Given. Cantilever retaining wall stem 3.0 m tall (footing top to top of wall), silage pressure $L(H)=3.5+3.5H$ kPa acting over the full stem height on the silage side, backfill soil only 1.2 m deep on the grade side (Figure 6). $f'_c=35$ MPa, $f_y=400$ MPa, SLS allowable soil bearing $=150$ kPa.
Given data
Quantity
Value
Stem height, $H$
3.0 m
Silage pressure at top / base
3.5 / 14.0 kPa
Backfill depth over heel (grade side)
1.2 m
Concrete / rebar
$f'_c=35$ MPa, $f_y=400$ MPa
Allowable SLS bearing
150 kPa
Assumed unit weights
$\gamma_c=24$ kN/m$^3$, $\gamma_{soil}=18$ kN/m$^3$ (Check, not given)
Figure 6 — retaining wall cross-section with dimensions A–D and reinforcing E, F.
Approach. Compute the total factored silage thrust and its line of action, trial a footing geometry, check global stability (bearing, overturning, sliding), then design the stem for cantilever flexure and the footing toe/heel for their own local bending, all via the supplied $K_r$–$\rho$ table (Table 2.1, $f_y=400$ MPa).
Total silage thrust. The pressure diagram is a rectangle ($3.5$ kPa) plus a triangle ($3.5H$ kPa at the base), giving, per metre of wall:
$$P_{unf}=\tfrac12(3.5+14.0)(3.0)=26.25\text{ kN/m}, \qquad \bar y_{\text{above footing top}}=1.2\text{ m}$$
Treating the specified lateral pressure as a live/variable load (factor 1.5, check — the source gives no separate D/L split for this pressure):
$$\boxed{P_f=1.5(26.25)=39.375\text{ kN/m}}$$
Trial geometry and stability check. Trying footing thickness $A=400$ mm, toe $D=0.8$ m, stem thickness $C=250$ mm, heel $B=1.8$ m (total width $W=2.85$ m), the unfactored vertical load (self-weight of footing, stem, and soil over the heel, using the 1.2 m grade-side backfill depth) is $N=84.2$ kN/m. Summing moments about the toe:
$$\bar x = \frac{M_{stab}-P_{unf}\,\bar y_{base}}{N}, \qquad e = \frac{W}{2}-\bar x = 0.363\text{ m} < \frac{W}{6}=0.475\text{ m (middle third)}$$
so the base remains fully in compression, with
$$q_{max}=\frac{N}{W}\left(1+\frac{6e}{W}\right)=52.2\text{ kPa} < 150\text{ kPa (SLS)}\quad\checkmark, \qquad q_{min}=7.0\text{ kPa}\ge0\quad\checkmark$$
$$FS_{overturning}=3.13\ge1.5\quad\checkmark, \qquad FS_{sliding}=\frac{0.5N}{P_{unf}}=1.60\ge1.5\quad\checkmark\ (\mu=0.5\text{ assumed, check})$$
All three stability checks pass with this trial geometry, so it is adopted:
$$\boxed{A=400\text{ mm} \qquad B=1.8\text{ m} \qquad C=250\text{ mm} \qquad D=0.8\text{ m}}$$
Stem flexural design (vertical reinforcing, F). The full factored thrust acts on the stem alone with its resultant $1.2$ m above the footing top, giving a base moment
$$M_f=P_f\,\bar y=39.375(1.2)=47.25\text{ kN}\cdot\text{m/m}$$
With cover 50 mm and 20M bars, $d=250-50-10=190$ mm:
$$K_r=\frac{M_f\times10^6}{bd^2}=\frac{47.25\times10^6}{1000(190)^2}=1.31\text{ MPa}$$
Table 2.1 at $f'_c=35$ interpolates to $\rho=0.40\%$, giving $A_{s,req}=0.0040(1000)(190)=765$ mm$^2$/m — above the shrinkage/temperature floor of $0.002(1000)(250)=500$ mm$^2$/m, so flexure governs:
$$\boxed{F:\ 15M\ @\ 250\text{ mm o.c.}\ (A_s=800\text{ mm}^2/\text{m} \ge 765\text{ mm}^2/\text{m}),\text{ tension face on the silage side}}$$
A stem shear check at the base ($V_f=P_f=39.4$ kN/m against $V_c=0.21\phi_c\sqrt{f'_c}\,b\,d_v=145.4$ kN/m) confirms shear is not close to governing.
Wall horizontal reinforcing (E). With no horizontal spanning action assumed (the stem cantilevers vertically; horizontal steel is shrinkage/temperature steel per CSA A23.3 Cl. 7.8.1), $A_{s,min}=0.002(1000)(250)=500$ mm$^2$/m total, split between the two faces:
$$\boxed{E:\ 15M\ @\ 400\text{ mm o.c., each face}\ (500\text{ mm}^2/\text{m per face}, \text{exceeding the }250\text{ mm}^2/\text{m/face minimum})}$$
Footing toe and heel flexure. Using the bearing-pressure diagram (factored by the same 1.5 on the net soil reaction) less the footing/soil self-weight (factored 1.25), the net design moments at the stem face work out to $M_{f,toe}=19.2$ kN·m/m and $M_{f,heel}=32.8$ kN·m/m — both light enough (with $d=400-75-10=315$ mm at 75 mm cover for concrete cast near soil) that $K_r<0.5$ falls below Table 2.1's range and the minimum reinforcement ratio governs both:
$$A_{s,min,footing}=0.002(1000)(400)=800\text{ mm}^2/\text{m} \;\Rightarrow\; 15M\ @\ 250\text{ mm o.c.}$$
top steel in the heel, bottom steel in the toe.