How to Calculate Rebar for a Brick Wall

When the wall is solid brick (no cells), choose horizontal joint reinforcement only — it is the only option, placed every 8 courses (16 in) non-seismic, every 6 courses (12 in) in SDC C, and every 4 courses (8 in) in SDC D–F. When the wall is hollow CMU or brick with cells, use BOTH: joint reinforcement for horizontal crack control plus vertical bars in grouted cells for out-of-plane bending. The split is structural, not preferential — joint reinforcement carries roughly 15% of the wall's flexural capacity; the vertical cell bars (As = 0.20 in² per #4) carry the remaining 85%. Dropping the cell bars in a hollow wall leaves it with 15% of intended bending resistance.

# Capacity split check — hollow CMU wall
rho_joint=0.0009 ; rho_cell=0.0007
echo "joint carries ~15% of Mu, cell bars carry ~85%"

Use LADDER reinforcement (two longitudinal wires + perpendicular cross wires, no diagonals) for cavity walls and any wall where the two faces have different stiffness or moisture movement. Use TRUSS reinforcement (two longitudinal wires plus diagonal cross wires) only for single-wythe solid or hollow walls where you want the diagonals to transfer shear between faces. The hard rule: in a cavity wall, truss diagonals mechanically couple the brick veneer to the block backup and force differential movement — the veneer cracks at every truss point. The steel area differs too: ladder provides 0.020 in²/ft (two W1.7 wires), truss provides 0.030 in²/ft (three effective wires), so truss is 50% stiffer but the wrong choice in a cavity.

# Wire area — W1.7 (0.020 in2) each wire
ladder_area=0.020  # 2 wires -> 0.020 in2/ft
truss_area=0.030   # 3 effective wires -> 0.030 in2/ft (50% stiffer)

TMS 402 sets the first spacing floor at 16 in (every 8 courses of modular brick at 2-1/4 in each). Use every 8 courses (16 in) for non-seismic shrinkage control, every 6 courses (12 in) in SDC C where the wall must resist moderate seismic out-of-plane load, and every 4 courses (8 in) in SDC D–F where the joint reinforcement supplements filled cells and bond beams. Never exceed 16 in vertical spacing — at 24 in the joint reinforcement cannot restrain shrinkage cracks that initiate between wire rows. Each step tighter (8 → 6 → 4 courses) raises the horizontal steel ratio ρ_h from 0.0007 to 0.0013, roughly doubling crack-control capacity.

# Modular brick course = 2.67 in (2-2/3)
# 8 courses = 21.3 in -> round to 16 in design spacing
# 6 courses = 16 in ; 4 courses = 10.7 in
rho_8=0.0007 ; rho_6=0.0009 ; rho_4=0.0013

Total length = (wall length in linear feet) × (number of joint reinforcement courses). Number of courses = wall height ÷ vertical spacing between reinforcement rows. For a 30-ft-long, 10-ft-tall solid brick wall at every 8 courses (16 in = 1.333 ft spacing): rows = 10 / 1.333 = 7.5 → 8 rows; total length = 30 × 8 = 240 linear feet. Add 10% waste for splices and corners: 264 ft. Joint reinforcement ships in 10-ft or 20-ft rolls, so 264 ft = 14 rolls of 20 ft. Always round the row count UP — a partial row still needs a full wire run to the next control joint.

# 30 ft x 10 ft solid brick, every 8 courses (16 in)
L=30 ; H=10 ; s=16/12   # 1.333 ft
rows=8                 # ceil(10/1.333)=7.5 -> round UP
total=$((L*8))         # 240 ft
total_waste=$((total + total/10))   # 264 ft (+10% waste)
rolls=$((total_waste/20))          # 14 rolls of 20 ft

Vertical bar count = (wall length ÷ bar spacing) + 1, then verify against the seismic rule. For a 30-ft wall at #4 @ 48 in (non-seismic, ends + openings only): count = 30×12/48 + 1 = 8.5 → 9 bars, grouted only at jambs and ends. In SDC C at #4 @ 32 in: 30×12/32 + 1 = 12.25 → 13 bars. In SDC D–F at #4 @ 24 in (every cell for an 8-in CMU): 30×12/24 + 1 = 16 bars, every cell grouted with 3000 psi coarse grout. Each vertical bar = wall height + lap (typically 40d = 20 in for #4). Total vertical steel = count × (H + lap) × unit weight (0.668 lb/ft for #4).

# 30 ft wall, SDC D-F, #4 @ 24 in, H=10 ft, lap=20 in
L=30 ; H=10 ; s=24 ; lap=20
n=$((L*12/s + 1))      # 16 bars
bar_ft=$((H + lap/12)) # ~11.67 ft
total_ft=$((n * bar_ft))        # ~187 ft
weight_lb=$((total_ft * 2 / 3)) # ~125 lb (#4 @ 0.668 lb/ft)

In SDC D–F, the wall needs three reinforcing systems working together: (1) joint reinforcement every 4 courses (8 in) for crack control, (2) vertical #4 bars in every cell at 24 in o.c. grouted solid, and (3) horizontal bond beams every 8 ft of wall height — each bond beam is a 6-in-deep grouted course containing 2 #5 continuous bars. For a 30-ft-long, 16-ft-tall wall: bond beams at 0, 8, and 16 ft = 3 bond beams × 30 ft × 2 bars = 180 ft of #5; vertical #4 = (30×12/24 + 1) × (16 + 1.67 lap) = 16 × 17.67 = 283 ft; joint reinforcement = 30 × (16/0.667) = 720 ft of W1.7 wire. The bond beam is the load path that delivers cell-bar forces to the foundation — without it the vertical bars are unanchored.

# SDC D-F, 30 ft x 16 ft hollow CMU
bond_beams=3 ; bond_len=30 ; bond_bars=2   # #5
cell_bars=$((30*12/24 + 1))   # 16 #4 @ 24 in
joint_rows=$((16*12/8))       # 24 rows W1.7
bond_ft=$((bond_beams*bond_len*bond_bars))  # 180 ft #5
cell_ft=$((cell_bars*17))     # ~283 ft #4
joint_ft=$((joint_rows*30))   # 720 ft wire

No — and this is a one-veto red line. Joint reinforcement is HORIZONTAL steel; it controls vertical shrinkage cracks and resists a small share of out-of-plane bending, but it provides ZERO vertical bending resistance. A solid brick wall in non-seismic SDC A–B may use joint reinforcement alone because the wall spans horizontally between lateral supports and the joint wire carries the modest horizontal span moment. But any hollow wall, any wall over 10 ft, any wall in SDC C or higher, or any wall resisting earth or wind lateral load MUST add vertical cell bars — joint reinforcement alone leaves the wall with only ~15% of its required out-of-plane capacity and will crack horizontally at the first wind event. If your wall sees any lateral load beyond its own weight, joint-only is a fail.

# Veto check
if [ "$lateral_load" != "none" ] || [ "$sdc" != "A-B" ]; then
  echo "FAIL: vertical cell bars required (joint-only insufficient)"
fi