U.S. Department of Transportation
Federal Highway Administration
1200 New Jersey Avenue, SE
Washington, DC 20590
202-366-4000


Skip to content U.S. Department of Transportation/Federal Highway AdministrationU.S. Department of Transportation/Federal Highway Administration

Bridges & Structures

LRFD Steel Girder SuperStructure Design Example

Steel Girder Design Example Design Step 3

Table of Contents

Design Step 3.1 - Obtain Design Criteria
Design Step 3.2 - Select Trial Girder Section
Design Step 3.3 - Compute Section Properties
Design Step 3.4 - Compute Dead Load Effects
Design Step 3.5 - Compute Live Load Effects
Design Step 3.6 - Combine Load Effects
Positive Moment Region:
Design Step 3.7 - Check Section Proportion Limits
Design Step 3.8 - Compute Plastic Moment Capacity
Design Step 3.9 - Determine if Section is Compact or Noncompact
Design Step 3.10 - Design for Flexure - Strength Limit State
Design Step 3.11 - Design for Shear
Design Step 3.12 - Design Transverse Intermediate Stiffeners
Design Step 3.14 - Design for Flexure - Fatigue and Fracture
Design Step 3.15 - Design for Flexure - Service Limit State
Design Step 3.16 - Design for Flexure - Constructibility Check
Design Step 3.17 - Check Wind Effects on Girder Flanges
Negative Moment Region:
Design Step 3.7 - Check Section Proportion Limits
Design Step 3.8 - Compute Plastic Moment Capacity
Design Step 3.9 - Determine if Section is Compact or Noncompact
Design Step 3.10 - Design for Flexure - Strength Limit State
Design Step 3.11 - Design for Shear
Design Step 3.12 - Design Transverse Intermediate Stiffeners
Design Step 3.14 - Design for Flexure - Fatigue and Fracture
Design Step 3.15 - Design for Flexure - Service Limit State
Design Step 3.16 - Design for Flexure - Constructibility Check
Design Step 3.17 - Check Wind Effects on Girder Flanges
Design Step 3.18 - Draw Schematic of Final Steel Girder Design

Design Step 3.1 - Obtain Design Criteria

The first design step for a steel girder is to choose the correct design criteria.

The steel girder design criteria are obtained from Figures 3-1 through 3-3 (shown below), from the concrete deck design example, and from the referenced articles and tables in the AASHTO LRFD Bridge Design Specifications (through 2002 interims). For this steel girder design example, a plate girder will be designed for an HL-93 live load. The girder is assumed to be composite throughout.

Refer to Design Step 1 for introductory information about this design example. Additional information is presented about the design assumptions, methodology, and criteria for the entire bridge, including the steel girder.

Span arrangement diagram consisting of a continuous two span bridge with expansion bearings at abutment 1 fixed bearings at pier and expansion bearings at abutment 2. Total structure length is 240 feet 0 inches with each span being 120 feet 0 inches between centerline of abutment and centerline of pier.

Figure 3-1 Span Configuration

Superstructure cross section consisting of a concrete deck with Jersey barriers on 5 steel girders which are spaced at 9 foot 9 inches each for a total of 39 foot 0 inches and with a 3 foot 11 and one quarter inch deck overhang on each side for a total deck width of 46 foot 10 and one half inches. The barriers are 3 foot 6 inches high and 1 foot 5 and one quarter inches wide. There are 2 lanes at 12 foot 0 inches each and 2 shoulders at 10 foot zero inches each.

Figure 3-2 Superstructure Cross Section

MathCad tip logo

Girder Spacing


Where depth or deflection limitations do not control the design, it is generally more cost-effective to use a wider girder spacing. For this design example, the girder spacing shown in Figure 3-2 was developed as a reasonable value for all limit states. Four girders are generally considered to be the minimum, and five girders are desirable to facilitate future redecking. Further optimization of the superstructure could be achieved by revising the girder spacing.


MathCad tip logo

Overhang Width

The overhang width is generally determined such that the moments and shears in the exterior girder are similar to those in the interior girder. In addition, the overhang is set such that the positive and negative moments in the deck slab are balanced. A common rule of thumb is to make the overhang approximately 0.35 to 0.5 times the girder spacing.

Figure 3 dash 3 shows a typical framing plan. There are five girders spaced at 9 feet 9 inches for a total of 39 feet 0 inches. The cross frames are spaced at 20 feet 0 inches from the centerline of bearing at the abutment to the centerline of pier for a total of 120 feet 0 inches. The framing plan is symmetrical about the centerline of pier.

Figure 3-3 Framing Plan

MathCad tip logo

Cross-frame Spacing

A common rule of thumb, based on previous editions of the AASHTO Specifications, is to use a maximum cross-frame spacing of 25 feet.

For this design example, a cross-frame spacing of 20 feet is used because it facilitates a reduction in the required flange thicknesses in the girder section at the pier.

This spacing also affects constructibility checks for stability before the deck is cured. Currently, stay-in-place forms should not be considered to provide adequate bracing to the top flange.

The following units are defined for use in this design example:

Formula: K = 1000lb Formula: kcf = Kips per cubic foot Formula: ksf = numerator (K) divided by denominator ( feet squared ) Formula: ksi = Kips per square inch

Design criteria:

Number of spans: Formula: N subscript spans = 2  
Span length: Formula: L subscript span = 120 feet  
Skew angle: Formula: Skew = 0deg  
Number of girders: Formula: N subscript girders = 5  
Girder spacing: Formula: S = 9 point 75 feet  
Deck overhang: Formula: S subscript overhang = 3 point 9375 feet  
Cross-frame spacing: Formula: L subscript b = 20 feet S6.7.4
Web yield strength: Formula: F subscript yw = 50ksi STable 6.4.1-1
Flange yield strength: Formula: F subscript yf = 50ksi STable 6.4.1-1
Concrete 28-day compressive strength: Formula: f prime subscript c = 4 point 0ksi S5.4.2.1 & STable C5.4.2.1-1
Reinforcement strength: Formula: f subscript y = 60ksi S5.4.3 & S6.10.3.7
Total deck thickness: Formula: t subscript deck = 8 point 5 inches  
Effective deck thickness: Formula: t subscript effdeck = 8 point 0 inches  
Total overhang thickness: Formula: t subscript overhang = 9 point 0 inches  
Effective overhang thickness: Formula: t subscript effoverhang = 8 point 5 inches  
Steel density: Formula: W subscript s = 0 point 490kcf STable 3.5.1-1
Concrete density: Formula: W subscript c = 0 point 150kcf STable 3.5.1-1
Additional miscellaneous dead load (per girder): Formula: W subscript misc = 0 point 015 Kips per foot  
Stay-in-place deck form weight: Formula: W subscript deckforms = 0 point 015ksf  
Parapet weight (each): Formula: W subscript par = 0 point 53 Kips per foot  
Future wearing surface: Formula: W subscript fws = 0 point 140kcf STable 3.5.1-1
Future wearing surface thickness: Formula: t subscript fws = 2 point 5 inches  
Deck width: Formula: w subscript deck = 46 point 875 feet  
Roadway width: Formula: w subscript roadway = 44 point 0 feet  
Haunch depth (from top of web): Formula: d subscript haunch = 3 point 5 inches  
Average Daily Truck Traffic (Single-Lane): Formula: ADTT subscript SL = 3000  

For this design example, transverse stiffeners will be designed in Step 3.12. In addition, a bolted field splice will be designed in Step 4, shear connectors will be designed in Step 5.1, bearing stiffeners will be designed in Step 5.2, welded connections will be designed in Step 5.3, cross-frames are described in Step 5.4, and an elastomeric bearing will be designed in Step 6. Longitudinal stiffeners will not be used, and a deck pouring sequence will not be considered in this design example.

Design factors from AASHTO LRFD Bridge Design Specifications:

Load factors:

STable 3.4.1-1 & STable 3.4.1-2

Limit State Load Factors
DC DW LL IM WS WL EQ
Strength I 1.25 1.50 1.75 1.75 - - -
Service II 1.00 1.00 1.30 1.30 - - -
Fatique - - 0.75 0.75 - - -

Table 3-1 Load Combinations and Load Factors

The abbreviations used in Table 3-1 are as defined in S3.3.2.

The extreme event limit state (including earthquake load) is generally not considered for a steel girder design.

Resistance factors:

S6.5.4.2

Resistance Factor
Type of Resistance Resistance Factor , Φ
For flexure Φf = 1.00
For shear Φv= 1.00
For axial compression Φc= 0.90

Table 3-2 Resistance Factors

MathCad tip logo

Multiple Presence Factors

Multiple presence factors are described in S3.6.1.1.2. They are already included in the computation of live load distribution factors, as presented in S4.6.2.2. An exception, however, is that they must be included when the live load distribution factor for an exterior girder is computed assuming that the cross section deflects and rotates as a rigid cross section, as presented in S4.6.2.2.2d.

Since S3.6.1.1.2 states that the effects of the multiple presence factor are not to be applied to the fatigue limit state, all emperically determined distribution factors for one-lane loaded that are applied to the single fatigue truck must be divided by 1.20 (that is, the multiple presence factor for one lane loaded). In addition, for distribution factors computed using the lever rule or based on S4.6.2.2.2d, the 1.20 factor should not be included when computing the distribution factor for one-lane loaded for the fatigue limit state. It should also be noted that the multiple presence factor still applies to the distribution factors for one-lane loaded for strength limit states.

Dynamic load allowance:

STable 3.6.2.1-1

Dynamic Load Allowance
Limit State Dynamic Load
Allowance, IM
Fatigue and Fracture
Limit State
15%
All Other Limit States 33%

Table 3-3 Dynamic Load Allowance

Dynamic load allowance is the same as impact. The term "impact" was used in previous editions of the AASHTO Specifications. However, the term "dynamic load allowance" is used in the AASHTO LRFD Bridge Design Specifications.

Design Step 3.2 - Select Trial Girder Section

Before the dead load effects can be computed, a trial girder section must be selected. This trial girder section is selected based on previous experience and based on preliminary design. For this design example, the trial girder section presented in Figure 3-4 will be used. Based on this trial girder section, section properties and dead load effects will be computed. Then specification checks will be performed to determine if the trial girder section successfully resists the applied loads. If the trial girder section does not pass all specification checks or if the girder optimization is not acceptable, then a new trial girder section must be selected and the design process must be repeated.

The span length from centerline of bearing at the abutment to the centerline at pier is 120 feet 0 inches and the beam projection is 8 inches. The girder is made up of three different sections. The girder has a constant 54 inch deep by one half inch thick web. The end of the first section is 84 feet 0 inches from the centerline of bearing at abutment. At the end of the first section, there is a bolted field splice. The first section has a top flange of 14 inches wide by five eights of an inch thick. The bottom flange for the first section is 14 inches wide by seven eights of an inch thick. The second section ends 108 feet 0 inches from the centerline of bearing at abutment or 24 feet 0 inches past the end of section one. The second section has a top flange of 14 inches wide by one and one quarter of an inch thick. The bottom flange for the second section is 14 inches wide by one and three eights of an inch thick. The third section ends 120 feet 0 inches from the centerline of bearing at abutment or 12 feet 0 inches past the end of section two. The third section has a top flange of 14 inches wide by two and one half inches thick. The bottom flange for the third section is 14 inches wide by two and three quarters of an inch thick. The girder has bearing stiffeners on both sides of the web at the centerline of bearing at abutment and the centerline of bearing at the pier.

Figure 3-4 Plate Girder Elevation

For this design example, the 5/8" top flange thickness in the positive moment region was used to optimize the plate girder. It also satisfies the requirements of S6.7.3. However, it should be noted that some state requirements and some fabricator concerns may call for a 3/4" minimum flange thickness. In addition, the AASHTO/NSBA Steel Bridge Collaboration Document "Guidelines for Design for Constructibility" recommends a 3/4" minimum flange thickness.

MathCad tip logo

Girder Depth

The minimum girder depth is specified in STable 2.5.2.6.3-1. An estimate of the optimum girder depth can be obtained from trial runs using readily available design software. The web depth may be varied by several inches more or less than the optimum without significant cost penalty.

MathCad tip logo

Web Thickness

A "nominally stiffened" web (approximately 1/16 inch thinner than "unstiffened") will generally provide the least cost alternative or very close to it. However, for web depths of approximately 50 inches or less, unstiffened webs may be more economical.

MathCad tip logo

Plate Transitions

A common rule of thumb is to use no more than three plates (two shop splices) in the top or bottom flange of field sections up to 130 feet long. In some cases, a single flange plate size can be carried through the full length of the field section.

MathCad tip logo

Flange Widths

Flange widths should remain constant within field sections. The use of constant flange widths simplifies construction of the deck. The unsupported length in compression of the shipping piece divided by the minimum width of the compression flange in that piece should be less than approximately 85.

MathCad tip logo

Flange Plate Transitions

It is good design practice to reduce the flange cross-sectional area by no more than approximately one-half of the area of the heavier flange plate. This reduces the build-up of stress at the transition.

The above tips are presented to help bridge designers in developing an economical steel girder for most steel girder designs. Other design tips are available in various publications from the American Institute of Steel Construction (AISC) and from steel fabricators.

Design Step 3.3 - Compute Section Properties

Since the superstructure is composite, several sets of section properties must be computed. The initial dead loads (or the noncomposite dead loads) are applied to the girder-only section. The superimposed dead loads are applied to the composite section based on a modular ratio of 3n or n, whichever gives the higher stresses.

S6.10.3.1

S6.10.3.1.1b

MathCad tip logo

Modular Ratio

As specified in S6.10.3.1.1b, for permanent loads assumed to be applied to the long-term composite section, the slab area shall be transformed by using a modular ratio of 3n or n, whichever gives the higher stresses.

Using a modular ratio of 3n for the superimposed dead loads always gives higher stresses in the steel section. Using a modular ratio of n typically gives higher stresses in the concrete deck, except in the moment reversal regions where the selection of 3n vs. n can become an issue in determining the maximum stress in the deck.

The live loads are applied to the composite section based on a modular ratio of n.

For girders with shear connectors provided throughout their entire length and with slab reinforcement satisfying the provisions of S6.10.3.7, stresses due to loads applied to the composite section for service and fatigue limit states may be computed using the composite section assuming the concrete slab to be fully effective for both positive and negative flexure.

S6.6.1.2.1 & S6.10.5.1

Therefore, for this design example, the concrete slab will be assumed to be fully effective for both positive and negative flexure for service and fatigue limit states.

For this design example, the interior girder controls. In general, both the exterior and interior girders must be considered, and the controlling design is used for all girders, both interior and exterior.

For this design example, only the interior girder design is presented. However, for the exterior girder, the computation of the live load distribution factors and the moment and shear envelopes are also presented.

For the design of an exterior girder, the composite section properties must be computed in accordance with S4.6.2.6.

The modular ratio is computed as follows:

Formula: W subscript c = 0 point 150 kcf

STable 3.5.1-1

Formula: f prime subscript c = 4 point 0ksi

S5.4.2.1 & STable C5.4.2.1-1

Formula: E subscript c = 33000 times ( W subscript c superscript 1 point 5) times square root of (f prime subscript c) Formula: E subscript c = 3834ksi

S5.4.2.4

Formula: E subscript s = 29000ksi

S6.4.1

Formula: n = numerator (E subscript s) divided by denominator (E subscript c) Formula: n = 7 point 6

Therefore, use n = 8.

In lieu of the above computations, the modular ratio can also be obtained from S6.10.3.1.1b. The above computations are presented simply to illustrate the process. Both the above computations and S6.10.3.1.1b result in a modular ratio of 8.

S6.10.3.1.1b

The effective flange width is computed as follows:

S4.6.2.6

For interior beams, the effective flange width is taken as the least of:

  1. One-quarter of the effective span length:

    Assume that the minimum, controlling effective span length equals approximately 60 feet (over the pier).

    Formula: Span subscript eff = 60 feet

    Formula: W subscript eff1 = numerator (Span subscript eff) divided by denominator (4) Formula: W subscript eff1 = 15 point 00 feet

  2. 12.0 times the average thickness of the slab, plus the greater of web thickness or one-half the width of the top flange of the girder:

    Formula: W subscript eff2 = 12 times t subscript effdeck + numerator (14 inches ) divided by denominator (2)

    Formula: W subscript eff2 = 8 point 58 feet

  3. The average spacing of adjacent beams:

    Formula: W subscript eff3 = S Formula: W subscript eff3 = 9 point 75 feet

    Therefore, the effective flange width is:

    Formula: W subscript effflange = min ( W subscript eff1 ,W subscript eff2 ,W subscript eff3 )

    Formula: W subscript effflange = 8 point 58 feet or

    Formula: W subscript effflange = 103 point 0 inches

Based on the concrete deck design example, the total area of longitudinal deck reinforcing steel in the negative moment region is computed as follows:

Formula: A subscript deckreinf = 2 times 0 point 31 inches squared times numerator (W subscript effflange) divided by denominator (5 inches )

Formula: A subscript deckreinf = 12 point 772 inches squared

MathCad tip logo

Slab Haunch

For this design example, the slab haunch is 3.5 inches throughout the length of the bridge. That is, the bottom of the slab is located 3.5 inches above the top of the web. For this design example, this distance is used in computing the location of the centroid of the slab. However, the area of the haunch is not considered in the section properties.

Some states and agencies assume that the slab haunch is zero when computing the section properties.

If the haunch depth is not known, it is conservative to assume that the haunch is zero. If the haunch varies, it is reasonable to use either the minimum value or an average value.

Based on the trial plate sizes shown in Figure 3-4, the noncomposite and composite section properties for the positive moment region are computed as shown in the following table. The distance to the centroid is measured from the bottom of the girder.

Positive Moment Region Section Properties
Section Area, A (Inches2) Centroid, d (Inches) A*d (Inches3) Io (Inches4) A*y2 (Inches4) Itotal (Inches4)
Girder only:
Top flange 8.750 55.188 482.9 0.3 7530.2 7530.5
Web 27.000 27.875 752.6 6561.0 110.5 6671.5
Bottom flange 12.250 0.438 5.4 0.8 7912.0 7912.7
Total 48.000 25.852 1240.9 6562.1 15552.7 22114.8
Composite (3n):
Girder 48.000 25.852 1240.9 22114.8 11134.4 33249.2
Slab 34.333 62.375 2141.5 183.1 15566.5 15749.6
Total 82.333 41.082 3382.4 22297.9 26700.8 48998.7
Composite (n):
Girder 48.000 25.852 1240.9 22114.8 29792.4 51907.2
Slab 103.000 62.375 6424.6 549.3 13883.8 14433.2
Total 151.000 50.765 7665.5 22664.1 43676.2 66340.3
Section ybotgdr (Inches) ytopgdr (Inches) ytopslab(Inches) Sbotgdr (Inches3) Stopgdr (Inches3) Stopslab(Inches3)
Girder only 25.852 29.648 --- 855.5 745.9 ---
Composite (3n) 41.082 14.418 25.293 1192.7 3398.4 1937.2
Composite (n) 50.765 4.735 15.610 1306.8 14010.3 4249.8

Table 3-4 Positive Moment Region Section Properties

Similarly, the noncomposite and composite section properties for the negative moment region are computed as shown in the following table. The distance to the centroid is measured from the bottom of the girder.

For the strength limit state, since the deck concrete is in tension in the negative moment region, the deck reinforcing steel contributes to the composite section properties and the deck concrete does not.

As previously explained, for this design example, the concrete slab will be assumed to be fully effective for both positive and negative flexure for service and fatigue limit states.

S6.6.1.2.1 & S6.10.5.1

Negative Moment Region Section Properties
Section Area, A (Inches2) Centroid, d (Inches) A*d (Inches3) Io (Inches4) A*y2 (Inches4) Itotal (Inches4)
Girder only:
Top flange 35.000 58.000 2030.0 18.2 30009.7 30027.9
Web 27.000 29.750 803.3 6561.0 28.7 6589.7
Bottom flange 38.500 1.375 52.9 24.3 28784.7 28809.0
Total 100.500 28.718 2886.2 6603.5 58823.1 65426.6
Composite (deck concrete using 3n):
Girder 100.500 28.718 2886.2 65426.6 8226.9 73653.5
Slab 34.333 64.250 2205.9 183.1 24081.6 24264.7
Total 134.833 37.766 5092.1 65609.7 32308.5 97918.3
Composite (deck concrete using n):
Girder 100.500 28.718 2886.2 65426.6 32504.5 97931.2
Slab 103.000 64.250 6617.8 549.3 31715.6 32264.9
Total 203.500 46.702 9503.9 65976.0 64220.1 130196.1
Composite (deck reinforcement only):
Girder 100.500 28.718 2886.2 65426.6 1568.1 66994.7
Deck reinf. 12.772 63.750 814.2 0.0 12338.7 12338.7
Total 113.272 32.668 3700.4 65426.6 13906.7 79333.4
Section ybotgdr (Inches) ytopgdr (Inches) ydeck (Inches) Sbotgdr (Inches3) Stopgdr (Inches3) Sdeck (Inches3)
Girder only 28.718 30.532 --- 2278.2 2142.9 ---
Composite (3n) 37.766 21.484 30.484 2592.8 4557.7 3212.1
Composite (n) 46.702 12.548 21.548 2787.8 10376.2 6042.3
Composite (rebar) 32.668 26.582 31.082 2428.5 2984.5 2552.4

Table 3-5 Negative Moment Region Section Properties

Design Step 3.4 - Compute Dead Load Effects

The girder must be designed to resist the dead load effects, as well as the other load effects. The dead load components consist of some dead loads that are resisted by the noncomposite section, as well as other dead loads that are resisted by the composite section. In addition, some dead loads are factored with the DC load factor and other dead loads are factored with the DW load factor. The following table summarizes the various dead load components that must be included in the design of a steel girder.

Dead Load Components
Resisted by Type of Load Factor
DC DW
Noncomposite section
  • Steel girder
  • Concrete deck
  • Concrete haunch
  • Stay-in-place deck forms
  • Miscellaneous dead load (including cross-frames, stiffeners, etc.)
 
Composite section
  • Concrete parapets
  • Future wearing surface

Table 3-6 Dead Load Components

For the steel girder, the dead load per unit length varies due to the change in plate sizes. The moments and shears due to the weight of the steel girder can be computed using readily available analysis software. Since the actual plate sizes are entered as input, the moments and shears are computed based on the actual, varying plate sizes.

For the concrete deck, the dead load per unit length for an interior girder is computed as follows:

Formula: W subscript c = 0 point 150 Kips per cubic foot Formula: S = 9 point 8 feet Formula: t subscript deck = 8 point 5 inches

Formula: DL subscript deck = W subscript c times S times numerator (t subscript deck) divided by denominator (12 inches per foot) Formula: DL subscript deck = 1 point 036 Kips per foot

For the concrete haunch, the dead load per unit length varies due to the change in top flange plate sizes. The moments and shears due to the weight of the concrete haunch can be computed using readily available analysis software. Since the top flange plate sizes are entered as input, the moments and shears due to the concrete haunch are computed based on the actual, varying haunch thickness.

For the stay-in-place forms, the dead load per unit length is computed as follows:

Formula: W subscript deckforms = 0 point 015 ksf Formula: S = 9 point 8 feet Formula: W subscript topflange = 14 inches

Formula: DL subscript deckforms = W subscript deckforms times ( S minus W subscript topflange )

Formula: DL subscript deckforms = 0 point 129 Kips per foot

For the miscellaneous dead load (including cross-frames, stiffeners, and other miscellaneous structural steel), the dead load per unit length is assumed to be as follows:

Formula: DL subscript misc = 0 point 015 Kips per foot

For the concrete parapets, the dead load per unit length is computed as follows, assuming that the superimposed dead load of the two parapets is distributed uniformly among all of the girders:

S4.6.2.2.1

Formula: W subscript par = 0 point 5 Kips per foot Formula: N subscript girders = 5

Formula: DL subscript par = W subscript par times numerator (2) divided by denominator (N subscript girders) Formula: DL subscript par = 0 point 212 Kips per foot

Although S4.6.2.2.1 specifies that permanent loads of and on the deck may be distributed uniformly among the beams, some states assign a larger percentage of the barrier loads to the exterior girders.

For the future wearing surface, the dead load per unit length is computed as follows, assuming that the superimposed dead load of the future wearing surface is distributed uniformly among all of the girders:

S4.6.2.2.1

Formula: W subscript fws = 0 point 140 kcf Formula: t subscript fws = 2 point 5 inches

Formula: w subscript roadway = 44 point 0 feet Formula: N subscript girders = 5

Formula: DL subscript fws = numerator (W subscript fws times numerator (t subscript fws) divided by denominator (12 inches per foot) times w subscript roadway) divided by denominator (N subscript girders) Formula: DL subscript fws = 0 point 257 Kips per foot

Since the plate girder and its section properties are not uniform over the entire length of the bridge, an analysis must be performed to compute the dead load moments and shears. Such an analysis can be performed using one of various computer programs.

MathCad tip logo

Need for Revised Analysis

It should be noted that during the optimization process, minor adjustments can be made to the plate sizes and transition locations without needing to recompute the analysis results. However, if significant adjustments are made, such that the moments and shears would change significantly, then a revised analysis is required.

The following two tables present the unfactored dead load moments and shears, as computed by an analysis computer program (AASHTO Opis software). Since the bridge is symmetrical, the moments and shears in Span 2 are symmetrical to those in Span 1.

Location in Span 1 Steel girder Concrete deck & haunches Other dead loads acting on girder alone Concrete parapets Future wearing surface
1.0L -421.5 -2418.3 -357.1 -436.1 -528.2
0.9L -244.0 -1472.0 -216.9 -255.0 -308.9
0.8L -107.2 -679.7 -99.9 -104.5 -126.6
0.7L -2.5 -43.1 -6.2 15.5 18.8
0.6L 73.6 436.6 64.4 104.9 127.1
0.5L 124.4 758.4 111.7 163.8 198.4
0.4L 150.0 922.4 135.8 192.2 232.7
0.3L 150.3 928.6 136.7 189.9 230.1
0.2L 125.5 776.9 114.3 157.2 190.4
0.1L 75.4 467.4 68.8 93.9 113.7
0.0L 0.0 0.0 0.0 0.0 0.0

Table 3-7 Dead Load Moments (Kip-feet)

Location in Span 1 Steel girder Concrete deck & haunches Other dead loads acting on girder alone Concrete parapets Future wearing surface
1.0L -16.84 -85.18 -12.65 -16.36 -19.82
0.9L -12.74 -72.52 -10.72 -13.82 -16.74
0.8L -10.06 -59.54 -8.78 -11.27 -13.65
0.7L -7.39 -46.55 -6.85 -8.73 -10.57
0.6L -5.29 -33.40 -4.91 -6.18 -7.49
0.5L -3.18 -20.24 -2.98 -3.63 -4.40
0.4L -1.08 -7.09 -1.04 -1.09 -1.32
0.3L 1.02 6.06 0.89 1.46 1.77
0.2L 3.12 19.22 2.83 4.00 4.85
0.1L 5.23 32.37 4.76 6.55 7.93
0.0L 7.33 45.53 6.70 9.10 11.02

Table 3-8 Dead Load Shears (kips)

Design Step 3.5 - Compute Live Load Effects

MathCad tip logo

LRFD Live Load

There are several differences between the live load used in Allowable Stress Design (ASD) or Load Factor Design (LFD) and the live load used in Load and Resistance Factor Design (LRFD). Some of the more significant differences are:

  • In ASD and LFD, the basic live load designation is HS20 or HS25. In LRFD, the basic live load designation is HL-93.

  • In ASD and LFD, the live load consists of either a truck load or a lane load and concentrated loads. In LRFD, the load consists of a design truck or tandem, combined with a lane load.

  • In ASD and LFD, the two concentrated loads are combined with lane load to compute the maximum negative live load moment. In LRFD, 90% of the effect of two design trucks at a specified distance is combined with 90% of the lane load to compute the maximum negative live load moment.

  • In ASD and LFD, the term "impact" is used for the dynamic interaction between the bridge and the moving vehicles. In LRFD, the term "dynamic load allowance" is used instead of "impact."

  • In ASD and LFD, impact is applied to the entire live load. In LRFD, dynamic load allowance is applied only to the design truck and design tandem.

For additional information about the live load used in LRFD, refer to S3.6 and C3.6.

The girder must also be designed to resist the live load effects. The live load consists of an HL-93 loading. Similar to the dead load, the live load moments and shears for an HL-93 loading can be obtained from an analysis computer program.

S3.6.1.2

Based on Table 3-3, for all limit states other than fatigue and fracture, the dynamic load allowance, IM, is as follows:

S3.6.2.1

Formula: IM = 0 point 33

The live load distribution factors for moment for an interior girder are computed as follows:

S4.6.2.2.2

First, the longitudinal stiffness parameter, Kg, must be computed:

S4.6.2.2.1

Formula: K subscript g = n times ( I + A times e subscript g squared )

Longitudinal Stiffness Parameter, Kg
  Region A
(Pos. Mom.)
Region B
(Intermediate)
Region C
(At Pier)
Weighted Average
Length (Feet) 84 24 12  
n 8 8 8  
I (Inches4) 22,114.8 34,639.8 65,426.6  
A (Inches2) 48.000 63.750 100.500  
eg (Inches) 36.523 35.277 35.532  
Kg (Inches4) 689,147 911,796 1,538,481 818,611
* Weighted average is estimated based on length of each region

Table 3-9 Longitudinal Stiffness Parameter

After the longitudinal stiffness parameter is computed, STable 4.6.2.2.1-1 is used to find the letter corresponding with the superstructure cross section. The letter corresponding with the superstructure cross section in this design example is "a."

If the superstructure cross section does not correspond with any of the cross sections illustrated in STable 4.6.2.2.1-1, then the bridge should be analyzed as presented in S4.6.3.

Based on cross section "a," STables 4.6.2.2.2b-1 and 4.6.2.2.2.3a-1 are used to compute the distribution factors for moment and shear, respectively.

S4.6.2.2.1

Check the range of applicability as follows:

STable 4.6.2.2.2b-1

Formula: 3 point 5 less than or equal to S less than or equal to 16 point 0

Formula: S = 9 point 75feet OK

Formula: 4 point 5 less than or equal to t subscript s less than or equal to 12 point 0

Formula: t subscript s = 8 point 0inches OK

Formula: 20 less than or equal to L less than or equal to 240

Formula: L = 120feet OK

Formula: N subscript b greater than or equal to 4

Formula: N subscript b = 5 OK

Formula: 10000 less than or equal to K subscript g less than or equal to 7000000

Formula: K subscript g = 818611 inches superscript 4 OK

For one design lane loaded, the distribution of live load per lane for moment in interior beams is as follows:

STable 4.6.2.2.2b-1

Formula: g subscript int_moment_1 = 0 point 06 + ( numerator (S) divided by denominator (14) ) superscript 0 point 4 ( numerator (S) divided by denominator (L) ) superscript 0 point 3 left bracket numerator (K subscript g) divided by denominator (12 point 0L times ( t subscript s ) cubed ) right bracket superscript 0 point 1

Formula: g subscript int_moment_1 = 0 point 472 lanes

For two or more design lanes loaded, the distribution of live load per lane for moment in interior beams is as follows:

STable 4.6.2.2.2b-1

Formula: g subscript int_moment_2 = 0 point 075 + ( numerator (S) divided by denominator (9 point 5) ) superscript 0 point 6 ( numerator (S) divided by denominator (L) ) superscript 0 point 2 left bracket numerator (K subscript g) divided by denominator (12 point 0 times L times ( t subscript s ) cubed ) right bracket superscript 0 point 1

Formula: g subscript int_moment_2 = 0 point 696 lanes

The live load distribution factors for shear for an interior girder are computed in a similar manner. The range of applicability is similar to that for moment.

STable 4.6.2.2.3a-1

For one design lane loaded, the distribution of live load per lane for shear in interior beams is as follows:

STable 4.6.2.2.3a-1

Formula: g subscript int_shear_1 = 0 point 36 + numerator (S) divided by denominator (25 point 0)

Formula: g subscript int_shear_1 = 0 point 750 lanes

For two or more design lanes loaded, the distribution of live load per lane for shear in interior beams is as follows:

STable 4.6.2.2.3a-1

Formula: g subscript int_shear_2 = 0 point 2 + numerator (S) divided by denominator (12) minus ( numerator (S) divided by denominator (35) ) superscript 2 point 0

Formula: g subscript int_shear_2 = 0 point 935 lanes

Since this bridge has no skew, the skew correction factor does not need to be considered for this design example.

S4.6.2.2.2e, S4.6.2.2.3c

This design example is based on an interior girder. However, for illustrative purposes, the live load distribution factors for an exterior girder are computed below, as follows:

S4.6.2.2.2

The distance, de, is defined as the distance between the web centerline of the exterior girder and the interior edge of the curb. For this design example, based on Figure 3-2:

Formula: d subscript e = 2 point 50 feet

Check the range of applicability as follows:

STable 4.6.2.2.2d-1

Formula: minus 1 point 0 less than or equal to d subscript e less than or equal to 5 point 5

Formula: d subscript e = 2 point 50feet OK

For one design lane loaded, the distribution of live load per lane for moment in exterior beams is computed using the lever rule, as follows:

STable 4.6.2.2.2d-1

Starting from left and going right. The figure shows an overhang, distance from the edge of deck to the first beam, of 3 feet 11 and one quarter inches and a beam spacing of 9 feet 9 inches. There is an assumed hinge at beam number 2. Again starting from left and going right. The parapet width is 1 foot 5 and one quarter inches wide. The distance from the face of the parapet to the first wheel load location is 2 feet 0 inches. The distance from the first wheel load to second wheel load is 6 feet. The distance from the second wheel load to beam number 2 is 4 feet 3 inches.

Figure 3-5 Lever Rule

Formula: g subscript ext_moment_1 = numerator (( 0 point 5) times ( 4 point 25 feet ) + ( 0 point 5) times ( 10 point 25 feet )) divided by denominator (9 point 75 feet )

Formula: g subscript ext_moment_1 = 0 point 744 lanes

Formula: Multiple_presence_factor = 1 point 20

Formula: g subscript ext_moment_1 = g subscript ext_moment_1 times Multiple_presence_factor

Formula: g subscript ext_moment_1 = 0 point 892 lanes (for strength limit state)

For two or more design lanes loaded, the distribution of live load per lane for moment in exterior beams is as follows:

STable 4.6.2.2.2d-1

Formula: e = 0 point 77 + numerator (d subscript e) divided by denominator (9 point 1) Formula: e = 1 point 045

Formula: g subscript ext_moment_2 = e times g subscript int_moment_2

Formula: g subscript ext_moment_2 = 0 point 727 lanes

The live load distribution factors for shear for an exterior girder are computed in a similar manner. The range of applicability is similar to that for moment.

STable 4.6.2.2.3b-1

For one design lane loaded, the distribution of live load per lane for shear in exterior beams is computed using the lever rule, as illustrated in Figure 3-5 and as follows:

STable 4.6.2.2.3b-1

Formula: g subscript ext_shear_1 = numerator (( 0 point 5) times ( 4 point 25 feet ) + ( 0 point 5) times ( 10 point 25 feet )) divided by denominator (9 point 75 feet )

Formula: g subscript ext_shear_1 = 0 point 744 lanes

Formula: Multiple_presence_factor = 1 point 20

Formula: g subscript ext_shear_1 = g subscript ext_shear_1 times Multiple_presence_factor

Formula: g subscript ext_shear_1 = 0 point 892 lanes (for strength limit state)

For two or more design lanes loaded, the distribution of live load per lane for shear in exterior beams is as follows:

STable 4.6.2.2.3b-1

Formula: e = 0 point 6 + numerator (d subscript e) divided by denominator (10) Formula: e = 0 point 850

Formula: g subscript ext_shear_2 = e times g subscript int_shear_2

Formula: g subscript ext_shear_2 = 0 point 795 lanes

In beam-slab bridge cross-sections with diaphragms or cross-frames, the distribution factor for the exterior beam can not be taken to be less than that which would be obtained by assuming that the cross-section deflects and rotates as a rigid cross-section. CEquation 4.6.2.2.2d-1 provides one approximate approach to satisfy this requirement. The multiple presence factor provisions of S3.6.1.1.2 must be applied when this equation is used.

S4.6.2.2.2d

Since this bridge has no skew, the skew correction factor does not need to be considered for this design example.

S4.6.2.2.2e, S4.6.2.2.3c

The following table presents the unfactored maximum positive and negative live load moments and shears for HL-93 live loading for interior beams, as computed using an analysis computer program. These values include the live load distribution factor, and they also include dynamic load allowance. Since the bridge is symmetrical, the moments and shears in Span 2 are symmetrical to those in Span 1.

Live Load Effects (for Interior Beams)
Location in Span 1 Maximum positive moment (K-ft) Maximum negative moment (K-ft) Maximum positive shear (kips) Maximum negative shear (kips)
1.0L 983 -2450 35.8 -131.4
0.9L 865 -1593 33.0 -118.5
0.8L 1006 -1097 32.1 -105.1
0.7L 1318 -966 33.5 -91.1
0.6L 1628 -966 37.1 -76.7
0.5L 1857 -968 42.5 -62.2
0.4L 1908 -905 49.6 -47.8
0.3L 1766 -777 61.0 -36.4
0.2L 1422 -583 76.6 -29.1
0.1L 836 -324 93.7 -28.7
0.0L 0 0 110.5 -33.8

Table 3-10 Live Load Effects (for Interior Beams)

The design live load values for HL-93 loading, as presented in the previous table, are computed based on the product of the live load effect per lane and live load distribution factor. These values also include the effects of dynamic load allowance. However, it is important to note that the dynamic load allowance is applied only to the design truck or tandem. The dynamic load allowance is not applied to pedestrian loads or to the design lane load.

S3.6.1, S3.6.2, S4.6.2.2

Design Step 3.6 - Combine Load Effects

After the load factors and load combinations have been established (see Design Step 3.1), the section properties have been computed (see Design Step 3.3), and all of the load effects have been computed (see Design Steps 3.4 and 3.5), the force effects must be combined for each of the applicable limit states.

For this design example, η equals 1.00. (For more detailed information about η, refer to Design Step 1.)

S1.3

Based on the previous design steps, the maximum positive moment (located at 0.4L) for the Strength I Limit State is computed as follows:

S3.4.1

Formula: LF subscript DC = 1 point 25

Formula: M subscript DC = 150 point 0K feet + 922 point 4K feet + 135 point 8K feet + 192 point 2K feet

Formula: M subscript DC = 1400 point 4 K feet

Formula: LF subscript DW = 1 point 50

Formula: M subscript DW = 232 point 7K feet

Formula: LF subscript LL = 1 point 75

Formula: M subscript LL = 1908K feet

Formula: M subscript total = LF subscript DC times M subscript DC + LF subscript DW times M subscript DW + LF subscript LL times M subscript LL

Formula: M subscript total = 5439 K feet

Similarly, the maximum stress in the top of the girder due to positive moment (located at 0.4L) for the Strength I Limit State is computed as follows:

Noncomposite dead load:

Formula: M subscript noncompDL = 150 point 0K feet + 922 point 4K feet + 135 point 8K feet

Formula: M subscript noncompDL = 1208 point 2 K feet

Formula: S subscript topgdr = 745 point 9 inches cubed

Formula: f subscript noncompDL = numerator ( minus M subscript noncompDL times ( numerator (12 inches ) divided by denominator ( feet ) )) divided by denominator (S subscript topgdr)

Formula: f subscript noncompDL = minus 19 point 44 ksi

Parapet dead load (composite):

Formula: M subscript parapet = 192 point 2K feet Formula: S subscript topgdr = 3398 point 4 inches cubed

Formula: f subscript parapet = numerator ( minus M subscript parapet times ( numerator (12 inches ) divided by denominator ( feet ) )) divided by denominator (S subscript topgdr) Formula: f subscript parapet = minus 0 point 68 ksi

Future wearing surface dead load (composite):

Formula: M subscript fws = 232 point 7K feet Formula: S subscript topgdr = 3398 point 4 inches cubed

Formula: f subscript fws = numerator ( minus M subscript fws times ( numerator (12 inches ) divided by denominator ( feet ) )) divided by denominator (S subscript topgdr) Formula: f subscript fws = minus 0 point 82 ksi

Live load (HL-93) and dynamic load allowance:

Formula: M subscript LL = 1908 K feet

Formula: S subscript topgdr = 14010 point 3 inches cubed

Formula: f subscript LL = numerator ( minus M subscript LL times ( numerator (12 inches ) divided by denominator ( feet ) )) divided by denominator (S subscript topgdr) Formula: f subscript LL = minus 1 point 63 ksi

Multiplying the above stresses by their respective load factors and adding the products results in the following combined stress for the Strength I Limit State:

S3.4.1

Formula: f subscript Str = ( LF subscript DC times f subscript noncompDL ) + ( LF subscript DC times f subscript parapet ) + ( LF subscript DW times f subscript fws ) + ( LF subscript LL times f subscript LL )

Formula: f subscript Str = minus 29 point 24 ksi

Similarly, all of the combined moments, shears, and flexural stresses can be computed at the controlling locations. A summary of those combined load effects for an interior beam is presented in the following three tables, summarizing the results obtained using the procedures demonstrated in the above computations.

Combined Effects at Location of Maximum Positive Moment
Summary of Unfactored Values:
Loading Moment (K-ft) fbotgdr (ksi) ftopgdr (ksi) ftopslab (ksi)
Noncomposite DL 1208 16.95 -19.44 0.00
Parapet DL 192 1.93 -0.68 -0.05
FWS DL 233 2.34 -0.82 -0.06
LL - HL-93 1908 17.52 -1.63 -0.67
LL - Fatigue 563 5.17 -0.48 -0.20
Summary of Factored Values:
Limit State Moment (K-ft) fbotgdr (ksi) ftopgdr (ksi) ftopslab (ksi)
Strength I 5439 57.77 -29.24 -1.33
Service II 4114 44.00 -23.06 -0.99
Fatigue 422 3.87 -0.36 -0.15

Table 3-11 Combined Effects at Location of Maximum Positive Moment

As shown in the above table, the Strength I Limit State elastic stress in the bottom of the girder exceeds the girder yield stress. However, for this design example, this value is not used because of the local yielding that occurs at this section.

Combined Effects at Location of Maximum Negative Moment
Summary of Unfactored Values (Assuming Concrete Not Effective):
Loading Moment (K-ft) fbotgdr (ksi) ftopgdr (ksi) fdeck (ksi)
Noncomposite DL -3197 -16.84 17.90 0.00
Parapet DL -436 -2.15 1.75 2.05
FWS DL -528 -2.61 2.12 2.48
LL - HL-93 -2450 -12.11 9.85 11.52
Summary of Unfactored Values (Assuming Concrete Effective):
Loading Moment (K-ft) fbotgdr (ksi) ftopgdr (ksi) fdeck (ksi)
Noncomposite DL -3197 -16.84 17.90 0.00
Parapet DL -436 -2.02 1.15 0.07
FWS DL -528 -2.44 1.39 0.08
LL - HL-93 -2450 -10.55 2.83 0.61
LL - Fatigue -406 -1.75 0.47 0.10
Summary of Factored Values:
Limit State Moment (K-ft) fbotgdr (ksi) ftopgdr (ksi) fdeck (ksi)
Strength I* -9621 -48.84 44.99 26.44
Service II** -7346 -35.01 24.12 0.94
Fatigue ** -305 -1.31 0.35 0.08

Legend:

*Strength I Limit State stresses are based on section properties assuming the deck concrete is not effective, and fdeck is the stress in the deck reinforcing steel.
**Service II and Fatigue Limit State stresses are based on section properties assuming the deck concrete is effective, and fdeck is the stress in the deck concrete.

Table 3-12 Combined Effects at Location of Maximum Negative Moment

Combined Effects at Location of Maximum Shear
Summary of Unfactored Values:
Loading Shear (kips)
Noncomposite DL 114.7
Parapet DL 16.4
FWS DL 19.8
LL - HL-93 131.4
LL - Fatigue 46.5
Summary of Factored Values:
Limit State Shear (kips)
Strength I 423.5
Service II 321.7
Fatigue 34.8

Table 3-13 Combined Effects at Location of Maximum Shear

Envelopes of the factored Strength I moments and shears are presented in the following two figures. Maximum and minimum values are presented, and values for both interior and exterior girders are presented. Based on these envelopes, it can be seen that the interior girder controls the design, and all remaining design computations are based on the interior girder.

Figure 3-6 Envelope of Strength I Moments. The figure shows a graph of factored moment with units of kip feet along the vertical axis versus distance with units of feet along the horizontal axis. Maximum and minimum values are presented, and values for both interior and exterior girders are presented. Based on these envelopes, it can be seen that the interior girder controls the design, and all remaining design computations are based on the interior girder.

Figure 3-6 Envelope of Strength I Moments

Figure 3-7 Envelope of Strength I Shears. The figure shows a graph of factored shear with units of kips along the vertical axis versus distance with units of feet along the horizontal axis. Maximum and minimum values are presented, and values for both interior and exterior girders are presented. Based on these envelopes, it can be seen that the interior girder controls the design, and all remaining design computations are based on the interior girder.

Figure 3-7 Envelope of Strength I Shears

Design Steps 3.7 through 3.17 consist of verifying the structural adequacy of critical beam locations using appropriate sections of the Specifications.

For this design example, two design sections will be checked for illustrative purposes. First, all specification checks for Design Steps 3.7 through 3.17 will be performed for the location of maximum positive moment, which is at 0.4L in Span 1. Second, all specification checks for these same design steps will be performed for the location of maximum negative moment and maximum shear, which is at the pier.

MathCad tip logo

Specification Check Locations

For steel girder designs, specification checks are generally performed using a computer program at the following locations:

  • Span tenth points

  • Locations of plate transitions

  • Locations of stiffener spacing transitions

However, it should be noted that the maximum moment within a span may not necessarily occur at any of the above locations.

The following specification checks are for the location of maximum positive moment, which is at 0.4L in Span 1, as shown in Figure 3-8.

A girder elevation showing the location of the maximum positive moment. The location of the maximum negative moment is 0 point 4 times the span length of 120 feet 0 inches which is 48 feet 0 inches from the centerline of bearing. The girder is symmetrical about the pier centerline.

Figure 3-8 Location of Maximum Positive Moment

Design Step 3.7 - Check Section Proportion Limits - Positive Moment Region

Several checks are required to ensure that the proportions of the trial girder section are within specified limits.

S6.10.2

The first section proportion check relates to the general proportions of the section. The flexural components must be proportioned such that:

S6.10.2.1

Formula: 0 point 1 less than or equal to numerator (I subscript yc) divided by denominator (I subscript y) less than or equal to 0 point 9

Formula: I subscript yc = numerator (0 point 625 inches times ( 14 inches ) cubed ) divided by denominator (12) Formula: I subscript yc = 142 point 9 inches superscript 4

Formula: I subscript y = numerator (0 point 625 inches times ( 14 inches ) cubed ) divided by denominator (12) + numerator (54 inches times ( numerator (1) divided by denominator (2) inches ) cubed ) divided by denominator (12) + numerator (0 point 875 inches times ( 14 inches ) cubed ) divided by denominator (12)

Formula: I subscript y = 343 point 6 inches superscript 4

Formula: numerator (I subscript yc) divided by denominator (I subscript y) = 0 point 416 OK

The second section proportion check relates to the web slenderness. For a section without longitudinal stiffeners, the web must be proportioned such that:

S6.10.2.2

Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) less than or equal to 6 point 77 times square root of ( numerator (E) divided by denominator (f subscript c)) less than or equal to 200

For the Strength I limit state at 0.4L in Span 1 (the location of maximum positive moment):

S6.10.3.1.4a

Formula: f subscript botgdr = 57 point 77 ksi (see Table 3-11 and explanation below table)

Formula: f subscript topgdr = minus 29 point 24 ksi (see Table 3-11)

Formula: t subscript topfl = 0 point 625 inches (see Figure 3-4)

Formula: D subscript web = 54 inches (see Figure 3-4)

Formula: t subscript botfl = 0 point 875 inches (see Figure 3-4)

Formula: Depth subscript gdr = t subscript topfl + D subscript web + t subscript botfl

Formula: Depth subscript gdr = 55 point 50 inches

Formula: Depth subscript comp = numerator ( minus f subscript topgdr) divided by denominator (f subscript botgdr minus f subscript topgdr) times Depth subscript gdr

C6.10.3.1.4a

Formula: Depth subscript comp = 18 point 65 inches

Formula: D subscript c = Depth subscript comp minus t subscript topfl

Formula: D subscript c = 18 point 03 inches

Formula: t subscript w = numerator (1) divided by denominator (2) inches (see Figure 3-4)

Formula: E = 29000ksi

S6.4.1

Formula: f subscript c = minus f subscript topgdr

Formula: f subscript c = 29 point 24 ksi

Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) = 72 point 1

Formula: 6 point 77 times square root of ( numerator (E) divided by denominator (f subscript c)) = 213 point 2

Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) less than or equal to 6 point 77 times square root of ( numerator (E) divided by denominator (f subscript c)) and Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) less than or equal to 200 OK

The third section proportion check relates to the flange proportions. The compression flanges on fabricated I-sections must be proportioned such that:

S6.10.2.3

Formula: b subscript f greater than or equal to 0 point 3 times D subscript c

Formula: b subscript f = 14 inches (see Figure 3-4)

Formula: D subscript c = 18 point 03 inches

Formula: 0 point 3 times D subscript c = 5 point 41 inches

Formula: b subscript f greater than or equal to 0 point 3 times D subscript c OK

According to C6.10.2.3, it is preferable for the flange width to be greater than or equal to 0.4Dc. In this case, the flange width is greater than both 0.3Dc and 0.4Dc, so this requirement is clearly satisfied.

C6.10.2.3

In addition to the compression flange check, the tension flanges on fabricated I-sections must be proportioned such that:

S6.10.2.3

Formula: numerator (b subscript t) divided by denominator (2 times t subscript t) less than or equal to 12 point 0

Formula: b subscript t = 14 inches (see Figure 3-4)

Formula: t subscript t = 0 point 875 inches (see Figure 3-4)

Formula: numerator (b subscript t) divided by denominator (2 times t subscript t) = 8 point 0 OK

Design Step 3.8 - Compute Plastic Moment Capacity - Positive Moment Region

For composite sections, the plastic moment, Mp, is calculated as the first moment of plastic forces about the plastic neutral axis.

S6.10.3.1.3

The width of the bottom flange is defined as b sub t and the bottom flange thickness is defined as t sub t. The web depth is defined as D sub w and the web thickness is defined as t sub w. The width of the top flange is defined as b sub c and the top flange thickness is defined as t sub c. The slab thickness is defined as t sub s and the slab width is defined as b sub s. The distance from the top of the slab to the plastic neutral axis is defined as Y. The tension force in the bottom flange is defined by P sub t and the tension in the web is defined as P sub w. The tension force in the top flange is defined by P sub c and the tension in the slab is defined as P sub s.

Figure 3-9 Computation of Plastic Moment Capacity for Positive Bending Sections

For the tension flange:

SAppendix A6.1

Formula: F subscript yt = 50ksi Formula: b subscript t = 14 inches Formula: t subscript t = 0 point 875 inches

Formula: P subscript t = F subscript yt times b subscript t times t subscript t Formula: P subscript t = 613 K

For the web:

Formula: F subscript yw = 50 point 0 ksi Formula: D subscript w = 54 inches Formula: t subscript w = 0 point 50 inches

Formula: P subscript w = F subscript yw times D subscript w times t subscript w Formula: P subscript w = 1350 K

For the compression flange:

Formula: F subscript yc = 50ksi Formula: b subscript c = 14 inches Formula: t subscript c = 0 point 625 inches

Formula: P subscript c = F subscript yc times b subscript c times t subscript c Formula: P subscript c = 438 K

For the slab:

Formula: f prime subscript c = 4 point 0ksi Formula: b subscript s = 103 inches Formula: t subscript s = 8 point 0 inches

Formula: P subscript s = 0 point 85 times f prime subscript c times b subscript s times t subscript s Formula: P subscript s = 2802 K

The forces in the longitudinal reinforcement may be conservatively neglected.

C6.10.3.1.3

Check the location of the plastic neutral axis, as follows:

SAppendix A6.1

Formula: P subscript t + P subscript w = 1963 K Formula: P subscript c + P subscript s = 3239 K

Formula: P subscript t + P subscript w + P subscript c = 2400 K Formula: P subscript s = 2802 K

Therefore, the plastic neutral axis is located within the slab.

Formula: Y = ( t subscript s ) times ( numerator (P subscript c + P subscript w + P subscript t) divided by denominator (P subscript s) )

STable A6.1-1

Formula: Y = 6 point 85 inches

Check that the position of the plastic neutral axis, as computed above, results in an equilibrium condition in which there is no net axial force.

Formula: Compression = 0 point 85 times f prime subscript c times b subscript s times Y

Formula: Compression = 2400 K

Formula: Tension = P subscript t + P subscript w + P subscript c

Formula: Tension = 2400 K OK

The plastic moment, Mp, is computed as follows, where d is the distance from an element force (or element neutral axis) to the plastic neutral axis:

STable A6.1-1

Formula: d subscript c = numerator ( minus t subscript c) divided by denominator (2) + 3 point 5 inches + t subscript s minus Y Formula: d subscript c = 4 point 33 inches

Formula: d subscript w = numerator (D subscript w) divided by denominator (2) + 3 point 5 inches + t subscript s minus Y Formula: d subscript w = 31 point 65 inches

Formula: d subscript t = numerator (t subscript t) divided by denominator (2) + D subscript w + 3 point 5 inches + t subscript s minus Y Formula: d subscript t = 59 point 08 inches

Formula: M subscript p = numerator (Y squared times P subscript s) divided by denominator (2 times t subscript s) + ( P subscript c times d subscript c + P subscript w times d subscript w + P subscript t times d subscript t )

Formula: M subscript p = 7419 K feet

Design Step 3.9 - Determine if Section is Compact or Noncompact - Positive Moment Region

The next step in the design process is to determine if the section is compact or noncompact. This, in turn, will determine which formulae should be used to compute the flexural capacity of the girder.

Where the specified minimum yield strength does not exceed 70.0 ksi, and the girder has a constant depth, and the girder does not have longitudinal stiffeners or holes in the tension flange, then the first step is to check the compact-section web slenderness provisions, as follows:

S6.10.4.1.1

Formula: numerator (2 times D subscript cp) divided by denominator (t subscript w) less than or equal to 3 point 76 times square root of ( numerator (E) divided by denominator (F subscript yc))

S6.10.4.1.2

Since the plastic neutral axis is located within the slab,

Formula: D subscript cp = 0 inches

Therefore the web is deemed compact. Since this is a composite section in positive flexure, the flexural resistance is computed as defined by the composite compact-section positive flexural resistance provisions of S6.10.4.2.2.

S6.10.4.1.2

For composite sections in positive flexure in their final condition, the provisions of S6.10.4.1.3, S6.10.4.1.4, S6.10.4.1.6a, S6.10.4.1.7, and S6.10.4.1.9 are considered to be automatically satisfied.

The section is therefore considered to be compact.

CFigure 6.10.4-1

Design Step 3.10 - Design for Flexure - Strength Limit State - Positive Moment Region

Since the section was determined to be compact, and since it is a composite section in the positive moment region, the flexural resistance is computed in accordance with the provisions of S6.10.4.2.2.

This is neither a simple span nor a continuous span with compact sections in the negative flexural region over the interior supports. (This will be proven in the negative flexure region computations of this design example.) Therefore, the nominal flexural resistance is determined using the following equation, based on the approximate method:

SFigure C6.10.4-1

S6.10.4.2.2a

Formula: M subscript n = 1 point 3 times R subscript h times M subscript y

All design sections of this girder are homogenous. That is, the same structural steel is used for the top flange, the web, and the bottom flange. Therefore, the hybrid factor, Rh, is as follows:

S6.10.4.3.1

Formula: R subscript h = 1 point 0

The yield moment, My, is computed as follows:

SAppendix A6.2

Formula: F subscript y = numerator (M subscript D1) divided by denominator (S subscript NC) + numerator (M subscript D2) divided by denominator (S subscript LT) + numerator (M subscript AD) divided by denominator (S subscript ST) Formula: M subscript y = M subscript D1 + M subscript D2 + M subscript AD

Formula: F subscript y = 50ksi

Formula: M subscript D1 = ( 1 point 25 times 1208K feet )

Formula: M subscript D1 = 1510 K feet

Formula: M subscript D2 = ( 1 point 25 times 192K feet ) + ( 1 point 50 times 233K feet )

Formula: M subscript D2 = 590 K feet

For the bottom flange:

Formula: S subscript NC = 855 point 5 inches cubed

Formula: S subscript LT = 1192 point 7 inches cubed

Formula: S subscript ST = 1306 point 8 inches cubed

Formula: M subscript AD = left bracket S subscript ST times ( F subscript y minus numerator (M subscript D1) divided by denominator (S subscript NC) minus numerator (M subscript D2) divided by denominator (S subscript LT) )right bracket times ( numerator (1 feet ) divided by denominator (12 inches ) )

Formula: M subscript AD = 2493 K feet

Formula: M subscript ybot = M subscript D1 + M subscript D2 + M subscript AD

Formula: M subscript ybot = 4592 K feet

For the top flange:

Formula: S subscript NC = 745 point 9 inches cubed

Formula: S subscript LT = 3398 point 4 inches cubed

Formula: S subscript ST = 14010 point 3 inches cubed

Formula: M subscript AD = S subscript ST times ( F subscript y minus numerator (M subscript D1) divided by denominator (S subscript NC) minus numerator (M subscript D2) divided by denominator (S subscript LT) )

Formula: M subscript AD = 27584 K feet

Formula: M subscript ytop = M subscript D1 + M subscript D2 + M subscript AD

Formula: M subscript ytop = 29683 K feet

The yield moment, My, is the lesser value computed for both flanges. Therefore, My is determined as follows:

SAppendix A6.2

Formula: M subscript y = min ( M subscript ybot , M subscript ytop )

Formula: M subscript y = 4592 K feet

Therefore, for the positive moment region of this design example, the nominal flexural resistance is computed as follows:

S6.10.4.2.2a

Formula: M subscript n = 1 point 3 times R subscript h times M subscript y

Formula: M subscript n = 5970 K feet

In addition, the nominal flexural resistance can not be taken to be greater than the applicable value of Mn computed from either SEquation 6.10.4.2.2a-1 or 6.10.4.2.2a-2.

S6.10.4.2.2a

Formula: D subscript p = Y Formula: D subscript p = 6 point 85 inches

S6.10.4.2.2b

Formula: D prime = beta times numerator (( d + t subscript s + t subscript h )) divided by denominator (7 point 5)

Formula: beta = 0 point 7 for Fy = 50 ksi

Formula: d = Depth subscript gdr Formula: d = 55 point 50 inches

Formula: t subscript s = 8 point 0 inches

Formula: t subscript h = 3 point 5 inches minus 0 point 625 inches Formula: t subscript h = 2 point 875 inches

Formula: D prime = beta times numerator (( d + t subscript s + t subscript h )) divided by denominator (7 point 5)

Formula: D prime = 6 point 19 inches

Formula: 5 times D prime = 30 point 97 inches

Therefore Formula: D prime less than or equal to D subscript p less than or equal to 5 times D prime

S6.10.4.2.2a

Formula: M subscript n = numerator (5 times M subscript p minus 0 point 85 times M subscript y) divided by denominator (4) + numerator (0 point 85 times M subscript y minus M subscript p) divided by denominator (4) times ( numerator (D subscript p) divided by denominator (D prime) )

Formula: M subscript n = 7326 K feet

Therefore, use Formula: M subscript n = 5970 K feet

The ductility requirement in S6.10.4.2.2b is checked as follows:

S6.10.4.2.2b

Formula: numerator (D subscript p) divided by denominator (D prime) = 1 point 1 Formula: numerator (D subscript p) divided by denominator (D prime) less than or equal to 5 OK

The factored flexural resistance, Mr, is computed as follows:

S6.10.4

Formula: phi subscript f = 1 point 00

S6.5.4.2

Formula: M subscript r = phi subscript f times M subscript n

Formula: M subscript r = 5970 K feet

The positive flexural resistance at this design section is checked as follows:

S1.3.2.1

Formula: Sigma eta subscript i times gamma subscript i times Q subscript i less than or equal to R subscript r

or in this case:

Formula: Sigma eta subscript i times gamma subscript i times M subscript i less than or equal to M subscript r

For this design example,

Formula: eta subscript i = 1 point 00

As computed in Design Step 3.6,

Formula: Sigma gamma subscript i times M subscript i = 5439K feet

Therefore

Formula: Sigma eta subscript i times gamma subscript i times M subscript i = 5439 K feet

Formula: M subscript r = 5970 K feet OK

MathCad tip logo

Available Plate Thicknesses

Based on the above computations, the flexural resistance is approximately 10% greater than the factored design moment, yielding a slightly conservative design. This degree of conservatism can generally be adjusted by changing the plate dimensions as needed.

However, for this design example, the web dimensions and the flange width were set based on the girder design requirements at the pier. In addition, the flange thicknesses could not be reduced any further due to limitations in plate thicknesses or because such a reduction would result in a specification check failure.

Available plate thicknesses can be obtained from steel fabricators. As a rule of thumb, the following plate thicknesses are generally available from steel fabricators:

3/16" to 3/4" - increments of 1/16"
3/4" to 1 1/2" - increments of 1/8"
1 1/2" to 4" - increments of 1/4"


Design Step 3.11 - Design for Shear - Positive Moment Region

Shear must be checked at each section of the girder. However, shear is minimal at the location of maximum positive moment, and it is maximum at the pier.

Therefore, for this design example, the required shear design computations will be presented later for the girder design section at the pier.

S6.10.7

It should be noted that in end panels, the shear is limited to either the shear yield or shear buckling in order to provide an anchor for the tension field in adjacent interior panels. Tension field is not allowed in end panels. The design procedure for shear in the end panel is presented in S6.10.7.3.3c.

S6.10.7.3.3c

Design Step 3.12 - Design Transverse Intermediate Stiffeners - Positive Moment Region

The girder in this design example has transverse intermediate stiffeners. Transverse intermediate stiffeners are used to increase the shear resistance of the girder.

As stated above, shear is minimal at the location of maximum positive moment but is maximum at the pier. Therefore, the required design computations for transverse intermediate stiffeners will be presented later for the girder design section at the pier.

S6.10.8.1

Design Step 3.14 - Design for Flexure - Fatigue and Fracture Limit State - Positive Moment Region

Load-induced fatigue must be considered in a plate girder design. Fatigue considerations for plate girders may include:

  1. Welds connecting the shear studs to the girder.
  2. Welds connecting the flanges and the web.
  3. Welds connecting the transverse intermediate stiffeners to the girder.

The specific fatigue considerations depend on the unique characteristics of the girder design. Specific fatigue details and detail categories are explained and illustrated in STable 6.6.1.2.3-1 and in SFigure 6.6.1.2.3-1.

For this design example, fatigue will be checked for the fillet-welded connection of the transverse intermediate stiffeners to the girder. This detail corresponds to Illustrative Example 6 in SFigure 6.6.1.2.3-1, and it is classified as Detail Category C' in STable 6.6.1.2.3-1.

S6.6.1

STable 6.6.1.2.3-1


SFigure 6.6.1.2.3-1

For this design example, the fillet-welded connection of the transverse intermediate stiffeners will be checked at the location of maximum positive moment. The fatigue detail is located at the inner fiber of the tension flange, where the transverse intermediate stiffener is welded to the flange. However, for simplicity, the computations will conservatively compute the fatigue stress at the outer fiber of the tension flange.

The fatigue detail being investigated in this design example is illustrated in the following figure:

Girder under a moment condition showing a detail of fillet welds between a tranverse intermediate stiffener and the bottom flange and web of the girder.

Figure 3-10 Load-Induced Fatigue Detail

The nominal fatigue resistance is computed as follows:

S6.6.1.2.5

Formula: ( Delta F ) subscript n = ( numerator (A) divided by denominator (N) ) superscript numerator (1) divided by denominator (3) greater than or equal to numerator (1) divided by denominator (2) ( Delta F ) subscript TH

for which:

STable 6.6.1.2.5-1

Formula: A = 44 point 0 times 10 superscript 8 ( ksi) cubed

Formula: N = ( 365) times ( 75) times n times ( ADTT) subscript SL

S6.6.1.2.5

Formula: n = 1 point 0

STable 6.6.1.2.5-2

Formula: ADTT subscript SL = 3000

Formula: N = ( 365) times ( 75) times n times ADTT subscript SL

Formula: N = 82125000

STable 6.6.1.2.5-3

Formula: Delta F subscript TH = 12 point 0 ksi

Formula: ( numerator (A) divided by denominator (N) ) superscript numerator (1) divided by denominator (3) = 3 point 77 ksi

Formula: numerator (1) divided by denominator (2) times Delta F subscript TH = 6 point 00 ksi

Formula: Delta F subscript n = max left bracket( numerator (A) divided by denominator (N) ) superscript numerator (1) divided by denominator (3) , numerator (1) divided by denominator (2) times Delta F subscript TH right bracket

S6.6.1.2.5

Formula: Delta F subscript n = 6 point 00 ksi

MathCad tip logo

Fatigue Resistance

CTable 6.6.1.2.5-1 can be used to eliminate the need for some of the above fatigue resistance computations. The above computations are presented simply for illustrative purposes.

The factored fatigue stress in the outer fiber of the tension flange at the location of maximum positive moment was previously computed in Table 3-11, as follows:

Formula: f subscript botgdr = 3 point 87 ksi

Formula: f subscript botgdr less than or equal to Delta F subscript n OK

In addition to the above fatigue detail check, fatigue requirements for webs must also be checked. These calculations will be presented later for the girder design section at the pier.

S6.10.6

Design Step 3.15 - Design for Flexure - Service Limit State - Positive Moment Region

The girder must be checked for service limit state control of permanent deflection. This check is intended to prevent objectionable permanent deflections due to expected severe traffic loadings that would impair rideability. Service II Limit State is used for this check.

The flange stresses for both steel flanges of composite sections must satisfy the following requirement:

S6.10.5

Formula: f subscript f less than or equal to 0 point 95F subscript yf

S6.10.5.2

The factored Service II flexural stress was previously computed in Table 3-11 as follows:

Formula: f subscript botgdr = 44 point 00 ksi Formula: f subscript topgdr = minus 23 point 06 ksi

Formula: F subscript yf = 50 point 0 ksi

Formula: 0 point 95 times F subscript yf = 47 point 50 ksi OK

In addition to the check for service limit state control of permanent deflection, the girder can also be checked for live load deflection. Although this check is optional for a concrete deck on steel girders, it is included in this design example.

Using an analysis computer program, the maximum live load deflection is computed to be the following:

S2.5.2.6.2

Formula: Delta subscript max = 1 point 43 inches

This maximum live load deflection is computed based on the following:

  1. All design lanes are loaded.
  2. All supporting components are assumed to deflect equally.
  3. For composite design, the design cross section includes the entire width of the roadway.
  4. The number and position of loaded lanes is selected to provide the worst effect.
  5. The live load portion of Service I Limit State is used.
  6. Dynamic load allowance is included.
  7. The live load is taken from S3.6.1.3.2.

S2.5.2.6.2

In the absence of other criteria, the deflection limit is as follows:

S2.5.2.6.2

Formula: Span = 120 feet

Formula: Delta subscript allowable = ( numerator (Span) divided by denominator (800) ) times ( numerator (12 inches ) divided by denominator ( feet ) )

Formula: Delta subscript allowable = 1 point 80 inches OK

Design Step 3.16 - Design for Flexure - Constructibility Check - Positive Moment Region

The girder must also be checked for flexure during construction. The girder has already been checked in its final condition when it behaves as a composite section. The constructibility must also be checked for the girder prior to the hardening of the concrete deck when the girder behaves as a noncomposite section.

S6.10.3.2

As previously stated, a deck pouring sequence will not be considered in this design example. However, it is generally important to consider the effects of the deck pouring sequence in an actual design because it will often control the design of the top flange in the positive moment regions of composite girders.

The investigation of the constructibility of the girder begins with the the noncompact section compression-flange slenderness check, as follows:

S6.10.4.1.4

Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) less than or equal to 12 point 0

Formula: b subscript f = 14 inches (see Figure 3-4)

Formula: t subscript f = 0 point 625 inches (see Figure 3-4)

Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) = 11 point 2

Therefore, the investigation proceeds with the noncompact section compression-flange bracing provisions of S6.10.4.1.9.

Formula: L subscript b less than or equal to L subscript p = 1 point 76 times r subscript t times square root of ( numerator (E) divided by denominator (F subscript yc))

S6.10.4.1.9

The term, rt, is defined as the radius of gyration of a notional section comprised of the compression flange of the steel section plus one-third of the depth of the web in compression taken about the vertical axis.

For the noncomposite loads during construction:

Formula: Depth subscript comp = 55 point 50 inches minus 25 point 852 inches

(see Figure 3-4 and Table 3-4)

Formula: Depth subscript comp = 29 point 65 inches

Formula: D subscript c = Depth subscript comp minus t subscript topfl

Formula: D subscript c = 29 point 02 inches Formula: numerator (D subscript c) divided by denominator (3) = 9 point 67 inches

Formula: b subscript c = 14 point 0 inches Formula: t subscript c = 0 point 625 inches

Formula: I subscript t = numerator (t subscript c times b subscript c cubed ) divided by denominator (12) + numerator ( numerator (D subscript c) divided by denominator (3) times t subscript w cubed ) divided by denominator (12) Formula: I subscript t = 143 point 0 inches superscript 4

Formula: A subscript t = ( t subscript c times b subscript c ) + ( numerator (D subscript c) divided by denominator (3) times t subscript w ) Formula: A subscript t = 13 point 6 inches squared

Formula: r subscript t = square root of ( numerator (I subscript t) divided by denominator (A subscript t)) Formula: r subscript t = 3 point 24 inches

Formula: E = 29000 ksi Formula: F subscript yc = 50 ksi

Formula: L subscript p = 1 point 76 times r subscript t times square root of ( numerator (E) divided by denominator (F subscript yc)) Formula: L subscript p = 11 point 46 feet

Formula: L subscript b = 20 point 0 feet

Therefore, the investigation proceeds with the noncomposite section lateral torsional buckling provisions of S6.10.4.2.6.

MathCad tip logo

Lateral Torsional Buckling

Lateral torsional buckling can occur when the compression flange is not laterally supported. The laterally unsupported compression flange tends to buckle out-of-plane between the points of lateral support. Because the tension flange is kept in line, the girder section twists when it moves laterally. This behavior is commonly referred to as lateral torsional buckling.

Lateral torsional buckling is generally most critical for the moments induced during the deck pouring sequence.

If lateral torsional buckling occurs, the plastic moment resistance, Mp, can not be reached.

Lateral torsional buckling is illustrated in the figure below.

This figure depicts a beam under lateral torsional buckling.

Figure 3-11 Lateral Torsional Buckling

The nominal flexural resistance of the compression flange is determined from the following equation:

S6.10.4.2.6a

Formula: F subscript n = R subscript b times R subscript h times F subscript cr

S6.10.4.2.4a

The load-shedding factor, Rb, is computed as follows:

S6.10.4.3.2

Formula: lamda subscript b = 4 point 64 for sections where Dc is greater than D/2

Formula: D subscript c = 29 point 02 inches

Formula: D = 54 point 0 inches Formula: numerator (D) divided by denominator (2) = 27 point 00 inches

Therefore Formula: lamda subscript b = 4 point 64

Check if Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) less than or equal to lamda subscript b times square root of ( numerator (E) divided by denominator (f subscript c))

Formula: D subscript c = 29 point 02 inches Formula: t subscript w = 0 point 50 inches

Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) = 116 point 1

Formula: E = 29000 ksi

Formula: f subscript c = 1 point 25 times ( 19 point 44 ksi)

Formula: f subscript c = 24 point 30 ksi

Formula: lamda subscript b times square root of ( numerator (E) divided by denominator (f subscript c)) = 160 point 3

Therefore: Formula: R subscript b = 1 point 0

For homogeneous section, Rh is taken as 1.0.

S6.10.4.3.1

Formula: R subscript h = 1 point 0

The critical compression-flange local buckling stress, Fcr, is computed as follows:

S6.10.4.2.4a

Formula: F subscript cr = numerator (1 point 904 times E) divided by denominator (( numerator (b subscript f) divided by denominator (2 times t subscript f) ) squared times square root of ( numerator (2 times D subscript c) divided by denominator (t subscript w)) ) less than or equal to F subscript yc without longitudinal web stiffeners

Formula: numerator (1 point 904 times E) divided by denominator (( numerator (b subscript f) divided by denominator (2 times t subscript f) ) squared times square root of ( numerator (2 times D subscript c) divided by denominator (t subscript w)) ) = 40 point 9 ksi Formula: F subscript yc = 50 point 0 ksi

Formula: F subscript cr = min left bracket numerator (1 point 904 times E) divided by denominator (( numerator (b subscript f) divided by denominator (2 times t subscript f) ) squared times square root of ( numerator (2 times D subscript c) divided by denominator (t subscript w)) ) , F subscript yc right bracket Formula: F subscript cr = 40 point 9 ksi

Therefore the nominal flexural resistance of the compression flange is determined from the following equation:

S6.10.4.2.4a

Formula: F subscript n = R subscript b times R subscript h times F subscript cr

Formula: F subscript n = 40 point 9 ksi

In addition, the nominal flexural resistance of the compression flange should not exceed the nominal flexural resistance based upon lateral-torsional buckling determined as follows:

S6.10.4.2.6a

Check if Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) less than or equal to lamda subscript b times square root of ( numerator (E) divided by denominator (F subscript yc))

Formula: D subscript c = 29 point 02 inches Formula: t subscript w = 0 point 5 inches

Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) = 116 point 1

Formula: lamda subscript b = 4 point 64 Formula: E = 29000 ksi Formula: F subscript yc = 50 point 0 ksi

Formula: lamda subscript b times square root of ( numerator (E) divided by denominator (F subscript yc)) = 111 point 7

Check if Formula: L subscript b less than or equal to L subscript r = 4 point 44 times square root of ( numerator (I subscript yc times d) divided by denominator (S subscript xc) times numerator (E) divided by denominator (F subscript yc))

Formula: I subscript yc = 142 point 9 inches superscript 4

Formula: d = 55 point 50 inches

Formula: S subscript xc = 745 point 9 inches cubed (see Table 3-4)

Formula: E = 29000 point 0 ksi

Formula: F subscript yc = 50 point 0 ksi

Formula: L subscript r = 4 point 44 times square root of ( numerator (I subscript yc times d) divided by denominator (S subscript xc) times numerator (E) divided by denominator (F subscript yc)) Formula: L subscript r = 29 point 06 feet

Formula: L subscript b = 20 point 0 feet

Therefore: Formula: M subscript n = C subscript b times R subscript b times R subscript h times M subscript y times left bracket 1 minus 0 point 5 ( numerator (L subscript b minus L subscript p) divided by denominator (L subscript r minus L subscript p) ) right bracket less than or equal to R subscript b times R subscript h times M subscript y

S6.10.4.2.6a

The moment gradient correction factor, Cb, is computed as follows:

S6.10.4.2.5a

Formula: C subscript b = 1 point 75 minus 1 point 05 times ( numerator (P subscript I) divided by denominator (P subscript h) ) + 0 point 3 times ( numerator (P subscript I) divided by denominator (P subscript h) ) squared less than or equal to K subscript b

Use: Formula: numerator (P subscript I) divided by denominator (P subscript h) = 0 point 5 (based on analysis)

Formula: 1 point 75 minus 1 point 05 times ( 0 point 5) + 0 point 3 times ( 0 point 5) squared = 1 point 30

Formula: K subscript b = 1 point 75

Therefore Formula: C subscript b = 1 point 30

Formula: M subscript y = ( 50 ksi) times 745 point 98 inches cubed Formula: M subscript y = 3108 K feet

S6.10.3.3.1

Formula: I subscript t = numerator (t subscript c times b subscript c cubed ) divided by denominator (12) Formula: I subscript t = 142 point 9 inches superscript 4

S6.10.4.2.6a

Formula: A subscript t = t subscript c times b subscript c Formula: A subscript t = 8 point 8 inches squared

Formula: r subscript t = square root of ( numerator (I subscript t) divided by denominator (A subscript t)) Formula: r subscript t = 4 point 04 inches

Formula: E = 29000 ksi Formula: F subscript yc = 50 ksi

Formula: L subscript p = 1 point 76 times r subscript t times square root of ( numerator (E) divided by denominator (F subscript yc)) Formula: L subscript p = 14 point 28 feet

Formula: L subscript b = 20 point 0 feet Formula: L subscript r = 29 point 06 feet

Formula: C subscript b times R subscript b times R subscript h times M subscript y times left bracket 1 minus 0 point 5 times ( numerator (L subscript b minus L subscript p) divided by denominator (L subscript r minus L subscript p) ) right bracket = 3258 K feet

Formula: R subscript b times R subscript h times M subscript y = 3108 K feet

Therefore Formula: M subscript n = R subscript b times R subscript h times M subscript y Formula: M subscript n = 3108 K feet

S6.10.4.2.6a

Formula: F subscript n = numerator (M subscript n) divided by denominator (S subscript xc) Formula: F subscript n = 50 point 0 ksi

Therefore, the provisions of SEquation 6.10.4.2.4a-2 control.

Formula: F subscript n = R subscript b times R subscript h times F subscript cr Formula: F subscript n = 40 point 9 ksi

The factored flexural resistance, Fr, is computed as follows:

S6.10.4

Formula: phi subscript f = 1 point 00

S6.5.4.2

Formula: F subscript r = phi subscript f times F subscript n Formula: F subscript r = 40 point 9 ksi

The factored construction stress in the compression flange is as follows:

Formula: f subscript c = 24 point 30 ksi (previously computed) OK

For the tension flange, the nominal flexural resistance, in terms of stress, is determined as follows:

S6.10.4.2.6b

Formula: F subscript n = R subscript b times R subscript h times F subscript yt

where: Formula: R subscript b = 1 point 0

S6.10.4.3.2b

Formula: R subscript h = 1 point 0

Formula: F subscript yt = 50 point 0 ksi

Formula: F subscript n = 50 point 0 ksi

The factored flexural resistance, Fr, is computed as follows:

S6.10.4

Formula: phi subscript f = 1 point 00

S6.5.4.2

Formula: F subscript r = phi subscript f times F subscript n

Formula: F subscript r = 50 point 0 ksi

The factored construction stress in the tension flange is as follows:

Formula: f subscript t = 1 point 25 times ( 16 point 95 ksi)

Formula: f subscript t = 21 point 19 ksi OK

Therefore, the girder design section at the location of maximum positive moment satisfies the noncomposite section flexural resistance requirements for construction loads based upon lateral torsional buckling for both the compression flange and the tension flange.

In addition, composite girders, when they are not yet composite, must satisfy the following requirement during construction:

S6.10.3.2.2

Formula: f subscript cw less than or equal to numerator (0 point 9 times E times alpha k) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) less than or equal to F subscript yw

for which:

Formula: E = 29000 ksi

Formula: alpha = 1 point 25 for webs without longitudinal stiffeners

Formula: D = 54 inches

Formula: D subscript c = 29 point 02 inches

Formula: k = 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared greater than or equal to 7 point 2 for webs without longitudinal stiffeners

Formula: 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared = 31 point 2

Formula: k = max left bracket 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared , 7 point 2 right bracket Formula: k = 31 point 2

Formula: t subscript w = numerator (1) divided by denominator (2) inches (see Figure 3-4)

Formula: numerator (0 point 9 times E times alpha k) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) = 87 point 15 ksi

Formula: F subscript yw = 50 point 0 ksi

Formula: min left bracket numerator (0 point 9 times E times alpha k) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) , F subscript yw right bracket = 50 point 0 ksi

Formula: f subscript cw = f subscript topgdr times ( numerator (D subscript c) divided by denominator (D subscript c + t subscript f) )

Formula: f subscript cw = minus 22 point 57 ksi OK

In addition to checking the nominal flexural resistance during construction, the nominal shear resistance must also be checked. However, shear is minimal at the location of maximum positive moment, and it is maximum at the pier.

Therefore, for this design example, the nominal shear resistance for constructibility will be presented later for the girder design section at the pier.

S6.10.3.2.3

Design Step 3.17 - Check Wind Effects on Girder Flanges - Positive Moment Region

As stated in Design Step 3.3, for this design example, the interior girder controls and is being designed.

S6.10.3.5

Wind effects generally do not control a steel girder design, and they are generally considered for the exterior girders only. However, for this design example, wind effects will be presented later for the girder design section at the pier.

Specification checks have been completed for the location of maximum positive moment, which is at 0.4L in Span 1.

C6.10.3.5.2 & C4.6.2.7.1

Now the specification checks are repeated for the location of maximum negative moment, which is at the pier, as shown in Figure 3-12. This is also the location of maximum shear.

A girder elevation showing the location of the maximum negative moment. The location of the maximum negative moment is 120 feet 0 inches from the centerline of bearing. The girder is symmetrical about the pier centerline.

Figure 3-12 Location of Maximum Negative Moment

Design Step 3.7 - Check Section Proportion Limits - Negative Moment Region

Several checks are required to ensure that the proportions of the trial girder section are within specified limits.

S6.10.2

The first section proportion check relates to the general proportions of the section. The flexural components must be proportioned such that:

S6.10.2.1

Formula: 0 point 1 less than or equal to numerator (I subscript yc) divided by denominator (I subscript y) less than or equal to 0 point 9

Formula: I subscript yc = numerator (2 point 75 inches times ( 14 inches ) cubed ) divided by denominator (12) Formula: I subscript yc = 628 point 8 inches superscript 4

Formula: I subscript y = numerator (2 point 75 inches times ( 14 inches ) cubed ) divided by denominator (12) + numerator (54 inches times ( numerator (1) divided by denominator (2) inches ) cubed ) divided by denominator (12) + numerator (2 point 5 inches times ( 14 inches ) cubed ) divided by denominator (12)

Formula: I subscript y = 1201 point 1 inches superscript 4

Formula: numerator (I subscript yc) divided by denominator (I subscript y) = 0 point 524 OK

The second section proportion check relates to the web slenderness. For a section without longitudinal stiffeners, the web must be proportioned such that:

S6.10.2.2

Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) less than or equal to 6 point 77 times square root of ( numerator (E) divided by denominator (f subscript c)) less than or equal to 200

At sections in negative flexure, using Dc of the composite section consisting of the steel section plus the longitudinal reinforcement is conservative.

C6.10.3.1.4a

MathCad tip logo

D c for Negative Flexure

At sections in negative flexure, using Dc of the composite section consisting of the steel section plus the longitudinal reinforcement, as described in C6.10.3.1.4a, removes the dependency of Dc on the applied loading, which greatly simplifies subsequent load rating calculations.

Formula: D subscript c = 32 point 668 inches minus 2 point 75 inches

(see Figure 3-4 and Table 3-5)

Formula: D subscript c = 29 point 92 inches

Formula: t subscript w = numerator (1) divided by denominator (2) inches (see Figure 3-4)

Formula: E = 29000ksi

S6.4.1

Formula: f subscript c = 48 point 84 ksi

Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) = 119 point 7

Formula: 6 point 77 times square root of ( numerator (E) divided by denominator (f subscript c)) = 165 point 0

Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) less than or equal to 6 point 77 times square root of ( numerator (E) divided by denominator (f subscript c)) and Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) less than or equal to 200 OK

The third section proportion check relates to the flange proportions. The compression flanges on fabricated I-sections must be proportioned such that:

S6.10.2.3

Formula: b subscript f greater than or equal to 0 point 3 times D subscript c

Formula: b subscript f = 14 inches (see Figure 3-4)

Formula: D subscript c = 29 point 92 inches

Formula: 0 point 3 times D subscript c = 8 point 98 inches

Formula: b subscript f greater than or equal to 0 point 3 times D subscript c OK

According to C6.10.2.3, it is preferable for the flange width to be greater than or equal to 0.4Dc. In this case, the flange width is greater than both 0.3Dc and 0.4Dc, so this requirement is clearly satisfied.

C6.10.2.3

In addition to the compression flange check, the tension flanges on fabricated I-sections must be proportioned such that:

S6.10.2.3

Formula: numerator (b subscript t) divided by denominator (2 times t subscript t) less than or equal to 12 point 0

Formula: b subscript t = 14 inches (see Figure 3-4)

Formula: t subscript t = 2 point 5 inches (see Figure 3-4)

Formula: numerator (b subscript t) divided by denominator (2 times t subscript t) = 2 point 8 OK

Design Step 3.8 - Compute Plastic Moment Capacity - Negative Moment Region

For composite sections, the plastic moment, Mp, is calculated as the first moment of plastic forces about the plastic neutral axis.

S6.10.3.1.3

The width of the bottom flange is defined as b sub c and the bottom flange thickness is defined as t sub c. The web depth is defined as D sub w and the web thickness is defined as t sub w. The width of the top flange is defined as b sub t and the top flange thickness is defined as t sub t. The area of the top transverse reinforcement is defined as A sub rt and of the bottom transverse reinforcement is defined as A sub rb. The distance from the bottom of the top flange to the plastic neutral axis is defined as Y. The tension force in the bottom flange is defined by P sub c and the tension in the web is defined as P sub w. The tension force in the top flange is defined by P sub t. The tension in the top slab reinforcement is defined as P sub rt and the tension in the bottom slab reinforcement is defined as P sub rb.

Figure 3-13 Computation of Plastic Moment Capacity for Negative Bending Sections

For the tension flange:

SAppendix A6.1

Formula: F subscript yt = 50ksi Formula: b subscript t = 14 inches Formula: t subscript t = 2 point 50 inches

Formula: P subscript t = F subscript yt times b subscript t times t subscript t Formula: P subscript t = 1750 K

For the web:

Formula: F subscript yw = 50 point 0 ksi Formula: D subscript w = 54 inches Formula: t subscript w = 0 point 50 inches

Formula: P subscript w = F subscript yw times D subscript w times t subscript w Formula: P subscript w = 1350 K

For the compression flange:

Formula: F subscript yc = 50ksi Formula: b subscript c = 14 inches Formula: t subscript c = 2 point 75 inches

Formula: P subscript c = F subscript yc times b subscript c times t subscript c Formula: P subscript c = 1925 K

For the longitudinal reinforcing steel in the top layer of the slab at the pier:

Formula: F subscript yrt = 60ksi

Formula: A subscript rt = 0 point 31 inches squared times ( numerator (103 inches ) divided by denominator (5 inches ) ) Formula: A subscript rt = 6 point 39 inches squared

Formula: P subscript rt = F subscript yrt times A subscript rt Formula: P subscript rt = 383 K

For the longitudinal reinforcing steel in the bottom layer of the slab at the pier:

Formula: F subscript yrb = 60ksi

Formula: A subscript rb = 0 point 31 inches squared times ( numerator (103 inches ) divided by denominator (5 inches ) ) Formula: A subscript rb = 6 point 39 inches squared

Formula: P subscript rb = F subscript yrb times A subscript rb Formula: P subscript rb = 383 K

Check the location of the plastic neutral axis, as follows:

SAppendix A6.1

Formula: P subscript c + P subscript w = 3275 K Formula: P subscript t + P subscript rb + P subscript rt = 2516 K

Formula: P subscript c + P subscript w + P subscript t = 5025 K Formula: P subscript rb + P subscript rt = 766 K

Therefore the plastic neutral axis is located within the web.

Formula: Y = ( numerator (D) divided by denominator (2) ) times ( numerator (P subscript c minus P subscript t minus P subscript rt minus P subscript rb) divided by denominator (P subscript w) + 1 )

STable A6.1-2

Formula: Y = 15 point 17 inches

Since it will be shown in the next design step that this section is noncompact, the plastic moment is not used to compute the flexural resistance and therefore does not need to be computed.

Design Step 3.9 - Determine if Section is Compact or Noncompact - Negative Moment Region

The next step in the design process is to determine if the section is compact or noncompact. This, in turn, will determine which formulae should be used to compute the flexural capacity of the girder.

Where the specified minimum yield strength does not exceed 70.0 ksi, and the girder has a constant depth, and the girder does not have longitudinal stiffeners or holes in the tension flange, then the first step is to check the compact-section web slenderness provisions, as follows:

S6.10.4.1.1

Formula: numerator (2 times D subscript cp) divided by denominator (t subscript w) less than or equal to 3 point 76 times square root of ( numerator (E) divided by denominator (F subscript yc))

S6.10.4.1.2

Since the plastic neutral axis is located within the web,

Formula: D subscript cp = D subscript w minus Y Formula: D subscript cp = 38 point 83 inches

Formula: numerator (2 times D subscript cp) divided by denominator (t subscript w) = 155 point 3 Formula: 3 point 76 times square root of ( numerator (E) divided by denominator (F subscript yc)) = 90 point 6

Therefore, the web does not qualify as compact. Since this is not a composite section in positive flexure, the investigation proceeds with the noncompact section compression-flange slenderness provisions of S6.10.4.1.4.

S6.10.4.1.4

Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) less than or equal to 12 point 0

Formula: b subscript f = 14 point 0 inches Formula: t subscript f = 2 point 75 inches

Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) = 2 point 5

Therefore, the investigation proceeds with the noncompact section compression-flange bracing provisions of S6.10.4.1.9.

Formula: L subscript b less than or equal to L subscript p = 1 point 76 times r subscript t times square root of ( numerator (E) divided by denominator (F subscript yc))

S6.10.4.1.9

The term, rt, is defined as the radius of gyration of a notional section comprised of the compression flange of the steel section plus one-third of the depth of the web in compression taken about the vertical axis.

Based on previous computations,

Formula: D subscript c = 29 point 92 inches Formula: numerator (D subscript c) divided by denominator (3) = 9 point 97 inches

Formula: b subscript c = 14 point 0 inches Formula: t subscript c = 2 point 75 inches

Formula: I subscript t = numerator (t subscript c times b subscript c cubed ) divided by denominator (12) + numerator ( numerator (D subscript c) divided by denominator (3) times t subscript w cubed ) divided by denominator (12) Formula: I subscript t = 628 point 9 inches superscript 4

Formula: A subscript t = ( t subscript c times b subscript c ) + ( numerator (D subscript c) divided by denominator (3) times t subscript w ) Formula: A subscript t = 43 point 5 inches squared

Formula: r subscript t = square root of ( numerator (I subscript t) divided by denominator (A subscript t)) Formula: r subscript t = 3 point 80 inches

Formula: L subscript p = 1 point 76 times r subscript t times square root of ( numerator (E) divided by denominator (F subscript yc)) Formula: L subscript p = 13 point 43 feet

Formula: L subscript b = 20 point 0 feet

Therefore, the investigation proceeds with the composite section lateral torsional buckling provisions of S6.10.4.2.5.

MathCad tip logo

Noncompact Sections

Based on the previous computations, it was determined that the girder section at the pier is noncompact. Several steps could be taken to make this a compact section, such as increasing the web thickness or possibly modifying the flange thicknesses to decrease the value Dcp. However, such revisions may not be economical.

Design Step 3.10 - Design for Flexure - Strength Limit State - Negative Moment Region

Since the section was determined to be noncompact and based on the computations in the previous design step, the nominal flexural resistance is computed based upon lateral torsional buckling.

S6.10.4.2.5

The nominal flexural resistance of the compression flange, in terms of stress, is determined from the following equation:

S6.10.4.2.5a

Formula: F subscript n = R subscript b times R subscript h times F subscript cr

S6.10.4.2.4a

The load-shedding factor, Rb, is computed as follows:

S6.10.4.3.2

Formula: lamda subscript b = 4 point 64 for sections where Dc is greater than D/2

Formula: D subscript c = 29 point 92 inches Formula: f subscript c = 48 point 84 ksi

Formula: D = 54 point 0 inches Formula: numerator (D) divided by denominator (2) = 27 point 00 inches

Therefore Formula: lamda subscript b = 4 point 64

Check if Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) less than or equal to lamda subscript b times square root of ( numerator (E) divided by denominator (f subscript c))

Formula: D subscript c = 29 point 92 inches Formula: t subscript w = 0 point 50 inches

Formula: numerator (2 times D subscript c) divided by denominator (t subscript w) = 119 point 7

Formula: lamda subscript b times square root of ( numerator (E) divided by denominator (f subscript c)) = 113 point 1

Therefore: Formula: R subscript b = 1 point 0

For homogeneous section, Rh is taken as 1.0.

S6.10.4.3.1

Formula: R subscript h = 1 point 0

The critical compression-flange local buckling stress, Fcr, is computed as follows:

S6.10.4.2.4a

Formula: F subscript cr = numerator (1 point 904 times E) divided by denominator (( numerator (b subscript f) divided by denominator (2 times t subscript f) ) squared times square root of ( numerator (2 times D subscript c) divided by denominator (t subscript w)) ) less than or equal to F subscript yc without longitudinal web stiffeners

Formula: numerator (1 point 904 times E) divided by denominator (( numerator (b subscript f) divided by denominator (2 times t subscript f) ) squared times square root of ( numerator (2 times D subscript c) divided by denominator (t subscript w)) ) = 779 point 0 ksi Formula: F subscript yc = 50 point 0 ksi

Formula: F subscript cr = min left bracket numerator (1 point 904 times E) divided by denominator (( numerator (b subscript f) divided by denominator (2 times t subscript f) ) squared times square root of ( numerator (2 times D subscript c) divided by denominator (t subscript w)) ) , F subscript yc right bracket Formula: F subscript cr = 50 point 0 ksi

Therefore the nominal flexural resistance of the compression flange is determined from the following equation:

S6.10.4.2.4a

Formula: F subscript n = R subscript b times R subscript h times F subscript cr

Formula: F subscript n = 50 point 0 ksi

In addition, the nominal flexural resistance of the compression flange should not exceed the nominal flexural resistance based upon lateral-torsional buckling determined as follows:

S6.10.4.2.5a

Check if Formula: L subscript b less than or equal to L subscript r = 4 point 44 times r subscript t times square root of ( numerator (E) divided by denominator (F subscript yc))

Formula: L subscript r = 4 point 44 times r subscript t times square root of ( numerator (E) divided by denominator (F subscript yc))

Formula: r subscript t = 3 point 80 inches Formula: E = 29000 ksi Formula: F subscript yc = 50 point 0 ksi

Formula: L subscript r = 33 point 9 feet Formula: L subscript b = 20 feet

Therefore:

Formula: F subscript n = C subscript b times R subscript b times R subscript h times F subscript yc times left bracket 1 point 33 minus 0 point 187 ( numerator (L subscript b) divided by denominator (r subscript t) ) times square root of ( numerator (F subscript yc) divided by denominator (E)) right bracket less than or equal to R subscript b times R subscript h times F subscript yc

The moment gradient correction factor, Cb, is computed as follows:

SC6.10.4.2.5a

Formula: C subscript b = 1 point 75 minus 1 point 05 times ( numerator (P subscript I) divided by denominator (P subscript h) ) + 0 point 3 times ( numerator (P subscript I) divided by denominator (P subscript h) ) squared less than or equal to K subscript b

Use: Formula: numerator (P subscript I) divided by denominator (P subscript h) = 0 point 5 (based on analysis)

Formula: 1 point 75 minus 1 point 05 times ( 0 point 5) + 0 point 3 times ( 0 point 5) squared = 1 point 30

Formula: K subscript b = 1 point 75

Therefore Formula: C subscript b = 1 point 30

Formula: C subscript b times R subscript b times R subscript h times F subscript yc times left bracket 1 point 33 minus 0 point 187 left bracket numerator (L subscript b times ( numerator (12 inches ) divided by denominator ( feet ) )) divided by denominator (r subscript t) right bracket times square root of ( numerator (F subscript yc) divided by denominator (E)) right bracket = 54 point 60 ksi

Formula: R subscript b times R subscript h times F subscript yc = 50 point 0 ksi

Therefore Formula: F subscript n = R subscript b times R subscript h times F subscript yc

Formula: F subscript n = 50 point 0 ksi

The factored flexural resistance, Fr, is computed as follows:

S6.10.4

Formula: phi subscript f = 1 point 00

S6.5.4.2

Formula: F subscript r = phi subscript f times F subscript n

Formula: F subscript r = 50 point 0 ksi

The negative flexural resistance at this design section is checked as follows:

S1.3.2.1

Formula: Sigma eta subscript i times gamma subscript i times Q subscript i less than or equal to R subscript r

or in this case:

Formula: Sigma eta subscript i times gamma subscript i times F subscript i less than or equal to F subscript r

For this design example,

Formula: eta subscript i = 1 point 00

As computed in Design Step 3.6, the factored Strength I Limit State stress for the compression flange is as follows:

Formula: Sigma gamma subscript i times F subscript i = 48 point 84ksi

Therefore

Formula: Sigma eta subscript i times gamma subscript i times F subscript i = 48 point 84 ksi

Formula: F subscript r = 50 point 00 ksi OK

For the tension flange, the nominal flexural resistance, in terms of stress, is determined as follows:

S6.10.4.2.5b

Formula: F subscript n = R subscript b times R subscript h times F subscript yt

where:

Formula: R subscript b = 1 point 0

S6.10.4.3.2b

Formula: R subscript h = 1 point 0

Formula: F subscript yt = 50 point 0 ksi

Formula: F subscript n = 50 point 0 ksi

The factored flexural resistance, Fr, is computed as follows:

S6.10.4

Formula: phi subscript f = 1 point 00

S6.5.4.2

Formula: F subscript r = phi subscript f times F subscript n

Formula: F subscript r = 50 point 0 ksi

The negative flexural resistance at this design section is checked as follows:

S1.3.2.1

Formula: Sigma eta subscript i times gamma subscript i times Q subscript i less than or equal to R subscript r

or in this case:

Formula: Sigma eta subscript i times gamma subscript i times F subscript i less than or equal to F subscript r

For this design example,

Formula: eta subscript i = 1 point 00

As computed in Design Step 3.6, the factored Strength I Limit State stress for the tension flange is as follows:

Formula: Sigma gamma subscript i times F subscript i = 44 point 99ksi

Therefore

Formula: Sigma eta subscript i times gamma subscript i times F subscript i = 44 point 99 ksi

Formula: F subscript r = 50 point 0 ksi OK

Therefore, the girder design section at the pier satisfies the flexural resistance requirements for both the compression flange and the tension flange.

Design Step 3.11 - Design for Shear - Negative Moment Region

Shear must be checked at each section of the girder. For this design example, shear is maximum at the pier.

S6.10.7

The first step in the design for shear is to check if the web must be stiffened. The nominal shear resistance of unstiffened webs of hybrid and homogeneous girders is:

S6.10.7.2

Formula: V subscript n = C times V subscript p

Formula: k = 5 point 0

S6.10.7.3.3a

Formula: numerator (D) divided by denominator (t subscript w) = 108 point 0

S6.10.7.3.3a

Formula: 1 point 10 times square root of ( numerator (E k) divided by denominator (F subscript yw)) = 59 point 2

Formula: 1 point 38 times square root of ( numerator (E k) divided by denominator (F subscript yw)) = 74 point 3

Therefore,

Formula: numerator (D) divided by denominator (t subscript w) greater than or equal to 1 point 38 times square root of ( numerator (E k) divided by denominator (F subscript yw))

Formula: C = numerator (1 point 52) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) times ( numerator (E k) divided by denominator (F subscript yw) )

Formula: C = 0 point 378

Formula: F subscript yw = 50 point 0 ksi Formula: D = 54 point 0 inches Formula: t subscript w = 0 point 5 inches

Formula: V subscript p = 0 point 58 times F subscript yw times D times t subscript w

S6.10.7.3.3a&c

Formula: V subscript p = 783 point 0 K

Formula: V subscript n = C times V subscript p

Formula: V subscript n = 295 point 9 K

The factored shear resistance, Vr, is computed as follows:

S6.10.7.1

Formula: phi subscript v = 1 point 00

S6.5.4.2

Formula: V subscript r = phi subscript v times V subscript n

Formula: V subscript r = 295 point 9 K

The shear resistance at this design section is checked as follows:

S1.3.2.1

Formula: Sigma eta subscript i times gamma subscript i times Q subscript i less than or equal to R subscript r

or in this case:

Formula: Sigma eta subscript i times gamma subscript i times V subscript i less than or equal to V subscript r

For this design example,

Formula: eta subscript i = 1 point 00

As computed in Design Step 3.6, the factored Strength I Limit State shear is as follows:

Formula: Sigma gamma subscript i times V subscript i = 423 point 5 K

Therefore

Formula: Sigma eta subscript i times gamma subscript i times V subscript i = 423 point 5 K

Formula: V subscript r = 295 point 9 K

Since the shear resistance of an unstiffened web is less than the actual design shear, the web must be stiffened.

MathCad tip logo

Nominally Stiffened Webs

As previously explained, a "nominally stiffened" web (approximately 1/16 inch thinner than "unstiffened") will generally provide the least cost alternative or very close to it. However, for web depths of approximately 50 inches or less, unstiffened webs may be more economical.

For this design example, transverse intermediate stiffeners are used and longitudinal stiffeners are not used. The transverse intermediate stiffener spacing in this design example is 80 inches. Therefore, the spacing of the transverse intermediate stiffeners does not exceed 3D. Therefore, the design section can be considered stiffened and the provisions of S6.10.7.3 apply.

S6.10.7.1

MathCad tip logo

Stiffener Spacing

The spacing of the transverse intermediate stiffeners is determined such that it satisfies all spacing requirement in S6.10.7 and such that the shear resistance of the stiffened web is sufficient to resist the applied factored shear.

First, handling requirements of the web are checked. For web panels without longitudinal stiffeners, transverse stiffeners must be used if:

S6.10.7.3.2

Formula: numerator (D) divided by denominator (t subscript w) greater than or equal to 150

Formula: D = 54 point 0 inches Formula: t subscript w = 0 point 5 inches

Formula: numerator (D) divided by denominator (t subscript w) = 108 point 0

Another handling requirement is that the spacing of transverse stiffeners, do, must satisfy the following:

S6.10.7.3.2

Formula: d subscript o less than or equal to D times left bracket numerator (260) divided by denominator (( numerator (D) divided by denominator (t subscript w) )) right bracket squared

Formula: D times left bracket numerator (260) divided by denominator (( numerator (D) divided by denominator (t subscript w) )) right bracket squared = 313 point 0 inches

Use Formula: d subscript o = 80 point 0 inches OK

This handling requirement for transverse stiffeners need only be enforced in regions where transverse stiffeners are no longer required for shear and where the web slenderness ratio exceeds 150. Therefore, this requirement must typically be applied only in the central regions of the spans of relatively deep girders, where the shear is low.

The nominal shear resistance of interior web panels of noncompact sections which are considered stiffened, as per S6.10.7.1, is as follows:

S6.10.7.3.3b

Check if Formula: f subscript u less than or equal to 0 point 75 times phi subscript f times F subscript y

The term, fu, is the flexural stress in the compression or tension flange due to the factored loading, whichever flange has the maximum ratio of fu to Fr in the panel under consideration.

Formula: f subscript u = 48 point 84 ksi (see Table 3-12)

Formula: 0 point 75 times phi subscript f times F subscript y = 37 point 5 ksi

Therefore,

Formula: f subscript u greater than or equal to 0 point 75 times phi subscript f times F subscript y

Formula: V subscript n = R times V subscript p times left bracket C + numerator (0 point 87 times ( 1 minus C)) divided by denominator ( square root of (1 + ( numerator (d subscript o) divided by denominator (D) ) squared ) ) right bracket greater than or equal to C times V subscript p

Formula: k = 5 + numerator (5) divided by denominator (( numerator (d subscript o) divided by denominator (D) ) squared ) Formula: k = 7 point 3

S6.10.7.3.3a

Formula: numerator (D) divided by denominator (t subscript w) = 108 point 0

S6.10.7.3.3a

Formula: 1 point 10 times square root of ( numerator (E k) divided by denominator (F subscript yw)) = 71 point 5

Formula: 1 point 38 times square root of ( numerator (E k) divided by denominator (F subscript yw)) = 89 point 7

Therefore,

Formula: numerator (D) divided by denominator (t subscript w) greater than or equal to 1 point 38 times square root of ( numerator (E k) divided by denominator (F subscript yw))

Formula: C = numerator (1 point 52) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) times ( numerator (E k) divided by denominator (F subscript yw) )

Formula: C = 0 point 550

The reduction factor applied to the factored shear, R, is computed as follows:

S6.10.7.3.3b

Formula: R = left bracket 0 point 6 + 0 point 4 times ( numerator (F subscript r minus f subscript u) divided by denominator (F subscript r minus 0 point 75 phi subscript f times F subscript y) ) right bracket

Formula: R = 0 point 637

Formula: V subscript p = 0 point 58 times F subscript yw times D times t subscript w

S6.10.7.3.3a&c

Formula: V subscript p = 783 point 0 K

Formula: R times V subscript p times left bracket C + numerator (0 point 87 times ( 1 minus C)) divided by denominator ( square root of (1 + ( numerator (d subscript o) divided by denominator (D) ) squared ) ) right bracket = 383 point 7 K

Formula: C times V subscript p = 430 point 7 K

Formula: V subscript n = max left bracket R times V subscript p times left bracket C + numerator (0 point 87 times ( 1 minus C)) divided by denominator ( square root of (1 + ( numerator (d subscript o) divided by denominator (D) ) squared ) ) right bracket , C times V subscript p right bracket

Formula: V subscript n = 430 point 7 K

The factored shear resistance, Vr, is computed as follows:

S6.10.7.1

Formula: phi subscript v = 1 point 00

S6.5.4.2

Formula: V subscript r = phi subscript v times V subscript n

Formula: V subscript r = 430 point 7 K

As previously computed, for this design example:

Formula: Sigma eta subscript i times gamma subscript i times V subscript i = 423 point 5 K

Formula: V subscript r = 430 point 7 K OK

Therefore, the girder design section at the pier satisfies the shear resistance requirements for the web.

Design Step 3.12 - Design Transverse Intermediate Stiffeners - Negative Moment Region

The girder in this design example has transverse intermediate stiffeners. Transverse intermediate stiffeners are used to increase the shear resistance of the girder. The shear resistance computations shown in the previous design step were based on a stiffener spacing of 80 inches.

S6.10.8.1

In this design example, it is assumed that the transverse intermediate stiffeners consist of plates welded to one side of the web. The required interface between the transverse intermediate stiffeners and the top and bottom flanges is described in S6.10.8.1.1.

The transverse intermediate stiffener configuration is assumed to be as presented in the following figure.

S6.10.8.1.1

There are two figures shown here. The first is a partial girder elevation at the pier. The elevation shows a bearing stiffener at the centerline of the pier and transverse intermediate stiffeners spaced at 6 feet 8 inches. There is a section A dash A thru the transverse stiffener. In the second figure Section A dash A, it show a one half inch thick web and a transverse intermediate stiffener with a thickness of one half of an inch and a width of 5 and one half inches.

Figure 3-14 Transverse Intermediate Stiffener

The first specification check is for the projecting width of the transverse intermediate stiffener. The width, bt, of each projecting stiffener element must satisfy the following:

S6.10.8.1.2

Formula: b subscript t greater than or equal to 2 point 0 + numerator (d) divided by denominator (30 point 0) and Formula: 16 point 0 times t subscript p greater than or equal to b subscript t greater than or equal to 0 point 25b subscript f

Formula: b subscript t = 5 point 5 inches

Formula: d = 59 point 25inches

Formula: t subscript p = 0 point 50 inches

Formula: b subscript f = 14 point 0 inches

Formula: b subscript t = 5 point 5 inches Formula: 2 point 0 + numerator (d) divided by denominator (30 point 0) = 3 point 98inches

Therefore, Formula: b subscript t greater than or equal to 2 point 0 + numerator (d) divided by denominator (30 point 0) OK

Formula: 16 point 0 times t subscript p = 8 point 0 inches

Formula: 0 point 25 times b subscript f = 3 point 5 inches

Therefore, Formula: 16 point 0 times t subscript p greater than or equal to b subscript t greater than or equal to 0 point 25 times b subscript f OK

The second specification check is for the moment of inertia of the transverse intermediate stiffener. This requirement is intended to ensure sufficient rigidity. The moment of inertia of any transverse stiffener must satisfy the following:

S6.10.8.1.3

Formula: I subscript t greater than or equal to d subscript o times t subscript w cubed times J

Formula: d subscript o = 80 point 0 inches Formula: t subscript w = 0 point 50 inches Formula: D = 54 point 0 inches

Formula: J = 2 point 5 times ( numerator (D) divided by denominator (d subscript o) ) squared minus 2 point 0 greater than or equal to 0 point 5

Formula: 2 point 5 times ( numerator (D) divided by denominator (d subscript o) ) squared minus 2 point 0 = minus 0 point 9

Therefore, Formula: J = 0 point 5

Therefore, Formula: d subscript o times t subscript w cubed times J = 5 point 0 inches superscript 4

Formula: I subscript t = numerator (t subscript p times b subscript t cubed ) divided by denominator (3) Formula: I subscript t = 27 point 7 inches superscript 4

Therefore, Formula: I subscript t greater than or equal to d subscript o times t subscript w cubed times J OK

The third specification check is for the area of the transverse intermediate stiffener. This requirement is intended to ensure sufficient area to resist the vertical component of the tension field. The area of any transverse stiffener must satisfy the following:

S6.10.8.1.4

Formula: A subscript s greater than or equal to left bracket 0 point 15 times B times numerator (D) divided by denominator (t subscript w) times ( 1 minus C) times ( numerator (V subscript u) divided by denominator (V subscript r) ) minus 18 right bracket times numerator (F subscript yw) divided by denominator (F subscript cr) times t subscript w squared

Formula: B = 2 point 4 for single plate stiffeners

Formula: D = 54 point 0 inches

Formula: t subscript w = 0 point 50 inches

Formula: C = 0 point 550

Formula: V subscript u = 423 point 5 K

Formula: V subscript r = 430 point 7 K

Formula: F subscript yw = 50 point 0 ksi

Formula: E = 29000 ksi

Formula: b subscript t = 5 point 5 inches

Formula: t subscript p = 0 point 5 inches Formula: F subscript cr = numerator (0 point 311 times E) divided by denominator (( numerator (b subscript t) divided by denominator (t subscript p) ) squared ) less than or equal to F subscript ys

Formula: numerator (0 point 311 times E) divided by denominator (( numerator (b subscript t) divided by denominator (t subscript p) ) squared ) = 74 point 5 ksi Formula: F subscript ys = 50 point 0 ksi

Therefore, Formula: F subscript cr = 50 point 0 ksi

Formula: left bracket 0 point 15 times B times numerator (D) divided by denominator (t subscript w) times ( 1 minus C) times ( numerator (V subscript u) divided by denominator (V subscript r) ) minus 18 right bracket times numerator (F subscript yw) divided by denominator (F subscript cr) times t subscript w squared = minus 0 point 2 inches squared

Therefore, the specification check for area is automatically satisfied.

Therefore, the transverse intermediate stiffeners as shown in Figure 3-13 satisfy all of the required specification checks.

Design Step 3.14 - Design for Flexure - Fatigue and Fracture Limit State - Negative Moment Region

For this design example, the nominal fatigue resistance computations were presented previously for the girder section at the location of maximum positive moment. Detail categories are explained and illustrated in STable 6.6.1.2.3-1 and SFigure 6.6.1.2.3-1.

S6.6.1

In addition to the nominal fatigue resistance computations, fatigue requirements for webs must also be checked. These checks are required to control out-of-plane flexing of the web due to flexure or shear under repeated live loading.

S6.10.6

S6.10.6.1

For this check, the live load flexural stress and shear stress resulting from the fatigue load must be taken as twice that calculated using the fatigue load combination in Table 3-1.

S6.10.6.2

As previously explained, for this design example, the concrete slab is assumed to be fully effective for both positive and negative flexure for fatigue limit states. This is permissible because the provisions of S6.10.3.7 were satisfied in Design Step 2.

S6.6.1.2.1

For flexure, the fatigue requirement for the web is as follows:

S6.10.6.3

If Formula: numerator (D) divided by denominator (t subscript w) less than or equal to 0 point 95 times square root of ( numerator (k times E) divided by denominator (F subscript yw)) then Formula: F subscript cf less than or equal to F subscript yw

Otherwise Formula: f subscript cf less than or equal to 0 point 9 k times E times ( numerator (t subscript w) divided by denominator (D) ) squared

Formula: D = 54 point 0 inches Formula: D subscript c = 29 point 92 inches

For the fatigue limit state at the pier (the location of maximum negative moment):

S6.10.3.1.4a

Formula: f subscript botgdr = ( minus 16 point 84 ksi) + ( minus 2 point 02 ksi) + ( minus 2 point 44ksi) + ( 2 times 0 point 75 times minus 1 point 75 ksi)

Formula: f subscript botgdr = minus 23 point 93 ksi

Formula: f subscript topgdr = ( 17 point 90 ksi) + ( 1 point 15 ksi) + ( 1 point 39ksi) + ( 2 times 0 point 75 times 0 point 47 ksi)

Formula: f subscript topgdr = 21 point 14 ksi

Formula: t subscript topfl = 2 point 50 inches (see Figure 3-4)

Formula: D subscript web = 54 inches (see Figure 3-4)

Formula: t subscript botfl = 2 point 75 inches (see Figure 3-4)

Formula: Depth subscript gdr = t subscript topfl + D subscript web + t subscript botfl

Formula: Depth subscript gdr = 59 point 25 inches

Formula: Depth subscript comp = numerator ( minus f subscript botgdr) divided by denominator (f subscript topgdr minus f subscript botgdr) times Depth subscript gdr

C6.10.3.1.4a

Formula: Depth subscript comp = 31 point 45 inches

Formula: D subscript c = Depth subscript comp minus t subscript botfl

Formula: D subscript c = 28 point 70 inches

Formula: k = 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared greater than or equal to 7 point 2 Formula: 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared = 31 point 9

Formula: k = max left bracket 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared , 7 point 2 right bracket Formula: k = 31 point 9

Formula: numerator (D) divided by denominator (t subscript w) = 108 point 0 Formula: 0 point 95 times square root of ( numerator (k times E) divided by denominator (F subscript yw)) = 129 point 1

Therefore, Formula: numerator (D) divided by denominator (t subscript w) less than or equal to 0 point 95 times square root of ( numerator (k times E) divided by denominator (F subscript yw))

Based on the unfactored stress values in Table 3-12:

Formula: f subscript cf = ( minus 16 point 84 ksi) + ( minus 2 point 02 ksi) + ( minus 2 point 44ksi) + ( 2 times 0 point 75 times minus 1 point 75 ksi)

Formula: f subscript cf = minus 23 point 93 ksi Formula: F subscript yw = 50 point 0 ksi

Therefore, Formula: f subscript cf less than or equal to F subscript yw OK

For shear, the fatigue requirement for the web is as follows:

S6.10.6.4

Formula: v subscript cf less than or equal to 0 point 58 times C times F subscript yw

Based on the unfactored shear values in Table 3-13:

Formula: V subscript cf = 114 point 7 K + 16 point 4 K + 19 point 8 K + ( 2 times 0 point 75 times 46 point 5 K)

Formula: V subscript cf = 220 point 7 K

Formula: D = 54 point 0 inches Formula: t subscript w = 0 point 50 inches

Formula: v subscript cf = numerator (V subscript cf) divided by denominator (D times t subscript w) Formula: v subscript cf = 8 point 17 ksi

Formula: C = 0 point 550 Formula: F subscript yw = 50 point 0 ksi

Formula: 0 point 58 times C times F subscript yw = 15 point 95 ksi

Therefore, Formula: v subscript cf less than or equal to 0 point 58 times C times F subscript yw OK

Therefore, the fatigue requirements for webs for both flexure and shear are satisfied.

Design Step 3.15 - Design for Flexure - Service Limit State - Negative Moment Region

The girder must be checked for service limit state control of permanent deflection. This check is intended to prevent objectionable permanent deflections due to expected severe traffic loadings that would impair rideability. Service II Limit State is used for this check.

S6.10.5

This check will not control for composite noncompact sections under the load combinations given in STable 3.4.1-1. Although a web bend buckling check is also required in regions of positive flexure at the service limit state according to the current specification language, it is unlikely that such a check would control in these regions for composite girders without longitudinal stiffeners since Dc is relatively small for such girders in these regions.

C6.10.5.1

The web must satisfy SEquation 6.10.3.2.2-1, using the appropriate value of the depth of the web in compression in the elastic range, Dc.

S6.10.5.1

Formula: f subscript cw less than or equal to numerator (0 point 9 times E times alpha k) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) less than or equal to F subscript yw

for which:

Formula: E = 29000 ksi

Formula: alpha = 1 point 25 for webs without longitudinal stiffeners

Formula: D = 54 point 0 inches

The factored Service II flexural stress was previously computed in Table 3-12 as follows:

Formula: f subscript botgdr = minus 35 point 01 ksi

Formula: f subscript topgdr = 24 point 12 ksi

Formula: Depth subscript gdr = 59 point 25 inches (see Figure 3-4)

Formula: Depth subscript comp = numerator ( minus f subscript botgdr) divided by denominator (f subscript topgdr minus f subscript botgdr) times Depth subscript gdr

Formula: Depth subscript comp = 35 point 08 inches

Formula: D subscript c = Depth subscript comp minus t subscript botfl

Formula: D subscript c = 32 point 33 inches

Formula: k = 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared greater than or equal to 7 point 2 for webs without longitudinal stiffeners

Formula: 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared = 25 point 1

Formula: k = max left bracket 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared , 7 point 2 right bracket Formula: k = 25 point 1

Formula: t subscript w = 0 point 5 inches (see Figure 3-4)

Formula: numerator (0 point 9 times E times alpha k) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) = 70 point 23 ksi

Formula: F subscript yw = 50 point 0 ksi

Formula: min left bracket numerator (0 point 9 times E times alpha k) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) , F subscript yw right bracket = 50 point 0 ksi

Formula: f subscript cw = f subscript botgdr times ( numerator (D subscript c) divided by denominator (D subscript c + t subscript f) )

Formula: f subscript cw = minus 32 point 27 ksi OK

In addition, the flange stresses for both steel flanges of composite sections must satisfy the following requirement:

Formula: f subscript f less than or equal to 0 point 95F subscript yf

As previously explained, for this design example, the concrete slab is assumed to be fully effective for both positive and negative flexure for service limit states.

The factored Service II flexural stress was previously computed in Table 3-12 as follows:

S6.10.5.1

Formula: f subscript botgdr = minus 35 point 01 ksi Formula: f subscript topgdr = 24 point 12 ksi

Formula: F subscript yf = 50 point 0 ksi

Formula: 0 point 95 times F subscript yf = 47 point 50 ksi OK

In addition to the check for service limit state control of permanent deflection, the girder can also be checked for live load deflection. Although this check is optional for a concrete deck on steel girders, it is included in this design example at the location of maximum positive moment.

S2.5.2.6.2

Design Step 3.16 - Design for Flexure - Constructibility Check - Negative Moment Region

The girder must also be checked for flexure during construction. The girder has already been checked in its final condition when it behaves as a composite section. The constructibility must also be checked for the girder prior to the hardening of the concrete deck when the girder behaves as a noncomposite section.

S6.10.3.2.2

The investigation of the constructibility of the girder begins with the the noncompact section compression-flange slenderness check, as follows:

S6.10.4.1.4

Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) less than or equal to 12 point 0

Formula: b subscript f = 14 inches (see Figure 3-4)

Formula: t subscript f = 2 point 75 inches (see Figure 3-4)

Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) = 2 point 5

In addition, composite girders, when they are not yet composite, must satisfy the following requirement during construction:

S6.10.3.2.2

Formula: f subscript cw less than or equal to numerator (0 point 9 times E times alpha k) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) less than or equal to F subscript yw

for which:

Formula: E = 29000 ksi

Formula: alpha = 1 point 25 for webs without longitudinal stiffeners

Formula: D = 54 point 0 inches

For the noncomposite loads during construction:

Formula: f subscript botgdr = 1 point 25 times ( minus 16 point 84 ksi)

Formula: f subscript botgdr = minus 21 point 05 ksi

Formula: f subscript topgdr = 1 point 25 times ( 17 point 90 ksi)

Formula: f subscript topgdr = 22 point 37 ksi

Formula: Depth subscript gdr = 59 point 25 inches (see Figure 3-4)

Formula: Depth subscript comp = numerator ( minus f subscript botgdr) divided by denominator (f subscript topgdr minus f subscript botgdr) times Depth subscript gdr

C6.10.3.1.4a

Formula: Depth subscript comp = 28 point 72 inches

Formula: D subscript c = Depth subscript comp minus t subscript botfl

Formula: D subscript c = 25 point 97 inches

Formula: k = 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared greater than or equal to 7 point 2 for webs without longitudinal stiffeners

Formula: 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared = 38 point 9

Formula: k = max left bracket 9 point 0 times ( numerator (D) divided by denominator (D subscript c) ) squared , 7 point 2 right bracket Formula: k = 38 point 9

Formula: t subscript w = 0 point 5 inches (see Figure 3-4)

Formula: numerator (0 point 9 times E times alpha k) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) = 108 point 83 ksi

Formula: F subscript yw = 50 point 0 ksi

Formula: min left bracket numerator (0 point 9 times E times alpha k) divided by denominator (( numerator (D) divided by denominator (t subscript w) ) squared ) , F subscript yw right bracket = 50 point 0 ksi

Formula: f subscript cw = f subscript botgdr times ( numerator (D subscript c) divided by denominator (D subscript c + t subscript f) )

Formula: f subscript cw = minus 19 point 03 ksi OK

In addition to checking the nominal flexural resistance in the web during construction, the nominal shear resistance in the web must also be checked as follows:

S6.10.3.2.3

Formula: V subscript n = C times V subscript p

Formula: C = 0 point 550

Formula: V subscript p = 783 point 0 K

Formula: V subscript n = 430 point 7 K

Formula: phi subscript v = 1 point 0

S6.5.4.2

Formula: V subscript r = phi subscript v times V subscript n

Formula: V subscript r = 430 point 7 K

Formula: V subscript u = ( 1 point 25 times 114 point 7 K) + ( 1 point 25 times 16 point 4 K) + ( 1 point 50 times 19 point 8 K)

Formula: V subscript u = 193 point 6 K OK

Therefore, the design section at the pier satisfies the constructibility specification checks.

Design Step 3.17 - Check Wind Effects on Girder Flanges - Negative Moment Region

As stated in Design Step 3.3, for this design example, the interior girder controls and is being designed.

S6.10.3.5

Wind effects generally do not control a steel girder design, and they are generally considered for the exterior girders only. However, for illustrative purposes, wind effects are presented below for the girder design section at the pier. A bridge height of greater than 30 feet is used in this design step to illustrate the required computations.

C6.10.3.5.2 & C4.6.2.7.1

S3.8.1.1

For noncompact sections, the stresses in the bottom flange are combined as follows:

S6.10.3.5.2

Formula: ( F subscript u + F subscript w ) less than or equal to F subscript r Formula: F subscript w = numerator (6 times M subscript w) divided by denominator (t subscript fb times b subscript fb squared )

Since the deck provides horizontal diaphragm action and since there is wind bracing in the superstructure, the maximum wind moment on the loaded flange is determined as follows:

C4.6.2.7.1

Formula: M subscript w = numerator (W times L subscript b squared ) divided by denominator (10)

Formula: L subscript b = 20 point 0 feet Formula: W = numerator ( eta times gamma times P subscript D times d) divided by denominator (2)

Formula: eta = 1 point 0

S1.3

MathCad tip logo

Strength Limit States for Wind on Structure

For the strength limit state, wind on the structure is considered for the Strength III and Strength V Limit States. For Strength III, the load factor for wind on structure is 1.40 but live load is not considered. Due to the magnitude of the live load stresses, Strength III will clearly not control for this design example (and for most designs). Therefore, for this design example, the Strength V Limit State will be investigated.

Formula: gamma = 0 point 40 for Strength V Limit State

STable 3.4.1-1

Assume that the bridge is to be constructed in Pittsburgh, Pennsylvania. The design horizontal wind pressure is computed as follows:

S3.8.1.2

Formula: P subscript D = P subscript B times ( numerator (V subscript DZ) divided by denominator (V subscript B) ) squared

Formula: P subscript B = 0 point 050 times ksf

STable 3.8.1.2.1-1

Formula: V subscript B = 100 MPH

Formula: V subscript DZ = 2 point 5 times V subscript o times ( numerator (V subscript 30) divided by denominator (V subscript B) ) times ln ( numerator (Z) divided by denominator (Z subscript o) )

S3.8.1.1

Formula: V subscript o = 12 point 0 MPH for a bridge located in a city

STable 3.8.1.1-1

Formula: V subscript 30 = 60 MPH assumed wind velocity at 30 feet above low ground or above design water level at bridge site

Formula: V subscript B = 100 MPH

S3.8.1.1

Formula: Z = 35 feet assumed height of structure at which wind loads are being calculated as measured from low ground or from water level

Formula: Z subscript o = 8 point 20 feet for a bridge located in a city

STable 3.8.1.1-1

Formula: V subscript DZ = 2 point 5 times V subscript o times ( numerator (V subscript 30) divided by denominator (V subscript B) ) times ln ( numerator (Z) divided by denominator (Z subscript o) )

S3.8.1.1

Formula: V subscript DZ = 26 point 1 MPH

Formula: P subscript D = P subscript B times ( numerator (V subscript DZ) divided by denominator (V subscript B) ) squared

S3.8.1.2.1

Formula: P subscript D = 0 point 00341 ksf

After the design horizontal wind pressure has been computed, the factored wind force per unit length applied to the flange is computed as follows:

C4.6.2.7.1

Formula: W = numerator ( eta times gamma times P subscript D times d) divided by denominator (2)

Formula: eta = 1 point 0

S1.3

Formula: gamma = 0 point 40 for Strength V Limit State

STable 3.4.1-1

Formula: P subscript D = 0 point 00341 ksf

Formula: d = 9 point 23 feet from bottom of girder to top of parapet

Formula: W = numerator ( eta times gamma times P subscript D times d) divided by denominator (2)

Formula: W = 0 point 00630 Kips per foot

Next, the maximum lateral moment in the flange due to the factored wind loading is computed as follows:

C4.6.2.7.1

Formula: M subscript w = numerator (W times L subscript b squared ) divided by denominator (10)

Formula: W = 0 point 00630 Kips per foot

Formula: L subscript b = 20 point 0 feet

Formula: M subscript w = 0 point 252 K feet

Finally, the flexural stress at the edges of the bottom flange due to factored wind loading is computed as follows:

S6.10.3.5.2

Formula: F subscript w = numerator (6 times M subscript w) divided by denominator (t subscript fb times b subscript fb squared )

Formula: M subscript w = 0 point 252 K feet

Formula: t subscript fb = 2 point 75 inches

Formula: b subscript fb = 14 point 0 inches

Formula: F subscript w = numerator (6 times M subscript w) divided by denominator (t subscript fb times b subscript fb squared )

Formula: F subscript w = 0 point 034 ksi

The load factor for live load is 1.35 for the Strength V Limit State. However, it is 1.75 for the Strength I Limit State, which we have already investigated. Therefore, it is clear that wind effects will not control the design of this steel girder. Nevertheless, the following computations are presented simply to demonstrate that wind effects do not control this design:

Formula: F subscript u = ( 1 point 25 times minus 16 point 84ksi) + ( 1 point 25 times minus 2 point 15ksi) + ( 1 point 50 times minus 2 point 61 ksi) + ( 1 point 35 times minus 12 point 11 ksi)

Formula: F subscript u = minus 44 point 00 ksi

Formula: F subscript w = minus 0 point 028 ksi

Formula: F subscript u + F subscript w = minus 44 point 03 ksi

Formula: F subscript r = 50 point 0 ksi

Therefore: Formula: ( F subscript u + F subscript w ) less than or equal to F subscript r OK

Therefore, wind effects do not control the design of this steel girder.

Design Step 3.18 - Draw Schematic of Final Steel Girder Design

Since all of the specification checks were satisfied, the trial girder section presented in Design Step 3.2 is acceptable. If any of the specification checks were not satisfied or if the design were found to be overly conservative, then the trial girder section would need to be revised appropriately, and the specification checks would need to be repeated for the new trial girder section.

The following is a schematic of the final steel girder configuration:

The span length from centerline of bearing at the abutment to the centerline at pier is 120 feet 0 inches and the beam projection is 8 inches. The girder is made up of three different sections. The girder has a constant 54 inch deep by one half inch thick web. The end of the first section is 84 feet 0 inches from the centerline of bearing at abutment. At the end of the first section, there is a bolted field splice. The first section has a top flange of 14 inches wide by five eights of an inch thick. The bottom flange for the first section is 14 inches wide by seven eights of an inch thick. The second section ends 108 feet 0 inches from the centerline of bearing at abutment or 24 feet 0 inches past the end of section one. The second section has a top flange of 14 inches wide by one and one quarter of an inch thick. The bottom flange for the second section is 14 inches wide by one and three eights of an inch thick. The third section ends 120 feet 0 inches from the centerline of bearing at abutment or 12 feet 0 inches past the end of section two. The third section has a top flange of 14 inches wide by two and one half inches thick. The bottom flange for the third section is 14 inches wide by two and three quarters of an inch thick. The girder has bearing stiffeners on both sides of the web at the centerline of bearing at abutment and the centerline of bearing at the pier. There are 5 and one half inch by one half inch transverse intermediate stiffeners spaced at 6 feet 8 inches in section 2 and 3. The transverse intermediate stiffeners are on one side of the web only, Interior side of the fascia girders.

Figure 3-15 Final Plate Girder Elevation

For this design example, only the location of maximum positive moment, the location of maximum negative moment, and the location of maximum shear were investigated. However, the above schematic shows the plate sizes and stiffener spacing throughout the entire length of the girder. Some of the design principles for this design example are presented in "tip boxes."

Design computations for a bolted field splice are presented in Design Step 4. Design computations and principles for shear connectors, bearing stiffeners, welded connections, and cross-frames are presented in Design Step 5. Design computations for an elastomeric bearing pad are presented in Design Step 6.

<< previous Contents next >>
Updated: 07/18/2013
Federal Highway Administration | 1200 New Jersey Avenue, SE | Washington, DC 20590 | 202-366-4000