A worked example — the FHWA simply supported deep beam, rerun under AASHTO LRFD 2024

Two 600 kip loads sit 108 in from the bearings, a shear span of 1.5 times the depth, so the whole beam is a D-region. The source compares it to a straddle bent over a skewed roadway. The deep-beam template takes the length, depth, width, bearing and load positions and plate sizes directly — in kip and inch, as in the source.

The source puts the top chord at mid-depth of an assumed 6 in compression block, 3 in below the top, and the tie 5 in above the bottom, so the truss is 64 in deep. To show both layouts, it carries the left load to its bearing through two panels, with a vertical tie BD at mid-shear-span, and the right load through one direct strut CF. The template places its intermediate station at the same half shear span, 54 in from the bearing; its extra diagonals were removed so that each half matches the source. The right half is a mechanism on its own. The source notes this is acceptable in STM, and the app reports it (one degree of freedom) and solves the load case as a lower-bound equilibrium. The forces are AB 506.3 · BC 1,012.5 · BD 600 · AD −785 · BE −785 · CF −1,176.9 · DE −506.2 · EF −1,012.5 kip, the source's Table 1-1 to within its rounding.

9 of 14 strength checks pass. The five that do not come from three places: the top strut EF and its two CCC nodes E and F — the deficiency the source also finds — the vertical tie BD, and node B. Anchorage is reported separately from the strength verdict, and both ends of the bottom tie fail it under the 2024 development equation. Steps ④–⑦ go through each.

The bottom tie BC carries 1,012.5 kip and needs 18.75 in²; 16-#10 in two layers give 20.27 in² (safety factor 1.08), the same margin as the source's φP_n = 1,097 kip. The vertical tie BD carries 600 kip and needs 11.11 in², as in the source. Nine sets of two #5 closed stirrups — 36 legs over the 48 in available length — give the source 11.16 in², just enough. The app takes a bar's area from its nominal diameter, π·d²/4 = 0.307 in² for #5, 1% below the ASTM nominal 0.31 in², so the same 36 legs give 11.05 in² and BD shows a safety factor of 0.99. The difference is the area convention, not the design.

Under AASHTO LRFD a strut is checked where it meets a node. On a strut-to-node interface ν = 0.85 − f′c/20 ksi = 0.60, so f_ce = 3.0 ksi with φ = 0.70. CF needs 11.68 in and has 12.30 in at node F (safety factor 1.05) — the source's 1,239 against 1,177 kip after it lengthened the plates from 12 to 14 in. With 12 in plates the app gives 0.97, the source's 1,138 < 1,177 kip. The top strut EF runs between two CCC nodes and bears on their back faces, where ν = 0.85. It needs 7.09 in, and the 6 in compression block leaves 6 in: a safety factor of 0.85, the source's 856 < 1,013 kip. The source makes up the difference with six #8 bars in the top strut (§5.6.4.4, 4.74 ≥ 4.67 in²). The app does not model strut reinforcement, so the NG stays.

ν by face, for f′c = 5 ksi. At node C the bearing face needs 5.10 of 14 in (2.74) and the strut interface 11.68 of 15.74 in (1.35); the source's first check, with 12 in plates, has 14.7 in and 1,482 kip. The app also checks the back face of the bonded #10 tie at C (1.16), which AASHTO §5.8.2.5.3b does not require. Nodes E and F are governed by EF's back face (0.85, step ⑤); their bearing faces have 3.33. Node B, at the lower end of the vertical tie, is a smeared node the source does not check. The app checks every node of a 2D model, and there the back face of tie BC needs 10.04 of 10.00 in, a safety factor of 0.996 listed as NG. Node A has 2.20 and node D 1.19.

The bottom tie ends in 90° hooks (#10, 2 in cover). The available length is the source's: at C, 8.4 in from the extended nodal zone to the plate, the 14 in plate and 5 in beyond it, less 2 in cover — 25.4 in; at A, where strut AD is steeper, 21.2 in. What changed is the required length. The source (8th edition) takes ℓdh = 38·d_b/60 · f_y/√f′c = 21.6 in and reduces it by the excess-reinforcement factor 0.92 to 19.9 in. The 2024 development equation gives 29.7 in for the same hook, and the app does not apply the excess-reinforcement reduction — safety factors 0.86 at C and 0.71 at A. The source checks only C.

Same beam, same truss, same forces. Under the 2024 edition the app reaches the source's own conclusion at the top strut — 0.85 on the back face of the CCC node. The other differences have known causes: the bar-area convention (BD), two checks the app makes that the source does not (node B and the tie back face at C), the strut reinforcement the app does not model (EF), and the new development equation (anchorage).