Section06B - Design Examples

8/8/2019 Section06B - Design Examples

1/68

MAR 2007 LRFD BRIDGE DESIGN DESIGN EXAMPLE 6-1

6.8 Tw o-Span

Plate Gi rd er

Des ign Examp le

This design example is taken from the MNDOT design manual. The

barriers and some of the details used by NDDOT are different and the

NDDOT details should be used where there is a difference. This example

illustrates the design of a two-span welded plate girder bridge with a 20

degree skew and is on a constant grade of 1.5%. The 304-0 long

bridge has two equal spans of 152-0. Specifically, this example

illustrates the detail design of a typical interior girder at the critical

sections in flexure (positive and negative) and shear for AASHTO HL-93

loading. Design of the stiffeners, end diaphragm, shear connectors, and

field splice is included. Fatigue is also checked at critical locations.

The superstructure consists of five girders spaced at 11-4 centers.

Girders are designed to act compositely with a 9 inch deck. A 1/2 inch

of wear is assumed and a deck thickness of 9 inches is used for

composite section properties. To simplify dead load computation the

wearing course is assumed to extend from outside of barrier to outside of

barrier.

The top flange thickness usually varies along the length of the girder. As

the top flange thickness varies along the girder line, the thickness of

concrete in the riser changes as well. For dead load purposes the

concrete portion of the riser is assumed to have a uniform thickness of

33/8 inches. This is arrived at by subtracting the thinnest top flange from

the thickest top flange and adding 1 inches. For section property

computations, a concrete riser thickness of 1 inches will be used. The

composite deck is assumed to have a unit weight of 0.150 kcf for dead

load computations and 0.145 kcf for elastic modulus computations.

The following material and geometric parameters are used in this

example:A.1 Mater ia l s

Concrete (deck and overlay)

Dead load unit weight = wc = 0.150 kcf

Compressive strength = fc = 4 ksi

Elastic modulus = Ec = 3644 ksi

Steel

Dead load unit weight = wst = 0.490 kcf

Yield strength = Fy = 50 ksi

Tensile strength = Fu = 70 ksi

Elastic modulus = Es = 29,000 ksi

Composite Section Properties[ 6 . 10 .3 .1 .1b ]


2/68


Short-term modular ratio = n = 8

Long-term modular ratio = 3n = 24

The geometry for the example is presented in Figures 6.9.1 through

6.9.4. Overall geometry (Figures 6.9.1 and 6.9.2) is presented in this

section. Girder geometry (Figures 6.9.3 and 6.9.4) is presented in the

next section where section properties are assembled.

A typical section for the bridge is shown in Figure 6.9.1. The deck is

supported on five lines of girders. The girders are spaced on 11-4

centers and the roadway is 48-0 wide (two 12-0 traffic lanes and two

12'-0" shoulders.) A Type F railing is provided on each side of the

bridge.

The framing of the superstructure is presented in Figure 6.9.2. The

structure has a 20 degree skew. Due to the span arrangement (2-span,

152-0/152-0), only a half-framing plan is provided. Rolled beam end-

diaphragms are located at the abutments. Cross-frame diaphragms are

located at other locations.

Fi

g

ure

6

Figur e 6 .9 .1


3/68


Figur e 6 .9 .2


4/68


A.2 De te rm i ne

Cross-Sect ion

Proper t i es

[ 2 . 5 .2 .6 .3 ]

The minimum depth of the steel girder is prescribed in LRFD Table

2.5.2.6.3-1 Traditional Minimum Depth of Constant Depth

Superstructures. For continuous span, steel I-beam structures the

depth of the I-beam or girder can be no less than 0.027 L. Mn/DOT uses

a preliminary depth of 0.033L.

Substituting 152 feet for L in the minimum depth ratios results in a

minimum girder height of 49 inches (LRFD) to 61 inches (Mn/DOT).

A member deeper than the minimum is usually the most economical.

Adequate clearance is assumed available for the example and a slightly

deeper section will be tried. Try a design with a 66 inch deep web.

Preliminary web and flange plate sizes for the girder are shown in Figure

6.9.3. The girder is symmetric about the pier with a 7/8 x 18 top

flange and a7

/8 x 20 bottom flange in the positive moment region. Inthe negative moment region, both flanges are 13/8 x 20 near the field

splice and 23/4 x 20 over the pier. For the web, a5/8 inch plate

thickness is used throughout.

The non-composite section properties of the girder are provided in Table

6.9.1.

Table 6.9.1 Non-Composite Section Properties

Parameter Positive Section*Negative

Section 1

**Negative

Section 2d (in) 67.75 68.75 71.50

A (in2) 74.50 96.25 151.25

I (in4) 52,106 77,399 145,024

yt (in) 34.66 34.38 35.75

yb (in) 33.09 34.38 35.75

St (in3) 1503 2252 4057

Sb (in3) 1575 2252 4057

* Negative Section 1 is from the field splice to the shop splice.

** Negative Section 2 is the section over the pier.


5/68


Figur e 6 .9 .3


6/68


Positive Moment Section Properties

The width of deck assumed to act compositely with the girder and resist

external loads is the smallest of three values: 1) of the effective span

length, 2) 12 times the deck thickness plus of the top flange width, or

3) the average spacing of adjacent beams. For section property

computations the deck thickness is reduced by inch to account for

wear.

[ 4 . 6 .2 .6 ]

1) 3191

12

4

1527.0 = in

2) 117182

1912 =+ in

3) 11-4 = 136 in

For this example, the controlling value is 117 inches. Using the modular

ratios provided earlier (n=8, 3n=24) results in a transformed deck widthof 14.63 inches for transient, short-term loads (n=8) and 4.88 inches for

permanent, long-term loads (n=24). The concrete riser or haunch has

an assumed thickness of 1 inches for section property computations.

Section B of Figure 6.9.4 contains a girder cross section with the primary

dimensions for section property computations identified.

[ 6 . 10 .3 .1 .1b ]

[ 6 . 10 .3 .1 .1c]

Negative Moment Section Properties

For negative moment regions, the section assumed effective in resisting

external loads is the steel girder section plus the reinforcement within an

effective width of the slab.

[ 6 . 10 .3 .7 ] In negative moment regions, the longitudinal reinforcing steel in the deck

is approximately 1% of the area of the deck. Two thirds of this steel is to

be placed in the top mat of reinforcement. A top mat with 20-#6 bars

and a bottom mat with 14-#5 bars is found to be satisfactory.

The top mat is located 3.63 inches from the top of the deck (based on 3

inches clear, for NDDOT this should be 2 clear) and the bottom mat is

located 1.94 inches from the bottom (based on 1 inch clear). See

Section C, Figure 6.9.4.


7/68


Figure 6 .9 .4

Table 6.9.2 contains composite girder section properties.

Table 6.9.2 Composite Section Properties

Parameter

Positive Section

N 3N

Positive

Section

for

Negative

Moment*

Negative

Section

1**

Negative

Section 2

****

Ac (in2) 209.50 119.50 87.6 109.39 164.38

Ic (in4) 131,943 98,515 70,371 96,048 165,867

Ycg (in) 19.04 29.90 39.10 40.06 42.93

Ytbeam

(in)8.54 19.40

28.6029.56 32.43

Ybbeam

(in)59.21 48.35

39.1539.19 39.07

Stbeam

(in3)15,443 5079

24603250 5114

Sbbeam

(in3)2229 2037

17982451 4246

* Positive Section for Negative Moment is used for splice design.

** Negative Section 1 is from the field splice to the flange butt splice.

*** Negative Section 2 is the section over the pier.


8/68


B. Select Load

Modi f i e rs

[ 1 . 3 .3 -1 .3 .5 ]

The following load multipliers will be used for this example.

= 1.00D

= 1.00R

= 1.00I

C. Select

Appl icable Load

Combinat ions and

Load Factors

[ 3 . 4 . 1 ]

The load combinations which will be considered for the design

example are identified below:

STRENGTH I - Standard HL-93 loading will be used. Primary applications

include:

maximum bottom flange stress in positive moment location maximum top and bottom flange stress in negative moment

locations

Mu = 1.25 DC + 1.75 LL

SERVICE II - Corresponds to the overload provisions in the AASHTO

Standard Specifications pertaining to yield control and to slip-critical

connections.

Mu = 1.0 (DC) + 1.3 LL

FATIGUE - .Checks to limit the potential for fatigue cracks to form in a

structure.

Mu = 0.75 LLrange

[ 3 . 4 . 2 ] CONSTRUCTION LOAD COMBINATIONDuring erection, the girder will need to resist stresses associated with the

steel section alone. In addition, the need for diaphragms or cross-frames

will be determined at this stage.

Mu = 1.25 DCtemp + 1.5 LLtemp

Due to the continuous configuration, maximum and minimum( ) load

factor values will be used.

p

Assume that traffic can be positioned anywhere between the F rail

barriers.

D. Calculat e Live

Load Force Ef fect s

[ 3 . 6 .1 ] [ 3 . 6 .2 ]

[ 4 . 6 .2 .2 ] Number of design lanes = =

12

484 lanes


9/68


Dynam ic Load

A l l owance

IM = 15% when evaluating fatigue and fracture

IM = 33% when evaluating all other limit states

a. Interior Beams with Concrete Decks

LRFD Table 4.6.2.2.1-1 lists common deck superstructure types that have

approximate live load distribution equations.

Di s t r i bu t i on

Factors f or

M o m e n t s

[ 4 . 6 .2 .2 .2b ]

A Type (a) superstructure describes the structural system used in this

example (cast-in-place concrete deck on steel beam supporting

components). Per LRFD Table 4.6.2.2.2b-1, the approximate distribution

equations can be used if a number of geometric parameters are within

the range of values used to arrive at the approximate equations. For a

Type (a) superstructure, the geometric constraints are:

[ C4.6 .2 .2 .1-1 ]

Type (a) Cross Section Range of Applicability Limits for Flexure

Parameter Design Example Minimum Maximum

Beam Spacing

(S)11.33 3.5 16.0

Slab Thickness

(ts)9.0 4.5 12

Number of

Beams (Nb)5 4 -

Span Length (L) 152 20 240

[ 4 . 6 .2 .2 .1 ] In addition to S, ts, and L, the distribution equations for live load moment

area also based on Kg. Kg is a longitudinal stiffness parameter defined in

LRFD Equation 4.6.2.2.1-1.

)eAI(nK2

gg +=

where n is the modular ratio, I is the noncomposite girder moment of

inertia, A is the noncomposite area of the girder and eg is the distance

between the centers of gravity of the noncomposite girder and the deck.

a.1 Positive Moment Region

For the positive moment region

I = 52,106 in4

A = 74.50 in2ts = 9.0 in

yt = 34.66 in

66.342

0.950.1

2riserconcrete ++=++=

t

s

gy

te = 40.66 in

= 1.402 x 106 in4)66.4050.74106,52(8K 2g +=


10/68


One design lane loaded:Distribution factor for moment

0.1

3s

g0.30.4

)(tL12

K

L

S

14

S0.06gM

+= [ Tab l e

4 .6 .2 .2 .2b-1]

0.1

3

60.30.4

(9)15212

1.402x10

152

11.33

14

11.330.06

+=

= 0.484 lanes/girder

Fatigue

[ 3 . 6 .1 .1 .2 ]

[ 3 . 6 .1 .4 ]

The design fatigue truck is a single lane vehicle, that does not include the

multiple presence factor. The tabulated approximate distribution factor

equations for moment include the multiple presence factors.

Consequently, when a designer is considering fatigue, the distribution

factor determined with the approximate equation for a single lane should

be divided by 1.20.

Distribution factor for fatigue moments

0.4031.2

0.484Mg f == lanes/girder

Two or more design lanes loaded

Distribution factor for moment

0.1

3s

g0.20.6

)(tL12

K

L

S

9.5

S

0.075gM

+=

[ Tab l e

4 .6 .2 .2 .2b-1]

0.1

3

60.20.6

(9)15212

1.402x10

152

11.33

9.5

11.330.075

+=


a.2 Negative Moment Region

The noncomposite section properties vary along the girder. The differing

I, A, and yt values impact the Kg term.

For the negative moment section over the pier

I = 145,024 in4

A = 151.25 in2

ts = 9.0 in

yt = 35.75 in

75.352

0.950.1

2riserconcrete ++=++=

t

s

gy

te = 41.75 in


11/68


= 3.269 x 106 in4)75.4125.151024,145(8K 2g +=

One design lane loaded


0.1

3s

g0.30.4

)(tL12

K

L

S

14

S0.06Mg

+= [ Tab l e

4 .6 .2 .2 .2b-1]

0.1

3

60.30.4

(9)15212

3.269x10

152

11.33

14

11.330.06

+=


Fatigue

Distribution factor for fatigue moment

0.4341.2

0.521gMf == lanes/girder



0.1

3s

g0.20.6

)(tL12

K

L

S

9.5

S0.075Mg

+=

[ Tab l e

4 .6 .2 .2 .2b-1]

0.1

3

60.20.6

(9)15212

3.269x10

152

11.33

9.5

11.33

0.075

+=


b. Exterior Beams

[ 4 . 6 .2 .2 .2d ] Table 4.6.2.2.2d-1 contains the approximate distribution factor equations

for exterior beams. Check the value of de to ensure that the

approximate distribution equations are valid.

de = distance from centerline of exterior girder to the gutterline

(see Figure 6.9.5)

de = 3.00 1.67 = 1.33 ft

which is greater than -1.0 and less than 5.5. The approximate equation

for two or more design lanes loaded can be used.

One design lane loaded

Use the lever rule and refer to Figure 6.9.5.


12/68


Exterior beam reaction or distribution factor is:

1.211.33

.67)]6(11.330.67)[(11.330.5gM

+=


Fatigue

Distribution factor for fatigue moment

1.2

0.811gMf = = 0.676 lanes/girder

Figur e 6 .9 .5


13/68


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 loaded.

[ Tab l e

4 .6 .2 .2 .2d-1]

eriorintgMegM =

where

1.9

33.177.0

1.9

d77.0e e +=+= = 0.916

which leads to:

740.0916.0gM = = 0.678 lanes/girder for positive moment

799.0916.0gM = = 0.732 lanes/girder for negative moment

Skewed Bridges

[ 4 . 6 .2 .2 .2e] The framing plan is skewed 20 degrees. There is no modification to themoments for skew until the skew angle is 30 degrees or greater.

Dis t r i but ion Factor for Shear

a. Interior Beams

Similar to flexure, in order to use the simplified distribution equations for

shear, geometric values for the bridge need to be within specific limits.

[ Tab l e

4 .6 .2 .2 .3a-1]

Type (a) Cross Section Range of Applicability Limits for Shear


Beam Spacing (S) 11.33 3.5 16.0

Slab Thickness(ts)

9.0 4.5 12

Number of Beams

(Nb)5 4 -


Pos. Mom. Long.

Stiffness (Kg)1.402x106 10,000 7.0x106

Neg. Mom. Long.

Stiffness (Kg)3.269x106 10,000 7.0x106

[ 4 . 6 .2 .2 .3a] All parameters for the design example are within permissible limits. Thesimplified equations for shear distribution can be used (Table 4.6.2.2.3a-

1 is used).

[ Tab l e 4 .6 .2 .2 .3a-1]

One Design Lane Loaded

25

33.1136.0

25

S0.36gV +=+= = 0.813 lanes/girder


14/68


Two or More Design Lanes Loaded

+=

+=

35

11.33-

12

33.112.0

35

S

12

S0.2gV

2


[ 4 . 6 .2 .2 .3b ] b. Exterior BeamsOne Design Lane Loaded

Use the lever rule, which results in the same factor that was computed

for flexure.

gV = 0.811 lanes/girder


e 0.73310

33.16.0

10

d0.6 e =+=+=

The exterior shear distribution factor for multiple lanes loaded is the

product of e and the interior girder factor.

1.0390.733gVeVg eriorint == = 0.762 lanes/girder

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

[ Tab l e

4 .6 .2 .2 .3c-1 ]

c. Skewed BridgesThere is a modification to the shear at the obtuse corners for bridges with

skewed lines of support. This example has a skew angle of 20 degrees.

Type (a) Cross Sections

Range of Applicability Limits for Skew Correction (Shear)


Skew Angle () 20 degrees 0 degrees 60 degrees

Beam Spacing

(S)11.33 3.5 16.0

Number of

Beams (Nb)5 4 -


)tan(K

tsL120.21.0CF

0.3

g

3

+=

(20)tan1.402x10

9152120.21.0CF

0.3

6

3

+=

= 1.07 lanes/girder at the abutment


15/68


(20)tan3.269x10

9152120.21.0CF

0.3

6

3

+=

= 1.06 lanes/girder at the pier

For simplicity, only the larger correction factor will be used to modify the

live load distribution factors for shear. The adjusted shear distribution

factors are:

Interior Girders

One design lane loaded = 0.813 x 1.07 = 0.870 lanes/girder

Two or more design lanes loaded =

1.039 x 1.07 = 1.112 lanes/girder

Exterior Girders

One design lane loaded = 0.811 x 1.07 = 0.868 lanes/girder

Two design lanes loaded

0.762 x 1.07 = 0.815 lanes/girder

Fatigue Shear

Interior Girder = 725.02.1

870.0gVf == lanes/girder

Exterior Girder = 723.02.1

868.0gVf == lanes/girder


16/68


Su m m a r y o f

Govern ing

D i s t r i bu t i on

Factors

Table 6.9.3 Distribution Factor Summary (lanes/girder)

Girder/Force Component One LaneMultiple

LaneControl

+ Moment 0.484 0.740 0.740

- Moment 0.521 0.799 0.799

Shear 0.870 1.112 1.112

+ Fatigue

Moment0.403 - 0.403

-Fatigue

Moment0.434 - 0.434

Interior

Girder

Fatigue Shear 0.725 - 0.725

+ Moment 0.811 0.678 0.811

- Moment 0.811 0.732 0.811

Shear 0.868 0.815 0.868

+ Fatigue

Moment0.676 - 0.676

-Fatigue

Moment0.676 - 0.676

Exterior

Girder

Fatigue Shear 0.723 - 0.723

E. Calcu lat e Force

Ef fects

Axial loads generated as a result of creep, shrinkage, and thermal

movements will not be considered for the design of the girders. These

loads are considered in the bearing and substructure design examples.

From this point forward only the design of an interior girder subject to

dead load and HL-93 live loads will be presented.

The unfactored bending moments and shear forces at different locations

along the girder are presented in Tables 6.9.5 through 6.9.12. Moments

and shears due to noncomposite loads (DC1 loads) are applied to a

continuous beam model with varying noncomposite section properties

(see Table 6.9.1). Moments and shears for DC2 loads are based on a

composite continuous beam model consisting of the steel girder plus the

concrete deck where a modular ratio of 3n is used for the section

properties. Moments and shears for live loads are based on a compositecontinuous beam model consisting of the steel girder plus the concrete

deck with a modular ratio of n. Table 6.9.4 presents the areas and

moments of inertias for the negative moment regions. The n and 3n

composite section properties for the positive moment regions are

provided in Table 6.9.2.


17/68


Table 6.9.4 Properties for DC2 and LL Moment Determination

Parameter

DC2 Negative Section

(3n)

Field Splice

Pier

LL Negative Section (n)

Field Splice

Pier

Ac (in2) 141.38 196.38 232.75 286.62

Ic (in4) 127,450 205,514 169,431 259,858

DC1 consists of the following loads: girder selfweight, concrete deck and

form loads. A 15% detail factor (based on the self weight of the girder)

is used to account for the dead load of connection and cross frame

elements. A 0.10 ksf load is considered during construction to account

for the weight of deck formwork.

==pierover-k/ft0.592

sectionnegative-k/ft0.377

sectionpositive-k/ft0.292

)15.1(490.0144

Aw beambeam

[ ]

k/ft1.410.1512

18

12

3.375

12

(9.5)11.33

0.150RiserAreaDeckAreaw deck

=

+=

+=

0.11311.33100.0w forms == k/ft

A 0.20 ksf temporary construction live loading is also considered. It is

assumed to be acting full length on a single span concurrent with wetconcrete placement. In Table 6.9.6, DCtemp consists of girder

selfweight, form load and one span of concrete

DC2 consists of long-term dead loads, barrier and future wearing surface

(FWS). The load factor for FWS would be 1.5 (DW in Table 3.4.1-1) and

for the barriers would be 1.25 (DC). Mn/DOT uses a FWS of 0.020 ksf

with a 1.25 load factor.

k/ft0.22711.33ksf0.020w

k/ft0.1755

2k/ft0.438w

fws

girders

barriersbarrier

==

==

The splice plate is located 108 feet from the abutment bearing,

approximately 0.71 of the span. This location was chosen as the nearest

even foot along the span to the dead load inflection point during the

initial sizing. During the design iteration, this dead load inflection point

moved to approximately 0.68 of the span. The splice location for the

example will remain at the assumed location of 0.71 of the span.


18/68


All of the DC1 loads presented in the example include the 0.010 ksf load

associated with the formwork. This increases the strength design loads

by 2% but greatly simplifies the calculations. The load is applied to the

noncomposite section but is removed from the composite section. The

actual stresses are also dependent on the pour sequence for the deck.

In the following tables, Girder Point 0.0 is the centerline of bearing at the

abutment. Girder Point 1.0 is centerline of bearing at the pier. Due to

the symmetry of the span arrangement, for most loads only data for

Girder Points 0.0 to 1.0 is provided. However, due to the unsymmetric

loading considered during construction, values are provided for both

spans in Table 6.9.6.

Table 6.9.5 Dead Load Bending Moments (unfactored)

DC1 Moment (K-FT) DC2 Moment (K-FT)

Girder

Point Girder

Slab

and

Stool

Forms TOTAL Barrier FWS TOTAL

0.1 198 954 77 1229 122 158 280

0.2 329 1582 127 2038 204 264 468

0.3 393 1884 151 2428 245 317 562

0.4 390 1860 150 2400 246 318 564

0.5 320 1511 121 1952 206 267 473

0.6 182 836 67 1085 125 163 288

0.7 -23 -165 -13 -201 5 6 11

0.71 * -47 -283 -23 -353 -10 -12 -22

0.8 -303 -1492 -120 -1915 -157 -203 -360

0.882

**-598 -2823 -227 -3648 -319 -413 -732

0.895**

*-650 -3054 -245 -3949 -347 -450 -797

0.9 -671 -3144 -253 -4068 -358 -464 -822

1.0 -1159 -5122 -412 -6693 -600 -778 -1378

* Field Splice

** Flange Butt Splice

*** First cross-frame off pier


19/68


20/68


Table 6.9.7 contains positive and negative live load moments due to

truck, lane, and truck train loading.

Table 6.9.8 lists the extreme positive and negative bending moments at

various girder points when the fixed axle fatigue truck is run across the

structural model.

Table 6.9.7 Live Load Design Moments Per Lane (unfactored)

Girder

Point

Truck

Pos

Mom

(K-

FT)

Lane

Pos

Mom

(K-

FT)

Truck

Neg

Mom

(K-

FT)

Lane

Neg

Mom

(K-

FT)

Truck

Train

(K-

FT)

Positive*

Mom

(K-FT)

Negative**

Mom

(K-FT)

0.1 867 558 -124 -109 1711 -274

0.2 1457 967 -246 -218 2905 -545

0.3 1796 1228 -369 -326 3617 -817

0.4 1923 1340 -492 -435 3898 -1089

0.5 1863 1305 -615 -543 3783 -1361

0.6 1642 1123 -738 -652 3307 -1634

0.7 1278 792 -860 -761 -1199 2492 -2120

0.71 1234 749 -873 -772 -1217 2390 -2152

0.8 816 342 -983 -898 -1370 1427 -2448

0.882 396 111 -1083-

1255-1534 638 -2966

0.895 348 87 -1100-

1335 -1626 550 -3148

0.9 329 78 -1106-

1369-1664 516 -3224

1.0 0 0 -1230-

2174-2459 0 -4900

*Positive Moment = [1.33 x Truck] + Lane

**Negative Moment = maximum of [1.33 x Truck]+Lane

or 0.9 x ([1.33 x Truck Train]+Lane)


21/68


Table 6.9.8 Live Load Fatigue Moments Per Lane (unfactored)

Girder Point

Truck Positive

Moment

(K-FT)

Truck Negative

Moment

(K-FT)

Fatigue Moment

Range

(K-FT)

0.1 788 -119 1043

0.2 1307 -236 1774

0.3 1625 -354 2276

0.4 1715 -472 2515

0.5 1654 -590 2581

0.6 1469 -708 2504

0.7 1126 -826 2245

0.71 1084 -838 2210

0.8 667 -943 1852

0.882 306 -1039 1547

0.895 268 -1055 1521

0.9 253 -1061 1511

1.0 0 -1180 1357

*Moment Range = 1.15 x [Positive Moment Negative Moment]

Table 6.9.9 presents the unfactored dead load shear forces at different

girder locations for different load components.


22/68


Table 6.9.9 Dead Load Shear (unfactored)

DC1 Shear (K) DC2 Shear (K)

Girder

Point Girder

Deck

and

Stool

Forms TOTAL Barrier FWS TOTAL

0.0 -15 -73 -6 -94 -9 -12 -21

0.1 -11 -52 -4 -67 -7 -9 -16

0.2 -6 -31 -2 -39 -4 -5 -9

0.3 -2 -9 -1 -12 -1 -2 -3

0.4 2 12 1 15 1 2 3

0.5 7 34 3 44 4 5 9

0.6 11 55 4 70 7 9 16

0.7 16 77 6 99 9 12 21

0.71 * 16 79 6 101 10 12 22

0.8 21 98 8 127 12 15 27

0.842*

*25 110 9 144 14 17 31

0.895*

**27 118 10 155 14 19 33

0.9 28 119 10 157 15 19 34

1.0 37 141 11 189 17 22 39

* Field Splice

** First transverse stiffener*** First cross-frame

Table 6.9.10 contains the dead load reactions at Girder Points 0.0 and

1.0. The reactions at Girder Point 1.0 are larger than the shear at Girder

point 1.0 because the reaction includes the load from both spans.

Table 6.9.10 Dead Load Reactions (unfactored)

Girder Point DC1 Reaction (KIPS) DC2 Reaction (KIPS)

0.0 95 22

1.0 378 79


23/68


Table 6.9.11 contains the live load shear extremes for various live load

components: truck and lane. Per LRFD Article 3.6.1.3.1, truck train

loading is not to be used for shear.

Table 6.9.11 Live Load Shear Per Lane (unfactored)

Girder

Point

Truck

Positive

Shear

(K)

Lane

Positive

Shear

(K)

Truck

Negative

Shear

(K)

Lane

Negative

Shear

(K)

Positive*

Shear

(K)

Negative*

Shear (K)

0.0 66 42 -8 -7 130 -18

0.1 57 32 -8 -8 108 -19

0.2 48 25 -13 -10 89 -27

0.3 39 18 -22 -13 70 -42

0.4 31 13 -31 -17 54 -58

0.5 24 8 -39 -23 40 -75

0.6 17 5 -47 -29 28 -92

0.7 11 3 -54 -36 18 -108

0.71 10 2 -55 -37 15 -110

0.8 6 1 -60 -45 9 -125

0.842 4 1 -63 -50 6 -134

0.895 2 0 -65 -53 3 -139

0.9 2 0 -65 -54 3 -140

1.0 0 0 -70 -63 0 -156

* = [1.33 x Truck] + Lane

Table 6.9.12 presents the live load reactions at the abutment (Girder

Point 0.0) and the pier (Girder Point 1.0). Similar to the dead load

reactions presented in Table 6.9.10, at the pier one can not simply take

the shear values to arrive at the reactions. The adjacent span influences

the magnitude of the reaction. Per LRFD Article 3.6.1.3.1 the truck train

loading needs to be considered for reactions at interior supports.

Table 6.9.12 Live Load Reactions (unfactored)

Girder Point

HL-93 Truck + Lane

Reaction

(KIPS)

HL-93 Truck Train + Lane

Reaction

(KIPS)0.0 130 -

1.0 221 273

The checks in this example begin with the strength checks on the

preliminary layout. Designers should be aware that deflections may

control the design. The deflection checks for this example are presented

in Section M.


24/68


F. I -Sect ions in

Flexur e

I nves t i ga te The

St reng t h L i m i t

S ta te

At the strength limit state the girder is designed to carry factored dead

and live loads. The resisting section in the positive moment regions is the

composite section. In the negative moment regions, resistance is

provided by the girder plus deck reinforcement composite section.

a. Maximum Positive Design MomentG.1 Determine

Max i mum Desi gn

M o m e n t s

At 0.4 (60.8) from each end, the maximum design positive flexure is:

Mu = 1.25(2400+564) + 1.75(3898) 0.740

= 3705 + 5048 = 8753 kip-ft

b. Maximum Negative Design MomentAt the pier, the maximum design negative flexure is:

Mu = 1.25(-6693-1378) + 1.75(-4900)0.799

= -10,089 6851 = -16,940 kip-ft

G.2 Cod e Check s a. General Proportions

To ensure that the section functions as an I section, the LRFD

specifications contain requirements on the proportions of the web,

flanges, and overall cross section.

The general proportions check is the following:

9.0I

I1.0

y

yc

where Iyc is the moment of inertia of the compression flange about a

vertical axis, and Iy is the moment of inertia of the girder about a vertical

axis.

[ 6 . 10 .2 .1 ] For the typical positive moment section this becomes:

3yc 18875.0

12

1I = = 425.3 in4

20875.012

1625.066

12

118875.0

12

1I 33y ++= = 1009.9 in4

42.09.1009

3.425

I

I

y

yc== OK

For the negative moment section over the pier this becomes:

3yc 2075.2

12

1I = = 1833.3 in4

333y 2075.2

12

1625.066

12

12075.2

12

1I ++= = 3668 in4


25/68


50.03668

3.1833

I

I

y

yc== OK

b. Web SlendernessTo ensure that lateral web deflections are not excessive, the following

check needs to be satisfied for webs without longitudinal stiffeners.

200f

E6.77

t

D2

cw

c

[ 6 . 10 .2 .2 ]

where, Dc is the depth of the web in compression in the elastic range, tw

is the thickness of the web, E is the modulus of the web steel and fc is

the stress in the compression flange due to the factored loading under

investigation.

[ C6.10.3 .1 .4a]

ftdtfcf

cfcD

+=

Section properties are from Tables 6.9.1 and 6.9.2, the moments are

from Tables 6.9.5 and 6.9.7 for the 0.4 Girder Point and the LL

Distribution factor is from Table 6.9.3. Start with the sum of the

compression stresses:

23.951503

1224001.25fDC1 =

= ksi

1.675079

1256425.1fDC2 =

= ksi

92.315,443

120.7438981.75f IMLL =

=+ ksi

Combining these terms produces:

29.5492.31.6723.95IMLLf2DCf1DCfcf =++=+++= ksi

Next compute the sum of the tension flange stresses:

22.861575

1224001.25DC1f =

= ksi

15.42037

1256425.1

DC2f =

= ksi

18.272229

120.7438981.75IMLLf =

=+ ksi

Combining these terms produces:19.5418.27.15422.86IMLLf2DCf1DCftf =++=+++= ksi

75.67875.066875.0d =++= in


26/68


tf = 0.875 in

Substituting back into the equation produces:

02.23875.075.6719.5454.29

54.29ftd

tfcf

cfcD =

+

=

+= in

The upper limit for the check is:

1.21229.54

29,0006.77

f

E6.77

c

== > 200, Use 200

The web slenderness check with values inserted becomes:

OK;002.7730.625

23.022

t

D2

w

c =

=

Table 6.9.13A Web Slenderness Check Along the Girder Unfactored

Moments

MomentsGirder Point

DC1 (K-FT) DC2 (K-FT) LL+IM (K-FT)

0.1 1229 280 1711

0.2 2038 468 2905

0.3 2428 562 3617

0.4 2400 564 3898

0.5 1952 473 3783

0.6 1085 288 3307

0.7 -201 11 2492

0.8 -1915 -360 -24480.9 -4068 -822 -3224

1.0 -6693 -1378 -4900

Table 6.9.13B Web Slenderness Check Along the Girder Section

Properties

Girder

Point

Sc

(in3)

Sc3n

(in3)

Scn

(in3)

St

(in3)

St3n

(in3)

Stn

(in3)tf (in)

0.1 1503 5079 15,443 1575 2037 2229 0.875

0.2 1503 5079 15,443 1575 2037 2229 0.875

0.3 1503 5079 15,443 1575 2037 2229 0.8750.4 1503 5079 15,443 1575 2037 2229 0.875

0.5 1503 5079 15,443 1575 2037 2229 0.875

0.6 1503 5079 15,443 1575 2037 2229 0.875

0.7 1503 5079 15,443 1575 2037 2229 0.875

0.8 2252 2451 2451 2252 3250 3250 1.38

0.9 4057 4246 4246 4057 5114 5114 2.75

1.0 4057 4246 4246 4057 5114 5114 2.75


27/68


Table 6.9.13C Web Slenderness Check Along the Girder Results

Girder

Point

Dc

(IN)2Dc/tw

cf

E77.6 Limit Check is

0.1 23.89 76.46 299.4 200.0 OK

0.2 23.76 76.03 232.2 200.0 OK0.3 23.51 75.24 212.2 200.0 OK

0.4 23.02 73.67 212.1 200.0 OK

0.5 21.98 70.35 232.1 200.0 OK

0.6 19.24 61.58 297.6 200.0 OK

0.7 1.37 4.40 1577.5 200.0 OK

0.8 34.48 110.3 208.8 200.0 OK

0.9 34.62 110.8 208.1 200.0 OK

1.0 34.57 110.6 164.7 164.7 OK

[ 6 . 10 .2 .3 ] c. Flange ProportionsFor the compression flange, a check is made to ensure that the web is

adequately restrained by the flange to control web bend buckling. The

width of the compression flange must be equal to at least 30 percent of

the depth of the web in compression.

[ 6 . 10 .2 .3 -1 ] cf D0.3b

where Dc is the same value computed in Table 6.9.13

Table 6.9.14 Compression Flange Proportion Check

Girder Point Dc (IN) 0.3 x Dc(IN)

bf (IN) Check

0.1* 23.89 7.17 18 OK

0.2 23.76 7.13 18 OK

0.3 23.51 7.05 18 OK

0.4 23.02 6.91 18 OK

0.5 21.98 6.60 18 OK

0.6 19.24 5.77 18 OK

0.7 1.37 0.41 18 OK

0.8 34.48 10.34 20 OK

0.9 34.62 10.39 20 OK1.0** 34.57 10.37 20 OK

* - controlling case for 18 flange

** - controlling case for 20 flange

For the tension flange, a simple check with flange width and flange

thickness is made to ensure that the flange will not distort excessively

during fabrication. The check also provides a measure of safety that the

flange has good proportions in the event of a stress reversal.


28/68


12.0t2

b

f

f

[ 6 . 10 .2 .3 -2 ]

where bf and tf of the tension flange are provided in Figure 6.9.3. Table

6.9.15 contains the results of the check for all of the Girder Points.

Table 6.9.15 Tension Flange Proportion Check

Girder Point bf (IN) tf (IN) bf/2tf

Check

12

ft2

fb

0.1 20 0.875 11.43 OK

0.2 20 0.875 11.43 OK

0.3 20 0.875 11.43 OK

0.4 20 0.875 11.43 OK

0.5 20 0.875 11.43 OK

0.6 20 0.875 11.43 OK

0.7 20 0.875 11.43 OK

0.8 20 1.375 7.27 OK

0.9 20 2.75 3.64 OK

1.0 20 2.75 3.64 OK

G.3 St r engt h L imi t

Sta t e Flexur a l

Resistance

[ 6 . 1 0 . 4 ]

a. Categorization of Flexural Resistance

The procedure for evaluating the flexural strength of a girder in

accordance with the LRFD Specifications is quite involved. To clarify the

steps involved, a flow chart is included in the commentary to LRFD Article

6.10.4. Figure 6.9.6 contains a copy of the flow chart.

The girder cross section has a constant web height. It is considered a

constant depth section. It also utilizes Grade 50 steel, has no holes in

the tension flange, and the web does not contain longitudinal stiffeners.

Consequently we can enter the flow chart at the upper left box.

With the two-span continuous girder, the nominal flexural resistance for

the positive moment section is potentially impacted by the negative

moment section. Thus, determine first the compactness of the negative

section, then check what portion of 6.10.4.2.2 (if any) can be followed.


29/68


Figur e 6 .9 .6


30/68


[ 6 . 10 .4 .1 .2 ] b. Negative Moment Section Compact-Section CheckFirst check the following equation to determine if the web is compact:

yw

cp

F

E3.76

t

D2

Figur e 6 .9 .7

Dcp is the depth of the web in compression at the plastic moment. The

location of the plastic neutral axis (PNA) can be determined with the

equations and figures contained in the Appendix to LRFD Section 6. The

figure in the appendix for Negative Bending Sections is provided in this

example as Figure 6.9.7.Pt = Force in the top flange = 50202.75 = 2750 kips

Pw = Force in the web = 50660.625 = 2063 kips

Pc = Force in the bottom flange = 50202.75 = 2750 kips

From Figure 6.9.4, Art = 8.84 in2 and Arb = 4.30 in2

Prt = Force in the top mat of rebar = 608.84 = 530.2 kips

Prb = Force in the bottom mat of rebar = 604.30 = 257.7 kips

With the force components known, the location of the plastic neutral axis

can be checked to see if it is located in the web or top flange of the

section. If the combined force in the web and bottom flange is greater

than the sum of the remaining force components the plastic neutral axis

is located in the web.

Pw + Pc = 2063 + 2750 = 4813 kips


31/68


Pt + Prb + Prt = 2750 + 257.7 + 530.2 = 3538 kips

The plastic neutral axis is in the web.

[ Tab l e A6 .1 -2 ] The location of the plastic neutral axis can be found with the equation inthe appendix for Case I.

+

=

1

P

PPPP

2

Dy

w

rtrbtc

+

= 1

2063

530.2257.727502750

2

66= 20.40 in

The depth of web in compression is the difference between D and y .

Dcp = D - y = 66.0 20.4 = 45.6 in

Returning to the web slenderness check in Article 6.10.4.1.2.

145.90.625

(45.6)2

t

D2

w

cp=

=

The maximum value for a compact web with 50 ksi steel is:

90.650

29,0003.76

F

E3.76

y

== < 145.9

Therefore, the web is not compact. Consequently, the section is

noncompact. Proceed down in the flow chart to the box with the heading

Article 6.10.4.1.4 and check the slenderness of the compression flange.12

t2

b

f

f

[ Equa t i on

6 .10.4 .1 .4-1 ]

1264.32.752

20=

The compression flange slenderness check is satisfied. Proceed to the

right in the flow chart to the box with the heading Article 6.10.4.1.9 and

determine the required bracing spacing for the bottom flange near the

pier.

ytpb

FEr1.76LL = [ Equa t i on

6 .10.4 .1 .9-1 ]

Lb is the length between compression flange brace points. If the flange

is not adequately braced, it may not be able to reach yield stress. rt is

the radius of gyration of the compression flange plus one third of the web

in compression.


32/68


183412

0.625

3

34.42

12

202.75I

33

rt =

+

= in4

620.6253

34.422.7520Art =+= in2

in5.4362

1834rt ==

Substitute rt into the equation to arrive at the lateral bracing limit for

plastic bending.

23050

29,0005.431.76Lp == in

From the framing plan, cross frames are located 16 feet away from the

centerline of the pier.192ft16Lb == in

This is less than Lp (230 inches), so adequate bracing is provided to

compute the flexural resistance of the section with the non-lateral

torsional buckling flexural resistance equations. Proceed to the right in

the flow chart to the box with the heading Article 6.10.4.2.4.

c. Negative Region Flexural Strength

Begin by checking the capacity of the compression flange.

[ 6 . 10 .4 .2 .4 ] crhbn FRRF =

where Rb is the load shedding factor, Rh is the hybrid factor, and the

equation for Fcr is provided in LRFD Article 6.10.4.2.4.

Ifc

bw

c

f

E

t

D2

then Rb is 1.0.[ 6 . 10 .4 .3 .2 ]

If the inequality is not satisfied a more refined equation must be used to

evaluate Rb. Within the inequality, Dc is the depth of the web in

compression in the elastic range and fc is the stress in the compression

flange due to factored loading.

in (from Table 6.9.13C)57.34Dc =

==2

66

2

D33 in

2

DDc > therefore b = 4.64


33/68


Now plug values into the inequality. First the left side:

=

=

625.0

57.342

t

D2

w

c 110.6

Now compute fc

48.98fc = (Table 6.9.13C)

The right side of the inequality becomes

98.48

000,2964.4

f

E

cb = = 112.9 > 110.6

Therefore, Rb = 1.0.

[ 6 . 10 .4 .3 .1a ] The cross section uses 50 ksi steel throughout, therefore Rh = 1.0

Now compute the maximum stress in the compression flange prior to

incorporating the Rb and Rh multipliers.

yc

w

c2

f

f

cr F

t

D2

t2

b

E1.904F

= [ 6 . 10 .4 .2 .4a ]

398.0

110.1(2.75)2

20

(29,000)1.9042

=

= ksi which is > 50 ksi, Fcr = 50 ksi

The maximum stress permitted in the compression flange can now be

computed.

Fn= (1)(1)(50) = 50.0 ksi

The compressive stress in the flange under factored loads has been

computed earlier as fc.

Fu = fc = 48.98 < 50.0 ksi

Therefore, the compression flange is okay.

[ 6 . 10 .4 .2 .4b ] Now check the tension flange.

The load shedding factor for the tension flange is:[ 6 . 10 .4 .3 .2 .b ] Rb = 1.0

[ 6 . 10 .4 .3 .1a ] The cross section uses 50 ksi steel throughout, therefore

Rh = 1.0

[ 6 . 10 .4 .2 .4b ] The yield strength of the tension flange is 50 ksi (Fyt)

The maximum stress permitted in the tension flange is


34/68


Fn = (1)(1)(50) = 50 ksi

The tension stress in the top flange under factored loads is

125114

(0.799)(4900)1.75

5114

(1378)1.25

3892

(6693)1.25ft

+

+

=

= 45.9 ksi

Fu = ft = 45.9 < 50.0 ksi

Therefore, the tension flange is okay.

The negative moment section has adequate flexural capacity.

[ 6 . 10 .4 .1 .23 ] d. Positive Moment Compact-Section Check

Now check the capacity of the positive moment section at the 0.4 Girder

Point. Return to the flow chart and enter it in the upper right hand

corner in the box with the heading Article 6.10.4.1.2. Check if the web is

compact. For the web to be compact the following inequality must be

satisfied:

yw

cp

F

E3.76

t

D2

To determine Dcp use Appendix A from Section 6. The figure for load

components for positive bending sections is presented in Figure 6.9.8.

Figur e 6 .9 .8


35/68


To simplify computations neglect the Prt and Prb terms.

Pc = Force in the top flange = 500.87518 = 787.5 kips

Pw = Force in the web = 500.62566 = 2063 kips

Pt = Force in the bottom flange = 500.87520 = 875 kips

Ps = Force in the slab = 0.854(9117+1.518) = 3672 kips

Begin by checking Case I (PNA in the web of the girder).

29382063875PP wt =+=+ kips

36725.787PP sc +=+ = 4459.5 > 2938

Therefore the PNA is not in the web.

Try Case II (PNA in the top flange)5.37255.7872063875PPP cwt =++=++ kips

< 3725.5 kips3672Ps =

Therefore the PNA is in the top flange. Use the equation in the appendix

to locate the position of the PNA in the top flange.

+

+

=

+

+

= 1

5.787

36728752063

2

875.01

P

PPP

2

ty

c

stwc =0.03 in

With the PNA located in the top flange the entire web is in tension. This

in turn implies that Dcp is zero and the inequality in Article 6.10.4.1.2 is

satisfied. For the positive moment section the web is compact.

The note on the flow chart for the asterisk states that for composite

sections in positive flexure, certain Articles are considered automatically

satisfied. Proceeding to the right from the box in the upper left-hand

corner we arrive at three boxes with asterisks. These boxes are

automatically satisfied. We proceed to the right and arrive at the box in

the upper right hand corner with the heading Article 6.10.4.2.1 or

6.10.4.2.2.

[ 6 . 10 .4 .2 .2 ] The span under consideration is continuous, with a negative flexural

region over the interior support that is noncompact. The LRFD

Specifications recognize the reduced capacity for load redistribution in

spans with noncompact sections. The positive flexural section is typically

not permitted to reach the plastic moment capacity (Mp) in such cases.


36/68


[ Equa t i on

6 .10.4 .2 .2a-3]

Compute the capacity with the Approximate Method. This method

requires the computation of My. The procedure for determining My is

presented in Appendix A6.2 of the LRFD Specifications.

n

AD

n3

2DC

nc

1DCy

S

M

S

M

S

MF ++=

Rearrange and solve for MAD, the additional moment required to reachyield in a flange.

=

n3

2DC

nc

1DCynAD

S

M

S

MFSM

The additional moment for the bottom flange is

=

2037

1256425.1

1575

12240025.1502229MAD

= 51,244 k-in = 4270 k-ft

And the additional moment for top flange is

=

5079

1256425.1

1503

12240025.150443,15MAD

= 376,534 k-in = 31,378 k-ft

A smaller additional moment is required to yield the bottom flange,

therefore MAD = 4270 k-ft

= 7975 k-ft427056425.1240025.1MMMM AD2DC1DCy ++=++=

The nominal flexural resistance based on LRFD Equation 6.10.4.2.2a-3

can now be computed.

= 10,367 k-ft797513.1MR3.1M yhn ==

The flexural resistance determined with Equation 3 can be no more than

the value computed for the case where the negative flexural section is

compact. The limiting value for this case is that computed with Equation

1 or 2. To determine which equation should be used, compute Dp and

D.

03.05.19yttD hsp ++=++= = 10.53 in

5.7

5.1975.677.0

5.7

ttd'D hs

++=

++= = 7.3 in[ Equa t i on

6 .10.4 .2 .2b-2]

Dp is greater than D and less than 5D, therefore Equation 2 is to be

used. Compute Mp using the Case II Equation in Appendix A6.1 Table

A6.1-1.

( )[ ] [ ]ttwwss2c2c

cp dPdPdPyty

t2

PM ++++

=


37/68


2

95.103.0

2

ttyd shs ++=++= = 6.03 in

2

6603.0875.0

2

Dytd cw +=+= = 33.85 in

)03.0875.0(662

875.0)yt(D

2

td c

tt ++=++= = 67.28 in

Substituting P, d, t, and y terms into the equation for Mp results in

( )[ ][ ]28.6787585.33206303.63672

03.0875.0)03.0(875.02

5.787M

22p

+++

+

=

= 151,166 k-in = 12,597 k-ft

Use LRFD Equation 6.10.4.2.2a-2 to compute the limiting flexural

resistance for the positive moment region.

+

=

'D

D

4

MM85.0

4

M85.0M5M

ppyypn

+

=

30.7

53.10

4

597,12797585.0

4

797585.0597,125Mn

= 11,954 k-ft

The 11,954 k-ft value from equation 2 is more than the 10,367 k-ft value

obtained with Equation 3. Therefore the flexural resistance of the

positive moment section is 10,367 k-ft.

74.0389875.1)564(25.1)2400(25.1Mu ++=

= 8753 < 10,367 k-ft

Therefore, the positive moment section has adequate flexural strength.

[ 6 . 10 .3 .2 ] e. Constructibility

The capacity of the girders should be evaluated during construction, prior

to composite action carrying the loads. For this example, the check

consists of placing selfweight and formwork on both spans, while deck

dead loads and a 20 psf construction live load is placed on one span.

Load factors for this check are based on the values provided in LRFD

Article 3.4.2, where 1.25 is used on dead loads and 1.5 is used on live

loads. The factored construction moment for the positive moment

section is:

temptemptempu LL5.1DC25.1M +=

)464(50.1)3425(25.1M tempu += = 4977 k-ft


38/68


We return to the flow chart to evaluate the constructability of the girder.

Enter the flow chart at the box in the lower left-hand column with the

heading Article 6.10.3.2.2.

The first check for constructability is to ensure that the bending stress

fcw in the web is not too large.

ksi50F

t

D

kE0.9f yw2

w

cw =

[ 6 . 10 .3 .2 .2 -1 ]

0.87534.66

52,106

124977

ty

I

Mf

f-t

tempucw

== = 38.72 ksi

Without longitudinal stiffeners in the web,

= 1.25

=

=

=

22

c 0.875-34.66

669

D

D9.0k 34.35

Substituting values into Equation 6.10.3.2.2-1 results in

ksi50ksi5.100

625.0

66

35.341.2529,0000.92

=

, set the limit at 50 ksi

, so the bending in the web is satisfactory.ksi50F72.38f ywcw ==

Proceed up in the flow chart to the box with the heading of Article

6.10.4.1.4.

The top flange satisfies the aspect ratio inequality in Article 6.10.4.1.4.

1229.10875.02

18

t2

b

f

f =

=

Proceed to the right in the flow chart to the box with the heading of

Article 6.10.4.1.9. Check if the bracing spacing is less than Lp. If so, thenon-lateral-torsional buckling equations are to be used to compute the

capacity of the section.

yctp

F

Er76.1L =

12

0.625

3

0.875-34.66

12

180.875I

33

rt

+

= = 425.5 in4


39/68


0.6253

0.875-34.66.875018Art += = 22.79 in2

in.324

22.79

425.5rt ==

Substituting values into the eq Lp produces:uation for

50

000,2932.476.1Lp = = 183 in

gin

by checking the following inequality to see if Equation 1 can be used:

From the framing plan, cross frames are spaced at 30 feet in the positive

moment region (Lb = 360 inches). Lb is greater than Lp. Therefore the

flexural resistance of the section is less than My and the lateral-torsional

buckling equations are used. Proceed down and to the right in the flow

chart to the box with the heading Article 6.10.4.2.5 or 6.10.4.2.6. Be

yb

w

c

F

E

t

D2

The left side of the inequality is

1.108625.0

)875.066.34(2

t

D2

w

c =

=

[ 6 . 10 .4 .2 .6 ]

Over half the web is in compression (top flange smaller than bottom

flange), therefore b = 4.64. The right side of the inequality becomes:

7.11150000,2964.4

FEy

b == > 108.1

Therefore, the inequality is satisfied and equation can be used:

yh

2

bycb

ychbn MR

L

d87.9

I

J772.0

L

IRCE14.3M

+

=

J is the Saint Venant torsional constant and can be found with:

3

tbtbtDJ

3tt

3ff

3w ++=

86.133

875.018875.020625.066 333=

++= in4

The weak axis moment of inertia of the top flange is:

43

yc in42512

875.018I =

=

Rh = 1.0 and d = 67.75 in


40/68


Due to the uncertainties associated with the construction loads, use amoment gradient correction factor (Cb) of 1.00.

Substituting values into the moment equation produces:

6.10.4 .2 .6a-1]

[ Equa t i on 2

n 360

75.6787.9

425

86.13772.0

360

4250.10.1000,2914.3M

+

=

kip-in = 5484 kip-ft810,65=

Compute the yield moment (My) and substitute in values to arrive at the

maximum flexural resistance.

,150751503500.1SFRMR xcyhyh === kip-in = 6262 k-ft

Use Mn = 5484 kip-ft

The flexural resistance of 5484 k-ft is greater than the factored moment

of 4977 k-ft. Use a cross frame spacing of 30 feet (NDDOT max 25).

H. I nves t i ga te t he

Serv i ce L imi t Sta t e

Overload provisions in past AASHTO design specifications controlled the

amount of permanent deflection and the performance of slip-critical

connections. The Service II load combination is used in a similar fashion

for LRFD designs.

Service II = 1.00(DC+DW)+1.3LL

[ 6 . 10 .5 .1 ] The narrative in LRFD Article 6.10.5.1 states that the web bending stress,

fcw, shall satisfy the LRFD Equation 6.10.3.2.2-1.

ksi50F

t

D

kE0.9f yw2

w

cw =

[ 6 . 10 .3 .2 .2 -1 ]

12

0.875)-(8.54

131,943

0.74(3898)1.30

0.875)-(19.40

98,515

(564)

0.875)-(34.66

52,106

(2400)

S

gM(3898)1.3

S

(564)

S

(2400)f

)n(web)n3(web(nc)webcw

++=

++=

= 22.6 ksi

There are no longitudinal stiffeners in the web, 25.1= . To arrive at a

conservative k value use the Dc value of the noncomposite section

35.340.875-34.66

669

D

D9.0k

22

c

=

=

=


41/68


Plug in the values to arrive at the limiting stress

ksi5.100

625.0

66

35.341.2529,0000.92

=

The upper limit is capped by the material strength of 50 ksi

fcw = 22.6 ksi 50 ksi, so the web bending stress is satisfactory.

H.1 Flang e Str ess

L i m i t a t i ons

For composite sections the stress in the flanges, ff, when subjected to

Service II load combinations must be less than 95 percent of the yield

strength of the flange. The noncompact web section over the pier limits

the stress in the flanges to Fy when evaluating strength load

combinations. The sections should readily pass this check due to the

smaller load factors associated with the Service II load combination.ksi47.5F0.95f yf =

[ 6 . 10 .5 .2 ] a. Positive Flexural Region

Top Flange

ksi47.5ksi23.41215,443

0.74(3898)1.3

5079

564

1503

2400ff =

++=

Bottom Flange

ksi5.47ksi.84112

2229

0.74(3898)1.3

2037

564

1575

2400ff =

++=

b. Negative Flexural Region

Top Flange

ksi47.5ksi35.0125114

0.799(4900)1.3

5114

1378

4057

6693ff =

++=

Bottom Flange

ksi47.5ksi38.1124246

0.799(4900)1.3

4246

1378

4057

6693ff =

++=

I . I n v e st i g a t e t h e

Fat igu e Lim i t S ta te

[ 6 . 6 .1 .2 .3 ]

Only details with fatigue resistance Category C or lower resistance need

to be evaluated during design. Details that are classified as Category B

and above no longer need to be checked.

To improve the fatigue resistance of steel superstructures, Mn/DOT

attaches cross frame connection plates and transverse stiffeners to the

tension flange with a bolted connection (NDDOT attaches connection

plates to flanges).


42/68


I . 1 Fa t i gue

Loading

[ 3 .6 .1 .4 ]

The HL-93 truck is used to generate the fatigue loads that are used to

evaluate different components of a design. For fatigue, the HL-93 truck

has a fixed rear axle spacing of 30 feet. In addition, a load factor of 0.75

is applied to calibrate the stresses to those observed in field studies. The

dynamic load allowance for fatigue loading is 15 percent. Distribution for

fatigue is equal to the one design lane loaded distribution, with the

multiple presence factor removed (if approximate equations are used for

one lane loaded).

[ 6 . 6 .1 .2 .2 ]

resistancefatiguenominalF)(

rangestressLoadLivef)(

0.75fatigueforfactorload

F)(f)(

n

n

=

=

==

I .2 Check Largest

St ress Rang e

Locat ion

The unfactored fatigue moments in Table 6.9.8 are multiplied by the

fatigue load factor (0.75) and the appropriate distribution factor to arrive

at the design moment ranges for fatigue. In Table 6.9.16 the stresses at

the top of the top flange are computed by dividing the design moment

range by the composite (n) section modulus.

For this example, the details with fatigue resistances less than B that

should be investigated for fatigue are: the shear studs attached to thetop flange and the web to connection plate/stiffener welds. Details

subject to stress ranges less than the infinite life fatigue threshold are

assumed to have infinite life. The factor accounts for the probability

that some vehicles larger than the HL-93 fatigue truck will cross the

bridge.

Designers should note that the fatigue distribution factor for the exterior

girder is significantly larger (0.676 versus 0.434) than that of the interior

girders.


43/68


Table 6.9.16 Fatigue Range (Truck Moments and Moment Range from

Table 6.9.8)

Fatigue

Loads Per Lane

Girder

Point

Truck

Positive

Moment

(K-FT)

Truck

Negative

Moment

(K-FT)

Moment*

Range

(K-FT)

Design**

Moment

Range

(K-FT)

Top

Stress

(KSI)

Bottom

Stress

(KSI)

0.1 788 -119 1043 315 0.24 1.70

0.2 1307 -236 1774 536 0.42 2.89

0.3 1625 -354 2276 688 0.53 3.70

0.4 1715 -472 2515 760 0.59 4.09

0.5 1654 -590 2581 780 0.61 4.20

0.6 1469 -708 2504 757 0.59 4.08

0.7 1126 -826 2245 679 0.53 3.66

0.71 1084 -838 2210 668 0.52 3.60

0.8 667 -943 1852 603 2.23 2.95

0.842 306 -1039 1547 504 1.86 2.47

0.895 268 -1055 1521 495 1.16 1.40

0.9 253 -1061 1511 492 1.15 1.39

1.0 0 -1180 1357 442 1.04 1.25

* Includes 15% Dynamic Load Allowance

**Girder Point 0.1 0.7: (Moment Range) x 0.75 x 0.403

Girder Point 0.71 1.0: (Moment Range) x 0.75 x 0.434

I .3 Top Flange

W i th Shear Studs

[ Tab le 6 .6 .1 .2 .3-1 ]

[ Tab le 6 .6 .1 .2 .5-3 ]

The top flange has welded shear studs that are a Category C detail.

Category C details have a constant amplitude fatigue threshold of 10.0

ksi. The shear connectors are attached to the top flange of the section.

From Table 6.9.16 the largest top flange stress range occurs at Girder

Point 0.71 (2.23 ksi). This value is below of the constant amplitude

fatigue threshold (5.0 ksi). Therefore, the shear studs are assumed to

have an infinite fatigue life.

I .4 St i f fen er To

W eb W e l d

Mn/DOT Detail B411 provides the cope detail for stiffeners andconnection plates. The connection to the tension flange is with a bolted

tab plate which is a Category B detail which need not be evaluated. The

stiffeners and connection plates are welded to the web of the girder. Per

Detail B411 this weld terminates approximately 3 inches above or below

the tension flange. The weld between the stiffener and web is classified

as a C fatigue detail that requires investigation.

[ Tab le 6 .6 .1 .2 .3-1 ]

[ Tab le 6 .6 .1 .2 .5-3 ]


44/68


The constant amplitude fatigue threshold for C details is 12.0 ksi and the

assumed infinite life fatigue threshold is 6.0 ksi. Reviewing Table 6.9.16

indicates that neither flange has a stress range over 4.20 ksi. The web

to stiffener welds are subject to a smaller stress range than the flanges.

By inspection, fatigue resistance is adequate.

Adequate resistance to load induced flexural fatigue is provided.

I . 5 Fa t i gue

Requ i rement s f o r

W eb

To control out-of-plane flexing of the web under repeated live loading the

following constraints are placed on webs.

Flexure Check

The following check compares the flange stress to a maximum value.

This assumes that the stress in the web due to flexure is approximately

the same as that found in the flange.

[ 6 . 10 .6 .3 ]

Ifyww FEk95.0

tD [ 6 . 10 .6 .3 -1 ]

the maximum compression flange stress (fcf) is Fyw (50 ksi). Otherwise

a more involved equation needs to be used. The value for k has been

previously computed as 34.35.

6.105625.0

66

t

D

w

==

6.1051.134

50

000,2935.3495.0

F

Ek95.0

yw

>=

=

Therefore, ywcf Ff

[ 6 . 10 .6 .2 ] The live load used for this check is twice that presented in Table 6.9.16.

In the positive moment region at the 0.4 Girder Point

ksi50.0ksi7.211215,443

7602

5079

564

1503

2400fcf =

++=

In the negative moment region at the 1.0 Girder Point

ksi50ksi7.28124246

442242461378

40576693fcf =

++=

[ 6 . 10 .6 .4 ] Shear Check

The computations for the resistance of the web in shear is based on the

following equation:[ Equa t i on 6 .10 .6 .4 -1 ]

ywcf F)C(58.0v


45/68


Where vcf is the maximum elastic shear stress in the web due to

unfactored permanent load and fatigue loading. C is defined in LRFD

Article 6.10.7.3.3a. It is the ratio of shear buckling stress to shear yield

strength.

Assume an unstiffened web.

6.105625.0

66

t

D

w

==

5

66

55

D

d

55k

22o

=

+=

+=

32.7450

0.5000,2938.1

F

kE38.1

yw

=

=

395.050

0.5000,29

625.0

66

52.1

F

kE

t

D

52.1C

2yw

2

w

=

=

=

500.3950.58F(C)0.58v ywcf = ksi5.11vcf

Table 6.9.17 Shear Fatigue

Fatigue

(per lane, no impact)Girder

Point

DC1

Shear(K)

DC2

Shear(K)

Minimum

Shear (K)

Maximum

Shear (K)

Vcf*

(K)

vcf**

(KSI)

0.0 94 21 -8 61 181 4.4

0.1 67 16 -8 52 140 3.4

0.2 39 9 -9 43 95 2.3

0.3 12 3 -17 35 53 1.3

0.4 -15 -3 -27 26 -47 -1.2

0.5 -44 -9 -34 20 -90 -2.2

0.6 -70 -16 -42 14 -132 -3.2

0.7 -99 -21 -50 8 -174 -4.2

0.711 -101 -22 -50 8 -177 -4.3

0.8 -127 -27 -56 4 -215 -5.2

0.842 -144 -31 -61 2 -241 -5.9

0.895 -155 -33 -62 2 -255 -6.2

0.9 -157 -34 -62 2 -258 -6.3

1.0 -189 -39 -67 0 -301 -7.3

* DC1+DC2+(Maximum Shear or Minimum Shear) x 0.725 x 0.75 x 2.0

** Vcf / (66 x 0.625)


46/68


47/68


6.105625.0

66

t

D

w

==

[ 6 . 10 .7 .3 .3a -7 ] The appropriate equation for C is selected based on how slender the webis:

3.7450

5000,2938.1F

kE38.1yw

== < 105.6

Therefore,

[ ]395.0

50

5000,29

6.105

52.1

F

kE

t

D

52.1C

2yw

2

w

=

=

=

The capacity of the unstiffened web is:4731196395.0Vn == kips

473VV nr == kips

J.1 Pier Sect io n Assume the critical section for shear is at Girder Point 1.0. Based on

Tables 6.9.3, 6.9.9, and 6.9.11, the factored shear force over the pier is:

kips5893042851.1121561.75)39(1891.25Vu =+=++= > nV

The resistance of an unstiffened web is less than the demand of 589 kips;

therefore transverse stiffeners are required near the pier.

[ 6 . 10 .7 .3 .2 ] Check the handling requirements. The maximum D/tw ratio permitted is

150. The web for the design example has a D/tw ratio of 105.6;

Equation 1 is satisfied. Investigate Equation 2. The maximum

transverse stiffener spacing that satisfies the handling check is:

in4002

105.6

26066

2

D/t

260D

w

=

=

To satisfy LRFD Article 6.10.7.1, the maximum spacing of transverse

stiffeners is three times the depth of the web:

in198(66)3D3 ==

Where required, transverse stiffeners can be spaced no further apartthan 198 inches.

Go to Article 6.10.7.3.3b to determine the required shear stiffener

spacing for a noncompact section.

Try a 8 foot spacing for transverse stiffeners (a cross frame is located at

16 foot and 32 foot distances from Girder Point 1.0).


48/68


Compute k for a stiffener spacing of 96 inches.

36.7

66

96

55

D

d

55k

22o

=

+=

+=

16.9050

36.7000,2938.1

F

kE38.1

yw

=

=

< 105.6

582.050

36.7000,29

625.0

66

52.1

F

kE

t

D

52.1C

2yw

2

w

=

=

=

For Girder Point 1.0, under factored loads, the compression flange has a

stress of 49.0 ksi, and a resistance of 50.0 ksi. Therefore, fu is greaterthan 75 percent of Fy and LRFD Equation 6.10.7.3.3b-2 is used to arrive

at the resistance of the stiffened web. Begin by computing the R and

Vp parameters:

[ 6 . 10 .7 .3 .3b -3 ]

632.05.3750

49504.06.0

F75.0F

fF4.06.0R

yfr

ur =

+=

+=

kips1196625.0665058.0tDF58.0V wywp ===

Substituting into Equation 2, one arrives at:

+

+=

+

+=

22o

pn

66

961

)582.01(87.0582.01196632.0

D

d1

)C1(87.0CVRV

[ 6 . 10 .7 .3 .3b -2 ]

= 595 kips

However, Vn should also be larger than CVp = 0.5821196 = 696 kips

The resistance of 696 kips is more than the demand of 589 kips. An 8foot stiffener spacing works over the pier. Next, determine where

transverse stiffeners can be dropped.

At 24 feet away from the pier, the shear demand is:

Vu = 1.25(144+31) + 1.75(134)1.112 = 480 kips

Which is slightly greater than the 473 kip resistance of the unstiffened

web.


49/68


The second cross-frame from the pier is the 0.789 point in the span. At

the 0.8 location the shear demand is:

436112.1)125(75.1)27127(25.1V 8.0 =++= kips

Therefore, provide stiffeners or cross frame connection plates to at least

32 feet away from the pier.

J .2 Abu tment

Sect ion

Begin by checking if the unstiffened web has adequate capacity.

For an unstiffened web the shear buckling coefficient (k) is equal to 5.

This leads to a C coefficient of 0.395 when using LRFD Equation

6.10.7.3.3a-7. Knowing Vp and C, the shear resistance for the end panel

can be computed.

[ 6 . 10 .7 .3 .3c ] Vn = CVp = 0.3951196 = 473 kips

At Girder Point 0.0 the shear demand is:

397253144)112.1()130(75.1)2194(25.1Vu =+=++= < 473 kips

The web has adequate capacity at the abutment without stiffeners.

J.3 Transverse

St i f fen er Des ign

[ 6 .10 .8 .1 ]

[ 6 . 7 . 3 ]

It will be shown that 3/8 x 5 stiffeners satisfy the code requirements for

transverse stiffeners, however, they are very thin. Ideally the size of the

stiffener should be coordinated with the cross frame connection plates.

Fabrication of the girder will be simplified if only one plate size and

thickness is welded to the web at non bearing locations. In addition,

transverse stiffeners and diaphragm connection plates should be detailedwith widths that are in quarter inch increments. This provides the

fabricator additional flexibility. They can either cut the stiffeners and

connection plates out of large mill plate or utilize standard flat bar stock.

Transverse stiffeners are required near the pier. Mn/DOTs Bridge Detail

B411 (Stiffener Details) address the constraints placed on stiffeners in

LRFD Article 6.10.8.1.1.

The dimensions of transverse stiffeners are required to fall within

geometric constraints based on section depth, flange width, and

projecting element thickness.

Begin with the projecting width constraint:

inches38.430

5.710.2

30

d0.2bt =+=+ [ 6 . 10 .8 .1 .2 -1 ]

The constraint based on flange width is:[ 6 . 10 .8 .1 .2 -2 ] inches0.52025.0b25.0bt0.16 ftp ==


50/68


Try a pair of 5-inch stiffeners. Each must be at least 3/8-inch thick, per

Mn/DOT Detail B402 or B407 (NDDOT uses single stiffeners).

In addition to good aspect ratios, the stiffeners must also have adequate

area and moment of inertia. Check the minimum required moment of

inertia to comply with LRFD Article 6.10.8.1.3.

5.0818.00.296

665.20.2

d

D5.2J

22

o

=

=

=

Therefore, J = 0.5.

The required stiffness of the stiffeners is:

Required in472.115.0625.096JtdI 33

wot ==

The stiffeners inertia taken about the center of the web is:

Actual 5.37625.10375.012

1I 3t == > 11.72 in4

Adequate stiffness is provided. Check to see if the area satisfies Article

6.10.8.1.4. Begin by determining Fcr.

5073.50

375.0

5

000,29311.0

t

b

E311.0F

22

p

t

cr =

=

= ; Fcr = 50 ksi

2wcryw

ru

ws t

F

F18

VV)C1(

tDB15.0A

With B = 1.0 for stiffener pairs:

8.4625.050

5018

696

589)582.01(

625.0

660.115.0A 2s =

The web provides adequate resistance, consequently, the required area

of the stiffeners is negative.

J.4 Bear ing

St i f fen er Des ign

[ 6 .10 .8 .2 .1 ]

For welded plate girders, bearing stiffeners are needed at both the

abutments and piers (for integral abutments a stiffener is not required at

the abutment).

Abutment Bearing

The reaction to be carried by the bearing stiffeners is:

kips397112.1)130(75.1)2194(25.1Bu =++=


51/68


Similar to transverse stiffeners, there are constraints on the geometry of

bearing stiffeners.

For a first try, assume the resistance bearing stress is approximately 35

ksi for 50 ksi stiffeners. Approximately 11.3 square inches of bearing

stiffener is required. Assume half is placed on each side (5.65 square

inches). Further, assume that the bearing stiffeners extend almost to the

outside edges of the narrower flange, which is the top flange and 18

inches in width near the abutment. Try a x 8 (Area = 6 in2) bearing

stiffener on each side of the web.

Begin by checking the projecting width.

67.850

000,2975.048.0

F

Et48.00.8b

yspt === OK[ Equa t i on

6 .10.8 .2 .2-1 ]

The bearing resistance check is based on the net area of steel in contactwith the flange. Assuming a 1 inch cope at the bottom of the stiffener

in accordance with the B411 detail.[ Equa t i on 6 .10.8 .2 .3-1 ]

48850)2)5.10.8(75.0(0.1FAB yspnbr === > 397 kips OK

The axial resistance of the bearing stiffener is found using the methods of

Article 6.9.2.1. Where restraint against buckling is provided in the plane

of the web and the effective length of the column is 75 percent of the

height of the web.

The stiffener will act like a column while supporting the bearing reaction.

The effective section consists of the stiffeners, plus 18(thickness of the

girder web) (see Figure 6.9.10).

[ 6 . 10 .8 .2 .4b ]

The area for this column is:A = 0.75 8.0 2 + 11.25 0.625 = 19.03 in2

The moment of inertia about the girder web is:

4.287625.1675.012

1625.05.10

12

1I 33 =+=

The radius of gyration is:

886.303.194.287

AIr ===

Check the width/thickness limits of Article 6.9.4.2

84.1050

000,2945.0

F

Ek67.10

75.0

0.8

t

b

y

==== OK


52/68


53/68


Therefore, use a pair of x 8 bearing stiffeners at the abutments.

Using the same design procedure, a pair of 1 x 9 bearing stiffeners

are adequate to carry the factored pier reaction of 1103 kips.

K. I nves t i ga te t he

Shear Connecto r

Design

LRFD Article 6.10.7.4.1 discusses the design of shear connectors.

Connectors are to be placed along the full length of the girder, including

negative moment regions, because the girder is designed as composite

for negative moment.

Shear connectors are designed to satisfy fatigue constraints after which a

strength check is performed. Assume that 7/8 inch diameter shear

connectors will be used.

The minimum transverse spacing for connectors is 4.0 stud diameters.

For 7/8 inch diameter studs, this translates into a minimum spacing of

31/2 inches. The minimum clear distance from a stud to the edge of a

flange is 1.0 inch. With a 18 inch top flange width, the maximum

number of stud spaces placed in a line across the flange is:

spaces3.45.3

875.0)1(218=

Five studs across the flange is permissible, but use 4 shear studs at each

location.

The pitch or longitudinal spacing of sets of 4 shear studs is dictated byLRFD Equation 6.10.7.4.1b-1.

MaxQV

IZnp

sr

r

where p is the pitch of the studs, n is the number of studs provided at a

location, Zr is the fatigue resistance of an individual connector, I is the

short-term moment of inertia, and Q is the first moment of the deck area

or the rebar area about the neutral axis of the short-term composite

section.

[ 6 . 1 0 . 1 0 . 2 - 1 ] The shear fatigue resistance of an individual connector is based on the

number of fatigue cycles anticipated. The resistance of a connector is

also no less than 2.75d2. This lower value corresponds to the resistance

for a stud subjected to approximately 26,200,000 cycles. Assume that

the connectors on the center girder are being designed. Also assume

that both traffic lanes generate fatigue loadings in the center girder.

Using 1.5 cycles per truck passage and a 75-year design life, 26.2 million

cycles is achieved with only 320 trucks per day in each direction.


54/68


cycles610x28.267532036525.1N ==

744.2Nlog28.45.34 ==

kips10.2)875.0(744.2dZ 22

r===

The inertia values are taken from Table 6.9.2. For the positive moment

region I = 131,943 in4 and for the negative moment region use an I of

96,048 in4 (value for the smaller negative moment section). Now

compute the Q values. For the positive moment region:

3in1914

2

91.58.549117

8

1

2

ststooltystbn

1Q =++=++=

And for the negative moment region.

248.430.484.8

94.130.437.584.8d avg_r =

+

+= in from bottom of deck

[ ] [ ]

3in464

4.251.529.564.308.84r_avgdstooltyrbArtAQ

=

+++=+++=

Knowing n, Zr, I, and Q leaves the pitch to be a function of the fatigue

shear force range. For the positive moment region

Maxsrsrsr

r

V

1.579

1914V

943,13110.24

QV

IZnp =

=

For the negative moment region the required pitch is

Maxsrsrsr

r

V

1739

464V

048,9610.24

QV

IZnp =

=


55/68


Table 6.9.18 Shear Connector Spacing Computations

Girder

Point

Negative

Shear

(K)

Positive

Shear

(K)

Fatigue*

Shear

gVf.

LL

Shear**

Range(K)

Max P

(positive)

Max P

(negative)

Max

P

limit

(in)

l***

0.0 -8 61 0.725 43 13.4 - 24

0.1 -8 52 0.725 38 15.4 - 24

0.2 -9 43 0.725 33 17.8 - 24

0.3 -17 35 0.725 33 17.8 - 24

0.4 -27 26 0.725 33 17.5 - 24

0.5 -34 20 0.725 34 17.0 - 24

0.6 -42 14 0.725 35 16.5 - 24

0.7 -50 8 0.725 36 16.0 - 24

0.8 -56 4 0.725 38 - 46.4 24

0.9 -62 2 0.725 40 - 43.5 24

1.0 -67 0 0.725 42 - 41.5 24

* Interior Girder Distribution Factor

** [maximum-minimum]1.15 0.75 gVf

*** The maximum limit for spacing of shear connectors is 24 inches perLRFD Article 6.10.7.4.1b

By inspection, the negative moment criteria is satisfied if sets of two

studs are placed on a 20 inch spacing.

K.1 De te rm i ne

Anchor S tuds a t

Con t ra f l exu re

Po i n t s [ 6 . 10 .10 .3 ]

To positively anchor the longitudinal reinforcment considered part of the

section in the negative moment region, additional studs need to be

placed in the vicinity of the dead load contraflexure points. The number

of additional studs, nac, required are:

r

srrac

Z

fAn

where Zr is the shear resistance of a single stud (2.10 kips), Ar is the

total area of longitudinal reinforcement within the effective flange width

in the negative moment area (13.14 in2) and fsr is the stress range inthe reinforcement. Using the composite section property information for

the smaller negative moment section contained in Table 6.9.2, the

section modulus for the rebar can be computed. The average height of

the rebar above the concrete riser is 4.25 inches.

[ ]3

r in27204.251.529.56

96,048S =

++=


56/68


The fatigue moment range at Girder Point 0.71 can be found in Table

6.9.8. It is 2210 k-ft. Multiplying by the load factor (0.75) and the

interior girder distribution factor of 0.725 results in a moment of

in-k14,4200.7250.75122210M ==

The stress in the rebar is

ksi30.52720

420,14

S

Mf

rsr ===

Plugging values into the equation, results in

2.3310.2

30.514.13

Z

fAn

r

srrac =

=

= studs Use 9 sets of 4 studs

They need to be placed within a length equal to 1/3 of the effective width

of the deck on each side of the contraflexure point. The effective deck

width is 117 inches, two-thirds of which is 78 inches. Dividing this

dimension by 8 (for 9 sets of studs) results in a spacing of 93/4 inches.

Place 9 sets of 4 studs at 9 inch centers near the contraflexure point.

K.2 St r engt h L imi t

Sta te [ 6 . 10 .10 .4 ]

In addition to anchorage studs and fatigue studs, adequate studs need to

be provided to ensure that the cross sections can generate the flexural

resistance computed earlier.

For positive moment areas, the lesser of the capacity of the deck or the

capacity of the steel section need to be provided on each side of the point

of maximum positive moment. The capacity of the deck is

kips35809117485.0tb'f85.0V sch ===

The capacity of the steel section is

[ ] kips3725875.018875.020625.06650

tbFtbFtDFV ffycttytwywh

=++=

++=

Provide resistance for 3580 kips on each side of the positive moment

peak location. The nominal resistance of a shear connector is

[ 6 . 10 .10 .4 .3 -1 ] uscccscn FAE'fA5.0Q =

kips366060.0FA&kips0.363605460.05.0 usc ====

Use a resistance of 36 kips for each shear stud at the strength limit state.

Each side of the positive moment peak requires

99.4kips/stud36

kips3580= shear studs, say 100 studs


57/68


For the negative moment region, each side of the pier must have

sufficient shear studs to develop the capacity of the longitudinal

reinforcement in the deck.

kips7886014.13FAV yrrh ===

studsshear22kips/stud36

kips788=

The final details for the shear studs need to satisfy all three constraints:

fatigue design, anchorage of negative reinforcement, and strength

design. After reviewing the constraints, the layout provided in Figure

6.9.13 satisfies all three constraints.

L . I nves t i ga te

th e Field Spl ice

Design

Several items need to be considered when locating and designing field

splices for steel girders. Typically, splices are located near inflection

points to minimize the flexural resistance required of the connection. In

addition, designers need to en

Section06B - Design Examples

Documents

Transcript of Section06B - Design Examples