How to Calculate Rebar in Retaining Wall

Read the design height H (top of stem to bottom of footing) and the retained soil unit weight γ (typical 120 pcf for granular backfill). Compute the active earth pressure coefficient Ka = (1 − sin φ) / (1 + sin φ); for φ = 30°, Ka = 0.33. The lateral load at the base is Pa = 0.5 × Ka × γ × H². This single number drives every downstream bar — skip it and the whole cage is mis-sized.

H=10 ; gamma=120 ; phi=30 ; ka=$(echo "scale=4; (1 - $(echo "scale=4; s($phi*3.14159/180)" | bc -l)) / (1 + $(echo "scale=4; s($phi*3.14159/180)" | bc -l))" | bc -l) ; Pa=$(echo "scale=1; 0.5*$ka*$gamma*$H*$H" | bc) ; echo "H=${H}ft | Ka=$ka | Pa=${Pa} lb/ft"

The stem acts as a vertical cantilever fixed at the footing. The design moment at the base is M = Pa × H / 3. Use Mu / (φ × d × fy) to estimate As per foot, then pick a bar size and spacing. The earth face takes the tension, so vertical bars land on the soil side. For a 10-ft wall at 1980 lb/ft, Mu ≈ 7920 lb·ft/ft → roughly #6 @ 12 in on the earth face.

Pa=1980 ; H=10 ; M=$(echo "scale=0; $Pa*$H/3" | bc) ; echo "M_base = ${M} lb-ft/ft" ; echo "As_est ≈ #6 @ 12in (earth face)"

Horizontal bars serve two roles: temperature/shrinkage control and distributing local pressure peaks. ACI 14.3.3 sets the minimum horizontal ratio at 0.0025 of the stem area for deformed bars ≤ #5 (Grade 60), and 0.0020 for #6 and larger. Place them on the exposed face, opposite the verticals. For a 12-in stem, As,min(horiz) = 0.0020 × 12 × 12 = 0.288 in²/ft → #5 @ 16 in.

t=12 ; rho_h=0.0020 ; As_min=$(echo "scale=3; $rho_h*12*$t" | bc) ; echo "As_min(horiz) = ${As_min} in2/ft -> #5 @ 16in"

The footing is a slab loaded upward by soil on the heel and downward on the toe. Heel steel (top mat, transverse bars) resists the heel cantilever moment from the backfill weight; toe steel (bottom mat, transverse bars) resists the toe cantilever from bearing pressure. Run main transverse bars perpendicular to the wall length; longitudinal distribution bars run parallel at ~18 in. The heel always needs more steel than the toe because soil weight dominates.

heel_L=4.0 ; toe_L=2.0 ; q_toe=2000 ; q_heel=500 ; M_heel=$(echo "scale=0; 0.5*120*$heel_L*$heel_L*12" | bc) ; echo "M_heel = ${M_heel} lb-in/ft -> #6 @ 9in" ; echo "M_toe  -> #6 @ 12in (lighter)"

Weep holes (3-in PVC at 8-ft spacing) drain hydrostatic pressure behind the stem — keep them clear of the vertical bar grid by offsetting bars 3 in around each hole. At the construction joint (typically at the top of the footing, where the stem is cast separately), leave the stem vertical bars projecting 40d–60d above the footing for lap splice. Clean and roughen the joint to 1/4-in amplitude before the stem pour, or the bond fails and the stem fixity is lost.

dia=0.75 ; lap=$(echo "$dia*40" | bc) ; echo "stem dowel projection = 40d = ${lap}in above footing" ; echo "weep hole = 3in PVC @ 8ft OC, offset bars 3in"

The stem moment peaks at the base and tapers parabolically to zero at the top — so the full vertical bar area is only needed near the base. Cut alternate bars at the height where the remaining bars still satisfy As ≥ Mu/(φ·d·fy). For a 10-ft wall, cut every other earth-face vertical at roughly 0.6H (6 ft) and the rest at 0.85H (8.5 ft). Extend every cut bar past the theoretical cutoff by the development length ld (≈ 30d for Grade 60 in 4000 psi).

H=10 ; cut1=$(echo "scale=1; 0.6*$H" | bc) ; cut2=$(echo "scale=1; 0.85*$H" | bc) ; ld=30 ; dia=0.75 ; echo "cut alternate bars at ${cut1}ft, extend +$(echo "$ld*$dia" | bc)in" ; echo "cut remaining at ${cut2}ft"

Confirm stem vertical ρ ≥ 0.0012 (min for walls, ACI 14.3.2), earth-face cover = 2 in (exposed to soil), exposed-face cover = 1.5 in, and footing cover = 3 in (against earth). Verify all stem dowel splices are Class B (1.3 × ld) unless staggered below 50%. Run the final As against Mu/(φ·d·fy); if provided As < 2× required As, the lap defaults to Class B.

cover_stem=2 ; cover_foot=3 ; class=B ; echo "cover: stem=${cover_stem}in | footing=${cover_foot}in | lap=Class ${class} (1.3*ld)"