Chapter 4 Compressibility and Consolidation of Soil 4

第四章 土的压缩与固结 Chapter 4 Compressibility and Consolidation of Soil 4. 1 Compaction Vs Consolidation 压实与固结 空气的体积减小 孔隙水的体积减小

What is consolidation? (什么是固结? ) § Consolidation is a time-related (时间性) process of increasing the density (增加密度) of a saturated soil by draining some of the water out of the voids. § Coarse-grained soils (粗粒土), such as sands and gravels, consolidate at a much faster rate than finegrained soils (细粒土) such as silts and clays.

The need of consolidation theory 固结理论的重要性 Most settlement in engineering is because of consolidation § Tilting of Piza Tower is due to differential settlement (不平均沉 降引起塔的倾斜) § Two important questions for consolidation problems (两 个 重 要问题) 1. How much ? (沉降有多少) 2. How long ? (时间有多久) Piza Tower, Italy

Examples Initial Loading condition (荷载) (初始状况) at time = T 0 Earth dam ( 土堤) Excess pore water pressure ( 超孔隙水压力) Static pore water pressure ( 静孔隙水压力)

4. 2 1 -D consolidation test (单向固结试验) Consolidation is a three-dimensional (三维) process, but the direction of flow of water is primarily vertical (主要在竖向) or one-dimensional (1 -D). ’z Simplify vy 简单化 vx vz (water flow) vz

(竖向荷载) Vertical settlement ( 竖向沉降) 侧向应变为零 土样本 Oedometer (固结仪) Load – settlement curve or 1 -D compression curve ( 单向压缩曲线)

1. Time – related consolidation (固结与时间关系) For each load increment (每一个荷载增量) hf ’f

2. 1 -D compression curve (单向压缩曲线) Compression curve Loading (加载 ) (压缩曲线) Unloading and Reloading ( 卸载与再加载) Re-compression curve ( 回弹曲线) where v is change in vertical strain (竖 向应变增量), h is the change in height for each loading increment (每次荷载增 量 的 高 度 改 变 ) and H 0 is the initial specimen height (样本初始高度)

(1)Coefficient of volume compressibility [体积压缩系数] (mv) No lateral strain (侧向应变为零) e 0 where vol is volumetric strain (体积 应变增量), ’v is change in vertical effective stress (应力增量) and e 0 is initial void ratio (初始孔隙比) e 1 ’v 0 ’v 1

Example (例子) How about a 100 -200

(2)Other compressibility parameters 其它压缩参数 Low compresibility Mid compresibility High compresibility

Oedometric modulus [侧限压缩模量] (Es) is defined as the ratio of vertical effective stress (’z) over vertical strain (z) under zero lateral strain condition , which reflects the capability of the soil resisting against compressive deformation, and shown as follows: From generalised Hooke’s law [广义虎克定律], elastic strains [弹性应变] in x, y and z directions are expressed as follows:

where E is Young’s modulus or elastic modulus [弹性模 量]. E represents the stess to strain ratio without confined latral strain. In 1 -D consolidation, ex = ey = 0，we get: soil is not ideal elastic material, above equation is only an approximate formula.

(3) e——log ’ curve 1). Compression Index [压缩指数] (Cc) e 0 e 1 e 2 e 3 ’v 0 ’v 1 2). Expansion Index [回弹指数] (Ce)

4. 3 Calculation formulae of soil compression with zero lateral strain 土的压缩量计算

4. 4 Settlement Calculation 沉降计算 Settlement found in soil due to loading may be divided into three categories: where St is total settlement (总 沉 降 ), Si is immediate settlement (瞬时沉降) – [elastic deformation, 弹性变形], Sc is primary consolidation (主固结沉降) – [change of ’, 有效应力改变] and Ss is secondary consolidation (次固结 沉降) – [time, 时间]

1。 Immediate settlement 瞬时沉降 § Immediate after loading (加荷後立即发生) § No volume change (体积不变) because time is too short to drain away pore-water § Vertical settlement induces lateral expansion (侧向变形) § If no lateral expansion, Si = 0 Similar to rubber (橡皮 )

§ Based on elastic theory (弹性理论 ) where p is the pressure at the bottom of the foundation (基底 压力), b is width of the foundation (基础的宽度) [shorter side of a rectangle (钜形的短边) or diameter of a circle (圆形 的直径)], E is Young’s modulus (弹性模量), is Poisson’s ratio (泊松比) and I is influence factor (影响系数).

2 。Primary consolidation § § 主固结沉降 Primary consolidation is the settlement due to the gradual dissipation of the pore water from the saturated soils or gradual increase in effective stress (有效应力的增加). It is calculated from the following two methods: 1. e : ’ curve 2. e : log ’ curve hf ’f

* e -- ’ curve method

Calculate stress of every layers Two situation for calculation load Consolidated under self-weight No-consolidated under self-weight Look for e 1 ie 2 i Calculate settlement of every layers

Calculate final settlement

4. 5 Effect of stress history 应力历史 where OCR is overconsolidation ratio (超固结比), ’v is current effective stress (现今有效应力) and ’vmax is maximum past effective stress (过去最 大有效应力) or pre-consolidation pressure [ ’c] (前期固结压力) § OCR = 1 Normally consolidated [NC] soil (正常固结土) § OCR > 1 Overconsolidated [OC] soil (超固结土)

* e-- log ’ curve method Overconsolidated Soil

In-situ recompression curve In-situ compression curve

Underconsolidated Soil

Example (例子) The unit weight of sand ( sand) is 20 k. N/m 3 and unit weight of clay ( clay) is 17 k. N/m 3. Calculate the consolidation settlement of the clay deposit after reclamation. Consolidation test results on a clay sample is shown in the following table: Sea level (海平面) 10 m Clay (粘土 ) 6 m Dense sand (密砂) Sand (砂土) 15 m Clay (粘土 ) 6 m Dense sand (密砂) After reclamation (填海后) Pressure (k. N/m 2) Height (mm) 0 19. 0 25 18. 8 50 18. 3 100 17. 6 200 16. 1 400 14. 5

(1) Consider clay deposit as a single layer and in the middle of clay layer: Before reclamation (填海前) 5 m Sand (砂土) 15 m After reclamation (填海后) Clay (粘土 ) 6 m Dense sand (密砂)

& Pressure, ’v (k. N/m 2) Height, H (mm) H/H 0 (H 0 = 19 mm) ’v (k. N/m 2) mv (m 2/k. N) 0 19. 0 25 18. 8 -0. 2 0. 011 25 4. 2 x 10 -4 50 18. 3 -0. 5 0. 026 25 1. 05 x 10 -3 100 17. 6 -0. 7 0. 037 50 7. 4 x 10 -4 200 16. 1 -1. 5 0. 079 100 7. 9 x 10 -4 400 14. 5 -1. 6 0. 084 200 4. 2 x 10 -4

(2) Divide the clay deposit into two layers 5 m Layer Depth of middle of layer (m) ’v 0 (k. N/m 2) ’v 1 (k. N/m 2) H (m) 1 1. 5 7. 2*1. 5 = 10. 8 20*5+10. 2*10 +10. 8 = 212. 8 3 2 4. 5 7. 2*4. 5 = 32. 4 20*5+10. 2*10 +32. 4 = 234. 4 3 Sand (砂土) 15 m 6 m Clay (粘土 ) Dense sand (密砂)

Example (例子) Pressure, ’v (k. N/m 2) Height, H (mm) H/H 0 (H 0 = 19 mm) ’v (k. N/m 2) mv (m 2/k. N) 0 19. 0 25 18. 8 -0. 2 0. 011 25 4. 2 x 10 -4 50 18. 3 -0. 5 0. 026 25 1. 05 x 10 -3 100 17. 6 -0. 7 0. 037 50 7. 4 x 10 -4 200 16. 1 -1. 5 0. 079 100 7. 9 x 10 -4 400 14. 5 -1. 6 0. 084 200 4. 2 x 10 -4

NC soil (正常固结土) e 0 e 1 e 2 e 3 ’ 1 ’ 0 Consider a soil layer of thickness H:

OC soil (超固结土) (i) If final stress ( ’ 1) pre-consolidation pressure ( ’c)

OC soil (超固结土) (ii) If final stress ( ’ 1) > pre-consolidation pressure ( ’c) e’ e’’

An oil tank of 10 m diameter which carries an uniform pressure of 100 k. N/m 2 is founded on a 2 m thick stiff clay deposit which rests on an incompressible dense sand layer. Calculate the settlement of the oil tank. The properties of clay are given as follows: sat = 19 k. N/m 3, Gs = 2. 7, ’c = 76 k. Pa, Cc = 0. 2 and Ce = 0. 05 10 m Oil Tank 100 k. N/m 2 Clay (粘土 ) Dense sand (密砂) 2 m

Consider the clay deposit as single layer and the stresses in the middle of the clay layer are Before building the oil tank After building the oil tank p = 100 k. N/m 2, r 0 = 5 m and z = 1 m Kr = 0. 992 [from 土力学 表 2 -5 p. 55]

4. 6 Terzaghi’s one-dimensional consolidation theory (太沙基-单向固结理论)

1 Assumptions (假设) 1)Soil is homogeneous (均质). 2)Soil is fully saturated (饱和). 3)Solid particles and the pore water are incompressible (土粒与水不能压 缩). 4)Flow of water and compression of soil are one-dimensional (单向). 5)Strains are small (小应变). 6)Darcy’s law is valid at all hydraulic gradients. 7） Coefficient of permeability (渗 透 糸 数 ) [k] and the coefficient of volume compressibility (体 积 压 缩 数 ) [mv] remain constant throughout the consolidation process. 8）There is a unique relationship between void ratio and effective stress.

Establish of equations

Find deformation of T time Find the time needed for a deformation