土力学地基基础课后答案_清华大学出版社(陈希哲第四版)

5.1(P236)

填土面 (1)求静止土压力值 干砂γ = 18 kN / m 3 = 36o

(2)求产生主动土压4 .0 m

力所需的位移 (3)求主动土压力值

(1) 1K o ≈ 1 sin ′ ≈ 1 sin 36 o = 0 . 412 1 1 2 Po = γ H K o = * 18 .0 * 4 .0 2 * 0 .412 = 59 .33 kN / m 2 2 (2) 密实砂土，产生主动土压力所需的位移约为墙高的0.5%,

即： ≈ H * 0 .5 % = 4 .0 * 0 .5 % = 0 .02 m = 20 .0 mm (3) K = tan2 (45o / 2) = tan2 (45o 36o / 2) = 0.2596 a

1 1 2 Pa = γ H K a = * 18 .0 * 4 .0 2 * 0 .2596 = 37 .38 kN / m 2 2

5.2(P237) 填土面 1γ = 18 kN / m 3

2

砂 土

= 36

o

2 . 0 m 地下水位面

γ sat = 21 .0 kN / m 3

(1)求静止土压力值 (2)求主动土压力值 (3)求墙后水压力值

2 .0 m

3 K o ≈ 1 sin ′ ≈ 1 sin 36 o = 0 . 412 (1) 根据 e o = γ h K o 得：

e o1 = 0 e o 2 = γ h1 K o = 18 * 2 . 0 * 0 . 412 = 14 . 832 kPa eo 3 = (γ h1 + γ w h2 ) K o = (18 * 2 .0 + 11 . 0 * 2 ) * 0 .412 = 23 .896 kPa

填土面 114 . 832

2 .0 m

22 .0 m

3 23 . 8961 1 Po = *14.832 * 2.0 + * (14.832 + 23.896) * 2.0 = 53.56kN / m 2 2

K a = tan 2 ( 45 o / 2 ) = tan 2 ( 45 o 36 o / 2 ) = 0 .2596 (2)

根据 e o = γ h K o 得：e a1 = 0 e a 2 = γ h1 K a = 18 * 2 . 0 * 0 . 2596 = 9 . 346 kPa ea 3 = (γ h1 + γ w h2 ) K a = (18 * 2 .0 + 11 .0 * 2 ) * 0 .2596 = 15 .057 kPa

填土面 19 . 346

2 .0 m

22 .0 m

3 15 . 057

1 1 Pa = * 9.346 * 2.0 + * (9.346 + 15.057) * 2.0 = 33.75kN / m 2 2

(3) Pw =

1 1 2 γ w hw = *10 * 2.0 2 = 20.0kN / m 2 2

5.3(P237) 根据 ε = β = 0 = 36 o

δ = 24o 查表5.1得： K a = 0.25

1 1 2 E a = γ H K a = * 18 . 0 * 4 . 0 2 * 0 . 252 = 36 . 288 kN / m 2 2

5.4(P237) 根据 ε = 0 = 30 o δ = 20o β = 12 o 查表得： K a = 0.35481 1 2 E a = γ H K a = * 17 .0 * 5 .0 2 * 0 .3548 = 75 .40 kN / m 2 2 Eax = 75.4 * cos 20o = 70.85kN / m Eay = 75.4 * sin 20o = 25.79kN / m

5.5(P237) 1.5m12oEay

Eax = 75.4 * cos 20o = 70.85kN / m

Eay = 75.4 * sin 20o = 25.79kN / m

5.0mEax

5/3 2.5m Ks = (G + Eay ) µ Eax [22* (1.5 + 2.5) *5* 0.5 + 25.79]* 0.4 = = 1.39 > 1.3 70.85 可以

G a + Eay b (22*1.0*5 / 2) *1/ 3 + (22*1.5*5 / 2) *(1.0 +1.5 / 2) Kt = = Eax 5 / 3 70.85*5 / 3

= 1.38 < 1.6 不可以

5.6(P237) 1 2 3 4

q = 20kPa中 γ 1 = 18.5kN / m3 砂 1 = 30 o 粗 砂 1γ 2 = 19.0kN / m 3= 35o

求主动土压 力和水压力 3.0m 3.0m 4.0m 地下水

γ 2sat = 20.0kN / m3

Ka1 = tan2 (45o 1 / 2) = tan2 (45o 30o / 2) = 0.333 Ka 2 = tan2 (45o 2 / 2) = tan2 (45o 35o / 2) = 0.271 p a 1 = 0 + q K a 1 = 20 . 0 * 0 . 333 = 6 . 666 kPa

p a 2 上 = ( q + γ 1 * h1 ) * K

a1

= ( 20 + 18 . 5 * 3 . 0 ) * 0 . 333 = 25 . 17 kPa

p a 2 下 = ( q + γ 1 * h1 ) * K a 2

= ( 20 + 18 . 5 * 3 . 0 ) * 0 . 271 = 20 . 46 kPap a 3 = ( q + γ 1 * h1 + γ

2 h 2 ) * Ka2

= ( 20 + 18 . 5 * 3 . 0 + 19 . 0 * 3 . 0 ) * 0 . 271 = 35 . 91 kPa′ p a 4 = ( q + γ 1 * h1 + γ 2 h2 + γ 2 h3 ) * K a 2

= ( 20 + 18 .5 * 3.0 + 19 .0 * 3.0 + 10 * 4.0) * 0.271 = 46 .75 kPa6 . 66

125 . 17

3.0m 3.0m

2 3 4

20 .4635 .91

4.0m46 . 75

1 E a 1 = * ( 6 . 666 + 25 . 17 ) * 3 . 0 = 47 . 75 kN / m 2 6.666 * 3.0 *1.5 + (25.17 6.666) * 3.0 *1.0 z p1 = = 1.791m 47.75

Ea2 z p2

1 = * ( 20 .46 + 35 .91) * 3 .0 = 84 .55 kN / m 2 20.46 * 3.0 *1.5 + (35.91 20.46) * 3.0 *1.0 = = 1.637m 84.55 1 = * ( 35 . 91 + 46 . 75 ) * 4 . 0 = 165 . 32 kN / m 2

E a3 z p3

35.91* 4.0 * 2.0 + (46.75 35.91) * 4.0 * 4.0 / 3 = = 2.087m 165.32

Ea = Ea1 + Ea 2 + Ea3 = 47.15 + 84.55 + 165.32 = 297.02kN / m1 1 2 E w = γ w h w = * 10 * 4 . 0 2 = 80 kN / m 2 2

1Ea1 = 47.75kN / m1.791m

3.0m

2Ea 2 = 84.55kN / m

3.0m

3

1.637Ea 3 = 165.32kN / m

4

2.087m

4.0mγ w hw = 40 .0 kPa

za

∑z E = ∑Eai ai

ai

=

(1.791+ 7.0) * 47.75+ (1.637+ 4.0) *84.55+ 2.087*165.32 297.02

= 4 . 18 m

5.7(P237)

1.0m6o20o 中 砂Eay

γ1 =17.0kN/ m3 1 = 30

7.0mo

Eax

1.5m3 细 γ 2 = 18.0kN / m 砂 = 20 o 5.0m 2

µ = 0 .4

(1)求作用在墙背上的土压力； (2)作用在墙前趾上的土压力， (3)验算挡土墙抗滑稳定性。

(1)作用在墙背上的土压力； 根据 ε = 20 o = 30 o δ = 15o β = 6 o 查表得： K a = 0.381

1 1 2 E a = γ H K a = * 17 .0 * 7 .0 2 * 0 .381 = 158 .69 kN / m 2 2Eay = 158.69 * sin(δ + ε ) = 158.69 sin 35o = 91.02kN / mEax = 158.69 * cos(δ + ε ) = 158.69 cos 35o = 129.99kN / m

(2)作用在墙前趾上的土压力

ε = tan 1[(5.0 1.0 7.0 tan20o ) / 7.0] = 11.72o根据 ε = 11 . 72 o β =0 = 20 o δ = 10o

ε

K 按式5.16计算得： p = 2.228

1.5m

1 1 2 E p = γ H K p = * 18 .0 * 1 .5 2 * 2 .228 = 45 .117 kN / m 2 2E py = 45.117 * sin(ε δ ) = 45.117 sin1.72o = 1.35kN / mE px = 45.117 * cos(ε δ ) = 45.117 cos1.72o = 45.10kN / m

(3)验算挡土墙抗滑稳定性。δ = 10o

ε ε

(G + Eay + Epy ) µ [24* (5.0 +1.0) * 7.0 / 2 + 91.02 +1.35]* 0.4 Ks = = = 2.81 Eax Epx 129.99 45.10

> 1 . 3 安全

5.8(P237) 根据 θ = 30 oN= c γ H = 20 o

查图5.48得：N = 0 . 026H= c 5 = = 12m γ N 16.0 * 0.026

5.9(P237)c 7 .0 N = = = 0 . 0389 γ H 18 . 0 * 10 . 0o 根据 = 20 N = 0 . 0389

查图5.48得：

θ = 35o

5.10(P237)

i = 1:1

γ = 18 kN / m 3 = 20 o c = 5.4kPa γ = 19 kN / m 3 = 16 oc = 10kPa

3.0m 3.0m 7.0m

6 4 2

8

6 4 2

第1个小土条,b=1.0m

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