movesaway, its velocity causes a change in pressure on the discharge surface.If the period (time interval) of the vortices leaving the surface is close tothat of the natural period of the structure, oscillations in the structure willbe induced by the pressure variations. With time these oscillations willincrease and shake a structure vigorously. The Tacoma Narrows Bridgefailure shown in Photo 2.1 is a dramatic example of the damage that vor-tex shedding can wreak. Tall chimneys and suspended pipelines are other

winddirection

vortex

vortex

directionof

oscillation

Figure 2.12: Vortices discharging from a steelgirder. As vortex speeds off, a reduction in pres-sure occurs, causing girder to move vertically.

Photo 2.1: Failure of the Tacoma Narrows Bridgeshowing the first section of the roadway as itcrashes into Puget Sound. The breakup of the nar-row, flexible bridge was produced by large oscilla-tion induced by the wind.

structures that are susceptible to wind-induced vibrations. To preventdamage to vibration-sensitive structures by vortex shedding, spoilers (seeFig. 2.13), which cause the vortices to leave in a random pattern, ordampers, which absorb energy, may be attached to the discharge surface.As an alternative solution, the natural period of the structure may bemodified so that it is out of the range that is sensitive to vortex shedding.The natural period is usually modified by increasing the stiffness of thestructural system.

For several decades after the Tacoma Narrows Bridge failure, design-ers added stiffening trusses to the sides of suspension bridge roadways tominimize bending of the decks (Photo 2.2). Currently designers use stiffaerodynamically shaped box sections that resist wind-induced deflec-tions effectively.

Structural Bracing Systems for Wind and Earthquake Forces

The floors of buildings are typically supported on columns. Under deadand live loads that act vertically downward (also called gravity load ),columns are loaded primarily by axial compression forces. Becausecolumns carry axial load efficiently in direct stress, they have relativelysmall cross sections—a desirable condition since owners want to maxi-mize usable floor space.

When lateral loads, such as wind or the inertia forces generated by anearthquake, act on a building, lateral displacements occur. These dis-


(a)

pipe

spoiler

spoiler

hanger

cable

(b)

pipe

hanger

cable

Figure 2.13: Spoilers welded to a suspenderpipe to change the period of vortices: (a) triangu-lar plate used as a spoiler; (b) spiral rod welded topipe used as spoiler.

Photo 2.2: Verrazano Narrows Bridge at theentrance to New York City harbor. This bridge,opened for traffic in 1964, joins Staten Island toBrooklyn. The photo shows the stiffening trussesat the level of the roadway that dampen wind-induced oscillations.


placements are zero at the base of the building and increase with height.Since slender columns have relatively small cross sections, their bendingstiffness is small. As a result, in a building with columns as the only sup-porting elements, large lateral displacements can occur. These lateral dis-placements can crack partition walls, damage utility lines, and producemotion sickness in occupants (particularly in the upper floors of multi-story buildings where they have the greatest effect).

To limit lateral displacements, structural designers often insert, atappropriate locations within the building, structural walls of reinforcedmasonry or reinforced concrete. These shear walls act in-plane as deepcantilever beam-columns with large bending stiffnesses several orders ofmagnitude greater than those of all the columns combined. Because oftheir large stiffness, shear walls often are assumed to carry all transverseloads from wind or earthquake into the foundation. Since the lateralloads act normal to the longitudinal axis of the wall, just as the shear actsin a beam, they are called shear walls (Fig. 2.14a). In fact, these walls mustalso be reinforced for bending along both vertical edges since they canbend in either direction. Figure 2.14b shows the shear and moment dia-grams for a typical shear wall.

Loads are transmitted to the walls by continuous floor slabs that actas rigid plates, termed diaphragm action (Fig. 2.14a). In the case of wind,

(a)

(c)

W1

W1

W2 W2W2W2

(b)

shear wall

Shear Diagram

elevation elevation

Moment Diagram

FF5

N

S

EW

FF4

FF3

FF2

FF1

(d)

FF5

FF4

FF3

FF2

FF1

Figure 2.14: Structural systems to resist lateralloads from wind or earthquake. (a) Reinforced con-crete shear wall carries all lateral wind loads.(b) Shear and moment diagrams for shear wall pro-duced by the sum of wind loads on the windwardand leeward sides of the building in (a). (c) Planof building showing position of shear walls andcolumns. (d) Cross-bracing between steel columnsforms a truss to carry lateral wind loads into thefoundations.

Photo 2.3: The cross-bracing, together with theattached columns and horizontal floor beams inthe plane of the bracing, forms a deep continu-ous, vertical truss that extends the full height ofthe building (from foundation to roof) and pro-duces a stiff, lightweight structural element fortransmitting lateral wind and earthquake forcesinto the foundation.

the floor slabs receive the load from air pressure acting on the exteriorwalls. In the case of earthquake, the combined mass of the floors andattached construction determines the magnitude of the inertia forces trans-mitted to the shear walls as the building flexes from the ground motion.

Shear walls may be located in the interior of buildings or in the exte-rior walls (Fig. 2.14c). Since only the in-plane flexural stiffness of the wallis significant, walls are required in both directions. In Figure 2.14c twoshear walls, labeled W1, are used to resist wind loads acting in the east-west direction on the shorter side of the building; four shear walls, labeledW2, are used to resist wind load, in the north-south direction, acting onthe longer side of the building.

In buildings constructed of structural steel, as an alternative to con-structing shear walls, the designer can add X-shaped or V-shaped cross-bracing between columns to form deep wind trusses, which are very stiffin the plane of the truss (Fig. 2.14d and Photo 2.3).

Equations to Predict Design Wind Pressures

Our primary objective in establishing the wind pressures on a building isto determine the forces that must be used to size the structural membersthat make up the wind bracing system. In this section we will discuss pro-cedures for establishing wind pressures using a simplified format basedon the provisions of the most recent edition of the ASCE Standard forMinimum Design Loads on Buildings. This edition of the ASCE standardcontains a major revision of the wind load provisions based on animproved understanding of a building’s response to wind forces.

If the mass density of air at 59°F (15°C) is substituted into Equation2.3a, the equation for the static wind pressure qs becomes

qs � 0.00256V 2 U.S. customary units (2.4a)

qs � 0.613V 2 SI units (2.4b)

where qs � static wind pressure, lb/ft2 (N/m2)V � basic wind speed, mph (m/s). Basic wind speeds, used to

establish the design wind force for particular locations inthe continental United States, are plotted on the map inFigure 2.15. These wind velocities are measured byanemometers located 33 ft (10 m) above grade in openterrain and represent wind speeds that have only a 2 percentprobability of being exceeded in any given year. Notice thatthe greatest wind velocities occur along the coast where thefriction between wind and water is minimal.

The static wind pressure qs given by Equation 2.4a or b is next mod-ified in Equation 2.5 by four empirical factors to establish the magnitudeof the velocity wind pressure qz at various elevations above ground level.



qz � 0.00256V 2IKzKztKd U.S. customary units (2.5a)

qz � 0.613V 2IKzKztKd SI units (2.5b)

Or using Equation 2.4a, we can replace the first two terms of Equation2.5 by qs to give

qz � qsIKzKztKd (2.6)

where qz � velocity wind pressure at height z above ground levelI � importance factor, which represents how essential a given

structure is to the community. For example, I � 1 foroffice buildings, but increases to 1.15 for hospitals, policestations, or other public facilities vital to the safety andwell-being of the community or whose failure might causelarge loss of life. For structures whose failure produces noserious economic loss or danger to the public, I reduces to0.87 or 0.77 if V exceeds 100 mph.

Kz � velocity exposure coefficient, which accounts for both theinfluence of height above grade and exposure conditions.Four exposure categories (A through D) considered are asfollows:A: Large city centers where 50 percent of buildings are

higher than 70 ft (21.3 m).

Figure 2.15: ASCE basic wind speed contourmap. Largest wind velocities occur along the east-ern and southeastern coasts of the United States.

B: Urban and suburban, or wooded areas with lowstructures.

C: Open terrain with scattered obstructions generally lessthan 30 ft (9.1 m) high.

D: Flat, unobstructed areas exposed to wind flowing overopen water for a distance of at least 1 mi (1.61 km).

Values of Kz are tabulated in Table 2.4 and showngraphically in Figure 2.16.


TABLE 2.4

Velocity Pressure Exposure Coefficient Kz

Height z above Ground Level Exposure

ft (m) A B C D

0–15 (0–4.6) 0.32 0.57 0.85 1.0320 (6.1) 0.36 0.62 0.90 1.0825 (7.6) 0.39 0.66 0.94 1.1230 (9.1) 0.42 0.70 0.98 1.1640 (12.2) 0.47 0.76 1.04 1.2250 (15.2) 0.52 0.85 1.09 1.2760 (18) 0.55 0.82 1.13 1.3170 (21.3) 0.59 0.89 1.17 1.3480 (24.4) 0.62 0.93 1.21 1.3890 (27.4) 0.65 0.96 1.24 1.40

100 (30.5) 0.68 0.99 1.26 1.43120 (36.6) 0.73 1.04 1.31 1.48140 (42.7) 0.78 1.09 1.36 1.52160 (48.8) 0.82 1.13 1.39 1.55180 (54.9) 0.86 1.17 1.43 1.58

0.0 0.5

Hei

ght (

ft)

Hei

ght (

m)

0.990.68 1.261.43

1.0Kz

A B C D

1.5 2.0

20

0

100

200

300

400

500

0

40

60

80

100

120

140

Figure 2.16: Variations of Kz.


Kzt � topographic factor, which equals 1 if building is located onlevel ground; for buildings located on elevated sites (top ofhills), Kzt increases to account for greater wind speed

Kd � wind direction factor, which accounts for the reducedprobability of maximum winds coming from any givendirection and for the reduced probability of the maximumpressure developing for any given wind direction (seeTable 2.5)

The final step for establishing the design wind pressure p is to mod-ify qz, given by Equation 2.5a or b, by two additional factors, G and Cp:

p � qzGCp (2.7)

where p � design wind pressure on a particular face of the buildingG � gust factor, which equals 0.85 for rigid structures; that is,

the natural period is less than 1 second. For flexiblestructures with a natural period greater than 1 second, aseries of equations for G are available in the ASCE standard.

Cp � external pressure coefficient, which establishes how afraction of the wind pressure (given by Equation 2.5a or b)is to be distributed to each of the four sides of the building(see Table 2.6). For the wind applied normal to the wall onthe windward side of the building Cp � 0.8. On theleeward side, Cp � 0.2 to 0.5. The minus signindicates a pressure acting outward from the face of thebuilding. The magnitude of Cp is a function of the ratio oflength L in the windward direction to length B in thedirection normal to the wind. The main wind bracingsystem must be sized for the sum of the wind forces on thewindward and leeward sides of the building. Finally, on thesides of the building perpendicular to the direction of thewind, where negative pressure also occurs, Cp � 0.7.

TABLE 2.5

Wind Directionality Factor Kd

Structural Type Kd

BuildingsMain wind force-resisting system 0.85Components and cladding 0.85

Chimneys, tanks, and similar structuresSquare 0.90Round 0.95

Trussed towersTriangular, square, rectangular 0.85All other cross sections 0.95

The wind pressure increases with height only on the windward side of abuilding where wind pressure acts inward on the walls. On the other threesides the magnitude of the negative wind pressure, acting outward, isconstant with height, and the value of Kz is based on the mean roof heighth. A typical distribution of wind pressure on a multistory building is shownin Figure 2.17. Example 2.6 illustrates the procedure to evaluate the windpressure on the four sides of a building 100 ft high.

Since wind can act in any direction, designers must also consideradditional possibilities of wind loading a building at various angles. Forhigh-rise buildings in a city—particularly those with an unusual shape—wind tunnel studies using small-scale models are often employed to deter-mine maximum wind pressures. For these studies, adjacent high-rise build-ings, which influence the direction of airflow, must be included. Modelsare typically constructed on a small platform that can be inserted into awind tunnel and rotated to determine the orientation of the wind that pro-duces the largest values of positive and negative pressure.


TABLE 2.6

External Pressure Coefficient Cp

Wall Pressure Coefficients Cp

Surface L/B Cp Use with

Windward wall All values 0.8 qz

Leeward wall 0–1 0.52 0.3 qh

4 0.2Side wall All values 0.7 qh

Notes:1. Plus and minus signs signify pressures acting toward and away from the surfaces,

respectively.2. Notations: B is the horizontal dimension of the building, in feet (meters) measured

normal to wind direction, and L is the horizontal dimension of the building in feet(meters), measured parallel to wind direction.

BL

wind

Figure 2.17: Typical wind load distribution on amultistory building.

L

qz GCp

qh GCp

qh GCp

qh GCp

wind

PLAN

B


Determine the wind pressure distribution on the four sides of an eight-story hotel located on level ground along the South Carolina coast—aregion known for frequent hurricanes. Consider the case of a strong windacting directly on face AB of the building in Figure 2.18a. The wind mapshown in Figure 2.15 indicates the building should be designed for abasic wind speed of V � 130 mph blowing off the ocean. Assume thebuilding is classified as stiff because its natural period is less than 1 s;therefore, the gust factor G equals 0.85. The importance factor I equals1.15. Since the building is located on level ground, Kzt � 1.

Solution

STEP 1 Compute the static wind pressure using Equation 2.4a:

qs � 0.00256VII � 0.00256(130)2 � 43.26 lb/ft2

STEP 2 Compute the magnitude of wind pressure on the windwardside at the top of the building, 100 ft above grade, using Equa-tion 2.5a. Since the wind blows off the ocean, exposure Dapplies.

Figure 2.18: Variation of wind pressure on sidesof buildings.

E X A M P L E 2 . 6

(a) (b)

L = 60�

A

B

C

D

A

B

C

D

E

F

G

100�

41.1 lb/ft2

35.94 lb/ft2 25.67 lb/ft2

35.94 lb/ft2

29.6 lb/ft2

leewardface

windwardface

B = 60�

wind =130 mph


I � 1.15

Kz � 1.43 (Fig. 2.16 or Table 2.4)

Kzt � 1 (level ground)

Kd � 0.85 (Table 2.5)

Substituting the above values into Equation 2.6 to determinethe design wind pressure at 100 ft above grade gives

qz � qsIKzKztKd

� 43.26(1.15)(1.43)(1)(0.85) � 60.4 lb/ft2

Note: To compute wind pressures at other elevations on thewindward side, the only factor that changes in the aboveequation is Kz, tabulated in Table 2.4. For example, at anelevation of 50 ft, Kz � 1.27 and qz � 53.64 lb/ft2.

STEP 3 Determine the design wind pressure on the windward faceAB, using Equation 2.7.

Gust factor G � 0.85, read Cp � 0.8 (from Table 2.6). Sub-stituting into Equation 2.7 produces

p � qzGCp � 60.4(0.85)(0.8) � 41.1 lb/ft2

STEP 4 Determine the wind pressure on the leeward side:

Cp � 0.5 (Table 2.6) and G � 0.85

p � qzGCp � 60.4(0.85)(0.5) � 25.67 lb/ft2

STEP 5 Compute the wind pressure on the two sides perpendicularto the wind:

Cp � 0.7 G � 0.85

p � qzGCp � 60.4(0.85)(0.7) � 35.94 lb/ft2

The distribution of wind pressures is shown in Figure 2.18b.



Section 2.5 Wind Load 55

Simplified Procedure: Wind Loads for Low-Rise Buildings

In addition to the procedure just discussed for computing wind loads, theASCE standard provides a simplified procedure to establish wind pres-sures on enclosed or partially enclosed low-rise buildings of regularshape whose mean roof height h does not exceed 60 ft (18.2 m) and towhich the following four conditions apply.

1. Floor and roof slabs (diaphragms) must be designed to act as rigidplates and connect to the main wind force-resisting system, whichmay include shear walls, moment frames, or cross-bracing.

2. Kzt � 1 (the building is not subject to topographical effects; i.e., theground is level, and no structures nearby create unusual windpatterns).

3. The building has an approximately symmetric cross section, and theroof slope u does not exceed 45°.

4. The building is classified as rigid; that is, its natural frequency isgreater than 1 Hz. (Most low-rise buildings with wind force-resisting systems, such as shear walls, moment frames, or cross-bracing, fall in this category.)

For such regular rectangular structures, the procedure to establish thedesign pressures follows:

1. Determine the wind velocity at the building site, using Figure 2.15.2. Establish the design wind pressure ps acting on the walls and roof

ps � lIps30 (2.8)

where ps30 is the simplified design wind pressure for exposure B, with h �30 ft and the importance factor I taken as 1.0 (see Table 2.7). If theimportance factor I differs from 1, substitute its value into Equation 2.8.

TABLE 2.7

Simplified Design Wind Pressure pS30 (lb/ft2) (Exposure B at h = 30 ft with I = 1.0)

Basic Wind Roof ZonesSpeed Angle Horizontal Pressures Vertical Pressures Overhangs(mph) (degrees) A B C D E F G H EOH GOH

0 to 5° 1 12.8 6.7 8.5 4.0 15.4 8.8 10.7 6.8 21.6 16.910° 1 14.5 6.0 9.6 3.5 15.4 9.4 10.7 7.2 21.6 16.915° 1 16.1 5.4 10.7 3.0 15.4 10.1 10.7 7.7 21.6 16.9

90 20° 1 17.8 4.7 11.9 2.6 15.4 10.7 10.7 8.1 21.6 16.925° 1 16.1 2.6 11.7 2.7 7.2 9.8 5.2 7.8 13.3 11.4

2 — — — — 2.7 5.3 0.7 3.4 — —30° to 45° 1 14.4 9.9 11.5 7.9 1.1 8.8 0.4 7.5 5.1 5.8

2 14.4 9.9 11.5 7.9 5.6 4.3 4.8 3.1 5.1 5.8

Load

Cas

e

For exposure C or D and for h other than 30 ft, the ASCE standard sup-plies an adjustment factor l, tabulated in Table 2.8.

The distribution of ps on the walls and roof for wind load in both thetransverse and longitudinal directions is shown in Figure 2.19. Each linein Table 2.7 lists the values of the uniform wind pressure for eight areasof a building’s walls and roof.

• Plus and minus signs signify pressures acting toward and awayfrom projected surfaces.

• Additional wind speeds are given in the ASCE standard.

These areas, shown in Figure 2.19, are labeled with circled letters (A toH). Table 2.7 contains values of ps30 for buildings subjected to 90-mphwinds; the complete standard provides data for winds varying from 85 to170 mph.

The value of a, which defines the extent of regions of greatest windpressure (see areas A, B, E, and F on the walls and roof in Fig. 2.19), isevaluated as 10 percent of the smaller horizontal dimension of the build-


Figure 2.19: Distribution of design wind pres-sures for the simplified method. See Table 2.7for the magnitude of the pressures in areas Athrough H. h � 60 ft. (From ASCE standard.)

H

G

FE

BMWFRS

direction beingevaluated

MWFRSdirection being

evaluated

Transverse

Referencecorner

2a

2a

Eaveheight

Eaveheight

Longitudinal

A

D

C

C

A

�

H

H

G

GF

E

�

Referencecorner

TABLE 2.8

Adjustment Factor Lfor Building Height and Exposure

Mean roof Exposureheight h (ft) B C D

15 1.00 1.21 1.4720 1.00 1.29 1.5525 1.00 1.35 1.6130 1.00 1.40 1.6635 1.05 1.45 1.7040 1.09 1.49 1.7445 1.12 1.53 1.7850 1.16 1.56 1.8155 1.19 1.59 1.8460 1.22 1.62 1.87

From ASCE standard.

Section 2.6 Earthquake Forces 57

ing or 0.4h, whichever is smaller (h is the mean height), but not less thaneither 4 percent of the least horizontal dimension or 3 ft (0.9 m). Noticethat the wind pressures are largest near the corners of walls and the edgesof roofs.

Example 2.7 illustrates the use of the simplified procedure to estab-lish the design wind pressures for the wind analysis of a 45-ft-high rec-tangular building.

Figure 2.15 indicates the velocity of the wind acting on the 45-ft-high,three-story building in Figure 2.20a is 90 mph. If exposure condition Capplies, determine the wind force transmitted to the building’s foundationsby each of the two large reinforced concrete shear walls that make up themain wind-resisting system. The walls located at the midpoint of each sideof the building have identical proportions. The importance factor I equals 1.

E X A M P L E 2 . 7

(a)

roof

3rdfloor

2ndfloor

15�

15�

15�

40�

34�

2a � 6�

AB

CF

shear wall

30�

V1

shear wall

NW

SE

Ps � 13 lb/ft2

Ps � 19.6 lb/ft2

E

DV2

18.8�

34�

40�

Note: Units of distributed load W are kips/ft.

2a � 6�

R � 4.2 kipsw � 0.147 kip/ft

w �0.0975 kip/ft

E

DA

RoofF

(b)

18.8�R � 8.4 kips

w � 0.294 kip/ftw �

0.195 kip/ft

3rd floor

18.8�R � 8.4 kips

w � 0.294 kip/ftw �

0.195 kip/ft

2nd floor

18.8�

D C

A B

(c)

R � 21 kips

V1 � 11.13 kips

V2 � 9.87 kips

30�

40�

V1 � 11.13 kips

(d )

4.45 kips

4.45 kips

2.23 kips15�

15�

15�

M1 = 300.6 kip•ft

Figure 2.20: Horizontal wind pressure analysisby the simplified method. (a) Wind pressure dis-tribution and details of the loaded structure;(b) wind forces applied by the exterior walls tothe edge of the roof and floor slabs; (c) plan viewof the resultant wind force and the reactions ofthe shear walls; (d) free body of the shear walllocated in plane ABDF showing the wind forcesapplied by the floor slabs and the reactions on base.


Solution

Compute the wind load transmitted from the wall on the windward side tothe roof and each floor slab. Assume each 1-ft-wide vertical strip of wallacts as a simply supported beam spanning 10 ft between floor slabs; there-fore, one-half of the wind load on the wall between floors is carried to theslabs above and below by the fictitious beam (see Fig. 2.20b).

STEP 1 Since the roof is flat, u � 0. For the simplified design windpressures ps30, read the top line in Table 2.7.

Region A: ps30 � 12.8 lb/ft2

Region C: ps30 � 8.5 lb/ft2

Note: There is need to compute the values of ps for zonesB and D because the building does not have a sloped roof.

STEP 2 Adjust ps30 for exposure C and a mean height of h = 45 ft.Read in Table 2.8 that adjustment factor l = 1.53. Computethe wind pressure ps � lIps30.

Region A: ps � 1.53(1)(12.8) � 19.584 round to 19.6 lb/ft2

Region C: ps � 1.53(1)(8.5) � 13.005 round to 13 lb/ft2

STEP 3 Compute the resultant wind forces transmitted from the exte-rior walls to the edge of the roof and floor slabs.

Load per foot, w, to roof slab (see Fig. 2.20b)

Load per foot, w, to second- and third-floor slabs

STEP 4 Compute the resultants of the distributed wind loads.

Roof slab:

R1 = 0.147 � 6 � 0.0975 � 34 � 4.197 round to 4.2 kips

Second and third floors:

R2 = 0.294 � 6 � 0.195 � 34 � 8.394 round to 8.4 kips

Total horizontal wind force � 4.2 � 8.4 � 8.4 � 21 kips

Region C: w � 15 �13

1000� 0.195 kip>ft

Region A: w � 15 �19.6

1000� 0.294 kip>ft

Region C: w � 15

2�

13

1000� 0.0975 kip>ft

Region A: w � 15

2�

19.6

1000� 0.147 kip>ft




STEP 5 Locate the position of the resultant. Sum moments about avertical axis through points A and F (see Fig. 2.20c).

At the level of the first floor slab

Since the variation of the pressure distribution is identical atall floor levels on the back of the wall, the resultant of allforces acting on the ends of the roof and floor slabs is locateda distance of 18.8 ft from the edge of the building (Fig. 2.20b).

STEP 6 Compute the shear force at the base of the shear walls. Sumthe moments of all forces about a vertical axis passing throughpoint A at the corner of the building (see Fig 2.20c).

�MA � 21 � 18.8 V2(40) and V2 � 9.87 kips Ans.

Compute V2: V2 � V1 � 21 kipsV1 � 21 9.87 � 11.13 kips Ans.

Note: A complete analysis for wind requires that the designerconsider the vertical pressures in zones E to H acting on theroof. These pressures are carried by a separate structural sys-tem, composed of the roof slabs and beams, to the columnsas well as to the shear walls. In the case of a flat roof, the windflowing over the roof produces upward pressures (uplift) thatreduce the axial compression in the columns.

2.6 Earthquake Forces

Earthquakes occur in many regions of the world. In certain locations wherethe intensity of the ground shaking is small, the designer does not haveto consider seismic effects. In other locations—particularly in regions nearan active geological fault (a fracture line in the rock structure), such asthe San Andreas fault that runs along the western coast of California—largeground motions frequently occur that can damage or destroy buildingsand bridges in large areas of cities (see Photo 2.4a and b). For example,San Francisco was devastated by an earthquake in 1906, before buildingand bridge codes contained seismic provisions.

The ground motions created by major earthquake forces cause build-ings to sway back and forth. Assuming the building is fixed at its base, thedisplacement of floors will vary from zero at the base to a maximum at

x � 18.797 ft round to 18.8 ft

4.197x � 0.882 13 2 � 3.315 16 � 34>2 2 Rx � ©F # x

the roof (see Fig. 2.21a). As the floors move laterally, the lateral bracingsystem is stressed as it acts to resist the lateral displacement of the floors.The forces associated with this motion, inertia forces, are a function ofboth the weight of the floors and attached equipment and partitions aswell as the stiffness of the structure. The sum of the lateral inertia forcesacting on all floors and transmitted to the foundations is termed the baseshear and is denoted by V (see Fig. 2.21b). In most buildings in which theweight of floors is similar in magnitude, the distribution of the inertiaforces is similar to that created by wind, as discussed in Section 2.6.

Although there are several analytical procedures to determine themagnitude of the base shear for which buildings must be designed, wewill only consider the equivalent lateral force procedure, described inthe ANSI/ASCE standard. Using this procedure, we compute the magni-tude of the base shear as

(2.8a)

but not to exceed

(2.8b)

and not less thanVmin � 0.044SDSIW (2.8c)

where W � total dead load of building and its permanent equipmentand partitions

Vmax �SDSW

R>I

V �SD1W

T 1R>I 2


�

(b)(a)

base

V = �Fi

F5

F4

F3

F2

F1

Figure 2.21: (a) Displacement of floors as build-ing sways; (b) inertia forces produced by motionof floors.

Photo 2.4: Earthquake damage to structures. (a) Hanshin Expressway collapsed during 1995 Hyogoken-Nanbu, Japan, earthquake. (b) Col-lapse of the Struve Sough Bridge. The severe shaking of the soil by the 1989 Loma Prieta Earthquake produced differential settlements of thefoundations supporting rows of columns that carried the roadway slab. This uneven settlement caused the columns that underwent the largestsettlements to transfer the weight of the bridge deck to adjacent columns whose settlement was smaller. The additional load, which had to betransferred into the column by shear stresses in the slab around the column’s perimeter, produced the punching shear failures shown.

(a) (b)


T � fundamental natural period of building, which can becomputed by the following empirical equation

T � Cthnx (2.9)

hn � the building height in feet (meters, above the base), Ct �0.028 (or 0.068 in SI units) for steel rigid frames (momentframes), Ct � 0.016 (0.044 SI) and x � 0.9 for reinforcedconcrete rigid frames, and Ct � 0.02 (0.055 SI) and x �0.75 for most other systems (for example, systems withbraced frames or structural walls). The natural period of abuilding (the time required for a building to go through onecomplete cycle of motion) is a function of the lateralstiffness and the mass of the structure. Since the base shearV is inversely proportional to the magnitude of the naturalperiod, it reduces as the lateral stiffness of the structuralbracing system increases. Of course, if the stiffness of thelateral bracing system is too small, lateral displacementsmay become excessive, producing damage to windows,exterior walls, and other nonstructural elements.

SD1 � a factor computed using seismic maps that shows intensityof design earthquake for structures with T � 1 s. Table 2.9gives the values for several locations.

SDS � a factor computed using seismic maps that shows intensity ofdesign earthquake at particular locations for structures withT � 0.2 s. See Table 2.9 for values at several locations.

R � response modification factor, which represents the ability of astructural system to resist seismic forces. This factor,which varies from 8 to 1.25, is tabulated in Table 2.10 forseveral common structural systems. The highest values areassigned to ductile systems; the lowest values, to brittlesystems. Since R occurs in the denominator of Equations2.8a and b, a structural system with a large value of R willpermit a large reduction in the seismic force the structuralsystem must be designed to support.

I � occupancy importance factor, which represents howessential a given structure is to the community. Forexample, I is 1 for office buildings, but increases to 1.5 for hospitals, police stations, or other public facilitiesvital to the safety and well-being of the community orwhose failure might cause large loss of life.

Note: The upper limit given by Equation 2.8b is required because Equa-tion 2.8a produces values of base shear that are too conservative for verystiff structures that have short natural periods. The ASCE standard alsosets a lower limit (Eq. 2.8c) to ensure that the building is designed for aminimum seismic force.

TABLE 2.9

Representative Values of SDS and SD1 atSelected Cities

City SDS, g SD1, g

Los Angeles, California 1.3 0.5Salt Lake City, Utah 1.2 0.5Memphis, Tennessee 0.83 0.27New York, New York 0.27 0.06

Note: Values of SDS and SD1 are based on theassumption that foundations are supported onrock of moderate strength. These values increasefor weaker soils with lower bearing capacity.

Distribution of Seismic Base Shear V to Each Floor Level

The distribution of the seismic base shear V to each floor is computed usingEquation 2.10.

(2.10)

where Fx � the lateral seismic force at level xwi and wx � deadweight of floor at levels i and xhi and hx � height from base to floors at levels i and x

k � 1 for T 0.5 s, 2 for T 2.5 s. For structureswith a period between 0.5 and 2.5 s, k isdetermined by linear interpolation between Tequal to 1 and 2 as

(2.11)

See Figure 2.22 for graphical representation of Equation 2.11.

Determine the design seismic forces acting at each floor of the six-storyoffice building in Figure 2.23. The structure of the building consists ofsteel moment frames (all joints are rigid) that have an R value of 8. The75-ft-tall building is located in a high seismic region with SD1 � 0.4g andSD1 � 1.0g for a building supported on rock, where g is the gravitationalacceleration. The deadweight of each floor is 700 kips.

SolutionCompute the fundamental period, using Equation 2.9:

T � Cthnx � 0.028(75)0.8 � 0.89 s

Assuming that the floor deadweight contains an allowance for the weightof columns, beams, partitions, ceiling, etc., the total weight W of the build-ing is

W � 700(6) � 4200 kips

k � 1 �T 0.5

2

Fx �wxhx

k

an

i�1wih i

k

V


TABLE 2.10

Values of R for Several Common Lateral Bracing Structural Systems

Description of Structural System R

Ductile steel or concrete frame with rigid joints 8Ordinary reinforced concrete shear walls 4Ordinary reinforced masonry shear wall 2

E X A M P L E 2 . 8

k = 1 +T – 0.5

2

0

0.5

1.01.21.5

2.0

0.890.5 1.5 2.52.0

T (seconds)

k (e

xpon

ent)

Figure 2.22: Interpolate for k value.

Section 2.8 Load Combinations 63

The occupancy importance factor I is 1 for office buildings. Compute thebase shear V using Equations 2.8a and c:

(2.8a)

but not more than

(2.8b)

and not less than

Vmin � 0.044SDSIW � 0.044 � 1.0 � 1 � 4200 � 184.8 kips (2.8c)

Therefore, use V � 236 kips.Computations of the lateral seismic force at each floor level are sum-

marized in Table 2.11. To illustrate these computations, we compute theload at the third floor. Since T � 0.89 s lies between 0.5 and 2.5 s, we mustinterpolate using Equation 2.11 to compute the k value (see Fig. 2.22):

�36,537

415,2621236 2 � 20.8 kips

F3rd floor �w3h3

k

an

i�1wihi

k

V

k � 1.0 �0.89 0.5

2� 1.2

k � 1 �T 0.5

2

Vmax �SDS

R>IW �1.0

8>1 14200 2 � 525 kips

V �SD1

T 1R>I 2W �0.4

0.89 18>1 2 14200 2 � 236 kips

�700 � 271.2

700 � 151.2 � 700 � 271.2 � 700 � 391.2 � 700 � 511.2 � 700 � 631.2 � 700 � 751.2 1236 2

15� 2nd floor

20 40 60force (kips)

80

3rd floor

4th floor

heig

ht (

ft)

5th floor

6th floor

roof

10.1

20.8

32.3

44.6

57.4

70.8

5 @

12�

= 6

0�

(a) (b)Figure 2.23:(a) Six-story building; (b) lateral load profile.


2.7 Other Loads

Snow load on roofs needs to be considered in cold regions. The designsnow load on a sloped roof is given by the ASCE standard as follows:

ps � 0.7CsCeCtIpg (2.12)

where pg � design ground snow load (for example, 40 lb/ft2 in Boston,25 lb/ft2 in Chicago)

Cs � roof slope factor (reduces from 1.0 as roof slope increases)Ce � exposure factor (0.7 in windy area and 1.3 in sheltered

areas with little wind)Ct � thermal factor (1.2 in unheated buildings and 1.0 in

heated buildings)I � importance factor

Flats roofs need to be properly drained to avoid the ponding of rainwater. The ASCE standard requires that each portion of the roof bedesigned to support the weight of all rainwater that could accumulate onit if the primary drainage system for that portion were blocked. If notproperly considered in design, rain loads may produce excessive deflec-tions of roof beams, producing an instability problem (called ponding),causing the roof to collapse.

When appropriate, other types of loading also need to be included inthe design of structures. These include soil pressures, hydrostatic pres-sures, thermally induced forces, among others.


Example 2.8 continues . . . TABLE 2.11

Computation of Seismic Lateral Forces Floor

Weight Height Floor wi (kips) hi ft wihi

k Fx (kips)

Roof 700 75 124,501 0.300 70.86th 700 63 100,997 0.243 57.45th 700 51 78,376 0.189 44.64th 700 39 56,804 0.137 32.33rd 700 27 36,537 0.088 20.82nd 700 15 18,047 0.043 10.1

V � a6

i�1Fi � 263a

6

i�1wihi

k � 415,262W � a6

i�1wi � 4200

wxhkx

a6

i�1wih

ki

Section 2.8 Load Combinations 65

2.8 Load Combinations

The forces (e.g., axial force, moment, shear) produced by various combi-nations of loads discussed need to be combined in a proper manner andincreased by a factor of safety (load factor) to produce the desired levelof safety. The combined load effect, sometimes called the required fac-tored strength, represents the minimum strength for which membersneed to be designed. Considering the load effect produced by the deadload D, live load L, roof live load Lr, wind load W, and earthquake load E,and snow load S, the ASCE standard requires that the following load com-binations be considered:

1.4D (2.13)

1.2D � 1.6L � 0.5Lr (2.14)

1.2D � 1.6Lr � 0.5L (2.15)

1.2D � 1.6W � 0.5L (2.16)

1.2D � 1.0E � 0.5L � 0.2S (2.17)

The load combination that produces the largest value of force representsthe load for which the member must be designed.

A column in a building is subject to gravity load only. Using the tributaryarea concept, the axial loads produced by the dead load, live load, and rooflive load are

PD � 90 kips

PL � 120 kips

PLr � 20 kips

What is the required axial strength of the column?

Solution1.4PD � 1.4(90) � 126 kips (2.13)

Therefore, the required axial load is 310 kips. In this case, the load com-bination in Equation 2.14 governs. However, if the dead load is signifi-cantly larger than the live loads, Equation 2.13 may govern.

E X A M P L E 2 . 9

1.2PD � 1.6PL � 0.5PLr �1.2(90) � 1.6(120) � 0.5(20) � 310 kips (2.14)

1.2PD � 1.6PLr � 0.5PL � 1.2(90) � 1.6(20) � 0.5(120) � 200 kips (2.15)

To determine the required flexural strength at one end of a beam in a con-crete frame, the moments produced by dead, live, and wind load are:

MD � 100 kip�ft

ML � 50 kip�ft

Mw � �200 kip�ft

where the minus sign indicates that the beam end is subject to counter-clockwise moment while the plus sign indicates clockwise moment. Boththe plus and minus signs are assigned to Mw because the wind load can acton the building in either direction. Compute the required flexural strengthfor both positive and negative bending.

SolutionNegative bending:

1.4MD � 1.4(100) � 140 kip�ft (2.13)

1.2MD � 1.6ML � 1.2(100) � 1.6(50) � 200 kip�ft (2.14)

1.2MD � 1.6Mw � 0.5ML � 1.2(100) � 1.6(200) � 0.5(50) � 465 kip�ft (governs) (2.16)

Positive bending: Load combinations from Equations 2.12 and 2.13 neednot be considered because both produce negative moments.

1.2MD � 1.6Mw � 0.5ML � 1.2(100) � 1.6(�200) � 0.5(50) � �175 kip�ft (2.16)

Therefore, the beam needs to be designed for a positive moment of 175kip�ft and a negative moment of 465 kip�ft.

Summary

• Loads that engineers must consider in the design of buildings andbridges include dead loads, live loads, and environmental forces—wind, earthquake, snow, and rain. Other types of structures such asdams, water tanks, and foundations must resist fluid and soilpressures, and for these cases specialists are often consulted toevaluate these forces.

• The loads that govern the design of structures are specified bynational and local building codes. Structural codes also specifyadditional loading provisions that apply specifically to constructionmaterials such as steel, reinforced concrete, aluminum, and wood.


E X A M P L E 2 . 1 0

• Since it is unlikely that maximum values of live load, snow, wind,earthquake, and so forth will act simultaneously, codes permit areduction in the values of loads when various load combinations areconsidered. Dead load, however, is never reduced unless it providesa beneficial effect.

• To account for dynamic effects from moving vehicles, elevators,supports for reciprocating machinery, and so forth, impact factorsthat increase the live load are specified in building codes.

• In zones where wind or earthquake forces are small, low-rise buildingsare initially proportioned for live and dead load, and then checkedfor wind or earthquake, or both, depending on the region; the designcan be easily modified as needed.

On the other hand, for high-rise buildings located in regionswhere large earthquakes or high winds are common, designers mustgive high priority in the preliminary design phase to select structuralsystems (for example, shear walls or braced frames) that resistlateral loads efficiently.

• Wind velocities increase with height above the ground. Values ofpositive wind pressures are given by the velocity pressure exposurecoefficient Kz tabulated in Table 2.4.

• Negative pressures of uniform intensity develop on three sides ofrectangular buildings that are evaluated by multiplying the magnitudeof the positive windward pressure at the top of the building by thecoefficients in Table 2.6.

• The wind bracing system in each direction must be designed to carrythe sum of the wind forces on the windward and leeward sides ofthe building.

• For tall buildings or for buildings with an unusual profile, wind tunnelstudies using instrumented small-scale models often establish themagnitude and distribution of wind pressures. The model must alsoinclude adjacent buildings, which influence the magnitude and thedirection of the air pressure on the building being studied.

• The ground motions produced by earthquakes cause buildings,bridges, and other structures to sway. In buildings this motion createslateral inertia forces that are assumed to be concentrated at eachfloor. The inertia forces are greatest at the top of buildings wherethe displacements are greatest.

• The magnitude of the inertia forces depends on the size of theearthquake, the weight of the building, the natural period of thebuilding, the stiffness and ductility of the structural frame, and thesoil type.

• Buildings with a ductile frame (that can undergo large deformationswithout collapsing) may be designed for much smaller seismicforces than structures that depend on a brittle structural system (forexample, unreinforced masonry).

Problems 67

P2.1. Determine the deadweight of a 1-ft-long segmentof the reinforced concrete ledger beam shown in FigureP2.1. Beam is constructed with normal weight concrete(see Table 2.1 for properties).

P2.2. Determine the deadweight of a 1-m-long segmentof the reinforced concrete girder in Figure P2.2 con-structed from lightweight concrete with a unit weight of16 kN/m3.

P2.3. Determine the deadweight of a 1-ft-long segmentof a typical 18-in-wide unit of a roof supported on a nom-inal 2 in � 16 in southern pine beam (the actual dimen-sions are in smaller). The -in plywood weighs 3 lb/ft2.

P2.4. Consider the floor plan shown in Figure P2.4.Compute the tributary areas for (a) floor beam B1,(b) girder G1, (c) girder G2, (d ) corner column C3, and(e) interior column B2.

P2.5. Refer to Figure P2.4 for the floor plan. Calculatethe influence areas for (a) floor beam B1, (b) girder G1,(c) girder G2, (d ) corner column C3, and (e) interiorcolumn B2.

34

12


30�

18�

18�

Inverted teeledger beam

12�

560 mm

180 mm

180 mm

180 mm

780 mm

114 mm

Prestress concretebridge girder

410 mm

190 mm

20� 20�

2� insulationthree ply felttar and gravel 3/4� plywood

15 1�2�1 1�2�

A

30� 25�

B1

C3

B2

G2G12 @ 10� � 20�

2 @ 10� � 20�

1 2 3

B

C

P2.1

P2.2

P2.3

P2.4

PROBLEMS

Problems 69

P2.6. The uniformly distributed live load on the floorplan in Figure P2.4 is 60 lb/ft2. Establish the loading formembers (a) floor beam B1, (b) girder G1, and (c) girderG2. Consider the live load reduction if permitted by theASCE standard.

P2.7. The elevation associated with the floor plan inFigure P2.4 is shown in Figure P2.7. Assume a live loadof 60 lb/ft2 on all three floors. Calculate the axial forcesproduced by the live load in column B2 in the third andfirst stories. Consider any live load reduction if permit-ted by the ASCE standard.

P2.8. A five-story building is shown in Figure P2.8. Fol-lowing the ASCE standard, the wind pressure along theheight on the windward side has been established asshown in Figure P2.8(c). Considering the windwardpressure in the east-west direction, use the tributary areaconcept to compute the resultant wind force at eachfloor level.

P2.9. The dimensions of a 9-m-high warehouse areshown in Figure P2.9. The windward and leeward windpressure profiles in the long direction of the warehouseare also shown. Establish the wind forces based on thefollowing information: basic wind speed � 40 m/s, windexposure category � B, Kd � 0.85, Kzt � 1.0, G � 0.85,and Cp � 0.8 for windward wall and 0.2 for leewardwall. Use the K8 values listed in Table 2.4. What is thetotal wind force acting in the long direction of thewarehouse?

20�24�

3 @

10�

= 3

0�

(a)

NA

1

3 @

30�

= 9

0�

4 @ 25� = 100�

2 3 4 5

B

C

D

(b) (c)

3 @

20�

= 6

0�

13

wind pressuresin lb/ft2

15

20

5 @

12�

= 6

0�

qzGCp qhGCp

40 m

9 m20 m

(not to scale)

P2.7

P2.8

P2.9

P2.8

P2.10. The dimensions of a gabled building are shownin Figure P2.10a. The external pressures for the windload perpendicular to the ridge of the building are shownin Figure P2.10b. Note that the wind pressure can acttoward or away from the windward roof surface. For theparticular building dimensions given, the Cp value forthe roof based on the ASCE standard can be determinedfrom Table P2.10, where plus and minus signs signifypressures acting toward and away from the surfaces,respectively. Where two values of Cp are listed, this indi-cates that the windward roof slope is subjected to either

positive or negative pressures, and the roof structureshould be designed for both loading conditions. TheASCE standard permits linear interpolation for the valueof the inclined angle of roof u. But interpolation shouldonly be carried out between values of the same sign.Establish the wind pressures on the building when posi-tive pressure acts on the windward roof. Use the follow-ing data: basic wind speed � 100 mi/h, wind exposurecategory � B, Kd � 0.85, Kzt � 1.0, G � 0.85, and Cp �0.8 for windward wall and –0.2 for leeward wall.


TABLE P2.10

Roof Pressure Coefficient Cp

Windward Leeward

Angle u 10 15 20 25 30 35 45 60 10 15 20Cp 0.9 0.7 0.4 0.3 0.2 0.2 0.0 0.01u* 0.5 0.5 0.6

0.0 0.2 0.2 0.3 0.4

(b)

(a)

48�

80�

wind

16�

16�

h

qhGCp

qhGCpqzGCp

qhGCp

�

P2.10

*u defined in Fig. P2.10

Problems 71

P2.11. Establish the wind pressures on the building inProblem P2.10 when the windward roof is subjected toan uplift wind force.

P2.12. (a) Determine the wind pressure distributionon the four sides of the 10-story hospital shown in Fig-ure P2.12. The building is located near the Georgiacoast where the wind velocity contour map in Figure2.15 of the text specifies a design wind speed of 140mph. The building, located on level flat ground, is clas-sified as stiff because its natural period is less than 1 s. Onthe windward side, evaluate the magnitude of the windpressure every 35 ft in the vertical direction. (b) Assum-ing the wind pressure on the windward side varies lin-early between the 35-ft intervals, determine the totalwind force on the building in the direction of the wind.Include the negative pressure on the leeward side.

P2.13. Consider the five-story building shown in Fig-ure P2.8. The average weights of the floor and roof are90 lb/ft2 and 70 lb/ft2, respectively. The values of SDS

and SD1 are equal to 0.9g and 0.4g, respectively. Sincesteel moment frames are used in the north-south direc-tion to resist the seismic forces, the value of R equals 8.Compute the seismic base shear V. Then distribute thebase shear along the height of the building.

P2.14. When a moment frame does not exceed 12 sto-ries in height and the story height is at least 10 ft, theASCE standard provides a simpler expression to com-pute the approximate fundamental period:

T � 0.1N

where N � number of stories. Recompute T with theabove expression and compare it with that obtained fromProblem P2.13. Which method produces a larger seis-mic base shear?

80�

140�

160�

wind140 mph

leeward

D

C

B

E

F

G

H

A

P2.12

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