How is the thrust composed for this wall?

Surcharge rectangle Ka q H plus soil triangle 0.5 Ka gamma H^2

Surcharge triangle plus soil rectangle

Why does the wall fail in overturning?

Resisting moment 368.75 is less than overturning moment 390 (FS = 0.95)

x_bar is negative (-0.094 m). What does this mean?

Resultant lies outside the base (beyond the toe), so the base cannot carry it

2078 Bhadra · 10 marks

Stability of wall with surcharge 25 kN/m²

Short answer

FS against overturning = 0.95, sliding = 0.48; resultant lies outside the base, causing tension. The wall is unsafe; redesign (e.g. wider base, shear key) is needed.

Check the stability of the retaining wall shown in figure below. Assume necessary conditions and take ultimate bearing capacity of the foundation soil as $250\ \text{kN/m}^2$ . [Figure: concrete gravity wall, $\gamma_c = 25\ \text{kN/m}^3$ , top width 0.5 m, height $H = 6.0$ m, base width 2.5 m, base friction $\delta_b = 20^\circ$ ; backfill $\gamma_t = 20\ \text{kN/m}^3$ , $\phi' = 30^\circ$ , $\delta_b = 20^\circ$ , with a surcharge $q = 25\ \text{kN/m}^2$ on the backfill.]

Assumptions (figure not drawn here): the back of the wall is vertical and the sloping face is on the front, so the section is a rectangle $0.5\times6$ m plus a triangle $2.0\times6$ m; Rankine's theory with horizontal backfill and a smooth back is used (conservative, ignores wall friction); passive resistance in front of the toe is neglected; base friction $\delta_b = 20^\circ$ ; $q_u = 250$ kN/m $^2$ with $FS \ge 3$ .

1. Active thrust

K_a = \tan^2(45^\circ-15^\circ) = 0.3333

Surcharge: $P_1 = K_a q H = 0.3333\times25\times6 = 50.0$ kN/m at 3.0 m.
Soil: $P_2 = \tfrac12 K_a\gamma H^2 = 0.5\times0.3333\times20\times36 = 120.0$ kN/m at 2.0 m.
$\sum H = 170.0$ kN/m; $M_o = 50\times3+120\times2 = 390.0$ kN m/m.

2. Wall weight (about the toe)

Part	Weight (kN/m)	Arm (m)	Moment
Rectangle 0.5 x 6 x 25	75.0	2.250	168.75
Triangle 0.5 x 2.0 x 6 x 25	150.0	1.333	200.00
Total	225.0		368.75

3. Checks

Overturning: $FS_{ot} = 368.75/390.0 = 0.95$ . This is less than 2 (even less than 1); the wall overturns.
Sliding: $FS_s = 225\times\tan20^\circ/170.0 = 0.48$ , less than 1.5; the wall slides.
Eccentricity: $\bar x = (368.75-390.0)/225 = -0.094$ m, so the resultant lies outside the base (negative), giving tension at the heel; $e = 1.34$ m $> B/6 = 0.42$ m.
Bearing: the resultant is beyond the toe, so the toe pressure is not valid and the available bearing area is lost. The pressure at the toe exceeds $q_u/3 = 83$ kN/m $^2$ .

Answer: the wall is unsafe against overturning ( $FS = 0.95$ ), sliding ( $FS = 0.48$ ) and bearing. Redesign: increase the base width. With the same section shape, $FS_{ot} = 2$ needs $B \approx 3.7$ m, and a shear key or wider base (about $0.6H$ to $0.7H$ for a gravity wall with surcharge) is needed for sliding; pressure at the toe must then be rechecked against $q_u/3$ .

✓ Answer checked by Claude Sonnet 5.5

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Stability of wall with surcharge 25 kN/m²

1. Active thrust

2. Wall weight (about the toe)

3. Checks

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