EGB373-TUT 08-Week 09- Rate of Consolidation- Answers - Colour

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Queensland University of Technology *

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373

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Mechanical Engineering

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Oct 30, 2023

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Q1 (i) What is the definition of H d in this equation? d v v H t c T  H d is the length of the average longest drainage path during consolidation (length of the drainage path) (a)In the case of double (two-way) drainage : H d =H c /2 (Where H c average height of the specimen) (b)In the case of single (one-way) drainage: H d =H c The average height during consolidation=(Initial height + Final height after consolidation)/2 As we deal with very small consolidation, you are allowed to use the initial height as the average height during consolidation if it is not enough information is given Then, H d =(Initial height/2) or (Initial height): depending on the drainage condition( double or single)

Q1. Continue… T v is a non dimensional time factor (no unit): For 50% consolidation, T 50 =0.197 For 90% consolidation, T 90 =0.848 d v v H t c T  What are the typical units of T v

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The average consolidation over the entire depth of clay at particular time as a percentage of the total primary consolidation of the layer, S P =S (∞) The average degree of consolidation, S(t)= The average settlement of the layer at time t S(∞)=Ultimate settlement of the layer from primary consolidation 100 ) ( ) (    S S U t Q1. Continue… What is the average degree of consolidation?

v w v m k C   ion, consolidat of t coefficien The ility compressib volume of t Coefficien m water of ht Unit weig ty permeabili of t Coefficien v w     k Q1. Continue… Mudstone has very low permeability, then why it has reasonably high coefficient of consolidation? Mudstone has low permeability (k) as well as very low compressibility (it is hard to compress by loading). Therefore, it has reasonably high coefficient of consolidation.

Final settlement (after many years) = 40mm Settlement after 3 years = 32 mm % 80 100 * 40 32 %   U 80 %  U 567 . 0  v T 2 d v v H t c T  ?  v c m 98 . 2 2 ) 04 . 0 0 . 3 ( ) 0 . 3 ( layer clay of thickess Average     m H d 49 . 1 2 98 . 2 2 thickness Average ) H ( path drainage longest drainage, way - two Since d    Q2. C H U% years, 3 after ion consolidat of degree Average U ( % ) 0 10 20 30 40 50 60 70 80 90 100 T v 0.000 0.008 0.031 0.071 0.126 0.196 0.287 0.403 0.567 0.848    v m m T M M U 2 0 2 exp 2 1        From the table

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567 . 0  v T 2 d v v H t c T  ?  v c m H d 49 . 1  Q2. Continue… years t 3  year m t H T c v v / 42 . 0 3 49 . 1 * 567 . 0 2 2 2    Coefficient of consolidation, C v

% 50  U ? 50 2     t H t c T d v v year m c v / 42 . 0 2  mm H d 10 2 20 drainage, way - two is Oedometer   Q2. Continue… C H year m c v / 42 . 0 2  If a 20mm thick soil sample taken from the same clay is tested on Oedometer, How long does it take to reach 50 % consolidation of the sample? ? 50  t % 50  U From the table U ( % ) 0 10 20 30 40 50 60 70 80 90 100 T v 0.000 0.008 0.031 0.071 0.126 0.196 0.287 0.403 0.567 0.848    v m m T M M U 2 0 2 exp 2 1        196 . 0  v T years Years c H T t v d v 5 2 2 50 10 667 . 4 42 . 0 1000 10 * 196 . 0            min 46 . 24 60 * 24 * 364 * 10 * 667 . 4 5 50    t

? c V 2   d v v H t c T hr m t H T c d v v / 000011 . 0 hr m 1 ) 0075 . 0 ( * 196 . 0 2 2 2 2    Q3. C H Sample thickness = 15 mm 50 % consolidation occurred in 1 hour h t U 1 % 50 50   U ( % ) 0 10 20 30 40 50 60 70 80 90 100 T v 0.000 0.008 0.031 0.071 0.126 0.196 0.287 0.403 0.567 0.848    v m m T M M U 2 0 2 exp 2 1        % 50 For  U From the table 196 . 0  v T m mm 0075 . 0 5 . 7 2 15 H drainage. way - two is Sample d    

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drainage) way - One ( ? H H c d 2    d v v H t c T v v c T t c H  m m H c 84 . 3 75 . 14 0653 . 0 10 0963 . 0     Q3. Continue…. ?  C H hr m c v / 000011 . 0 2  Final settlement (after many years) = 35mm Settlement after 10 years = 10 mm % 57 . 28 100 * 35 10   U U years, 10 after ion consolidat of degree Average U ( % ) 0 10 20 30 40 50 60 70 80 90 100 T v 0.000 0.008 0.031 0.071 0.126 0.196 0.287 0.403 0.567 0.848    v m m T M M U 2 0 2 exp 2 1        % 57 . 28 For  U From the table 0653 . 0  v T year year m hr m c v / m 0.0963 / 365 24 000011 . 0 / 000011 . 0 2 2 2     

Q4. mm H C 20  kPa 100    0 50 100 150 200 0 20 40 60 80 100 120 Time (minutes) Excess pore w ater pressure (kPa) Excess pore water pressure at A, after loading From the definition of degree of consolidation, at any given point at any given time, t 100 1 ' ' 0 1 0 1 ' 1 0 0 ) , (                                  e e t z u u e e e e U kPa 100   

Q4. mm H C 20  kPa 100    0 50 100 150 200 0 20 40 60 80 100 120 Time (minutes) Excess pore w ater pressure (kPa) Excess pore water pressure at A, after loading Consider excess pore water pressure,u e(z,t) at A, (z = 20 mm) after 100 mm (t =100 min) of the application of stress,  = 100 kPa % 72 100 100 28 1 100 1 ) , (                      e t z u U kPa u e 28  kPa u e 28  Degree of consolidation at A , after 100 min of loading(z= 20mm, t = 100 min), U (20,100)

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Q4. Conti.. mm H C 20  kPa 100    Degree of consolidation at A , after 100 min of loading (z= 20mm, t = 100 min), U (20,100) % 72 ) , (  t z U drainage) way - (One 20 mm H H c d   1 20 20 H z A, at d    z/H d   v m m z T M H Mz M U 2 0 exp sin 2 1              figure) the (From 6 . 0   v T 2 d v v H t c T  H d = 20 mm t =100 min year year t H T C d v v / m 26 . 1 / m 364 24 60 100 1000 20 0.6 2 2 2 2                  