AASHTO -Prestressed Beams AASHTO Examples

MAR 2007 P r e s t r e s se d I -  

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-1

This example illustrates the design of a pretensioned I-Beam for a two

B e a m D e si g n

span bridge without without skew. skew.

Ex a m p l e

Mn/DOT 72" beams.

The 130'-0" spans are supported with

Mn/DOT standard details and drawings for

diaphragms (B406, B812), railings (Fig. 5-397.117), and beams (Fig. 5397.517) are to be used with this example. This example contains the design of a typical interior beam at the critical sections in positive flexure, shear, and deflection. The superstructure superstructure consists of six beams spaced at 9'-0" centers. centers. A typical transverse transverse superstructure superstructure section section is provided in Figure 5.7.2.1.

A framing plan is provided in Figure 5.7.2.2.

The

roadway section is composed of two 12' traffic lanes and two 12' shoulders. A Type F railing is provided on each side side of the bridge and a 9" composite concrete concrete deck is used. used. End diaphragms (B812) (B812) are used used at each end of the bridge and interior diaphragms (B406) are used at the interior third points and at the pier.

F ig ig u r e 5 . 7 . 2 . 1  

MAR 2007

LRFD BRIDGE DESIGN

F ig ig u r e 5 . 7 . 2 . 2

DESIGN EXAMPLE 5-2

MAR 2007 A.

Materials

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-3

The modulus of elasticity for high strength concrete suggested by ACI Committee 363 is used for the the beam concrete. concrete.

The composite deck is

assumed to have a unit weight of 0.150 kcf for dead load computations and 0.145 kcf kcf for elastic modulus computations. computations. The beam concrete is assumed to have a unit weight of 0.155 kcf for dead load computations. The material and geometric parameters used in the example are shown in Table 5.7.2.1: Table 5.7.2.1 Material Properties Material Parameter

  e   t   e   r   c   n   o    C

Prestressed Beam

Deck

′ at transfer f ci

7 ksi

---

f c at 28 days

8.0 ksi

4 ksi

Eci at trans ransffer

)

′ + 1000 f ci

---

= 4347 ksi 1265 ⋅ f c′ + 1000

33,000 ⋅ (0.145)1.5⋅ f c′

= 4578 ksi

= 3644 ksi

fy for rebar rebar

60 ksi

60 ksi

fpu for for stra strand nd

270 270 ksi ksi

-----

Es for rebar rebar

29,000 29,000 ksi

29,000 29,000 ksi

Ep for strand strand

28,500 28,500 ksi

---

Ec at 28 day days s

   l   e   e   t    S

(1265 ⋅

Strand type

0.6 inch diameter 270 ksi, low relaxation

---

B. Determine

The beams are designed to act compositely with the deck on simple

Cross-Section

spans. The deck consists consists of a 7 inch thick concrete concrete slab with a 2 inch

Properties for a

wearing course. course.

Typical I nterior

beams are designed assuming the full 9 inches of thickness is placed in a

B e am

single pour. A 1 /2 inch of wear is assumed. A thickness of 81 /2 inches is

For simplicity and in order to be conservative, conservative, the

used for composite section section properties. The haunch or or stool is assumed assumed to have an average thickness of 2 1 /2 inches for dead load computations and 11 /2 inches for section property computations. [4.6.2.6.1]

The effective flange width, b e , is the smallest of: 1 1) 1 /4 x Effective Span Length = ⋅ 130 ⋅ 12 = 390.0 in 4 2) 12 x Slab Thickness + 1 /2 Top Flange Width

= 12 ⋅ 8.5 +

30 = 117.0 in 2

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-4

3) Average Beam Spacing = 108.0 in

GOVERNS

The modular ratio of the deck concrete to the beam concrete is: n=

Ecdeck 3644 = = 0.796 Ecbeam 4578

This results in a transformed effective flange width of: b etrans = 0.796 ⋅ (108.0) = 86.0 in Properties for an interior beam are given in Table 5.7.2.2. Table 5.7.2.2 Cross-Section Properties Parameter

Non-composite Section

Composite Section

Height of section, h

72 in

82.0 in

Deck thickness

---

8.5 in

Average stool thickness

---

Effective flange width, be

---

Area, rea, A

786 786 in 2

1553 in 2

Momen Momentt of inerti inertia, a, I

547,92 547,920 0 in 4

1,235 ,235,0 ,000 00 in 4

Centroidal axis height, y

35.60 in

56.29 in

Bottom section modulus, Sb

15,390 in 3

21,940 in 3

Top section modulus, St

15,050 in 3

48,040 in 3  (beam concrete) 60,350 in 3  (deck concrete)

Top of prestressed beam

15,050 in 3

78,610 in 3

1.5 in (section properties) 2.5 in (dead load) 108.0 in (deck concrete) 86.0 in (beam concrete)

C. Sh ear Forces

Three load combinations will be considered; Strength I, Service I, and

and Bending

Service III. As a result of of the simple span configuration, configuration, only maximum maximum

Moments

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-5

γ p  values need to be considered. Load effects related to settlement, thermal effects, water load, or stream pressure will not be considered. It shall be assumed that traffic can be positioned anywhere between the barriers. Number of design lanes =

distance between barriers 48 = =4 design lane width 12

[3.6.2]

Dynamic load allowance IM = 33%

[ 4 .6 .2 . 2 ]

1. Determine Live Load Distribution Factors

Designers should note that the approximate distribution factor equations include the multiple presence factors. [4.6.2.2.2]

Distribution Factor for Moment – Interior Beams

LRFD Table 4.6.2.2.1-1 lists the common deck superstructure types for which approximate live load distribution equations have been assembled. The cross section for this design example is Type (k). To ensure that the approximate distribution equations can be used, several parameters need to be checked. 1) 3.5 ft ≤ beam spacing = 9.0 ft ≤ 16.0 ft

OK

2) 3.5 in ≤ slab thickness = 8.5 in ≤ 12.0 in

OK

3) 20 ft ≤ span length = 130 ft ≤ 240 ft

OK

4) 4 ≤ number of beams = 6

OK

The distribution factor equations use a K g  factor that is defined in LRFD Article 4.6.2.2.1.

η=

E c beam Ec deck

=

4578 = 1.256 3644

e g = ( deck centroid ) − ( beam centroid ) = 77.75 − 35.60 = 42.15 in 2 2 K g = η ⋅ I + A ⋅ (e g ) = 1.256 ⋅ 547,920 + 786 ⋅ (42.15) = 2.442  x 10 6

One design lane loaded:

⎛  S  ⎞ gM = 0.06 + ⎜ ⎟ ⎝ 14 ⎠

0.4

⎛ 9.0 ⎞ gM = 0.06 + ⎜ ⎟ ⎝  14  ⎠

⎛ S ⎞ ⋅⎜ ⎟ ⎝ L  ⎠

0.4

0.3

⎛  90  ⎞ ⋅⎜ ⎟ ⎝ 130 ⎠

⎛   ⎞ K ⎟ ⋅⎜ ⎜ 12 ⋅ L ⋅ t 3 ⎟ s  ⎠ ⎝  0.3

0.1

⎛  2.442 x 106  ⎞ ⎟ ⋅⎜ ⎜ 12 ⋅ 130 ⋅ 8.53 ⎟ ⎝   ⎠

0.1

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-6

gM  = 0.473  lanes/beam

Two or more design lanes loaded:

⎛  S  ⎞ gM = 0.075 + ⎜ ⎟ ⎝ 9.5 ⎠

0.6

⎛ S ⎞ ⋅⎜ ⎟ ⎝ L ⎠

0.2

⎛ 9.0 ⎞ gM = 0.075 + ⎜ ⎟ ⎝ 9.5 ⎠

0.6

⎛  90  ⎞ ⋅⎜ ⎟ ⎝ 130 ⎠

⎛  K g  ⎞ ⎟ ⋅⎜ ⎜ 12 ⋅ L ⋅ t 3 ⎟ s  ⎠ ⎝ 

0.2

0.1

⎛  2.442 x 106  ⎞ ⎟ ⋅⎜ ⎜ 12 ⋅ 130 ⋅ 8.53 ⎟ ⎝   ⎠

0.1

gM  = 0.698  lanes/beam

[4.6.2.2.2d]

Distribution Factor for Moment - Exterior Beams

LRFD Table 4.6.2.2.2d-1 contains the approximate distribution factor equations for exterior beams.

Type (k) cross-sections have a deck

dimension check to ensure that the approximate equations are valid. The distance from the inside face of barrier to the centerline of the fascia beam is defined as de . For the example this distance is: de = 24 − (2.5 − 9.0) = 1.50 ft

The check to use the approximate equations is:

−1.0 ft ≤ de = 1.50 ft ≤ 5.5 ft One design lane loaded:

F ig ig u r e 5 . 7 .2 . 3

OK

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-7

The lever rule shall be used to determine the live load distribution factor for one lane. The fascia beam live load distribution factor is found by summing reactions about the first interior beam: W1 = W2 = 0.5 lanes

⎛ W1 ⋅ L 1 + W2 ⋅ L 2  ⎞ ⎛ 0.5 ⋅ 8.5 + 0.5 ⋅ 2.5 ⎞ ⎟⎟ = 1.2 ⋅ ⎜ ⎟ S 9.0 ⎝   ⎠ ⎝   ⎠

gM = 1.2 ⋅ ⎜⎜

gM  = 0.733  lanes/beam

Two or more design lanes loaded: The distribution factor is equal to the factor “e” multiplied by the interior girder distribution factor for two or more lanes de = 24.0 − 22.5 = 1.5 ft

⎛  d  ⎞ ⎛ 1.5 ⎞ e = 0.77 + ⎜⎜ e ⎟⎟ = 0.77 + ⎜ ⎟ = 0.935 ⎝ 9.1 ⎠ ⎝ 9.1 ⎠ gM = e ⋅ gint = 0.935 ⋅ 0.698 = 0.653  lanes/beam

[4.6.2.2.2e]

Skew Factor No correction is necessary for a skew angle of zero.

[4.6.2.2.3]

Distribution Factor for Shear – Interior Beams

[ 4 . 6 .2 . 2 .3 a ]

LRFD Table 4.6.2.2.3a-1 can be used. One design lane loaded:

⎛  S  ⎞ ⎛  9.0  ⎞ ⎟ = 0.36 + ⎜ ⎟ = 0.720  lanes/beam ⎝ 25.0 ⎠ ⎝ 25.0 ⎠

gV = 0.36 + ⎜

Two or more design lanes loaded: 2

2

⎛  S  ⎞ ⎛  S  ⎞ ⎛ 9.0 ⎞ ⎛ 9.0 ⎞ ⎟−⎜ ⎟ = 0.884  lanes/beam ⎟ − ⎜ ⎟ = 0.2 + ⎜ ⎝ 12 ⎠ ⎝ 35 ⎠ ⎝  12  ⎠ ⎝  35  ⎠

gV = 0.2 + ⎜

[4.6.2.2.3b]

Distribution Factor for Shear – Exterior Beams One Design Lane Loaded: The lever rule shall be used which results in the same factor that was computed for flexure and is equal to 0.733 lanes/beam Two or more design lanes loaded:

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-8

⎛ d  ⎞ ⎛ 1.5 ⎞ e = 0.6 + ⎜ e ⎟ = 0.6 + ⎜ ⎟ = 0.750 ⎜ 10 ⎟ ⎝  10  ⎠ ⎝   ⎠

The exterior beam shear distribution factor for two or more design lanes is determined by modifying the interior distribution factor: gV = e ⋅ gint = 0.750 ⋅ 0.884 = 0.663  lanes/beam

[ 4 . 6 .2 . 2 .3 c ]

Skew Factor No correction is necessary for a skew angle of zero.

[2.5.2.6.2]

Distribution Factor for Deflection

[ T a b l e 3 . 6 . 1 .1 .1 . 2 - 1 ]

The distribution factor for checking live load deflections assumes that the entire cross section participates in resisting the live load. The minimum Multiple Presence Factor (MPF) used by Mn/DOT when checking live load deflection is 0.85. The deflection distribution factor is: gD =

(# of lanes) ⋅ (MPF ) 4 ⋅ 0.85 = = 0.567  lanes/beam (# of beam lines ) 6

Table 5.7.2.3 contains a summary of the live load distribution factors. Table 5.7.2.3 Distribution Factor Summary (lanes per beam) Loading   r   o  m    i   r   a   e   t   e    B   n    I

  r   o    i   m   r   a   e   t   e   x   B    E

[ 1.3.3 – 1.3.5]

Flexure

Shear

One Design Lane

0.473

0.720

Two or More Design Lanes

0.698

0.884

Deflection

0.567

-

One Design Lane

0.733

0.733

Two or More Design Lanes

0.653

0.663

Deflection

0.567

-

2. Load Modifiers The following load modifiers will be used for this example: Strength

Service

Fatigue

Ductility

ηD  

1.0

1.0

1.0

Redundancy

ηR  

1.0

1.0

1.0

Importance

ηI  

1.0

n/a

n/a

η = ηD ⋅ ηR ⋅ ηI

1.0

1.0

1.0

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-9

3. Dead and and Live Load Summary Summary

(

)

Beam Selfweight = (786 / 144) ⋅ 0.155 kip/ft 3 = 0.846 kip/ft Stool Weight = (2.5 ft ) ⋅ (0.208 ft ) ⋅ 0.150 kip/ft 3 = 0.078 kip/ft Deck Weight = (9.0 ft ) ⋅ (0.75 ft ) ⋅ 0.150 kip/ft 3 = 1.013 kip/ft Future Wearing Surface = 0.020 kip/ft 2 ⋅ (48 ft ) ⋅ (1 / 6 ) = 0.160 kip/ft Barrier Weight = 2 ⋅ (0.439 kip/ft ) ⋅ (1 / 6 ) = 0.146 kip/ft Diaphragm Weight ≅ (9.0) ⋅ (2 ) ⋅ (0.0103 ) + 0.0149

⎛ 17 ⎞ ⎛ 0.5 ⎞ + 2 ⋅ (4.17) ⋅ ⎜ ⎟ ⋅ ⎜ ⎟ ⋅ (0.490) = 0.561 kip ⎝ 12 ⎠ ⎝  12  ⎠ The bending moments and shears for the dead and live loads were obtained with a line girder model of the bridge. They are summarized summarized in Tables 5.7.2.4 and 5.7.2.5. Table 5.7.2.4 Shear Force Summary (kips/beam) Critical 0.1 Strand 0.2 0.3 0.4 0.5 Brg Brg Trans Shear Span Dev Span Span Span Span Load Type/Combination Type/Combination CL Face Point Point Point Point Point Point Point Point (0.0') (0.63') (2.38') (5.8') (13.0') (13.6') (26.0') (39.0') (52.0') (65.0')

  s    d   a   o    L    d   a   e    D

  s    d   a   o    L   e   v    i    L

Selfweight

55

54

53

50

44

44 44

33

22

11

0

Stool

5

5

5

5

4

4

3

2

1

0

Deck

66

65

63

60

53

52

40

26

13

0

FWS

10

10

10

9

8

8

6

4

2

0

Barrier

9

9

9

9

8

8

6

4

2

0

Diaphragms

1

1

1

1

1

1

1

1

0

0

Total

146

144

141

134

118

113

89

59

29

0

Uniform Lane

37

36

35

34

30

30

24

18

13

9

Truck with DLA

79

78

77

75

70

70

62

53

45

36

Total

116

114

112

109

100

100

86

71

58

45

386

380

372

358

323

316

262

198

138

79

262

258

253

243

218

213

175

130

87

45

239

235

231

221

198

193

158

116

75

36

Strength I Load Comb (1.25 ⋅ DL + 1.75 ⋅ LL) Service I Load Comb (1.00 ⋅ DL + 1.00 ⋅ LL) Service III Load Comb (1.00 ⋅ DL + 0.80 ⋅ LL)

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-10

Table 5.7.2.5 Bending Moment Summary (kip-ft/beam) Load Type/Combination Type/Combination

Brg CL

Brg Face

Trans Point

(0.0') (0.63') (2.38')

  s    d   a   o    L    d   a   e    D

  s    d   a   o    L   e   v    i    L

Critical Shear Point (5.8')

0.1 Span

Strand Dev

0.2 Span

0.3 Span

0.4 Span

0.5 Span

Point Point Point Point Point* Point (13.0') (13.6') (26.0') (39.0') (52.0') (65.0')

Selfweight

0

34

128

305

643

670

1144

1501

1716

1787

Stool

0

3

12

28

59

62

105

138

158

165

Deck

0

41

154

365

770

802

1370

1798

2054

2140

Diaphragms

0

0

1

3

7

8

15

22

24

24

Total DC1

0

78

297

701

1479

1542

2634

3459

3952

4116

Barrier

0

6

22

53

111

116

197

259

296

308

FWS

0

7

24

58

122

127

216

284

324

338

Total DC2

0

13

46

111

233

243

413

543

620

646

Total (DC1+DC2)

0

91

343

812

1712

1785

3047

4002

4572

4762

Uniform Lane

0

18

68

161

340

354 354

604

793

906

944

Truck with DLA

0

39

145

343

719

749

1265

1638

1857

1912

Total

0

57

213

504

1059

1103

1869

2431

2763

2856

0

214

802

1897

3993

4162

7080

9257

0

148

556

1316

2771

2888

4916

6433

7335

7618

0

137

513

1215

2559

2667

4542

5947

6782

7047

   1    C    D

   2    C    D

Strength I - Load Comb (1.25 ⋅ DL + 1.75 ⋅ LL) Service I - Load Comb (1.00 ⋅ DL + 1.00 ⋅ LL) Service III – Load Comb (1.00 ⋅ DL + 0.80 ⋅ LL)

10,550 10,951

* Drape point for strands.

D.

Design

Prestressing

Typically the tension at the bottom of the beam at midspan dictates the required level of prestressing. 1. Estimate Required Prestress Service III load combination shall be used Bottom of beam stress:

⎛ M  ⎞ ⎛ M  ⎞ ⎛ M ⋅ 0.8 ⎞ ⎟ = ⎜ DC1 ⎟ + ⎜⎜ DC2 ⎟⎟ + ⎜⎜ LL ⎜ Sgb ⎟ ⎝  S cb  ⎠ ⎝  S cb  ⎠⎟ ⎝   ⎠ ⎛ 4116 ⋅ 12 ⎞ ⎛ 646 ⋅ 12 ⎞ ⎛ 2856 ⋅ 12 ⋅ 0.8 ⎞ = ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = 4.81 ksi ⎝  15,390  ⎠ ⎝  21,940  ⎠ ⎝  21,940  ⎠

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-11

As a starting point, the total prestress losses will be assumed to be 30%. This results in an effective prestress of f pe = 0.75 ⋅ f pu ⋅ (1 − 0.30) = 0.75 ⋅ 270 ⋅ 0.70 = 141.8 ksi Strands are are typically placed placed on a 2" grid.

The bottom flange flange of a 72"

beam can hold a maximum of 48 strands. The centroid of a 48 strand pattern would be

⎡ Σ (# of strands) ⋅ (γ of strands)⎤ ⎥ (total # of strands) ⎣ ⎦

y str = ⎢

⎡10 ⋅ (2 + 4 + 6 ) + (4 ⋅ 8 ) + 2 ⋅ (3 + 5 + 7 + 9 + 11 + 13 + 15 )⎤ =⎢ ⎥ = 5.79 in 48 ⎣ ⎦ Using the centroid of this group as an estimate of the strand pattern eccentricity results in e 48 = y g − 5.79 = 35.60 − 5.79 = 29.81 in The area of a 0.6" diameter 7-wire strand is 0.217 in2 The axial compression produced by the prestressing strands is P = A s ⋅ f pe = (# of strands) ⋅ (0.217 ) ⋅ (141.8 ) The internal moment produced by the prestressing strands is Mp / s = A s ⋅ f pe ⋅ e 48 = (# of strands) ⋅ 0.217 ⋅ 141.8 ⋅ 29.81 The allowable tension after losses = 0.19 ⋅

f c′ = 0.19 ⋅

8 = 0.54 ksi

This moment and the axial compression from the prestress must reduce the bottom flange tension from 4.81 ksi tension to a tension of 0.54 ksi or Required f pe = 4.81 − 0.54 = 4.27 ksi Using the fact that f pe =

P M + A S

One can estimate the required number of strands:

MAR 2007

LRFD BRIDGE DESIGN ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ 4.27 ⎢ ⎥  ⎞ ⎥ 1 29 . 81 ⎢ ⎛  ⎜ ⎟ ⎢⎜ A + S ⎟⎥ gb  ⎠ ⎥⎦ ⎢⎣ ⎝  g = (0.217 + 141.8)

DESIGN EXAMPLE 5-12

⎡ ⎤ ⎢ ⎥ ⎢ ⎥ 4.27 ⎢ ⎥  ⎞ ⎥ 1 29 . 81 ⎢ ⎛  ⎜ ⎢ ⎜ 786 + 15,390 ⎟⎟ ⎥ ⎢⎣ ⎝   ⎠ ⎥⎦ = 43.2 strands (0.217 + 141.8)

A strand pattern with 44 strands should be tried. After reviewing Bridge Details Part II Figure 5-397.517, a 44 strand draped strand pattern pattern was selected. selected. Also, the drape drape points were chosen chosen to be at 0.40L = 52.0 ft from the centerline of bearing locations. locations. trial strand pattern is shown in Figure 5.7.2.4. The properties of this strand pattern at midspan are:

⎡10 ⋅ (2 + 4 + 6) + 2 ⋅ (3 + 5 + 7 + 8 + 9 + 11 + 13) ⎤ ⎥ = 5.27 in 44 ⎣ ⎦

y strand = ⎢

e strand = y b − y strand = 35.60 − 5.27 = 30.33 in Section Modulus at the strand pattern centroid is S gps =

Ig e strand

=

547,920 = 18,065 in3 30.33

F ig ig u r e 5 . 7 .2 . 4

The

MAR 2007 [5.9.5]

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-13

2. Prestress Losses Prestress losses are computed using the refined method.

[5.9.5.4.4b]

Initial Relaxation Loss It shall be assumed that the prestress is transferred 18 hours after stressing t=

18 = 0.75 days 24

f pj = 0.75 ⋅ f pu = 0.75 ⋅ 270 = 202.50 ksi

[5.9.5.2.3]

Δf pR1 =

⎞ f pj log (24 ⋅ t ) ⎛  ⋅ ⎜⎜ − 0.55⎟⎟ ⋅ f pj 40 ⎝ f pγ  ⎠

Δf pR1 =

log (24 ⋅ 0.75) ⎛  202.50  ⎞ ⋅⎜ − 0.55 ⎟ ⋅ 202.50 = 1.80 ksi 40 ⎝ 0.9 ⋅ 270  ⎠

Elastic Shortening Loss The alternative equation presented in the LRFD C5.9.5.2.3a shall be used. 2

Δf pES =

A ps f pbt ⋅ Ig + em A g − em Mg A g

(

2

)

A ps Ig + em A g +

A g Ig E ci Ep

A ps = (# of strands) ⋅ (strand area) = 44 ⋅ 0.217 = 9.55 in2 f pbt = f pj − Δf pR1 = 202.50 − 1.80 = 200.70 ksi em = e strand = 30.33 in A g Ig E ci Ep

=

786 (547,920) (4347) = 65,687,764 in6 28,500

2 2 A ps Ig + em A g = 9.55 [547,920 + (30.33) (786 )] = 12,137,748 in6

Δf pES = [5.9.5.4.2]

200.70 ⋅ (12,137,748) − 30.33 (1787 ) (12 ) (786 ) = 24.73 ksi 12,137,748 + 65,687,764

Shrinkage Loss Use an average humidity for North Dakota of 70%.

Δf pSR = 17.0 − (0.150 ⋅ H) = 17.0 − (0.150 ⋅ 73) = 6.05 ksi [5.9.5.4.3]

Creep Loss Non-composite dead load moment excluding selfweight MDC 1− SW = (4116 − 1787) = 2329  kip-ft

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-14

Composite dead load moment, MDC 2 = 646 kip-ft

⎛   ⎞ ⎛  y − y g + e strand ⎞ e ⎟ Δfcdp = ⎜⎜MDC1− SW ⋅ strand ⎟ + ⎜MDC 2 ⋅ cg ⎟ ⎜ ⎟ I I g cg ⎝   ⎠ ⎝   ⎠ ⎛  30.33  ⎞ ⎛ 56.29 − 35.60 + 30.33 ⎞ = ⎜⎜ 2329.12 ⋅ ⎟⎟ + ⎜⎜ 646.12 ⋅ ⎟⎟ 547,920 ⎠ ⎝  1,235,000 ⎝   ⎠ = 1.87 ksi f cgp =

P ⋅e M Pi + i strand − sw Ag S gps S gps

Pi = f pj − Δf pR1 − Δf pES A ps = (202.50 − 1.80 − 24.73) (9.55) = 1681 kips f cgp =

1681 1681 (30.33) 1787 (12 ) + − = 3.77 ksi 786 18,065 18065

Δf pCR = 12 ⋅ f cgp − 7 ⋅ Δf cdp = (12 ⋅ 3.77) − (7 ⋅ 1.87) = 32.15 ksi [ 5 . 9 .5 . 4 .4 c ]

Relaxation Loss After Transfer

Δf pR2 = 0.30 ⋅ 20 − 0.4 ⋅ Δf pES − 0.2 ⋅ Δf pSR + Δf pCR = 0.30 ⋅ [20 − 0.4 ⋅ 24.73 − 0.2 ⋅ (6.05 + 32.15)] = 0.74 ksi [ 5 .9 .5 . 1 ]

Total Losses

ΔTL = Δf pES + Δf pSR + Δf pCR + Δf pR2 = 24.73 + 6.05 + 32.15 + 0.74 = 63.67 ksi f pe = f pj − ΔTL = 202.50 − 63.67 = 138.83 ksi prestress loss percentage =

[ 5 .9 .4 . 1 ]

ΔTL f pj

⋅ 100 =

63.67 ⋅ 100 = 31.4% 202.50

3. Stresses at Transfer (compression (compression +, tension tension -) Stress Limits for P/S Concrete at Release Compression in the concrete is limited to:

′ = 0.60 ⋅ 7.0 = 4.20 ksi 0.60 ⋅ f ci Tension in the concrete is limited to: The minimum of − 0.0948 ⋅

′ = −0.0948 ⋅ 7.0 = −0.25 ksi f ci or

Tension limit = -0.20 ksi

−0.20 ksi

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-15

Check Release Stresses at Drape Point (0.40 Point of Span) Pi ⋅ e strand = 1681 ⋅ 30.33 = 50,985  kip-in

⎛  P  ⎞ ⎛ P ⋅ e strand ⎞ ⎛ 1681 ⎞ ⎛ 50,985 ⎞ ⎟=⎜ ⎟⎟ Top stress due to P/S = ⎜ i ⎟ − ⎜ i ⎟ − ⎜⎜ ⎜ A g ⎟ ⎜ S gt ⎟ ⎝   ⎠ ⎝   ⎠ ⎝  786  ⎠ ⎝ 15,050 ⎠ = −1.25 ksi ⎛  P  ⎞ ⎛ 1681 ⎞ ⎛ 50,985 ⎞ P ⋅e ⎜⎜ ⎟⎟ Bottom Stress due to P/S = ⎜ i + i strand ⎟ = ⎜ ⎜ Ag ⎟ ⎝  786  ⎠⎟ + ⎝  S 15 , 390  ⎠ gb ⎝   ⎠ = 5.45 ksi Selfweight moment at drape point = Msw0.40 = 1716  kip-ft

⎛ M  ⎞ ⎛ 1716 ⋅ 12 ⎞ Top stress due to selfweight = ⎜ sw0.40 ⎟ = ⎜⎜ ⎜ Sgt ⎟ ⎝  15,050  ⎠⎟⎟ = 1.37 ksi ⎝   ⎠ ⎛ M  ⎞ ⎛ 1716 ⋅ 12 ⎞ = −1.34 ksi Bottom stress due to selfweight = ⎜ sw0.40 ⎟ = ⎜⎜ ⎜ S gb ⎟ ⎝  15,390 ⎟⎟⎠ ⎝   ⎠ Top stress at drape point = −1.25 + 1.37 = 0.12 ksi < −0.20 ksi

OK

Bottom stress at drape point = 5.45 − 1.34 = 4.11 ksi < 4.20 ksi

OK

Check Release Stresses at End of Beam The strands need to be draped to raise the eccentricity of the prestress force and limit the potential for cracking the top of the beams. Centroid of strand pattern at the end of the beams:

⎡10 ⋅ (2 + 4 + 6 ) + 2 ⋅ (8 + 57 + 59 + 61 + 63 + 65 + 67 )⎤ ⎥ = 20.0 in 44 ⎣ ⎦

y strand = ⎢

The eccentricity of the strand pattern is: e strand = y b − y strand = 35.60 − 20.00 = 15.60 in The internal prestress moment is: Pi ⋅ e strand = 1681 ⋅ 15.60 = 26,224  kip-in

⎛  P  ⎞ ⎛ P ⋅ e  ⎞ ⎛ 1681 ⎞ ⎛ 26,224 ⎞ − ⎜⎜ Top stress at end = ⎜ i ⎟ − ⎜ i strand ⎟ = ⎜ ⎟⎟ ⎜ A g ⎟ ⎜ S gt ⎟ ⎝  786  ⎠⎟ ⎝  15 , 050  ⎠ ⎝   ⎠ ⎝   ⎠ = 0.40 ksi < −0.20 ksi

OK

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-16

⎛  P  ⎞ ⎛ P ⋅ e  ⎞ ⎛ 1681 ⎞ ⎛ 26,224 ⎞ − ⎜⎜ ⎟⎟ Bottom stress at end = ⎜ i ⎟ − ⎜ i strand ⎟ = ⎜ ⎜ Ag ⎟ ⎜ Sgb ⎟ ⎝  786  ⎠⎟ ⎝  15 , 390  ⎠ ⎝   ⎠ ⎝   ⎠ = 3.84 ksi < 4.20 ksi

OK

For simplicity, the stresses were checked at the end of the beam assuming the full prestress force was was effective.

The check could have

been made at the transfer point (60 strand diameters away from the end of the beam). [ 5 .9 .4 . 2 ]

4. Stresses at Service Loads (compression (compression +, tension tension -) Stress Limits for P/S Concrete After All Losses Compression in the concrete is limited to (Service I Load Combination): 0.45 ⋅ f c′ = 0.45 ⋅ 8.0 = 3.60 ksi (for prestress and permanent loads) Check the bottom stress at end of beam and the top stress at midspan against this limit. 0.40 ⋅ f c′ = 0.40 ⋅ 8.0 = 3.20 ksi (for live load and 1 /2 of prestress and permanent loads) Check the top stress at midspan against this limit. 0.60 ⋅ Q w ⋅ f c′ = 0.60 ⋅ 1.0 ⋅ 8.0 = 4.80 ksi (for live load, prestress, permanent loads, and transient loads) Check the top stress at midspan against this limit. Tension in the concrete is limited lim ited to (Service III Load L oad Combination):

− 0.19 ⋅ f c′ = −0.19 ⋅ 8.0 = −0.54 ksi Check the bottom stress at midspan against this limit. Check Stresses at Midspan After Losses: Let Pe = A ps ⋅ f pe = 9.55 ⋅ 138.83 = 1326 kips Bottom stress

⎛ MDC 1 ⎞ ⎛ MDC 2  ⎞ ⎛ MLL ⋅ 0.8 ⎞ ⎛  Pe  ⎞ ⎛ Pe ⋅ e strand ⎞ ⎟+⎜ ⎟−⎜ ⎟ ⎟ ⎜ ⎟⎟ − ⎜ =⎜ ⎜ S ⎜ Scb ⎟ + ⎝  ⎜ S gb ⎟ ⎝  ⎜ ⎟ ⎜ ⎟ A S cb gb  ⎠ ⎝  g  ⎠ ⎝   ⎠ ⎝   ⎠  ⎠ ⎛ 4116 ⋅ 12 ⎞ ⎛ 646 ⋅ 12 ⎞ ⎛ 2856 ⋅ 12 ⋅ 0.8 ⎞ ⎛ 1326 ⎞ ⎛ 1326 ⋅ 30.33 ⎞ = ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ − ⎜ ⎟⎟ ⎟ − ⎜⎜ ⎝  15,390  ⎠ ⎝  21,940  ⎠ ⎝  21,940  ⎠ ⎝  786  ⎠ ⎝  15,390  ⎠ OK = −0.51 ksi < −0.54 ksi

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-17

Top stress due to all loads

⎛  P  ⎞ ⎛ P ⋅ e  ⎞ ⎛ MDC 1 ⎞ ⎛ MDC 2 + MLL  ⎞ ⎟+⎜ ⎟ = ⎜⎜ e ⎟⎟ − ⎜⎜ e strand ⎟⎟ + ⎜⎜ ⎟ ⎜ ⎟ A S S S g gt gt gtc ⎝   ⎠ ⎝   ⎠ ⎝   ⎠ ⎝   ⎠ ⎛ 1326 ⎞ ⎛ 1326 ⋅ 30.33 ⎞ ⎛ 4116 ⋅ 12 ⎞ ⎡ (646 + 2856 ) ⋅ 12 ⎤ =⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎢ ⎟ − ⎜⎜ ⎥ 78,610 ⎝  786  ⎠ ⎝  15,050  ⎠ ⎝  15,050  ⎠ ⎣ ⎦ = 2.83 ksi < 4.80 ksi

OK

By inspection, top stress due to prestress and permanent loads is less than 3.60 ksi.

Also by inspection, top stress due to live load plus

1

 /2 prestress and permanent loads is less than 3.20 ksi.

Check the Compression Stresses at End of Beam After Losses Bottom flange stress due to prestress and permanent loads Pe Ag [5.5.4]

+

Pe ⋅ e strand S

=

1326 1326 ⋅ 15.60 + = 3.03 ksi < 3.60 ksi 786 15,390

OK

5. Flexure – Strength Limit State Resistance factors at the strength limit state are:

φ = 1.00  for flexure and tension φ = 0.90  for shear and torsion φ = 1.00  for tension in steel in anchorage zones Strength I design moment is 10,951 kip-ft at midspan. Rectangular beam behavior shall be checked by comparing the tensile capacity of the strands and the compression capacity of the deck. Tensile strength of 44 strands is: 44 ⋅ 0.217 ⋅ 270 = 2578 kips Maximum compressive force generated by the deck is 0.85 ⋅ f c′ ⋅ b ⋅ t s = 0.85 ⋅ 4.0 ⋅ 108 ⋅ 8.5 = 3121 kips The strands have less capacity than the deck, so assume a rectangular cross section.

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-18

From previous calculations, distance to strand centroid from bottom of the beam is: y strand = 5.27 in [5.7.3.1.1]

⎛  ⎜ ⎝ 

f py  ⎞

243 ⎞ ⎟ = 2 ⋅ ⎛  1.04 − ⎜ ⎟ = 0.280 f pu ⎠⎟ 270 ⎠ ⎝  dp = ( beam height ) + stool + deck − y strand

k = 2 ⋅ ⎜1.04 −

= 72 + 1.5 + 8.5 − 5.27 = 76.73 in [5.7.3.1.1-4]

⎡ ⎤ ⎢ ⎥ (Aps ⋅ f pu ) ⎢ ⎥ c=⎢ ⎥ f pu  ⎞ ⎥ ⎢ ⎛  ⎜ ⎟ ⎢ ⎜ 0.85 ⋅ f c′ ⋅ β1 ⋅ b ⋅ k ⋅ A ps ⋅ d ⎟ ⎥ p  ⎠ ⎥⎦ ⎢⎣ ⎝  ⎛   ⎞ ⎜ ⎟ 9.55 ⋅ 270 ⎜ ⎟ = 8.02 in = ⎜ 270 ⎟ ⎜ 0.85 ⋅ 4.0 ⋅ 0.85 ⋅ 108 + 0.28 ⋅ 9.55 ⋅ ⎟ 76.73 ⎠ ⎝  ⎛  c  ⎞⎟ 8.02  ⎞ ⎛  = 270 ⋅ ⎜1 − 0.28 ⋅ fps = f pu = ⎜1 − k ⋅ ⎟ = 262.10 ksi ⎜ ⎟ d 76 . 73 ⎝   ⎠ p  ⎠ ⎝  a = β1 ⋅ c = 0.85 ⋅ 8.02 = 6.82 in Internal lever arm between compression and tension flexural force components: dp −

a 6.82 = 76.73 − = 73.32 in 2 2

Mn = A ps ⋅ f ps ⋅ 73.32 = 9.55 ⋅ 262.10 ⋅ 73.32 = 18,524  kip-in

= 15,294  kip-ft φ Mn = 1.0 ⋅ 15,294 = 15,294  kip-ft > Mu = 10,951 kip-ft [5.7.3.3.1]

OK

6. Limits of Reinforcement Maximum Reinforcement The depth of the flexural compressive block is compared to the depth of the steel centroid to verify adequate ductility. c 8.02 = = 0.10 < 0.42 dp 76.73

OK

MAR 2007 [5.7.3.3.2]

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-19

Minimum Reinforcement f c′ = 0.24 ⋅

f r = 0.24 ⋅ f peb =

=

8.0 = 0.68 ksi

Pe P ⋅e + e strand Ag S gb 1326 1326 ⋅ 30.33 + = 4.30 ksi 786 15,390

⎛ S cgb  ⎞ − 1⎟ Mcr = (f r + f peb ) ⋅ S cgb − MDC 1 ⋅ ⎜ ⎜ S gb ⎟ ⎝   ⎠ ⎛ 21,940  ⎞ = (0.68 + 4.30) ⋅ 21,940 − (4116 ⋅ 12 ) ⋅ ⎜⎜ − 1⎟⎟ = 88,240  kip-in ⎝ 15,390 ⎠ = 7353 kip-ft 1.2 ⋅ Mcr = 1.2 ⋅ 7353 = 8824 < 15,294 kip-ft provided

E.

Design

R ei n f o r ce m e n t f o r Shear

OK

1. Vertical Shear Design Determine d v and Critical Section for Shear A theta angle of 26 degrees degrees shall be assumed. assumed. If the theta angle ends up flatter, the critical section will move towards midspan and the shear

[5.8]

demand will reduce. The effective effective shear shear depth depth dv shall be determ determined ined at the critic critical al section section for shear. dv = dp −

a 2

The effective shear depth is no less than: dv ≥ 0.72 ⋅ h = 0.72 ⋅ (72 + 1.5 + 8.5) = 59.04 in The internal face is assumed to be at the inside edge of the 15 inch sole plate. The critical section section will be at least 66.54 inches (59.04 + 15 / 2)  or 5.55 feet away from the centerline of bearing. Find the centroid centroid of the prestressing strands at this location: l ocation: The centroid of the prestressing strands is at:

⎛  5.55 ft.  ⎞ ⎟⎟ ⋅ (y end − y drape ) y str @ dv = y end − ⎜⎜ ⎝ 0.40 ⋅ span ⎠ ⎛  5.55  ⎞ = 20 − ⎜ ⎟ ⋅ (20 − 5.27) = 18.43 in ⎝ 0.40 ⋅ 130 ⎠

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-20

With this approximation to the strand centroid, dp can be computed: dp = h − y str @ dv = (72 + 1.5 + 8.5) − 18.43 = 63.57 in From the flexural strength computations, a = 6.82 in a 6.82 = 60.16 in dv = dp − = 63.57 − 2 2 But the effective shear depth dv need not be less than dv ≥ 0.72 ⋅ h = 59.04 in or dv ≥ 0.9 de = 0.9 dp = 0.9 (63.57 ) = 57.21 in Therefore take dv = 60.16 in Based on the assumed θ , the critical section location is: dcritv = 7.5 + 0.5 ⋅ dv ⋅ cot (θ) = 7.5 + 0.5 ⋅ (60.16 ) ⋅ cot (26 )

= 69.17 in = 5.8 ft  

GOVERNS

or dcritv = 7.5 + dv = 7.5 + 60.16 = 67.63 in Determine Shear Stress From Table 5.7.2.4 the Strength I design shear at 5.8 ft is Vu = 358 kips The amount of force carried by the draped strands at their effective prestress level is: P12d = 12 ⋅ 0.217 ⋅ 138.83 = 361.5 kips The inclination of the draped strands is:

⎡ (62 − 8) / 12 ⎤ φ = arctan ⎢ ⎥ = 4.887 deg rees ⎣ 52.63 ⎦ The vertical prestress component is: Vp = P12d ⋅ sin (φ) = 361.5 ⋅ sin (4.887 ) = 30.8 kips The maximum shear capacity of the section Vn = 0.25 ⋅ f c′ ⋅ dv ⋅ b v + Vp = 0.25 ⋅ 8.0 ⋅ 60.16 ⋅ 6.0 + 30.8 = 752.7 kips The maximum design shear the section can have is:

φ v ⋅ Vn = 0.90 ⋅ 752.7 = 677.4 kips >> 358 kips

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-21

The shear stress on the section is:

⎛ Vu − φ v − Vp  ⎞ ⎛ 358 − 0.90 ⋅ 30.8 ⎞ ⎟ ⎜ v=⎜ ⎟ = 1.017 ksi ⎜ φ v ⋅ b v ⋅ dv ⎟ = ⎝  0.90 ⋅ 6 ⋅ 60.16  ⎠ ⎝   ⎠ The ratio of the shear stress to the compressive strength is: v 1.017 = = 0.127 f c′ 8.0 Determine Longitudinal Strain

x

It shall be assumed that minimum transverse reinforcement will be provided in the cross section. A ps shall first be determined. determined. Note that A ps computed computed here is different different than the Aps computed earlier. This Aps  includes only the area of prestressing steel found on the flexural tension side of the member. Near the end of the beam, A ps must also be reduced for development Development length ld

ld

is:

2 ⎛   ⎞ = K ⎜ f ps − f pe ⎟ db 3 ⎝   ⎠ 2 ⎡ ⎤ = 1.6 ⎢262.10 − (138.83)⎥ (0.6) = 162.8 in 3 ⎣ ⎦

Transfer length l tr

l tr

is:

= 60 ⋅ db = 60 (0.6) = 36.0 in

At the critical section dcritv = 69.14 in   from the beam end, the strand development fraction Fdev =

=

f pe f pu

+

f   ⎞ dcritv − l tr ⎛  ⎜1 − pe ⎟ ⎜ ⎟ l d − l tr ⎝  f pu ⎠

138.83 69.14 − 36.0 ⎛  138.83 ⎞ + ⎜1 − ⎟ = 0.641 270 162.8 − 36.0 ⎝  270  ⎠

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-22

The flexural tension side of the member is defined as: hcomp 2

=

82.0 = 41.0 in 2

At dcritv none of the draped strands fall on the flexural tension side. Therefore, A ps = (# straight str.) (strand area) ( Fdev )

= (32) (0.217) (0.641) = 4.451 in2 [5.8.3.4.3]

Equation 5.8.3.4.2-1 shall be used to compute the strain:

⎡ Mu ⎤ ⎢ d + 0.5 ⋅ (Vu − Vp ) ⋅ cot (θ) − A ps ⋅ f po ⎥ ⎥ εx = ⎢ v ⎢ ⎥ 2 ⋅ (Ep ⋅ A ps ) ⎢ ⎥ ⎣ ⎦ ⎡ 1897.12 ⎤ ⎢ 60.16 + 0.5 ⋅ (358 − 30.8) ⋅ cot (26 ) − (4.451 ⋅ 0.70 ⋅ 270 )⎥ =⎢ ⎥ 2 ⋅ (28,500 ⋅ 4.451) ⎢ ⎥ ⎣⎢ ⎦⎥ ⎛ − 127.4  ⎞ = ⎜⎜ ⎟⎟ = −0.000502 ⎝ 253,707 ⎠ Because the value is negative, equation three should be investigated with the additional concrete term: From Figure 5.4.6.1, A c = 431 in2

⎡ Mu ⎤ ⎢ d + 0.5 ⋅ (Vu − Vp ) ⋅ cot (θ) − A ps ⋅ f po ⎥ ⎥ εx = ⎢ v ⎢ ⎥ ( ) 2 ⋅ Ec ⋅ Ac + Ep ⋅ A ps ⎢ ⎥ ⎣ ⎦ ⎡ 1897.12 ⎤ ⎢ 60.16 + 0.5 ⋅ (358 − 30.8) ⋅ cot (26) − (4.451 ⋅ 0.70 ⋅ 270)⎥ =⎢ ⎥ 2 ⋅ (4578 ⋅ 431 + 28,500 ⋅ 4.451) ⎢ ⎥ ⎢⎣ ⎥⎦ ⎛ − 127.4  ⎞ = ⎜⎜ ⎟⎟ = −0.000030 ⎝ 4,199,943 ⎠ With the strain and shear stress to f c′   ratio determined, interpolate to find β and θ in LRFD Table 5.8.3.4.2-1.

θ  = 23.2 degrees β  = 2.90

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-23

Since the original assumption for θ   (26 degrees) does not match the computed angle, do an iteration assuming θ = 23.2 degrees.

Because

new angle is flatter than original assumption, do not revise location of critical section for shear (conservative). For θ = 23.2 degrees, ε x = −0.000019 Then with

v = 0.126 and ε x = −0.000019 , interpolate to get: f c′

θ = 23.4  degrees (close enough to assumed angle) β = 2.89 The concrete contribution: Vc = 0.0316 ⋅ β ⋅

f c′ ⋅ b v ⋅ dv = 0.0316 ⋅ 2.896 ⋅ 8 ⋅ 6 ⋅ 60.16 = 93.2 kips

The required steel contribution is: Vu

Vs = Vn − Vc − Vp =

− Vc − Vp =

φv

358 − 93.2 − 30.8 = 273.8 kips 0.90

The required spacing of double leg #13 stirrups: s=

A v ⋅ f y ⋅ dv ⋅ cot (θ) 2 ⋅ 0.20 ⋅ 60 ⋅ 60.16 ⋅ cot (23.4) = = 12.2 in Vs 273.8

Double leg stirrups at a 12 inch spacing at the end of the beam shall be provided. A v = 040 in2 / ft  

Vs = 277.9 kips

Other sections are investigated similarly. [5.8.4]

2. Interface Shear Transfer Top flange width b v = 30 in The Strength I vertical shear at the critical shear section due to all superimposed loads is: Vu = 1.25 (9 + 9) + 1.75 (109) = 213.3 Interface shear force is: Vh =

Vu de

=

Vu dp −

a 2

=

213.3 = 3.55  kip/in 60.16

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-24

Required nominal interface design shear is: Vn =

Vh

φv

=

3.55 = 3.94  kip/in 0.90

The interface area per 1 inch length of beam is: A cv = 30 ⋅ 1 = 30.0 in2 /in The upper limits on nominal interface shear are: 0.2 ⋅ f c′ ⋅ A cv = 0.2 ⋅ 4 ⋅ 30.0 = 24.0 kip/in > 3.94 kip/in

OK

and 0.8 ⋅ A cv = 0.8 ⋅ 30.0 = 24.0 kip/in > 3.94 kip/in

OK

The amount of interface shear carried by cohesion shall be determined. A note on the Bridge Details II Fig. 5-397.517 requires the top flanges of the beam to be roughened. roughened. Consequently use a cohesion (c) of 0.100 ksi and a friction factor ( μ ) of 1.0. The nominal interface shear resistance is: Vn = c A cv + μ (A vf  f γ + Pc ) Pc = 0.0 kip The required interface shear steel is: A vf  =

[ En g . 5 . 8 .4 . 1 - 4 ]

V − c A cv 3.94 − 0.1 (30.0) = = 0.016 in2 /in = 0.19 in2 /ft μ ⋅ f y 1.0 ⋅ 60

The minimum shear steel that needs to be provided is: A vf min =

0.05 ⋅ b v 0.05 ⋅ 30 = = 0.025 in2 /in = 0.30 in2 /ft f y 60

The minimum requirement controls. Vertical shear reinforcement A v = 0.40 in2 /ft > 0.30 in2 /ft at the critical section for shear. shear. Therefore, no additional reinforcement reinforcement is required required for interface shear. Other sections are investigated similarly.

MAR 2007 [ 5 .8 .3 . 5 ]

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-25

3. Minimum Longitudinal Longitudinal Reinforcement Reinforcement Requirement Requirement A diagonal section shall be checked with a crack starting at the inside edge of the bearing sole plate. Straight strands cross the crack at: sole plate + y 32 ⋅ cot (θ ) = 15 + 4.25 ⋅ cot (23.4) = 24.82 in The transfer length for 0.6" strands is: 60 ⋅ ds = 60 ⋅ 0.6 = 36 in To find the tensile capacity of the straight strands at the crack interpolate: Tr = f pe ⋅ A ps ⋅

length to crack 24.82 = 138.83 ⋅ 32 ⋅ 0.217 ⋅ = 664.6 kips transfer length 36

The force to carry is: ⎛ V  ⎞ T = ⎜⎜ u − 0.5 ⋅ Vs − Vp ⎟⎟ ⋅ cot (θ)

⎝ ϕ v

 ⎠

⎛  358  ⎞ =⎜ − 0.5 ⋅ 277.9 − 30.8 ⎟ ⋅ cot (23.4) ⎝ 0.90  ⎠ = 526.9 kips < 664.6 kips

OK

F . D e s ig n

Bursting Reinforcement

Pretensioned

To prevent cracking in the beam end due to the transfer of the

Anchorag e Zone

prestressing force from the strands to the concrete, bursting steel needs

R ei n f o r ce m e n t

to be provided in the anchorage zone.

[5.10.10.1]

A load factor of 1.0 and lateral force component of 4% shall be used to determine the required amount of steel. The factored design bursting force is: Pb = 1.0 ⋅ 0.04 ⋅ Pi = 1.0 ⋅ 0.04 ⋅ 1681 = 67.2 kips The amount of resisting reinforcement is determined using a steel stress f s of 20 ksi: P 67.2 As = b = = 3.36 in2 f s 20

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-26

This steel shall be located at the end of the beam within a distance of: h 72 = = 18.0 in 4 4 The number of #16 double legged stirrups necessary to provide this area is: As 3.36 = = 5.4 2 ⋅ Ab 2 ⋅ 0.31 The first set of stirrups is located 2 inches from the end of the beam. Provide

six

( 2 + 5 ⋅ 3 = 17 [5.10.10.2]

sets

of

#16

stirrups

spaced

at

3

inch

centers.

in < 18 in)

Confinement Reinforcement Reinforcement is required at the ends of the beam to confine the prestressing steel steel in the bottom flange. G1303E and G1607E bars (see Figure 5.7.2.5) will be placed at a maximum spacing of 6 inches out to 1.5d from the ends of the beam. For simplicity in detailing and ease of tying the reinforcement, the vertical shear reinforcement shall be spaced with the confinement reinforcement in this area. 1.5 d = 1.5 (72) = 108.0 in

G. Determ ine

Camber Due to Prestressing and Dead Load Deflection

Ca m b e r a n d

Using the PCI handbook (Figure 4.10.13 of the 3 rd  Edition), the camber

Deflection

due to prestress can can be found. The centroid of the the prestressing has has an

[2.5.2.6.2] [3.6.1.3.2]

eccentricity emid of 30.33 inches at midspan. At the end of the beams the eccentricity e e is 15.60 inches. E is the initial concrete modulus

[5.7.3.6.2]

(4347 ksi), Po equals the prestress force just after transfer (1681 kips). The drape points points are at 0.4 of the span. The span length length is 130.0 feet. Using the equation for the two point depressed strand pattern: e′ = emid − e e = 30.33 − 15.60 = 14.73 in

Δ ps

Po e e L2 Po e′ ⎛ L2 a2  ⎞ ⎜ − ⎟ = + ⎜8 8 EI EI ⎝  6  ⎠⎟ 1681(15.60))((130 ⋅ 12)2 1681(14.73) ⎡ (130 ⋅ 12)2 (0.4 ⋅ 130 ⋅ 12)2 ⎤ = + ⎢ − ⎥ 8 (4347) (547,920) 4347(547,920) ⎢ 8 6 ⎥

⎣

= 5.84 in

⎦

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-27

Downward deflection due to selfweight 0.846 (130 ⋅ 12)4 5⋅ 5 ⋅ w ⋅ L4 12 Δ sw = = = 2.28 in 384 ⋅ E ⋅ I 384 ⋅ 4347 ⋅ 547,920 Camber at release Δ rel = Δ ps − Δ sw = 5.84 − 2.28 = 3.56 in To estimate camber at the time of erection the deflection components are multiplied by standard standard PCI handbook multipliers. rd

the 3

(See Figure 4.6.3 of

Edition.) They are:

Release to Erection Multipliers: Prestress = 1.80 Selfweight = 1.85 Camber and selfweight deflection values at erection are: 1.80 ⋅ 5.84 = 10.51 in

Prestress: Selfweight:

1.85 ⋅ (− 2.28) = −4.22 in

Diaphragm DL: Deck and stool DL: Parapet:

−0.03 in −2.84 in −0.17 in

The values to be placed in the camber diagram on the beam plan sheet are arrived at by combining the values above.  “Initial Total Camber” 10.51 − 4.22 − 0.03 = 6.26 in

say 61 /4 in

 “Est. Dead Load Deflection” 2.84 + 0.17 = 3.01 in

say 3 in

 “Est. Residual Camber” 61 /4 – 3 = 3 1 /4 in

Live Load Deflection The deflection of the bridge is checked when subjected to live load and compared against the limiting values of L /800 for vehicle only bridges and L

 /1000 for bridges with bicycle or pedestrian traffic.

Deflection due to lane load is:

Δ lane

0.64 ⎡ 4 ⎤ ⎛  5 ⋅ w ⋅ L4  ⎞ ⎢ 5 ⋅ 12 ⋅ (130 ⋅ 12) ⎥ ⎟= =⎜ = 0.73 in ⎜ 384 ⋅ E ⋅ I ⎟ ⎢⎢ 384 ⋅ 4578 ⋅ 1,235,000 ⎥⎥ ⎝   ⎠ ⎣⎢ ⎦⎥

MAR 2007

LRFD BRIDGE DESIGN

DESIGN EXAMPLE 5-28

Deflection due to a truck with dynamic load allowance is found using hand computations or computer tools to be:

Δ truck = 1.46 in Two deflections are computed and compared to the limiting values; that of the truck alone and that of the lane load plus 25% of the truck. Both deflections need to be adjusted with the distribution factor for deflection.

Δ1 = DFΔ ⋅ Δ truck = 0.567 ⋅ 1.46 = 0.83 in Δ 2 = DFΔ ⋅ (Δlane + 0.25 ⋅ Δ truck ) = 0.567 ⋅ (0.73 + 0.25 ⋅ 1.46 ) = 0.62 in There is no bicycle or pedestrian traffic on the bridge, the deflection limit is: L 130 ⋅ 12 = = 1.95 in >> than Δ1 or Δ 2 800 800

H . D e t a i l in in g I t e m s

OK

Approximate weight of each beam is: L ⋅ A ⋅ y = 131.25 ft ⋅

786 in2 144 in2 /ft 2

⋅

0.155 kips ft 3

⋅

1 ton = 55.5 tons 2 kips

Initial prestress force at jacking is: 44 ⋅ 0.217 ⋅ 0.75 ⋅ 270 = 1933 kips

Figure 5.7.2.5 shows the detailed beam sheet for the bridge.

MAR 2007

LRFD BRIDGE DESIGN

Figure 5.7.2.5

DESIGN EXAMPLE 5-29