Connections, Splices, Cross-Frames, and Bracing

Engineering story

A steel bridge is what its joints let it be

A rolled beam ships from the mill up to 50–60 ft long. A welded plate girder up to about 130 ft. Beyond that, every steel bridge is a collection of shorter members joined in the field with splices, and every parallel-girder system is pinned together laterally with cross-frames and lateral bracing that keep the girders plumb, share wind load across the deck, and prevent flange lateral-torsional buckling before composite action develops. AASHTO §6.13 devotes 40+ pages to these details for a reason: 80 % of bridge fatigue cracks initiate at a connection, and every steel bridge failure since the Silver Bridge (1967) has traced back to a joint.

Chapter objectives

What you will be able to do

Learning objectives

By the end of this chapter you will be able to:

1Distinguish bearing-type, slip-critical, and pretensioned high-strength bolted connections and their permissible use per §6.13.2.
2Compute nominal bolt shear and slip resistance per §6.13.2.7 and §6.13.2.8.
3Design a bolted field splice for a plate-girder cross section — web splice for shear and moment, flange splice for tension and compression.
4Design a full-penetration groove weld splice per §6.13.3, including AWS D1.5 nondestructive testing requirements.
5Compute cross-frame spacing for a composite steel bridge and design its diagonals as concentric axial-force members per §6.9 and §6.8.
6Design a top-flange lateral bracing system for a phased-construction girder subject to wind uplift before composite action develops.
7Detail gusset plates for concentric truss and cross-frame connections per §6.14.2.8.

11.1 — Connection families

Bolts, welds, and why we still use both

AASHTO LRFD §6.13.1

Modern bridges use high-strength bolts (ASTM F3125 Grade A325 or A490) for field connections and full-penetration groove welds for shop connections. Bolts are used in the field because they need no power, are inspectable by torque or turn-of-the-nut, and can be pretensioned to create slip-critical friction connections that never move. Field welds are limited to secondary elements because achieving AWS D1.5 quality outdoors is difficult and the fatigue category of a field weld is inferior to a properly installed bolted joint.

Fig. 11.1Bolt loading. (a) Shear — the bolt shank resists the force acting across the ply interface. (b) Tension — the bolt is pulled along its axis; external load first reduces the clamping pretension before lengthening the bolt.

11.2 — Bolt resistance

Shear, bearing, tension, and slip

AASHTO LRFD §6.13.2.7, §6.13.2.8, §6.13.2.9

Nominal shear resistance per bolt (threads excluded from shear plane):

R_n \;=\; 0.48 \, A_b \, F_{ub} \, N_s

(11.1)

$A_b$: nominal bolt cross-section [in²]
$F_{ub}$: specified minimum tensile strength (120 ksi A325, 150 ksi A490)
$N_s$: number of shear planes

If threads are in the shear plane, use 0.38 instead of 0.48. Factored capacity is $\phi_s R_n$ with $\phi_s = 0.80$ .

Bearing resistance on plate at a bolt hole (§6.13.2.9):

R_n \;=\; 2.4 \, d \, t \, F_u \qquad (\text{clear end distance} \ge 2d)

(11.2)

$d$: bolt diameter [in]
$t$: thickness of connected plate [in]
$F_u$: tensile strength of plate [ksi]

Slip resistance (Class B clean mill-scale, standard holes):

R_n \;=\; K_h \, K_s \, N_s \, P_t

(11.3)

$K_h$: hole-size factor (1.0 standard, 0.85 oversize)
$K_s$: surface condition factor (0.50 Class B, 0.33 Class A)
$P_t$: minimum required bolt pretension [kip]

11.3 — Field splices of plate girders

Web splice for shear + moment, flange splice for direct force

AASHTO LRFD §6.13.6

A field splice occurs where two girder segments meet on the falsework or on temporary shoring. AASHTO §6.13.6.1.4 requires the splice to be designed for the greater of the factored force at that section and 75 % of the yield capacity of the smaller member on either side. The 75 % rule guarantees the splice never becomes the weak link, so that plastic redistribution — if it ever occurs — happens in the girder, not in the splice.

Fig. 11.2Bolted field splice of a plate girder. Inner and outer splice plates on each flange and on both faces of the web transfer axial force, shear, and moment across the joint without a single field weld.

Flange splice — direct force method. The flange resists tension or compression proportional to the moment couple:

F_{fl} \;=\; \dfrac{M_u}{h_{eff}}

(11.4)

$F_{fl}$: flange direct force at splice [kip]
$M_u$: factored moment at splice, or 0.75·M_p if smaller
$h_{eff}$: distance between flange centroids [in]

n_b \;\ge\; \dfrac{F_{fl}}{\phi_s R_n}

(11.5)

$n_b$: number of bolts required on one side of the splice

Web splice — resist shear plus the web share of moment. Bolts are grouped in a rectangular pattern; the elastic vector combination governs:

R_v \;=\; \dfrac{V_u}{n_b}, \qquad R_h \;=\; \dfrac{M_{web} \, y_i}{\sum y_i^2}

(11.6)

R_{max} \;=\; \sqrt{R_v^2 + R_h^2} \;\le\; \phi_s R_n

(11.7)

$R_v, R_h$: vertical (shear) and horizontal (bending) force per bolt [kip]
$y_i$: distance from bolt group centroid to bolt row
$M_{web}$: share of moment resisted by web = M · (I_w / I_g)

Fig. 11.3Alternative shop-welded splice. Complete-joint-penetration groove welds on both flanges and the web develop the full cross-sectional yield strength; NDT (UT for tension flanges, MT for surfaces) is required by AWS D1.5.

11.4 — Cross-frames and diaphragms

Keeping the girders plumb and sharing load

AASHTO LRFD §6.7.4

In an I-girder bridge, adjacent girders are tied together by transverse cross-frames — a truss made of angles or WT sections. Their four structural jobs are:

Prevent lateral-torsional buckling of the compression flange before composite action develops.
Distribute wind load on the exterior girder to the interior girders through the deck.
Maintain the geometry of the girder line during erection and deck placement.
Provide load paths in curved girders where the primary load is torsion.

Fig. 11.4Three cross-frame families. The X-type (a) is the most common; the K-type (b) is preferred when interior clearance to the deck soffit is tight; the Warren-type (c) uses only diagonals and is used for shallow girders.

Fig. 11.5Cross-frame plan. Cross-frames are placed at each support and at intermediate points such that unbraced flange length

L_b

keeps the compression flange safely below its LTB length limit during construction.

Maximum cross-frame spacing during construction is governed by the noncomposite compression-flange LTB limit and by the AASHTO rule of thumb:

L_b \;\le\; \min\!\Bigl(\; 25\;\text{ft},\;\; L_p = 1.0\, r_t \sqrt{E / F_{yc}}\;\Bigr)

(11.8)

$L_b$: unbraced length of compression flange [ft]
$r_t$: radius of gyration of the flange plus 1/3 of the compression web area [in]
$F_{yc}$: compression-flange yield strength [ksi]

The diagonal force in an X-type cross-frame of height $h_{cf}$ and width $s_g$ under lateral force $W$ is:

F_{diag} \;=\; \dfrac{W}{2\cos\alpha}, \qquad \alpha \;=\; \arctan\!\left(\dfrac{h_{cf}}{s_g}\right)

(11.9)

Fig. 11.6Figure 11.6. Free-body diagram of a cross-frame. A lateral wind force

W

at the top chord is resolved into axial tension in one diagonal and compression in the other; the chord axial force closes the couple.

Fig. 11.7Gusset plate detail. Force lines of all three members must intersect at a common working point to avoid inducing local moments in the gusset; §6.14.2.8 sets the Whitmore-section rule for checking gusset tension.

11.5 — Lateral bracing

A horizontal truss built into the flanges

AASHTO LRFD §6.7.5

For long-span or curved girder bridges, top-flange lateral bracing (a horizontal truss between the top flanges, with diagonals in the plane of the deck) is added during construction to (a) resist wind load on the girders before the deck is cast, and (b) prevent lateral flange buckling during launch or lift. Once the deck cures composite, the deck itself becomes the diaphragm and the lateral bracing becomes redundant.

Fig. 11.8Plan of top-flange lateral bracing during construction. The X-pattern in the plane of the top flange resists wind and any construction lateral load until the deck is fully composite.

11.6 — Worked example 1

Bolted flange splice of a composite plate girder

AASHTO LRFD §6.13.6

Problem statement

A composite steel plate-girder bridge (Ch. 8 example) has a field splice at the point of contraflexure of the interior span. Design the tension bottom-flange splice for the negative-moment case (deck in tension, bottom flange in compression at the splice).

Given

Bottom flange $b_f = 18\;\text{in},\; t_f = 1.25\;\text{in},\; F_y = 50\;\text{ksi}$
Girder depth $h_{eff} = 66\;\text{in}$ (flange centroid to flange centroid)
Factored moment at splice $M_u = 4{,}200\;\text{kip-ft}$
0.75 M_p limitLarger than M_u ⇒ M_u governs
BoltsASTM F3125 A325, 7/8 in. Ø, threads excluded, standard holes, Class B slip surface

Required

Compute the flange direct force and required number of bolts on one side of the splice. Verify slip resistance for Service II and bearing / shear for Strength I.

Step 1 — Flange direct force.

Formula

F_{fl} \;=\; \dfrac{M_u}{h_{eff}}

Substitute

F_{fl} \;=\; \dfrac{4{,}200 \times 12}{66}

Result

F_{fl} \;=\; 764\;\text{kip}

Step 2 — Bolt properties.

Formula

A_b \;=\; \dfrac{\pi}{4}\,(7/8)^2

Substitute

A_b \;=\; \dfrac{\pi}{4}(0.875)^2

Result

A_b \;=\; 0.601\;\text{in}^2

Step 3 — Strength I shear per bolt. Two shear planes (outer + inner cover plate):

Formula

R_n \;=\; 0.48\, A_b\, F_{ub}\, N_s

Substitute

R_n \;=\; 0.48\,(0.601)(120)(2)

Result

R_n \;=\; 69.2\;\text{kip/bolt}, \;\; \phi_s R_n = 0.80\,(69.2) = 55.4\;\text{kip}

Formula

n_b \;=\; \dfrac{F_{fl}}{\phi_s R_n}

Substitute

n_b \;=\; \dfrac{764}{55.4}

Result

n_b \;=\; 13.8 \;\Rightarrow\; \text{use 16 bolts (4 rows × 4 columns) on each side}

Step 4 — Bearing on flange at bolt hole.

Formula

R_n \;=\; 2.4\, d\, t\, F_u

Substitute

R_n \;=\; 2.4\,(0.875)(1.25)(65)

Result

R_n \;=\; 171\;\text{kip/bolt} \;\;\text{(governs neither)}

Step 5 — Slip check, Service II. Service II moment = $M_{s2} = 2{,}950\;\text{kip-ft}$ , flange service force = 536 kip. Slip resistance of one bolt:

Formula

R_{slip} \;=\; K_h\, K_s\, N_s\, P_t

Substitute

R_{slip} \;=\; 1.0\,(0.50)(2)(39)

Result

R_{slip} \;=\; 39\;\text{kip/bolt}

Formula

n_b^{slip} \;=\; \dfrac{536}{39}

Result

= 13.7 \;\Rightarrow\; 16 \text{ bolts still adequate} \quad \checkmark

Both Strength I and Service II are satisfied with 16 A325 bolts per side. Adopt this splice.

Fig. 11.9Figure 11.9. Splice detail for the worked example — outer and inner cover plates on the bottom flange with a symmetric 4 × 3 bolt group on each side of the joint.

Final section detailing (from computed A_s)

Bottom-flange bolted field splice — composite steel plate girder

Location	A_s required	Bars provided	Spacing / detail
Bolts	n = 13.8 (Strength I), 13.7 (Service II slip)	16 – 7/8 in. Ø ASTM F3125 A325 per side (4 × 4 pattern)	3 in. pitch, 3 in. gage, 1.5 in. edge distance
Outer cover plate	Area ≥ tension flange area / 2 with holes deducted	18 in. × 1/2 in. plate, F<sub>y</sub> = 50 ksi	Extends 6 in. beyond outermost bolt each side
Inner cover plates (pair)	Area ≥ remaining tension flange area / 2	Two – 7 in. × 1/2 in. plates, F<sub>y</sub> = 50 ksi	One each side of the web centerline; 3 in. gage
Faying surface	Class B, µ = 0.50	Blast-cleaned to SSPC-SP10, unpainted or with Class-B coating	Applied within 30 days of assembly
Bolt installation	Pretensioned per §6.13.2.1.2	Turn-of-the-nut method, 1/2 turn from snug tight	Verify with calibrated wrench on 10 % of bolts

The splice is designed for the actual factored moment; 0.75·M_p would govern if M_u were smaller. Service II slip check is required by §6.10.9.3 for composite girders in the negative-moment region because slip after deck cracking can produce non-recoverable deformation.

11.7 — Worked example 2

Cross-frame diagonal for a 4-girder composite bridge

AASHTO LRFD §6.7.4, §6.9.4

Problem statement

For the same 130-ft composite bridge as Ch. 10 Example 1, size the diagonal of an intermediate X-type cross-frame subject to wind on the exterior girder before the deck cures composite.

Given

Girder spacing $s_g = 8\;\text{ft}$
Cross-frame height $h_{cf} = 5\;\text{ft}$
Cross-frame spacing25 ft o.c.
Wind pressure (§3.8) $p_w = 0.050\;\text{ksf}$ (on 6 ft exposed depth)
Diagonal materialASTM A709 Grade 50, angle section, F_y = 50 ksi

Required

Compute the factored lateral wind force per cross-frame, resolve into the diagonal axial force, and select an equal-leg angle that meets the AASHTO §6.9.4 compression rules.

Step 1 — Tributary wind at one cross-frame.

Formula

W \;=\; 1.4\, p_w \, h_{exp} \, s_{cf}

Substitute

W \;=\; 1.4\,(0.050)(6)(25)

Result

W \;=\; 10.5\;\text{kip}

Step 2 — Diagonal angle geometry.

Formula

\alpha \;=\; \arctan\!\left(\dfrac{h_{cf}}{s_g}\right)

Substitute

\alpha \;=\; \arctan(5/8)

Result

\alpha \;=\; 32.0°, \; \cos\alpha = 0.848, \; L_{diag} = \sqrt{5^2+8^2} = 9.43\;\text{ft}

Step 3 — Diagonal axial force.

Formula

F_{diag} \;=\; \dfrac{W}{2\cos\alpha}

Substitute

F_{diag} \;=\; \dfrac{10.5}{2\,(0.848)}

Result

F_{diag} \;=\; 6.2\;\text{kip (design for both tension and compression)}

Step 4 — Try L 4 × 4 × 3/8. Properties: $A_g = 2.86\;\text{in}^2,\; r_z = 0.782\;\text{in},\; r_x = r_y = 1.23\;\text{in}$ . Effective length K = 1.0, unbraced length = 9.43 ft = 113 in. Governing slenderness:

Formula

\dfrac{K L}{r_z}

Substitute

\dfrac{1.0 \times 113}{0.782}

Result

= 145 \;<\; 200 \quad \checkmark

Step 5 — Nominal compressive resistance (§6.9.4).

Formula

P_e \;=\; \dfrac{\pi^2 E}{(KL/r)^2} A_g

Substitute

P_e \;=\; \dfrac{\pi^2 (29{,}000)}{(145)^2}\,(2.86)

Result

P_e \;=\; 39.0\;\text{kip}

Formula

P_o \;=\; F_y \, A_g

Substitute

P_o \;=\; 50\,(2.86)

Result

P_o \;=\; 143\;\text{kip}

$P_e / P_o = 0.273 < 0.44$ , so use the elastic (inelastic) branch:

Formula

P_n \;=\; 0.877\, P_e

Substitute

P_n \;=\; 0.877\,(39.0)

Result

P_n \;=\; 34.2\;\text{kip},\;\; \phi_c P_n = 0.95\,(34.2) = 32.5\;\text{kip} \;>\; 6.2\;\text{kip} \quad \checkmark

Step 6 — Connection to gusset. Use 2 – 3/4 in. Ø A325 bolts at each end; capacity per bolt $\phi_s R_n = 30.5\;\text{kip} \times 1 = 30.5\;\text{kip}$ — one bolt covers demand, but AASHTO §6.13.2.6 requires a minimum of two bolts at each end for load-bearing angle connections.

Final section detailing (from computed A_s)

Intermediate cross-frame diagonal — 4-girder composite steel bridge

Location	A_s required	Bars provided	Spacing / detail
Cross-frame diagonal	F<sub>diag</sub> = 6.2 kip (± tension/compression), KL/r ≤ 200	L 4 × 4 × 3/8, A709 Grade 50, KL/r = 145	9.43 ft long, single-angle, connected by 2 – 3/4 in. Ø A325 bolts each end
Cross-frame chord (top & bottom)	chord force from wind moment couple	L 5 × 5 × 3/8, one on each chord	8 ft long horizontal, welded to gusset plate
Gusset plate	Whitmore-section tension, §6.14.2.8	3/8 in. thick × 12 in. wide, Grade 50	Bolted to girder web with 4 – 7/8 in. Ø A325
Cross-frame spacing	L<sub>b</sub> ≤ 25 ft (§6.10.1.6, construction stage)	25 ft o.c. between piers, 15 ft near supports	One cross-frame at every support and at every field splice
Bolt pretension	snug-tight acceptable for cross-frames per §6.13.2.1.2	Snug-tight installation	Pretensioned installation not required for cross-frame diagonals

Selected angle carries the wind demand with a 5:1 factor of safety on capacity. The KL/r = 145 is well below the 200 limit and provides erection stiffness. Two bolts per end are the AASHTO minimum for a load-bearing single-angle connection.

11.8 — Guided practice

Web splice bolt group of a plate girder

A plate girder web is 72 in. deep × 1/2 in. thick, G50 steel. The splice must resist $V_u = 340\;\text{kip}$ and $M_{web} = 1{,}200\;\text{kip-ft}$ . Use 7/8 in. Ø A325 bolts (threads excluded). Propose a rectangular bolt group and verify the extreme bolt with Eqs. 11.6–11.7.

Expected result

Try 2 columns × 12 rows @ 4 in. pitch (bolt group 44 in. deep). Extreme

y_{max} = 22\;\text{in}

;

\sum y_i^2 \approx 2\cdot(4^2+8^2+\ldots+22^2)\cdot 2 \approx 4224\;\text{in}^2

. Then

R_v = 14.2, R_h = 75.0

R_{max} = 76.3\;\text{kip}

— exceeds 55 kip capacity. Add columns or rows and iterate.

11.9 — Mini design challenge

Complete field splice and cross-frame system for a 3-span composite bridge

Fig. 11.10Design-challenge geometry — three-span composite steel bridge with two field splices per span and cross-frames at 20 ft o.c.

Deliver:

A complete bolted field splice (flange and web) at the point of maximum positive moment in Span 1.
A cross-frame diagonal design at the exterior girder line.
A top-flange lateral bracing plan for the construction (non-composite) stage.
A gusset-plate detail for one cross-frame connection.
Bolt schedule, splice plate schedule, and welded-shop-fabrication notes.
A one-page design memo and marked shop drawings.

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11.10 — Chapter summary

What you leave with

Bearing-type, slip-critical, and pretensioned bolted-connection categories, and the AASHTO §6.13.2 resistance equations for each.
Bolted field-splice design: 75 %-of-yield rule, direct force method for flanges, and the elastic-vector web check.
Cross-frame families and the maximum spacing that keeps the compression flange stable in construction.
Top-flange lateral bracing as a temporary horizontal truss until the deck goes composite.
Gusset-plate detailing under the Whitmore-section rule.

Section 2

Fully Worked Examples

Complete AASHTO LRFD solutions with knowns, assumptions, step calculations, verification, and design commentary. Difficulty rises from basic to consulting-grade.

Worked Example 1

Bolt shear capacity — slip-critical A325 in standard holes

Basic

Problem

Compute total slip-resistance R_r of the group.

Step-by-Step

R_{n} = K_{h}\cdot K_{s}\cdot N_{s}\cdot P_{t} = 1.0\cdot 0.50\cdot 1\cdot 39

Result

R_{n} = 19.5 kip/bolt

R_{r} = 8\cdot \phi \cdot R_{n}; \phi = 1.00 for slip

Result

R_{r} = 156 kip

Design Verification

For slip-critical, φ = 1.00 (φ ≠ resistance factor for strength). Strength-limit shear check must also be satisfied.

Discussion

Never use faying-surface reduction K_h = 0.85 for oversize holes on primary members — always specify standard holes for slip-critical.

Worked Example 2

Block shear rupture of a bolted splice plate

Intermediate

Problem

Compute block-shear resistance R_n.

Step-by-Step

R_{1} = 0.58\cdot 65\cdot 3.8 + 1.0\cdot 65\cdot 1.2 = 143.3 + 78

Result

R_{1} = 221 kip

R_{2} = 0.58\cdot 50\cdot 5.5 + 78 = 159.5 + 78

Result

R_{2} = 238 kip

Design Verification

Block shear is often the hidden governor of bolted-connection design. Rupture path controls when F_u·A_nv < F_y·A_gv.

Discussion

Check block shear on both plates AND the connected member. A splice plate can be block-shear critical even when bolt shear passes with margin.

Worked Example 3

Fatigue check on a welded transverse stiffener detail

Intermediate

Problem

Verify infinite-life fatigue per §6.6.1.2.

Step-by-Step

ADTT_{SL}\cdot (75\cdot 365) = 1200\cdot 27{,}375 = 3.29e7 > 10^{7} cycles \rightarrow \ \text{Infinite Life required.}

\gamma \cdot \Delta f = 1.75\cdot 6.0

Result

10.5 ksi

Design Verification

Infinite-life design requires γ·Δf ≤ (ΔF)_TH. Finite-life design uses cycle-count and A/N³ curves — more permissive but more analysis.

Discussion

Category C' (transverse stiffener to web) is more common than C. Confirm detail category from Table 6.6.1.2.3-1; a wrong category off by one letter can halve or double allowable Δf.

Worked Example 4

Fillet-weld group loaded eccentrically in shear

Intermediate

Problem

Compute the resultant stress at the critical weld end and check against φF_nw.

Step-by-Step

A_w = 2\,(8) = 16\ \text{in};\quad I_p \approx \tfrac{1}{12}(2)(8)^{3} + 2(8)(3)^{2} = 85.3 + 144 = 229.3\ \text{in}^{3}

f_v = \dfrac{P}{A_w} = \dfrac{40}{16} = 2.50\ \text{kip/in}

Design Verification

Utilization 24.3/33.6 = 0.72. If eccentricity doubles (e = 8 in), f_R climbs above the limit — a good reminder that torsion, not direct shear, usually governs.

Discussion

The elastic vector method is conservative because it ignores weld ductility. The instantaneous-center method reduces required weld length by ~15–25% but requires tabulated coefficients (AISC Table 8-4) or software.

Worked Example 5

Splice-plate net-section tension (Whitmore + shear-lag)

Advanced

Problem

Check the splice plate for gross yielding, net-section fracture (with shear-lag U), and confirm P_u is safe.

Step-by-Step

A_g = w\,t = 8\,(0.375) = 3.00\ \text{in}^{2};\ \ \phi P_y = 0.95\,F_y\,A_g = 0.95\,(50)(3.00) = 142.5\ \text{kip}

A_n = (w - d_h)\,t = (8-1.00)(0.375) = 2.625\ \text{in}^{2}

Design Verification

Shear-lag governs because the plate transfers load through only one line — U drops to 0.875. Adding a second line at 3-in gage restores U ≈ 1.0 and rebalances the check.

Discussion

The block-shear check (§6.13.4) must also be run — for single-line splices it usually governs before net section. Always compute all three (yield, net fracture, block shear) and take the minimum.

Section 3

Guided Practice

Complete the missing steps. Use Hints for AASHTO article pointers and setup logic before revealing the full step. Submit at the end to send your work to your instructor.

Guided Problem 1

Slip-critical bolted splice

A325 7/8-in bolts in double shear on a Class B slip surface. Design $M_u/S$ at splice = 950 kip. Standard holes ( $K_h = 1.0$ ), $K_s = 0.50$ , $N_s = 2$ .

Step 1

Minimum bolt tension $P_t$ for 7/8-in A325 (kip).

Answer

Step 2

Slip resistance per bolt $R_n = K_h K_s N_s P_t$ (kip).

Answer

Step 3

Bolts required at slip (Service II). $\phi = 1.0$ . $n = 950/39$ (round up).

Answer

Step 4

Corresponding rows if 2-across pattern, 12 rows minimum? #bolts (round up).

Answer

Guided Problem 2

Cross-frame design load — curved bridge

Interior K-frame between two 66-in-deep girders. Radius $R = 700\ \text{ft}$ , span between cross-frames $L_b = 20\ \text{ft}$ , girder factored moment $M_u = 3200\ \text{k-ft}$ .

Step 1

Approximate lateral flange force $H = M_u L_b/(D R)$ (kip). $D = 66/12\ \text{ft}$ .

Answer

Step 2

Half force per K-frame diagonal (kip).

Answer

Step 3

Angle diagonal length if depth 60 in and horizontal 60 in (in).

Answer

Step 4

Axial demand $P = H/\cos(45^{\circ})$ (kip).

Answer

Guided Problem 3

Fillet weld to develop shear

Two continuous ¼-in fillet welds attach a 6-in stiffener to a girder web resisting $V = 40\ \text{kip}$ along 8-in length. E70 electrode.

Step 1

Throat $= 0.707 \cdot \text{leg}$ (in).

Answer

Step 2

Nominal shear strength per inch: $F_{exx}\cdot0.6\cdot\text{throat} = 70(0.6)(0.177)$ (kip/in).

Answer

Step 3

$\phi R_n$ both welds over 8 in (kip). $\phi = 0.80$ .

Answer

Step 4

Minimum weld length ratio $L/\text{leg}$ for a 1/4-in weld.

Answer

Guided Problem 4

Field splice — flange plate design

Bottom-flange splice: flange 16×1 in ( $A_g = 16\ \text{in}^{2}$ ), $M_u = 2800\ \text{k-ft}$ at splice, $d = 60\ \text{in}$ .

Step 1

Flange force $F = M_u \cdot 12/d$ (kip).

Answer

Step 2

Required $\phi F_y A_{pl}$ for splice plates, $\phi = 1.0$ , $F_y = 50\ \text{ksi}$ (in² of plate).

Answer

Step 3

Try 2 plates 16×0.375 in each side. $A = 2(16)(0.375)$ (in²).

Answer

Step 4

Bolts req'd for 560 kip with 30-kip slip capacity (round up).

Answer

Section 4

Independent Practice

Every problem randomizes its inputs. Work each step, submit for immediate feedback, request new values to practice again.

Practice 1

Slip resistance per bolt

K_h

Kh = 0.7000000000000001 -

K_s

Ks = 0.30000000000000004 -

N_s

Ns = 2 -

P_t (kip)

Pt = 49 kip

Step 1K_h K_s N_s P_t.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 2

Bolts to carry a factored splice force

Splice force

F = 1950 kip

R_n per bolt

Rn = 24 kip

Step 1Ceil(F/R_n).

Answer (-)

Randomized inputs, symbolic grading (±2%).

Practice 3

Fillet weld capacity (double weld)

Leg

leg = 0.3125 in

Length

L = 17 in

Step 1throat = 0.707·leg.

Answer (in)

Step 22·70·0.6·throat·L (kip).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 4

Block shear (simple approximation)

A_nt

Ant = 3.7 in²

A_nv

Anv = 7.800000000000001 in²

F_u

Fu = 70 ksi

Step 1F_u·A_nt + 0.6·F_u·A_nv.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 5

Cross-frame V-load (curved girder)

M_u

M = 3800 k-ft

L_b (ft)

Lb = 26 ft

Girder depth (ft)

D = 4.5 ft

Radius (ft)

R = 500 ft

Step 1M·L_b/(D·R) (kip).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 6

Diagonal axial from lateral force

H = 5 kip

diag angle (deg)

th = 52 deg

Step 1H/cos(θ).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 7

Bolt bearing on ply

d_b

d = 1 in

t_ply

t = 0.9375 in

F_u

Fu = 86 ksi

Step 12.4·d·t·F_u.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 8

Weld group polar moment (2-side)

Length

L = 7 in

spacing

d = 5 in

Step 1≈ (L³/6) + L·(d/2)² ×2.

Answer (in⁴/in)

Randomized inputs, symbolic grading (±2%).

Practice 9

Development of shear stud connector (steel-concrete)

A_sc

As = 1.1 in²

f′_c

fc = 4 ksi

Step 10.5·A_sc·√(f′_c·E_c). E_c ≈ 1820·√f′_c.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 10

Splice plate area required

Flange force

F = 325 kip

F_y

Fy = 55 ksi

Step 1A = F/(F_y).

Answer (in²)

Randomized inputs, symbolic grading (±2%).

Practice 11

Anchor rod tension (pot bearing)

Uplift T

Ttot = 185 kip

# rods

n = 5 -

Step 1T/n (kip).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 12

Girder lateral force from truck impact (rough)

Axle

P = 16 kip

Step 15%·P.

Answer (kip)

Randomized inputs, symbolic grading (±2%).