Tải bản đầy đủ - 0 (trang)
16 Example - LRFD for Wide-Flange Column in a Multistory Rigid Frame

# 16 Example - LRFD for Wide-Flange Column in a Multistory Rigid Frame

Tải bản đầy đủ - 0trang

12.148

SECTION TWELVE

be found by proportion from those previously calculated. Thus, the moment at the support

is

210 ϫ 0.247

MLL ϭ Ϫ

ϭ Ϫ172 ft-kips

0.0302

and the moment at midspan is

MLL ϭ

110 ϫ 0.247

ϭ 90 ft-kips

0.302

Impact. For a 30-ft span, impact is taken as 30%.

At midspan, MI ϭ 0.30 ϫ 90 ϭ 27 ft-kips.

At supports, MI ϭ 0.30(Ϫ172) ϭ Ϫ52 ft-kips.

Total Floorbeam Moments. The design moments previously calculated are summarized in

Table 12.78.

Properties of Floorbeam Sections. For stress computations, an effective width so of the

deck plate is assumed to act as the top flange of member III. For determination of so , the

effective spacing of floorbeams sƒ is taken equal to the actual spacing, 180 in. The effective

span le , with the floorbeam ends considered fixed, is taken as 0.7 ϫ 30 ϫ 12 ϭ 252 in.

Hence,

sƒ 180

ϭ

ϭ 0.715

le

252

From Table 4.6 for this ratio,

so

ϭ 0.53, and so ϭ 0.53 ϫ 180 ϭ 95 in

(Fig. 12.63b and c).

The neutral axis of the floorbeam sections at midspan and supports can be located by

taking moments of component areas about middepth of the web. This computation and those

for moments of inertia and section moduli are given in Table 12.79.

Floorbeam Stresses. These are determined for the total moments in Table 12.78 with the

section properties given in Table 12.79. Calculations for the stresses at midspan and the

supports are given in Table 12.80. Since the stresses are well within the allowable, the

floorbeam sections are satisfactory.

TABLE 12.78 Moments, ft-kips, in Floorbeam

with Orthotropic-Plate Flange

Midspan

Supports

MDL

MLL

MI

Total M

104

0

90

27

221

Ϫ712

Ϫ52

Ϫ224

BEAM AND GIRDER BRIDGES

12.149

TABLE 12.79 Floorbeam Moments of Inertia and Section Moduli

(a) At midspan

Material

A

Deck 95 ϫ 3⁄8

Web 21 ϫ 3⁄8

Bottom flange 10 ϫ 1⁄2

35.6

7.9

5.0

48.5

d

10.69

381

4,070

Io

I

4,070

290

580

580

4,940

Ϫ6.73 ϫ 327 ϭ Ϫ2,210

INA ϭ 2,730

290

Ϫ10.75

Ϫ54

327

d ϭ 327 / 48.5 ϭ 6.73 in

Distance from neutral axis to:

Top of deck plate ϭ 10.50 ϩ 0.375 Ϫ 6.73 ϭ 4.15 in

Bottom of rib ϭ 10.50 ϩ 0.50 ϩ 6.73 ϭ 17.73 in

Section moduli

Top of deck plate

Bottom of rib

St ϭ 2,730 / 4.15 ϭ 658 in3

Sb ϭ 2,730 / 17.73 ϭ 154 in3

(b) At supports, gross section

Material

A

Deck 95 ϫ 3⁄8

Web 18 ϫ 3⁄8

Bottom flange 10 ϫ 1⁄2

35.6

6.8

5.0

47.4

d

9.19

327

3,010

Io

I

3,010

180

430

3,620

Ϫ5.93 ϫ 281 ϭ Ϫ1,670

Gross INA ϭ 1,950

180

Ϫ9.25

Ϫ46

430

281

dg ϭ 281 / 47.4 ϭ 5.93 in

Distance from neutral axis to:

Bottom of rib ϭ 9 ϩ 0.50 ϩ 5.93 ϭ 15.43 in

Section modulus, bottom of rib

Sb ϭ 1,950 / 15.43 ϭ 126 in3

(c) At supports, net section

Material

A

Gross section

Top-flange holes

Bottom-flange holes

Web holes

Ϫ10.2

Ϫ1.0

Ϫ2.3

9.19

Ϫ9.25

Ϫ94

47.4

33.9

dnet ϭ 196 / 33.9 ϭ 5.77 in

d

Io

281

9

196

I

3,620

Ϫ860

Ϫ90

Ϫ120

Ϫ860

Ϫ90

Ϫ120

2,550

Ϫ5.77 ϫ 196 ϭ Ϫ1,130

Net INA ϭ 1,420

12.150

SECTION TWELVE

TABLE 12.79 Floorbeam Moments of Inertia and Section Moduli (Continued )

Distance from neutral axis to:

Top of deck plate ϭ 9 ϩ 0.375 Ϫ 5.77 ϭ 3.61

Section modulus, top of deck plate

St ϭ 1,420 / 3.61 ϭ 392 in3

Floorbeam Shears. For maximum shear, the truck wheels are placed in each design lane

as indicated in Fig. 12.64. The 16-kip wheels are placed over the floorbeam. A 4-kip wheel

is located 14 ft away on each of the adjoining rib spans. Thus, with the floorbeams assumed

acting as rigid supports for the ribs, the wheel load is 16.5 kips, as for maximum floorbeam

moment. (The effects of floorbeam flexibility can be determined as for bending moments.)

This loading produces a simple-beam reaction of 41.8 kips. It also causes end moments

of Ϫ202 and Ϫ86, which induce a reaction of (Ϫ86 ϩ 202) / 30 ϭ 3.9 kips. Hence, the

maximum live-load reaction and shear equal

VLL ϭ 41.8 ϩ 3.9 ϭ 45.7 kips

Shear due to impact is

VI ϭ 0.30 ϫ 45.7 ϭ 13.7 kips

MAXIMUM FLOORBEAM SHEARS,

KIPS

VDL

VLL

VI

Total V

13.9

45.7

13.7

73.3

Allowable shear stress in the web for Grade 50W steel is 17 ksi. Average shear stress in

the web is

ƒv ϭ

73.3

ϭ 10.9 Ͻ 17 ksi

18 ϫ 3⁄8

Transverse stiffeners are not required.

Flange-to-Web Welds. The web will be connected to the deck plate and the bottom flange

by a fillet weld on opposite sides of the web. These welds must resist the horizontal shear

between flange and web. For the weld to the 10 ϫ 1⁄2-in bottom flange, the minimum size

TABLE 12.80 Bending Stresses in Member III

At midspan:

Top of deck plate

Bottom flange

At supports:

Top of deck plate

Bottom flange

ƒb ϭ 221 ϫ 12 / 658 ϭ 4.03 ksi (compression)

ƒb ϭ 221 ϫ 12 / 154 ϭ 17.2 Ͻ 27 ksi (tension)

ƒb ϭ 224 ϫ 12 / 392 ϭ 6.9 ksi (tension)

ƒb ϭ 224 ϫ 12 / 126 ϭ 21.4 Ͻ 27 ksi

BEAM AND GIRDER BRIDGES

12.151

FIGURE 12.64 Positions of truck wheels for maximum shear in floorbeam.

fillet weld permissible with a 1⁄2-in plate, 1⁄4 in, may be used. Shear, however, governs for

the weld to the deck plate.

For computing the shear v, kips per in, between web and deck plate, the total maximum

shear V is 73.3 kips and the moment of inertia of the floorbeam cross section I is 1,950 in.4

The static moment of the deck plate is

Q ϭ 35.6(4.15 Ϫ 0.19) ϭ 141 in3

Hence, the shear to be carried by the welds is

VQ 73.3 ϫ 141

ϭ

ϭ 5.30 kips per in

I

1,950

The allowable stress on the weld is 18.9 ksi. So the allowable load per weld is 18.9 ϫ

0.707 ϭ 13.4 kips per in, and for two welds, 26.7 kips per in. Therefore, the weld size

required is 5.30 / 26.7 ϭ 0.20 in. Use 1⁄4-in fillet welds.

Floorbeam Connections to Girders. Since the bottom flange of the floorbeam is in compression, it can be connected to the inner web of each box girder with a splice plate of the

same area. Use a 10 ϫ 1⁄2-in plate, shop-welded to the girder and field-bolted to the floorbeam. With A325 7⁄8-in-dia. high-strength bolts in slip-critical connections with Class A

surfaces, the allowable load per bolt is 9.3 kips. If the capacity of the 10 ϫ 1⁄2-in flange is

developed at the allowable stress of 27 ksi, the number of bolts required in the connection

is 27 ϫ 5 / 9.3 ϭ 15. Use 16.

The deck plate is spliced to the girder with 7⁄8-in-dia. high-strength bolts. To meet girder

requirements, the pitch may vary from 3 to 51⁄2 in (Fig. 12.60d ). But the bolts also must

transmit the tensile forces from the deck plate to the girder when the plate acts as the top

flange of member III. The shear in the bolts from the girder compression is perpendicular

to the shear from the floorbeam tension. Hence, the allowable load per bolt decreases from

9.3 to 9.3 ϫ 0.707 ϭ 6.6 kips. With an average tensile stress in the deck plate of 6.2 ksi,

and a net area after deduction of holes of 35.6 Ϫ 10.2 ϭ 25.4 in2, the plate carries a tensile

force of 25.4 ϫ 6.2 ϭ 158 kips. Thus, to transmit this force, 158 / 6.6 ϭ 24 bolts are needed.

If a pitch of 3 in is used in the 95-in effective width of the plate, 31 bolts are provided. Use

a 3-in pitch for 4 ft on each side of every floorbeam.

The web connection to the girder must transmit both vertical shear, V ϭ 73.3 kips, and

bending moment. The latter can be computed from the stress diagram for the cross section

(Fig. 12.65a).

12.152

SECTION TWELVE

FIGURE 12.65 (a) Bending stresses in floorbeam at supports. (b)

Bolted web connection of floorbeam to girder.

M ϭ 1⁄2 ϫ 3⁄8(4.23 ϫ 3.07 ϫ 2.05 ϩ 20.7 ϫ 14.93 ϫ 9.95) ϭ 581 in-kips

Assume that the connection will be made with two rows of six bolts each, on each side

of the connection centerline (Fig. 12.65b). The polar moment of inertia of these bolts can

be computed as the sum of the moments of inertia about the x (horizontal) and y (vertical)

axes.

Ix ϭ 4(1.52 ϩ 4.52 ϩ 7.52) ϭ 315

Iy ϭ 12(1.5)2

ϭ 27

ϭ

J ϭ 342

Load per bolt due to shear is

Pv ϭ

73.3

ϭ 6.1 kips

12

Load on the outermost bolt due to moment is

Pm ϭ

581 ϫ 7.63

ϭ 12.95 kips

342

The vertical component of this load is

Pv ϭ

12.95 ϫ 1.5

ϭ 2.5 kips

7.63

and the horizontal component is

Ph ϭ

12.95 ϫ 7.5

ϭ 12.7 kips

7.63

The total load on the outermost bolt is the resultant

P ϭ ͙(6.1 ϩ 2.5)2 ϩ 12.72 ϭ 15.3 Ͻ 2 ϫ 9.3

For the web connection plates, try two plates 171⁄2 ϫ 5⁄16 in. They have a net moment of

inertia

BEAM AND GIRDER BRIDGES

Iϭ2

12.153

(5⁄16)17.53

Ϫ 50 ϭ 228 in4

12

To transmit the 581-in-kip moment in the web, they carry a bending stress of

ƒb ϭ

581 ϫ 8.75

ϭ 22.3 Ͻ 27 ksi

228

The assumed plates are therefore satisfactory if Grade 50W steel is used.

12.15.4

Design of Deck Plate

The deck plate is to be made of Grade 50W steel. This steel has a yield strength Fy ϭ

50 ksi for the 3⁄8-in deck thickness.

Stresses. Bending stresses in the deck plate as the top flange of ribs (member II), floorbeams (member III), and girders (member IV) are relatively low. Combining the stresses of

members II and IV yields 4.75 ϩ 9.73 ϭ 14.48 Ͻ 1.25 ϫ 27 ksi.

Deflection. The thickness of deck plate to limit deflection to 1⁄300 of the rib spacing can

be computed from Eq. 11.72. For a 16-kip wheel, assumed distributed over an area of

26 ϫ 12 ϭ 312 in2, the pressure, including 30% impact, is

p ϭ 1.3 ϫ 16 / 312 ϭ 0.0667 ksi

Required thickness with rib spacing a ϭ e ϭ 12 in is

t ϭ 0.07 ϫ 12(0.0667)1/3 ϭ 0.341 Ͻ 0.375 in

The 3⁄8-in deckplate is satisfactory.

12.16

CONTINUOUS-BEAM BRIDGES

Articles 12.1 and 12.3 recommended use of continuity for multispan bridges. Advantages

over simply supported spans include less weight, greater stiffness, smaller deflections, and

fewer bearings and expansion joints. Disadvantages include more complex fabrication and

erection and often the costs of additional field splices.

Continuous structures also offer greater overload capacity. Failure does not necessarily

occur if overloads cause yielding at one point in a span or at supports. Bending moments

are redistributed to parts of the span that are not overstressed. This usually can take place

in bridges because maximum positive moments and maximum negative moments occur with

loads in different positions on the spans. Also, because of moment redistribution due to

yielding, small settlements of supports have no significant effects on the ultimate strength

of continuous spans. If, however, foundation conditions are such that large settlements could

While analysis of continuous structures is more complicated than that for simple spans,

design differs in only a few respects. In simple spans, maximum dead-load moment occurs

at midspan and is positive. In continuous spans, however, maximum dead-load moment

occurs at the supports and is negative. Decreasing rapidly with distance from the support,

the negative moment becomes zero at an inflection point near a quarter point of the span.

positive, with a maximum about half the negative moment at the supports.

12.154

SECTION TWELVE

As for simple spans, live loads are placed on continuous spans to create maximum stresses

at each section. Whereas in simple spans maximum moments at each section are always

positive, maximum live-load moments at a section in continuous spans may be positive or

negative. Because of the stress reversal, fatigue stresses should be investigated, especially in

the region of dead-load inflection points. At interior supports, however, design usually is

governed by the maximum negative moment, and in the midspan region, by maximum positive moment. The sum of the dead-load and live-load moments usually is greater at supports

than at midspan. Usually also, this maximum is considerably less than the maximum moment

in a simple beam with the same span. Furthermore, the maximum negative moment decreases

rapidly with distance from the support.

The impact fraction for continuous spans depends on the length L, ft, of the portion of

the span loaded to produce maximum stress. For positive moment, use the actual loaded

length. For negative moment, use the average of two adjacent loaded spans.

Ends of continuous beams usually are simply supported. Consequently, moments in threespan and four-span continuous beams are significantly affected by the relative lengths of

interior and exterior spans. Selection of a suitable span ratio can nearly equalize maximum

positive moments in those spans and thus permit duplication of sections. The most advantageous ratio, however, depends on the ratio of dead load to live load, which, in turn, is a

function of span length. Approximately, the most advantageous ratio for length of interior

to exterior span is 1.33 for interior spans less than about 60 ft, 1.30 for interior spans between

When composite construction is advantageous (see Art. 12.1), it may be used either in

the positive-moment regions or throughout a continuous span. Design of a section in the

positive-moment region in either case is similar to that for a simple beam. Design of a

section in the negative-moment regions differs in that the concrete slab, as part of the top

flange, cannot resist tension. Consequently, steel reinforcement must be added to the slab to

resist the tensile stresses imposed by composite action.

Additionally, for continuous spans with a cast-in-place concrete deck, the sequence of

concrete pavement is an important design consideration. Bending moments, bracing requirements, and uplift forces must be carefully evaluated.

12.17

ALLOWABLE-STRESS DESIGN OF BRIDGE WITH

CONTINUOUS, COMPOSITE STRINGERS

The structure is a two-lane highway bridge with overall length of 298 ft. Site conditions

require a central span of 125 ft. End spans, therefore, are each 86.5 ft (Fig. 12.66a). The

typical cross section in Fig. 12.66b shows a 30-ft roadway, flanked on one side by a 21-inwide barrier curb and on the other by a 6-ft-wide sidewalk. The deck is supported by six

rolled-beam, continuous stringers of Grade 36 steel. Concrete to be used for the deck is

Class A, with 28-day strength ƒЈc ϭ 4,000 psi and allowable compressive stress ƒc ϭ 1,600

psi. Loading is HS20-44. Appropriate design criteria given in Sec. 11 will be used for this

structure.

Concrete Slab. The slab is designed to span transversely between stringers, as in Art. 12.2.

A 9-in-thick, two-course slab will be used. No provision will be made for a future 2-in

wearing course.

Stringer Loads. Assume that the stringers will not be shored during casting of the concrete

slab. Then, the dead load on each stringer includes the weight of a strip of concrete slab

plus the weights of steel shape, cover plates, and framing details. This dead load will be

referred to as DL and is summarized in Table 12.81.

### Tài liệu bạn tìm kiếm đã sẵn sàng tải về

16 Example - LRFD for Wide-Flange Column in a Multistory Rigid Frame

Tải bản đầy đủ ngay(0 tr)

×