By The ASCE Task Committee On Design Criteria For Composite Structures in Steel and Concrete ABSTRACT: This Commentary Complements The "Proposed Specification For Struc
By The ASCE Task Committee On Design Criteria For Composite Structures in Steel and Concrete ABSTRACT: This Commentary Complements The "Proposed Specification For Struc
By The ASCE Task Committee On Design Criteria For Composite Structures in Steel and Concrete ABSTRACT: This Commentary Complements The "Proposed Specification For Struc
1.0 SCOPE
This specification (ASCE: "Proposed Specification for Structural Steel
Beams with Web Openings" 1992) is based on the design procedures and
guidelines presented by Darwin (1990) for both composite and noncom-
posite beams with web openings. Darwin (1990) drew heavily on design
guidelines presented by Redwood and Shrivastava (1980) for noncomposite
sections, as well as a wide range of other references, many of which are
cited in this commentary. Steel sections must meet the requirements of a
compact section for reasons explained in section 4.1 of this commentary.
The specification is formulated in terms of load and resistance factor
design because the capacity of beams at web openings can be accurately
expressed in terms of strength, but not in terms of stresses at stages below
ultimate.
2.0 DESIGN
Several design procedures have been developed for structural steel beams
with web openings (Cho and Redwood 1986; Clawson and Darwin 1980;
Darwin and Donahey 1988; Darwin and Lucas 1990; Donoghue 1982; Kuss-
man and Cooper 1976; Redwood 1968, 1971; Redwood and Poumbouras
1984; Redwood and Shrivastava 1980; Redwood and Wong 1982; Wang et
al. 1975). Most of the techniques have been developed for specific types of
members, such as noncomposite beams with reinforced openings (Kussman
and Cooper 1976; Redwood 1971; Redwood and Shrivastava 1980; Wang
et al. 1975) or composite beams with unreinforced openings (Cho and Red-
Note. Discussion open until May 1, 1993. To extend the closing date one month,
a written request must be filed with the A S C E Manager of Journals. The manuscript
for this paper was submitted for review and possible publication on February 11,
1992. This paper is part of the Journal of Structural Engineering, Vol. 118, No. 12,
December, 1992. ©ASCE, ISSN 0733-9445/92/0012-3325/$1.00 + $.15 per page.
Paper No. 3436.
3325
tions that accurately match experimental results such as detailed for non-
composite beams by Bower (1968), Clawson and Darwin (1980), Congdon
and Redwood (1970), Cooper and Snell (1972), Cooper et al. (1977), Lupien
and Redwood (1978), Redwood et al. (1978), and Redwood and Mc-
Cutcheon (1968), and for composite beams by Cho (1982), Clawson and
Darwin (1982), Donahey and Darwin (1988), Granade (1968), Redwood
and Poumbouras (1983), Redwood and Wong (1982), and "Structural In-
vestigation" (1984). The method must include the required resistance fac-
tors.
The procedure presented next provides a single technique that applies to
all types of structural steel beams with web openings, i.e., composite and
noncomposite members, reinforced and unreinforced openings. These pro-
cedures presented are generally easier to apply than the earlier techniques
and provide for more efficient designs (Lucas and Darwin 1990).
" / \3 1 ~Ui
Noncomposite Beams
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Mm = Mp - FyAAs (^ + e (3)
1 1—*y—H
u 1
t '
0
f /§ Openingi 1
rl h
'04
1 ^q Steel Section t
1
u
\ H/H
(a)
-K-A
2
< Ar
1 •w
(b) (C)
3327
Composite Beams
Fig. 2 illustrates stress diagrams for composite sections in pure bending.
In each case, the force in the concrete Pc, is limited to the lower of the
concrete compressive strength, the shear connector capacity, or the yield
strength of the net steel section
Pc s 0.85/# e f e (6a)
Pc^NQ„ (66)
PcsT = FyAm (6c)
where/c = concrete compressive strength;
be = effective width of concrete slab (Load 1986b);
te = effective thickness of concrete slab ( = ts for solid slab; = t's for
slab with ribs perpendicular to steel beam; = (t's + ts)l2 for slab
with ribs parallel to steel beam);
3328
A_r
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~\ /lHM,
7
— a
• — - ^
(b) (c)
FIG. 2. Composite Sections in Pure Bending: (a) Neutral Axis above Top of Flange;
(b) Neutral Axis in Flange; (c) Neutral Axis in Web
d AAse
Mm = T - + + ts (7)
2 Asn
where LAS = hj„, - 2Ar;
e = opening eccentricity; for composite sections, positive when the center
line of the opening is above the center line of the steel section; and
a = depth of concrete compression block = Pcl(0.&5f'cbe) for solid slabs
and ribbed slabs for which a < t's.
If a > t's, as it can be for ribbed slabs with longitudinal ribs, the term (r,
- a/2) in Commentary Eq. (7) must be replaced with the appropriate
expression for the distance between the top of the steel flange and the
centroid of the concrete force.
If Pc < T [Commentary Eqs. (6a) or (6b)], the plastic neutral axis (PNA)
is in either the flange or the web of the top tee, based on whether
Pc + 2FyAf \ T (8a)
or
Pc + 2FyAf < T (86)
where Af = the flange area = bftf.
Commentary Eqs. (8a) and (8b) are derived from the inequalities Pc +
FyAf > Fy(Am - Af) and Pc + FyAf < Fy(Asn - Af), respectively, which
determine whether the force in the concrete, Pc, and the force in the flange,
FyAf, is greater than or less than the force that can be developed in the
steel section below the flange.
If Commentary Eq. (8a) governs, the PNA is in the flange (Fig. 2b) at
a distance x = (T — Pc)/(2bfFy) from the top flange. In this case
d k.Ase - bfx'
Mm = V + P, (9)
2 A...
If Commentary Eq. (&b) governs, the PNA is in the web [Fig. 2(c)] at a
distance x = (As„ - 2A/)/(2f,„) - PJ(2Fytw) + tf from the top of the
flange. In this case
Commentary Eq. (11) is always conservative for Am < As. The values of
Mpc can be conveniently obtained from part 4 of the AISC Load and Resis-
tance Factor Design Manual (1986a).
For beams in which the PNA in the unperforated member is located below
the top of the flange and Pc > Pcmin = Fy[(3/4)twd - AAS], the value of
Mm may be approximated by Commentary Eq. (12)
Top
Tea
Bottom
Tea
High
Moment
End
0.85 fe 0.85 fc
. W, -r Knf,
/ t 4
\ r\ L, dh
Top
TM
Bottom
Tea
High
Moment
End
ment increases, the PNA moves further from the outside, thus reducing the
effective moment arm of both the normal stresses in the web and the re-
inforcement. The effect of this movement is handled by modifying s, in the
calculation of v only. The actual value of st should be used to calculate (x
in Specification Eq. (3) (ASCE: "Proposed" 1992).
The design criteria presented in 4.0 are based on both theoretical con-
siderations and experimental observations. Many of the criteria were de-
veloped for noncomposite beams (Redwood and Shrivastava 1980) and ex-
tended as appropriate to composite beams (Darwin 1990). The criteria help
insure that the limit states can be obtained upon which the design formulas
are based.
3333
validated experimentally only for compact sections (Darwin and Lucas 1990;
Lucas and Darwin 1990).
in excess of 0.7d.
P = F A < Fy t" a°
r y r
~ 2V3
Ar = cross-sectional area of reinforcement above or below the opening.
The required strength of a weld within each extension is
Rwr = $0FyAr (16)
The factor 2 in Commentary Eq. (15) is used because the reinforcement is
in tension on one side of the opening and in compression on the other end
when the tee is subjected to shear (Figs. 3 and 4). Within the extensions,
the reinforcement must be anchored to provide the full yield strength of
the bars since the expressions for M,n are based on this assumption.
The terms 2Pr in Commentary Eq. (15) and FyAr in Commentary Eq.
(16) are multiplied by (J>0 to convert these forces into equivalent factored
loads. The weld is then designed to resist the factored load, Rwr, with a
value of (J> = 0.75 (Load 1986b). The result is a design that is consistent
with the Load and Resistance Factor Design Specification for Structural Steel
Buildings (1986b).
The criteria for placing the reinforcement on one side of the web limit
the reductions in strength caused by out-of-plane deflections caused by
eccentric loading of the reinforcement (Lupien and Redwood 1978). The
limitations on the area of reinforcement, Ar, and the aspect ratio of the
opening, a0lh0, represent the extreme values that have been tested. The
limitation on sjtw is primarily empirical. The limitation on MJ{Vud) restricts
the use of unsymmetrical reinforcement to regions subject to some shear
loading. For regions subjected to pure bending or negligibly low shear, the
out-of-plane deflections of the web can be severe. Under shear, the lateral
3336
The provisions of 4.12 should not be construed to restrict the use of flange
cover plates, or a thickened concrete slab in composite beams, which are
considered to be modifications of the unperforated section.
4.13.a
There is strong experimental evidence to suggest that the concrete slab
improves the shear strength at a web opening, even in regions of negative
bending. However, since no tests have actually been carried out for openings
in negative moment regions, the traditional approach, ignoring the contri-
bution of the concrete slab, is taken for the design of web openings in
negative moment regions of composite members.
4.13.b
Slabs tend to crack both transversely and longitudinally in the vicinity of
web openings. The minimum slab reinforcement ratio of 0.0025 is used to
limit crack width and improve the postcrack strength of the slab in the
vicinity of a web opening (Donahey and Darwin 1986, 1988).
4.13.C
At failure, a significant amount of bridging (lifting of the slab from the
steel section) tends to occur between the low moment end of the opening
and a point past the high moment end of the opening in the direction of
increasing moment (Donahey and Darwin 1986, 1988). The required min-
imum number of shear connectors in the direction of increasing moment
limits bridging, although the studs do not directly enter into the calculation
of member strength at the opening. The minimum of two studs per foot
(0.3 m) applies to the total number of studs. If this criteria is already satisfied
by normal stud requirements, additional shear connectors are not needed.
4.13.d
This requirement recognizes that a composite beam with adequate strength
at a web opening may not provide adequate capacity during construction,
when it must perform as a noncomposite member.
4.14 Fatigue
Web openings are not recommended for members that will be subjected
to significant cyclic or fatigue loading. This is due to both a lack of exper-
imental data and a number of specific considerations. However, most mem-
bers in buildings are not subject to a large enough number of cycles of
sufficient amplitude to require design for fatigue.
A web opening in a steel beam can create severe stress concentrations,
especially near the corners of an opening (Clawson and Darwin 1980). The
magnitude of the stresses are dependent upon several factors, including: (1)
The geometry and location of the opening; (2) workmanship; (3) corner
radius; and (4) loading (Frost and Leffler 1971). Reinforcement of an open-
ing with bars welded to the web can also increase local stresses. High lo-
calized stresses under fatigue (cyclic) load conditions can lead to crack
3337
4.15 Deflections
Web openings reduce the local moment of inertia of beams, which results
in an increase in the maximum deflection. Openings also result in a local
decrease in the shear stiffness, which leads to deflections through the length
of the opening. The first effect is often greatest when the opening is located
in a region of high moment, the latter when the opening is in a region of
high shear. The effects of the opening on member deformation must be
considered by the engineer. However, in most cases, the effect of a single
web opening on deflection is small. A review of analysis methods that
account for the effects of web openings on beam deflection is presented by
Darwin (1990).
Given
A W16 x 31 composite section with Fy = 36 ksi has been selected to
support a service live load of 1.0 kips/ft and a service dead load of 0.65
kips/ft. The beam span and spacing are 30 ft and 10 ft, respectively. The
total slab thickness is 6.25 in. The slab is placed on a composite metal deck
with a rib depth of 3 in., an average rib width of 6 in., and a rib spacing of
12 in. The ribs are oriented perpendicular to the beam center line {be = 90
in.). Lightweight concrete (f'c = 3.5 ksi [115 pcf]) is used. A total of 26
3/4 in. diameter X 5 in. headed studs (Qn = 19.8 kips) have been selected.
Shored construction is specified. [Note: The basic beam design is identical
to example 1, pages 4-9 through 4-12 of the AISC Load and Resistance
Factor Design Manual (1986a).]
A 10 in. x 20 in. rectangular opening is required at the quarter point of
the span (Fig. 5). Determine if the opening can be placed in the selected
beam. If not, modify the design to allow the same beam depth to be used.
Solution
A. Loads
Factored load = 1.2(0.65) + 1.6(1.0) = 2.38 kips/ft.
^Opening 4 Support
30'-0"
Ar
FIG. 5. Opening Location
3338
C. Section Properties
As = 9.12 in.2, d = 15. 1 in., tw = 0.275 in., bf = 5.525 in., tf = 0.440
in., ts = 6.25 in., te = t's -- 3.25 in., andfr„ = 90 in.
D. Preliminary Calculations
Preliminary calculations indicate that the selected beam must be modified
to accept the required opening. It should be noted that a number of options
are available, including: (1) Use of a heavier section; (2) use of higher
strength steel, and (3) use of reinforcement at the opening. For this example,
the second and third options will be investigated. Final selection will be
dependent upon material and fabrication costs.
Opening <5
3.25 in.
/ \ / \ t 7- 3.00 in.
2.94 in.
10 in. _&.
2.94 in.
p c
"' ~ 3 ~ 3\V3 J 3\ V3
+ Vc = 84 + Vc
139
+ 139 | 6.25
1.7(3.5)(90),
= 3336 in.-kip
Maximum Shear Capacity (Section 3.3)
Bottom tee:
Vpt = Fytwst/V3 = 50(0.275)(2.94)/V3 = 23.3 kips
Using Specification Eq. (2) with |x = 0 and v = 6.80:
3340
M„ V„
+ < 1.0
3
2,410 17.9
+ = 0.61 + 0.35 = 0.96 O.K.
0.85(3,336) 0.85(30.0)
3— 12.94 h.
Mil
M„ 2,410
= 8.5 < 20 O.K.
y„rf 17.9(15.88)
Maximum Moment Capacity
AA, = h0tw - 2Ar = 10(0.275) - 2(0.75) = 1.25 in.2
and
As„ = As - M.s = 9.12 - 1.25 = 7.87 in.2
Use Commentary Eqs. (6a), (6b), and (6c) to calculate the force in the
concrete:
Pc < 0.85/^,4 = 0.85(3.5)(90)(3.25) = 870 kips
Pc ^ NQn = 7(19.8) = 139 kips controls
Pc<T' = FyAsn = 36(7.87) = 283 kips
Again, since Pc < T and Pc + 2FyAf = 139 + 2(36)(5.525)(0.440) = 314
kips > T" [Commentary Eq. (8a)], the PNA is in the flange and Mm is given
by Commentary Eq. (9)
where
T - Pr 283 - 139
x = = 0.362 in.
2bfFy 2(5.525)(36)
Therefore
139
139 6.25
1.7(3.5)(90)/
3343
Ft a
p r = FA
y r < y 'V-°
~ 2V3
Therefore,
0 2 0
Pr = 36(0.75) = 27.0 kips * ^ C ^ ) = 57.2 kips
Bottom tee:
^ = ^ = 36(0.275^(2.94) = 168k .ps
With reinforcement at the bottom tee, (JL is nonzero (Specification Eq. 3).
_ 2Prdr + Pchdh - Pcldi _ 2(27.0)(2.75) + 0 - 0
^" V^, 16.8(2.94)
For reinforced openings, v is initially given by v = a 0 /s, = a0/[sr - A r /
(26,)]. Therefore, v = 20/{2.94 - 0.75/[2(5.525)]} = 6.96. Using Specifi-
cation Eq. (2) with JJL = 3.00 and v = 6.96:
V6 + (JL V6 4- 3.00
V V
- = VT^ * = 6 * ^ / 3 (16-8^ = 10'5 ^
Top tee:
p = Fytwst =36(0.275)(2.94) = ps
' "W vJ—
As with the unreinforced opening, Pch must be found before calculating \i.
Since Specification Eq. (4c) gives Pch = K,AM = Fy(A„,/2 - etw) = 36(7.87/
2 - 0(0.275)) = 141, Specification Eq. (Ab) again provides the initial limit
on the concrete force with Pch - 139 kips. The value for |x [Specification
Eq. (3)] is therefore
2Prdr + Pchdh - Pc,d, = 2(27.0)(2.75) + 139(5.99) - 119(3.22)
^ Vpts, 16.8(2.94)
= 12.1
Check (V6 + |x)/(v + V3):
V6 + n, V6 + 12.1
£ = -F= = 1.68 > 1.0
v + V3 6.96 + V 3
Fcft is therefore limited by the combined yield capacity of the top flange
and the reinforcement [Specification Eq. (9)]
Pch =£ Fy[tf(bf - tw) + Ar] = 36[0.440(5.525 - 0.275) + 0.75]
= 110 kips
3344
d = = 6 25 = 6 04in
" '• " ilk ' - 1.7(3.5)(90) - '
di
= '• - < ;+ uKte = 6 - 2 5 -3-25 + UiBjwr3-17 in
-
fx [Specification Eq. (3)] is now given by
= 2Prdr + Pchdh - Pcld, = 2(27.0)(2.75) + 110(6.04) - 90(3.17)
^ Vplst 16.8(2.94)
= 10.7
Finally, v = ajst = 6.80 for use in Specification Eq. (8)
y =
- I^ = S = 26 4kipS
'
Check to see if Vm, < Vm,(sh) [Specification Eq. (10a)]
Vmt{sh) = Vpt + 0.llVfcAvc = 16.8 + 0.1lV53(3)(6.25)(3.25)
= 29.4 kips O.K.
Total Shear Capacity
4>0M,J \4>0Vm
2,410 17.9
-- 0.80 + 0.18 = 0.98 O.K.
0.85(3,053) + 0.85(36.9)
Reinforcement details (Section 4.12):
V3A V3(0.75)
Extension a _ _ r = _____ = 2 . 3 6 in.
ACKNOWLEDGMENTS
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To convert To Multiply by
ft m 0.305
in. mm 25.4
in.-kip m-N 112.98
kip kN 4.448
kip/ft kN/m 14.59
ksi N/mm2 6.895
APPENDIX V. NOTATION
The notation listed here includes only those symbols used in the Com-
mentary that are not defined in the Nomenclature section of the Specification.
The section number in parentheses after the definition of a symbol refers to
the section in this Commentary where the symbol is first defined.
3349