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Bridges & Structures

LRFD Steel Girder SuperStructure Design Example

Bolted Field Splice Design Example Design Step 4

Table of Contents

Design Step 4.1 - Obtain Design Criteria
Design Step 4.2 - Select Girder Section as Basis for Field Splice Design
Design Step 4.3 - Compute Flange Splice Design Loads
Design Step 4.4 - Design Bottom Flange Splice
Design Step 4.5 - Design Top Flange Splice
Design Step 4.6 - Compute Web Splice Design Loads
Design Step 4.7 - Design Web Splice
Design Step 4.8 - Draw Schematic of Final Bolted Field Splice Design

Design Step 4.1 - Obtain Design Criteria

This splice design example is based on AASHTO LRFD Bridge Design Specifications (through 2002 interims). The design methods presented throughout the example are meant to be the most widely used in general bridge engineering practice.

The first design step is to identify the appropriate design criteria. This includes, but is not limited to, defining material properties, identifying relevant superstructure information, and determining the splice location.

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 splice.

Presented in Figure 4-1 is the final steel girder configuration as designed in Design Step 3. Included in Figure 4-1 is the bolted field splice location. This location was determined using the criteria presented in the narrative below.

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 4-1 Plate Girder Elevation

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

Formula: K = 1000lb Formula: kcf = Kips per cubic foot Formula: ksi = Kips per square inch

For relatively long girders, field splices are generally required to reduce the girder shipping length. The location of the field splice is generally based on economy and includes the following considerations:

  1. Field splices are generally located to provide girder segment lengths that do not exceed the allowable girder shipping length. The allowable girder shipping length is often a function of the shipping route from the fabrication site to the construction site.

  2. The Specifications recommends locating splices near points of dead load contraflexure.

    S6.13.6.1.4a

  3. Field splices are generally located where total moment in the girder is relatively small. This minimizes the required splice plate thicknesses and the required number of bolts.

In Design Step 1.1, the steel properties of the girder were defined. These properties will be used for the splice plates as well.

Yield Strength: Formula: F subscript y = 50 ksi STable 6.4.1-1
Tensile Strength: Formula: F subscript u = 65 ksi
For Specifications equations requiring the flange yield strength:
Flange Yield Strength: Formula: F subscript yf = 50 ksi

Plate Dimensions of the Left Girder (reference Design Step 3.18):

Web Thickness: Formula: t subscript w = 0 point 50 inches
Web Depth: Formula: D = 54 inches
Top Flange Width: Formula: b subscript fltL = 14 inches
Top Flange Thickness: Formula: t subscript fltL = 0 point 625 inches
Bottom Flange Width: Formula: b subscript flbL = 14 inches
Bottom Flange Thickness: Formula: t subscript flbL = 0 point 875 inches

Plate Dimensions of the Right Girder (reference Design Step 3.18):

Web Thickness: Formula: t subscript w = 0 point 50 inches
Web Depth: Formula: D = 54 inches
Top Flange Width: Formula: b subscript fltR = 14 inches
Top Flange Thickness: Formula: t subscript fltR = 1 point 25 inches
Bottom Flange Width: Formula: b subscript flbR = 14 inches
Bottom Flange Thickness: Formula: t subscript flbR = 1 point 375 inches

Splice Bolt Properties:

Bolt Diameter: Formula: d subscript bolt = 0 point 875 inches
Bolt Hole Diameter:(for design purposes) Formula: d subscript hole = 1 point 0 inches S6.13.2.5 & S6.8.3
Bolt Tensile Strength: Formula: Fu subscript bolt = 120 ksi S6.4.3.1

Concrete Deck Properties (reference Design Step 3.3):

Effective Slab Thickness: Formula: t subscript seff = 8 inches
Modular Ratio: Formula: n = 8
Haunch Depth (measured from top of web): Formula: d subscript haunch = 3 point 5 inches
Effective Flange Width: Formula: W subscript eff = 103 inches

Based on the concrete deck design example and as illustrated in Figure 2-18, the area of longitudinal deck reinforcing steel in the negative moment region is computed as follows:

For the top steel: Formula: A subscript deckreinftop = ( 0 point 31 inches squared ) times numerator (W subscript eff) divided by denominator (5 inches )
Formula: A subscript deckreinftop = 6 point 386 inches squared
For the bottom steel: Formula: A subscript deckreinfbot = ( 0 point 31 inches squared ) times numerator (W subscript eff) divided by denominator (5 inches )
Formula: A subscript deckreinfbot = 6 point 386 inches squared

Resistance Factors: S6.5.4.2

Flexure: Formula: phi subscript f = 1 point 0
Shear: Formula: phi subscript v = 1 point 0
Axial Compression: Formula: phi subscript c = 0 point 90
Tension, fracture in net section: Formula: phi subscript u = 0 point 80
Tension, yielding in gross section: Formula: phi subscript y = 0 point 95
Bolts bearing on material: Formula: phi subscript bb = 0 point 80
A325 and A490 bolts in shear: Formula: phi subscript s = 0 point 80
Block shear: Formula: phi subscript bs = 0 point 80

Design Step 4.2 - Select Girder Section as Basis for Field Splice Design

Where a section changes at a splice, the smaller of the two connected sections shall be used in the design. Therefore, the bolted field splice will be designed based on the left adjacent girder section properties. This will be referred to as the Left Girder throughout the calculations. The girder located to the right of the bolted field splice will be designated the Right Girder.

S6.13.6.1.1

Design Step 4.3 - Compute Flange Splice Design Loads

Girder Moments at the Splice Location:

Based on the properties defined in Design Step 3 (Steel Girder Design), any number of commercially available software programs can be used to obtain the design dead and live loads at the splice. For this design example, the AASHTO Opis software was used. A summary of the unfactored moments at the splice from the initial trial of the girder design are listed below. The live loads include impact and distribution factors.

Loads Moments
Dead Loads: Noncomposite: Formula: M subscript NDL = minus 51 point 8 K feet
Composite: Formula: M subscript CDL = 15 point 5 K feet
Future Wearing Surface: Formula: M subscript FWS = 18 point 8 K feet
Live Loads: HL-93 Positive: Formula: M subscript PLL = 1307 point 8 K feet
HL-93 Negative: Formula: M subscript NLL = minus 953 point 3 K feet
Fatigue Positive: Formula: M subscript PFLL = 394 point 3 K feet
Fatigue Negative: Formula: M subscript NFLL = minus 284 point 0 K feet

Typically, splices are designed for the Strength I, Service II, and Fatigue Limit States. The load factors for these limit states are shown in Table 4-1:

S6.13.6

  Load Factors
Strength I Service II Fatigue
Load γmax γmin γmax γmin γmax γmin
DC 1.25 0.90 1.00 1.00 - -
DW 1.50 0.65 1.00 1.00 - -
LL 1.75 1.75 1.30 1.30 0.75 0.75

STable 3.4.1-1

STable 3.4.1-2

Table 4-1 Load Factors

Flange Stress Computation Procedure:

As previously mentioned, the applicable limit states for the splice design are Strength I, Service II, and Fatigue. The stresses corresponding to these limit states will be computed at the midthickness of the top and bottom flanges. The appropriate section properties and load factors for use in computing stresses are described below. Where necessary, refer to the signs of the previously documented design moments.

S6.13.6

Strength I Limit State:

At the strength limit state, the section properties for flexural members with holes in the tension flange shall be computed using an effective flange area.

S6.10.3.6

Case 1: Dead Load + Positive Live Load

For this case, stresses will be computed using the effective top flange area for the noncomposite dead load, and the effective bottom flange area for the composite dead load, future wearing surface, and live load. The minimum load factor is used for the DC dead loads (noncomposite and composite) and the maximum load factor is used for the future wearing surface. The composite dead load and future wearing surface act on the 3n- or the n-composite slab section, whichever gives the higher stresses, and the live load acts on the n-composite slab section.

STable 3.4.1-2

S6.10.3.1.1b

Case 2: Dead Load + Negative Live Load

For this case, stresses will be computed using the effective top flange area for all loads. The future wearing surface is excluded and the maximum load factor is used for the DC dead loads. The live load acts on the composite steel girder plus longitudinal reinforcement section. The composite dead load is applied to this section as well, as a conservative assumption for simplicity and convenience, since the net effect of the live load is to induce tension in the slab. The reinforcing steel in the deck that is used corresponds to the negative moment deck reinforcement shown in Figure 2-18.

Service II Limit State:

S6.13.6.1.4a

Case 1: Dead Load + Positive Live Load

For this case, stresses will be computed using the gross steel section. The future wearing surface is included and acts, along with the composite dead load, on the 3n- or n-composite slab section, whichever gives the higher stresses. The live load acts on the n-composite slab section.

Case 2: Dead Load + Negative Live Load

S6.10.3.1.1c

For this case, stresses will be computed using the gross steel section. The future wearing surface is excluded. The composite dead load acts on the 3n- or n-composite slab section, whichever gives the larger stresses. The live load acts on the n-composite slab section.

Fatigue Limit State:

C6.13.6.1.4a

Case 1: Positive Live Load

For this case, stresses will be computed using the gross steel section. The live load acts on the n-composite slab section.

Case 2: Negative Live Load

S6.10.3.1.1c

For this case, stresses will be computed using the gross steel section. The live load acts on the n-composite slab section.

Section Properties:

Effective Flange Areas:

S6.13.6.1.4c & S6.10.3.6

Formula: A subscript e = A subscript n + beta times A subscript g less than or equal to A subscript g

SEquation 6.10.3.6-1

For holes equal to or less than 1.25 inches in diameter:

Formula: beta = ( numerator (A subscript n) divided by denominator (A subscript g) ) times left bracket( numerator ( phi subscript u times F subscript u) divided by denominator ( phi subscript y times F subscript yf) ) minus 1 right bracket greater than or equal to 0 point 0

The effective area of the bottom flange of the steel girder is as follows:

Formula: A subscript g = t subscript flbL times b subscript flbL Formula: A subscript g = 12 point 25 inches squared

The net area of the bottom flange of the steel girder is defined as the product of the thickness of the flange and the smallest net width. The net width is determined by subtracting from the width of the flange the sum of the widths of all holes in the assumed failure chain, and then adding the quantity s2 /4g for each space between consective holes in the chain. Since the bolt holes in the flanges are lined up transverse to the loading direction, the governing failure chain is straight across the flange (i.e., s2 /4g is equal to zero).

S6.8.3

The net area of the bottom flange of the steel girder now follows:

Formula: A subscript n = ( b subscript flbL minus 4 times d subscript hole ) times t subscript flbL Formula: A subscript n = 8 point 75 inches squared

Formula: beta = ( numerator (A subscript n) divided by denominator (A subscript g) ) times left bracket( numerator ( phi subscript u times F subscript u) divided by denominator ( phi subscript y times F subscript yf) ) minus 1 right bracket Formula: beta = 0 point 07

With the gross and net areas identified, along with beta, the effective tension area of the bottom flange can now be computed as follows:

Formula: A subscript e = A subscript n + beta times A subscript g

Formula: A subscript e = 9 point 58 inches squared

Check:

Formula: A subscript e = 9 point 58 inches squared < Formula: A subscript g = 12 point 25 inches squared OK

Effective bottom flange area: Formula: A subscript ebot = 9 point 58 inches squared

Similar calculations determine the effective tension area for the top flange of the steel girder:

Effective top flange area: Formula: A subscript etop = 6 point 84 inches squared

The transformed effective area of the concrete flange of the steel girder is now determined. This requires the modular ratio as follows:

Ac= Effective Slab Width x tseff

Modular Ratio

where:

Effective Slab Width: Formula: W subscript eff = 103 inches

Modular Ratio: Formula: n = 8

For the n-composite beam:

Formula: A subscript c = numerator (W subscript eff) divided by denominator (n) times t subscript seff Formula: A subscript c = 103 point 00 inches squared

For the 3n-composite beam:

Formula: A subscript c3n = numerator (W subscript eff) divided by denominator (3n) times t subscript seff Formula: A subscript c3n = 34 point 33 inches squared

The section properties for the Left Girder are calculated with the aid of Figure 4-2 shown below:

This figure shows the girder and the slab, which will aid in the calculation of the section properties. Area 1 is the transverse reinforcement in the top layer of the slab. Area 2 is the transverse reinforcement in the bottom layer of the slab. Area 3 is the area of the top flange. Area 4 is the area of the web. Area 5 is the area of the bottom flange. Y bar is defined as the distance from neutral axis to the bottom of the bottom flange.

Figure 4-2 Girder, Slab and Longitudinal Reinforcement

The following tables contain the section properties for the left (i.e., smaller) girder section at the splice location. The properties in Table 4-2 are based on the gross area of the steel girder, and these properties are used for computation of stresses for the Service II and Fatigue Limit States. The properties in Tables 4-3 and 4-4 are based on the effective top flange and effective bottom flange of the steel girder, respectively, and these properties are used for computation of stresses for the Strength I Limit State.

Gross 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 ybotmid (Inches) ytopmid (Inches) Sbotweb (Inches3) Sbotmid (Inches3) Stopmid (Inches3) Stopweb (Inches3)
Girder only 25.414 29.336 885.4 870.2 753.8 762.0
Composite (3n) 40.644 14.106 1218.7 1205.5 3473.7 3552.4
Composite (n) 50.327 4.423 1329.7 1318.2 15000.3 16140.8

Table 4-2 Section Properties Based on Gross Steel Section

Section Properties - Effective Top Flange Area
Section Area, A (Inches2) Centroid, d (Inches) A*d (Inches3) Io(Inches4) A*y2 (Inches4) Itotal (Inches4)
Girder only:
Top flange 6.840 55.188 377.5 0.3 6384.5 6384.8
Web 27.000 27.875 752.6 6561.0 283.3 6844.3
Bottom flange 12.250 0.438 5.4 0.8 7173.1 7173.9
Total 46.090 24.636 1135.5 6562.1 13840.9 20402.9
Deck Steel:
Girder 46.090 24.636 1135.5 20402.9 3009.2 23412.2
Top Steel 6.386 63.438 405.1 0.0 6027.1 6027.1
Bottom Steel 6.386 60.313 385.2 0.0 4863.3 4863.3
Total 58.862 32.716 1925.7 20402.9 13899.7 34302.6
Composite (3n):
Girder 46.090 24.636 1135.5 20402.9 11963.5 32366.4
Slab 34.333 62.375 2141.5 183.1 16060.1 16243.2
Total 80.423 40.747 3277.0 20586.1 28023.6 48609.7
Composite (n):
Girder 46.090 24.636 1135.5 20402.9 31330.6 51733.5
Slab 103.000 62.375 6424.6 549.3 14019.7 14569.0
Total 149.090 50.708 7560.1 20952.3 45350.2 66302.5
Section ybotmid (Inches) ytopmid (Inches) Sbotmid (Inches3) Stopmid (Inches3)
Girder only 24.198 30.552   843.2 667.8  
Deck Steel 32.279 22.471   1062.7 1526.5  
Composite (3n) 40.309 14.441   1205.9 3366.2  
Composite (n) 50.271 4.479   1318.9 14802.1  

Table 4-3 Section Properties Using Effective Top Flange Area of Steel Girder

Section Properties - Effective Bottom Flange Area
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 6781.3 6781.6
Web 27.000 27.875 752.6 6561.0 7.5 6568.5
Bottom flange 9.580 0.438 4.2 0.6 6937.8 6938.5
Total 45.330 27.348 1239.7 6561.9 13726.7 20288.6
Deck Steel:
Girder 45.330 27.348 1239.7 20288.6 2611.1 22899.7
Top Steel 6.386 63.438 405.1 0.0 5186.8 5186.8
Bottom Steel 6.386 60.313 385.2 0.0 4111.7 4111.7
Total 58.102 34.938 2030.0 20288.6 11909.6 32198.2
Composite (3n):
Girder 45.330 27.348 1239.7 20288.6 10329.9 30618.4
Slab 34.333 62.375 2141.5 183.1 13638.4 13821.5
Total 79.663 42.444 3381.2 20471.7 23968.3 44440.0
Composite (n):
Girder 45.330 27.348 1239.7 20288.6 26816.1 47104.7
Slab 103.000 62.375 6424.6 549.3 11801.7 12351.0
Total 148.330 51.671 7664.3 20837.9 38617.8 59455.7
Section ybotmid (Inches) ytopmid (Inches)   Sbotmid (Inches3) Stopmid (Inches3)  
Girder only 26.911 27.839   753.9 728.8  
Deck Steel 34.501 20.249   933.3 1590.1  
Composite (3n) 42.007 12.743   1057.9 3487.3  
Composite (n) 51.233 3.517   1160.5 16906.8  

Table 4-4 Section Properties Using Effective Bottom Flange Area of Steel Girder

Strength I Limit State Stresses - Dead Load + Positive Live Load:

The section properties for this case have been calculated in Tables 4-3 and 4-4. The stresses at the midthickness of the flanges are shown in Table 4-6, which immediately follows the sample calculation presented below.

A typical computation for the stresses occurring at the midthickness
of the flanges is presented in the example below. The stress in the bottom flange of the girder is computed using the 3n-composite section for the composite dead load and future wearing surface,
and the n-composite section for the live load:

Formula: f = numerator (M) divided by denominator (S)

Noncomposite DL:

Stress at the midthickness:

Formula: f = f subscript botgdr subscript 1

Noncomposite DL Moment:

Formula: M subscript NDL = minus 51 point 8 K feet

Section Modulus (girder only), from Table 4-3:

Formula: S subscript botgdr subscript 1 = 843 point 1 inches cubed

Stress due to the noncomposite dead load:

Formula: f subscript botgdr subscript 1 = numerator (M subscript NDL) divided by denominator (S subscript botgdr subscript 1) Formula: f subscript botgdr subscript 1 = minus 0 point 74 ksi

Composite DL:

Stress at the midthickness:

Formula: f = f subscript botgdr subscript 2

Composite DL Moment:

Formula: M subscript CDL = 15 point 5 K feet

Section Modulus (3n-composite), From Table 4-4:

Formula: S subscript botgdr subscript 2 = 1057 point 9 inches cubed

Stress due to the composite dead load:

Formula: f subscript botgdr subscript 2 = numerator (M subscript CDL) divided by denominator (S subscript botgdr subscript 2) Formula: f subscript botgdr subscript 2 = 0 point 18 ksi

Future Wearing Surface:

Stress at the midthickness:

Formula: f = f subscript botgdr subscript 3

FWS Moment:

Formula: M subscript FWS = 18 point 8 K feet

Section Modulus (3n-composite), From Table 4-4:

Formula: S subscript botgdr subscript 3 = 1057 point 9 inches cubed

Stress due to the composite dead load:

Formula: f subscript botgdr subscript 3 = numerator (M subscript FWS) divided by denominator (S subscript botgdr subscript 3) Formula: f subscript botgdr subscript 3 = 0 point 21 ksi

Positive Live Load:

Stress at the midthickness:

Formula: f = f subscript botgdr subscript 4

Live Load Moment:

Formula: M subscript PLL = 1307 point 8 K feet

Section Modulus (n-composite), From Table 4-4:

Formula: S subscript botgdr subscript 4 = 1160 point 5 inches cubed

Stress due to the positive live load:

Formula: f subscript botgdr subscript 4 = numerator (M subscript PLL) divided by denominator (S subscript botgdr subscript 4) Formula: f subscript botgdr subscript 4 = 13 point 52 ksi

The preceding stresses are now factored by their respective load factors to obtain the final factored stress at the midthickness of the bottom flange for this load case. The applicable load factors for this case were discussed previously.

STable 3.4.1-1 & STable 3.4.1-2

Formula: f subscript botgdr = ( 0 point 90 times f subscript botgdr subscript 1 + 0 point 90 times f subscript botgdr subscript 2 + 1 point 50 times f subscript botgdr subscript 3 + 1 point 75 times f subscript botgdr subscript 4 )

Formula: f subscript botgdr = 23 point 48 ksi

The stresses at the midthickness of the top flange for this load case are computed in a similar manner. The section properties used to obtain the stresses in the top flange are also from Tables 4-3 and 4-4.

The top and bottom flange midthickness stresses are summarized in Table 4-5, shown below.

Strength I - Dead Load + Positive Live Load
Summary of Unfactored Values
Loading Moment (K-ft) fbotmid(ksi) ftopmid(ksi)
Noncomposite DL -51.80 -0.74 0.93
Composite DL 15.50 0.18 -0.05
FWS DL 18.80 0.21 -0.06
Live Load - HL-93 1307.80 13.52 -0.93
Summary of Factored Values
Limit State
Strength I 2284.18 23.48 -0.93

Table 4-5 Strength I Flange Stresses for Dead + Pos. LL

The computation of the midthickness flange stresses for the remaining load cases are computed in a manner similar to what was shown in the sample calculation that preceded Table 4-5.

Strength I Limit State - Dead Load + Negative Live Load:

The computed stresses in the following table require the use of section properties from Table 4-3.

Strength I - Dead Load + Negative Live Load
Summary of Unfactored Values
Loading Moment (K-ft) fbotmid(ksi) ftopmid(ksi)
Noncomposite DL -51.80 -0.74 0.93
Composite DL 15.50 0.18 -0.12
Live Load - HL-93 -953.30 -10.76 7.49
Summary of Factored Values
Limit State
Strength I -1713.65 -19.54 14.13

Table 4-6 Strength I Flange Stresses for Dead + Neg. LL

Service II Limit State - Dead Load + Positive Live Load:

The computed stresses in the following table require the use of section properties from Table 4-2.

Service II - Dead Load + Positive Live Load
Summary of Unfactored Values
Loading Moment (K-ft) fbotmid(ksi) ftopmid(ksi)
Noncomposite DL -51.80 -0.71 0.82
Composite DL 15.50 0.15 -0.05
FWS 18.80 0.19 -0.06
Live Load - HL-93 1307.80 11.91 -1.05
Summary of Factored Values
Limit State
Service II 1682.64 15.10 -0.65

Table 4-7 Service II Flange Stresses for Dead + Pos. LL

Service II Limit State - Dead Load + Negative Live Load:

The computed stresses in the following table require the use of section properties from Table 4-2.

Service II - Dead Load + Negative Live Load
Summary of Unfactored Values
Loading Moment (K-ft) fbotmid(ksi) ftopmid(ksi)
Noncomposite DL -51.80 -0.71 0.82
Composite DL 15.50 0.14 -0.01
Live Load - HL-93 -953.30 -8.68 0.76
Summary of Factored Values
Limit State
Service II -1275.59 -11.85 1.80

Table 4-8 Service II Flange Stresses for Dead + Neg. LL

Fatigue Limit State - Positive Live Load:

The computed stresses in the following table require the use of section properties from Table 4-2.

Fatigue - Positive Live Load
Summary of Unfactored Values
Loading Moment (K-ft) fbotmid(ksi) ftopmid(ksi)
Live Load-Fatigue 394.30 3.59 -0.32
Summary of Factored Values
Limit State
Fatigue 295.73 2.69 -0.24

Table 4-9 Fatigue Flange Stresses for Positive LL

Fatigue Limit State - Negative Live Load:

The computed stresses in the following table require the use of section properties from Table 4-2.

Fatigue - Negative Live Load
Summary of Unfactored Values
Loading Moment (K-ft) fbotmid(ksi) ftopmid(ksi)
Live Load-Fatigue -284.00 -2.59 0.23
Summary of Factored Values
Limit State
Fatigue -213.00 -1.94 0.17

Table 4-10 Fatigue Flange Stresses for Negative LL

Fatigue Limit State:

The computed stresses in the following table require the use of section properties from Table 4-2.

Fatigue - Live Load
Summary of Unfactored Values
Loading Moment (K-ft) fbotweb(ksi) ftopweb(ksi)
Live Load-Pos 394.3 3.56 -0.29
Live Load-Neg -284.00 -2.56 0.21
Summary of Factored Values
Limit State
Pos Fatigue 295.73 2.67 -0.22
Neg Fatigue -213.00 -1.92 0.16

Table 4-11 Fatigue Web Stresses for Positive and Negative Live Load

A summary of the factored stresses at the midthickness of the top and bottom flanges for the Strength I, Service II, and Fatigue limit states are presented below in Tables 4-12 through 4-14. Table 4-14 also contains the top and bottom web fatigue stresses.

  Stress (ksi)
Limit State Location Dead + Pos. LL Dead + Neg. LL
Strength I Bottom Flange 23.48 -19.54
Top Flange -0.93 14.13

Table 4-12 Strength I Flange Stresses

  Stress (ksi)
Limit State Location Dead + Pos. LL Dead + Neg. LL
Service II Bottom Flange 15.10 -11.85
Top Flange -0.65 1.80

Table 4-13 Service II Flange Stresses

  Stress (ksi)
Limit State Location Positive LL Negative LL
Fatigue Bottom Flange 2.69 -1.94
Top Flange -0.24 0.17
Bottom of Web 2.67 -1.92
Top of Web -0.22 0.16

Table 4-14 Fatigue Flange and Web Stresses

Strength I Minimum Design Force - Controlling Flange:

S6.13.6.1.4c

The next step is to determine the minimum design forces for the controlling flange of each load case (i.e., positive and negative live load). By inspection of Table 4-12, it is obvious that the bottom flange is the controlling flange for both positive and negative live load for the Strength I Limit State.

The minimum design force for the controlling flange, Pcu, is taken equal to the design stress, Fcf, times the smaller effective flange area, Ae,on either side of the splice. When a flange is in compression, the effective compression flange area shall be taken as Ae= Ag.

S6.10.3.6

The calculation of the minimum design force is presented below for the load case of dead load with positive live load.

The minimum design stress for the controlling (bottom) flange is computed as follows:

Formula: F subscript cf = numerator (( Vertical Bar numerator (f subscript cf) divided by denominator (R subscript h) Vertical Bar + alpha times phi subscript f times F subscript yf )) divided by denominator (2) greater than or equal to 0 point 75 times alpha times phi subscript f times F subscript yf

SEquation 6.13.6.1.4c-1

where:

Maximum flexural stress due to the factored loads at the midthickness of the controlling flange at the point of splice (from Table 4-12): Formula: f subscript cf = 23 point 48 ksi

Hybrid girder reduction factor.

For homogeneous girders: Formula: R subscript h = 1 point 0

Flange stress reduction factor: Formula: alpha = 1 point 0

Resistance factor for flexure (Design Step 4.1): Formula: phi subscript f = 1 point 0

Minimum yield strength of the flange: Formula: F subscript yf = 50 ksi

Formula: F subscript cf subscript 1 = numerator (( Vertical Bar numerator (f subscript cf) divided by denominator (R subscript h) Vertical Bar + alpha times phi subscript f times F subscript yf )) divided by denominator (2)

Formula: F subscript cf subscript 1 = 36 point 74 ksi

Compute the minimum required design stress:

Formula: F subscript cf subscript 2 = 0 point 75 times alpha times phi subscript f times F subscript yf

Formula: F subscript cf subscript 2 = 37 point 50 ksi

The minimum design stress for the bottom flange for this load case is:

Formula: F subscript cf = max ( F subscript cf subscript 1 , F subscript cf subscript 2 )

Formula: F subscript cf = 37 point 50 ksi

The minimum design force now follows:

Formula: P subscript cu = F subscript cf times A subscript e

The gross area of the bottom flange is:

Formula: A subscript flbL = b subscript flbL times t subscript flbL

Formula: A subscript flbL = 12 point 25 inches squared

Since the bottom flange force for this load case is a tensile force, the effective area will be used. This value was computed previously to be:

Formula: A subscript ebot = 9 point 58 inches squared

Therefore:

Formula: P subscript cu = F subscript cf times A subscript ebot

Formula: P subscript cu = 359 point 25 K

Table 4-15 presents the minimum design forces for the Strength I Limit State for both the positive and negative live load cases.

  Strength I Limit State Controlling Flange
Load Case Location fcf (ksi) Fcf (ksi) Area (in2) Pcu (kips)
Dead + Pos. LL Bot. Flange 23.48 37.5 9.58 359.25
Dead + Neg. LL Bot. Flange -19.54 37.5 12.25 459.38

Table 4-15 Controlling Flange Forces

In the above table, the design controlling flange force (Pcu) is a compressive force for negative live load.

Strength I Minimum Design Force - Noncontrolling Flange:

S6.13.6.1.4c

The next step is to determine the minimum design forces for the noncontrolling flange of each load case (i.e., positive and negative live load). By inspection of Table 4-12, the top flange is the noncontrolling flange for both positive and negative live load for the Strength I Limit State.

The minimum design force for the noncontrolling flange, Pncu, is taken equal to the design stress, Fncf, times the smaller effective flange area, Ae,on either side of the splice. When a flange is in compression, the effective compression flange area shall be taken as Ae= Ag.

S6.10.3.6

The calculation of the minimum design force is presented below for the load case of dead load with positive live load.

The minimum design stress for the noncontrolling (top) flange is computed as follows:

Formula: F subscript ncf = R subscript cf times Vertical Bar numerator (f subscript ncf) divided by denominator (R subscript h) Vertical Bar greater than or equal to 0 point 75 times alpha times phi subscript f times F subscript yf

SEquation 6.13.6.1.4c-2

where:

Maximum flexural stress due to the factored loads at the midthickness of the noncontrolling flange at the point of splice concurrent with fcf (see Table 4-12): Formula: f subscript ncf = minus 0 point 93 ksi

Controlling flange design stress: Formula: F subscript cf = 37 point 50 ksi

Controlling flange actual stress: Formula: f subscript cf = 23 point 48 ksi

Controlling flange stress ratio:

Formula: R subscript cf = Vertical Bar numerator (F subscript cf) divided by denominator (f subscript cf) Vertical Bar Formula: R subscript cf = 1 point 60

Hybrid girder reduction factor: Formula: R subscript h = 1 point 00

Therefore:

Formula: F subscript ncf subscript 1 = R subscript cf times Vertical Bar numerator (f subscript ncf) divided by denominator (R subscript h) Vertical Bar

Formula: F subscript ncf subscript 1 = 1 point 49 ksi

Compute the minimum required design stress:

Formula: F subscript ncf subscript 2 = 0 point 75 times alpha times phi subscript f times F subscript yf

Formula: F subscript ncf subscript 2 = 37 point 50 ksi

The minimum design stress in the top flange is:

Formula: F subscript ncf = max ( F subscript ncf subscript 1 , F subscript ncf subscript 2 )

Formula: F subscript ncf = 37 point 50 ksi

The minimum design force now follows:

Formula: P subscript ncu = F subscript ncf times A subscript e

For the positive live load case, the top flange is in compression. The effective compression flange area shall be taken as:

Formula: A subscript e = A subscript g

SEquation 6.10.3.6-2

Formula: A subscript g = t subscript fltL times b subscript fltL Formula: A subscript g = 8 point 75 inches squared

Therefore:

Formula: P subscript ncu = F subscript ncf times A subscript g

Formula: P subscript ncu = 328 point 13 K (compression)

Table 4-16 presents the minimum design forces for the Strength I Limit State for both the positive and negative live load cases.

  Strength I Limit State Noncontrolling Flange
Load Case Location fncf (ksi) Fncf (ksi) Area (in2) Pncu (kips)
Dead + Pos. LL Top Flange -0.93 37.5 8.75 328.13
Dead + Neg. LL Top Flange 14.13 37.5 6.84 256.50

Table 4-16 Noncontrolling Flange Forces

In the above table, the design noncontrolling flange force (Pncu) is a compressive force for positive live load.

Service II Limit State Flange Forces:

S6.13.6.1.4c

Per the Specifications, bolted connections for flange splices are to be designed as slip-critical connections for the service level flange design force. This design force shall be taken as the Service II design stress, Fs, multiplied by the smaller gross flange area on either side of the splice.

Fs is defined as follows:

Formula: F subscript s = numerator (f subscript s) divided by denominator (R subscript h)

SEquation 6.13.6.1.4c-4

fs = maximum flexural Service II stress at the midthickness of the flange under consideration.

The factored Service II design stresses and forces are shown in Table 4-17 below.

  Service II Limit State
Load Case Location Fs (ksi) Agross (in2) Ps (kips)
Dead + Pos. LL Bot. Flange 15.10 12.25 184.98
Top Flange -0.65 8.75 -5.69
Dead + Neg. LL Bot. Flange -11.85 12.25 -145.16
Top Flange 1.80 8.75 15.75

Table 4-17 Service II Flange Forces

It is important to note here that the flange slip resistance must exceed the larger of: (1) the Service II flange forces or (2) the factored flange forces from the moments at the splice due to constructibility (erection and/or deck pouring sequence). However, in this design example, no special erection procedure is prescribed and, per the Introduction in Design Step 1, the deck is placed in a single pour. Therefore, the constructibility moment is equal to the noncomposite dead load moment shown at the beginning of this design step. By inspection, the Service II Limit State will control for checking of slip-critical connections for the flanges and the web in this example.

S3.4.2

Fatigue Limit State Stresses:

C6.13.6.1.4c

The final portion of this design step is to determine the range of the stresses at the midthickness of both flanges, and at the top and bottom of the web for the Fatigue Limit State. The ranges are calculated below and presented in Table 4-18.

A typical calculation of the stress range for the bottom flange is shown below.

From Tables 4-9 and 4-10, the factored stresses at the midthickness of the bottom flange are:

Case 1 - Positive Live Load:

Formula: f subscript spos = 2 point 69 ksi

Case 2 - Negative Live Load:

Formula: f subscript sneg = minus 1 point 94 ksi

The stress range is determined by:

Formula: Delta f = Vertical Bar f subscript spos Vertical Bar + Vertical Bar f subscript sneg Vertical Bar

Formula: Delta f = 4 point 63 ksi

  Fatigue Limit State
Stress Range (ksi)
Location Df (ksi)
Bottom Flange 4.63
Top Flange 0.41
Bottom of Web 4.59
Top of Web 0.38

Table 4-18 Fatigue Stress Ranges

Design Step 4.4 - Design Bottom Flange Splice

Splice Plate Dimensions:

The width of the outside plate should be at least as wide as the width of the narrowest flange at the splice. Therefore, try a 7/16" x 14" outside splice plate with two 1/2" x 6" inside splice plates. Include a 1/2" x 14" fill plate on the outside. Figure 4-3 illustrates the initial bottom flange splice configuration.

This sketch shows the splice plates at the bottom flange. The bottom flange on the left side is 14 inches wide by seven eights of an inch thick and the bottom flange on the right is 14 inches wide by one and three eights of an inch thick. The outside splice plate is 14 inches wide by seven sixteenths of an inch thick. The fill plate is 14 inches wide by one half of an inch thick. There are two inside splice plates that measure 6 inches by one half of an inch thick.

Figure 4-3 Bottom Flange Splice

The dimensions of the elements involved in the bottom flange splice from Figure 4-3 are:

Thickness of the inside splice plate: Formula: t subscript in = 0 point 50 inches

Width of the inside splice plate: Formula: b subscript in = 6 inches

Thickness of the outside splice plate: Formula: t subscript out = 0 point 4375 inches

Width of the outside splice plate: Formula: b subscript out = 14 inches

Thickness of the fill plate: Formula: t subscript fill = 0 point 50 inches

Width of the fill plate: Formula: b subscript fill = 14 inches

If the combined area of the inside splice plates is within ten percent of the area of the outside splice plate, then both the inside and outside splice plates may be designed for one-half the flange design force.

C6.13.6.1.4c

Gross area of the inside and outside splice plates:

Inside:

Formula: A subscript gross_in = 2 times t subscript in times b subscript in

Formula: A subscript gross_in = 6 point 00 inches squared

Outside:

Formula: A subscript gross_out = t subscript out times b subscript out

Formula: A subscript gross_out = 6 point 13 inches squared

Check:

Formula: ( 1 minus numerator (A subscript gross_in) divided by denominator (A subscript gross_out) ) times 100 % = 2 point 04 %

The combined areas are within ten percent.

If the areas of the inside and outside splice plates had differed by more than ten percent, the flange design force would be proportioned to the inside and outside splice plates. This is calculated by multiplying the flange design force by the ratio of the area of the splice plate under consideration to the total area of the inner and outer splice plates.

C6.13.6.1.4c

Yielding and Fracture of Splice Plates:

S6.13.6.1.4c

Case 1 - Tension:

S6.13.5.2

At the Strength Limit State, the design force in the splice plates subjected to tension shall not exceed the factored resistances for yielding, fracture, and block shear.

From Table 4-15, the Strength I bottom flange tension design force is:

Formula: P subscript cu = 359 point 25 K

The factored tensile resistance for yielding on the gross section is:

Formula: P subscript r = phi subscript y times P subscript ny

SEquation 6.8.2.1-1

Formula: P subscript r = phi subscript y times F subscript y times A subscript g

Formula: F subscript y = 50 ksi (Design Step 4.1)

Formula: phi subscript y = 0 point 95 (Design Step 4.1)

For yielding of the outside splice plate:

Formula: A subscript g = A subscript gross_out

Formula: P subscript r = phi subscript y times F subscript y times A subscript g

Formula: P subscript r = 290 point 94 K

The outside splice plate takes half of the design load:

Formula: numerator (P subscript cu) divided by denominator (2) = 179 point 63 K

Formula: P subscript r = 290 point 94 K > Formula: numerator (P subscript cu) divided by denominator (2) = 179 point 63 K OK

For yielding of the inside splice plates:

Formula: A subscript g = A subscript gross_in

Formula: P subscript r = phi subscript y times F subscript y times A subscript g

Formula: P subscript r = 285 point 00 K

The inside splice plate takes half of the design load:

Formula: numerator (P subscript cu) divided by denominator (2) = 179 point 63 K

Formula: P subscript r = 285 point 00 K > Formula: numerator (P subscript cu) divided by denominator (2) = 179 point 63 K OK

The factored tensile resistance for fracture on the net section is:

Formula: P subscript r = phi subscript u times P subscript nu

SEquation 6.8.2.1-2

Formula: P subscript r = phi subscript u times F subscript u times A subscript n times U

Formula: F subscript u = 65 ksi (Design Step 4.1)

Formula: phi subscript u = 0 point 80 (Design Step 4.1)

Formula: U = 1 point 0

S6.13.5.2

To compute the net area of the splice plates, assume four 7/8" bolts across the width of the splice plate.

The net width shall be determined for each chain of holes extending across the member along any transverse, diagonal or zigzag line. This is determined by subtracting from the width of the element the sum of the width of all holes in the chain and adding the quantity s2/4g for each space between consecutive holes in the chain. For non-staggered holes, such as in this design example, the minimum net width is the width of the element minus the number of bolt holes in a line straight across the width.

S6.8.3

For fracture of the outside splice plate:

The net width is:

Formula: b subscript n_out = b subscript out minus 4 times d subscript hole

Formula: d subscript hole = 1 point 0 inches (Design Step 4.1)

Formula: b subscript n_out = 10 point 00 inches

The nominal area is determined to be:

Formula: A subscript n_out = b subscript n_out times t subscript out

Formula: A subscript n_out = 4 point 38 inches squared

The net area of the connecting element is limited to 0.85 Ag:

S6.13.5.2

Formula: A subscript n less than or equal to 0 point 85 times A subscript g

Formula: A subscript gross_out = 6 point 13 inches squared

Formula: A subscript n_out = 4 point 38 inches squared < Formula: 0 point 85 times A subscript gross_out = 5 point 21 inches squared OK

Formula: P subscript r = phi subscript u times F subscript u times A subscript n_out times U

Formula: P subscript r = 227 point 50 K

The outside splice plate takes half of the design flange force:

Formula: P subscript r = 227 point 50 K > Formula: numerator (P subscript cu) divided by denominator (2) = 179 point 63 K OK

For fracture of the inside splice plates:

The net width is:

Formula: b subscript n_in = b subscript in minus 2 times d subscript hole

Formula: b subscript n_in = 4 point 00 inches

The nominal area is determined to be:

Formula: A subscript n_in = 2 ( b subscript n_in times t subscript in ches )

Formula: A subscript n_in = 4 point 00 inches squared

The net area of the connecting element is limited to 0.85 Ag:

S6.13.5.2

Formula: A subscript n less than or equal to 0 point 85 times A subscript g

Formula: A subscript gross_in = 6 point 00 inches squared

Formula: A subscript n_in = 4 point 00 inches squared < Formula: 0 point 85 times A subscript gross_in = 5 point 10 inches squared OK

Formula: P subscript r = phi subscript u times F subscript u times A subscript n_in times U

Formula: P subscript r = 208 point 00 K

The inside splice plates take half of the design flange force:

Formula: P subscript r = 208 point 00 K > Formula: numerator (P subscript cu) divided by denominator (2) = 179 point 63 K OK

Case 2 - Compression:

S6.13.6.1.4c

From Table 4-15, the Strength I bottom flange compression design force is:

Formula: P subscript cu = 459 point 38 K

This force is distributed equally to the inside and outside splice plates.

The factored resistance of the splice plate is:

Formula: R subscript r = phi subscript c times F subscript y times A subscript s

SEquation 6.13.6.1.4c-3

Formula: phi subscript c = 0 point 90 (Design Step 4.1)

For yielding of the outside splice plate:

Formula: A subscript s = A subscript gross_out

Formula: R subscript r_out = phi subscript c times F subscript y times A subscript s

Formula: R subscript r_out = 275 point 63 K

Formula: R subscript r_out = 275 point 63 K > Formula: numerator (P subscript cu) divided by denominator (2) = 229 point 69 K OK

For yielding of the inside splice plates:

Formula: A subscript s = A subscript gross_in

Formula: R subscript r_in = phi subscript c times F subscript y times A subscript s

Formula: R subscript r_in = 270 point 00 K

Formula: R subscript r_in = 270 point 00 K > Formula: numerator (P subscript cu) divided by denominator (2) = 229 point 69 K OK

Block Shear:

S6.13.6.1.4c

S6.13.5.2

S6.13.4

All tension connections, including connection plates, splice plates and gusset plates, shall be investigated to ensure that adequate connection material is provided to develop the factored resistance of the connection. Block shear rupture will usually not govern the design of splice plates of typical proportion. However, the block shear checks are carried out here for completeness.

From Table 4-15, the Strength I bottom flange tension design force is:

Formula: P subscript cu = 359 point 25 K

To determine the appropriate block shear equation:

If Formula: A subscript tn greater than or equal to 0 point 58 times A subscript vn then: Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript y times A subscript vg + F subscript u times A subscript tn )

SEquation 6.13.4-1

Otherwise: Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript u times A subscript vn + F subscript y times A subscript tg )

SEquation 6.13.4-2

where, from Design Step 4.1:

Minimum yield strength of the connected material: Formula: F subscript y = 50 ksi

Minimum tensile strength of the connected material: Formula: F subscript u = 65 ksi

Resistance factor for block shear: Formula: phi subscript bs = 0 point 80

Outside Splice Plate:

Failure Mode 1:

A bolt pattern must be assumed prior to checking an assumed block shear failure mode. An initial bolt pattern for the bottom flange splice, along with the first assumed failure mode, is shown in Figure 4-4. The outside splice plate will now be checked for block shear.

This sketch shows the outside splice plate for the bottom flange being checked for block shear failure. The distance from the centerline of the splice to the centerline of the first row of vertical bolts is one point 9375 inches. The plate width of 14 inches and an edge distance of 1 point 5 inches. There are three vertical rows of bolts with a 3 inch spacing. There are four horizontal rows of bolts with a 3 inch spacing between the first and second rows, a 5 inch spacing between the second and third rows and a 3 inch spacing between the third and fourth rows. There are 1 inch diameter bolt holes in the plate.

Figure 4-4 Outside Splice Plate - Failure Mode 1

Applying the factored resistance equations presented previously to the outside splice plate for Failure Mode 1:

Gross area along the plane resisting shear stress:

Formula: A subscript vg = left bracket 2 times ( 3 point 00 inches ) + 1 point 50 inches right bracket times t subscript out

Formula: A subscript vg = 3 point 28 inches squared

Net area along the plane resisting shear stress:

Formula: A subscript vn = left bracket 2 times ( 3 point 00 inches ) + 1 point 50 inches minus 2 point 5 times d subscript hole right bracket times t subscript out

Formula: A subscript vn = 2 point 19 inches squared

Gross area along the plane resisting tension stress:

Formula: A subscript tg = left bracket 2 times ( 3 point 00 inches ) + 5 point 00 inches + 1 point 50 inches right bracket times t subscript out

Formula: A subscript tg = 5 point 47 inches squared

Net area along the plane resisting tension stress:

Formula: A subscript tn = left bracket left bracket 2 times ( 3 point 00 inches ) + 5 point 00 inches + 1 point 50 inches right bracket minus 3 point 5 times d subscript hole right bracket times t subscript out

Formula: A subscript tn = 3 point 94 inches squared

To determine which equation should be applied to calculate the factored resistance:

Formula: A subscript tn = 3 point 94 inches squared > Formula: 0 point 58 times A subscript vn = 1 point 27 inches squared

Therefore, use SEquation 6.13.4-1:

Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript y times A subscript vg + F subscript u times A subscript tn )

Formula: R subscript r = 280 point 88 K

Check:

Formula: R subscript r = 280 point 88 K > Formula: numerator (P subscript cu) divided by denominator (2) = 179 point 63 K OK

Failure Mode 2:

See Figure 4-5 for Failure Mode 2:

This sketch shows the outside splice plate for the bottom flange being checked for block shear failure. The distance from the centerline of the splice to the centerline of the first row of vertical bolts is one point 9375 inches. The plate width of 14 inches and an edge distance of 1 point 5 inches. There are three vertical rows of bolts with a 3 inch spacing. There are four horizontal rows of bolts with a 3 inch spacing between the first and second rows, a 5 inch spacing between the second and third rows and a 3 inch spacing between the third and fourth rows. There are 1 inch diameter bolt holes in the plate. There are two failure planes. The first plane is vertically thru the top two bolts closest to the centerline of the field splice and then horizontal thru the second row of horizontal bolts. The second plane is vertically thru the bottom two bolts closest to the centerline of the field splice and then horizontal thru the third row of horizontal bolts.

Figure 4-5 Outside Splice Plate - Failure Mode 2

Applying the factored resistance equations presented previously to the outside splice plate for Failure Mode 2:

Gross area along the plane resisting shear stress:

Formula: A subscript vg = 2 left bracket 2 times ( 3 point 00 inches ) + 1 point 50 inches right bracket times t subscript out

Formula: A subscript vg = 6 point 56 inches squared

Net area along the plane resisting shear stress:

Formula: A subscript vn = 2 left bracket 2 times ( 3 point 00 inches ) + 1 point 50 inches minus 2 point 5 times d subscript hole right bracket times t subscript out

Formula: A subscript vn = 4 point 38 inches squared

Gross area along the plane resisting tension stress:

Formula: A subscript tg = 2( 3 point 00 inches + 1 point 50 inches ) times t subscript out

Formula: A subscript tg = 3 point 94 inches squared

Net area along the plane resisting tension stress:

Formula: A subscript tn = 2 left bracket ( 3 point 00 inches + 1 point 50 inches ) minus 1 point 5d subscript hole right bracket times t subscript out

Formula: A subscript tn = 2 point 63 inches squared

To determine which equation should be applied to calculate the factored resistance:

Formula: A subscript tn = 2 point 63 inches squared > Formula: 0 point 58 times A subscript vn = 2 point 54 inches squared

Therefore, use SEquation 6.13.4-1:

Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript y times A subscript vg + F subscript u times A subscript tn )

Formula: R subscript r = 288 point 75 K

Check:

Formula: R subscript r = 288 point 75 K > Formula: numerator (P subscript cu) divided by denominator (2) = 179 point 63 K OK

Inside Splice Plates:

The inside splice plates will now be checked for block shear. See Figure 4-6 for the assumed failure mode:

This sketch shows the inside splice plate for the bottom flange being checked for block shear failure. There are two inside cover plates and there is a 1 inch gap between the plates. The distance from the centerline of the splice to the centerline of the first row of vertical bolts is one point 9375 inches. The plates each have a 6 inches width for a total plate width of 13 inches including the 1 inch gap. Both plates have an edge distance of 1 point 5 inches. There are three vertical rows of bolts with a 3 inch spacing in both plates. There are four horizontal rows of bolts, 2 in each plate, with a 3 inch spacing. There are also 1 inch diameter bolt holes in the plate. There are two failure planes. The first plane is vertically thru the top two bolts closest to the centerline of the field splice and then horizontal thru the second row of horizontal bolts. The second plane is vertically thru the bottom two bolts closest to the centerline of the field splice and then horizontal thru the third row of horizontal bolts.

Figure 4-6 Inside Splice Plates - Block Shear Check

Applying the factored resistance equations presented previously to the inside splice plates for the assumed failure mode:

Gross area along the plane resisting shear stress:

Formula: A subscript vg = 2 left bracket 2 times ( 3 point 00 inches ) + 1 point 50 inches right bracket times t subscript in

Formula: A subscript vg = 7 point 50 inches squared

Net area along the plane resisting shear stress:

Formula: A subscript vn = 2 left bracket left bracket 2 times ( 3 point 00 inches ) + 1 point 50 inches right bracket minus 2 point 5 times d subscript hole right bracket times t subscript in

Formula: A subscript vn = 5 point 0 inches squared

Gross area along the plane resisting tension stress:

Formula: A subscript tg = 2( 3 point 00 inches + 1 point 50 inches ) times t subscript in

Formula: A subscript tg = 4 point 50 inches squared

Net area along the plane resisting tension stress:

Formula: A subscript tn = 2 left bracket ( 3 point 00 inches + 1 point 50 inches ) minus 1 point 5d subscript hole right bracket times t subscript in

Formula: A subscript tn = 3 point 00 inches squared

To determine which equation should be applied to calculate the factored resistance:

Formula: A subscript tn = 3 point 0 inches squared > Formula: 0 point 58 times A subscript vn = 2 point 90 inches squared

Therefore, use SEquation 6.13.4-1:

Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript y times A subscript vg + F subscript u times A subscript tn )

Formula: R subscript r = 330 point 00 K

Check:

Formula: R subscript r = 330 point 00 K > Formula: numerator (P subscript cu) divided by denominator (2) = 179 point 63 K OK

Girder Bottom Flange:

The girder bottom flange will now be checked for block shear. See Figure 4-7 for the assumed failure mode:

This sketch shows the outside splice plate for the bottom flange being checked for block shear failure. The distance from the centerline of the splice to the centerline of the first row of vertical bolts is one point 9375 inches. The plate width of 14 inches and an edge distance of 1 point 5 inches. There are three vertical rows of bolts with a 3 inch spacing. There are four horizontal rows of bolts with a 3 inch spacing between the first and second rows, a 5 inch spacing between the second and third rows and a 3 inch spacing between the third and fourth rows. There are 1 inch diameter bolt holes in the plate. There are two failure planes. The first plane is vertically thru the top two bolts farthest from the centerline of the field splice and then horizontally thru the first and second rows of horizontal bolts. The second plane is vertically thru the bottom two bolts farthest from the centerline of the field splice and then horizontally thru the third and fourth rows of horizontal bolts.

Figure 4-7 Bottom Flange - Block Shear Check

Applying the factored resistance equations presented previously to the bottom flange for the assumed failure mode:

Gross area along the plane resisting shear stress:

Formula: A subscript vg = 4 left bracket 2 times ( 3 point 00 inches ) + 1 point 75 inches right bracket times t subscript flbL

Formula: A subscript vg = 27 point 13 inches squared

Net area along the plane resisting shear stress:

Formula: A subscript vn = 4 left bracket left bracket 2 times ( 3 point 00 inches ) + 1 point 75 inches right bracket minus 2 point 5 times d subscript hole right bracket times t subscript flbL

Formula: A subscript vn = 18 point 38 inches squared

Gross area along the plane resisting tension stress:

Formula: A subscript tg = 2( 3 point 00 inches ) times t subscript flbL

Formula: A subscript tg = 5 point 25 inches squared

Net area along the plane resisting tension stress:

Formula: A subscript tn = 2 left bracket ( 3 point 00 inches ) minus 1 point 0d subscript hole right bracket times t subscript flbL

Formula: A subscript tn = 3 point 50 inches squared

To determine which equation should be applied to calculate the factored resistance:

Formula: A subscript tn = 3 point 50 inches squared < Formula: 0 point 58 times A subscript vn = 10 point 66 inches squared

Therefore, use SEquation 6.13.4-2:

Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript u times A subscript vn + F subscript y times A subscript tg )

Formula: R subscript r = 764 point 19 K

Check:

Formula: R subscript r = 764 point 19 K > Formula: P subscript cu = 359 point 25 K OK

It should be noted that although the block shear checks performed in this design example indicate an overdesign, the number of bolts cannot be reduced prior to checking shear on the bolts and bearing at the bolt holes. These checks are performed in what follows.

Flange Bolts - Shear:

Determine the number of bolts for the bottom flange splice plates that are required to develop the Strength I design force in the flange in shear assuming the bolts in the connection have slipped and gone into bearing. A minimum of two rows of bolts should be provided to ensure proper alignment and stability of the girder during construction.

The Strength I flange design force used in this check was previously computed (reference Table 4-15):

Formula: P subscript cu = 459 point 38 K

The factored resistance of an ASTM A325 7/8" diameter high-strength bolt in shear must be determined, assuming the threads are excluded from the shear planes. For this case, the number of bolts required to provide adequate shear strength is determined by assuming the design force acts on two shear planes, known as double shear.

The nominal shear resistance is computed first as follows:

Formula: R subscript n = ( 0 point 48 times A subscript b times F subscript ub times N subscript s )

SEquation 6.13.2.7-1

where:

Area of the bolt corresponding to the nominal diameter: Formula: A subscript b = numerator (pi) divided by denominator (4) times d subscript bolt squared

Formula: A subscript b = 0 point 60 inches squared

Specified minimum tensile strength of the bolt from Design Step 4.1: Formula: F subscript ub = Fu subscript bolt

Formula: F subscript ub = 120 ksi

Number of shear planes per bolt: Formula: N subscript s = 2

Formula: R subscript n = 2 times ( 0 point 48 times A subscript b times F subscript ub )

Formula: R subscript n = 69 point 27 K

The factored shear resistance now follows:

Formula: R subscript u = phi subscript s times R subscript n

Formula: phi subscript s = 0 point 80 (Design Step 4.1)

Formula: R subscript u = 55 point 42 K

When bolts carrying loads pass through fillers 0.25 inches or more in thickness in axially loaded connections, including girder flange splices, either:

S6.13.6.1.5

The fillers shall be extended beyond the gusset or splice material and shall be secured by enough additional bolts to distribute the total stress in the member uniformly over the combined section of the member and the filler.

or

The fillers need not be extended and developed provided that the factored resistance of the bolts in shear at the Strength Limit State, specified in Article 6.13.2.2, is reduced by an appropriate factor:

In this design example, the reduction factor approach will be used. The reduction factor per the Specifications is:

Formula: R = left bracket numerator (( 1 + gamma )) divided by denominator (( 1 + 2 gamma )) right bracket

SEquation 6.13.6.1.5-1

where: Formula: gamma = numerator (A subscript f) divided by denominator (A subscript p)

Sum of the area of the fillers on the top and bottom of the connected plate:

Formula: A subscript f = b subscript fill t subscript fill

Formula: A subscript f = 7 point 00 inches squared

The smaller of either the connected plate area (i.e., girder flange) or the sum of the splice plate areas on the top and bottom of the connected plate determines Ap.

Bottom flange area:

Formula: b subscript flbL = 14 inches

Formula: t subscript flbL = 0 point 875 inches

Formula: A subscript p1 = ( b subscript flbL ) times ( t subscript flbL )

Formula: A subscript p1 = 12 point 25 inches squared

Sum of splice plate areas is equal to the gross areas of the inside and outside splice plates:

Formula: A subscript gross_in = 6 point 00 inches squared Formula: A subscript gross_out = 6 point 13 inches squared

Formula: A subscript p2 = A subscript gross_in + A subscript gross_out

Formula: A subscript p2 = 12 point 13 inches squared

The minimum of the areas is:

Formula: A subscript p = min ( A subscript p1 , A subscript p2 )

Formula: A subscript p = 12 point 13 inches squared

Therefore:

Formula: gamma = numerator (A subscript f) divided by denominator (A subscript p) Formula: gamma = 0 point 58

The reduction factor is determined to be:

Formula: R subscript fill = left bracket numerator (( 1 + gamma )) divided by denominator (( 1 + 2 gamma )) right bracket Formula: R subscript fill = 0 point 73

To determine the total number of bolts required for the bottom flange splice, divide the applied Strength I flange design force by the reduced allowable bolt shear strength:

Formula: R = R subscript u times R subscript fill

Formula: R = 40 point 57 K

The number of bolts required per side is:

Formula: N = numerator (P subscript cu) divided by denominator (R) Formula: N = 11 point 32

The minimum number of bolts required on each side of the splice to resist the maximum Strength I flange design force in shear is twelve.

Flange Bolts - Slip Resistance:

Bolted connections for flange splices shall be designed as slip-critical connections for the Service II flange design force, or the flange design force from constructibility, whichever governs. In this design example, the Service II flange force controls (see previous discussion in Design Step 4.3).

S 6.13.6.1.4c

When checking for slip of the bolted connection for a flange splice with inner and outer splice plates, the slip resistance should always be determined by dividing the flange design force equally to the two slip planes regardless of the ratio of the splice plate areas. Slip of the connection cannot occur unless slip occurs on both planes.

C6.13.6.1.4c

From Table 4-17, the Service II bottom flange design force is:

Formula: P subscript s = 184 point 98 K

The factored resistance for slip-critical connections is:

Formula: R subscript r = R subscript n

SEquation 6.13.2.2-1

Formula: R subscript n = K subscript h K subscript s times N subscript s times P subscript t

SEquation 6.13.2.8-1

Determine the factored resistance per bolt assuming a Class B surface condition for the faying surface, standard holes (which are required per S6.13.6.1.4a) and two slip planes per bolt:

Class B surfaces are unpainted blast-cleaned surfaces and blast-cleaned surfaces with Class B coatings.

S6.13.2.8

Additionally:

Number of slip planes per bolt: Formula: N subscript s = 2

Minimum required bolt tension: Formula: P subscript t = 39 point 0 K

STable 6.13.2.8-1

Hole size factor: Formula: K subscript h = 1 point 0

STable 6.13.2.8-2

Surface condition factor for Class B surface conditions: Formula: K subscript s = 0 point 50

STable 6.13.2.8-3

Formula: R subscript r = K subscript h K subscript s times N subscript s times P subscript t Formula: R subscript r = 39 point 00 K

The minimum number of bolts required to prevent slip is:

Formula: N = numerator (P subscript s) divided by denominator (R subscript r) Formula: N = 4 point 74

Use:

Formula: N = 5 bolts < N = 12 bolts determined previously to satisfy the bolt shear requirements.

Therefore, the number of bolts required for the bottom-flange splice is controlled by the bolt shear requirements. Arrange the bolts in three rows of four bolts per line with no stagger.

MathCad tip logo

Friction Coefficient Selection

Weathering steel can be blasted for a Class B surface. Also, for painted steel, most inorganic zinc (IOZ) primers provide a Class B surface.

Flange Bolts - Minimum Spacing:

S6.13.2.6.1

The minimum spacing between centers of bolts in standard holes shall be no less than three times the diameter of the bolt.

Formula: d subscript bolt = 0 point 875 inches (Design Step 4.1)

Formula: s subscript min = 3 times d subscript bolt

Formula: s subscript min = 2 point 63 inches

For this example, Formula: s = 3 point 00 inches (see Figures 4-4 thru 4-7)

The minimum spacing requirement is satisfied.

Flange Bolts - Maximum Spacing for Sealing:

S6.13.2.6.2

The maximum spacing of the bolts is limited to prevent penetration of moisture in the joints.

For a single line adjacent to a free edge of an outside plate or shape (for example, the bolts along the edges of the plate parallel to the direction of the applied force):

Formula: s less than or equal to ( 4 point 0 + 4 point 0 times t) less than or equal to 7 point 0

where:

Thickness of the thinner outside plate or shape: Formula: t subscript out = 0 point 4375 inches

Maximum spacing for sealing:

Formula: 4 point 0 inches + 4 point 0 times t subscript out = 5 point 75 inches Formula: 5 point 75 inches less than or equal to 7 point 00 inches

Formula: s less than or equal to 5 point 75 inches OK

Next, check for sealing along the free edge at the end of the splice plate. The bolts are not staggered, therefore the applicable equation is:

Formula: s less than or equal to ( 4 point 00 + 4 point 00 times t) less than or equal to 7 point 00

Maximum spacing along the free edge at the end of the splice plate (see Figures 4-4 thru 4-7):

Formula: s subscript end = 5 point 00 inches

Maximum spacing for sealing:

Formula: 4 point 00 inches + 4 point 00 times t subscript out = 5 point 75 inches

Formula: s subscript end less than or equal to 5 point 75 inches OK

Therefore the requirement is satisfied.

Flange Bolts - Maximum Pitch for Stitch Bolts:

S6.13.2.6.3

The maximum pitch requirements are applicable only for mechanically fastened built-up members and will not be applied in this example.

Flange Bolts - Edge Distance:

S6.13.2.6.6

Minimum:

The minimum required edge distance is measured as the distance from the center of any bolt in a standard hole to an edge of the plate.

For a 7/8" diameter bolt measured to a sheared edge, the minimum edge distance is 1 1/2".

STable 6.13.2.6.6-1

Referring to Figures 4-4 thru 4-7, it is clear that the minimum edge distance specified for this example is 1 1/2" and thus satisfies the minimum requirement.

Maximum:

The maximum edge distance shall not be more than eight times the thickness of the thinnest outside plate or five inches.

Formula: 8 times t less than or equal to 5 point 00 inches

where:

Formula: t = t subscript out

Formula: t subscript out = 0 point 4375 inches

The maximum edge distance allowable is:

Formula: 8 times t subscript out = 3 point 50 inches

The maximum distance from the corner bolts to the corner of the splice plate or girder flange is equal to (reference Figure 4-7):

Formula: square root of (( 1 point 50 inches ) squared + ( 1 point 75 inches ) squared ) = 2 point 30 inches

and satisfies the maximum edge distance requirement.

Formula: 2 point 30 inches less than or equal to 3 point 50 inches OK

Flange Bolts - Bearing at Bolt Holes:

S6.13.2.9

Check bearing of the bolts on the connected material under the maximum Strength I Limit State design force. The maximum Strength I bottom flange design force from Table 4-15 is:

Formula: P subscript cu = 459 point 38 K

The design bearing strength of the connected material is calculated as the sum of the bearing strengths of the individual bolt holes parallel to the line of the applied force.

The element of the bottom flange splice that controls the bearing check in this design example is the outer splice plate.

To determine the applicable equation for the calculation of the nominal resistance, the clear distance between holes and the clear end distance must be calculated and compared to the value of two times the nominal diameter of the bolt. This check yields:

Formula: d subscript bolt = 0 point 875 inches (Design Step 4.1)

Formula: 2 times d subscript bolt = 1 point 75 inches

For the bolts adjacent to the end of the splice plate, the edge distance is 1 1/2". Therefore, the clear end distance between the edge of the hole and the end of the splice plate:

Formula: d subscript hole = 1 point 0 inches (Design Step 4.1)

Formula: L subscript c subscript 1 = 1 point 50 inches minus numerator (d subscript hole) divided by denominator (2)

Formula: L subscript c subscript 1 = 1 point 00 inches

The center-to-center distance between bolts in the direction of the force is three inches. Therefore, the clear distance between edges of adjacent holes is computed as:

Formula: L subscript c subscript 2 = 3 point 00 inches minus d subscript hole

Formula: L subscript c subscript 2 = 2 point 00 inches

For standard holes, where either the clear distance between holes or the clear end distance is less than twice the bolt diameter:

Formula: R subscript n = 1 point 2 times L subscript c times t times F subscript u

SEquation 6.13.2.9-2

For the outside splice plate:

Thickness of the connected material: Formula: t subscript out = 0 point 4375 inches

Tensile strength of the connected material (Design Step 4.1): Formula: F subscript u = 65 ksi

The nominal resistance for the end row of bolt holes is computed as follows:

Formula: R subscript n subscript 1 = 4 times ( 1 point 2 times L subscript c subscript 1 times t subscript out times F subscript u )

Formula: R subscript n subscript 1 = 136 point 50 K

The nominal resistance for the remaining bolt holes is computed as follows:

Formula: R subscript n subscript 2 = 8 times ( 1 point 2 times L subscript c subscript 2 times t subscript out times F subscript u )

Formula: R subscript n subscript 2 = 546 point 00 K

The total nominal resistance of the bolt holes is:

Formula: R subscript n = R subscript n subscript 1 + R subscript n subscript 2

Formula: R subscript n = 682 point 50 K

Formula: phi subscript bb = 0 point 80 (Design Step 4.1)

Formula: R subscript r = phi subscript bb times R subscript n

Formula: R subscript r = 546 point 00 K

Check:

Formula: numerator (P subscript cu) divided by denominator (2) = 229 point 69 K < Formula: R subscript r = 546 point 00 K OK

Fatigue of Splice Plates:

S6.6.1

Check the fatigue stresses in the base metal of the bottom flange splice plates adjacent to the slip-critical connections. Fatigue normally does not govern the design of the splice plates, and therefore, an explicit check is not specified. However, a fatigue check of the splice plates is recommended whenever the combined area of the inside and outside flange splice plates is less than the area of the smaller flange at the splice.

From Table 4-18, the factored fatigue stress range at the midthickness of the bottom flange is:

Formula: Delta f subscript fact = 4 point 63 ksi

For load-induced fatigue considerations, each detail shall satisfy:

Formula: gamma times ( Delta f ) less than or equal to ( Delta F ) subscript n

SEquation 6.6.1.2.2-1

where:

Load factor for the fatigue load combination: Formula: gamma = 0 point 75

Force effect, live load stress range due to the passage of the fatigue load:

Formula: gamma ( Delta f ) = Delta f subscript fact

Nominal fatigue resistance:

( Delta F ) subscript n 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) times Delta F subscript TH

SEquation 6.6.1.2.5-1

The fatigue detail category under the condition of Mechanically Fastened Connections for checking the base metal at the gross section of high-strength bolted slip-resistant connections is Category B.

STable 6.6.1.2.3-1

The parameters required for the determination of the nominal fatigue resistance are as follows: Formula: N = ( 365) times ( 75) times n times ( ADTT) subscript SL

SEquation 6.6.1.2.5-2

For Fatigue Category B: Formula: A = 120 times 10 superscript 8

STable 6.6.1.2.5-1

For a span length greater than 40.0 feet and at a location near the interior support, the number of stress range cycles per truck passage: Formula: n = 1 point 5

STable 6.6.1.2.5-2

Single-lane ADTT, from Design Step 3.1: Formula: ADTT subscript SL = 3000

Constant-amplitude fatigue threshold: Formula: Delta F subscript TH = 16 ksi

STable 6.6.1.2.5-3

Therefore:

Formula: N = 123187500

Determine the nominal fatigue resistance:

Condition 1:

Formula: Delta F subscript n = ( numerator (A) divided by denominator (N) ) superscript numerator (1) divided by denominator (3) Formula: Delta F subscript n = 4 point 60ksi

Condition 2:

Formula: Delta F subscript n = numerator (1) divided by denominator (2) times Delta F subscript TH Formula: Delta F subscript n = 8 point 00 ksi (governs)

Check that the following is satisfied:

Formula: Delta f subscript fact less than or equal to ( Delta F ) subscript n

Formula: Delta f subscript fact = 4 point 63 ksi < Formula: Delta F subscript n = 8 point 00 ksi OK

Control of Permanent Deflection - Splice Plates:

S6.10.5.2

A check of the flexural stresses in the splice plates at the Service II Limit State is not explicitly specified in the specifications. However, whenever the combined area of the inside and outside flange splice plates is less than the area of the smaller flange at the splice (which is the case for the bottom flange splice in this example), such a check is recommended.

The maximum Service II flange force in the bottom flange is taken from Table 4-17:

Formula: P subscript s = 184 point 98 K

The following criteria will be used to make this check. The equation presented is for both steel flanges of composite section:

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

SEquation 6.10.5.2-1

where:

Elastic flange stress caused by the factored loading: f subscript f

Specified minimum yield strength of the flange (Design Step 4.1): Formula: F subscript yf = 50 ksi

The flange force is equally distributed to the inner and outer splice plates due to the areas of the flanges being within 10 percent of each other:

Formula: P = numerator (P subscript s) divided by denominator (2) Formula: P = 92 point 49 K

The resulting stress in the outside splice plate is:

Formula: A subscript gross_out = 6 point 13 inches squared

Formula: f_out = numerator (P) divided by denominator (A subscript gross_out) Formula: f_out = 15 point 10 ksi

Formula: f_out = 15 point 10 ksi < Formula: 0 point 95 times F subscript yf = 47 point 50 ksi OK

The resulting stress in the inside splice plates is:

Formula: A subscript gross_in = 6 point 00 inches squared

Formula: f_in = numerator (P) divided by denominator (A subscript gross_in) Formula: f_in = 15 point 42 ksi

Formula: f_in = 15 point 42 ksi < Formula: 0 point 95 times F subscript yf = 47 point 50 ksi OK

Design Step 4.5 - Design Top Flange Splice

The design of the top flange splice is not included in this design example (for the sake of simplicity and brevity). However, the top flange splice is designed using the same procedures and methods presented in this design example for the bottom flange splice.

Design Step 4.6 - Compute Web Splice Design Loads

S6.13.6.1.4b

Web splice plates and their connections shall be designed for shear, the moment due to the eccentricity of the shear at the point of splice, and the portion of the flexural moment assumed to be resisted by the web at the point of the splice.

Girder Shear Forces at the Splice Location:

Based on the girder properties defined in Design Step 3 (Steel Girder Design), any number of commercially available software programs can be used to obtain the design dead and live loads at the splice. For this design example, the AASHTO Opis software was used. A summary of the unfactored shears at the splice from the initial trial of the girder design are listed below. The live loads include impact and distribution factors.

Loads Shears
Dead Loads: Noncomposite: Formula: V subscript NDL = minus 60 point 8 K
Composite: Formula: V subscript CDL = minus 8 point 7 K
Future Wearing Surface: Formula: V subscript FWS = minus 10 point 6 K
Live Loads: HL-93 Positive: Formula: V subscript PLL = 14 point 5 K
HL-93 Negative: Formula: V subscript NLL = minus 91 point 1 K
Fatigue Positive: Formula: V subscript PFLL = 5 point 0 K
Fatigue Negative: Formula: V subscript NFLL = minus 33 point 4 K

Web Moments and Horizontal Force Resultant:

C6.13.6.1.4b

Because the portion of the flexural moment assumed to be resisted by the web is to be applied at the mid-depth of the web, a horizontal design force resultant, Huw, must also be applied at the mid-depth of the web to maintain equilibrium. The web moment and horizontal force resultant are applied together to yield a combined stress distribution equivalent to the unsymmetrical stress distribution in the web. For sections with equal compressive and tensile stresses at the top and bottom of the web (i.e., with the neutral axis located at the mid-depth of the web), Huw will equal zero.

In the computation of the portion of the flexural moment assumed to be resisted by the web, Muw, and the horizontal design force resultant, Huw, in the web, the flange stresses at the midthickness of the flanges are conservatively used. This allows use of the same stress values for both the flange and web splices, which simplifies the calculations. It is important to note that the flange stresses are taken as signed quantities in determining Muw and Huw (positive for tension; negative for compression).

C6.13.6.1.4b

The moment, Muv, due to the eccentricity of the design shear, Vuw, is resisted solely by the web and always acts about the mid-depth of the web (i.e., horizontal force resultant is zero). This moment is computed as:

Formula: M subscript uv = V subscript uw times e

where e is defined as the distance from the centerline of the splice to the centroid of the connection on the side of the joint under consideration. For this design example:

S6.13.6.1.4b

Formula: e = 1 point 9375 inches + numerator (3 point 00 inches ) divided by denominator (2) (Reference Figure 4-8)

Formula: e = 3 point 44 inches

The total web moment for each load case is computed as follows:

Formula: M subscript total = M subscript uw + M subscript uv

In general, and in this example, the web splice is designed under the conservative assumption that the maximum moment and shear at the splice will occur under the same loading condition.

Strength I Limit State:

Design Shear:

S6.13.6.1.4b

For the Strength I Limit State, the girder web factored shear resistance is required when determining the design shear. Assume an unstiffened web at the splice location.

S6.10.7.2

Formula: phi subscript v = 1 point 00 (Design Step 4.1)

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

SEquation 6.10.7.1-1

Formula: V subscript n = C times V subscript p

SEquation 6.10.7.2-1

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

SEquation 6.10.7.2-2

where:

Ratio of shear buckling stress to the shear yield strength, C, is dependent upon the ratio of D/tw in comparison to:

S6.10.7.3.3a

1 point 10 times square root of ( numerator (E k) divided by denominator (F subscript yw)) and 1 point 38 times square root of ( numerator (E k) divided by denominator (F subscript yw))

And:

Formula: k = 5 point 0

S6.10.7.2

Modulus of Elasticity: Formula: E = 29000 ksi

Specified minimum yield strength of the web (Design Step 4.1): Formula: F subscript yw = F subscript y

Formula: F subscript yw = 50 ksi

From Figure 4-1:

Web Depth: Formula: D = 54 inches

Thickness of the web: Formula: t subscript w = 0 point 50 inches

Compare:

Formula: numerator (D) divided by denominator (t subscript w) = 108 point 00

to the values for:

Formula: 1 point 10 times square root of ( numerator (E k) divided by denominator (F subscript yw)) = 59 point 24 and Formula: 1 point 38 times square root of ( numerator (E k) divided by denominator (F subscript yw)) = 74 point 32

Based on the computed value of D/tw, use the following equation to determine C:

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) )

SEquation 6.10.7.3.3a-7

Formula: C = 0 point 38

The nominal shear resistance is computed as follows:

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

Formula: V subscript p = 783 point 00 K

Formula: V subscript n = C times V subscript p

Formula: V subscript n = 295 point 91 K

The factored shear resistance now follows:

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

Formula: V subscript r = 295 point 91 K

At the strength limit state, the design shear, Vuw, shall be taken as:

If Vu < 0.5 Vr, then:

SEquation 6.13.6.1.4b-1

Formula: V subscript uw = 1 point 5 times V subscript u

Otherwise:

SEquation 6.13.6.1.4b-2

Formula: V subscript uw = numerator (V subscript u + V subscript r) divided by denominator (2)

The shear due to the Strength I loading at the point of splice, Vu, is computed from the girder shear forces at the splice location listed at the beginning of this design step.

For the Strength I Limit State, the factored shear for the positive live load is:

Formula: V subscript upos = 0 point 90 times ( V subscript NDL + V subscript CDL ) + 1 point 75 times V subscript PLL

Formula: V subscript upos = minus 37 point 17 K

For the Strength I Limit State, the factored shear for the negative live load is:

Formula: V subscript uneg = 1 point 25 times ( V subscript NDL + V subscript CDL ) + 1 point 50 times V subscript FWS + 1 point 75 times V subscript NLL

Formula: V subscript uneg = minus 262 point 20 K (controls)

Therefore:

Formula: V subscript u = Vertical Bar V subscript uneg Vertical Bar

Since Vu exceeds one-half of Vr:

Formula: V subscript uw = numerator (V subscript u + V subscript r) divided by denominator (2)

SEquation 6.13.6.1.4b-2

Formula: V subscript uw = 279 point 05 K

Web Moments and Horizontal Force Resultants:

Case 1 - Dead Load + Positive Live Load:

For the loading condition with positive live load, the controlling flange was previously determined to be the bottom flange. The maximum elastic flexural stress due to the factored loads at the midthickness of the controlling flange, fcf, and the design stress for the controlling flange, Fcf,were previously computed for this loading condition. From Table 4-15:

Formula: f subscript cf = 23 point 48 ksi

Formula: F subscript cf = 37 point 50 ksi

For the same loading condition, the concurrent flexural stress at the midthickness of the noncontrolling (top) flange, fncu, was previously computed. From Table 4-16:

Formula: f subscript ncf = minus 0 point 93 ksi

Therefore, the portion of the flexural moment assumed to be resisted by the web is computed as:

Formula: M subscript w = numerator (t subscript w times D squared ) divided by denominator (12) times Vertical Bar R subscript h times F subscript cf minus R subscript cf times f subscript ncf Vertical Bar

CEquation 6.13.6.1.4b-1

where:

The hybrid girder reduction factor: Formula: R subscript h = 1 point 00

The ratio Rcf is computed as follows:

Formula: R subscript cf = Vertical Bar numerator (F subscript cf) divided by denominator (f subscript cf) Vertical Bar Formula: R subscript cf = 1 point 60

Web thickness: Formula: t subscript w = 0 point 50 inches

Web depth: Formula: D = 54 inches

Compute the portion of the flexural moment to be resisted by the web:

Formula: M subscript w_str_pos = numerator (t subscript w times D squared ) divided by denominator (12) times Vertical Bar R subscript h times F subscript cf minus R subscript cf times f subscript ncf Vertical Bar times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript w_str_pos = 394 point 73 K feet

The total web moment is:

Formula: V subscript uw = 279 point 05 K Formula: e = 3 point 44 inches

Formula: M subscript tot_str_pos = M subscript w_str_pos + ( V subscript uw times e ) times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript tot_str_pos = 474 point 66 K feet

Compute the horizontal force resultant (the variables included in this equation are as defined for Mw_str_pos):

Formula: H subscript w_str_pos = numerator (t subscript w times D) divided by denominator (2) times ( R subscript h times F subscript cf + R subscript cf times f subscript ncf )

CEquation 6.13.6.1.4b-2

Formula: H subscript w_str_pos = 486 point 20 K

The above value is a signed quantity, positive for tension and negative for compression.

Case 2 - Dead Load + Negative Live Load:

Similarly, for the loading condition with negative live load, the controlling flange was determined to be the bottom flange. For this case the stresses were previously computed. From Table 4-15:

Formula: f subscript cf = minus 19 point 54 ksi

Formula: F subscript cf = minus 37 point 50 ksi

For the noncontrolling (top) flange, the flexural stress at the midthickness of the flange, from Table 4-16:

Formula: f subscript ncf = 14 point 13 ksi

The ratio, Rcf, is computed as follows:

Formula: R subscript cf = Vertical Bar numerator (F subscript cf) divided by denominator (f subscript cf) Vertical Bar Formula: R subscript cf = 1 point 92

Therefore:

Formula: M subscript w_str_neg = numerator (t subscript w times D squared ) divided by denominator (12) times Vertical Bar R subscript h times F subscript cf minus R subscript cf times f subscript ncf Vertical Bar times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript w_str_neg = 654 point 25 K feet

The total web moment is:

Formula: V subscript uw = 279 point 05 K Formula: e = 3 point 44 inches

Formula: M subscript tot_str_neg = M subscript w_str_neg + ( V subscript uw times e ) times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript tot_str_neg = 734 point 19 K feet

Compute the horizontal force resultant:

Formula: H subscript w_str_neg = numerator (t subscript w times D) divided by denominator (2) times ( R subscript h times F subscript cf + R subscript cf times f subscript ncf )

Formula: H subscript w_str_neg = minus 140 point 16 K

The above value is a signed quantity, positive for tension, and negative for compression.

Service II Limit State:

Design Shear:

As a minimum, for checking slip of the web splice bolts, the design shear shall be taken as the shear at the point of splice under the Service II Limit State, or the shear from constructibility, whichever governs. In this design example, the Service II shear controls (see previous discussion in Design Step 4.3).

S6.13.6.1.4b

The elastic shears due to the unfactored loads at the point of the splice are listed at the beginning of this design step.

For the Service II Limit State, the factored shear for the positive live load is (ignore future wearing surface):

Formula: V subscript ser_pos = 1 point 00 times V subscript NDL + 1 point 00 times V subscript CDL + 1 point 30 times V subscript PLL

Formula: V subscript ser_pos = minus 50 point 65 K

For the Service II Limit State, the factored shear for the negative live load is (include future wearing surface):

Formula: V subscript ser_neg = 1 point 00 times V subscript NDL + 1 point 00 times V subscript CDL + 1 point 00 times V subscript FWS + 1 point 30 times V subscript NLL

Formula: V subscript ser_neg = minus 198 point 53 K (governs)

Therefore:

Formula: V subscript w_ser = Vertical Bar V subscript ser_neg Vertical Bar

Web Moments and Horizontal Force Resultants:

The web design moment and horizontal force resultant are computed using CEquation 6.13.6.1.4b-1 and CEquation 6.13.6.1.4b-2, modified for the Service II Limit State as follows:

C6.13.6.1.4b

Formula: M subscript w_ser = numerator (t subscript w times D squared ) divided by denominator (12) times Vertical Bar f subscript s minus f subscript os Vertical Bar Formula: H subscript w_ser = numerator (t subscript w times D) divided by denominator (2) times ( f subscript s + f subscript os )

In the above equations, fs is the maximum Service II midthickness flange stress for the load case considered (i.e., positive or negative live load). The Service II midthickness flange stress in the other flange, concurrent with fs, is termed fos.

Case 1 - Dead Load + Positive Live Load:

The maximum midthickness flange flexural stress for the load case with positive live load moment for the Service II Limit State occurs in the bottom flange. From Table 4-13:

Formula: f subscript s_bot_pos = 15 point 10 ksi

Formula: f subscript os_top_pos = minus 0 point 65 ksi

Therefore, for the load case of positive live load:

Formula: M subscript w_ser_pos = numerator (t subscript w times D squared ) divided by denominator (12) times Vertical Bar f subscript s_bot_pos minus f subscript os_top_pos Vertical Bar times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript w_ser_pos = 159 point 47 K feet

The total web moment is:

Formula: V subscript w_ser = 198 point 53 K Formula: e = 3 point 44 inches

Formula: M subscript tot_ser_pos = M subscript w_ser_pos + ( V subscript w_ser times e ) times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript tot_ser_pos = 216 point 34 K feet

Compute the horizontal force resultant:

Formula: H subscript w_ser_pos = numerator (t subscript w times D) divided by denominator (2) times ( f subscript s_bot_pos + f subscript os_top_pos )

Formula: H subscript w_ser_pos = 195 point 08 K

The above value is a signed quantity, positive for tension, and negative for compression.

Case 2 - Dead Load + Negative Live Load:

The maximum midthickness flange flexural stress for the load case with negative live load moment for the Service II Limit State occurs in the bottom flange. From Table 4-13:

Formula: f subscript s_bot_neg = minus 11 point 85 ksi

Formula: f subscript os_top_neg = 1 point 80 ksi

Therefore:

Formula: M subscript w_ser_neg = numerator (t subscript w times D squared ) divided by denominator (12) times Vertical Bar f subscript s_bot_neg minus f subscript os_top_neg Vertical Bar times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript w_ser_neg = 138 point 21 K feet

The total web moment is:

Formula: V subscript w_ser = 198 point 53 K Formula: e = 3 point 44 inches

Formula: M subscript tot_ser_neg = M subscript w_ser_neg + ( V subscript w_ser times e ) times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript tot_ser_neg = 195 point 08 K feet

Compute the horizontal force resultant:

Formula: H subscript w_ser_neg = numerator (t subscript w times D) divided by denominator (2) times ( f subscript s_bot_neg + f subscript os_top_neg )

Formula: H subscript w_ser_neg = minus 135 point 68 K

The above value is a signed quantity, positive for tension, and negative for compression.

Fatigue Limit State:

Fatigue of the base metal adjacent to the slip-critical connections in the splice plates may be checked as specified in STable 6.6.1.2.3-1 using the gross section of the splice plates and member. However, the areas of the web splice plates will often equal or exceed the area of the web to which it is attached (the case in this design example). Therefore, fatigue will generally not govern the design of the splice plates, but is carried out in this example for completeness.

C6.13.6.1.4a

Design Shear:

For the Fatigue Limit State, the factored shear for the positive live load is:

Formula: V subscript fat_pos = 0 point 75 times V subscript PFLL

Formula: V subscript fat_pos = 3 point 75 K

For the Fatigue Limit State, the factored shear for the negative live load is:

Formula: V subscript fat_neg = 0 point 75 times V subscript NFLL

Formula: V subscript fat_neg = minus 25 point 05 K

Web Moments and Horizontal Force Resultants:

The portion of the flexural moment to be resisted by the web and the horizontal force resultant are computed from equations similar to CEquations 6.13.6.1.4b-1 and 6.13.6.1.4b-2, respectively, with appropriate substitutions of the stresses in the web caused by the fatigue-load moment for the flange stresses in the equations. Also, the absolute value signs are removed to keep track of the signs. This yields the following equations:

Formula: M subscript w = numerator (t subscript w times D squared ) divided by denominator (12) times ( f subscript botweb minus f subscript topweb )

Formula: H subscript w = numerator (t subscript w times D) divided by denominator (2) times ( f subscript botweb + f subscript topweb )

Case 1 - Positive Live Load:

The factored stresses due to the positive live load moment for the Fatigue Limit State at the top and bottom of the web, from Table 4-14, are:

Formula: f subscript topweb_pos = minus 0 point 22 ksi

Formula: f subscript botweb_pos = 2 point 67 ksi

Therefore:

Formula: M subscript w_fat_pos = numerator (t subscript w times D squared ) divided by denominator (12) times ( f subscript botweb_pos minus f subscript topweb_pos ) times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript w_fat_pos = 29 point 26 K feet

The total web moment is:

Formula: V subscript fat_pos = 3 point 75 K Formula: e = 3 point 44 inches

Formula: M subscript tot_fat_pos = M subscript w_fat_pos + ( V subscript fat_pos times e ) times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript tot_fat_pos = 30 point 34 K feet

Compute the horizontal force resultant:

Formula: H subscript w_fat_pos = numerator (t subscript w times D) divided by denominator (2) times ( f subscript botweb_pos + f subscript topweb_pos )

Formula: H subscript w_fat_pos = 33 point 08 K

The above value is a signed quantity, positive for tension, and negative for compression.

Case 2 - Negative Live Load:

The factored stresses due to the negative live load moment for the Fatigue Limit State at the top and bottom of the web, from Table 4-14, are:

Formula: f subscript botweb_neg = minus 1 point 92 ksi

Formula: f subscript topweb_neg = 0 point 16 ksi

Therefore:

Formula: M subscript w_fat_neg = numerator (t subscript w times D squared ) divided by denominator (12) times ( f subscript botweb_neg minus f subscript topweb_neg ) times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript w_fat_neg = minus 21 point 06 K feet

The total web moment is:

Formula: V subscript fat_neg = minus 25 point 05 K Formula: e = 3 point 44 inches

Formula: M subscript tot_fat_neg = M subscript w_fat_neg + ( V subscript fat_neg times e ) times ( numerator (1) divided by denominator (12 inches per foot) )

Formula: M subscript tot_fat_neg = minus 28 point 24 K feet

Compute the horizontal force resultant:

Formula: H subscript w_fat_neg = numerator (t subscript w times D) divided by denominator (2) times ( f subscript botweb_neg + f subscript topweb_neg )

Formula: H subscript w_fat_neg = minus 23 point 76 K

The above value is a signed quantity, positive for tension, and negative for compression.

Design Step 4.7 - Design Web Splice

Web Splice Configuration:

Two vertical rows of bolts with sixteen bolts per row will be investigated. The typical bolt spacings, both horizontally and vertically, are as shown in Figure 4-8. The outermost rows of bolts are located 4 1/2" from the flanges to provide clearance for assembly (see the AISC Manual of Steel Construction for required bolt assembly clearances). The web is spliced symmetrically by plates on each side with a thickness not less than one-half the thickness of the web. Assume 5/16" x 48" splice plates on each side of the web. No web fill plate is necessary for this example.

This figure shows an elevation view of the splice. The gap between the sections of the girders is three-eights of an inch. The vertical and horizontal edge distances are both 1 point 50 inches. The distance from the centerline of the splice to the centerline of the first row of vertical bolts is one point 9375 inches. The horizontal row spacing and the vertical row spacing are both 3 inches. The distance from the bottom of the top flange to the edge of the splice plate is 3 inches.

Figure 4-8 Web Splice

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Web Splice Design

It is recommended to extend the web splice plates as near as practical to the full depth of the web between flanges without impinging on bolt assembly clearances. Also, two vertical rows of bolts in the web on each side of the splice is considered a standard minimum. This may result in an overdesigned web splice, but is considered good engineering practice.

Web Bolts - Minimum Spacing:

S6.13.2.6.1

This check is only dependent upon the bolt diameter, and is therefore satisfied for a three inch spacing per the check for the flange bolts from Design Step 4.4.

Web Bolts - Maximum Spacing for Sealing:

S6.13.2.6.2

The maximum spacing of the bolts is limited to prevent penetration of moisture in the joints.

For a single line adjacent to a free edge of an outside plate or shape (for example, the bolts along the edges of the plate parallel to the direction of the applied force):

Formula: s less than or equal to ( 4 point 0 + 4 point 0 times t) less than or equal to 7 point 0

where:

Thickness of the thinner outside plate or shape, in this case the web plate: Formula: t subscript wp = 0 point 3125 inches

Maximum spacing for sealing:

Formula: 4 point 0 inches + 4 point 0 times t subscript wp = 5 point 25 inches Formula: 5 point 25 inches less than or equal to 7 point 00 inches

Formula: 3 point 0 inches less than or equal to 5 point 25 inches OK

S6.13.2.6.3

Web Bolts - Maximum Pitch for Stitch Bolts:

The maximum pitch requirements are applicable only for mechanically fastened built-up members and will not be applied in this example.

Web Bolts - Edge Distance:

S6.13.2.6.6

Minimum:

The minimum required edge distance is measured as the distance from the center of any bolt in a standard hole to an edge of the plate.

For a 7/8" diameter bolt measured to a sheared edge, the minimum edge distance is 1 1/2".

STable 6.13.2.6.6-1

Referring to Figure 4-8, it is clear that the minimum edge distance specified for this example is 1 1/2" and thus satisfies the minimum requirement.

Maximum:

The maximum edge distance shall not be more than eight times the thickness of the thinnest outside plate or five inches.

Formula: 8 times t less than or equal to 5 point 00 inches

where:

Formula: t = t subscript wp

Formula: t subscript wp = 0 point 3125 inches

The maximum edge distance allowable is:

Formula: 8 times t subscript wp = 2 point 50 inches

The maximum distance from the corner bolts to the corner of the splice plate or girder flange is equal to (reference Figure 4-8):

Formula: square root of (( 1 point 50 inches ) squared + ( 1 point 50 inches ) squared ) = 2 point 12 inches

and satisfies the maximum edge distance requirement.

Formula: 2 point 12 inches less than or equal to 2 point 50 inches OK

Web Bolts - Shear:

Calculate the polar moment of inertia, Ip, of the bolt group on each side of the centerline with respect to the centroid of the connection. This is required for determination of the shear force in a given bolt due to the applied web moments.

Formula: I subscript p = numerator (n times m) divided by denominator (12) left bracket s squared times ( n squared minus 1 ) + g squared times ( m squared minus 1 ) right bracket

CEquation 6.13.6.1.4b-3

where:

Number of vertical rows of bolts: Formula: m = 2

Number of bolts in one vertical row: Formula: n = 16

Vertical pitch: Formula: s = 3 point 00 inches

Horizontal pitch: Formula: g = 3 point 00 inches

The polar moment of inertia is:

Formula: I subscript p = numerator (n times m) divided by denominator (12) left bracket s squared times ( n squared minus 1 ) + g squared times ( m squared minus 1 ) right bracket

Formula: I subscript p = 6192 point 00 inches squared

The total number of web bolts on each side of the splice, assuming two vertical rows per side with sixteen bolts per row, is:

Formula: N subscript b = 32

Strength I Limit State:

Under the most critical combination of the minimum design shear, moment and horizontal force, it is assumed that the bolts in the web splice have slipped and gone into bearing. The shear strength of an ASTM A325 7/8" diameter high-strength bolt in double shear, assuming the threads are excluded from the shear planes, was computed in Design Step 4.4 for Flange Bolts - Shear:

Formula: R subscript u = 55 point 42 K

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Threads in the Shear Plane

Since the bolt shear strength for both the flange and web splices is based on the assumption that the threads are excluded from the shear planes, an appropriate note should be placed on the drawings to ensure that the splice is detailed to exclude the bolt threads from the shear planes.

Case 1 - Dead Load + Positive Live Load:

The following forces were computed in Design Step 4.6:

Formula: V subscript uw = 279 point 05 K

Formula: M subscript tot_str_pos = 474 point 66 K feet

Formula: H subscript w_str_pos = 486 point 20 K

The vertical shear force in the bolts due to the applied shear force:

Formula: P subscript v_str = numerator (V subscript uw) divided by denominator (N subscript b)

Formula: P subscript v_str = 8 point 72 K

The horizontal shear force in the bolts due to the horizontal force resultant:

Formula: P subscript H_str_pos = numerator (H subscript w_str_pos) divided by denominator (N subscript b)

Formula: P subscript H_str_pos = 15 point 19 K

Determine the horizontal and vertical components of the bolt shear force on the extreme bolt due to the total moment in the web:

Formula: P subscript Mv = numerator (M subscript total times x) divided by denominator (I subscript p)

and

Formula: P subscript Mh = numerator (M subscript total y) divided by denominator (I subscript p)

For the vertical component:

Formula: x = numerator (g) divided by denominator (2) Formula: x = 1 point 50 inches

For the horizontal component:

Formula: y = numerator (15 times s) divided by denominator (2) Formula: y = 22 point 50 inches

Calculating the components:

Formula: P subscript Mv_str_pos = numerator (M subscript tot_str_pos times ( x)) divided by denominator (I subscript p) times ( 12 inches per foot )

Formula: P subscript Mv_str_pos = 1 point 38 K

Formula: P subscript Mh_str_pos = numerator (M subscript tot_str_pos ( y)) divided by denominator (I subscript p) times ( 12 inches per foot )

Formula: P subscript Mh_str_pos = 20 point 70 K

The resultant bolt force for the extreme bolt is:

P subscript r_str_pos = square root of ( ( P subscript v_str + P subscript Mv_str_pos ) squared + ( P subscript H_str_pos + P subscript Mh_str_pos ) squared )

Formula: P subscript r_str_pos = 37 point 29 K

Case 2 - Dead Load + Negative Live Load:

The following forces were computed in Design Step 4.6:

Formula: V subscript uw = 279 point 05 K

Formula: M subscript tot_str_neg = 734 point 19 K feet

Formula: H subscript w_str_neg = minus 140 point 16 K

The vertical shear force in the bolts due to the applied shear force:

Formula: P subscript v_str = numerator (V subscript uw) divided by denominator (N subscript b)

Formula: P subscript v_str = 8 point 72 K

The horizontal shear force in the bolts due to the horizontal force resultant:

Formula: P subscript H_str_neg = numerator (Vertical Bar H subscript w_str_neg Vertical Bar) divided by denominator (N subscript b)

Formula: P subscript H_str_neg = 4 point 38 K

Determine the horizontal and vertical components of the bolt shear force on the extreme bolt due to the total moment in the web:

Calculating the components:

Formula: P subscript Mv_str_neg = numerator (M subscript tot_str_neg times ( x)) divided by denominator (I subscript p) times ( 12 inches per foot )

Formula: P subscript Mv_str_neg = 2 point 13 K

Formula: P subscript Mh_str_neg = numerator (M subscript tot_str_neg ( y)) divided by denominator (I subscript p) times ( 12 inches per foot )

Formula: P subscript Mh_str_neg = 32 point 01 K

The resultant bolt force is:

P subscript r_str_neg = square root of ( ( P subscript v_str + P subscript Mv_str_neg ) squared + ( P subscript H_str_neg + P subscript Mh_str_neg ) squared )

Formula: P subscript r_str_neg = 37 point 98 K

The governing resultant bolt force is:

Formula: P subscript r_str = max ( P subscript r_str_pos , P subscript r_str_neg )

Formula: P subscript r_str = 37 point 98 K

Check:

Formula: P subscript r_str = 37 point 98 K < Formula: R subscript u = 55 point 42 K OK

Service II Limit State:

The factored slip resistance, Rr, for a 7/8" diameter high-strength bolt in double shear for a Class B surface and standard holes was determined from Design Step 4.4 to be:

Formula: R subscript r = 39 point 00 K

Case 1 - Dead Load + Positive Live Load:

The following forces were computed in Design Step 4.6:

Formula: V subscript w_ser = 198 point 53 K

Formula: M subscript tot_ser_pos = 216 point 34 K feet

Formula: H subscript w_ser_pos = 195 point 08 K

The vertical shear force in the bolts due to the applied shear force:

Formula: P subscript s_ser = numerator (V subscript w_ser) divided by denominator (N subscript b)

Formula: P subscript s_ser = 6 point 20 K

The horizontal shear force in the bolts due to the horizontal force resultant:

Formula: P subscript H_ser_pos = numerator (H subscript w_ser pos) divided by denominator (N subscript b minus )

Formula: P subscript H_ser_pos = 6 point 10 K

Determine the horizontal and vertical components of the bolt shear force on the extreme bolt due to the total moment in the web:

For the vertical component:

Formula: x = 1 point 50 inches

Formula: P subscript Mv_ser_pos = numerator (M subscript tot_ser_pos times ( x)) divided by denominator (I subscript p) times ( 12 inches per foot )

Formula: P subscript Mv_ser_pos = 0 point 63 K

For the horizontal component:

Formula: y = 22 point 50 inches

Formula: P subscript Mh_ser_pos = numerator (M subscript tot_ser_pos ( y)) divided by denominator (I subscript p) times ( 12 inches per foot )

Formula: P subscript Mh_ser_pos = 9 point 43 K

The resultant bolt force is:

P subscript r_ser_pos = square root of ( ( P subscript s_ser + P subscript Mv_ser_pos ) squared + ( P subscript H_ser_pos + P subscript Mh_ser_pos ) squared )

Formula: P subscript r_ser_pos = 16 point 97 K

Case 2 - Dead Load + Negative Live Load:

The following forces were computed in Design Step 4.6:

Formula: V subscript w_ser = 198 point 53 K

Formula: M subscript tot_ser_neg = 195 point 08 K feet

Formula: H subscript w_ser_neg = minus 135 point 68 K

The vertical shear force in the bolts due to the applied shear force:

Formula: P subscript s_ser = numerator (V subscript w_ser) divided by denominator (N subscript b)

Formula: P subscript s_ser = 6 point 20 K

The horizontal shear force in the bolts due to the horizontal force resultant:

Formula: P subscript H_ser_neg = numerator (Vertical Bar H subscript w_ser neg Vertical Bar) divided by denominator (N subscript b minus )

Formula: P subscript H_ser_neg = 4 point 24 K

Determine the horizontal and vertical components of the bolt shear force on the extreme bolt due to the total moment in the web:

For the vertical component:

Formula: P subscript Mv_ser_neg = numerator (M subscript tot_ser_neg times ( x)) divided by denominator (I subscript p) times ( 12 inches per foot )

Formula: P subscript Mv_ser_neg = 0 point 57 K

For the horizontal component:

Formula: P subscript Mh_ser_neg = numerator (M subscript tot_ser_neg ( y)) divided by denominator (I subscript p) times ( 12 inches per foot )

Formula: P subscript Mh_ser_neg = 8 point 51 K

The resultant bolt force is:

P subscript r_ser_neg = square root of ( ( P subscript s_ser + P subscript Mv_ser_neg ) squared + ( P subscript H_ser_neg + P subscript Mh_ser_neg ) squared )

Formula: P subscript r_ser_neg = 14 point 43 K

The governing resultant bolt force is:

Formula: P subscript r_ser = max ( P subscript r_ser_pos , P subscript r_ser_neg )

Formula: P subscript r_ser = 16 point 97 K

Check:

Formula: P subscript r_ser = 16 point 97 K < Formula: R subscript r = 39 point 00 K OK

Thirty-two 7/8" diameter high-strength bolts in two vertical rows on each side of the splice provides sufficient resistance against bolt shear and slip.

Shear Yielding of Splice Plates:

S6.13.6.1.4b

Check for shear yielding on the gross section of the web splice plates under the Strength I design shear force, Vuw:

Formula: V subscript uw = 279 point 05 K

The factored resistance of the splice plates is taken as:

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

SEquation 6.13.5.3-1

Formula: R subscript n = 0 point 58 times A subscript g times F subscript y

SEquation 6.13.5.3-2

The gross area of the web splice is calculated as follows:

Number of splice plates: Formula: N subscript wp = 2

Thickness of plate: Formula: t subscript wp = 0 point 3125 inches

Depth of splice plate: Formula: d subscript wp = 48 inches

Formula: A subscript gross_wp = N subscript wp times t subscript wp times d subscript wp

Formula: A subscript gross_wp = 30 point 00 inches squared

From Design Step 4.1:

Specified minimum yield strength of the connection element: Formula: F subscript y = 50 ksi

Resistance factor for shear: Formula: phi subscript v = 1 point 0

The factored shear resistance is then:

Formula: R subscript r = phi subscript v times ( 0 point 58) times ( A subscript gross_wp ) times ( F subscript y )

Formula: R subscript r = 870 point 00 K

Check:

Formula: V subscript uw = 279 point 05 K < Formula: R subscript r = 870 point 00 K OK

Fracture and Block Shear Rupture of the Web Splice Plates:

S6.13.6.1.4b

Strength I Limit State checks for fracture on the net section of web splice plates and block shear rupture normally do not govern for plates of typical proportion. These checks are provided in this example for completeness.

From Design Step 4.6, the factored design shear for the Strength I Limit State was determined to be:

Formula: V subscript uw = 279 point 05 K

Fracture on the Net Section:

C6.13.4

Investigation of critical sections and failure modes, other than block shear, is recommended, including the case of a net section extending across the full plate width, and, therefore, having no parallel planes. This may be a more severe requirement for a girder flange or splice plate than the block shear rupture mode.

For this case, the areas of the plate resisting tension are considered to be zero.

Formula: A subscript tg = 0 point 0 inches squared Formula: A subscript tn = 0 point 0 inches squared

Therefore, the factored resistance is:

Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript u times A subscript vn + F subscript y times A subscript tg )

SEquation 6.13.4-2

Formula: phi subscript bs = 0 point 80 (Design Step 4.1)

where the net area resisting shear:

Formula: A subscript vn = N subscript wp times ( d subscript wp minus N subscript fn times d subscript hole ) times t subscript wp

Number of web plates: Formula: N subscript wp = 2

Depth of the web plate: Formula: d subscript wp = 48 inches

Number of bolts along one plane: Formula: N subscript fn = 16

Thickness of the web plate: Formula: t subscript wp = 0 point 3125 inches

From Design Step 4.1:

Specified minimum yield strength of the connected material: Formula: F subscript y = 50 ksi

Specified minimum tensile strength of the connected material: Formula: F subscript u = 65 ksi

Diameter of the bolt holes: Formula: d subscript hole = 1 point 0 inches

Net area resisting shear:

Formula: A subscript vn = N subscript wp times ( d subscript wp minus N subscript fn times d subscript hole ) times t subscript wp

Formula: A subscript vn = 20 point 00 inches squared

Avn of the splice plates to be used in calculating the fracture strength of the splice plates cannot exceed eighty-five percent of the gross area of the plates:

S6.13.5.2

Formula: A subscript 85 = 0 point 85 times A subscript gross_wp

Formula: A subscript gross_wp = 30 point 00 inches squared

Formula: A subscript 85 = 25 point 50 inches squared > Formula: A subscript vn = 20 point 00 inches squared OK

The factored resistance is then:

Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript u times A subscript vn )

Formula: R subscript r = 603 point 20 K > Formula: V subscript uw = 279 point 05 K OK

Block Shear Rupture Resistance:

S6.13.4

Connection plates, splice plates and gusset plates shall be investigated to ensure that adequate connection material is provided to develop the factored resistance of the connection.

Determine the applicable equation:

If Formula: A subscript tn greater than or equal to 0 point 58 times A subscript vn then:

Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript y times A subscript vg + F subscript u times A subscript tn )

SEquation 6.13.4-1

otherwise:

Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript u times A subscript vn + F subscript y times A subscript tg )

SEquation 6.13.4-2

This figure shows an elevation view of the web splice and the mode of block shear failure. The vertical and horizontal edge distances are both 1 point 50 inches. The horizontal row spacing and the vertical row spacing are both 3 inches. The failure mode is vertically thru the vertical row of blots closest to the centerline of the splice and horizontally thru the top two horizontal bolts.

Figure 4-9 Block Shear Failure Mode - Web Splice Plate

Gross area along the plane resisting shear stress:

Formula: A subscript vg = N subscript wp times ( d subscript wp minus 1 point 50 inches ) times t subscript wp

Formula: A subscript vg = 29 point 06 inches squared

Net area along the plane resisting shear stress:

Formula: A subscript vn = N subscript wp times left bracket d subscript wp minus 1 point 50 inches minus 15 point 50 times ( d subscript hole ) right bracket times t subscript wp

Formula: A subscript vn = 19 point 38 inches squared

Gross area along the plane resisting tension stress:

Formula: A subscript tg = N subscript wp times ( 1 point 50 inches + 3 point 0 inches ) times t subscript wp

Formula: A subscript tg = 2 point 81 inches squared

Net area along the plane resisting tension stress:

Formula: A subscript tn = N subscript wp times left bracket 1 point 50 inches + 3 point 0 inches minus 1 point 5 times ( d subscript hole ) right bracket times t subscript wp

Formula: A subscript tn = 1 point 88 inches squared

Identify the appropriate block shear equation:

Formula: A subscript tn = 1 point 88 inches squared < Formula: 0 point 58 times A subscript vn = 11 point 24 inches squared

Therefore, SEquation 6.13.4-2 is the governing equation:

Formula: R subscript r = phi subscript bs times ( 0 point 58 times F subscript u times A subscript vn + F subscript y times A subscript tg )

Formula: R subscript r = 696 point 85 K

Check:

Formula: V subscript uw = 279 point 05 K < Formula: R subscript r = 696 point 85 K OK

Flexural Yielding of Splice Plates:

S6.13.6.1.4b

Check for flexural yielding on the gross section of the web splice plates for the Strength I Limit State due to the total web moment and the horizontal force resultant:

Formula: f = numerator (M subscript Total) divided by denominator (S subscript pl) + numerator (H subscript uw) divided by denominator (A subscript gross_wp) less than or equal to phi subscript f times F subscript y

where:

Resistance factor for flexure (Design Step 4.1): Formula: phi subscript f = 1 point 0

Section modulus of the web splice plate:

Formula: S subscript pl = numerator (1) divided by denominator (6) times A subscript gross_wp times d subscript wp

Formula: S subscript pl = 240 point 00 inches cubed

Case 1 - Dead Load + Positive Live Load:

Formula: M subscript tot_str_pos = 474 point 66 K feet

Formula: H subscript w_str_pos = 486 point 20 K

Formula: f subscript str_pos = numerator (M subscript tot_str_pos) divided by denominator (S subscript pl) times ( 12 inches per foot ) + numerator (H subscript w_str_pos) divided by denominator (A subscript gross_wp)

Formula: f subscript str_pos = 39 point 94 ksi

Formula: f subscript str_pos = 39 point 94 ksi < Formula: phi subscript f times F subscript y = 50 ksi OK

Case 2 - Dead Load + Negative Live Load:

Formula: M subscript tot_str_neg = 734 point 19 K feet

Formula: H subscript w_str_neg = minus 140 point 16 K

Formula: f subscript str_neg = numerator (M subscript tot_str_neg) divided by denominator (S subscript pl) times ( 12 inches per foot ) + numerator (Vertical Bar H subscript w_str_neg Vertical Bar) divided by denominator (A subscript gross_wp)

Formula: f subscript str_neg = 41 point 38 ksi

Formula: f subscript str_neg = 41 point 38 ksi < Formula: phi subscript f times F subscript y = 50 ksi OK

Control of Permanent Deflection - Splice Plates:

S6.10.5.2

Check the maximum normal stress on the gross section of the web splice plates for the Service II Limit State due to the total web moment and horizontal force resultant:

Formula: f = numerator (M subscript Total) divided by denominator (S subscript pl) + numerator (H subscript w) divided by denominator (A subscript gross_wp) less than or equal to 0 point 95 times F subscript y

where:

Formula: S subscript pl = 240 point 00 inches cubed

Formula: A subscript gross_wp = 30 point 00 inches squared

Case 1 - Dead Load + Positive Live Load:

Formula: M subscript tot_ser_pos = 216 point 34 K feet

Formula: H subscript w_ser_pos = 195 point 08 K

Formula: f subscript ser_pos = numerator (M subscript tot_ser_pos) divided by denominator (S subscript pl) times ( 12 inches per foot ) + numerator (H subscript w_ser_pos) divided by denominator (A subscript gross_wp)

Formula: f subscript ser_pos = 17 point 32 ksi

Formula: f subscript ser_pos = 17 point 32 ksi < Formula: 0 point 95 times F subscript y = 47 point 50 ksi OK

Case 2 - Dead Load + Negative Live Load:

Formula: M subscript tot_ser_neg = 195 point 08 K feet

Formula: H subscript w_ser_neg = minus 135 point 68 K

Formula: f subscript ser_neg = numerator (M subscript tot_ser_neg) divided by denominator (S subscript pl) times ( 12 inches per foot ) + numerator (Vertical Bar H subscript w_ser_neg Vertical Bar) divided by denominator (A subscript gross_wp)

Formula: f subscript ser_neg = 14 point 28 ksi

Formula: f subscript ser_neg = 14 point 28 ksi < Formula: 0 point 95 times F subscript y = 47 point 50 ksi OK

Web Bolts - Bearing Resistance at Bolt Holes:

S6.13.2.9

Since the girder web thickness is less than twice the thickness of the web splice plates, the girder web will control for the bearing check.

Check the bearing of the bolts on the connected material for the Strength I Limit State assuming the bolts have slipped and gone into bearing. The design bearing strength of the girder web at the location of the extreme bolt in the splice is computed as the minimum resistance along the two orthogonal shear failure planes shown in Figure 4-10. The maximum force (vector resultant) acting on the extreme bolt is compared to this calculated strength, which is conservative since the components of this force parallel to the failure surfaces are smaller than the maximum force.

This figure shows an elevation view of the girder with a close-up of the end of the girder, the bottom flange, the splice bolt holes and the shear places for bearing. There are two bolt holes shown, both are shown vertically and hole 1 is closest to the bottom flange. The horizontal distance from the end of the girder to the bolt holes is defined as Lc1. The vertical distance between the two bolt holes is defined as Lc2. The shear planes are shown horizontally from the end of the girder to the first bolt hole and then vertically between the bolt holes.

Figure 4-10 Bearing Resistance - Girder Web

To determine the applicable equation for the calculation of the nominal bearing resistance, the clear distance between holes and the clear end distance must be calculated and compared to the value of two times the nominal diameter of the bolt. This check yields:

S6.13.2.9

Formula: d subscript bolt = 0 point 875 inches (Design Step 4.1)

Formula: 2 times d subscript bolt = 1 point 75 inches

The edge distance from the center of the hole to the edge of the girder is taken as 1.75". Therefore, the clear distance between the edge of the hole and the edge of the girder is computed as follows:

S6.13.2.6.6

Formula: L subscript c subscript 1 = 1 point 75 inches minus numerator (d subscript hole) divided by denominator (2)

Formula: d subscript hole = 1 point 0 inches (Design Step 4.1)

Formula: L subscript c subscript 1 = 1 point 25 inches

The center-to-center distance between adjacent holes is 3". Therefore, the clear distance between holes is:

Formula: L subscript c subscript 2 = 3 point 00 inches minus d subscript hole

Formula: L subscript c subscript 2 = 2 point 00 inches

For standard holes, where either the clear distance between holes is less than 2.0d, or the clear end distance is less than 2.0d:

Formula: R subscript n = 1 point 2 times L subscript c times t times F subscript u

SEquation 6.13.2.9-2

From Design Step 4.1:

Thickness of the connected material: Formula: t subscript w = 0 point 50 inches

Tensile strength of the connected material: Formula: F subscript u = 65 ksi

The nominal bearing resistance at the extreme bolt hole is as follows:

Formula: R subscript n = 1 point 2 times L subscript c subscript 1 times t subscript w times F subscript u

Formula: R subscript n = 48 point 75 K

The factored bearing resistance is:

Formula: R subscript r = phi subscript bb times R subscript n

Formula: phi subscript bb = 0 point 80 (Design Step 4.1)

Formula: R subscript r = 39 point 00 K

The controlling minimum Strength I resultant bolt force was previously computed:

Formula: P subscript r_str = 37 point 98 K < Formula: R subscript r = 39 point 00 K OK

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Bearing Resistance at Web Bolt Holes

Should the bearing resistance be exceeded, it is recommended that the edge distance be increased slightly in lieu of increasing the number of bolts or thickening the web.

Fatigue of Splice Plates:

For load-induced fatigue considerations, each detail shall satisfy:

Formula: gamma times ( Delta f ) less than or equal to ( Delta F ) subscript n

SEquation 6.6.1.2.2-1

Fatigue is checked at the bottom edge of the splice plates, which by inspection are subject to a net tensile stress.

The normal stresses at the bottom edge of the splice plates due to the total positive and negative fatigue-load web moments and the corresponding horizontal force resultants are as follows:

Formula: f = numerator (M subscript total) divided by denominator (S subscript pl) + numerator (H subscript w) divided by denominator (A subscript gross_wp)

From previous calculations:

Formula: S subscript pl = 240 point 00 inches cubed

Formula: A subscript gross_wp = 30 point 00 inches squared

Case 1 - Positive Live Load:

From Design Step 4.6:

Formula: M subscript tot_fat_pos = 30 point 34 K feet

Formula: H subscript w_fat_pos = 33 point 08 K

Formula: f subscript fat_pos = numerator (M subscript tot_fat_pos) divided by denominator (S subscript pl) times ( 12 inches per foot ) + numerator (H subscript w_fat_pos) divided by denominator (A subscript gross_wp)

Formula: f subscript fat_pos = 2 point 62 ksi

Case 2 - Negative Live Load:

From Design Step 4.6:

Formula: M subscript tot_fat_neg = minus 28 point 24 K feet

Formula: H subscript w_fat_neg = minus 23 point 76 K

Formula: f subscript fat_neg = numerator (M subscript tot_fat_neg) divided by denominator (S subscript pl) times ( 12 inches per foot ) + numerator (H subscript w_fat_neg) divided by denominator (A subscript gross_wp)

Formula: f subscript fat_neg = minus 2 point 20 ksi

The total fatigue-load stress range at the bottom edge of the web splice plates is therefore:

Formula: gamma Delta f = Vertical Bar f subscript fat_pos Vertical Bar + Vertical Bar f subscript fat_neg Vertical Bar

Formula: gamma Delta f = 4 point 82 ksi

From Design Step 4.4, the fatigue resistance was determined as:

Formula: Delta F subscript n = 8 point 00 ksi

The fatigue check is now completed as follows:

Formula: gamma Delta f = 4 point 82 ksi < Formula: Delta F subscript n = 8 point 00 ksi OK

Design Step 4.8 - Draw Schematic of Final Bolted Field Splice Design

Figure 4-11 shows the final bolted field splice as determined in this design example.

This figure shows an elevation view of the final field splice. The distance from the centerline of the splice to the centerline of the first row of vertical bolts is one point 9375 inches. The web is 54 inches deep by one half inch thick. The web splice plate is 48 inches by five sixteenths of an inch thick. The vertical and horizontal edge distances for the web splice plate are both 1 point 50 inches. The horizontal row spacing and the vertical row spacing for the web splice plate are both 3 inches. There are a total of 4 vertical rows and 16 horizontal rows in the web splice plate.The bottom flange on the left side is 14 inches wide by seven eights of an inch thick and the bottom flange on the right is 14 inches wide by one and three eights of an inch thick. The outside splice plate is 14 inches wide by seven sixteenths of an inch thick. The fill plate is 14 inches wide by one half of an inch thick. There are two inside splice plates that measure 6 inches by one half of an inch thick. There are two horizontal spaces at 3 inches in the bottom flange, for a total of 12 bolts, on each side of the centerline of splice. The distance from the top of the bottom flange to the bottom row of bolts in the web splice plate is 4 point 5 inches.All the bolts are seven eights inch in diameter ASTM A325.

Figure 4-11 Final Bolted Field Splice Design

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Updated: 07/18/2013
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