美国钢结构设计手册第七章十三十四节

7.13 LRFD FOR COMPOSITE BEAM WITH UNIFORM LOADS

The typical floor construction of a multistory building is to have composite framing. The floor consists of 31⁄4-in-thick lightweight concrete over a 2-in-deep steel deck. The concrete weighs 115 lb/ft3and has a compressive strength of 3.0 ksi. An additional 30% of the dead load is assumed for equipment load during construction. The deck is to be supported onsteel beams with stud shear connectors on the top flange for composite action (Art. 7.12).

Unshored construction is assumed. Therefore, the beams must be capable of carrying their own weight, the weight of the concrete before it hardens, deck weight, and construction loads. Shear connectors will be3⁄4 in in diameter and 31⁄2 in long. The floor system should be investigated for

vibration, assuming a damping ratio of 5%.

FIGURE 7.6 Seven locations of the plastic neutral axis used for determining the strength of a composite beam. (a) For cases 6 and 7, the PNA lies in the web. (b) For cases 1 through 5, the PNA lies in the steel flange.

A typical beam supporting the deck is 30 ft long. The distance to adjacent beams is 10 ft. Ribs of the deck are perpendicular to the beam. Uniform dead loads on the beam are construction, 0.50 kips per ft, plus 30% for equipment loads, and superimposed load, 0.25 kips per ft. Uniform live load is 0.50 kips per ft.

TABLE 7.3 Qn for Partial Composite Design(kips)

ƒ†C(n) compressive force at location (n). Beam Selection. Initially, a beam of A36 steel that can support the construction loads is selected. It is assumed to weigh 26 lb /ft. Thus the beam is to be designed for a service dead load of 0.5×1.3+0.026=0.676 kips per ft.

Factored load=0.676*1.4=0.946 kips per ft

Factored moment = Mu=0.946×302/8=106.5 kip-ft

The plastic section modulus required therefore is

Z=Mu106.5⨯123==39.4in φFy0.9⨯36

34Use a W16 ×26 (Z =44.2 inand moment of inertia I =301 in).

The beam should be cambered to offset the deflection due to a dead load of 0.50 ＋0.026 =0.526 kips per ft.

5⨯0.526⨯304⨯123

Camber ==1.1in 384⨯29,000⨯301

Camber can be specified on the drawings as 1 in.

Strength of Fully Composite Section.

Next, the composite steel section is designed to support the total loads. The live load may be reduced in accordance with area supported (Art. 7.9). The reduction factor is R = 0.0008(300－150) =0.12. Hence the reduced live load is 0.5(1 － 0.12) = 0.44 kips per ft. The factored load is the larger of the following:

1.2(0.50 + 0.25 + 0.026) × 1.6 ＋0.44= 1.635 kips per ft

1.4(0.5 + 0.25 + 0.026) =1.086 kips per ft

Hence the factored moment is

Mu=1.635⨯30/8=183.9kip-ft

The concrete-flange width is the smaller of b = 10 ×12 = 120 in or b = 2(30 ×12⁄8) =90 in (governs).

The compressive force in the concrete C is the smaller of the values computed from Eqs. (7.24) and (7.25). 2

Cc=0.85fc'Ac=0.85⨯3⨯90⨯3.25=745.9kips

Ct=AsFy=7.68×36=276.5 kips (governs)

The depth of the concrete compressive-stress block (Fig. 7.5) is

a=C27605==1.205in '0.85fcb0.85⨯3.0⨯90

Since Cc>Ct,the plastic neutral axis will line in the concrete slab (case 3, Art.7.12). The distance between the compression and tension forces on the W16 ×26 (Fig.7.5d) is

e =0.5d + 5.25 － 0.5a

= 0.5 × 15.69 ＋ 5.25－ 0.5 ×1.205 =12.493 in

The design strength of the W16 × 26 is

φMn=0.85Cte=0.85×276.5×12.493/12=244.7 kip-ft >183.9 kip-ft—OK

Partial Composite Design. Since the capacity of the full composite section is more than required, a partial composite section may be satisfactory. Seven values of the composite section (Fig. 7.6) are calculated as follows, with the flange area Af= 5.5 ×0.345 = 1.898in.

1.Full composite:

2∑Qn=AsFy= 276.5 kips

φMn= 276.5 kips

2.Plastic neutral axis ∆Af=Af/4 = 0.4745 in below the top of the top flange. From Table

7.3,∑Qn=AsFy-2∆AfFy

∑Qn=276.5 －2 × 0.4745 ×36 = 242.3

a =242.3/(0.85 × 3.0 × 90) = 1.0558 in

e = 15.69/2 × 5.25 － 1.0558/2 = 12.567 in

Mn=242.3 × 12.567 ＋0.5(276.5－242.3)

×(15.69 － 0.345276.5-242.3)= 3,312 kip-in 2⨯1.898⨯36

φMn= 0.85 × 3312/12 ? 234.6 kip-ft

3.PNA ∆Af=Af/2=0.949 in below the top of the top flange:

∑Qn=208.2 kips

φMn= 224.0 kip-ft

4. PN∆Af=3Af/4 =1.4235 in below the top of the top flange:

∑Qn= 174.0 kips

φMn= 212.8 kip-ft

5. PNA at the bottom of the top flange (∆Af=Af):

∑Qn=139.9 kips

φMn=201.0 kip-ft

6. Plastic neutral axis within the web.

∑Qnis the average of items 5 and 7. (See Table 7.3.) ∑Qn= (139.9 ? 69.1)/2 ? 104.5 kips

φMn=186.4 kip-ft

7. ∑Qn=0.25 ? 276.5 ? 69.1 kips

φMn=166.7 kip-ft

From the partial composite values 2 to 7, value 6 is just greater than Mu= 183.9 kip-ft. The AISC ‘‘Manual of Steel Construction’’ includes design tables for composite beams that greatly simplify the calculations. For example, the table for the W16 × 26, grade 36, composite beam gives φMnfor the seven positions of the PNA and for several values of the distance Y2(in) from the concrete compressive force C to the top of the steel beam. For the preceding example,

Y2=Ycon－a/2 (7.31)

where Ycon= total thickness of floor slab, in

a=depth of the concrete compressive-stress block, in

From the table for case 6,

a=∑Qn=104 kips. 104=0.453 in 0.85⨯3.0⨯90

Substitution of a andYcon=5.25 in in Eq. (7.31) gives

Y2=5.25－0.453/2 =5.02 in

The manual table gives the corresponding moment capacity for case 6 and Y2= 5.02 in as φMn=186 kip-ft > 183.9 kip-ft—OK

The number of shear studs is based on C=104.5 kips. The nominal strength Qn of one stud is given by Eq. (7.28). For a 3⁄4-in stud, with shearing area Asc= 0.442 inand tensile strength 2

Fu=60 ksi, the limiting strength is AscFu= 0.442× 60 = 26.5 kips. With concrete unit weight w=115 lb/ftand compressive strength fc'=3.0 ksi, and modulus of elasticityEc= 2136 ksi, the nominal strength given by Eq. (7.28) is

Qn=0.5 ×0.442 33.0⨯2136= 17.7 kips

The number of shear studs required is 2 × 104.5/17.7 =11.8. Use 12. The total number of metal deck ribs supported on the steel beam is 30. Therefore, only one row of shear studs is required, and no reduction factor is needed.

Deflection Calculations. Deflections are calculated based on the partial composite properties of the beam. First, the properties of the transformed full composite section (Fig. 7.7) are determined. The modular ratioEsEn is n = 29,000/2136 = 13.6. This is used to determine the transformed concrete area A1= 3.25 × 90/13.6 = 21.52 in2. The area of the W16 × 26 is 7.68 in2, and its moment of inertia Is= 301 in. The location of the elastic neutral axis is determined by taking moments of the transformed concrete area and the steel area about the top of the concrete slab: X=421.52⨯3.25/2+7.68(0.5⨯15.69+5.25)=4.64 in 21.52+7.68

The elastic transformed moment of inertia for full composite action is

90⨯3.2533.25215.69Itr=+21.52(4.64-)+7.68(+5.25-4.64)2+301=1065in4 13.6⨯1222

Since partial composite construction is used, the effective moment of inertia is determined from Ieff=301+(106-530)0.45/27.65=77.07in4

Ieffis used to calculate the immediate deflection under service loads (without long-term effects). For long-term effect on deflections due to creep of the concrete, the moment of inertia is reduced to correspond to a 50% reduction in Ec. Accordingly, the transformed moment of inertia with full composite action and 50% reduction in Ec isItr= 900.3inand is based on a modular ratio 2n = 4

27.2. The corresponding transformed concrete area is A1=10.76 in

. 2

FIGURE 7.7 Transformed section of a composite beam.

The reduced effective moment of inertia for partial composite construction with long- term effect is determined from Eq. (7.32):