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


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

Bridges & Structures

LRFD Steel Girder SuperStructure Design Example

Miscellaneous Steel Design Example Design Step 5

Table of Contents

Design Step 5.1 - Design Shear Connectors
Design Step 5.2 - Design Bearing Stiffeners
Design Step 5.3 - Design Welded Connections
Design Step 5.4 - Design Cross-frames 21

(It should be noted that Design Step 5.4 presents a narrative description rather than design computations.)

Design Step 5 consists of various design computations associated with the steel girder but not necessarily required to design the actual plates of the steel girder. Such miscellaneous steel design computations include the following:

  1. Shear connectors
  2. Bearing stiffeners
  3. Welded connections
  4. Diaphragms and cross-frames
  5. Lateral bracing
  6. Girder camber

For this design example, computations for the shear connectors, a bearing stiffener, a welded connection, and a cross-frame will be presented. The other features must also be designed, but their design computations are not included in this design example.

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

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

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 design features included in this design step.

Design Step 5.1 - Design Shear Connectors

Since the steel girder has been designed as a composite section, shear connectors must be provided at the interface between the concrete deck slab and the steel section to resist the interface shear. For continuous composite bridges, shear connectors are normally provided throughout the length of the bridge. In the negative flexure region, since the longitudinal reinforcement is considered to be a part of the composite section, shear connectors must be provided.

Studs or channels may be used as shear connectors. For this design example, stud shear connectors are being used throughout the length of the bridge. The shear connectors must permit a thorough compaction of the concrete to ensure that their entire surfaces are in contact with the concrete. In addition, the shear connectors must be capable of resisting both horizontal and vertical movement between the concrete and the steel.

S6.10.7.4.1

The following figure shows the stud shear connector proportions, as well as the location of the stud head within the concrete deck.

S6.10.7.4.1a

A partial section through the deck, showing one beam with three 7 eights inch diameter shear connectors 6 inches in height. The middle shear connector is located at the centerline of beam and the first and third shear connectors 5 inches from the centerline of beam. The total deck thickness is 8 and one half inches and a 3 and one half inch dimension from the bottom of deck to the bottom of the top flange. Dimensions vertically are shown for values A, B and C. These values are given in Table 5-1, Shear Connector Embedment. 'A' represents the distance from the bottom of slab to the top of the top flange. 'B' represents the distance from the top of shear connectors to the bottom of slab, while 'C' represents the distance from top of slab to top of shear connectors.

Figure 5-1 Stud Shear Connectors

Shear Connector Embedment
Flexure Region A B C
Positive 2.875" 3.125" 5.375"
Intermediate 2.25" 3.75" 4.75"
Negative 1.00" 5.00" 3.50"

Table 5-1 Shear Connector Embedment

Formula: MathCad tip logo

Shear Connector Layout

It is common to use several stud shear connectors per transverse row along the top flange of the girder. The number of shear connectors per transverse row will depend on the top flange width. Refer to S6.10.7.4.1c for transverse spacing requirements.

Formula: MathCad tip logo

Shear Connector Length

The stud shear connector length is commonly set such that its head is located near the middle of the deck slab. Refer to S6.10.7.4.1d for shear connector embedment requirements.

The ratio of the height to the diameter of a stud shear connector must not be less than 4.0. For this design example, the ratio is computed based on the dimensions presented in Figure 5-1, as follows:

S6.10.7.4.1a

Formula: Height subscript stud = 6 point 0 inches

Formula: Diameter subscript stud = 0 point 875 inches

Formula: numerator (Height subscript stud) divided by denominator (Diameter subscript stud) = 6 point 86 OK

The pitch of the shear connectors must be determined to satisfy the fatigue limit state as specified in S6.10.7.4.2 and S6.10.7.4.3, as applicable. The resulting number of shear connectors must not be less than the number required to satisfy the strength limit states as specified in S6.10.7.4.4.

The pitch, p, of the shear connectors must satisfy the following equation:

S6.10.7.4.1b

Formula: p less than or equal to numerator (n times Z subscript r times I) divided by denominator (V subscript sr times Q)

The parameters I and Q are based on the short-term composite section and are determined using the deck within the effective flange width.

In the positive flexure region:

Formula: n = 3 (see Figure 5-1)

Formula: I = 66340 point 3 inches superscript 4 (see Table 3-4)

S6.10.3.1.1b

Formula: Q = left bracket numerator (( 8 point 0 inches ) times ( 103 point 0 inches )) divided by denominator (8) right bracket times ( 62 point 375 inches minus 50 point 765 inches )

Formula: Q = 1195 point 8 inches cubed

In the positive flexure region, the maximum fatigue live load shear range is located at the abutment. The factored value is computed as follows:

Formula: V subscript sr = 0 point 75 times ( 41 point 45 K + 5 point 18 K)

Formula: V subscript sr = 34 point 97 K

(see live load analysis computer run)

Formula: Z subscript r = alpha times d squared greater than or equal to numerator (5 point 5 times d squared ) divided by denominator (2)

S6.10.7.4.2

Formula: N = 82125000 (see Design Step 3.14 at location of maximum positive flexure)

S6.6.1.2.5

Formula: alpha = 34 point 5 minus 4 point 28 times log( N)

Formula: alpha = 0 point 626

Formula: d = 0 point 875 in

Formula: alpha times d squared = 0 point 48 Formula: numerator (5 point 5 times d squared ) divided by denominator (2) = 2 point 11

Therefore, Formula: Z subscript r = 2 point 11 K

Formula: p = numerator (n times Z subscript r times I) divided by denominator (V subscript sr times Q) Formula: p = 10 point 04 inches

In the negative flexure region:

Formula: n = 3 (see Figure 5-1)

In the negative flexure region, the parameters I and Q may be determined using the reinforcement within the effective flange width for negative moment, unless the concrete slab is considered to be fully effective for negative moment in computing the longitudinal range of stress, as permitted in S6.6.1.2.1. For this design example, I and Q are assumed to be computed considering the concrete slab to be fully effective.

SC6.10.7.4.1b

Formula: I = 130196 point 1 inches superscript 4 (see Table 3-5)

Formula: Q = left bracket numerator (( 8 point 0 inches ) times ( 103 point 0 inches )) divided by denominator (8) right bracket times ( 64 point 250 inches minus 46 point 702 inches )

Formula: Q = 1807 point 4 inches cubed

Formula: V subscript sr = 0 point 75 times ( 0 point 00 K + 46 point 53 K)

Formula: V subscript sr = 34 point 90 K

(see Table 3-1 and live load analysis computer run)

Formula: Z subscript r = alpha times d squared greater than or equal to numerator (5 point 5 times d squared ) divided by denominator (2)

S6.10.7.4.2

Formula: Z subscript r = 2 point 11 K (see previous computation)

Formula: p = numerator (n times Z subscript r times I) divided by denominator (V subscript sr times Q) Formula: p = 13 point 07 inches

Therefore, based on the above pitch computations to satisfy the fatigue limit state, use the following pitch throughout the entire girder length:

Formula: p = 10 inches

Formula: MathCad tip logo

Shear Connector Pitch

The shear connector pitch does not necessarily have to be the same throughout the entire length of the girder. Many girder designs use a variable pitch, and this can be economically beneficial.

However, for this design example, the required pitch for fatigue does not vary significantly over the length of the bridge. Therefore, a constant shear connector pitch of 10 inches will be used.

In addition, the shear connectors must satisfy the following pitch requirements:

S6.10.7.4.1b

Formula: p less than or equal to 24 inches OK

Formula: p greater than or equal to 6 times d

Formula: d = 0 point 875 inches Formula: 6 times d = 5 point 25 inches OK

For transverse spacing, the shear connectors must be placed transversely across the top flange of the steel section and may be spaced at regular or variable intervals.

Stud shear connectors must not be closer than 4.0 stud diameters center-to-center transverse to the longitudinal axis of the supporting member.

S6.10.7.4.1c

Formula: 4 times d = 3 point 50 inches

Formula: Spacing subscript transverse = 5 point 0 inches (see Figure 5-1) OK

In addition, the clear distance between the edge of the top flange and the edge of the nearest shear connector must not be less than 1.0 inch.

Formula: Distance subscript clear = numerator (14 inches ) divided by denominator (2) minus 5 inches minus numerator (d) divided by denominator (2) (see Figure 5-1)

Formula: Distance subscript clear = 1 point 56 inches OK

S6.10.7.4.1d

The clear depth of concrete cover over the tops of the shear connectors should not be less than 2.0 inches, and shear connectors should penetrate at least 2.0 inches into the deck. Based on the shear connector penetration information presented in Table 5-1, both of these requirements are satisfied.

For the strength limit state, the factored resistance of the shear connectors, Qr, is computed as follows:

S6.10.7.4.4

Formula: Q subscript r = phi subscript sc times Q subscript n

S6.10.7.4.4a

Formula: phi subscript sc = 0 point 85

S6.5.4.2

The nominal shear resistance of one stud shear connector embedded in a concrete slab is computed as follows:

S6.10.7.4.4c

Formula: Q subscript n = 0 point 5 times A subscript sc times square root of (f prime subscript c times E subscript c) less than or equal to A subscript sc times F subscript u

Formula: A subscript sc = pi times numerator (d squared ) divided by denominator (4) Formula: A subscript sc = 0 point 601 inches squared

Formula: f prime subscript c = 4 point 0 ksi (see Design Step 3.1)

S5.4.2.1

Formula: E subscript c = 3834 ksi (see Design Step 3.3)

S5.4.2.4

Formula: F subscript u = 60 point 0 ksi

S6.4.4

Formula: 0 point 5 times 0 point 601 times square root of (4 point 0 times 3834) = 37 point 21 K

Formula: 0 point 601 times 60 point 0 = 36 point 06 K

Therefore, Formula: Q subscript n = 36 point 06 K

Formula: Q subscript r = phi subscript sc times Q subscript n

Therefore, Formula: Q subscript r = 30 point 65 K

The number of shear connectors provided between the section of maximum positive moment and each adjacent point of 0.0 moment or between each adjacent point of 0.0 moment and the centerline of an interior support must not be less than the following:

S6.10.7.4.4a

Formula: n = numerator (V subscript h) divided by denominator (Q subscript r)

The total horizontal shear force, Vh, between the point of maximum positive moment and each adjacent point of 0.0 moment is equal to the lesser of the following:

S6.10.7.4.4b

Formula: V subscript h = 0 point 85 times f prime subscript c times b times t subscript s

or

Formula: V subscript h = F subscript yw times D times t subscript w + F subscript yt times b subscript t times t subscript t + F subscript yc times b subscript f times t subscript f

where Formula: f prime subscript c = 4 point 0 ksi (see Design Step 3.1)

S5.4.2.1

Formula: b = 103 point 0 inches (see Design Step 3.3)

Formula: t subscript s = 8 point 0 inches (see Design Step 3.1)

Formula: F subscript yw = 50 ksi (see Design Step 3.1)

STable 6.4.1-1

Formula: D = 54 inches (see Design Step 3.18)

Formula: t subscript w = 0 point 50 inches (see Design Step 3.18)

Formula: F subscript yt = 50 ksi (see Design Step 3.1)

STable 6.4.1-1

Formula: b subscript t = 14 inches (see Design Step 3.18)

Formula: t subscript t = 0 point 875 inches (see Design Step 3.18)

Formula: F subscript yc = 50 ksi (see Design Step 3.1)

STable 6.4.1-1

Formula: b subscript f = 14 inches (see Design Step 3.18)

Formula: t subscript f = 0 point 625 inches (see Design Step 3.18)

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

Formula: F subscript yw times D times t subscript w + F subscript yt times b subscript t times t subscript t + F subscript yc times b subscript f times t subscript f = 2400 K

Therefore, Formula: V subscript h = 2400 K

Therefore, the number of shear connectors provided between the section of maximum positive moment and each adjacent point of 0.0 moment must not be less than the following:

S6.10.7.4.4a

Formula: n = numerator (V subscript h) divided by denominator (Q subscript r)

Formula: n = 78 point 3

The distance between the end of the girder and the location of maximum positive moment is approximately equal to:

Formula: L = 48 point 0 feet (see Table 3-7)

Similarly the distance between the section of the maximum positive moment and the point of dead load contraflexure is approximately equal to:

Formula: L = 83 point 6 feet minus 48 point 0 feet (see Table 3-7)

Formula: L = 35 point 6 feet

Using a pitch of 10 inches, as previously computed for the fatigue limit state, and using the minimum length computed above, the number of shear connectors provided is as follows:

Formula: n = 3 times numerator (L times ( 12 inches per foot )) divided by denominator (p)

Formula: L = 35 point 6 feet Formula: p = 10 inches

Formula: n = 128 point 2 OK

For continuous span composite sections, the total horizontal shear force, Vh, between each adjacent point of 0.0 moment and the centerline of an interior support is equal to the following:

S6.10.7.4.4b

Formula: V subscript h = A subscript r times F subscript yr

where Formula: A subscript r = 12 point 772 inches squared (see Design Step 3.3)

Formula: F subscript yr = 60 ksi (see Design Step 3.1)

Formula: V subscript h = A subscript r times F subscript yr

Formula: V subscript h = 766 K

Therefore, the number of shear connectors provided between each adjacent point of 0.0 moment and the centerline of an interior support must not be less than the following:

S6.10.7.4.4a

Formula: n = numerator (V subscript h) divided by denominator (Q subscript r)

Formula: n = 25 point 0

The distance between the point of dead load contraflexure and the centerline of the interior support is approximately equal to:

Formula: L = 120 feet minus 83 point 6 feet (see Table 3-7)

Formula: L = 36 point 4 feet

Using a pitch of 10 inches, as previously computed for the fatigue limit state, the number of shear connectors provided is as follows:

Formula: n = 3 times numerator (L times ( 12 inches per foot )) divided by denominator (p) Formula: p = 10 inches

Formula: n = 131 point 0 OK

Therefore, using a pitch of 10 inches for each row, with three stud shear connectors per row, throughout the entire length of the girder satisfies both the fatigue limit state requirements of S6.10.7.4.1 and S6.10.7.4.2 and the strength limit state requirements of S6.10.7.4.4.

Therefore, use a shear stud spacing as illustrated in the following figure.

An elevation of a girder from the centerline of abutment to the centerline of pier, which is the point of symmetry. Identified in the elevation is the shear connector spacing. It says 144 spaces at 10 inches equal 120 feet 0 inches. Below the spacing callout is a note saying '3 stud shear connectors per row'.

Figure 5-2 Shear Connector Spacing

Design Step 5.2 - Design Bearing Stiffeners

Bearing stiffeners are required to resist the bearing reactions and other concentrated loads, either in the final state or during construction.

For plate girders, bearing stiffeners are required to be placed on the webs at all bearing locations and at all locations supporting concentrated loads.

Therefore, for this design example, bearing stiffeners are required at both abutments and at the pier. The following design of the abutment bearing stiffeners illustrates the bearing stiffener design procedure.

The bearing stiffeners in this design example consist of one plate welded to each side of the web. The connections to the web will be designed to transmit the full bearing force due to factored loads and is presented in Design Step 5.3.

S6.10.8.2.1

The stiffeners extend the full depth of the web and, as closely as practical, to the outer edges of the flanges.

Each stiffener will either be milled to fit against the flange through which it receives its reaction or attached to the flange by a full penetration groove weld.

The following figure illustrates the bearing stiffener layout at the abutments.

Two sketches, first a Partial Girder Elevation at Abutment and second, Section A-A below. The partial elevation is located at the centerline of abutment with Section A-A taken, looking down through the bearing stiffener. Section A-A shows a partial web length with a thickness of one half inch. Bearing stiffeners are on each side of the web 5 and one half inches wide by 11 sixteenths thick.

Figure 5-3 Bearing Stiffeners at Abutments

Formula: MathCad tip logo

Bearing Stiffener Plates

Bearing stiffeners usually consist of one plate connected to each side of the web. This is generally a good starting assumption for the bearing stiffener design. Then, if this configuration does not provide sufficient resistance, two plates can be used on each side of the web.

The projecting width, bt, of each bearing stiffener element must satisfy the following equation. This provision is intended to prevent local buckling of the bearing stiffener plates.

S6.10.8.2.2

Formula: b subscript t less than or equal to 0 point 48 times t subscript p times square root of ( numerator (E) divided by denominator (F subscript ys))

Formula: t subscript p = numerator (11) divided by denominator (16) inches (see Figure 5-3)

Formula: E = 29000 ksi

S6.4.1

Formula: F subscript ys = 50 ksi

STable 6.4.1-1

Formula: 0 point 48 times t subscript p times square root of ( numerator (E) divided by denominator (F subscript ys)) = 7 point 95 inches

Formula: b subscript t = 5 point 5 inches (see Figure 5-3) OK

The bearing resistance must be sufficient to resist the factored reaction acting on the bearing stiffeners. The factored bearing resistance, Br, is computed as follows:

S6.10.8.2.3

Formula: B subscript r = phi subscript b times A subscript pn times F subscript ys

Formula: phi subscript b = 1 point 00

S6.5.4.2

Part of the stiffener must be clipped to clear the web-to-flange weld. Thus the area of direct bearing is less than the gross area of the stiffener. The bearing area, Apn, is taken as the area of the projecting elements of the stiffener outside of the web-to-flange fillet welds but not beyond the edge of the flange. This is illustrated in the following figure:

Section through the girder showing bearing stiffeners on each side of the web. The total width of each stiffener plate is 5 and one half inches. The top and bottom of each plate is clipped 1 inch by 1 inch to clear the girder fillet welds. This leaves 4 and one half inches bearing width on the flanges.

Figure 5-4 Bearing Width

Formula: b subscript brg = b subscript t minus 1 point 0 inches Formula: b subscript brg = 4 point 5 inches

Formula: A subscript pn = 2b subscript brg times t subscript p Formula: A subscript pn = 6 point 19 inches squared

Formula: F subscript ys = 50 ksi

Formula: B subscript r = phi subscript b times A subscript pn times F subscript ys

Formula: B subscript r = 309 point 4 K

The factored bearing reaction at the abutment is computed as follows, using load factors as presented in STable 3.4.1-1 and STable 3.4.1-2 and using reactions obtained from a computer analysis run:

Formula: Reaction subscript Factored = ( 1 point 25 times 68 point 7 K) + ( 1 point 50 times 11 point 0 K) + ( 1 point 75 times 110 point 5 K)

Formula: Reaction subscript Factored = 295 point 8 K

Therefore, the bearing stiffener at the abutment satisfies the bearing resistance requirements.

The final bearing stiffener check relates to the axial resistance of the bearing stiffeners. The factored axial resistance is determined as specified in S6.9.2.1. The radius of gyration is computed about the midthickness of the web, and the effective length is taken as 0.75D, where D is the web depth.

S6.10.8.2.4

S6.10.8.2.4a

For stiffeners consisting of two plates welded to the web, the effective column section consists of the two stiffener elements, plus a centrally located strip of web extending not more than 9tw on each side of the stiffeners. This is illustrated in the following figure:

S6.10.8.2.4b

A section looking down through the web showing bearing stiffeners on each side. The thickness of web is one half inch and the thickness of the bearing plates are 11 sixteenths. The bearing plates are also 5 and one half inches wide. A dimension extending along the web from the centerline of stiffener in each direction indicates a strip of web equal to 9 times the web thickness. This dimension is equal to 4 and one half inches.

Figure 5-5 Bearing Stiffener Effective Section

Formula: P subscript r = phi subscript c times P subscript n

S6.9.2.1

Formula: phi subscript c = 0 point 90

S6.5.4.2

Formula: lamda = ( numerator (k times I) divided by denominator (r subscript s times pi) ) squared times numerator (F subscript y) divided by denominator (E)

S6.9.4.1

Formula: kI = ( 0 point 75) times ( 54 inches )

S6.10.8.2.4a

Formula: I subscript s = numerator (left bracket 0 point 6875 inches times ( 11 point 5 inches ) cubed right bracket + left bracket 8 point 3125 inches times ( 0 point 5 inches ) cubed right bracket) divided by denominator (12)

S6.10.8.2.4b

Formula: I subscript s = 87 point 22 inches superscript 4

Formula: A subscript s = ( 0 point 6875 inches times 11 point 5 inches ) + ( 8 point 3125 inches times 0 point 5 inches )

S6.10.8.2.4b

Formula: A subscript s = 12 point 06 inches squared

Formula: r subscript s = square root of ( numerator (I subscript s) divided by denominator (A subscript s))

Formula: r subscript s = 2 point 69 inches

Formula: F subscript y = 50ksi

Formula: lamda = ( numerator (kI) divided by denominator (r subscript s times pi) ) squared times numerator (F subscript y) divided by denominator (E)

S6.9.4.1

Formula: lamda = 0 point 0396

Therefore, Formula: lamda less than or equal to 2 point 25

Therefore, Formula: P subscript n = 0 point 66 superscript lamda subscript F subscript y times A subscript s

S6.9.4.1

Formula: P subscript n = 593 point 3 K

Formula: P subscript r = phi subscript c times P subscript n

S6.9.2.1

Formula: P subscript r = 533 point 9 K

Formula: Reaction subscript Factored = 295 point 8 K

Therefore, the bearing stiffener at the abutment satisfies the axial bearing resistance requirements.

The bearing stiffener at the abutment satisfies all bearing stiffener requirements. Therefore, use the bearing stiffener as presented in Figures 5-3 and 5-4.

Design Step 5.3 - Design Welded Connections

Welded connections are required at several locations on the steel superstructure. Base metal, weld metal, and welding design details must conform to the requirements of the ANSI/AASHTO/AWS Bridge Welding Code D1.5.

For this design example, two fillet welded connection designs will be presented using E70 weld metal:

  1. Welded connection between the bearing stiffeners and the web.
  2. Welded connection between the web and the flanges.

For the welded connection between the bearing stiffeners and the web, the fillet weld must resist the factored reaction computed in Design Step 5.2.

S6.13.3

Formula: Reaction subscript Factored = 295 point 8 K

Assume a fillet weld thickness of 1/4 inches.

Formula: Thickness subscript Weld = 0 point 25 inches

Formula: MathCad tip logo

Fillet Weld Thickness

In most cases, the minimum weld thickness, as specified in Table 5-2, provides a welded connection that satisfies all design requirements. Therefore, the minimum weld thickness is generally a good starting point when designing a fillet weld.

The resistance of the fillet weld in shear is the product of the effective area and the factored resistance of the weld metal. The factored resistance of the weld metal is computed as follows:

S6.13.3.2.4b

Formula: R subscript r = 0 point 6 times phi subscript e2 times F subscript exx

Formula: phi subscript e2 = 0 point 80

S6.5.4.2

Formula: F subscript e70 = 70 ksi

SC6.13.3.2.1

Formula: R subscript r = 0 point 6 times phi subscript e2 times F subscript e70

S6.13.3.2.4b

Formula: R subscript r = 33 point 60 ksi

The effective area equals the effective weld length multiplied by the effective throat. The effective throat is the shortest distance from the joint root to the weld face.

S6.13.3.3

Formula: Length subscript Eff = 4 times ( 54 inches minus 2 inches ) Formula: Length subscript Eff = 208 point 0 inches

Formula: Throat subscript Eff = numerator (Thickness subscript Weld) divided by denominator ( square root of (2) ) Formula: Throat subscript Eff = 0 point 177 inches

Formula: Area subscript Eff = Length subscript Eff times Throat subscript Eff Formula: Area subscript Eff = 36 point 77 inches squared

The resistance of the fillet weld is then computed as follows:

S6.13.3.2.4b

Formula: Resistance = R subscript r times Area subscript Eff

Formula: Resistance = 1235 K OK

For material 0.25 inches or more in thickness, the maximum size of the fillet weld is 0.0625 inches less than the thickness of the material, unless the weld is designated on the contract documents to be built out to obtain full throat thickness.

S6.13.3.4

For the fillet weld connecting the bearing stiffeners to the web, the bearing stiffener thickness is 11/16 inches and the web thickness is 1/2 inches. Therefore, the maximum fillet weld size requirement is satisfied.

The minimum size of fillet welds is as presented in Table 5-2. In addition, the weld size need not exceed the thickness of the thinner part joined.

S6.13.3.4

Minimum Size of Filet Welds
Base Metal Thickness of
Thicker Part Joined(T)
(inches)
Minimum Size of
Filet Weld
(inches)
T ≤ 3/4 1/4
T > 3/4 5/16

STable 6.13.3.4-1

Table 5-2 Minimum Size of Fillet Welds

In this case, the thicker part joined is the bearing stiffener plate, which is 11/16 inches thick. Therefore, based on Table 5-2, the minimum size of fillet weld is 1/4 inch, and this requirement is satisfied.

The minimum effective length of a fillet weld is four times its size and in no case less than 1.5 inches. Therefore, this requirement is also satisfied.

Since all weld design requirements are satisfied, use a 1/4 inch fillet weld for the connection of the bearing stiffeners to the web.

S6.13.3.5

For the welded connection between the web and the flanges, the fillet weld must resist a factored horizontal shear per unit length based on the following equation:

S6.13.3

Formula: v = numerator (V times Q) divided by denominator (I)

This value is greatest at the pier, where the factored shear has its highest value.

The following computations are for the welded connection between the web and the top flange. The welded connection between the web and the bottom flange is designed in a similar manner.

The shear is computed based on the individual section properties and load factors for each loading, as presented in Design Steps 3.3 and 3.6:

For the noncomposite section, the factored horizontal shear is computed as follows:

Formula: V subscript Noncomp = ( 1 point 25 times 114 point 7 K)

Formula: V subscript Noncomp = 143 point 4 K

Formula: Q subscript Noncomp = ( 14 inches times 2 point 5 inches ) times ( 58 point 00 inches minus 28 point 718 inches )

Formula: Q subscript Noncomp = 1024 point 9 inches cubed

Formula: I subscript Noncomp = 65426 point 6 inches superscript 4

Formula: v subscript Noncomp = numerator (V subscript Noncomp times Q subscript Noncomp) divided by denominator (I subscript Noncomp) Formula: v subscript Noncomp = 2 point 25 numerator (K) divided by denominator ( inches )

For the composite section, the factored horizontal shear is computed as follows:

Formula: V subscript Comp = ( 1 point 25 times 16 point 4 K) + ( 1 point 50 times 19 point 8 K) + ( 1 point 75 times 131 point 4 K)

Formula: V subscript Comp = 280 point 1 K

Formula: Q subscript Comp = ( 14 inches times 2 point 5 inches ) times ( 58 point 00 inches minus 32 point 668 inches )

Formula: Q subscript Comp = 886 point 6 inches cubed

Formula: I subscript Comp = 79333 point 4 inches superscript 4

Formula: v subscript Comp = numerator (V subscript Comp times Q subscript Comp) divided by denominator (I subscript Comp) Formula: v subscript Comp = 3 point 13 numerator (K) divided by denominator ( inches )

Based on the above computations, the total factored horizontal shear is computed as follows:

Formula: v subscript Total = v subscript Noncomp + v subscript Comp

Formula: v subscript Total = 5 point 38 numerator (K) divided by denominator ( inches )

Assume a fillet weld thickness of 5/16 inches.

Formula: Thickness subscript Weld = 0 point 3125 inches

The resistance of the fillet weld in shear is the product of the effective area and the factored resistance of the weld metal. The factored resistance of the weld metal was previously computed as follows:

S6.13.3.2.4b

Formula: R subscript r = 0 point 6 times phi subscript e2 times F subscript e70 Formula: R subscript r = 33 point 60 ksi

The effective area equals the effective weld length multiplied by the effective throat. The effective throat is the shortest distance from the joint root to the weld face. In this case, the effective area is computed per unit length, based on the use of one weld on each side of the web.

S6.13.3.3

Formula: Throat subscript Eff = numerator (Thickness subscript Weld) divided by denominator ( square root of (2) ) Formula: Throat subscript Eff = 0 point 221 inches

Formula: Area subscript Eff = 2 times Throat subscript Eff Formula: Area subscript Eff = 0 point 442 numerator ( inches squared ) divided by denominator ( inches )

The resistance of the fillet weld is then computed as follows:

S6.13.3.2.4b

Formula: Resistance = R subscript r times Area subscript Eff

Formula: Resistance = 14 point 85 numerator (K) divided by denominator ( inches ) OK

For material 0.25 inches or more in thickness, the maximum size of the fillet weld is 0.0625 inches less than the thickness of the material, unless the weld is designated on the contract documents to be built out to obtain full throat thickness.

S6.13.3.4

For the fillet weld connecting the web to the flanges, the web thickness is 0.5 inches, the minimum flange thickness is 0.625 inches, and the maximum flange thickness is 2.75 inches. Therefore, the maximum fillet weld size requirement is satisfied.

The minimum size of fillet welds is as presented in Table 5-2. In addition, the weld size need not exceed the thickness of the thinner part joined.

S6.13.3.4

In this case, the thicker part joined is the flange, which has a minimum thickness of 0.625 inches and a maximum thickness of 2.75 inches. Therefore, based on Table 5-2, the minimum size of fillet weld is 5/16 inch, and this requirement is satisfied.

The minimum effective length of a fillet weld is four times its size and in no case less than 1.5 inches. Therefore, this requirement is also satisfied.

Since all weld design requirements are satisfied, use a 5/16 inch fillet weld for the connection of the web and the top flange. The welded connection between the web and the bottom flange is designed in a similar manner.

S6.13.3.5

Load-induced fatigue must be considered in the base metal at a welded connection. Fatigue considerations for plate girders may include:

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

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

S6.6.1.2.5

In Design Step 3.14 for the positive moment region, the fatigue check is illustrated for the fillet-welded connection of the transverse intermediate stiffeners to the girder. This procedure must be considered for the base metal at welded connections.

Additional weld connection requirements are presented in S6.13.3 and in ANSI/AASHTO/AWS Bridge Welding Code D1.5.

Design Step 5.4 - Design Cross-frames

Diaphragms and cross-frames may be placed at the following locations along the bridge:

  • At the end of the structure
  • Across interior supports
  • Intermittently along the span

S6.7.4.1

Formula: MathCad tip logo

Diaphragm or Cross-frame Spacing

A common rule of thumb, based on previous editions of the AASHTO Specifications, is to use a maximum diaphragm or cross-frame spacing of 25 feet. Based on C6.7.4.1, the arbitrary requirement for a 25 foot maximum spacing has been replaced by a requirement for a rational analysis that will often result in the elimination of fatigue-prone attachment details.

For this design example, cross-frames are used at a spacing of 20 feet. The 20-foot spacing in this design example facilitates a reduction in the required flange thicknesses in the girder section at the pier.

The need for diaphragms or cross-frames must be investigated for:

  • All stages of assumed construction procedures
  • The final condition
Formula: MathCad tip logo

Difference Between Diaphragms

and Cross-frames

The difference between diaphragms and cross-frames is that diaphragms consist of a transverse flexural component, while cross-frames consist of a transverse truss framework.

Both diaphragms and cross-frames connect adjacent longitudinal flexural components.

When investigating the need for diaphragms or cross-frames and when designing them, the following must be considered:

  • Transfer of lateral wind loads from the bottom of the girder to the deck and from the deck to the bearings
  • Stability of the bottom flange for all loads when it is in compression
  • Stability of the top flange in compression prior to curing of the deck
  • Distribution of vertical dead and live loads applied to the structure

Diaphragms or cross-frames can be specified as either:

  • Temporary - if they are required only during construction
  • Permanent - if they are required during construction and in the bridge's final condition

At a minimum, the Specifications require that diaphragms and cross-frames be designed for the following:

  • Transfer of wind loads according to the provisions of S4.6.2.7
  • Applicable slenderness requirements in S6.8.4 or S6.9.3

In addition, connection plates must satisfy the requirements of S6.6.1.3.1.

Formula: MathCad tip logo

Cross-frame Types

K-type cross-frames are as shown in Figure 5-6, while X-type cross-frames have an X-shape configuration of angles or structural tees rather than a K-shape configuration of angles or structural tees.

A common rule of thumb is to use K-type cross-frames when the aspect ratio (that is, the ratio of the girder spacing to the girder depth) is greater than about 1.5 to 1 and to use X-type cross-frames when the aspect ratio is less than 1.5 to 1.

For this design example, cross-frames will be used.

Girder spacing: Formula: S = 9 point 75 feet (see Figure 3-2)

Girder depth: Formula: D = 4 point 9375 feet (see Figure 3-15)

(maximum value)

Aspect ratio: Formula: numerator (S) divided by denominator (D) = 1 point 97

Therefore, use K-type cross-frames.

The geometry of a typical K-type cross-frame for an intermediate cross-frame is illustrated in Figure 5-6.

As illustrated in Figure 5-6, the intersection of the centroidal axes of the two diagonals coincides with the centroidal axis of the bottom strut. In addition, the intersection of the centroidal axis of each diagonal and the centroidal axis of the top strut coincides with the vertical centerlines of the girders.

Partial cross section through the superstructure showing the deck with a barrier on two girders with a cross-frame between the girders. The spacing of the girders is 9 feet 9 inches center to center. Also, the center of the cross frame is dimensioned at 4 feet 10 and one half inches. The cross frame is identified as either steel angle or tee, typical.

Figure 5-6 K-Type Cross-frame

Based on previous computations in Design Step 3.17 for the negative moment region, the unfactored wind load is computed as follows:

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

C4.6.2.7.1

Formula: eta = 1 point 0

S1.3

Formula: gamma = 1 point 40 (for Strength III Limit State)

STable 3.4.1-1

Formula: P subscript D = 0 point 00529 times ksf (see Design Step 3.17)

Formula: d = 4 point 9375 feet (maximum value)

Formula: W = numerator ( eta times gamma times P subscript D times d) divided by denominator (2) Formula: W = 0 point 0183 Kips per foot

The horizontal wind force applied to the brace point may then be computed as specified in C4.6.2.7.1, as follows:

Formula: P subscript w = W times L subscript b

C4.6.2.7.1

Formula: W = 0 point 0183 Kips per foot

Formula: L subscript b = 20 feet

Formula: P subscript w = W times L subscript b Formula: P subscript w = 0 point 366 K

For the design of the cross-frame members, the following checks should be made using the previously computed wind load:

  • Slenderness
  • Axial compression
  • Flexure about the major axis
  • Flexure about the minor axis
  • Flexure and axial compression
<< previous Contents next >>
Updated: 07/18/2013
Federal Highway Administration | 1200 New Jersey Avenue, SE | Washington, DC 20590 | 202-366-4000