Simulate Soil Consolidatio编程代写

Simulate Soil Consolidation

Due date 11:59PM Friday 28 May April 2021 (local

Sydney time)

You are tasked to write the parts needed for the simulation of soil consolidation.

Changes to description

This document is being updated:

Students should refer to the Ed website for any clari�cations, or changes to being made to the

assignment description.

A brief list of changes is Documented here.

9 May 2021

CLAY_COMPRESSIBILITY_RATE is 0.001. It is a constant dening water units moved per 1kN per

hour (no longer 100kN)

A message to all students about Academic Integrity

This is an assignment and sta are not permitted to give specic guidance on your code, or how to

solve the specic problem. That is the purpose of the assessment that you are required to perform to

achieve the grade.

You may ask clarication questions about the assignment description. This is often necessary to

implement functionality that is otherwise ambiguous.

The assignment description is not intended to be complete and you can conrm your assumptions in

a form of a question. In asking the question you should be quoting the description you are asking

about.

If you have a question to Ask on Ed please search before asking. However, remember that you

should not be posting any assignment code publicly, as this would constitute academic

dishonesty.

Also, do not wait too long before starting. This assignment needs time and sustained eort.

Background

Consolidation is the gradual changes in volume of a partly or fully saturated soil when subject to a

sustained load. The changes are mainly due to the removal of gases, uids and organic matter from

the soil. We simplify our model to consider this matter as water.

A sample of soil shows the particles arranged with voids between them.

Soil can be composed of many minerals, primary silica, but clay, sand, shale, rock. Some of these

have more water content than others. Water is present within soil and we can assume water is an

incompressible uid, where any pressure applied will cause the water to move to a lower pressure.

The water is eectively squeezed out of the soil very slowly.

Soil consolidation has a huge impact on the planning and construction of buildings throughout

history. The leaning tower of Pisa is a great example to showcase the importance of soil consolidation

[https://www.geoengineer.org/education/web-class-projects/ce-179-geosystems-engineeringdesign/assignments/the-tilt-of-the-tower-of-pisa-why-and-how

https://www.youtube.com/watch?v=nK4oDD-4CeE

The purpose of the simulation is to determine:

how long consolidation will take for a given load and placement over a soil conguration

how much water is displaced during consolidation

what are the changes in Height of the soil after consolidation

Simulation Input and Output

functionality for le input/output provided for you

The simulation will require information about the nature of the soil, the load, and the parameters

used to simulate. These are read from a le using the three command line arguments.

$ simulate.py <simulation parameters file path> <soil data file path> <output file pat

Your program will read in a le for the simulation parameters from argument 1

Your program will read in a le for the soil geometry and composition from argument 2

Your program will write to a �le for the results of the simulation from argument 3

For example:

$ simulate.py tests/params_example1.in tests/soil_example1.in sim_results_p1_s1.txt

Modelling of the problem

The modelling of soil consolidation in this assignment makes the following assumptions:

soil particles have no air

soil particles initially have a capacity to hold water and is considered full

for example, if clay can hold 40% water, then at the beginning of the simulation, the clay

particle holds 40% water.

water is an incompressible uid

water moving out of a particle will cause the particle to compress

any amount of water removed from a particle cannot be reintroduced

soil particle categorisation is limited to Clay and Shale and only relevant to the initial conditions

bedrock is an incompressible Particle and will always provide an equal and opposite reactive

force

The void particle is a simple characterisation of representing a lower pressure area and it is

assume to have no volume or capacity. A sand column could be represented as a lower

pressure region, but as a void it has no capacity.

We model a particle of soil.

soil consists of a mixture of solid matter (aggregate) and water.

soil particle has a capacity to hold water. This is dictated by the soil type and the water capacity

value [0,1]

soil particle has pressure acting on it.

Initially the soil particle is at a rest state, in equilibrium with its neighbours.

adding force to the soil will cause change in pressure and result in a movement of water

assume a particle is 1 unit wide and 1 unit deep (square)

We model the movement of water.

water is an incompressible uid.

the entire soil mass is considered as a body of uid where the pressure is even throughout.

water reaching a void in the soil Will cause it to leak out (sink). Void are explicitly dened in

every simulation.

pressure acting on the soil will cause the equivalent water mass to leave the exit points of the

soil.

There should always be a leak point, place for water to be displaced.

If soil has no leak points, the pressure will continue to rise and no water movement is

possible until there is a break in those soil barriers such as bedrock. This building of

pressure and breaking is not considered in the model or the simulation.

water can move up into a void, against gravity. This is to reduce complexity of the assignment.

We model the rate at which water moves out of soil:

it is constant (for simplicity)

Water volume moved per 1kN per hour

CLAY_COMPRESSIBILITY_RATE = 0.001

SHALE_COMPRESSIBILITY_RATE = 0.0005

More water cannot be removed from a particle:

if there is 0.04 water in a particle of clay, and 100kN is applied over one hour.

Water moved is 100,000 x 0.001 = 0.1 . 0.1 > 0.04 , and we expect the particle

of clay will have 0.00 units of water.

File formats

Applicable to parse_sim_parameters and parse_soil_data

The le format contains pairs of Labels and Values , as well as Comments .

# comments

label1

value(s) for label1

label2

value(s) for label2

label3

value(s) for label3

# comments

...

Comments

A comment is identied when the # symbol rst appears in the line after any whitespace.

# a valid comment

# also a valid comment

### valid comment

-- # INVALID comment

NOT # a # comment

Comments can appear after or before a label/value(s) pair.

# comment1

# more of comment1

label1

value(s) for label1

# comment2

label2

value(s) for label2

# comment3

label3

value(s) for label3

# comment4

# more comments...

Labels and Values

A label is the text used to identify the parameter. e.g. Load weight . A value is information for that

label. e.g. 1000

Load weight

1000

Once a label has been identied, the value(s) are always in next lines that follow. A blank line is used

to separate denitions of labels and their values.

Labels can appear at any point within the le. There is no speci�c ordering.

Label matching and identication

Each le format has a well dene set of labels.

All label matching is case insensitive.

Duplicate labels

Labels appearing more than once are permitted, however, only the last value is used.

For example:

Load weight

1000

Load weight

800

During the reading of the le, the program should print a message to the console stderr Warning

label Load weight defined twice. Using last value of 800

Simulation Parameters le format

Applicable to parse_sim_parameters

Load location, width

Load weight

<value>

Load type

<value>

Load timing

<value>

Load custom data

<time1, load1>, <time2, load2>, <time3, load3>, ...

Format details

Load location, width

2 integers, representing columns starting and how many columns wide

comma separated

must each be positive integers from 0 to 100

Load weight

1 integer representing kilonewtons of weight

must be a positive integer between 0 and 1,000,000

Load type

1 string - Constant or Linear or Custom

Optional, default is Constant .

Load timing

1 integer representing the number of hours the load will apply for Linear

not used for Constant or Custom Load types

must be a positive integer between 0 and 1,000,000

Load custom data

pairs of integer values representing the weight of the load at the given time

Comma separated in the order <hour>, <load weight>

Used when Custom Load type is dened. Otherwise ignored.

Must have an even number of values (a whole number of pairs)

de�nes zero and positive integers

time values dened incrementally

no duplicate time values

Example le

Load location, width

0, 1

Load weight

100

Load type

Linear

Load timing

100

Load location, width

1, 3

Load weight

3860

Load type

Linear

Load timing

Load custom data

1,10, 25,50, 75,55, 4,100, 5,150

Soil data le format

Soil width, depth

Soil keys

<key1>, <description1>, <key2>, <description2>, ...

Soil data

... ...

... <key> ...

... ...

The keys describe a symbol to use for a particular soil category. The width of the symbol is 1

character.

The key description is a string of any size

Example

Soil width, depth

30, 20

Soil keys

C,Clay, H,Shale, B,Bedrock, V,Void

Soil data

A special particle is the void. Void is represented by a V . It represents a space which will allows water

to move to (lower pressure).

Format details

Applicable to parse_soil_data

Soil width, depth

2 integers representing the number of columns and the number of layers (rows) of soil data to

comma separated

must each be positive integers from 0 to 10,000

Soil keys

pairs of character,string for representing a soil particle type by a single character

comma separated

must have an even number of values

must de�ne single character followed by a string

no duplicate values (key or description)

Soil data

string(s)

must have the appropriate number of columns and rows previously de�ned

must be characters de�ned in the Soil keys

all rows are equal length

all columns are equal length

Simulation parameters

Applicable to check_simulation_data, calculate_applicable_load, calculate_current_load,

simulation_start

Load location, width

Where and how wide the load is placed.

The location is the column integer, where 0 is the left most column.

The width is the side of the load

For example:

Load location = 1 Load width = 1

In this document, we will represent load with L to illustrate the load location and width. However,

the symbol L will not be present in the �le formats, or simulation.

BCCV

BCCB

BBBB

Load location = 2 Load width = 1

BCCV

BCCB

BBBB

Most of the problems will focus on a single load point, where Load width =1

Distributed load

The weight of the load implies a force. The width of the load implies a distribution of force over area,

pressure.

For example: Load location = 1 Load width = 3

LLL

BCCCV

BCCCB

BBBBB

When simulating, the load will only a�ect the body of water which are under load

For example: Load location = 0 Load width = 3

LLL

BCCCV

BCCCB

BBBBB

Load on Column 0 has no e�ect in this model Load on Column 1 has e�ect in this model Load on

Column 2 has e�ect in this model

For simpli�cation, you may assume that where a column is not all bedrock, the load will a�ect it.

For example: Load location = 1 Load width = 3

LLL

BCCCV

BBCBB

BCCCB

BBBBB

All columns a�ected

Hint: You should focus on load width of 1 to simulate �rst before moving to distributed loads

Load type and timing

Constant load is �xed. Every hour will impose a force of Load weight at the load point(s).

Linear. At time zero, there is zero load. At time Load timing , there will be 100% of the Load weight

at the load point(s). For example, if Load weight=100 and Load timing = 10 . It would take 10

hours for the load to reach 100kN.

t=0, load=0

t=1, load=10

t=2, load=20

t=3, load=30

t=4, load=40

t=5, load=50

t=6, load=60

t=7, load=70

t=8, load=80

t=9, load=90

t=10, load=100

This a�ects the rate of consolidation.

Custom load are user speci�ed times and load weights to be used in the simulation. For example, the

user may specify:

t=0, load=100

t=1, load=120

t=2, load=180

t=3, load=200

t=4, load=150

t=5, load=110

The user may alternatively specify sparse time sequences:

t=0, load=0

t=1, load=500

t=6, load=1300

t=14, load=2000

t=15, load=1500

t=23, load=400

For the simulation, you may assume that the times not speci�ed will use the same value as the

previous time.

t=0, load=0

t=1, load=500

t=2-5, load=500 # this is implicit

t=6, load=1300

t=7-13, load=1300 # this is implicit

t=14, load=2000

t=15, load=1500

t=16-22, load=1300 # this is implicit

t=23, load=400

t=24-end of simulation, 400 # this is implicit

The only error checking you need for custom data is where there are missing pairs of time/load

values. Otherwise, you may assume that the custom data is correct.

A group of connected soil particles as a body of water

Particles of clay that are adjacent will share the same pressure and push of water throughout.

Therefore, the simulation requires knowing which particles constitute a group, or a body of water.

Example: 1 body of water. All clay particles C are adjacent to all other clay C particles. A pressure

acting on any clay particle will be transmissible throughout the body of water.

BCCCCCCCCCV

BCCCBCCBCBB

BBBBBBBBBBB

Example: 3 bodies of water. There are bedrock particles creating a barrier between the clay particles.

BCCCBCCCCCB

BCCCBCCBBBB

BCCCBCCBCBB

BBBBBBBBBBB

Important:

You do not have to calculate a body of water

You do not have to consider the case where a load is applied to 2 or more bodies of water.

This is calculated for you using the function

leak_points, fluid_body_particles = find_leak_points(start_particle)

Every hour of the simulation the body of water will lose an amount of water related to the load

applied.

The formula for calculating the water moved in 1 hour relates to how much load is applied at the

time, how much is applicable to the soil (exclude bedrock), and the rate of water movement out of

soil particles (constant).

water_moved = CLAY_COMPRESSIBILITY_RATE x load_applicable

Consistency checking between parameters and soil

Having both parameters and soil data, we can test whether these conditions are satis�ed before

proceeding.

Load over soil

Load geometry must be positioned entirely over soil (Load geometry cannot be overhanging the

edge)

Example of not permitted:

VCC

BCC

Load not overhanging

Load must be placed over compressible soil particles (e.g. Clay)

Example of not permitted:

LLL

VCC

BCC

Simulation examples

Example 1

Consider a column of clay soil with a force applied vertically.

Let the load L be 100kN

The water volume and capacity for clay is 0.4 in this example.

The rate of water movement from clay is 0.001 per hour for every 1kN of force.

The sides of the column are empty, meaning water can leak there in�nitely

There are 4 leak points in this example Void (V) particles

Initial conditions can be represented as so:

VCV

BBB

Initial conditions, height of middle column = 3

0 1 0

1 1 1

Initial conditions, as numeric values of water capacity:

i 0.4 i

0 0 0

where i is in�nite

Initial conditions, as numeric values of water volume:

0 0.4 0

0 0 0

After 1 hour

There is 0.1 of water removed per hour on the entire soil mass. The soil pressure is even throughout,

and 0.1 water move to the 4 leak points.

Water leaked = 0.1 / 4 = 0.025

capacity (always the same)

i 0.4 i

0 0 0

water volume

0.025 0.35 0.025

0 0 0

0.1 water units removed from the soil in this hour 0.1 total water units removed since t=0

After 2 hours

water volume

0.050 0.3 0.050

0 0 0

0.1 water units removed from the soil in this hour 0.2 total water units removed since t=0

After 3 hours

water volume

0.075 0.25 0.075

0 0 0

0.1 water units removed from the soil in this hour 0.3 total water units removed since t=0

After 4 hours

water volume

0.100 0.20 0.100

0 0 0

0.1 water units removed from the soil in this hour 0.4 total water units removed since t=0

After 5 hours

water volume

0.125 0.15 0.125

0 0 0

0.1 water units removed from the soil in this hour 0.5 total water units removed since t=0

After 6 hours

water volume

0.150 0.10 0.150

0 0 0

0.1 water units removed from the soil in this hour 0.6 total water units removed since t=0

After 7 hours

water volume

0.175 0.05 0.175

0 0 0

0.1 water units removed from the soil in this hour 0.7 total water units removed since t=0

After 8 hours

water volume

0.200 0.00 0.200

0 0 0

0.1 water units removed from the soil in this hour 0.8 total water units removed since t=0

Final result

height of middle column = 2.2

0 0.6 0

1 1 1

0.8 units of water has been removed from the soil.

Example 2

So long as there is a lower force area, water can be pushed out.

Consider two columns of clay soil surrounded by bedrock with a vertical force applied on the left

column only.

Let the load L be 100kN

The water volume and capacity for clay is 0.4 in this example.

The rate of water movement from clay is 0.001 per hour for every 1kN of force.

There is one leak point in this example Void (V) particle. water �ows toward this point.

Initial conditions can be represented as so:

BCBCV

BCBCB

BCCCB

BBBBB

Initial conditions

5 columns:

height of column 1 = 4

height of column 2 = 4

height of column 3 = 4

height of column 4 = 4

height of column 5 = 3

Heights:

4 4 4 4 3

1 1 1 1 0

1 1 1 1 1

Initial conditions, as numeric values of water capacity:

0 0.4 0 0.4 i

0 0.4 0 0.4 0

0 0.4 0.4 0.4 0

0 0 0 0 0

Initial conditions, as numeric values of water volume:

0 0.4 0 0.4 0

0 0.4 0.4 0.4 0

0 0 0 0 0

After 1 hour

capacity (always the same)

0 0.4 0 0.4 0

0 0.4 0.4 0.4 0

0 0 0 0 0

Water volume decreases by 0.1 from all particles sharing this force. 7 particles: 0.1 / 7 = 0.014

0 0.386 0 0.386 0.1

0 0.386 0 0.386 0

0 0.386 0.386 0.386 0

0 0 0 0 0

0.1 water units removed from the soil in this hour 0.1 total water units removed since t=0

After 2 hours

water volume

0 0.371 0 0.371 0.2

0 0.371 0 0.371 0

0 0.371 0.371 0.371 0

0 0 0 0 0

0.1 water units removed from the soil in this hour 0.2 total water units removed since t=0

After 3 hours

water volume

0 0.357 0 0.357 0.3

0 0.357 0 0.357 0

0 0.357 0.357 0.357 0

0 0 0 0 0

0.1 water units removed from the soil in this hour 0.3 total water units removed since t=0

...

After 28 hours

water volume

0 0 0 0 2.8

0 0 0 0 0

0.1 water units removed from the soil in this hour 2.8 total water units removed since t=0

Final result

height of column 2 = 2.8

0.6

height of column 3 = 3.6 (cavity collapses)

0.6

height of column 4 = 2.8

0.6

2.8 units of water has been removed from the soil.

Example 3

This example will demonstrate the load linearly increasing to the maximum weight.

Consider a column of clay soil with a force applied vertically.

Let the load L be 2450kN

Let the load timing be 10 hours

The water volume and capacity for clay is 0.4 in this example.

The rate of water movement from clay is 0.001 per hour for every 1kN of force.

There are 3 leak points in this example Void (V) particles

Initial conditions can be represented as so:

BCCCCV

BBBBBB

Initial conditions t=0

5 columns:

height of column 1 = 4

height of column 2 = 4

height of column 3 = 4

height of column 4 = 4

height of column 5 = 4

height of column 6 = 1

Heights:

4 4 4 4 4 1

1 1 1 1 1 0

1 1 1 1 1 1

Initial conditions, as numeric values of water capacity:

0 0.4 0.4 0.4 0.4 i

0 0 0 0 0 0

Initial conditions, as numeric values of water volume:

0 0.4 0.4 0.4 0.4 0

0 0 0 0 0 0

After 1 hour t=1

capacity (always the same)

0 0.4 0.4 0.4 0.4 i

0 0 0 0 0 0

Force applied at this time is: (1/10) * 2450 = 245

Based on the formula: water_moved = CLAY_COMPRESSIBILITY_RATE x load_applicable

water_moved = ( 0.001 * 245 ) = 0.245

Water volume decreases by 0.245 from all particles sharing this force.

12 particles: 0.245 / 12 = 0.020416667 loss each

0 0.380 0.380 0.380 0.380 0.0817

0 0 0 0 0 0

0.245 water units removed from the soil in this hour 0.245 total water units removed since t=0

After 2 hours

Force applied at this time is: (2/10) * 2450 = 490

water_moved = ( 0.1 * 490 ) / 100 = 0.49

Water volume decreases by 0.490 from all particles sharing this force.

12 particles: 0.490 / 12 = 0.040833333 loss each

0 0.339 0.339 0.339 0.339 0.245

0 0 0 0 0 0

0.490 water units removed from the soil in this hour 0.735 total water units removed since t=0

After 3 hours

Force applied at this time is: (3/10) * 2450 = 735

water_moved = ( 0.001 * 735 ) = 0.735

Water volume decreases by 0.735 from all particles sharing this force.

12 particles: 0.735 / 12 = 0.06125 loss each

0 0.278 0.278 0.278 0.278 0.49

0 0 0 0 0 0

0.735 water units removed from the soil in this hour 1.47 total water units removed since t=0

After 4 hours

Force applied at this time is: (4/10) * 2450 = 980

water_moved = ( 0.001 * 980 ) = 0.98

Water volume decreases by 0.98 from all particles sharing this force.

12 particles: 0.98 / 12 = 0.081666667 loss each

0 0.196 0.196 0.196 0.196 0.816

0 0 0 0 0 0

0.98 water units removed from the soil in this hour 2.45 total water units removed since t=0

After 5 hours

Force applied at this time is: (5/10) * 2450 = 1225

water_moved = ( 0.001 * 1225 ) = 1.225

Water volume decreases by 1.225 from all particles sharing this force.

12 particles: 1.225 / 12 = 0.102083333 loss each

0 0.094 0.094 0.094 0.094 0.816

0 0 0 0 0 0

1.225 water units removed from the soil in this hour 3.675 total water units removed since t=0

After 6 hours

Force applied at this time is: (6/10) * 2450 = 1470

water_moved = ( 0.001 * 1470 ) = 1.47

Water volume decreases by 1.47 from all particles sharing this force.

12 particles: 1.47 / 12 = 0.1225 loss each

0 0.000 0.000 0.000 0.000 0.816

0 0.000 0.00 0.000 0.000 0.816

0 0 0 0 0 0

1.125 water units removed from the soil in this hour (remainder) 4.8 total water units removed since

t=0

Final result

heights:

Column 1 2 3 4 5 6

--------------------------------

4, 2.8, 2.8, 2.8, 2.8, 1

4.8 units of water has been removed from the soil in 6 hours.

Example 4

This example will demonstrate the custom load.

Consider the scenario where sand columns are used to consolidate faster:

Load custom data is: 0,200, 15,550, 25,977, 99,1489

The water volume and capacity for clay is 0.4 in this example.

The rate of water movement from clay is 0.001 per hour for every 1kN of force.

There are 16 leak points in this example Void (V) particles

There are 39 soil particles in this example Clay (C)

There is only one body of water

Sand columns are emulated as void and are positioned between clay columns.

Initial conditions can be represented as so:

Final result

15.6 units of water has been removed from the soil in 33 hours.

Your tasks

Please see the following functions for you to complete. They describe the necessary input and output.

[ ] Complete the code necessary to load and parse the simulation parameters �le.

def parse_sim_parameters(file_obj, config):

''' input: open file object

output:

set the config variables load_location, load_width, load_weight, load_type, load_timing, load

close the file_obj

return True on success, otherwise False

'''

[ ] Complete the code necessary to load and parse the soil data �le

def parse_soil_data(file_obj, config):

''' input: open file object

output:

set the config variables soil_width, soil_depth, soil_key_desc, soil_data

close the file_obj

return True on success, otherwise False

'''

[ ] Complete the code for the summarise_sim_data() and summarise_soil_data() functions

def summarise_sim_data(config):

''' print the summary of the simulation parameter data

input: config variables load_location, load_width, load_weight, load_type, load_timing, load_cus

output: list of strings

'''

def summarise_soil_data(config):

''' print the summary of the soil data

input: config variables soil_width, soil_depth, soil_key_desc, soil_data

output: list of strings

'''

See basic simulation scenario for the output string formats.

[ ] Complete the code for the functions of class sim_results

[ ] Complete the code for the calculate_heights() function

def calculate_heights(particles2d):

''' calculate the height of each column

input: 2D grid of particles (list of lists)

output: a list of heights for each column

'''

[ ] Complete the code for the get_bedrock_columns function

def get_bedrock_columns(particles2d):

''' returns a list of boolean values to represent whether the column is entirely bedrock

input: 2D grid of particles (list of lists)

output: a list of bool, where the list length is the number of columns

example:

BCCS

BBBB

[ True, False, False, False ]

'''

[ ] Complete the code for the check_simulation_data function

def check_simulation_data(config):

'''

check the simulation parameters and soil data are compatible

- load location must be within the columns of soil defined

- load width cannot overhang last soil column

input: config variables for simulation parameters and soil

output: on success, return True, otherwise return False

'''

[ ] Complete the code for the calculate_applicable_load function

def calculate_applicable_load(config, particles2d, current_load):

'''

Calculate how much of the load will be applied based on whether there are bedrock columns. When t

Formula for your idea:

load = load * ( #non-bedrock-cols / width + #bedrock-cols / width )

input:

current_load, the number of kN for the given time instance (externally calculated based o

config data with Load location and dimensions.

particle2d - 2D grid of particles at present

output: the kN (single float) applied to the body of water

'''

[ ] Complete the code for the calculate_current_load function

def calculate_current_load(config, hours_passed):

'''

caclulate the amount of weight to be applied at hours_passed time.

- Where the load type is constant. config.load_weight is returned.

- Where the load type is linear, a calculation is needed based on

hours_passed and load_timing. If the hours_passed exceeds load_timing,

then the full load_weight is used

- Where the load type is custom, the calculation follows the pairs

of time,load values in config.load_custom_data. config.load_custom_data

is assumed to be in time sorted order with no duplicates. The

intermediate values of custom data use the last known time's load value.

If the hours_passed is before all time/load pairs, then the load is zero.

input:

config information config.load_weight, config.load_type, config.load_timing, config.load_custom_d

hours_passed - representing the current time in the simulation. must be a positive integer

output:

on success, the load applied (single float) is returned (without considering the soil information

on failure, -1.0 is returned

'''

[ ] Complete the partially completed code for the simulation_start() function. Refer to the

sca�old code provided.

Testing

Sta� will run tests on your code. The functions will be called to evalute the output to your functions.

You are to ensure the correct data is returned or set in memory as described by the function.

Test loading the con�guration �le for simulation parameters

Test loading the con�guration �le for soil data

Test for checking the simulation data is consistent

Test for summarising the simulation con�guration

Test for calculating the load for a given time

Test for calculating the load applicable

Some testing �les and output date are provided for you. Please refer to the tests directory in the

workspace on Ed.

How do I begin?

It is always good to start with the simplest possible simulation to run. Small number of particles, very

well de�ned input and outcome. Pick a simple scenario. Here is a suggested one.

Create a simple con�guration

The con�guration �les:

basic_params.in

Load location, width

0, 1

Load weight

100

Load type

Constant

basic_soil.in

Soil width, depth

1, 4

Soil keys

C,Clay, B,Bedrock, V,Void

Expectations

Can you predict the expected outcome?

C becomes

C ---> C

V ---> C

B ---> B

Height should be 0.6 + 0.6 + 0 + 1 = 2.2

Time taken?

1 clay has 0.4 water

total water for 2 clay is thus 0.8 water

CLAY_COMPRESSIBILITY_RATE = 0.001 hence it would take 8 hours

Code for the basic simulation scenario

No need to write code for:

simulate.py

summarise_sim_data()

Expected to return this list of strings

[ 'Load location, width: 0, 1',

'Load weight: 100',

'Load type: constant',

'Load timing: -1',

'Load custom data: [0, 0]' ]

summarise_soil_data()

Expected to return this list of strings

[ "Soil width, depth: 1, 4",

"Soil keys and description:",

" C - Clay",

"Soil data:",

"['C']",

"['V']",

"['B']" ]

Easy implementation for the basic simulation scenario

check_simulation_data() - always returns True

calculate_applicable_load() - always returns current_load

calculate_current_load() - always returns config.load_weight

sim_loader.py

parse_sim_parameters()

config.load_location = 0

config.load_width = 1

config.load_weight = 100

config.load_type = "Constant"

return True

parse_soil_data()

config.soil_width = 1

config.soil_depth = 4

config.soil_keys = [ 'C','Clay', 'B','Bedrock', , 'V','Void']

config.soil_data = [ ['C'], ['C'], ['V'], ['B'] ]

return True

sim_particles2d.py

get_bedrock_columns() - always returns [ False ]

calculate_heights()

height = 0

height += (1 - particles2d[0][0].capacity) + particles2d[0][0].water

height += (1 - particles2d[1][0].capacity) + particles2d[1][0].water

# for particles2d[2][0], skip adding height of void particle

# for particles2d[3][0], always add height 1 for bedrock particle

height += 1

return [height]

sim_results.py init() - use suggested variables get_consolidation_time() - returns

self.consolidation_time get_total_water_removed() - returns self.total_water_removed

get_height_data() - returns self.heights_data get_load_data() - returns self.load_data

Hardest part

simulate.py

simulation_start()

Run the code:

python3 simulate.py tests/basic_params.in tests/basic_soil.in results.txt

Check your results

The results �le should look like this:

sim_param: basic_params.in

soil_data: basic_soil.in

Consolidation time: 8

What next?

Now you can pick which part to increase in complexity.

How about checking the �le format values for this basic simulation scenario are correct or

rejected? (test your code!)

How about changing the number of clay particles from 2 --> 3 ? but ideally how can you

change it from 2 --> N particles!

How about a di�erent shape of soil?

How about a di�erent load location?

How about changing from Constant to Linear Load Type?

etc.

Take small steps. Test as you go using the previous as well as new scenarios. Use scenarios where

you already know or can con�rm the outcome!

Code Submission

Your code submission must be made via Ed in this lesson.

To make a submission, you will need to press the "Mark" button.

You may submit as many times as you wish without penalty.

You are able to view previous submissions from the code submission section.

Every submission you make includes the README.md and the python .py �les

The following rules apply to your code submission:

Only the �le simulate.py will be started by the automarker

Only the last submission will be graded.

Only the �nal README.md �le of the last submission will be graded by your tutor

Submission after the due date will incur a late penalty as described in the unit of study outline.

Your submission must be able to compile and run within the Ed environment provided.

After each submission, the marking system will automatically check your code against the public test

cases.

The Python version is that which is presently being used on Ed system:

$ python3 --version

Python 3.9.2

Please ensure you carefully follow the assignment description. Your program output must exactly

match the output shown in the examples.

Late is late!

The computer does not discriminate about what is late. There are two �les for submission

all python �les .py

README.md

They are both in your Ed workspace. Make sure these are the ones for your submission.

Changes to these �les after the deadline will incur a penalty for both (not individually).

Marking criteria

This assignment is 20% of your �nal course grade.

The grade you will receive is based on the number of test cases passed as well as manual grading by

your tutor.

We have provided you with some sample test cases but these do not test all the functionality

described in the assignment. It is important that you thoroughly test your own code.

Automatic tests 15 / 20

These are marked by a computer.

There are public test cases

There are hidden test cases

There are private test cases

Manual grading 5 / 20

The manual grading from your tutor will consider the style, layout and comments you provide in

your code. Your tutor will not be debugging or running your code against any test cases.

edit the README.md �le to describe the parts you have completed and the general structure of

your code. Maximum 1000 words. See sca�old �le directory in Ed.

Explain how each of these modules and their functions are used as part of the simulation:

sim_con�g.py

sim_loader.py

sim_particle.py

sim_particles2d.py

sim_results.py

simulate.py

Justify the choice of testing tool and type of tests that you have conducted to verify your

work. Additional Python �les and logs that were created to run these tests must be

included in your submission.

in your python code �les .py:

Please ensure your code is comprehensible for the purposes of marking. We recommend

that your code adheres to the PEP 8 style guide https://www.python.org/dev/peps/pep-

0008/

Please use comments or docstrings for sections of code to explain the section's purpose.

e.g. the following block of code will "search for", "calculate", "get input for options" etc.

Academic declaration

By submitting this assignment you declare the following:

I declare that I have read and understood the University of Sydney Academic Dishonesty and

Plagiarism in Coursework Policy, and except where speci�cally acknowledged, the work contained in

this assignment or project is my own work, and has not been copied from other sources or been

previously submitted for award or assessment.

I understand that failure to comply with the the Academic Dishonesty and Plagiarism in Coursework

Policy, can lead to severe penalties as outlined under Chapter 8 of the University of Sydney By-Law

1999 (as amended). These penalties may be imposed in cases where any signi�cant portion of my

submitted work has been copied without proper acknowledgement from other sources, including

published works, the internet, existing programs, the work of other students, or work previously

submitted for other awards or assessments.

I realise that I may be asked to identify those portions of the work contributed by me and required to

demonstrate my knowledge of the relevant material by answering oral questions or by undertaking

supplementary work, either written or in the laboratory, in order to arrive at the �nal assessment

mark.

I acknowledge that the School of Computer Science, in assessing this assignment, may reproduce it

entirely, may provide a copy to another member of faculty, and or communicate a copy of this

assignment to a plagiarism checking service or in house computer program, and that a copy of the

assignment may be maintained by the service or the School of Computer Science for the purpose of

future plagiarism checking.

Warning: Any attempts to deceive or disrupt the marking system will result in an immediate zero for

the entire assignment. Negative marks can be assigned if you do not properly follow the assignment

description, or your code is unnecessarily or deliberately obfuscated.

请加QQ：99515681 或邮箱：99515681@qq.com WX：codehelpSimulate Soil Consolidatio编程代写

Simulate Soil Consolidation

Due date 11:59PM Friday 28 May April 2021 (local

Sydney time)

You are tasked to write the parts needed for the simulation of soil consolidation.

Changes to description

This document is being updated:

Students should refer to the Ed website for any clari�cations, or changes to being made to the

assignment description.

A brief list of changes is documented here.

9 May 2021

CLAY_COMPRESSIBILITY_RATE is 0.001. It is a constant dening water units moved per 1kN per

hour (no longer 100kN)

A message to all students about Academic Integrity

This is an assignment and sta are not permitted to give specic guidance on your code, or how to

solve the specic problem. That is the purpose of the assessment that you are required to perform to

achieve the grade.

You may ask clarication questions about the assignment description. This is often necessary to

implement functionality that is otherwise ambiguous.

The assignment description is not intended to be complete and you can conrm your assumptions in

a form of a question. In asking the question you should be quoting the description you are asking

about.

If you have a question to ask on Ed please search before asking. However, remember that you