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SOLUTIONS MANUAL For

Principles of Foundation Engineering Ninth Edition

BRAJA M. DAS

NAGARATNAM SIVAKUGAN

Contents Chapter 2............................................................................................................................. 1 Chapter 3........................................................................................................................... 11 Chapter 5........................................................................................................................... 23 Chapter 6........................................................................................................................... 31 Chapter 7........................................................................................................................... 49 Chapter 8........................................................................................................................... 59 Chapter 9........................................................................................................................... 69 Chapter 10......................................................................................................................... 83 Chapter 11......................................................................................................................... 89 Chapter 12......................................................................................................................... 91 Chapter 13....................................................................................................................... 113 Chapter 14....................................................................................................................... 123 Chapter 15....................................................................................................................... 125 Chapter 16....................................................................................................................... 131 Chapter 17....................................................................................................................... 141 Chapter 18....................................................................................................................... 153 Chapter 19....................................................................................................................... 171

Chapter 2 2.1

From Eq. (2.18),

Gs ρw 2450 2.80 × 1000 ; e = 0.0571 = = 1+ e 0.925 1+ e

ρd =

From Eq. (2.6), Porosity, n =

2.2

e 0.0571 = = 0.054 1 + e 1 + 0.0571

From Eq. (2.13), the dry density ρd =

ρ 2060 = = 1786.6 kg/m3 1 + w 1 + 0.153

From Eq. (2.18), ρd = e=

Gs ρw 1+ e

Gs ρw 2.70 × 1000 −1 = − 1 = 0.511 ρd 1786.6

Once saturated, from Eq. (2.19), ρsat =

2.3

(Gs + e ) ( 2.70 + 0.511) × 1000 = 2125.1 kg/m 3 ρw = (1 + e) (1 + 0.511)

Let’s consider a 1-m2 area in plan. The initial volume of this soil is V = 1 × 0.5 = 0.5 m3. Volume of the solids is Vs. e = 0 .9 =

0.5 − Vs ; Vs = 0.2632 m3 Vs

Ws = 0.2632 × 2.68 × 9.81 = 6.919 kN The new volume after compaction = 1 × 0.455 = 0.455 m3 The dry unit weight, γd = From Eq. (2.12), γd =

6.919 = 15.21 kN/m3 0.455

Gs γw 1+ e

SOURCE: Browsegrades.net

Gs γ w 2.68 × 9.81 −1 = − 1 = 0.729 γd 15.21

From Eq. (2.13), γ = (15.21)(1 + 0.20) = 18.25 kN/m 3

2.4

At the compacted road base, the weight of solids, Ws = 120,000 × 19.5 = 2,340,000 kN At the borrow pit, the dry unit weight, γd = Volume of the pit, V =

γ 17.5 = = 16.13 kN/m 3 1 + w 1 + 0.085

2,340,000 = 145,080 m3 16.13

The moisture content has to be increased from 8.5% (at the borrow pit) to 14.0% (at the road base). The quantity of water to add, 2,340,000 × (0.14 ‒ 0.085) = 128,700 kN Volume of water to be added =

2.5

γd =

γ 110.4 = = 99.9 lb/ft 3 1 + w 1 + 0.105

γd =

Gs γ w 2.65 × 62.4 ; e= − 1 = 0.655 1+ e 99.9

From Eq. (2.23), Dr =

2.6

2.7

128,700 = 13,119.3 m 3 9.81

0.870 − 0.655 × 100 = 60.6% 0.870 − 0.515

a. A-1-a

c. A-3

b. A-1-b

d. A-7-6

Soil A: % of gravel = 50, % of sand = 13, % of fines = 37 D10 = 0.035 mm, D30 = 0.061 mm, D60 = 9.8 mm Cu = 280; Cc = 0.02 LL = 58, PL = 34, PI = 24 plots below the A-line; hence, silt The soil can be described as poorly (gap) graded sandy silty gravel with a group symbol of GM.

SOURCE: Browsegrades.net

Soil B: % of gravel = 24, % of sand = 69, % of fines = 7 D10 = 0.17 mm, D30 = 0.82 mm, D60 = 2.6 mm Cu = 15.3; Cc = 1.5 LL = 42, PL = 22, PI = 20 plots above the A-line; hence, clay The soil can be described as well graded clayey gravelly sand with a group symbol of SW-SM. Soil C: % of gravel = 1, % of sand = 99, % of fines = 0 D10 = 0.7 mm, D30 = 1.2 mm, D60 = 1.6 mm Cu = 2.3; Cc = 1.3 The soil can be described as poorly (uniformly) graded sand with a group symbol of SP. Soil D: % of gravel = 0, % of sand = 12, % of fines = 88 LL = 75, PL = 31, PI = 44 plots above the A-line; hence, clay. The soil can be described as sandy clay of high plasticity with a group symbol of CH.

2.8

The head loss from the reservoir to the ditch, ∆h = 38 − 28 = 10.0 m The length of the sand seam in the direction of the flow, L = 200 cos10 = 203.1 m The hydraulic gradient, i = 10.0 203.1 = 0.0492 By Darcy’s law [Eq. (2.35)], v = ( 2.6 × 10 −5 m/s)( 0.0492 ) = 0.128 × 10 −5 m/s The cross section of the sand seam through which the flow takes place is

1.0 × 500.0 = 500.0 m2 The flow rate = (0.128 × 10 −5 m/s)(500.0 m 2 ) = 64.0 × 10 −5 m 3/s Volume of water flowing into the ditch per day: = (64.0 × 10 −5 m 3/s)(24 )( 3600) = 55.3 m 3

2.9

For the flow net shown in Figure P2.9, Nf = 3 and Nd = 10 Total head loss from right to left, hmax = 5.0 m The flow rate is given by [Eq. (2.46)] q = khmax

 3 = (1.5 × 10−5 )(5.0)  = 2.25 × 10−5 m 3/s/m length Nd  10 

= ( 2.25 × 10− 5 )(50.0)( 24)( 3600 m 3/day) = 97.2 m 3 /day

SOURCE: Browsegrades.net

2.10

On top of the soft clay layer (i.e., at 10 m depth), initially: σ ′ = 1 × 17.0 + 9( 20 − 9.81) = 108.7 kN/m 2

After the water table is lowered, σ ′ = 3 × 17.0 + 7( 20 − 9.81) = 122.3 kN/m 2

By lowering the water table, the effective stress has increased by (122.3 – 108.7) = 13.6 kN/m2

2.11

The soil below the water table can be assumed to be fully saturated (i.e., S = 1).

e = wGs = 0.25 × 2.70 = 0.675 The saturated unit weight can be computed as

γsat =

(Gs + e)γw (2.70 + 0.675) × 9.81 = = 19.8 kN/m3 1+ e 1 + 0.675

At a depth of 5 m into the sandy clay, σ = 4 × 9.81 + 5 × 19.8 = 138.3 kN/m2 u = 9 × 9.81 = 88.3 kN/m2 σ ′ = σ − u = 138.3 − 88.3 = 50 kN/m2

2.12

Refer to the figure.

a. The compression index Cc is given by [Eq. (2.53)],

Cc =

e1 − e2 1.10 − 0.85 = = 0.441 log σ 2′ − log σ1′ log 240 − log 65 SOURCE: Browsegrades.net

b. Let the void ratio at 460 kN/m2 pressure be e3 .

 460  e1 − e3 = Cc (log 460 − log 65) = 0.441 × log  = 0.375  65  e3 = 1.10 − 0.375 = 0.725

2.13

a. The clay is below the water table and, hence, is saturated. The initial void ratio eo can be determined as eo = wG s = 0.225 × 2.72 = 0.612 The saturated unit weight is determined as

γsat =

(Gs + e)γw ( 2.72 + 0.612)(9.81) = 20.3 kN/m3 = 1+ e 1 + 0.612

The effective overburden stress at the middle of the clay is

σ o′ = 2 × 17.0 + 3( 20.2 − 9.81) + 1.5( 20.3 − 9.81) = 80.3 kN/m2 < 110.0 kN/m2 Since the preconsolidation pressure is greater than the current overburden pressure, the clay is overconsolidated. The overconsolidation ratio

OCR = 110.0 80.3 = 1.37 b. The 2-m-high compacted fill imposes a surcharge of 2 × 20 = 40 kN/m2 (i.e., ∆σ ′ = 40.0 kN/m2 , σo′ = 80.3 kN/m2 , and σ c′ = 110.0 kN/m2 ) Since σ o′ + ∆σ ′ > σ c′ , the consolidation settlement can be computed from Eq. (2.69) as Sp = =

 σ′  C H  σ ′ + ∆σ'  Cs H  log c  + c log o ′ ′ 1 + eo σ 1 + e σ o o c     0.06 × 3000  110.0  0.52 × 3000  80.3 + 40.0  log log +  1 + 0.612 1 + 0.612 110  80.3   

= 52.9 mm

2.14

For U = 75%, Tv = 0.477 (Table 2.12)

Tv =

cv t H dr2

From two-way (doubly drained) to one-way (singly drained), Hdr is doubled. For the same U and, hence, the same Tv, this would increase the time fourfold. Therefore, it will take 4t years. SOURCE: Browsegrades.net

2.15

a. The clay layer with one-way drainage has Hc of 6.0 m. After one year,

Tv =

cvt 0.0014 × 365 × 24 × 3600 = = 0.123 ; settlement, Sc(t) = 160 mm H2 6002

From Table 2.12, U = 39.6%

Sc ( t ) Sc (max)

; Sc (max) = 160 0.396 = 404 mm

When t = 2 years, Tv = 0.246. From Table 2.12, U = 55.8%. Consolidation settlement during the first two years is 0.558 × 404 = 225.4 mm b. The initial effective overburden stress at the middle of the clay is σo′ = 1.5 × 17.0 + 0.5(18.5 − 9.81) + 3.0(19.0 − 9.81) = 57.4 kN/m2

∆σ ′ = 3 × 19 = 57 kN/m2 Sc =

 σ o′ + ∆σ ′  Cc H  log 1 + eo σ o′  

404 =

2.16

Cc × 6000  57.4 + 57.0  log ; Cc = 0.41 1 + 0.810 57.4  

a. For the clay layer, assuming S = 100% below the water table, eo = 0.45 × 2.70 = 1.215

γsat =

(Gs + e)γw (2.70 + 1.215) × 9.81 = 17.3 kN/m3 = 1+ e 1 + 1.215

The initial effective overburden stress at the middle of the clay layer is

σo′ = 1 × 16 + 1(19.0 − 9.81) + 1.5(17.3 − 9.81) = 36.4 kN/m2 With OCR = 1.5, the preconsolidation pressure σ c′ = 1.5 × 36.4 = 54.6 kN/m2 When the fill is placed, it imposes a surcharge of ∆σ ′ = 20 × 1.5 = 30.0 kN/m2 From Eq. (2.69), Sc = =

 σ′  C H  σ ′ + ∆σ'  Cs H  log c  + c log o 1 + eo  σ o′  1 + eo  σ c′  0.08 × 3000  54.6  0.65 × 3000  66.4  log log +  = 93.9 mm 1 + 1.215 1 + 1.215  36.4   54.6  SOURCE: Browsegrades.net

b. Including the fill load and the warehouse load, ∆σ ′ = 30 + 40 = 70 kN/m2

Sc = =

 σ′  C H  σ ′ + ∆σ'  Cs H  log c  + c log o 1 + eo  σo′  1 + eo  σ c′  0.08 × 3000  54.6  0.65 × 3000  106.4  log log +  1 + 1.215 1 + 1.215  36.4   54.6 

= 274.2 mm Consolidation settlement due to the warehouse alone is 274.2 – 93.9 = 180.3 mm

2.17

The direct shear test data are plotted in the figure. From the failure envelope, c′ = 6.5 kN/m2 and ϕ' = tan‒1(0.4292) = 23.2°

2.18

a. σ3′ = 100 kN/m2 and ∆σ f = 260 kN/m2 Therefore, σ1′ = σ 3′ + ∆σ f = 360 kN/m 2

φ′  φ′    From Eq. (2.91), σ1′ = σ 3′ tan 2  45 +  + 2c′ tan  45 +  2 2   For normally consolidated clays, c′ = 0; hence,

φ′   360 = 100 tan 2  45 +  ; φ ′ = 34.4 ° 2 

SOURCE: Browsegrades.net

b. For the second specimen, σ3′ = 200 kN/m2

34.4   2 σ1′ = 200 tan 2  45 +  = 719.5 kN/m 2  

∆σ f = 719.5 − 200 = 519.5 kN / m 2

2.19

a. In normally consolidated clay, c′ = 0 For the first specimen (consolidated drained test), using Eq. (2.91),

φ′   260 + 150 = 410 = 150 tan 2  45 +  ; φ ′ = 27.7° 2  In the second specimen (consolidated undrained test), applying the same value of φ ′ in Eq. (2.91),

σ3 = 150 kN/m 2 and ∆σ f = 115 kN/m 2 σ1 = σ 3 + ∆σ f = 265 kN/m 2

σ3′ = 150 − u f and σ1′ = 265 − u f

φ′  27.7    σ1′ = σ 3′ tan 2  45 +  = σ 3′ tan 2  45 +  = 2.737σ 3′ 2 2    265 − u f = (150 − u f ) × 2.737; u f = 83.8 kN/m2

b. From Eq. (2.96), A f =

2.20

uf ∆σ f

83.8 = 0.73 115

At failure the pore water pressure is uf, σ3 = 100 kN/m2 and σ1 = 207 kN/m2.

σ3′ = 100 − u f and σ1′ = 207 − u f Substituting for σ3′ and σ1′ in Eq. (2.91),

φ′  φ′    σ1′ = σ 3′ tan 2  45 +  + 2c′ tan 45 +  2 2  

(207 − u f ) = (100 − u f )tan 2  45 + 262  + 2 × 10 tan 45 + 262  







uf = 51.9 kN/m2 SOURCE: Browsegrades.net

2.21

The following table can be prepared from the given data, and the Mohr circles are plotted as shown in the figure. Sample No.

(kN/m )

1 2 3

100 200 350

σ3 2

(∆σ d ) f

(∆ud ) f

(kN/m ) 88.2 138.5 232.1

(kN/m ) 57.4 123.7 208.8

σ3′

σ1

σ1′

(kN/m )

(kN/m2)

188.2 338.5 582.1

42.6 76.3 141.2

130.8 214.8 373.3

The failure envelope is drawn tangent to the Mohr circles in the figure and, from measurements, c′ = 10.0 kN/m2 and ϕ' = 24.7°

2.22

The unconfined compressive strength of the clay specimen qu = 2cu = 90 kN/m2 Cross sectional area of the specimen =

π 4

× 752 = 4417.9 mm2

Maximum load the specimen can carry = 90 × 4417.9 × 10−6 = 0.398 kN = 398 N Weight of one steel plate = 1.5 × 9.81 N = 7.358 N Therefore, number of plates that can be stacked on the specimen = 398 7.358 = 54 With qu = 90 kN/m2 (see Table 2.14), it is a medium clay (consistency).

2.23

a. From Figure P2.7, D10 = 0.7 mm, D15 = 0.9 mm, D30 = 1.2 mm, D50 = 1.4 mm, D60 = 1.6 mm, and D85 = 2.05 mm

D60 1.6 D302 1 .2 2 = = 2.3 ; Cc = = = 1.3 D10 × D60 0.7 × 1.6 D10 0.7 From Eq. (2.87), φ ′ = 26 + (10 × 0.8) + (0.4 × 2.3) + 1.6 log(1.4) = 35.2° Cu =

SOURCE: Browsegrades.net

D   2.05  b. From Eq. (2.89), a = 2.101 + 0.097  85  = 2.101 + 0.097  = 2.322  0 .9   D15  From Eq. (2.90), b = 0.845 – 0.398a = 0.845 – 0.398×2.322 = ‒ 0.0792

From Eq. (2.88),

1  1   −1   = tan   = 36.8°  ae + b   2.322 × 0.61 − 0.0792 

φ ′ = tan −1 

SOURCE: Browsegrades.net

Chapter 3 3.1

Do = 3.0 in.; Di = 3.0 ‒ 2 × 0.065 = 2.870 in. From Eq. (3.3), the area ratio is given by AR (%) =

3.2

Do2 − Di2 3.02 − 2.870 2 × 100 = × 100 = 9.3% Di2 2.870 2

Assume the wall thickness is t (mm). Therefore, the outer diameter is (75 + 2t) mm. For the samples to be undisturbed, AR < 10%. From Eq. (3.3), (75 + 2t )2 − 752 AR = × 100 ≤ 10 752 2

 75 + 2t   75 + 2t    ≤ 1.1 ;   ≤ 1.0488; t ≤ 1.83 mm  75   75 

The maximum possible wall thickness is 1.83 mm. 3.3 Depth from ground surface (m)

cu (kN/m2)

OCR

N60

[Eq. (3.8b)]

σ′o (MN/m2)

[Eq. (3.9)]

3.0

76.5

1 [(1.5)(16.5) + (1.5)(19 − 9.81)] = 0.03854 1000

5.51

4.5

107.3

0.03854 +

1 (1.5)(16.8 − 9.81) = 0.0490 1000

6.46

6.0

107.3

0.0490 +

1 (1.5)(16.8 − 9.81) = 0.0595 1000

5.65

7.5

116.8

0.0595 +

1 (1.5)(16.8 − 9.81) = 0.07 1000

5.48

9.0

126.0

0.07 +

1 (1.5)(16.8 − 9.81) = 0.0805 1000

5.35

Average cu = 106.78 kN/m2

Average OCR = 5.69

SOURCE: Browsegrades.net

3.4 Depth from ground surface N60 (m) 3.0 5 4.5 8 6.0 8 7.5 9 9.0 10 a See Problem 3.3

(kN/m2) 38.54 49.0 59.5 70.0 80.5

Depth (m)

σ o′ (kN/m2 )

σ o′ a

OCR [Eq. (3.10)]

σ c′ (kN/m2 )

7.52 9.47 8.0 7.46 7.2

235 376 376 423 470

[Eq. (3.11)]

3.5 18 × 1.5 = 27 18 × 3.0 = 54 18 × 4.5 = 81 18 × 6.0 = 108 108 + (1.5)(20.2 – 9.81) = 123.6 123.6 + (1.5)(20.2 – 9.81) = 139.2

1.5 3.0 4.5 6.0 7.5 9.0

 1  Eq. (3.13) : CN =    (σ o′ pa ) Depth (m)

N60

σ o′ (kN/m 2 )

0.5

(N1 )60 a

12 11 10 8 12 12

1.5 6 27 1.92 3.0 8 54 1.36 4.5 9 81 1.11 6.0 8 108 0.96 7.5 13 123.6 0.9 9.0 14 139.2 0.85 a Rounded off to nearest whole number

3.6

CN =

2 1 + (σ o′ pa )

Depth (m)

N60

σ′o a (kN/m2)

(N1)60 b

1.5 3.0 4.5 6.0 7.5 9.0

6 8 9 8 13 14

27 54 81 108 123.6 139.2

1.57 1.3 1.1 0.96 0.89 0.84

9 10 10 8 12 12

a b

See Problem 3.5 Rounded off to nearest whole number

SOURCE: Browsegrades.net

3.7

From Problem 3.5, the average value of ( N1 )60 =

1 (12 + 11 + 10 + 8 + 12 + 12) = 10.83 ≈ 11 6

Eq. (3.31b): φ ′ = 15.4( N1 )60 + 20 = (15.4)(11) + 20 = 33°

3.8

3.9

From Problem 3.5, Depth (m)

σ′o (kN/m2)

pa (kN/m2)

1.5 3.0 4.5 6.0 7.5 9.0

27 54 81 108 123.6 139.2

100 100 100 100 100 100

φ ′ (deg) [Eq. (3.30)] 6 34.7 8 34.9 9 34.0 8 31.4 13 34.9 14 34.9 Average φ ′= 34.1° ≈ 34°° N60

Eq. (3.29): φ ′ = 27.1 + 0.3( N1 )60 − 0.00054[( N1 )60 ]2 Depth (m) 1.5 3.0 4.5 6.0 7.5 9.0

N60

φ ′ (deg)

[Eq. (3.29)] 12 30.6 11 30.4 10 30.05 8 29.47 12 30.6 12 30.6 Average φ ′ = 29.93° ≈ 30.45°°

3.10 Depth (m)

σ′o (kN/m2)

pa (kN/m2)

N60

Dr (%) [Eq. (3.22)]

1.5 3.0 4.5 6.0 7.5 9.0

27 54 81 108 123.6 139.2

100 100 100 100 100 100

6 8 9 8 13 14

50.6 51.8 49.7 43.2 52.8 52.7

Average Dr = 50.13% ≈ 50%

SOURCE: Browsegrades.net

0.5

3.11

 1  Eq. (3.13): CN =   ; ( N1 )60 = C N N60  (σo′ pa ) 0 .5

Depth (m) 1.5 3.0 4.5 6.0 7.5 9.0

σ′o (kN/m2) 27 54 81 108 123.6 139.2

pa (kN/m2) 100 100 100 100 100 100

N60

1.08( N1 )60

6 8 9 8 13 14

1.92 1.36 1.11 0.96 0.9 0.85

12.44 11.75 10.78 8.29 12.64 12.85

 1.08( N1 ) 60  Dr (%) =   60   [Eq. (3.28)] 45.3 44.3 42.4 37.2 45.9 46.3 Average = 43.56% ≈ 44%

3.12 CN γavg Depth (m) (kN/m3) Liao & Whitman Skempton [Eq. (3.13)] [Eq. (3.14)] 3.5 18.0 1.27 1.23 8.8 18.0 0.80 0.78 12.4 18.5 0.66 0.62 18.5 18.5 0.54 0.46 23.6 19.0 0.48 0.37 26.9 19.0 0.45 0.33 a

3.13

(N1 )60 a

N60

Liao & Whitman Skempton [Eq. (3.13)] [Eq. (3.14)]

12 8 15 19 21 26

15 6 10 10 10 12

15 6 9 9 8 9

Rounded off to nearest whole number

Eq. (3.13):

 1  CN =    σo′ pa 

0.5

  1 =   (25 × 110) 2000

0.5

= 0.85

( N 1 ) 60 = C N N 60 = ( 0.85)( 28) ≈ 24 a. From Eq. (3.21):

  σ′  Dr = 12.2 + 0.75222 N 60 + 2311 − 711OCR − 779  o  − 50Cu2   pa   

0.5

  25 × 110  2 = 2.2 + 0.75222 × 28 + 2311 − 711 × 1 − 779  − (50)3.2   2000    = 71.4%

0.5

SOURCE: Browsegrades.net

b. From Eq. (3.22):     N 60   Dr =    17 + 24  σ o′    p     a   

0. 5

    28   =   17 + 24  25 × 110       2000     

0 .5

= 0.748 or 74.8%

c. From Eq. (3.23): 1 .7    0.06    N 60  0.23 +  D50   1     Dr =  ×  9 σ o′ pa      

0 .5

1 .7    0.06    28 0.23 +    1 1 .2     = ×   9  ( 25 × 110 ) 2000     

0 .5

= 0.510 or 51.0%

d. From Eq. (3.24): C P = 60 + 25 log D50 = 60 + 25 log 1.2 = 62 5000 = 1 .28 100

C A = 1 .2 + 0 .05 log

COCR = OCR1.8 = 1  ( N1 )60  Dr =    CPC ACOCR  e.

0.5

  24 =   (62)(1.28)(1) 

0.5

= 0.55 or 55%

From Eq. (3.28): (N )  Dr =  1 60   60 

0 .5

 24  =   60 

0 .5

= 0.632 or 63.2%

SOURCE: Browsegrades.net

3.14 Friction angle (deg) CN σ′o Depth N60 (N1)60 (m) (kN/m2) Eq. (3.13) Eq. (3.29) Eq. (3.30) Eq. (3.31b) 2.0 37.0 17 1.644 27.9 35.1 43.6 40.7 3.5 64.8 23 1.243 28.6 35.2 44.1 41.0 5.0 92.5 26 1.040 27.0 34.8 43.3 40.4 6.5 120.3 28 0.912 25.5 34.4 42.4 39.8 8.0 148.0 29 0.822 23.8 33.9 41.3 39.2

3.15

a. At 25 ft depth, σo′ = (10)(111.5) + (15)(125.0 − 62.4) = 2054.0 lb/ft 2 From Eq. (3.13),

 1  CN =    σo′ pa 

0.5

  1 =   2054 2000 

0.5

= 0.99

(N1)60 = CN N60 = 0.99 × 26 ≈ 26

b. From Eq. (3.28), (N )  Dr =  1 60   60 

0 .5

 26  =   60 

0. 5

= 0.658 or 65.8%

c. From Eq. (3.30), −1 

 N 60 φ ′ = tan   12.2 + 20.3(σo′ pa )

0.34

−1 

 26 = tan   12.2 + 20.3(2054 2000)

0.34

= 42.7° d. From Eq. (3.32), Es = 10N 60 pa

Es = (10)(26)( 2000) = 520,000 lb/ft 2

SOURCE: Browsegrades.net