Retaining Walls

Example 1

Retaining walls (Rankine active force) — solve for active thrust — Retaining Walls

a cantilever retaining wall holding back sandy backfill Given active earth pressure coefficient (K_a) = 0.3400; soil unit weight (gamma) = 16.5000 kN/m^3; wall height (H) = 9.0000 m, determine the active thrust (P_a) in kN/m.

Given

$active earth pressure coefficient (K_a) = 0.3400$
$soil unit weight (gamma) = 16.5000 kN/m^3$
$wall height (H) = 9.0000 m$

Find

active thrust (P_a), in kN/m

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except P_a is given, so isolate P_a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 1 — schematic for Retaining walls (Rankine active force) — solve for active thrust — Retaining Walls

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that P_a stands alone on the left-hand side.
Step 3 — List the givens: active earth pressure coefficient (K_a) = 0.3400, soil unit weight (gamma) = 16.5000 kN/m^3, wall height (H) = 9.0000 m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$P_{a} = 227.2\ \text{kN/m}$
Step 6 — Check: returning P_a = 227.2 kN/m to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

P_{a} = 227.2\ \text{kN/m}

Why the other options are there

454.4 — kept a factor of two that cancels in the correct rearrangement.
113.6 — dropped that same factor in the other direction.
249.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Example 2

Retaining walls (Rankine active force) — solve for active earth pressure coefficient — Retaining Walls (2)

a gravity retaining wall supporting a highway cut Given soil unit weight (gamma) = 17.5000 kN/m^3; wall height (H) = 5.0000 m; active thrust (P_a) = 488.0 kN/m, determine the active earth pressure coefficient (K_a).

Given

$soil unit weight (gamma) = 17.5000 kN/m^3$
$wall height (H) = 5.0000 m$
$active thrust (P_a) = 488.0 kN/m$

Find

active earth pressure coefficient (K_a)

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except K_a is given, so isolate K_a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 2 — schematic for Retaining walls (Rankine active force) — solve for active earth pressure coefficient — Retaining Walls (2)

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that K_a stands alone on the left-hand side.
Step 3 — List the givens: soil unit weight (gamma) = 17.5000 kN/m^3, wall height (H) = 5.0000 m, active thrust (P_a) = 488.0 kN/m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$K_{a} = 2.2309$
Step 6 — Check: returning K_a = 2.2309 to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

K_{a} = 2.2309

Why the other options are there

4.4617 — kept a factor of two that cancels in the correct rearrangement.
1.1154 — dropped that same factor in the other direction.
2.4539 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Example 3

Retaining walls (Rankine active force) — solve for wall height — Retaining Walls (3)

retaining walls designed for active earth pressure from granular fill Given active earth pressure coefficient (K_a) = 0.2600; soil unit weight (gamma) = 18.0000 kN/m^3; active thrust (P_a) = 295.0 kN/m, determine the wall height (H) in m.

Given

$active earth pressure coefficient (K_a) = 0.2600$
$soil unit weight (gamma) = 18.0000 kN/m^3$
$active thrust (P_a) = 295.0 kN/m$

Find

wall height (H), in m

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except H is given, so isolate H symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 3 — schematic for Retaining walls (Rankine active force) — solve for wall height — Retaining Walls (3)

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that H stands alone on the left-hand side.
Step 3 — List the givens: active earth pressure coefficient (K_a) = 0.2600, soil unit weight (gamma) = 18.0000 kN/m^3, active thrust (P_a) = 295.0 kN/m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$H = 11.2280\ \text{m}$
Step 6 — Check: returning H = 11.2280 m to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

H = 11.2280\ \text{m}

Why the other options are there

22.4560 — kept a factor of two that cancels in the correct rearrangement.
5.6140 — dropped that same factor in the other direction.
12.3508 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Example 4

Retaining walls (Rankine active force) — solve for active thrust (case 2) — Retaining Walls (4)

a cantilever retaining wall holding back sandy backfill Given active earth pressure coefficient (K_a) = 0.3100; soil unit weight (gamma) = 18.0000 kN/m^3; wall height (H) = 10.0000 m, determine the active thrust (P_a) in kN/m.

Given

$active earth pressure coefficient (K_a) = 0.3100$
$soil unit weight (gamma) = 18.0000 kN/m^3$
$wall height (H) = 10.0000 m$

Find

active thrust (P_a), in kN/m

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except P_a is given, so isolate P_a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 4 — schematic for Retaining walls (Rankine active force) — solve for active thrust (case 2) — Retaining Walls (4)

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that P_a stands alone on the left-hand side.
Step 3 — List the givens: active earth pressure coefficient (K_a) = 0.3100, soil unit weight (gamma) = 18.0000 kN/m^3, wall height (H) = 10.0000 m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$P_{a} = 279.0\ \text{kN/m}$
Step 6 — Check: returning P_a = 279.0 kN/m to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

P_{a} = 279.0\ \text{kN/m}

Why the other options are there

558.0 — kept a factor of two that cancels in the correct rearrangement.
139.5 — dropped that same factor in the other direction.
306.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Example 5

Retaining walls (Rankine active force) — solve for active earth pressure coefficient (case 2) — Retaining Walls (5)

a gravity retaining wall supporting a highway cut Given soil unit weight (gamma) = 18.5000 kN/m^3; wall height (H) = 3.0000 m; active thrust (P_a) = 38.0000 kN/m, determine the active earth pressure coefficient (K_a).

Given

$soil unit weight (gamma) = 18.5000 kN/m^3$
$wall height (H) = 3.0000 m$
$active thrust (P_a) = 38.0000 kN/m$

Find

active earth pressure coefficient (K_a)

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except K_a is given, so isolate K_a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 5 — schematic for Retaining walls (Rankine active force) — solve for active earth pressure coefficient (case 2) — Retaining Walls (5)

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that K_a stands alone on the left-hand side.
Step 3 — List the givens: soil unit weight (gamma) = 18.5000 kN/m^3, wall height (H) = 3.0000 m, active thrust (P_a) = 38.0000 kN/m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$K_{a} = 0.4565$
Step 6 — Check: returning K_a = 0.4565 to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

K_{a} = 0.4565

Why the other options are there

0.9129 — kept a factor of two that cancels in the correct rearrangement.
0.2282 — dropped that same factor in the other direction.
0.5021 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Example 6

Retaining walls (Rankine active force) — solve for wall height (case 2) — Retaining Walls (6)

retaining walls designed for active earth pressure from granular fill Given active earth pressure coefficient (K_a) = 0.4200; soil unit weight (gamma) = 19.5000 kN/m^3; active thrust (P_a) = 425.0 kN/m, determine the wall height (H) in m.

Given

$active earth pressure coefficient (K_a) = 0.4200$
$soil unit weight (gamma) = 19.5000 kN/m^3$
$active thrust (P_a) = 425.0 kN/m$

Find

wall height (H), in m

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except H is given, so isolate H symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 6 — schematic for Retaining walls (Rankine active force) — solve for wall height (case 2) — Retaining Walls (6)

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that H stands alone on the left-hand side.
Step 3 — List the givens: active earth pressure coefficient (K_a) = 0.4200, soil unit weight (gamma) = 19.5000 kN/m^3, active thrust (P_a) = 425.0 kN/m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$H = 10.1875\ \text{m}$
Step 6 — Check: returning H = 10.1875 m to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

H = 10.1875\ \text{m}

Why the other options are there

20.3750 — kept a factor of two that cancels in the correct rearrangement.
5.0937 — dropped that same factor in the other direction.
11.2062 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Example 7

Retaining walls (Rankine active force) — solve for active thrust (case 3) — Retaining Walls (7)

a cantilever retaining wall holding back sandy backfill Given active earth pressure coefficient (K_a) = 0.3200; soil unit weight (gamma) = 19.0000 kN/m^3; wall height (H) = 7.0000 m, determine the active thrust (P_a) in kN/m.

Given

$active earth pressure coefficient (K_a) = 0.3200$
$soil unit weight (gamma) = 19.0000 kN/m^3$
$wall height (H) = 7.0000 m$

Find

active thrust (P_a), in kN/m

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except P_a is given, so isolate P_a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 7 — schematic for Retaining walls (Rankine active force) — solve for active thrust (case 3) — Retaining Walls (7)

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that P_a stands alone on the left-hand side.
Step 3 — List the givens: active earth pressure coefficient (K_a) = 0.3200, soil unit weight (gamma) = 19.0000 kN/m^3, wall height (H) = 7.0000 m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$P_{a} = 149.0\ \text{kN/m}$
Step 6 — Check: returning P_a = 149.0 kN/m to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

P_{a} = 149.0\ \text{kN/m}

Why the other options are there

297.9 — kept a factor of two that cancels in the correct rearrangement.
74.4800 — dropped that same factor in the other direction.
163.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Example 8

Retaining walls (Rankine active force) — solve for active earth pressure coefficient (case 3) — Retaining Walls (8)

a gravity retaining wall supporting a highway cut Given soil unit weight (gamma) = 16.5000 kN/m^3; wall height (H) = 10.0000 m; active thrust (P_a) = 331.0 kN/m, determine the active earth pressure coefficient (K_a).

Given

$soil unit weight (gamma) = 16.5000 kN/m^3$
$wall height (H) = 10.0000 m$
$active thrust (P_a) = 331.0 kN/m$

Find

active earth pressure coefficient (K_a)

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except K_a is given, so isolate K_a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 8 — schematic for Retaining walls (Rankine active force) — solve for active earth pressure coefficient (case 3) — Retaining Walls (8)

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that K_a stands alone on the left-hand side.
Step 3 — List the givens: soil unit weight (gamma) = 16.5000 kN/m^3, wall height (H) = 10.0000 m, active thrust (P_a) = 331.0 kN/m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$K_{a} = 0.4012$
Step 6 — Check: returning K_a = 0.4012 to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

K_{a} = 0.4012

Why the other options are there

0.8024 — kept a factor of two that cancels in the correct rearrangement.
0.2006 — dropped that same factor in the other direction.
0.4413 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Example 9

Retaining walls (Rankine active force) — solve for wall height (case 3) — Retaining Walls (9)

retaining walls designed for active earth pressure from granular fill Given active earth pressure coefficient (K_a) = 0.3300; soil unit weight (gamma) = 18.0000 kN/m^3; active thrust (P_a) = 123.0 kN/m, determine the wall height (H) in m.

Given

$active earth pressure coefficient (K_a) = 0.3300$
$soil unit weight (gamma) = 18.0000 kN/m^3$
$active thrust (P_a) = 123.0 kN/m$

Find

wall height (H), in m

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except H is given, so isolate H symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 9 — schematic for Retaining walls (Rankine active force) — solve for wall height (case 3) — Retaining Walls (9)

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that H stands alone on the left-hand side.
Step 3 — List the givens: active earth pressure coefficient (K_a) = 0.3300, soil unit weight (gamma) = 18.0000 kN/m^3, active thrust (P_a) = 123.0 kN/m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$H = 6.4354\ \text{m}$
Step 6 — Check: returning H = 6.4354 m to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

H = 6.4354\ \text{m}

Why the other options are there

12.8708 — kept a factor of two that cancels in the correct rearrangement.
3.2177 — dropped that same factor in the other direction.
7.0789 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Example 10

Retaining walls (Rankine active force) — solve for active thrust (case 4) — Retaining Walls (10)

a cantilever retaining wall holding back sandy backfill Given active earth pressure coefficient (K_a) = 0.3000; soil unit weight (gamma) = 17.5000 kN/m^3; wall height (H) = 8.0000 m, determine the active thrust (P_a) in kN/m.

Given

$active earth pressure coefficient (K_a) = 0.3000$
$soil unit weight (gamma) = 17.5000 kN/m^3$
$wall height (H) = 8.0000 m$

Find

active thrust (P_a), in kN/m

Start with the thinking

The governing relation printed in this handbook section is Retaining walls (Rankine active force).
Everything except P_a is given, so isolate P_a symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Retaining walls must resist the active lateral earth force from backfill soil computed with the Rankine coefficient.

Figure 10 — schematic for Retaining walls (Rankine active force) — solve for active thrust (case 4) — Retaining Walls (10)

Step-by-step solution

Step 1 — State the governing relation:
$P_a = \tfrac{1}{2} K_a \gamma H^2$
Step 2 — Rearrange the relation so that P_a stands alone on the left-hand side.
Step 3 — List the givens: active earth pressure coefficient (K_a) = 0.3000, soil unit weight (gamma) = 17.5000 kN/m^3, wall height (H) = 8.0000 m.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$P_{a} = 168.0\ \text{kN/m}$
Step 6 — Check: returning P_a = 168.0 kN/m to
$P_a = \tfrac{1}{2} K_a \gamma H^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

P_{a} = 168.0\ \text{kN/m}

Why the other options are there

336.0 — kept a factor of two that cancels in the correct rearrangement.
84.0000 — dropped that same factor in the other direction.
184.8 — rounded an intermediate value before the final step.

Reference: FE Handbook — Retaining Walls

Handbook notes for this section

Core formulas for this FE topic

Worked exam-style examples

Retaining Walls

Handbook notes for this section

Core formulas for this FE topic

FSoverturning=MRFS overturning = M RFSoverturning=MR

FSsliding=/FSsliding = /FSsliding=/

Vhtan⁡d+BCa+Pp^ V h \tan d + BC a + PpVhtand+BCa+Pp

FSsliding=Pacos⁡aFS sliding = Pa \cos aFSsliding=Pacosa

FSbearingcapacity=qULTFS bearing capacity = qULTFSbearingcapacity=qULT

qtoe=Bc1+Bmq toe = B c_{1} + B mqtoe=Bc1​+Bm

e=2−eoe = 2 -e oe=2−eo

e=eccentricitye = eccentricitye=eccentricity

B=widthofbaseB = width of baseB=widthofbase

MR=resistingmomentMR = resisting momentMR=resistingmoment

MO=overturningmomentMO = overturning momentMO=overturningmoment

FR=resistingforcesFR = resisting forcesFR=resistingforces

FD=drivingforcesFD = driving forcesFD=drivingforces

V=verticalforcesV = vertical forcesV=verticalforces

δ=k1ϕ2\delta = k_{1}\phi_{2}δ=k1​ϕ2​

Ca=k2C2Ca = k_{2}C_{2}Ca=k2​C2​

FS=factorofsafetyagainstslopeinstabilityFS = factor of safety against slope instabilityFS=factorofsafetyagainstslopeinstability

=TFF/TMOB= TFF/TMOB=TFF/TMOB

TFF=availableshearingresistancealongslipsurfaceTFF = available shearing resistance along slip surfaceTFF=availableshearingresistancealongslipsurface

= cLS + WM cos αS tan φNeeds technical review

TMOB=mobilizedshearforcealongslipsurfaceTMOB = mobilized shear force along slip surfaceTMOB=mobilizedshearforcealongslipsurface

= WM sin αSNeeds technical review

LS=lengthofassumedplanarslipsurfaceLS = length of assumed planar slip surfaceLS=lengthofassumedplanarslipsurface

WM=weightofsoilaboveslipsurfaceWM = weight of soil above slip surfaceWM=weightofsoilaboveslipsurface

αS = angle of assumed slip surface with respect to horizontalNeeds technical review

2.00mm(No.10)≤50−−−−−−−−−−2.00 mm (No. 10) \le 50 - - - - - - - - - -2.00mm(No.10)≤50−−−−−−−−−−

0.425mm(No.40)≤30≤50≥51−−−−−−−−0.425 mm (No. 40) \le 30 \le 50 \ge 51 - - - - - - - -0.425mm(No.40)≤30≤50≥51−−−−−−−−

0.075mm(No.200)≤15≤25≤10≤35≤35≤35≤35≥36≥36≥36≥360.075 mm (No. 200) \le 15 \le 25 \le 10 \le 35 \le 35 \le 35 \le 35 \ge 36 \ge 36 \ge 36 \ge 360.075mm(No.200)≤15≤25≤10≤35≤35≤35≤35≥36≥36≥36≥36

LIQUIDLIMIT−−≤40≥41≤40≥41≤40≥41≤40≥41LIQUID LIMIT - - \le 40 \ge 41 \le 40 \ge 41 \le 40 \ge 41 \le 40 \ge 41LIQUIDLIMIT−−≤40≥41≤40≥41≤40≥41≤40≥41

PLASTICITYINDEX∗6maxNP≤10≤10≥11≥11≤10≤10≥11≥11PLASTICITY INDEX * 6 m ax NP \le 10 \le 10 \ge 11 \ge 11 \le 10 \le 10 \ge 11 \ge 11PLASTICITYINDEX∗6maxNP≤10≤10≥11≥11≤10≤10≥11≥11

soil.Example:A−6withGI=15islesssuitablethanA−6withGI=10.soil. Example: A-6 with GI = 15 is less suitable than A-6 with GI = 10.soil.Example:A−6withGI=15islesssuitablethanA−6withGI=10.

GI=(F−35)[0.2+0.005(LL−40)]+0.01(F−15)(PI−10)GI = ( F - 35 ) [ 0.2 + 0.005 ( L L - 40 ) ] + 0.01 ( F - 15 ) ( PI - 10 )GI=(F−35)[0.2+0.005(LL−40)]+0.01(F−15)(PI−10)

where:F=PercentpassingNo.200sieve,expressedasawholenumber.Thispercentagewhere: F = Percent passing No. 200 sieve, expressed as a whole number. This percentagewhere:F=PercentpassingNo.200sieve,expressedasawholenumber.Thispercentage

LL=LiquidlimitLL = Liquid limitLL=Liquidlimit

PI=PlasticityindexPI = Plasticity indexPI=Plasticityindex

IfthecomputedvalueofGI<0,thenuseGI=0.If the computed value of GI < 0, then use GI = 0.IfthecomputedvalueofGI<0,thenuseGI=0.

COARSE−GRAINEDGravelsCleanGravelsCu≥4and1≤Cc≤3DGWWell−gradedgravelECOARSE-GRAINED Gravels Clean Gravels Cu \ge 4 and 1 \le Cc \le 3D GW Well-graded gravelECOARSE−GRAINEDGravelsCleanGravelsCu≥4and1≤Cc≤3DGWWell−gradedgravelE

Worked exam-style examples