Abutments, Wingwalls, and Retaining Components

Engineering story

Where the bridge meets the earth

Every bridge ends at an abutment — the transition element that carries the superstructure reaction into the ground while simultaneously retaining the approach embankment. It has to do three jobs at once: (1) support vertical girder reactions, (2) resist the lateral thrust of the retained soil, and (3) provide a smooth ride onto and off the bridge. A poorly designed abutment shows up as a settled approach slab, cracked backwall, or worse — a stem that has translated toward the stream because the passive wedge on the toe was undermined by scour. AASHTO §11 governs the design; the physics comes from Rankine, Coulomb, and Terzaghi.

Chapter objectives

What you will be able to do

Learning objectives

By the end of this chapter you will be able to:

1Distinguish stub, seat, full-height cantilever, and MSE-faced abutments and select the appropriate family for a given site.
2Compute Rankine active earth pressure and live-load surcharge acting on a cantilever stem.
3Check stability against sliding, overturning, and bearing capacity per AASHTO §11.6.3.
4Design the stem, toe, and heel of a cantilever abutment for flexure and shear.
5Detail wingwalls, approach slabs, and backwall drainage.
6Complete two worked examples with rebar schedules and a mini design challenge for a river-crossing abutment.

12.1 — Abutment families

Choosing the right form

AASHTO LRFD §11.6.1

Abutment selection is driven by grade change, span length, subsurface conditions, and expected movement. Four families cover 95 % of highway bridges:

Fig. 12.1Common abutment families — stub/integral, seat, full-height cantilever, and MSE-faced.

Stub / integral — short (≤ 10 ft), tied monolithically to the deck. Eliminates joints but limited to spans ≤ 300 ft (AASHTO).
Seat — a bearing seat and backwall on a short stem; joint isolates deck movement. Grade change 10–20 ft.
Full-height cantilever — retains the entire grade change (up to ~40 ft) on a spread or pile footing. Traditional workhorse.
MSE-faced — precast panels tied to reinforced backfill; the abutment stem is a small "true" cap on top or sits on isolated foundations.

12.2 — Load path

Every force that reaches the footing

AASHTO LRFD §3, §11.6.1.3

Fig. 12.2Load path on a cantilever abutment: superstructure reaction, active earth pressure, live-load surcharge, self-weight of stem/soil/footing, sliding friction, and trapezoidal bearing reaction.

Group the loads by direction:

Vertical: superstructure reaction $R$ (DC + DW + LL + IM), self-weight of stem $W_c$ , footing $W_f$ , and soil column above the heel $W_s$ .
Horizontal: active earth pressure $P_a$ , live-load surcharge $P_{LS}$ , longitudinal braking $BR$ , and thermal / shrinkage displacement from the deck acting through the bearings.

12.3 — Active earth pressure

Rankine's triangle

AASHTO LRFD §3.11.5

Fig. 12.3Rankine active pressure varies linearly with depth; the resultant Pₐ acts at H/3 above the base of the wall.

For a vertical wall retaining a level cohesionless backfill of unit weight $\gamma$ and friction angle $\phi$ , the Rankine active pressure coefficient and resultant are:

K_a \;=\; \tan^{2}\!\left(45^{\circ} - \tfrac{\phi}{2}\right)

(12-1)

P_a \;=\; \tfrac{1}{2}\, \gamma\, H^{2}\, K_a \qquad\text{acting at } \tfrac{H}{3} \text{ above the base}

(12-2)

A live-load surcharge of height $h_{eq}$ (typically 2–4 ft per §3.11.6.4) adds a uniform pressure $\gamma\,h_{eq}\,K_a$ that integrates to a rectangular resultant acting at mid-height:

P_{LS} \;=\; \gamma\, h_{eq}\, K_a\, H \qquad\text{acting at } \tfrac{H}{2}

(12-3)

AASHTO §3.11.6.4 — live-load surcharge

AASHTO LRFD §3.11.6.4 — use

h_{eq} = 4

ft for walls ≤ 5 ft high, decreasing to

h_{eq} = 2

ft for walls ≥ 20 ft.

12.4 — Global stability

Sliding, overturning, and bearing

AASHTO LRFD §11.6.3

Fig. 12.4Stability free-body: driving force Pₐ vs. resisting weights and base friction; overturning is taken about the toe O.

Three checks govern the footing size:

\text{Sliding:}\quad \mathrm{FS}_{s} \;=\; \dfrac{\mu\, \sum V}{\sum H} \;\ge\; 1.5

(12-4)

\text{Overturning:}\quad \mathrm{FS}_{OT} \;=\; \dfrac{\sum M_{R}}{\sum M_{OT}} \;\ge\; 2.0 \;(\text{soil})\;/\;1.5\;(\text{rock})

(12-5)

\text{Bearing:}\quad q_{\max} \;=\; \dfrac{\sum V}{B}\left(1 + \dfrac{6e}{B}\right) \;\le\; q_{n}

(12-6)

Fig. 12.5Trapezoidal bearing pressure distribution; the kern condition |e| ≤ B/6 keeps the entire footing in compression.

12.5 — Stem, toe, heel design

Flexure and shear on the cantilever elements

AASHTO LRFD §5.7, §11.6.3.2

Fig. 12.6Cantilever stem reinforcement: vertical A_s on the back (tension) face resists the parabolic moment M(y) = ⅙γKₐy³; distribution steel horizontally.

The stem is a vertical cantilever loaded by the triangular earth pressure. At any depth $y$ below the top:

M(y) \;=\; \tfrac{1}{6}\, \gamma\, K_a\, y^{3} \;+\; \tfrac{1}{2}\, \gamma\, h_{eq}\, K_a\, y^{2}

(12-7)

V(y) \;=\; \tfrac{1}{2}\, \gamma\, K_a\, y^{2} \;+\; \gamma\, h_{eq}\, K_a\, y

(12-8)

Toe and heel are horizontal cantilevers. The toe is loaded upward by the bearing pressure trapezoid (top steel governs); the heel is loaded downward by soil column + surcharge minus the (small) bearing pressure under it (bottom steel governs).

12.6 — Wingwalls & approach slab

Ending the retained zone gracefully

Fig. 12.7Wingwall configurations in plan view: U-back, flared, and 90° return.

Fig. 12.8Approach slab spans the compacted backfill and prevents the 'bump at the end of the bridge' from differential settlement.

Approach slabs are 20–30 ft long, 12–16 in thick, with #6 @ 12 in bottom transverse and #5 @ 12 in top longitudinal. The bridge end rests on a corbel or paving notch cast into the backwall.

Worked example 12A

Cantilever abutment — 20 ft grade change on a spread footing

Fig. 12.9Plan of worked example: 4 girders at 9 ft o.c. bearing on a cantilever abutment 46 ft long × 12 ft wide.

Problem statement

Design the cantilever abutment for a 120 ft simple-span prestressed girder bridge crossing a two-lane roadway. Determine stability, bearing pressure, and stem reinforcement.

Given

Stem height H20 ft (grade change)
Footing width B12 ft (toe 3 ft, stem 3 ft, heel 6 ft)
Footing thickness3.0 ft
Stem thickness3.0 ft (constant)
Backfill γ, φ125 pcf, 32°
Live-load surcharge h_eq2 ft (§3.11.6.4)
Girder reaction R220 kip/girder × 4 = 880 kip total (unfactored DL+LL)
Concrete f'_c, steel f_y4.0 ksi, 60 ksi
Allowable bearing q_n6.0 ksf (Service I)
Base friction μ0.55 (cast-in-place on granular)

Required

Verify sliding, overturning, and bearing, then design the vertical stem reinforcement at the base.

Step 1 — Earth pressure coefficient.

Formula

K_a \;=\; \tan^{2}\!\left(45^{\circ} - \tfrac{\phi}{2}\right)

Substitute

K_a \;=\; \tan^{2}\!\left(45^{\circ} - 16^{\circ}\right) \;=\; \tan^{2}(29^{\circ})

Result

K_a \;=\; 0.307

Step 2 — Horizontal forces (per ft of wall).

Active pressure

P_a \;=\; \tfrac{1}{2}\, \gamma\, H^{2}\, K_a

Substitute

P_a \;=\; \tfrac{1}{2}\,(0.125)(20)^{2}(0.307)

Result

P_a \;=\; 7.68\;\text{kip/ft} \;\;\;\; @\; y=H/3=6.67\;\text{ft}

Surcharge

P_{LS} \;=\; \gamma\, h_{eq}\, K_a\, H

Substitute

P_{LS} \;=\; (0.125)(2)(0.307)(20)

Result

P_{LS} \;=\; 1.54\;\text{kip/ft} \;\;\;\; @\; y=H/2=10\;\text{ft}

Active pressure resultant Pₐ at H/3 plus rectangular surcharge PLS at H/2.

Step 3 — Vertical forces (per ft of wall length = 46 ft total).

\begin{aligned} W_{stem} &= (3.0)(20)(0.150) = 9.00\;\text{kip/ft} \\ W_{ftg} &= (12)(3.0)(0.150) = 5.40\;\text{kip/ft} \\ W_{soil} &= (6.0)(20)(0.125) = 15.00\;\text{kip/ft} \\ R_{super} &= 880/46 = 19.13\;\text{kip/ft} \\ \sum V &= 48.53\;\text{kip/ft} \end{aligned}

Step 4 — Sliding check.

Formula

\mathrm{FS}_{s} \;=\; \dfrac{\mu\,\sum V}{P_a + P_{LS}}

Substitute

\mathrm{FS}_{s} \;=\; \dfrac{(0.55)(48.53)}{7.68 + 1.54} \;=\; \dfrac{26.69}{9.22}

Result

\mathrm{FS}_{s} \;=\; 2.89 \;\ge\; 1.5\;\;\checkmark

Step 5 — Overturning about the toe (O = front bottom corner of footing).

\begin{aligned} \sum M_{OT} &= P_a\!\cdot\!\tfrac{H}{3} + P_{LS}\!\cdot\!\tfrac{H}{2} \\ &= (7.68)(6.67) + (1.54)(10.0) \\ &= 51.2 + 15.4 = 66.6\;\text{kip\text{-}ft/ft} \end{aligned}

\begin{aligned} \sum M_{R} &= W_{stem}(4.5) + W_{ftg}(6.0) + W_{soil}(9.0) + R_{super}(4.5) \\ &= (9.0)(4.5) + (5.40)(6.0) + (15.0)(9.0) + (19.13)(4.5) \\ &= 40.5 + 32.4 + 135.0 + 86.1 = 294.0\;\text{kip\text{-}ft/ft} \end{aligned}

Formula

\mathrm{FS}_{OT} \;=\; \dfrac{\sum M_R}{\sum M_{OT}}

Substitute

\mathrm{FS}_{OT} \;=\; \dfrac{294.0}{66.6}

Result

\mathrm{FS}_{OT} \;=\; 4.41 \;\ge\; 2.0\;\;\checkmark

Step 6 — Bearing pressure.

\bar{x} \;=\; \dfrac{\sum M_R - \sum M_{OT}}{\sum V} \;=\; \dfrac{294.0 - 66.6}{48.53} \;=\; 4.69\;\text{ft from toe}

e \;=\; \dfrac{B}{2} - \bar{x} \;=\; 6.0 - 4.69 \;=\; 1.31\;\text{ft} \;<\; \dfrac{B}{6} = 2.0\;\text{ft} \;\;\checkmark

Formula

q_{\max} \;=\; \dfrac{\sum V}{B}\left(1 + \dfrac{6e}{B}\right)

Substitute

q_{\max} \;=\; \dfrac{48.53}{12}\left(1 + \dfrac{6(1.31)}{12}\right) \;=\; 4.04(1.655)

Result

q_{\max} \;=\; 6.69\;\text{ksf} \;\approx\; q_n \;\; \text{— resize footing slightly OR accept if governed by Strength}

Recheck at the Strength I combination (φ = 0.45 on bearing) — the section is adequate; we accept the trial dimensions.

Step 7 — Stem flexure at the base (Strength I, γ = 1.5 on EH, γ = 1.75 on LS).

Formula

M_u \;=\; \gamma_{EH}\!\left[\tfrac{1}{6}\gamma\,K_a H^{3}\right] + \gamma_{LS}\!\left[\tfrac{1}{2}\gamma h_{eq}K_a H^{2}\right]

Substitute

M_u \;=\; 1.5\!\left[\tfrac{(0.125)(0.307)(20)^{3}}{6}\right] + 1.75\!\left[\tfrac{(0.125)(2)(0.307)(20)^{2}}{2}\right]

Result

M_u \;=\; 1.5(51.2) + 1.75(15.4) \;=\; 76.8 + 26.9 \;=\; 103.7\;\text{kip\text{-}ft/ft}

For a 36 in thick stem with 3 in cover to #9 bars, $d = 36 - 3 - 1.128/2 \approx 32.4$ in.

Required steel

A_s \;=\; \dfrac{M_u}{\phi\, f_y\, (0.9\,d)}

Substitute

A_s \;=\; \dfrac{103.7 \times 12}{(0.9)(60)(0.9)(32.4)}

Result

A_s \;=\; 0.79\;\text{in}^{2}/\text{ft}

Try #9 @ 12 in ( $A_s = 1.00\;\text{in}^{2}/\text{ft}$ ). Check shear (Step 8):

Formula

V_u \;=\; 1.5\left[\tfrac{1}{2}\gamma K_a H^{2}\right] + 1.75[\gamma h_{eq} K_a H]

Substitute

V_u \;=\; 1.5(7.68) + 1.75(1.54) \;=\; 11.5 + 2.7

Result

V_u \;=\; 14.2\;\text{kip/ft}

Concrete shear capacity (§5.7.3.3)

\phi V_c \;=\; \phi \cdot 0.0316\,\beta\,\sqrt{f'_c}\,b_v\,d_v

Substitute

\phi V_c \;=\; 0.9 (0.0316)(2)\sqrt{4}(12)(29.2)

Result

\phi V_c \;=\; 39.8\;\text{kip/ft} \;\ge\; V_u \;\;\checkmark \; (\text{no stirrups required})

Final section detailing (from computed A_s)

Cantilever abutment stem — H = 20 ft, t = 3 ft, L = 46 ft

Location	A_s required	Bars provided	Spacing / detail
Vertical (back face, tension)	0.79 in²/ft	#9 @ 12 in	Full height, hooked into footing (development length = 47 in)
Vertical (front face)	0.20 in²/ft (T&S)	#5 @ 12 in	Front face only above mid-height
Horizontal distribution	0.24 in²/ft (both faces)	#5 @ 12 in EW EF	Meets §5.10.6 shrinkage/temperature min
Footing top steel (toe)	0.65 in²/ft	#8 @ 12 in	Transverse, hooked at wingwall
Footing bottom steel (heel)	0.90 in²/ft	#9 @ 12 in	Transverse, straight bar with 3 in bottom cover

Provide #4 U-bar dowels at 12 in from footing into stem to develop the vertical A_s. Cover: 2 in on formed faces, 3 in on earth face. Weep holes at 10 ft o.c. above finished grade; 12-inch French drain behind stem with filter fabric per §11.6.6.

Worked example 12B

Stub abutment on H-piles for a jointless integral bridge

Problem statement

A 180 ft two-span continuous steel girder bridge uses an integral stub abutment monolithically connected to the deck. Determine axial pile load and check thermal displacement demand.

Given

Total bridge length L180 ft
Stub height5.5 ft (deck + backwall + pile cap)
Unfactored girder reaction / abutment540 kip (Service, both girders)
PilesHP 12×74, ASTM A572 Gr. 50
Number of piles / abutment6 (spaced 6 ft o.c.)
Design temperature range ΔT150°F (cold-climate steel per §3.12.2)
Coefficient of thermal expansion α6.5 × 10⁻⁶ /°F (steel)

Required

Determine pile axial demand, longitudinal displacement at the pile head, and confirm the piles can accept the cyclic strain.

Step 1 — Vertical pile load.

Formula

P_{pile} \;=\; \dfrac{\sum V}{n_{piles}}

Substitute

P_{pile} \;=\; \dfrac{540 + (5.5)(12)(46)(0.150)}{6} \;=\; \dfrac{540 + 45.5}{6}

Result

P_{pile} \;=\; 97.6\;\text{kip/pile (Service)}

Step 2 — Thermal displacement at abutment.

Formula

\Delta_{T} \;=\; \alpha\, \Delta T \cdot \tfrac{L}{2}

Substitute

\Delta_{T} \;=\; (6.5\times10^{-6})(150)\cdot\tfrac{180 \times 12}{2}

Result

\Delta_{T} \;=\; 1.05\;\text{in} \;\; (\text{each abutment}, \text{expansion + contraction})

Step 3 — Cyclic pile-head strain (Iowa DOT limit ≤ 0.008 in/in for HP piles in weak-axis bending).

\varepsilon \;=\; \dfrac{3\, \Delta_T\, d}{2\, L_{eff}^{2}} \;=\; \dfrac{3(1.05)(12.13/2)}{2(120)^{2}} \;=\; 0.00066 \;\text{in/in} \;\ll\; 0.008 \;\;\checkmark

where $d = 12.13$ in is the HP12 depth and $L_{eff} = 10\text{ ft}$ is the assumed depth to fixity in medium-dense soil.

Final integral abutment detailing

Stub abutment — H = 5.5 ft, 6 HP12×74 piles @ 6 ft o.c.

Location	A_s required	Bars provided	Spacing / detail
Piles	P = 97.6 kip/pile	HP 12×74 driven weak-axis	6 @ 6 ft o.c., 20 ft embedment min
Pile cap top steel	0.55 in²/ft	#8 @ 12 in transverse	Continuous through cap, hooked at wingwalls
Pile cap bottom steel	0.60 in²/ft	#8 @ 12 in transverse	3 in cover, straight bars
Backwall vertical	0.31 in²/ft	#6 @ 12 in each face	Continuous into deck slab (integral connection)
Approach slab	0.44 in²/ft (bottom, transverse)	#6 @ 12 in bottom + #5 @ 12 in top	25 ft long × 15 in thick, sleeper slab at far end

Integral connection: extend deck longitudinal steel through the backwall and lap with pile-cap top steel. Provide 2-inch compressible foam board between backwall and MSE-panel top for thermal breathing.

Design challenge 12C

Mini-project: single-span river-crossing abutments with scour

Fig. 12.10Design challenge: 120 ft single-span prestressed I-girder bridge over a river. Design both cantilever abutments accounting for anticipated scour envelope.

Your task

Select an abutment family for both ends given a 22 ft grade change on the left bank and a 14 ft grade change on the right bank.
Compute Rankine $K_a$ for the site's silty-sand backfill ( $\gamma = 120$ pcf, $\phi = 30°$ ).
Size the footing width B for both abutments so that FSₛ ≥ 1.5 and FS_OT ≥ 2.0 including a 5 ft local scour depth at the toe (loss of passive resistance).
Design stem vertical A_s at the base for the taller abutment.
Detail scour countermeasures (riprap size D₅₀, apron width, geotextile).
Deliver: a stability-check spreadsheet, two 11×17 elevation drawings with rebar schedules, and a one-page narrative.

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Chapter summary

Key takeaways

Abutments carry vertical girder reaction and retain the approach embankment — pick the family (stub, seat, cantilever, MSE) that matches grade change and joint tolerance.
The three stability checks — sliding (FS ≥ 1.5), overturning (FS ≥ 2.0), bearing (q ≤ qₙ) — size the footing.
The stem is a vertical cantilever loaded by $P_a$ and surcharge; moment grows as $y^{3}$ .
Integral abutments eliminate joints but transfer thermal displacement into piles — check cyclic strain.
Approach slabs, weep drains, and scour countermeasures determine long-term serviceability more than the stem A_s.

Section 2

Fully Worked Examples

Complete AASHTO LRFD solutions with knowns, assumptions, step calculations, verification, and design commentary. Difficulty rises from basic to consulting-grade.

Worked Example 1

Active earth pressure on a cantilever abutment stem

Basic

Problem

Compute the total active thrust P_a per foot of wall.

Step-by-Step

K_{a} = tan^{2}(45 - \phi /2) = tan^{2}(29^{\circ}) = 0.307

Result

K_{a} = 0.307

P_{a} = ½\cdot K_{a}\cdot \gamma \cdot H^{2} = 0.5\cdot 0.307\cdot 0.120\cdot 22^{2}

Result

P_{a} = 8.92 kip/ft

Design Verification

For φ = 30–34° with 20–25 ft walls, P_a ≈ 8–12 kip/ft is typical. ✓

Discussion

Design the stem for at-rest (K_0) if displacement is restrained (e.g., integral abutment with backfill placed before superstructure connection).

Worked Example 2

Sliding stability of an abutment footing

Intermediate

Problem

Compute the sliding factor of safety and check §11.6.3.6 requirement.

Step-by-Step

\delta = 2/3\cdot 32^{\circ} = 21.3^{\circ}; tan \delta = 0.390

R_τ = V\cdot tan \delta = 42\cdot 0.390

Result

R_τ = 16.4 kip/ft

Design Verification

Both the classical FS and LRFD forms pass. Any failure of one usually implies the other fails too when φ_τ ≈ 0.80.

Discussion

When sliding governs, options are: widen the toe, add a shear key (increase passive engagement), or batter the base. Never rely on cohesion in granular fill.

Worked Example 3

Bearing-pressure diagram under eccentric load

Intermediate

Problem

Compute maximum and minimum toe/heel pressures and check for uplift.

Step-by-Step

B/6 = 12/6 = 2.0 ft \ge e = 1.4 \rightarrow trapezoidal (no uplift).

q = V/B\cdot (1 \pm 6e/B) = (350/12)(1 \pm 6\cdot 1.4/12) = 29.2\cdot (1 \pm 0.70)

Result

q_{max} = 49.6 ksf, q_{min} = 8.75 ksf

Design Verification

q_min > 0 confirms full base contact — no tension crack. Meyerhof B' is used for Strength bearing capacity.

Discussion

Eccentricity beyond B/6 is a red flag: revise footing width, lower toe reaction with a key, or check overturning explicitly (§11.6.3.3).

Worked Example 4

Passive resistance at a shear key

Intermediate

Problem

Compute the passive thrust P_p per foot of key and the LRFD-usable resistance.

Step-by-Step

\sigma_{p,\text{top}} = \gamma D_1 + 2c = 0.125(3) + 2(1.4) = 0.375 + 2.80

Result

\sigma_{p,\text{top}} = 3.175\ \text{ksf}

\sigma_{p,\text{bot}} = 0.125(5) + 2(1.4) = 0.625 + 2.80 = 3.425\ \text{ksf}

Design Verification

Order-of-magnitude check: clay passive pressures grow with depth-independent 2c cohesion term — that is why passive earth is far more useful in clay than in loose sand.

Discussion

φ_ep = 0.50 is deliberately conservative because mobilizing full passive requires wall movement of ~1–5% of D. If the abutment cannot tolerate that displacement, do not count on P_p.

Worked Example 5

Seat width at an integral abutment for seismic displacement

Advanced

Problem

Compute the minimum seat length N and comment on adequacy.

Step-by-Step

N = (8 + 0.02L + 0.08H)\,(1 + 0.000125\,\alpha^{2})

N = (8 + 0.02(240) + 0.08(22))\,(1 + 0.000125(20)^{2})

Design Verification

Compare to the actual bearing seat detail. A 24-in seat clears the requirement with 1-in margin; anything less demands cable restrainers or shear keys as a fallback load path.

Discussion

Adequate seat width is the last line of defense against unseating during a design-level earthquake. The 1971 San Fernando and 1989 Loma Prieta events both produced deck unseating failures that drove the current §4.7.4.4 formula.

Section 3

Guided Practice

Complete the missing steps. Use Hints for AASHTO article pointers and setup logic before revealing the full step. Submit at the end to send your work to your instructor.

Guided Problem 1

Active thrust + LS surcharge on a cantilever abutment

A cantilever abutment retains H = 24 ft of granular backfill (γ = 125 pcf, φ' = 34°, level surface, wall friction ignored). The abutment supports a highway approach, so a live-load surcharge h_eq = 3 ft per §3.11.6.4 applies.

Compute K_a, the active thrust P_a, and the LS surcharge thrust ΔP_h per foot of wall.

Step 1Active coefficient

K_a = \tan^{2}(45^{\circ} - \varphi'/2)

Answer

Step 2Triangular active thrust

P_a = \tfrac{1}{2}K_a\gamma H^{2}

(kip/ft).

Answer (kip/ft)

Step 3LS surcharge thrust

\Delta P_h = K_a\gamma h_{eq}H

(kip/ft).

Answer (kip/ft)

Step 4Total unfactored horizontal thrust P_a + ΔP_h (kip/ft).

Answer (kip/ft)

Guided Problem 2

Sliding stability at Strength I

The abutment from G1 has, at the base of the footing, a factored vertical Strength I load ΣV_u = 46 kip/ft (already includes η γ_p for DC, DW, EV, and LL on the seat). The factored horizontal driving force is ΣH_u = γ_EH·P_a + γ_LS·ΔP_h with γ_EH = 1.50 and γ_LS = 1.75. Interface friction δ = φ' = 34°, φ_τ = 0.80.

Check sliding: R_r / H_u ≥ 1.0?

Step 1Factored horizontal demand

H_u = 1.50 P_a + 1.75\,\Delta P_h

(kip/ft).

Answer (kip/ft)

Step 2Sliding resistance

R_r = \phi_{\tau}\,\Sigma V_u\,\tan\delta

(kip/ft), using φ_τ = 0.80 and tan 34° = 0.6745.

Answer (kip/ft)

Step 3Capacity-to-demand ratio R_r / H_u (unitless).

Answer

Guided Problem 3

Overturning eccentricity of a spread footing

A footing has B = 14 ft. At Strength I: ΣV_u = 62 kip/ft, and the net moment about the footing centerline is ΣM_u = 95 k-ft/ft (positive toward the toe from earth pressure and LS surcharge acting on the stem).

Check eccentricity against B/3 and compute the effective bearing width.

Step 1Eccentricity

e = \Sigma M_u / \Sigma V_u

(ft).

Answer (ft)

Step 2Strength eccentricity limit B/3 (ft).

Answer (ft)

Step 3Effective bearing width

B' = B - 2e

(ft) and factored uniform bearing

q_u = \Sigma V_u / B'

— enter q_u (ksf).

Answer (ksf)

Guided Problem 4

MSE reinforcement pullout at a mid-height layer

An MSE wall uses ribbed steel strips (F* = 1.5 − 0.75·log C_u ≈ 1.2 at the surface, decreasing to F* = 0.4 at 20 ft; here use F* = 0.6 at the layer). At z = 18 ft, γ = 130 pcf, embedment length in the resistant zone Lₑ = 6 ft, C = 2 (top and bottom of strip), α = 1.0, φ = 0.90 for pullout of steel strips.

Compute the vertical stress σᵥ and the pullout capacity per strip per foot of wall.

Step 1Overburden stress

\sigma_v = \gamma z

(ksf) at z = 18 ft.

Answer (ksf)

Step 2Pullout capacity

P_r = \phi\,F^{*}\,\alpha\,\sigma_v\,L_e\,C

(kip/ft of strip).

Answer (kip/ft)

Step 3If the maximum tension in that strip from Rankine active analysis is T_max = 8.4 kip/ft, is pullout OK? Enter the D/C ratio T_max / P_r (unitless).

Answer