π2 π =
2 2
2
29 000 29 000
83 82 40 74 ,
( )
,
. .
Equation (E3.2) yields the critical stress
Fcr=( .0 65842 40 74/ . )×42 27 28= . ksi The nominal strength Pn of the HSS 7 × 7 is given by
Pn=F Acr g=27 28 7 59 207. × . = kip The design strength = ϕcPn = 0.9 × 207 = 186 kip > 17.6 kip.
3.3 DESIGN OF MEMBERS SUBJECT TO BENDING
It should be noted that compression portions of beams subjected to bending may have buckling problems similar to compression members. Referring to Figures 3.10 through 3.12, we note that the segment of the beam above the neutral axis in the positive member region, acts in a manner similar to a column. Unless there are closely spaced lateral restraints, we can expect that there is some criti-cal moment, Mcr, analogous to the critical load, Per, at which compression members buckle laterally.
The characteristic mode of buckling for an I-shaped section consists in part due to St. Venant’s torsion characterized by the polar moment of inertia, J, and warping of the beam flanges as delin-eated by the warping moment of inertia Cw. The polar moment of inertia also called St. Venant tor-sion constant, J, is typically expressed in units of in.4, while Cw, in units of in.6
Shapes that are intended to be used primarily as beams are generally proportioned so that the moment of inertia about the major principal axis is considerably larger than that about the minor principal axis. This is done to produce shapes that make economical beams. As a result, they are rela-tively weak in resistance to torsion and to bending about the minor axis, and if not held in-line by floor
Positive bending regions (a)
(b)
(c) Points of contraflexure
FIGURE 3.10 Lateral torsional buckling of beams: (a) continuous beam with a uniformly distributed load;
(b) bending moment diagram showing positive bending regions; (c) point of contraflexure.
Gravity Systems for Steel Buildings 129
construction or by bracing, they may become unstable under load. The instability manifests itself as a sidewise bending accompanied by twist and is called lateral buckling or later-torsional buckling.
It is not always easy to decide whether a beam has adequate support against lateral buckling.
Embedment of the top flange in a concrete slab provides support except when the beam is a cantilever in which the compression flange is at the bottom. A completely encased beam is laterally braced no matter which flange is in compression. Wood flooring spiked to nailing strips fastened to the top flange should furnish lateral support. Corrugated sheet-metal roofs are sometimes attached to purlins by metal straps or clips. It is questionable whether such connections provide dependable lateral support.
Lateral bracing must be adequate to hold the braced beam in position. Thus, stiffness as well as strength is required. As a general rule, bracing will be adequate if each lateral brace is designed for 2% of the compressive force in the flanges of the beam it braces. This rule of thumb is based on
Point of contraflexure, M = 0;
shown symbolically as a pin.
If beam is continuous, point of contraflexure shifts when load distributions change
W
Continuous (a) beam
(b)
Negative bending
Positive bending
(d)
X X
(c)
C T
FIGURE 3.11 Lateral buckling of beams: (a) continuous beam, uniformly distributed load; (b) bending moment diagram; (c) portion of positive bending region; (d) lateral-torsional buckling of I beam.
130 Structural Analysis and Design of Tall Buildings: Steel and Composite Construction
Conceptually, a wide flange beam with a moment of inertia about the principal axis considerably larger than about the minor axis is similar to a strip with a narrow rectangular cross section. If bent in its plane by couples M applied at the ends (Figure 3.13), the beam may become unstable and at a certain critical value of M buckle sideways as shown in the figure. Such lateral buckling accompa-nied by twist may be of practical importance and should be considered in the case of beams without lateral support if the flexural rigidity of the beam in the plane of bending is very large in comparison with its lateral rigidity.
3.3.1 ComPaCt, nonComPaCt, and Slender SeCtIonS
There are two main categories of beams, compact and noncompact. Compact beams by virtue for their special controls on their geometry are particularly stable. Compact section criteria are based on the yield strength of steel, the type of cross section, the ratios of width to thickness of the ele-ments of cross section. Members, whether they are rolled shapes or shapes made from plates, are compact if they fulfill certain criteria. To meet all criteria, the member must cope with certain stringent limits, of bf/tf and h/tw ratios, have unsupported length of compression flange less than Lc and be bent about its major axis. Nonfulfillment of any of the compact criteria will degrade the member to noncompact. If the width to thickness ratios exceed the limiting width to thickness ratios of noncompact elements, then it is classified as a slender element section.
3.3.2 flexural deSIGn of doubly SymmetrIC ComPaCt I-SHaPed memberS
and CHannelS bent about tHeIr major axIS
The design flexural strength, ϕbMn shall be determined using plastic section modulus Mn and ϕb = 0.9.
The nominal flexural strength, Mn, shall be the lower value obtained according to the limit states of yielding (plastic moment) and lateral-torsional buckling.
(a)
Lateral restraint prevents translation Lb
Compression flange
(b)
Compression flange
Not effective Effective
Compression flange
Effective
FIGURE 3.12 Concept of Lb, Length between points that are either braced against lateral displacement of compression flange or braced against twist of the cross section: (a) lateral restraints limit effective length of compression flange; (b) effective lateral restraint must prevent translation of compression flange.
Gravity Systems for Steel Buildings 131
1. Yielding
Mn=Mp=F Zy x
where
Fy is the specified minimum yield stress of the type of steel being used, ksi Zx is the plastic section modulus about the x-axis, in.3
2. Laterial-torsional buckling
(a) When Lb≤ Lp, the limit state of lateral-torsional buckling does not apply.
(b) When Lp < Lb≤ Lr
M C M M F S L L
L L M
x
n b p p y b p
r p
= − − − p
−
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎢
⎤
⎦⎥
⎥≤ ( 0 7. )
(c) When Lb > Lr
Mn=F Scr x≤Mp
where L is the length between points that are either braced against lateral
displace-y
M M z
(a)
h
(b) x
X L
z
M y
z
b
(c)
FIGURE 3.13 Buckling of narrow deep beams: (a) simply supported beam subject to moments M in the vertical plane; (b) plan showing lateral displacement of beam; (c) section showing lateral and torsional displacements.
132 Structural Analysis and Design of Tall Buildings: Steel and Composite Construction
E is the modulus of elasticity of steel = 29,000 ksi J is the torsional constant, in.4
Sx is the elastic section modulus taken about the x-axis, in.3 The limiting lengths Lp and Lr are determined as follows:
L r E
For a doubly symmetric I-shape: c = 1 For a channel: c h I
= o C 2
y w
where ho is the distance between the flange centroids, in.
The formulas given above apply to
1. Doubly symmetric I-shaped members and channels bent about their major axis, having compact webs and compact flanges
All current ASTM A6 W, S, M, C and MC shapes except W21 × 48, W14 × 99, W14 × 90, W12 × 65, W10 × 12, W8 × 31, W8 × 10, W6 × 15, W6 × 15, W6 × 9, W6 × 8.5, and M4 × 6 have compact flanges for Fy = 50 ksi; all current ASTM A6 W, S, M, HP, C, and MC shapes have compact webs at Fy≤ 65 ksi.
2. Beams and girders that are restrained against rotation about their longitudinal axis at the supports
The highest possible nominal flexural strength is the plastic moment, Mn = Mp. Being able to use this value in design represents the optimum use of the steel. In order to attain Mp the beam cross section must be compact and the member must be laterally braced. Compactness depends on the flange and web plate width-to-thickness ratios. When compact conditions are not met, the available nominal flexural strength diminishes. For laterally braced beams, the plastic moment region extends over the range of plate width-thickness ratios λ terminating at λr. This is the range where the section is noncompact. Beyond λr the section is a slender-element section.
These three ranges are illustrated in Figure 3.14 for the case of rolled wide-flange members for the limit state of flange local buckling. The curve in that figure shows the relationship between the flange width–thickness ratio bf/2tf and the nominal flexural strength Mn.
Gravity Systems for Steel Buildings 133
The basic relationship between the nominal flexural strength, Mn, and the unbraced length, Lb, for the limit state of lateral-torsional buckling is shown in Figure 3.15 for a compact section [W27 × 84Fy = 50 ksi] subjected to uniform bending, Cb = 1.0.
There are four principal zones defined on the basic curve by the lengths Lpd, Lp, and Lr. Note that Lp is the limiting unbraced length needed to reach Mp.
Elastic lateral-torsional buckling will occur when the unbraced length is greater than Lr. The range of inelastic lateral-torsional buckling is a straight line between the defined limits Mp at Lp and 0.7FySx
at L. Buckling strength in the elastic region is given by AISC 360-05 Equations F2-3 and F2-4 for
Mp
Compact flange
Slender flange
0
Slenderness, A = bt/2tf Nominal flexural strength, Mn
0
0.38 E
Fy 1.0 E
Fy γpf γrf
Noncompact flange 0.7FySx
FIGURE 3.14 Nominal flexural strength as a function of flange width–thickness ratio of rolled I-shapes.
Mp
Plastic design Nominal flexural stength, Mn
Basic strength × Cb Cb= 1.0 (Basic strength)
0
0.7Fy Sy
Mp
Inelastic
LTB Elastic LTB (lateral torsion buckling) Unbraced length, Lb
0 LpdLp Lr
FIGURE 3.15 Nominal flexural strength as a function of unbraced length and moment gradient.
134 Structural Analysis and Design of Tall Buildings: Steel and Composite Construction
Shown in Figure 3.16 is the beam area for the calculation of radius of gyration rt required for determining Lp and Lr.
3.3.3 deSIGn examPleS, memberS SubjeCt to bendInG and SHear
3.3.3.1 General Comments
According to the AISC Specification, the nominal capacity Mp (in kip) of a steel section in flexure is equal to the plastic moment:
Mp=ZFy
where
Z is the plastic section modulus (in.3) Fy is the steel yield strength (ksi)
However, this applies only when local or lateral torsional buckling of the compression flange is not a governing criterion. The nominal capacity Mp is reduced when the compression flange is not braced laterally for a length that exceeds the limiting unbraced length for full plastic bending capacity Lp. Also, the nominal moment capacity is less than Mp when the ratio of the compression-element width to its thickness exceeds limiting slenderness parameters for compact sections. The same is true for the effect of the ratio of web depth to thickness.
In addition to strength requirements for design of beams, serviceability is important. Deflection limitations defined by local codes or standards of practice must be maintained in selecting member sizes. Dynamic properties or the beams are also important design parameters in determining the vibration behavior of floor systems for various uses.
The factored shear strength, Vv of the web of wide-flange sections shall be verified with the design shear strength, especially if large concentrated loads occur near the supports. For LRFD, the AISC Specification requires that the factored shear Vv (kip) not exceed ϕVn
Vv φv nV
Note that ϕv = 0.9 for all cases except for webs of wide flanges with h t/wʺ 2 24. E F/ y. The nominal shear strength Vn (kip) is calculated as follows:
Vn= 0 6. F A Cyw w v
where
Fyw is the yield strength of the web, ksi Aw is the web area, in.
Comp. flange y
d x x
y
d3 rt= ry for shade area
FIGURE 3.16 Beam section showing area for the calculation of rt, the radius of gyration used in determin-ing Lp and Lr.
Gravity Systems for Steel Buildings 135
Cv is the web shear coefficient determined as follows:
1. For h t/wʺ 1 10. k E Fv / yw or hltwʺ 2 34. E F/ yw for rolled I-shaped members (kv = 5),
where a is the distance between transverse stiffeners.
The reader is referred to Chapter G, of ASCE 360-05 specifications for further require-ments of shear design.
3.3.3.2 Simple-Span Beam, Braced Top Flange Given
A 40-ft simply supported wide flange beam of Grade 50 steel with Fy = 50 ksi. The beam supports open web joists spaced at 4 ft center to center with 2-in. deep metal deck and a 3-in. thick normal weight concrete topping. For purposes of beam design, assume unfactored uniformly distributed dead load of 2.5 kip/ft (including self weight of beam) and a live load of 1.1 kip/ft.
Required
Design the beam for flexure while limiting the live load deflection to L/360 40 12 . . 360 1 34
= × = in
Comment on cambering of beams. Also, design for shear.
Solution
136 Structural Analysis and Design of Tall Buildings: Steel and Composite Construction Although the top flange of the beam is not continually braced, the presence of closely spaced joists attached to the beam, allows us to consider the top flange as fully braced. Therefore, the required plastic modulus Zx, is determined as follows.
The factored moment Mu may not exceed the design strength of ϕMn = ϕZxFy
The moment of inertia of W24 × 94 is 2700 in.4 and the modulus of elasticity for steel is 29,000 ksi.
Hence, the deflection due to live load is
ΔL= L = × × × in.
The immediate dead load deflection due to the weight of concrete topping is important in main-taining reasonably level floors. If the calculated load deflection multiplied by a factor equal to 0.75 (to account for some fixity at simply supported connections) exceeds ¾-in., typical practice in build-ing construction is to camber the beams. The purpose is to achieve a commercially acceptable level floor and to prevent ponding of wet concrete during placement.
In the present example, deflection due to a dead load of 2.5 kip/ft is
ΔL= × × × in
This is considerably larger than 0.75 in., the cut of deflection value above which camber is recom-mended. Therefore, cambering of the example beam at midspan for 75% of the calculated deflection is recommended.
Required camber = 0.75 × 1.84 = 1.38 in., say 138in.
3.3.3.2.1 Shear Design
For review of the shear capacity of the section, the depth/thickness ratio of the web for W24 × 94 is
h
The design shear strength is given by
ϕVn = 0.9 × 0.6 × Fy× AwCv
Note Cv = 1.0
φVn=0 9 0 6 50 23 6 0 395 252. × . × × . × . = kip
Gravity Systems for Steel Buildings 137
The factored shear force near the support is
Vu=4 76×40= kip< kip, OK
2 95 2 252
. .
As illustrated in this example, it usually is not necessary to review shear capacity of typical floor beams, subject to uniformly distributed loads.
3.3.3.3 Simple-Span Beam, Unbraced Top Flange Given
ASTM A992, Grade 50 steel beam with a simple span of 30 ft carrying a transfer column with Pu = 70 kip at midspan. No floor deck is present on either side of the beam. The beam top flange is therefore unbraced for the entire length. At center span, the transfer column is not capable of later-ally bracing the top flange; see Figure 3.17 for beam schematics.
Required
Design a W18 wide flange beam to carry the given loads. Neglect self-weight of beam.
Solution
Factored moment at midspan, Mu=70×30= kip ft
4 525 . From the Beam Table 3-16 of the Steel Manual, we note that W18 × 71 as the lightest beam with ϕMp = 548 kip ft exceeding the required Mu = 525 kip ft. This beam would have worked if the beam had
1. Lateral bracing of top flange at a spacing not exceeding Lp (Lp = 6.0 ft for W18 × 71 from Beam Table 3-16)
2. A torsional bracing at a distance not exceeding Lr, the rotation of the beam (Lr = 19.6 ft for W18 × 71) from the same Table
For our W18 × 71, Lp = 30 ft. is much greater than 19.6 ft, the maximum distance between beam bracings for developing full plastic capacity. Hence, its nominal bending strength, Mn will be less than Mp = FyZx.
Therefore, calculations to find the lowest-weight, larger-size beam must be continued. In prac-tice, however, this trial-and-error process can be eliminated by using beam-selector charts in the AISC Manual. These charts give the beam design moment corresponding to unbraced length for various rolled sections. See AISC manual Tables 3-10 through 3-11.
A sample design table for Mu = ϕMn ranging from 450 kip-ft to 600 kip-ft is shown in Figure 3.18.
For ϕMn > 525 kip-ft-and L = 30 ft, the figure indicates that a W18 × 106 of A992 steel satisfies the criteria. As a check, the following calculation is made with the properties of the W18 × 106.
Transfer girder, unbraced top
flange for entire span
138 Structural Analysis and Design of Tall Buildings: Steel and Composite Construction
Moment, Mu = φMn (3 kip−ft increments)
540
FIGURE 3.18 Moment Mu versus unbraced length.
Gravity Systems for Steel Buildings 139
C = 1.0 for doubly symmetric I-shape Sx = 204 in.3
Observe, for convenience we assumed Cb = 1.0 in the calculations above.
In fact, exact calculations will show Cb = 1.14.
Therefore, improved value of φMn 1 14 577 85 658 75 kip-ft
Although any type of steel cross section may be used as a tension member, in practice the selection is influenced by the type of connections required at the ends. The nominal tensile strength Pn is based on both the yield criteria over the gross section, and the fracture criteria based on effective net area. Thus,
Pn=F Ay g (yield criteria based on gross area, )