Download - HOMEWORK 2 - courses.ce.metu.edu.trcourses.ce.metu.edu.tr/.../2017/12/CE363_FALL_2017_HW2_Solutions.pdfHomework solutions will be posted to the course website on December 26, 2017,

CE 363 Soil Mechanics Middle East Technical University

Homework 2 Department of Civil Engineering

1

HOMEWORK 2

Homework 2 is due on December 26, 2017, Tuesday at 16:59.

CE363 Homeworks are to be submitted to the “CE363 Homework box” in Soil

Mechanics Lab.

Soil Mechanics lab door is locked every day at 17:00 and homeworks cannot be

submitted under the door.

Homework solutions will be posted to the course website on December 26, 2017,

Tuesday at 17:00.

Unless otherwise stated, use γwater = 10 kN/m3

Question 1 (a:15%, b:10%)

A very wide granular embankment that is 2-m-thick, having unit weight of 20 kN/m3, is placed

over the ground surface of a soft (normally consolidated) clay deposit that is 3-m thick. Below the

clay layer, there is impervious rock, and water table is at the ground surface of soft clay. Saturated

unit weight of clay is 18 kN/m3, it is fully saturated and its initial average moisture content is 23%,

Gs=2.67, compression index is 0.35 and coefficient of consolidation is 6 m2/year.

a) Plot time versus consolidation settlement of clay layer until average degree of

consolidation is 95%. Note that horizontal axis should be time (in months), and vertical

axis should be settlement (vertical axis should increase in downward direction).

b) After the fill is placed, at the time when the ground surface settlement is measured as 15

cm,

- what is the average degree of consolidation of the clay layer?

- what is the degree of consolidation at 2 m depth below the top of the clay?

- what is the total pore water pressure and excess pore water pressure at 2 m depth below

the top of clay?



2

EMBANKMENT

IMPERVIOUS ROCK

N.C SOFT CLAY

2 m

3 m

𝛥𝜎 = 20 𝑘𝑁 𝑚3⁄ ∗ 2 𝑚 = 40 𝑘𝑃𝑎 (𝑑𝑢𝑒 𝑡𝑜 𝑣𝑒𝑟𝑦 𝑤𝑖𝑑𝑒 𝑒𝑚𝑏𝑎𝑛𝑘𝑚𝑒𝑛𝑡)

𝜎0′ = 1.5 𝑚 ∗ (18 𝑘𝑁 𝑚3⁄ − 10 𝑘𝑁 𝑚3⁄ ) = 12 𝑘𝑃𝑎 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑙𝑦 𝑎𝑡 𝑡ℎ𝑒 𝑚𝑖𝑑 − 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑙𝑎𝑦 𝑙𝑎𝑦𝑒𝑟)

𝜎𝑓′ = 12 𝑘𝑃𝑎 + 40 𝑘𝑃𝑎 = 52 𝑘𝑃𝑎 (𝑓𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑚𝑖𝑑 − 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑙𝑎𝑦 𝑙𝑎𝑦𝑒𝑟)

Since 𝑤 ∗ 𝐺𝑠 = 𝑆𝑟 ∗ 𝑒,

𝑒0 = 𝑤0 ∗ 𝐺𝑠 = 23% ∗ 2.67 = 0.614 (𝑆𝑟 = 1, 𝑓𝑢𝑙𝑙𝑦 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑)

Given, 𝐶𝑐 = 0.35 & the soil is normally consolidated,

𝑆𝑐 =𝐶𝑐

1 + 𝑒0∗ 𝐻 ∗ log (

𝜎𝑓′

𝜎0′) =

0.35

1 + 0.614∗ 3 ∗ log (

52

12) = 0.414 𝑚 = 41.4 𝑐𝑚

a) Consolidation settlement vs time graph is asked. Given,

𝑐𝑣 = 6 𝑚2 𝑦𝑒𝑎𝑟⁄ = 0.5 𝑚2/𝑚𝑜𝑛𝑡ℎ

𝑇𝑣 =𝑐𝑣 ∗ 𝑡

𝐻2, 𝑤ℎ𝑒𝑟𝑒 𝐻 (𝑜𝑟 𝑑, 𝐷) = 3 𝑚 (𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑙𝑎𝑦𝑒𝑟 𝑖𝑠 ℎ𝑎𝑙𝑓 𝑐𝑙𝑜𝑠𝑒𝑑. )

𝑇𝑣 𝑐𝑎𝑛 𝑏𝑒 𝑟𝑒𝑙𝑎𝑡𝑒𝑑 𝑤𝑖𝑡ℎ 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑠:

𝑓𝑜𝑟 𝑈 < 60 % → 𝑇𝑣 =𝜋

4∗ 𝑈2

𝑓𝑜𝑟 𝑈 > 60% → 𝑇𝑣 = −0.933 ∗ log(1 − 𝑈) − 0.085

𝐻𝑒𝑛𝑐𝑒, 𝑡 → 𝑇𝑣 → 𝑈 → 𝑆

𝐹𝑜𝑟 𝑖𝑛𝑠𝑡𝑎𝑛𝑐𝑒,

𝐴𝑡 𝑡 = 3 𝑚𝑜𝑛𝑡ℎ𝑠, 𝑇𝑣 =0.5 𝑚2 𝑚𝑜𝑛𝑡ℎ⁄ ∗ 3 𝑚𝑜𝑛𝑡ℎ𝑠

9 𝑚2= 0.167



3

𝑈 = √4 ∗ 𝑇𝑣

𝜋= √

4 ∗ 0.167

3.14= 0.461 = 46.1 % (𝑈 < 60%)

𝑆 = 41.4 𝑐𝑚 ∗ 46.1 % = 19.1 𝑐𝑚

𝐴𝑡 𝑡 = 12 𝑚𝑜𝑛𝑡ℎ𝑠, 𝑇𝑣 =0.5 𝑚2 𝑚𝑜𝑛𝑡ℎ⁄ ∗ 12 𝑚𝑜𝑛𝑡ℎ𝑠

9 𝑚2= 0.667

𝑈 = 1 − 10−(𝑇𝑣+0.085)/0.933 = 1 − 10−0.667+0.085

0.933 = 0.844 = 84.4 % (𝑈 > 60%)

𝑆 = 41.4 𝑐𝑚 ∗ 84.4 % = 34.9 𝑐𝑚

Similarly,

t (months) Tv U Settlement (cm)

0 0.000 0.000 0.0

0.1 0.006 0.084 3.5

0.5 0.028 0.188 7.8

1 0.056 0.266 11.0

2 0.111 0.376 15.6

3 0.167 0.461 19.1

4 0.222 0.532 22.0

5 0.278 0.595 24.6

6 0.333 0.644 26.7

7 0.389 0.689 28.6

8 0.444 0.729 30.2

9 0.500 0.764 31.6

10 0.556 0.794 32.9

11 0.611 0.821 34.0

12 0.667 0.844 34.9

13 0.722 0.864 35.8

14 0.778 0.881 36.5

15 0.833 0.896 37.1

16 0.889 0.910 37.7

17 0.944 0.921 38.2

18 1.000 0.931 38.6

19 1.056 0.940 38.9

20 1.111 0.948 39.3

20.3 1.129 0.950 39.4



4

b) 𝑊ℎ𝑒𝑛 𝑆 = 15 𝑐𝑚, 𝑈 =15 𝑐𝑚

41.4 𝑐𝑚= 36.2 % 𝒂𝒗𝒆𝒓𝒂𝒈𝒆 𝒅𝒆𝒈𝒓𝒆𝒆 𝒐𝒇 𝒄𝒐𝒏𝒔𝒐𝒍𝒊𝒅𝒂𝒕𝒊𝒐𝒏

𝑇𝑣 = 𝜋

4∗ (36.2 %)2 = 0.103

𝐷𝑒𝑝𝑡ℎ 𝑓𝑟𝑜𝑚 𝑡𝑜𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑙𝑎𝑦 = 2 𝑚, 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑟𝑎𝑖𝑛𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 3 𝑚

𝑈𝑧 = 0.14 , 𝒅𝒆𝒈𝒓𝒆𝒆 𝒐𝒇 𝒄𝒐𝒏𝒔𝒐𝒍𝒊𝒅𝒂𝒕𝒊𝒐𝒏 𝑎𝑡 2 𝑚 𝑑𝑒𝑝𝑡ℎ 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑙𝑎𝑦

0

5

10

15

20

25

30

35

40

45

0 2 4 6 8 10 12 14 16 18 20 22

Set

tlem

ent

(cm

)

Time (months)



5

𝑈𝑧 =𝑢𝑖𝑒 − 𝑢𝑒

𝑢𝑖𝑒, 𝑤ℎ𝑒𝑟𝑒 𝑢𝑖𝑒 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑟𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝑢𝑒 = 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑟𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝑢𝑖𝑒 = 𝛥𝜎 = 40 𝑘𝑃𝑎

𝐹𝑜𝑟 𝑈𝑧 = 0.14 = 40 − 𝑢𝑒

40,

𝑢𝑒 = 34.4 𝑘𝑃𝑎, 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑟𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 2 𝑚 𝑑𝑒𝑝𝑡ℎ 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑙𝑎𝑦

𝐵𝑒𝑓𝑜𝑟𝑒 𝑒𝑚𝑏𝑎𝑛𝑘𝑚𝑒𝑛𝑡 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛, 𝑢0 = 2 𝑚 ∗ 10 𝑘𝑁 𝑚3⁄ = 20 𝑘𝑃𝑎

𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑟𝑒 𝑤𝑎𝑡𝑒𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑤ℎ𝑒𝑛 𝑆 = 15 𝑐𝑚 , 𝑢𝑓 = 20 𝑘𝑃𝑎 + 34.4 𝑘𝑃𝑎 = 54.4 𝑘𝑃𝑎

Question 2 (a: 14%; b: 5 %)

Following results are obtained in a series of CU triaxial tests performed on saturated samples of

Ankara clay.

a) Calculate the effective stress-based modified shear strength parameters, intercept a' and

slope angle ' , by filling up the following table and plotting the modified Mohr-Coulomb

failure envelopes on the chart provided below.

Test Number 1 2

Confining Pressure

(kPa) 40 80

Deviatoric Stress

(kPa) 120 170

Pore water pressure

at failure (kPa) 15 25

'3,f (kPa) 25 55

'1,f (kPa) 145 225

'1,f+'3,f)/2 (kPa) 85 140

'1,f-'3,f)/2 (kPa) 60 85

From the figure above: a' = 21.4 kPa & ' = 24.4o

sin(𝜙′) = tan(𝛼′) → sin(𝜙′) = tan(24.4) → 𝜙′ = 27°

𝑐′ =𝑎′

cos(𝜙′)=

21.4

cos(27)= 24 𝑘𝑃𝑎

y = 0.4545x + 21.364

0

50

100

150

200

250

0 50 100 150 200 250

(1'-

3')

/2

(1'+3')/2



6

b) Vane shear test is used to determine the undrained shear strength of the same clay in Part

(a). If the clay is known to have a sensitivity ratio of 3 and undrained shear strength of

cu=35 kPa, estimate the peak and the residual torques. The shear vane used in the test has

a diameter of 75 mm and length of 150 mm.

)62

(32 DLD

cT u

𝑇𝑝𝑒𝑎𝑘 = 𝜋 ∗ 35 ∗ ((0.0752𝑥0.15

2) + (

0.0753

6)) = 54.1 𝑁. 𝑚

𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜 = 𝑇𝑝𝑒𝑎𝑘

𝑇𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙= 3 → 𝑇𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 = 18.0 𝑁. 𝑚

Question 3 (4%)

A silty clay specimen was placed in a triaxial cell and sheared as part of an unconfined

compression test. If it is sheared at deviator stress of 50 kN/m2,

Draw the Mohr circle corresponding to failure state. Estimate the followings:

i. Confining stress, σ3………0………. ...kPa........

ii. Major principal stress, σ1……50…………. .kPa..........

iii. Unconfined compressive strength, qu………50………. .kPa.

iv. Undrained shear strength, cu……25…………. ..kPa.........

0

10

20

30

40

50

60

0 10 20 30 40 50 60

t(k

Pa)

(kPa)



7

Question 4 (a: 7%, b: 3%)

The one dimensional consolidation test data for a soil sample is given below:

Effective stress (kPa) 25 50 100 150 200 250 300

void ratio 0.96 0.93 0.89 0.84 0.78 0.71 0.65

The pre-consolidation pressure of the soil sample is 𝟏𝟏𝟓 𝒌𝑷𝒂 and initial void ratio is given as

eo=0.98.

a) Determine the coefficient of volume of compressibility, compression index, and reload

(recompression) index assuming that the initial effective stress on the sample is 155 kPa

and the final effective stress will be 245 kPa.

𝑚𝑉 =1

1+𝑒0∗ (

∆𝑒

∆𝜎) =

1

1+0.84∗ (

0.84−0.71

250−150) = 7.06 ∗ 10−4𝑚2/𝑘𝑁 (e0 for σ = 150 kPa)

𝐶𝑟 =(𝑒0 − 𝑒𝜎𝑝)

log σ 𝑝 − log 𝑝𝑜=

∆𝑒

log 𝜎𝑝/𝑝𝑜=

(0.93 − 0.89)

log (10050

)= 0.133

The closest void ratio data for stress level of 155 kPa and 245 kPa are 150 kPa and 250 kPa in the

table given above. Therefore Cc is calculated based on these stress levels. Alternatively you may

determine void ratio values for 155 kPa and 245 kPa and calculate Cc.

0.50

0.60

0.70

0.80

0.90

1.00

10 60 110 160 210 260 310 360

e

σ' (kPa)

e-σ'



8

𝐶𝑐 =(𝑒𝜎𝑝 − 𝑒)

log 𝜎 − log 𝜎𝑝=

∆𝑒

log 𝜎/𝜎𝑝=

(0.84 − 0.71)

log (250150

)= 0.586

b) Calculate the consolidation settlement of a 5 m-thick clay layer in part (a) if initial and final

effective stresses at mid-depth are 𝟓𝟎 𝒌𝑷𝒂 𝒂𝒏𝒅 𝟐𝟎𝟎 𝒌𝑷𝒂 respectively.

𝑆𝑐 =𝐶𝑟

(1 + 𝑒𝑜)∗ 𝐻 ∗ log (

𝜎𝑓

𝜎𝑖) +

𝐶𝑐

(1 + 𝑒𝑜)∗ 𝐻 ∗ log (

𝜎𝑓

𝜎𝑖)

𝑆𝑐 =0.133

(1 + 0.98)∗ 5 ∗ log (

115

50) +

0.586

(1 + 0.98)∗ 5 ∗ log (

200

115)

𝑆𝑐 = 0.121 + 0.356 = 0.477 𝑚 = 47.7 𝑐𝑚

Question 5 (a:4%, b:4%, c:4%)

A direct shear apparatus has 𝟔𝟑 𝒎𝒎 sample diameter. Following data were recorded at failure

during the tests caried out over the samples of a clayey sand using this device.

Do not perform area correction

𝑵 (𝒌𝑵) 𝑻 (𝒌𝑵) 𝝈 (𝒌𝑷𝒂) 𝝉 (𝒌𝑷𝒂)

Test 1 0.15 0.15

Test 2 0.45 0.36

a) Complete the table and plot the failure envelope of the tested soil. When plotting, use an

appropriate scaling for the axes and clearly indicate the stresses defining the envelope.

σ =N

A=

150 N

3117.2 𝑚𝑚2= 0.048 𝑀𝑃𝑎 = 48 𝑘𝑃𝑎

τ =T

A=

150 N

3117.2 𝑚𝑚2= 0.048 𝑀𝑃𝑎 = 48 𝑘𝑃𝑎

σ =N

A=

450 N

3117.2 𝑚𝑚2= 0.1444 𝑀𝑃𝑎 = 144.4 𝑘𝑃𝑎

τ =T

A=

360 N

3117.2 𝑚𝑚2= 0.1155 𝑀𝑃𝑎 = 115.5 𝑘𝑃𝑎



9

𝑵 (𝒌𝑵) 𝑻 (𝒌𝑵) 𝝈 (𝒌𝑷𝒂) 𝝉 (𝒌𝑷𝒂)

Test 1 0.15 0.15 48 48

Test 2 0.45 0.36 144.4 115.5

b) Calculate the shear strength parameters of the clayey sand.

∅′ = 𝑎𝑟𝑐𝑡𝑎𝑛 (∆𝑦

∆𝑥) = 𝑎𝑟𝑐𝑡𝑎𝑛 (

∆𝑦

∆𝑥) = 𝑎𝑟𝑐𝑡𝑎𝑛 (

115.5 − 48

144.4 − 48) = 𝑎𝑟𝑐𝑡𝑎𝑛 (

67.5

96.4)

∅ ≅ 35 °

𝜏 = 𝑐′ + 𝜎′ ∗ 𝑡𝑎𝑛∅′

115.5 = 𝑐′ + 144.4 ∗ 𝑡𝑎𝑛35

𝑐′ ≅ 14.4 𝑘𝑃𝑎

c) At what shear stress would this soil fail if the normal stress is 𝟖𝟓 𝒌𝑷𝒂.

𝜏 = 𝑐′ + 𝜎′ ∗ 𝑡𝑎𝑛∅′

𝜏 = 14.4 + 85 ∗ 𝑡𝑎𝑛35 = 73.9 𝑘𝑃𝑎

0

20

40

60

80

100

120

140

0 20 40 60 80 100 120 140 160

τ

(kP

a)

σ (kPa)



10

Question 6 (a:6%, b:4%)

During a consolidated undrained triaxial test on a saturated clay sample, failure is reached at a

principle stress difference of 385 kPa. The consolidation (all-round) pressure was 250 kPa and the

pore pressure was measured at failure as 145 kPa.

a) Plot the total and effective stress Mohr circles corresponding to the stress state at failure.

Use an appropriate scale when drawing and clearly indicate the necessary data to define

the circles.

Given

∆𝜎 = 385 𝑘𝑃𝑎

𝜎3 = 250 𝑘𝑃𝑎

𝑢 = 145 𝑘𝑃𝑎

𝜎3 = 250 𝑘𝑃𝑎

𝜎1 = 𝜎3 + ∆𝜎 = 250 + 385 = 635 𝑘𝑃𝑎

𝜎´1 = 𝜎1 − 𝑢 = 635 − 145 = 490 𝑘𝑃𝑎

𝜎´3 = 𝜎3 − 𝑢 = 250 − 145 = 105 𝑘𝑃𝑎

b) If the effective angle of internal shear of this clay is known to be 30°, calculate the effective

cohesion parameter.

𝑠𝑖𝑛𝜙´ =

𝜎´1 − 𝜎´3

2𝜎´1 + 𝜎´3

2+

𝑐´𝑡𝑎𝑛𝜙´

𝑠𝑖𝑛30 =

490 − 1052

490 + 1052

+𝑐´

𝑡𝑎𝑛30

Solving for 𝑐´ = 50.5 𝑘𝑃𝑎

0

50

100

150

200

250

300

350

400

450

500

-200 -100 0 100 200 300 400 500 600 700 800 900 1000

t(k

Pa)

, ' (kPa)

TSCESC

105 kPa 250 kPa

u

635 kPa

c´=50.5 kPa



11

Question 7 (a:12%, b:4%, c:4%)

Consider the retaining wall shown in the figure below:

a) Using Rankine’s theory of lateral earth pressure calculate and plot the active and passive

earth pressure distributions at the back and in front of the wall respectively.

𝐾𝑎𝑠𝑜𝑖𝑙−1 =1 − 𝑠𝑖𝑛∅′

1 + 𝑠𝑖𝑛∅′=

1 − 𝑠𝑖𝑛27

1 + 𝑠𝑖𝑛27= 0.375

𝐾𝑎𝑠𝑜𝑖𝑙−2 =1 − 𝑠𝑖𝑛∅′

1 + 𝑠𝑖𝑛∅′=

1 − 𝑠𝑖𝑛30

1 + 𝑠𝑖𝑛30= 0.333

𝐾𝑝𝑠𝑜𝑖𝑙−2 =1 + 𝑠𝑖𝑛∅′

1 − 𝑠𝑖𝑛∅′=

1 + 𝑠𝑖𝑛30

1 − 𝑠𝑖𝑛30= 3

Calculate the active pressure (z=depth from ground surface)

𝜎ℎ′ = 𝜎𝑣

′ ∗ 𝐾𝑎 − 2 ∗ 𝑐′ ∗ √𝐾𝑎


′ ∗ 𝐾𝑝 + 2 ∗ 𝑐′ ∗ √𝐾𝑝

𝐴𝑡 𝑧 = 0 𝑚; 𝜎ℎ′ = 18 ∗ 0 ∗ 0.375 − 2 ∗ 10 ∗ √0.375 = −12.25 𝑘𝑃𝑎

𝐴𝑡 𝑧 = 4 𝑚; 𝜎ℎ′ = 18 ∗ 4 ∗ 0.375 − 2 ∗ 10 ∗ √0.375 = 14.75 𝑘𝑃𝑎

𝐴𝑡 𝑧 = 4 𝑚; 𝜎ℎ′ = 18 ∗ 4 ∗ 0.333 − 2 ∗ 0 ∗ √0.333 = 24 𝑘𝑃𝑎

𝐴𝑡 𝑧 = 7𝑚; 𝜎ℎ′ = ((18 ∗ 4 + (21 − 10) ∗ 3) ∗ 0.333 − 2 ∗ 0 ∗ √0.333 = 35 𝑘𝑃𝑎

4 m

3 m

Soil-1

c´=10 kPa

ϕ´=27°

γ=18 kN/m3

Soil-2

c´=0 kPa

ϕ´=30°

γsat=21 kN/m3



12

Calculate the passive pressure (z=depth from ground surface)

𝐴𝑡 𝑧 = 0 𝑚; 𝜎ℎ′ = (21 − 10) ∗ 0 ∗ 3 + 2 ∗ 0 ∗ √3 = 0 𝑘𝑃𝑎

𝐴𝑡 𝑧 = 3 𝑚; 𝜎ℎ′ = (21 − 10) ∗ 3 ∗ 3 + 2 ∗ 0 ∗ √3 = 99 𝑘𝑃𝑎

4 m

3 m

12.25 kPa

14.75 kPa

24 kPa

35 kPa 30 kPa30 kPa 99 kPa

b) Calculate the depth of tension crack at the back of the wall.


′ ∗ 𝐾𝑎 − 2 ∗ 𝑐′ ∗ √𝐾𝑎 = 0

𝜎ℎ′ = 0 = 18 ∗ 𝑧 ∗ 0.375 − 2 ∗ 10 ∗ √0.375

𝑧 =12.25

18 ∗ 0.375= 1.81 𝑚

c) Calculate the total passive resistance in front of the wall.

Total passive resistance (including water pressure)

𝐹𝑝 = (99 ∗ 3 ∗1

2) = 148.5 𝑘𝑁/𝑚

𝐹𝑤 = 30 ∗ 3 ∗1

2= 45 𝑘𝑁/𝑚

Active Pressure

Distribution Passive Pressure

Distribution

Hydrostatic Pressure

Distribution

Hydrostatic Pressure

Distribution