How to Calculate Rebar in Odd Shapes

Decompose the odd shape into primitive rectangles and triangles, run the standard grid formula (n = length ÷ spacing + 1) on each unit, then deduplicate any bars that fall on a shared boundary between two units. For an L-shape split into a vertical leg (t × H) and a horizontal leg ((W − t) × t), the two rectangles share the corner column of bars — summing both counts double-counts that column, so subtract one shared set. The decomposition keeps the math on rectangles (where the +1 rule is exact) and avoids hand-fitting bars to an angled edge.

Split the L into two rectangles: a vertical leg t wide × H tall, and a horizontal leg (W − t) wide × t tall. Bars in each direction: vertical-leg vertical bars = ceil(t × 12 / s) + 1, horizontal-leg bottom bars = ceil((W − t) × 12 / s) + 1. The two rectangles share the corner line of bars where they meet — subtract that shared count once. For a 20×20 ft L with an 8 ft leg thickness at #4 @ 12 in: vertical-leg vertical bars = ceil(8×12/12)+1 = 9; horizontal-leg bars = ceil(12×12/12)+1 = 13; total ≈ 32 bars after dedup, vs 48 for a full 20×20 bounding box — a 33% saving.

Do NOT use the average width for every bar. Bar length steps linearly from the long base B1 to the short base B2. Bar count n = ceil(H × 12 / s) + 1; the i-th bar (from 0 to n−1) has length L_i = B1 − (B1 − B2) × (i / (n−1)). Sum the stepped lengths for total steel. For a trapezoid with B1 = 24 ft, B2 = 12 ft, H = 16 ft at #4 @ 12 in: n = ceil(16×12/12)+1 = 17 bars; lengths step from 24 ft down to 12 ft; total ≈ 306 ft. The average-width method (18 ft × 17) gives 306 ft here only when n is odd and the midpoint is exact — but at a 2:1 base ratio with even n it underestimates by ~12%, so always step.

For a cut corner (chamfer), extend the orthogonal grid to the bounding box and trim the two bars that cross the corner to the chamfer line — the offcuts become the corner bars themselves, so waste is near zero. For a curved (arc) edge, terminate the straight grid bars at the arc and add fan-shaped radial bars along the arc at the same spacing; each radial bar is a unique length read from the arc geometry. The rule: keep the orthogonal grid in the field (where bending is uniform) and only custom-fit the edge bars (where the geometry forces it). Never bend the entire grid to follow a curve — it triples labor for no structural gain.

Lay a standard orthogonal grid over the bounding rectangle of the polygon, then clip every bar to the polygon boundary. Bar count each way = ceil(bounding_dim × 12 / s) + 1, then discard the portion of each bar that falls outside the polygon. For a hexagon with 14 ft bounding radius at #4 @ 12 in: grid = 29 × 29 = 841 intersections; only ~65% fall inside the hexagon, so ~547 effective bars plus edge trims. Waste runs 18–25% on polygons, which is acceptable because labor stays low (straight runs, no custom bending). For polygons under 8 ft, switch to a radial layout to cut the waste.

Draw the odd shape to scale, then overlay the rebar grid as a hatch or array. AutoCAD's HATCH command reports the enclosed area instantly, and the ARRAYRECT command lays the grid so you can count bars that fall inside the boundary by selecting them. The workflow: (1) draw the outline, (2) ARRAYRECT the bar spacing in both directions over the bounding box, (3) TRIM bars to the outline, (4) LIST or QSELECT to count the surviving bars and sum their lengths. This eliminates the hand-decomposition error and gives an exact piece count for the fabricator's cut list in under 5 minutes per shape.