FLEXURAL SYSTEMS: B E A M S

Structure Systems & Structure Behavior

INTRODUCTION TO STRUCTURAL CONCEPTS

SKELETON STRUCTURES • Axial Systems

• Beams

• Frames

• Arches

• Cable-supported Structures

SURFACE STRUCTURES • Membranes: beams, walls

• Plates: slabs

• Hard shells

• Soft shells: tensile membranes

• Hybrid tensile surface systems: tensegrity

SPACE FRAMES

LATERAL STABILITY OF STRUCTURES

LIN

E E

LE

ME

NT

SS

UR

FA

CE

E

LE

ME

NT

S

FLEXURAL STRUCTURE

SYSTEMS

FLEXURAL-AXIAL STRUCTURE SYSTEMS

TENSILE MEMBERS

COMPRESSIVE

MEMBERS

BEAMS

BEAM-COLUMN

MEMBERS

FRAMES

TENSILE MEMBRANES

PLATES

MEMBRANE FORCES

SOFT SHELLS

SLABS, MEMBRANE BENDING and TWISTING

AXIAL STRUCTURE

SYSTEMS

SHELLS RIGID SHELLS

The Parthenon, Acropolis, Athens, 448 B.C., Ictinus and Callicrates

tower cranes are

stationary vs.

mobile cranes

Incheon International Airport , Seoul, 2001, Fentress Bradburn Arch

Steel Tree House, 2005, Tahoe Donner, CA, Joel Sherman

tizio table lamp,

Richard Sapper, 1972

project by Eric Owen Moss Architects (EOMA)

Guthrie Theater , Minneapolis, 2006, Jean Nouvel, 2006,

INSTITUTE OF CONTEMPORARY ART, Boston Harbor, 2006, Diller Scofidio &

Renfro of New York, 2006

MAXXI Art Museum,

Rome, Italy, Zaha Hadid,

2010

The Tampa Museum of Art. Tampa, 2010,

Stanley Saitowitz Office / Natoma

Architects Inc., San , Walter P Moore,

Lufthansa Reception Building, Hamburg, 2000, Renner Hainke Wirth Architects

Maxxi, the new museum of contemporary art, Rome, Italy, Zaha Hadid, 2009

Fallingwater, Ohiopyle, (Bear Run), Pennsylvania, 1937, Frank Lloyd Wright

Maxxi, the new museum of contemporary art, Rome, Italy, 2009, Zaha Hadid

Maxxi, the new museum of contemporary art, Rome, Italy, 2009, Zaha Hadid

Maxxi, the new museum of contemporary

art, Rome, Italy, 2009, Zaha Hadid

Oslo Opera House, Norway, 2007, .Snohetta

Oslo Opera House, Norway, 2007, .Snohetta

Shenzhen Stock Exchange HQ, 2007-, OMA- Rem Koolhaas

Shenzhen Stock Exchange HQ, 2007-,

OMA- Rem Koolhaas

American Indian Museum, Washington DC, 2004, architect Johnpaul Jones

American Indian Museum, Washington DC, 2004, architect Johnpaul Jones

Nelson Mandela Bay Stadium, Port Elizabeth,

South Africa, 2009, GMP Architect (Berlin),

Schlaich Bergermann and Partner

Beams constitute FLEXURAL SYSTEMS. The frame

element in SAP2000 is used to model axial truss members as well as

beam-column behavior in planar and three-dimensional skeletal

structures. In contrast to truss structures, the joints may not be hinged

but rigid. The loads may not be applied at the nodes but along the

members causing a member behavior much more complicated than for

trusses.

Beams cannot transfer loads directly to the boundaries as axial

members do, they must bend in order to transmit external forces to the

supports. The deflected member shape is usually caused by the

bending moments.

Beams are distinguished in shape (e.g. straight, tapered, curved), cross-

section (e.g. rectangular, round, T-, or I-sections), material (e.g.

homogeneous, mixed, composite), and support conditions (simple,

continuous, fixed). Depending on their span-to-depth ratio (L/t) beams

are organized as shallow beams with L/t > 5 (e.g. rectangular solid, box,

or flanged sections), deep beams (e.g. girder, trusses), and wall beams

(e.g. walls, trusses, frames).

TABLE B.3

ASTM standard reinforcing bars

Nominal Dimensions

Bar Sizea (SI)b

Diameter

in mm

Cross-Sect.

Area

in2 mm2

Weight Mass

lbs/ft kg/m

#3 #10 0.375 9.5 0.11 71 0.376 0.560

#4 #13 0.500 12.7 0.20 129 0.668 0.944

#5 #16 0.625 15.9 0.31 199 1.043 1.552

#6 #19 0.750 19.1 0.44 284 1.502 2.235

#7 #22 0.875 22.2 0.60 387 2.044 3.042

#8 #25 1.000 25.4 0.79 510 2.670 3.973

#9 #29 1.128 28.7 1.00 645 3.400 5.060

#10 #32 1.270 32.3 1.27 819 4.303 6.404

#11 #36 1.410 35.8 1.56 1006 5.313 7.907

#14 #43 1.693 43.0 2.25 1452 7.650 11.380

#18 #57 2.257 57.3 4.00 2581 13.600 20.240

REBARS

Approximate allowable stresses: the allowable stress design is used as a first

simplified structural design approach (see Projects 2010)

Compress.

St. N/mm2

(MPa)

Tensile Stress

N/mm2

(MPa)

Flexural Str.

N/mm2

(MPa)

Shear Stress

N/mm2

(MPa)

Steel, A36 (≈Q235) 150 150 150 100

Rebars, A615Gr60 (≈HRB400) Fy = 360

Concrete, 4000 psi (≈C30 ) 7 0.7 12 0.5

Masonry 3 0.2 5 0.2

Wood 10 4 8 1

Dead loads Live loads Snow loads Wind loads

kN/m2 kN/m2 kN/m2 kN/m2

Floors 4.00 3.00 # #

Roofs 2.00 1.00 1.00 #

Walls # # # 1.00

Typical preliminary vertical and horizontal design loads (see Projects 2010)

It is apparent that loads cause a beam to deflect. External loads initiate the

internal forces: shear and moment (disregarding axial forces and torsion),

deflection must be directly dependent on shear and moment.

Typical beams are of the shallow type where deflection is generally

controlled by moments. In contrast, the deflection of deep beams is

governed by shear.

In the following discussion it is helpful to treat moment and beam deflection

as directly related. Since the design of beams is primarily controlled by

bending, emphasis is on the discussion of moments rather than shear.

BEAMS may not only be the common

• planar beams

• spatial beams (e.g. folded plate, shell beams , corrugated sections

• space trusses.

They may be not only the typical rigid beams but may be flexible

beams such as

• cable beams.

The longitudinal profile of beams may be shaped as a funicular form

in response to a particular force action, which is usually gravity

loading; that is, the beam shape matches the shape of the moment

diagram to achieve constant maximum stresses.

Beams may be part of a repetitive grid (e.g. parallel or two-way joist

system) or may represent individual members; they may support

ordinary floor and roof structures or span a stadium; they may form a

stair, a bridge, or an entire building. In other words, there is no limit to

the application of the beam principle.

Calder mobile, Hirschorn Museum, Washington, 1935

Barcelona chair, 1929, Mies van der Rohe

Breuer chair, 1928

Wassily chair, 1925,

Marcel Breuer

Atrium, Germanisches Museum, Nuremberg, Germany

Shanghai Stadium, 1997, Weidlinger Assoc.

Shanghai Stadium, 1997,

Weidlinger Assoc.

Residence, Aspen,

Colorado, 2004,

Voorsanger & Assoc.,

Weidlinger Struct. Eng.

Petersbogen shopping center, Leipzig, 2001, HPP Hentrich-Petschnigg

Petersbogen shopping center, Leipzig,

2001, HPP Hentrich-Petschnigg

TU Munich, Germany

TU Munich

Auditorium Maximum, TU Munich, 1994, Rudolf Wienands

Potsdammer Platz, Berlin,

1998, Richard Rogers

Sobek House, Stuttgart, 2000,

Werner Sobek

Tokyo International Forum, Tokyo,

Japan, 1996, Rafael Vinoly Arch. and

Kunio Watanabe Eng

Petersbogen shopping center, Leipzig, 2001, HPP Hentrich-Petschnigg

Ski Jump Berg

Isel, Innsbruck,

2002, Zaha

Hadid

Dresdner Bank, Verwaltungszentrum, Leipzig, 1997, Engel und Zimmermann Arch.

Ningbo downtown, 2002,

Qingyun Ma

Hotel, Oberhof, Thueringen, Germany

National Gallery of Art, East Wing, Washington, 1978, I.M. Pei

Everson Museum, Syracuse, NY, 1968, I. M. Pei

Herbert F. Johnson Museum of Art, Cornell University, 1973, I. M. Pei

Hirshorn Museum, Washington, 1974, Gordon Bunshaft/ SOM

Hamburg Ferry Terminal, 1993, W.Alsop & J.Stoermer Arch

Atrium, Germanisches Museum, Nuremberg, Germany

Pedestrian bridge Nuremberg

Cologne/Bonn Airport, Germany, 2000, Helmut Jahn Arch., Ove Arup USA Struct. Eng.

Incheon International Airport, Seoul, S. Korea, 2001, Fentress Bradburn Arch.

Library University of Halle, Germany

Shanghai-Pudong International Airport, Paul Andreu principal architect, Coyne et

Bellier structural engineers

Berlin

CUMT, Xuzhou, 2005

Documentation Center Nazi Party Rally Grounds, Nuremberg, 2001, Guenther Domenig

Architect

German Museum of Technology, Berlin,

2001, Helge Pitz and Ulrich Wolff Architects

College for Basic Studies , Sichuan University, Chengdu, 2002

Super C, RWTH Aachen, Germany, 2008, Fritzer + Pape

World Trade Center proposal, New York, 2002, Rafael Vinoly

The building as a vertical cantilever beam

Eiffel Tower, Paris, 1889,

Gustave Eiffel

Jin Mao Tower, Shanghai,

1999, SOM

The FRAME ELEMENT for Flexural Systems

FLEXURAL SYSTEMS: BEAMS

BEHAVIOR of BEAMS

FLEXURAL SYSTEMS: shallow beams, deep beams

BEAM TYPES

LIVE LOAD ARRANGEMENT

EFFECT of SPAN

LOAD TYPES and LOAD ARRANGEMENTS

MOMENT SHAPE

DESIGN of BEAMS

• steel

• concrete

FLOOR and ROOF FRAMING STRUCTURES

BEHAVIOR of BEAMS

Beams, generally, must be checked for the primary structural

determinants of bending, shear, deflection, possibly load effects of

bearing, and lateral stability.

Usually short beams are governed by shear, medium-span

beams by flexure, and long-span beams by deflection. The

moment increases rapidly with the square of the span (L2), thus the

required member depth must also correspondingly increase so that the

stresses remain within the allowable range.. The deflection, however,

increases with the span to the fourth power (L4), clearly indicating that

with increase of span deflection becomes critical.

On the other hand, with decrease of span or increase of beam depth (i.e.

increase of depth-to-span ratio), the effect of shear must be taken into

account, which is a function of the span (L) and primarily dependent on

the cross-sectional area of the beam (A). Deflections in the elastic range

are independent of material strength and are only a function of the

stiffness EI, while shear and bending are dependent on the material

strength.

The direction, location, and nature of the loads as well as the member

shape and curvature determine how the beam will respond to force

action.

In this context it is assumed that the beam material obeys Hooke’s law

and that for shallow beams a linear distribution of stresses across the

member depth holds true.

For deep beams other design criteria must be developed.

Only curved beams of shallow cross-section that makes them only

slightly curved (e.g. arches) can be treated as straight beams using

linear bending stress distribution.

Furthermore it is assumed that the beam will act only in simple

bending and not in torsion; hence, there will be no unsymmetrical

flexure.

The condition of symmetrical bending occurs for doubly symmetrical

shapes (e.g. rectangular and W shapes), when the static loads are

applied through the centroid of their cross-section, which is typical

for most cases in building construction.

SHALLOW BEAMS

The general form of the flexure formula: fb = Mc / I = M/S

Where I is defined as Moment of Inertia, a section that measures the size and

"spread-outness" of a section with respect to an axis.

Tables for standard steel and timber sections list two values for moment of

inertia

A strong axis value called Ixx, for the section bending in its strongest

orientation.

A weak axis value called Iyy, for the section bending in its weakest

orientation.

The general definition of section modulus: S = I/c

Where c is the distance from the neutral axis to the extreme fiber of the

section.

Section modulus is also defined in terms of strong axis and weak axis

properties: Sxx = Ixx / cxx , Syy = Iyy / cyy

CONTOURS of BENDING STRESS

CONTOURS of SHEAR STRESS

General Form of the Flexure Formula

For non-rectangular sections, there is a more general derivation of the

flexure formula.

Shear causes a racking deformation, inducing diagonal tension and compression on mutually

perpendicular axes.

Shear failure in beams may manifest itself in several forms

• Diagonal cracking (concrete).

• Diagonal buckling (thin plates in steel beams).

• Horizontal cracking (timber).

In beams, the shearing stresses are maximum at the neutral axis because this is where the

tension and compression resultants of the unbalanced moment create the greatest horizontal

sliding action.

Since maximum bending stresses occur at the extreme edge of a beam section while

maximum shear stresses occur at the neutral axis, shear and bending stresses can be

considered separately in design. They are uncoupled.

SHEAR IN BEAMS

BEAM TYPES: the Effect of Support Conditions

Beams can be supported at one point requiring a fixed support joint

(e.g. cantilever beams), at two points (e.g. simple beams, overhanging

beams), and at several points (e.g. continuous beams). Beams may be

organized according to their support types as follows:

• simple beams

• cantilever beams

• overhanging beams

• hinge-connected cantilever beams

• fixed-end beams

• continuous beams

BEAM TYPES

A.

B.

C.

D.

E.

F.

SIMPLE BEAMS

OVERHANGING BEAMS: SINGLE-CANTILEVER BEAMS

OVERHANGING BEAMS: DOUBLE-CANTILEVER BEAMS

2-SPAN CONTINUOUS BEAMS

3-SPAN CONTINUOUS BEAMS

HINGE-CONNECTED BEAMS

FIXED BEAMS

G

The effect of different boundary types (pin, hinge,

overhang, fixity, continuity, and free end) on the behavior

of beams is investigated using the typical uniform

loading conditions. It is known that a uniform load

generates a parabolic moment diagram with a maximum

moment of Mmax = wL2/8 at midspan. It is shown in the

subsequent discussion how the moment diagram is

affected by the various boundary conditions. In the

following drawing the movement of the moment diagram

is demonstrated in relation to various beam types.

moving the supports

MEMBER ORIENTATION Is defined by local coordinate system

Each part of the structure (e.g. joint,

element) has its own LOCAL

coordinate system 1-2-3.

The joint local coordinate system is

normally the same as the global X-Y-Z

coordinate system.

For the elements, one of the element

local axes is determined by the

geometry of the individual element;

the orientation of the remaining two

axes is defined by specifying a single

angle of rotation.

For frame elements, for example, the

local axis 1 is always the

longitudinal axis of the element

with the positive direction from I to

J. The default orientation of the local

1-2 plane in SAP is taken to be

vertical (i.e. parallel to the Z-axis). The

local 3-axis is always horizontal (i.e.

lies in the X-Y plane).

Typical: Moment 3-3, Shear 2-2

STEEL MEMBER PROPERTIES

CONCRETE MEMBER PROPERTIES

DESIGN

Modeling Steel Members using

SAP2000 (see also Appendix A)

SAP2000 assumes by default that

frame elements (i.e., beams and

columns) are laterally unsupported

for their full length. But beams are

generally laterally supported by the floor

structure (Fig. 4.1). Therefore, assume

an unsupported length of say Lb = 2 ft

for preliminary design purposes, or

when in doubt, take the spacing

between the filler beams. For example,

for a beam span of, L = 24 ft, assume

an unbraced length ratio about the

minor axis of Lb /L = 2 ft/24 ft = 0.083,

or say 0.1; that is, take the minor

direction unbraced length as 10% of the

actual span length.

SIMPLE and CONTINUOUS FLOOR BEAMS

American Indian Museum, Washington DC, 2004, architect Johnpaul Jones

Wanli University, Ningbo

Atrium, Germanisches Museum, Nuremberg, Germany

William J. Clinton Presidential Center, Little Rock, AR, 2004, Polshek Partnership

Steel Tree House, Tahoe Donner, 2008, Joel Sherman

Fallingwater, Pittsburgh,

1937, Frank Llyod Wright

Guthrie Theatre, Minneapolis, 2006, Jean Nouvel, Ericksen & Roed

Phaeno Science Center, 2005, Wolfsburg, Zaha Hadid, Adams Kara Taylor

OVERHANGING BEAMS

Usually, cantilever beams are natural extensions of beams; in other words, they are

formed by adding to the simple beam a cantilever at one end or both ends, which

has a beneficial effect since the cantilever deflection counteracts the field deflection,

or the cantilever loads tend to lift up the beam loads. The beam is said to be of

double curvature, hence it has positive and negative moments. It is obvious that at

the point of contraflexure or the inflection point (where the moment changes signs)

the moment must be zero.

For demonstration purposes, a symmetrical overhanging beam with double

cantilevers of 0.35L span has been chosen. The negative cantilever moments at

each support are equal to

-Ms = w(.35L)0.35L/2 ≈ wL2/16 = M/2

The cantilever moments must decrease in a parabolic shape, in response to the

uniform load, to a maximum value at midspan because of symmetry of beam

geometry and load arrangement. We can visualize the moment diagram for the

simple beam to be lifted up to the top of the support moments that are caused by

the loads on the cantilever portion (i.e. moment diagrams by parts in contrast to

composite M-diagrams). Therefore, the maximum field moment, Mf , must be

equal to the simple beam moment, M, reduced by the support moment Ms.

+Mf = M – Ms = wL2/8 – wL2/16 = wL2/16 = M/2

In general, with increase of span, the simply supported beam concept becomes

less efficient because of the rapid increase in moment and deflection that is

increase in dead weight. The magnitude of the bending stresses is very much

reduced by the cantilever type of construction as the graphical analysis

demonstrates. The maximum moment in the symmetrical double cantilever beam

is only 17% of that for the simple beam case for the given arrangement of supports

and loading! Often this arrangement is used to achieve a minimal beam depth for

conditions where the live load, in comparison to the dead load, is small so that the

effect of live load arrangement becomes less critical. As the cantilever spans

increase, the cantilever moments increase, and the field moment between the

supports decreases. When the beam is cantilevered by one-half of the span, the

field moment at midspan is zero because of symmetry and the beam can be

visualized as consisting of two double-cantilever beams. For this condition the

maximum moment is equal to that of a simple span beam.

A powerful design concept is demonstrated by the two balanced, double-cantilever

structures carrying a simply supported beam; this balanced cantilever beam

concept is often used in bridge construction. It was applied for the first time on

large scale to the 1708-ft span Firth of Forth Rail Bridge in Scotland, 1890. The

form of the balancing double-cantilever support structures is in direct response to

the force flow intensity, in other words, the shape of the trusses conforms to that

of the moment diagram.

Firth of Forth Bridge (1708 ft), Scotland, 1890, John Fowler and Benjamin Baker

DOUBLE CANTILEVER

STRUCTURES

Gerber beam

Nelson Mandela Bay Stadium, Port Elizabeth, South Africa, 2009, GMP Architect

(Berlin), Schlaich Bergermann and Partner

Sporthalle, Ulster, 1994,

S. Camenzind & M.

Graefensteiner, Geilinger

Stahlbau

International Terminal, San Francisco International Airport, 2001, SOM

TGV Station, Paris-Roissy, 1994,

Paul Andreu/ Peter Rice.

LIVE LOAD ARRANGEMENT

DEAD LOAD (D)

LIVE LOAD 1 (L1)

LIVE LOAD 2 (L2)

LIVE LOAD 3 (L3)

LIVE LOAD 4 (L4)

PATTERN LOADING

In contrast to simply supported beams, for continuous beams and overhanging

beams the arrangement of the live loads must be considered in order to determine

the maximum beam stresses. Typical live load layouts are shown in the following

figure. For example with respect to the critical bending moments of a 3-span

continuous beam:

• to determine the maximum field moment at mid-span of the center beam, the

dead load case together with live load case L2 should be considered

• to determine the maximum field moments of the exterior beams, the dead load

case together with L3 should be taken,

• to determine the maximum interior support moment, the dead load case with

L4 should be used.

For the preliminary design of a continuous roof beam, the uniform gravity loading

may be assumed to control the design. It would be questionable to consider a

critical live load arrangement for flat roofs where the snow does not follow such

patterns, assuming constant building height and no effect of parapets, that is ,

assuming areas do not collect snow. Furthermore, the roof live loads are often

relatively small in comparison to the dead load, as is the case in concrete

construction, so the effect of load placement becomes less pronounced. Therefore,

the beam moment usually used for the design is based on the first interior support

and is equal to,

M = wL2/10

This moment should also cover the effect of possible live load arrangement during

construction at the interior column supports.

D, L1

L2

L3

L4

COMB1 (D + L1)

COMB2 (D + L2)

COMB3 (D + L3)

COMB4 (D + L4)

EFFECT OF SPAN

A.

B.

C.

D.

E.

F.

SIMPLE BEAMS

OVERHANGING BEAMS: SINGLE-CANTILEVER BEAMS

OVERHANGING BEAMS: DOUBLE-CANTILEVER BEAMS

2-SPAN CONTINUOUS BEAMS

3-SPAN CONTINUOUS BEAMS

HINGE-CONNECTED BEAMS

FIXED BEAMS

G

LOAD TYPES and LOAD ARRANGEMENTS

Beam loads can be arranged symmetrically and asymmetrically. Remember, for

symmetrical beams with symmetrical loading, the reactions can be determined

directly – each reaction carries one-half of the total beam load.

Notice, the asymmetrical single load on a simple beam in Table A14 top, can be

treated as a symmetrical load case plus a rotational load case. In other words,

asymmetry of loading clearly introduces the effect of rotation upon the

supports.

Beam loads can consist of concentrated loads, line loads, and any combination of

the two. Line loads usually are uniformly or triangularly distributed; occasionally they

are of curvilinear shape. The various types of loads acting on a simple beam for

symmetrical conditions by keeping the total beam load W constant are shown in the

following drawing. We may conclude the following from the figure with respect to the

shapes of the shear force and bending moment diagrams:

• The shear is constant between single loads and translates vertically at the loads.

• The shear due to a uniform load varies linearly (i.e. first-degree curve).

• The shear due to a triangular load varies parabolically (i.e. second-degree curve).

• The moment varies linearly between the single loads (i.e. first-degree curve).

• The moment due to a uniform load varies parabolically (i.e. second-degree curve).

• The moment due to a triangular load represents a cubic parabola (i.e. third degree curve).

LOAD ARRANGEMENT

LOAD TYPES

MOMENT SHAPE

For general loading conditions, it is extremely helpful to derive the shape of the

moment diagram by using the funicular cable analogy.

The single cable must adjust its suspended form to the respective transverse

loads so that it can respond in tension. Under single loads, for example, it takes the

shape of a string or funicular polygon, whereas under distributed loading, the polygon

changes to a curve and, depending on the type of loading, takes familiar geometrical

forms, such as a second- or third-degree parabola. For a simple cable, the cable sag at

any point is directly proportional to the moment diagram or an equivalent beam on the

horizontal projection carrying the same load. In a rigid beam, the moments are resisted

by bending stiffness, while a flexible cable uses its geometry to resist rotation in pure

tension.

The various cases in the figure demonstrate how helpful it is to visualize the

deflected shape of the cable (i.e. cable profile) as the shape of the moment

diagram.

The effect of overhang, fixity, or continuity can easily be taken into account by lifting up

the respective end of the moment diagram.

FUNICULAR CABLE ANALOGY

Load Types and Boundary Conditions

I.

J.

K.

L.

M.

N.

O.

A.

B.

C.

D

E.

F.

G.

H.

18 kft

12 k

6 k6 k

4 k 4 k 4 k

2 k/ft

2 k/ft

2 k/ ft 0.5 k/ft

12 kft

1.5k/ft

1 kft/ft

1 k/ft

.

12 kft

1 k/ft

1 k/ft1 k/ft

1 k/ft 1 k/ft

1 k/ft

I.

J.

K.

L.

M.

N.

O.

A.

B.

C.

D

E.

F.

G.

H.

84 kN

42 kN

28 kN

42 kNm

4 kNm/m

14 kN/m

.

14 kN/m

14 kN/m

14 kN/m

14 kN/m

14 kN/m

14 kN/m

28 kN/m

28 kN/m

28 kN/m

28 kN/m

42 kNm 63 kNm

7 kN/m

21 kN/m

42 kN

28 kN 28 kN

2 m

2 m

M = 20.84 ft-k

P = 97.87 k

2.5 ft-k

- 2.5 ft-k

Mt = 10 ft-k

Mt M

R ∆

- 2.5 ft

2.5 ft-k

W8 x 10 W14 x 30

W8 x 10

W8 x 10

W16 x 31

W14 x 30

15.88"

7.89"7.99"

7.99"/2

4"

13.84"7.89"

5.95"

5.95"/2

2.98"

a.

b.

c.

8'8' 8'

8' 16'

12'12'

DESIGN of BEAMS

In steel design, for the condition where a given member stress is

checked that is the member input is known just assign the section to the

member. But, when the member has to be designed the Automatic Steel

Selection Feature in SAP will pick up the most economical member

available from a list that has been pre-selected, i.e. for the conditions

where the members are not known and an efficient solution must be

found, more sections for the selection process have to be stored.

The design results are based on default SAP2000 assuming, that the

frame elements (i.e. beams and columns) are laterally unsupported for

their full length. But beams are generally laterally supported by the floor

structure. Therefore assume an unsupported length of say Lb = 2 ft for

preliminary design purposes, or when in doubt, take the spacing

between the filler beams (e.g. as 33% of the actual beam span). For

example, for a beam span of L = 24 ft assume an unbraced length ratio

about the minor axis of Lb/L = 2 ft/ 24 ft = 0.083 or say of 0.1, that is

taking the minor direction unbraced length as 10% of the actual span

length.

The stress ratios in SAP represent the DEMAND/CAPACITY ratios as

reflected by the various colors ranging from gray to red.

Concrete frame elements can have the area of longitudinal and

shear reinforcing steel automatically chosen for a selected section

according to the selected design code.

For normal loading conditions the program has built-in default loading

combinations for each design code. For other special loading conditions

the user must define design loading combinations. K-factors are calculated

for concrete frame members, which are defined as type column under the

frame section definition, reinforcement.

In concrete design you must define the frame section as a beam or

column! Beams are not designed for axial forces. Treat one-way slabs as

shallow, one-foot wide beam strips.

In contrast to steel design, where SAP selects the least weight section

from a list that has been pre-selected, in concrete design the area of the

bars depends on the concrete section that is the STEEL RATIO (As/bd) or in

SAP on the REBAR PERCENTAGE, As/bh.

4"

18

"

be

= 63“

bw

= 10“bw

= 10“a. b.

h =

18

"

be

b.

POSITIVE MOMENT @ MID-SPANNEGATIVE MOMENT @ SUPPORT4

"

bw

= 10“

a.

d =

15

.5"

d =

14

.5"

h =

18

"b

e

b.

POSITIVE MOMENT @ MID-SPANNEGATIVE MOMENT @ SUPPORT

4"

bw

= 10“

a.

d =

15

.5"

d =

14

.5"

Low REBAR PERCENTAGE BM18x30

Typical REBAR PERCENTAGE BM16x28

High REBAR PERCENTAGE BM14x24

Critical stirrup spacing: s = (1/0.061)0.22 = 3.61 in > ≈ 3 in

BM 14 x 24 in

BM 14 x 20 in

TBM 24 in deep

PRESTRESSED CONCRETE

BEAM Load balancing

Ps cosθ

Ps

e = 12"

Ps

L = 32'

18"

30"

wD = 2 k/ft

wL= 1.0 k/ft

wp

FLOOR and ROOF FRAMING STRUCTURES Whereas typical wood beams are rectangular solid sections, steel beams for

floor or roof framing in building construction are the common rolled sections,

cover-plated W-sections, open web steel joists, trusses, castellated beams, stub

girders, plate girders, and tapered and haunched-taper beams. In cast-in-place

concrete construction the beams form an integral part of the floor framing

system. With respect to gravity loading they constitute T-sections (or L-sections

for spandrel beams) with respect to positive bending along the midspan region,

but only rectangular sections for negative bending close to the supports.

Simple rectangular sections or inverted T-sections are also typical for precast

concrete construction, where the slab may rest on the beams without any

continuous interaction.

There are numerous framing arrangements and layouts possible depending on

the bay proportions, column layout, span direction, beam arrangement, framing

floor openings, etc. A typical floor framing bay is shown to demonstrate the

nature of load flow (i.e. hierarchy of members), and beam loading arrangements.

It is shown how the load flows (and the type of loads it generates) from the floor

deck (i.e. 1-ft slab strips) to the beams (or joists), to the girders, columns, and

finally to the foundations.

FLOOR-ROOF FRAMING SYSTEMS

Horizontal gravity force flow

Kaifeng, Xiangguo Si temple complex, Kaifeng

Beilin Bowuguan (Forest of Stelae Museum), Kaifeng

3 Sp @ 8' = 24'

25

'

BM

BM

BM

BM

GI

GI

Beam design: The beam carries the following uniform load assuming the beam

weight included in the floor dead load.

w = wD + wL = 8(0.080) + 8(0.080)0.96 = 0.64 + 0.61 = 1.25 k/ft

The maximum moment is, Mmax = wL2/8 = 1.25(25)2/8 = 97.66 ft-k

The required section modulus is,

Sx = Mx/Fb = Mx/0.66Fy = 97.66(12)/0.66(36) = 49.32 in.3

Try W18x35, Sx = 57.6 in.3, Ix = 510 in.4 (W460 x 52)

The maximum live load deflection is within the allowable limits as shown,

ΔL = 5wL4/(384EI) = 5(0.61/12)(25 x 12)4/ (384(29000)510) = 0.36 in.

≤ L/360 = 25(12)/360 = 0.83 in.

Girder design:

The girder weight is for this preliminary design approach ignored, it will have almost

no effect upon the design of the beam. The girder must support the following reaction

forces of the beams,

P = [0.080 + 0.080(0.8)](25 x 8) = 28.80 k

The maximum moment is, Mmax = PL/3 = 28.80(24)/3 = 230.40 ft-k

The required section modulus is,

Sx = Mx/0.66Fy = 230.40(12)/0.66(36) = 116.36 in.3

Try W18 x 71, Sx = 131 in.3

Notice, SAP uses a reduction factor of 0.96 therefore yielding a W18 x 76.

In ETABS, when floor elements are

modeled with plate bending

capacity (e.g. DECK section for

steel framing), vertical uniform floor

loads are automatically converted

to line loads on adjoining beams or

point loads on adjacent columns

thereby evading the tedious task of

determining the tributary loads on

the floor beams as in SAP.

LOAD MODELING

A typical floor structure layout with a stair opening is investigated in the

following figure in order to study asymmetrical loading conditions in addition

to setting up beam loading. The floor deck spans in the short direction

perpendicular to the parallel beams that are 8 ft (2.44 m) apart, as indicated

by the arrows. Visualize the deck to act between the beams as parallel, 1-ft

(0.31 m) wide, simply supported beam panels or as joists spaced 1 ft

apart that transfer one-half of the deck loads to the respective supporting

beams. The contributing floor area each beam must support is shaded and

identified in the figure; it is subdivided into parallel load strips that cause a

uniform line load on the parallel beams. However, beam B7 is positioned on

an angle and hence will have to carry a triangular tributary area. The

loading diagrams with numerical values are given for the various beams as

based on a hypothetical load of 100 psf (4.79 kPa) including the beam

weight; this load is also used for the stair area, but is assumed on the

horizontal projection of the opening.

Beam B1 is supported by beam B2 framing the opening; its reaction causes

single loads on B2 and G2. Beam B2, in turn, rests on beams B3 and B4; its

reactions are equal to the single loads acting on these two beams. Since

most of the beams are supported by the interior girders, their reactions cause

single load action on the girders, as indicated for G1, where the beam

reactions from the other side are assumed to be equal to the ones for B5; the

girder weight is ignored.

L

S

45 deg.

wS/2

M

M

M

Mw

S/2

wS

/3

(wS/3)(3 - (s/L)2)/2

BM1

BM2

BM1

BM2

G 2

BM3

G 1

BM

2

BM

1

BM

1

BM

2

G 3

BM

5

G 4

3 Sp @ 7' = 21' 21'

3 S

p @

7' =

21

'2

1'

G 1

BM3

BM4

BM

5

BM

5

BM

5

BM

5

BM

5

8' 8' 8'

24'

P P

20

'

R R

In concrete design you must define the frame section as a beam or column! Beams are

not designed for axial forces. Treat one-way slabs as shallow, one-foot wide beam strips.

• Define material and the concrete section (e.g. rectangular, T-section).

• For the design of beams enter the top and bottom concrete cover in the text edit boxes. If

you want to specify top and bottom longitudinal steel, enter reinforcement area for the section

in the appropriate text edit box, otherwise leave values of zero for SAP2000 to calculate

automatically the amount of reinforcing required.

• First the load cases must be defined such as, D (dead load), L1 (live load 1), L2 (live

load 2), L3 (live load 3), etc. according to the number of live load arrangements.

• Click Define, then Analysis Cases and the load cases occur, highlight load case and click

Modify/Show Case to check whether the load case is OK. In case of load factor design

change the scale factor with the load factor (e.g. 1.2 D, 1.6 L).

• Then define load combinations such as for a continuous beam for D + L: go to

combinations and click Add New Combo button and define such as COMB1 (D + L1),

COMB2 (D + L2), COMB3 (D + L3), etc. Change Scale Factor for combined load action

such as 0.75(D + L + W or E)

Check the strength reduction factors in Options/Preferences.

• Assign C L E A R M E M B E R LE N G T H S, select member (click on member)

then click on Assign, then Frame, then End Offsets: (total beam length -clear span, or

support width of girder, for example)/2, then Offset Lengths

• Click Analysis and check results

• Click Design, then Concrete Frame Design then Select Design Combos and select combos,

then click Start Design/Check of Structure. Start Design/ Check of Structure button, then

select member, then right click, then choose ReDesign button, then check under Element

Type: NonSway (for beams

and laterally braced columns), or Sway Ordinary (for ordinary frames, laterally non braced

columns). Click on member, then click right button of the mouse to obtain the Concrete

Design Information, then highlight the critical location (e.g. support and center-span for

longitudinal reinforcing, or support for shear reinforcing), then click Details to obtain the

maximum moment and shear reinforcement areas which are displayed for the governing

design combination by default

EXAMPLE 6.4: Design of concrete floor framing

A 6-story concrete frame office building consists of 30 x 34-ft (9.14 x 13.11-m) bays

with the floor framing shown in Fig. 6.13 The 6.5-in (165-mm) concrete slab

supports 5 psf (0.24 kPa) for ceiling and floor finish, a partition of 20 psf (0.96 kPa),

as well as a live load of 80 psf (3.83 kPa). The girders are 24 in (610 mm) high and

16 in wide (406 mm), whereas the beams have the same depth but are 12 in (305

mm) wide. Investigate a typical interior beam. The beam dead load is 1.81 klf (26.41

kN/m) and the reduced live load is 0.85 klf (12.40 kN/m).

Use a concrete strength of fc' = 4000 psi (28 MPa ), fy = fys = 60 ksi (414 MPa ) and

a concrete cover of 2.5 in.(63.5 mm).

1. Treat the typical interior span of the continuous beam as a fixed beam using

the net span.

2. Model the intermediate floor beam (i.e. beam between column lines) as a

continuous three-span beam fixed at the exterior supports. Consider live load

arrangement.

3. Use the equivalent rigid-frame method by modeling the beam along the column

lines as a continuous three-span beam to be framed into 18 x 18-in. columns and

to form a continuous frame, where the ends of the 12-ft columns are assumed

fixed. Consider live load arrangement.

4. Model six structural bays to design the beams using ETABS and then export

the floor framing to SAFE to design the floor slab. . For this preliminary

investigation, establish live load patterns for the design of the intermediate

beams only, that is not for the beams along the column lines.

GI

GI

BM

BM

BM

16/24

16/24

12

/24

12

/ 24

12/2

4

18"x18"

15' 15'

34

'

34'

16"

24

"

6.5

"

11

A1

lnet

= 34 - 16/12 = 32.67'.

lnet

= 34 - 18/12 = 32.50'

12'

12'

hn

et = 1

2 -

24/1

2 =

10'

hn

et = 1

2 -

24

/12 =

10

'

1

A1

1

A1

EQUIVALENT RIGID FRAME METHOD

THREE-SPAN CONTINUOUS BEAM

FIXED BEAM

COMB1 = D + DS + L1

COMB2 = D + DS + L2

COMB3 = D + DS + L3

Floor Beam Grids

The floor framing systems discussed till now consisted of one-dimensional

resisting beams, in other words, the loads were carried by single beams in one-

directional fashion. However, when beams intersect loads may be transferred in

two or more directions as is the case for beam grids.

First let us investigate various cross beam layouts for floor framing shown in Fig.

7.22. The two left cases identify on directional beams, where either the short

beams are supported by the long beam (left case) or the long beams are

supported by the short beam, hence the structures are statically determinate.

However, in the two other cases the beams are continuous and support each

other; together they share the load and disperse the load in two-directional

fashion, which makes the analysis statically indeterminate.

Ls

LL =

2L

S

a. b. c. d.

PP P P

0.25P0.5P

0.5P

0.5P

0.25P

0.25P

0.25P

0.06P

0.06P

0.44P0.44P

LL = L

S = L

0.5P

The effect of beam continuity

a.

b.

c. d.

rectangular and skew beam grids