The drawing says 4-20 mm top bars. The column is 400 mm square with 8-20 mm verticals, three to a face. The cage is tied, the pour is booked, and the second and third top bars of the crossing beam will not go in: they meet the column's middle vertical head-on. Four bars were drawn; two will sit where the drawing put them. The fixer springs the other two around the vertical, or taps them down until they slide under the girder steel.
QA writes "top bars not per plan." True, but ask what "per plan" looks like at that joint and nobody can say, because no sheet draws it. Nothing in the set draws the 400 mm box where both beams, the column verticals, the ties and the slab bars arrive together. Add the depths up and the congestion stops being a mystery: 40 mm cover, a 10 mm stirrup, then one beam's top bars and the other's beneath them is 40 + 10 + 20 + 20 = 90 mm of cover and steel in the beam top, with the 120 mm slab sitting over it.
Every threshold quoted in this article comes from NSCP 2015, the Philippine structural code whose parent document is ACI 318, and steel is Grade 415 throughout, the local designation. Where a provision carries a condition or an exception that decides the outcome, the condition is stated in words rather than left to be inferred.
Sections are drawn one member at a time. A beam section is a truthful drawing of the beam and a silent lie about the joint: it cannot show what arrives from the perpendicular direction.
Both beams carry the same d. Two 500 mm beams with 40 mm cover, 10 mm stirrups and 20 mm top bars both leave the design sheet at d = 440 mm. Where they cross, one set must pass under the other: somebody loses 20 mm, and the drawings do not say who.
The spacing rule is checked in the wrong place. NSCP 2015 sets clear spacing between parallel bars in a layer at the greatest of 25 mm, one bar diameter and four-thirds of the nominal maximum aggregate size; that is checked inside the beam section, where it passes easily. Nobody checks the gap to a column vertical, because the code rule does not reach it: those are not parallel bars in a layer. Say so before a reviewer does. What you apply there is a constructability criterion, not a clause: the largest stone has to pass, or the joint honeycombs.
Column 400 × 400 mm, 8-20 mm verticals (three per face), 10 mm ties, 40 mm cover. Girder B1 and beam B2 both 300 × 500 mm with 4-20 mm top bars at the support, 10 mm stirrups, 40 mm cover. Slab 120 mm with 10 mm top bars at 200 mm, 20 mm cover. fc' = 28 MPa, Grade 415 steel (fy = 415 MPa), nominal maximum aggregate size 20 mm as the design value. Bar unit masses 3.85 kg per metre for 25 mm and 2.47 kg per metre for 20 mm. B2 was designed for Mu = 185 kN-m taken at the face of support, not at the column centreline.
Four assumptions carry the arithmetic, and each of them belongs on the sheet. The joint is interior and B2's top bars run through it; had B2 terminated at the joint on hooks or heads, both the plan geometry below and the joint-dimension rule under Fix B would change, because anchorage length would govern instead. The four top bars are assumed evenly spread at the 60 mm pitch the clear width allows, and a fixer nudging a bar, or the ordinary placing tolerance on the column cage, is the same order of magnitude as the 10 mm gap the whole clash turns on. The vertical drop of B2 beneath B1 is taken as exactly one bar diameter, bar hard against bar. And 185 kN-m is a face-of-support moment: reading it at the centreline instead typically moves the demand by several percent, the same order as the 2.7% shortfall reported below, so the sign of that shortfall depends on where the moment was taken.
4/3 × 20 = 26.7 mm, so the governing value is the greatest of 25, 20 and 26.7 = 26.7 mm. The aggregate leg governs, not the 25 mm floor everybody remembers.
Know what that leg rests on before you lean on it. The four-thirds figure is the inverse of the rule limiting nominal maximum aggregate size to three quarters of the minimum clear spacing between bars, and that limit is expressly waivable: it does not apply where the licensed design professional judges that workability and methods of consolidation are such that the concrete can be placed without honeycombs or voids. A reviewer willing to sign that judgement takes 26.7 mm off the table and puts the 25 mm floor back, and the whole Step 2 conclusion moves with it. State the aggregate size on the drawing and state whether the waiver has been exercised.
Clear width inside the stirrups = 300 - 2(40) - 2(10) = 200 mm. Four bars take 4 × 20 = 80 mm, leaving 120 mm over three gaps: 120 ÷ 3 = 40.0 mm clear, and 40.0 ≥ 26.7, so four fit. Five would take 100 mm, leaving 100 mm over four gaps: 100 ÷ 4 = 25.0 mm, short of 26.7 by 1.7 mm, so five 20 mm bars do not fit. That 1.7 mm is the entire margin, and it is exactly what the waiver in Step 1 reaches.
Beam bar centres from the beam's side face: 40 + 10 + 10 = 60 mm, at a pitch of 40 + 20 = 60 mm, giving 60, 120, 180 and 240 mm (240 + 60 = 300 mm, the full width). B2 is centred on the column, so shift by (400 - 300) ÷ 2 = 50 mm: the centres become 110, 170, 230, 290. Column verticals sit 40 + 10 + 10 = 60 mm from each face, so 60 and 340, and the middle bar at (60 + 340) ÷ 2 = 200 mm lands on the column centreline, which is also the beam centreline. Take each bar as its centre ± 10 mm:
Ten millimetres, against a 20 mm stone: two of the four bars cannot pass. Four on the drawing, two at the column.
B1 is the girder and therefore primary: top bar centre 40 + 10 + 10 = 60 mm from the top, d = 500 - 60 = 440 mm as designed. B2's bars pass beneath in contact, centre 60 + 20 = 80 mm, so d = 500 - 80 = 420 mm. One bar diameter, every crossing.
One diameter is the optimistic figure, because it assumes bar hard against bar. A chair, a spacer, a tie wire crossing under the bar or a deformation lug seating on a lug all add to the drop, push d below 420 mm, and make the shortfall in Step 5 worse rather than better.
As = 4 × 314.16 = 1256.6 mm², As fy = 1256.6 × 415 = 521,506 N, and a = 521,506 ÷ (0.85 × 28 × 300) = 521,506 ÷ 7,140 = 73.04 mm, so a ÷ 2 = 36.52 mm. Tension control at the real depth: beta1 = 0.85 for fc' = 28 MPa, c = 73.04 ÷ 0.85 = 85.9 mm, strain = 0.003 × (420 - 85.9) ÷ 85.9 = 0.003 × 3.888 = 0.0117, above 0.005, so phi = 0.90.
Lever arm as designed = 440 - 36.52 = 403.48 mm; as built = 420 - 36.52 = 383.48 mm; ratio = 383.48 ÷ 403.48 = 0.9504, about a 5 percent loss. Mn = 521,506 × 403.48 = 210,417,000 N-mm = 210.4 kN-m, phiMn = 189.4 kN-m as designed; 521,506 × 383.48 = 199,987,000 N-mm = 200.0 kN-m, phiMn = 180.0 kN-m as built. Against Mu = 185 kN-m, short by 5.0 kN-m, or 2.7%.
2.7% is not a collapse, but it is not a rounding error and placing tolerance does not absorb it. Tolerances on effective depth are a band about the detailed position; this 20 mm sits underneath that band, and tolerance stacks on top.
The same four bars, tied as two bundles of two, clear of the middle vertical: bundle 1 at 110 and 130 (occupying 100 to 140), bundle 2 at 270 and 290 (260 to 300).
NSCP 2015 allows up to four bars in a bundle, tied in contact, enclosed by stirrups or ties, and treated as one bar of the equivalent diameter for spacing and cover. A two-bar bundle carries no development-length increase; a three-bar bundle takes 20 percent and a four-bar bundle 33 percent. Extra steel: none.
That equivalent diameter reaches the cover check as well as the spacing check: cover on a bundle is at least the equivalent diameter, and need not exceed 50 mm. Here the 40 mm cover beats the 28.3 mm equivalent, so Fix A passes on cover too. A reader carrying 25 mm cover, or stepping up to three-bar bundles, fails that leg silently, because no beam section drawing shows it.
What Fix A does not do is repair the capacity. The steel area is still 1256.6 mm² and the effective depth is still 420 mm, so phiMn stays at 180.0 kN-m against Mu = 185 kN-m: the 2.7% shortfall Step 5 just identified survives the bundling untouched. Fix A has to be combined with Fix C, or with a larger section.
It is also needed twice. B1 is likewise 300 × 500 mm with 4-20 mm top bars centred on the same 400 mm column, and the column carries a mid-face vertical on every face, so B1's bars 2 and 3 meet verticals with the same 10 mm gaps in the perpendicular direction. The clash exists in both directions and the fix is needed in both: bundling B2 alone leaves the girder exactly where it was.
2-32 instead of 4-20 gives As = 2 × 804.25 = 1608.5 mm², and 1608.5 ÷ 1256.6 = 1.28, so 28% more steel. Only B2 changes to 32 mm bars; B1 keeps its 4-20. Re-run Step 3: the bars occupy 100 to 132 and 268 to 300, leaving 190 - 132 = 58 mm to the middle vertical. The drop grows with the bar, but only the lower bar grew: B1's top bar still has its underside at 60 + 10 = 70 mm, so B2's 32 mm bars centre at 70 + 16 = 86 mm and d = 500 - 86 = 414 mm. Every check in this option runs at 414 mm.
The real catch is the joint. Where the beam longitudinal reinforcement extends through the joint, a special moment frame under NSCP 2015 sets a minimum joint dimension along those bars of 20 bar diameters for Grade 415, flat, with no increase for higher grades. A 400 mm column with 20 mm bars is exactly 20 × 20 = 400 mm, at the limit; 32 mm bars would need 640 mm. At an exterior joint, where the bars terminate in hooks or heads rather than passing through, that rule does not govern and anchorage length does. Deepening the column to 640 mm removes the special-moment-frame joint-dimension problem for Fix B, and the intermediate-frame joint provisions must be checked separately rather than assumed to follow from the special-frame result.
Give B2 d = 420 mm from the start. Required As for Mu = 185 kN-m, by one iteration: assume a = 75 mm, lever arm 420 - 37.5 = 382.5 mm, As = 185,000,000 ÷ (0.90 × 415 × 382.5) = 185,000,000 ÷ 142,864 = 1294.9 mm². Check: a = 1294.9 × 415 ÷ 7,140 = 537,383.5 ÷ 7,140 = 75.26 mm, close enough to stop. So about 1295 mm² is needed against 1256.6 supplied.
Confirm the phi that was assumed in solving for it. At 1295 mm² with d = 420 mm, c = 75.26 ÷ 0.85 = 88.5 mm and the strain = 0.003 × (420 - 88.5) ÷ 88.5 = 0.003 × 3.746 = 0.0112, well past 0.005, so phi = 0.90 holds and the area solved for stands.
2-25 plus 2-20 (2 × 490.87 + 2 × 314.16 = 1610.1 mm²) covers it, and covers it heavily: 1610.1 ÷ 1295 = 1.24, 24% above what the moment asks for. A tighter selection, 1-25 plus 3-20, gives 490.87 + 942.48 = 1433 mm², still clears 1295 mm² and costs materially less. The figure below is therefore a consequence of the bar selection, not of the deficiency being repaired.
Over a 2.0 m support-bar length (common practice, not a code figure): 2 × 2.0 × 3.85 + 2 × 2.0 × 2.47 = 15.4 + 9.9 = 25.3 kg against 4 × 2.0 × 2.47 = 19.8 kg, an extra 5.5 kg, about ₱265 per joint at an illustrative ₱48/kg.
Carry one inconsistency knowingly: a mixed layer shifts the centroid. With B1's bar underside at 70 mm, a 25 mm bar in B2 centres at 82.5 mm and a 20 mm bar at 80 mm, so the centroid of 2-25 plus 2-20 is (2 × 490.87 × 82.5 + 2 × 314.16 × 80) ÷ 1610.1 = 81.5 mm from the top, and the effective depth is about 418.5 mm, not the 420 mm the area was solved for. It is small here only because the selected area sits so far above the required area; trim the selection toward 1295 mm² and the slip has to be iterated out.
Fix C repairs the capacity, not the fit: one of the four bars still lands on the column centreline, so Fix A's geometry is needed as well.
Close it with a joint plan detail at each typical column type with the critical gaps dimensioned, a note nominating the primary direction, and a bending schedule carrying the bundles or revised sizes and the reduced d.
Re-running a capacity check at a corrected effective depth takes two minutes, and you will do it many times across one floor plate. The reinforced-concrete calculators on the RHCES web tools page take b, d, fc', fy and the bar layout and return phiMn, so the designed and the real d sit side by side.
Normally the girder: the direction with the larger support moment, or the deeper beam where depths differ. Where both are identical, pick the one where 20 mm hurts less and write it down. What matters is that somebody picks in an office, with the calculation in front of them.
The bundling provisions are general and are not switched off in seismic frames, and a two-bar bundle carries no development-length penalty. The open question is the special-moment-frame joint dimension, written in terms of bar diameter: a reviewer may judge the bundle on its equivalent diameter, 28.3 mm here, which would demand 20 × 28.3 = 566 mm of column. Ask before you detail it.
Recompute d from the centroid. With four bars at 80 mm from the top and a fifth at 80 + 10 + 25 + 10 = 125 mm, the centroid is (4 × 80 + 125) ÷ 5 = 445 ÷ 5 = 89 mm, so d = 411 mm, not 431 mm.