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Friday, November 18, 2011

Experiment to determine the centre of pressure on a partially submerged plane surface



OBJECTIVE OF EXPERIMENT:


To determine the centre of pressure on a partially submerged plane surface


EQUIPMENT SET – UP:


Hydraulics Bench F1 – 10 and Hydrostatic Pressure Apparatus F1 – 12.







SUMMARY OF THEORY
   
 Partial immersion:
    
mgL = [(ρg/2)](by²)[(a+d)-y/3]
    
m/y² = [(ρb)/(2L)][a+d]- [(ρb)/2L](y/3)




PROCEDURE:


1. Place the quadrant on the two dowel pins and using the clamping screw, fasten to the balance arm. Measure a, L, depth d and width b, of the quadrant end face. With the Perspex tank on the bench, position the balance arm of the knife edges    (pivot). Hang the balance pan from the end of the balance arm. Connect a length of hose from the drain cock to the sump and a length from the bench feed to the triangular aperture on the top of the Perspex tank. Level the tank using the adjustable feet and spirit level. Move the counter balance weight until the balance arm is horizontal.

2. Close the drain cock and admit water until the level reaches the bottom edge of the quadrant. Place a weight on the balance pan, slowly adding water into the tank until the balance arm is horizontal. Record the water level on the quadrant and the weight on the balance pan.

3. Fine adjustment of the water level can be achieved by overfilling and slowly draining, using the stop cock.

4. Repeat the above for each increment of weight until the water level reaches the top of the quadrant end face. Then remove each increment of weight noting weights and water levels until the weights have been removed.




RESULTS AND CALCULATIONS:



FILLING TANK
DRAINING TANK
AVERAGE

Weight
m
(g)
Height of water y (mm)
Weight
m
(g)
Height of water y (mm)
m
(g)
y
(mm)
m/y²
50
43
200
44
50
43.5
1892.25
0.026
100
64
150
63
100
63.5
4032.25
0.025
150
79
100
78
150
78.5
6162.25
0.024
200
93
50
93
200
93
8649.00
0.023

Therefore, graph m/y² against as shown:


From the graph


P1 = x : 30
        y : 0.0265


P2 = x : 50
        y : 0.025


My = (y2 – y1) / (x2 – x1)


      = (0.025 – 0.0265) / (50 -80)


      = 0.000033


C   = 0.029




CONCLUSION
From this experiment, we can conclude that center of pressure can be determine by the graph parameters.

Experiment Hydrometer Test


Description of Test 

This method describes the quantitative determination of the distribution of particle sizes in soils. The distribution of particle sizes larger than 71 um is determined by a sedimentation process, using a hydrometer to secure the necessary data. 

The hydrometer test is an application of Stokes Law, which in essence states that larger particles fall more quickly in a suspending fluid, while finer particles remain in suspension longer. The time at which the hydrometer readings are taken determines the size of particle remaining in suspension, while the reading on the hydrometer determines the amount of that size. 

Several assumptions are made about particles shape and other test conditions, so the results are somewhat approximate. The sieve portion and hydrometer portion of the test may not exactly line up. 

The method as presented, assumes a particle specific gravity of 2.65. For most purposes this will be sufficiently accurate even though S.G.'s may be somewhat lower or higher. If further refinement is required, additional corrections may be found in the reference. 

Results are used to indicate whether the soil is frost susceptible and for comparing soils from different areas or strata. 

Because the sample size is small, take extra care to obtain representative material. Considerable care should also be taken in all weighing and liquid volume measurements. 

The sample must be completely dispersed and remain dispersed throughout the test. 

Be sure the dispersing agent is not more than one month old. Also make sure the stirring paddle is not badly worn. Some soil (like heavy clays) tends to coagulate and form curds and then settle quickly giving false readings. If you see any evidence of coagulation, you must re-run the test. Reducing the sample size to 25 g sometimes helps this problem.

Basic objective of the test

Distribution of soil particles having sizes less than 75 micron (Fine Grained soils) is often determined by a sedimentation process using a hydrometer to obtain the necessary data such as the borderline between clay and silt. Using this test the GSD or grain size distribution for soils containing appreciable amount of fines is obtained.



APPARATUS AND MATERIALS 

Equipment Required 

Balance - sensitive to 0.01 g. 

Stirring apparatus - mechanically operated, with an electrical motor able to turn a vertical shaft at a speed of not less than 10,000 r.p.m. without load. The shaft shall have a stirring paddle made of metal, plastic, or hard rubber. The paddle shall be not less than 19.0 mm or more than 38.1 mm above the bottom of the dispersion cup. For details of the paddle and dispersion cup see Figure 205-10-1. 

Hydrometer - graduated to read in grams per litre of suspension and conforming to requirements for hydrometer 152 H in ASTM Specification E 100. 

Sedimentation Cylinder - glass, 457 mm in height, 63.5 mm in diameter and marked for 1000 ml volume. 
Thermometer - accurate to 0.5o C. 

Sieves – full set of Canadian metric sieves

Water bath or constant room temperature – to maintain the soil suspension at or near 20°C during the analysis. If a room can be controlled at a constant temperature no water bath is necessary. 

Beaker - 250 ml. 

Timing device - a watch or clock with a second hand. 

Oven or hot plate.


Materials Required 

Dispersing agent - prepare a solution of sodium hexametaphosphate (sometimes called sodium metaphosphate) in distilled or demineralized water. Use 40 g of sodium hexametaphosphate/litre of solution. Make new solutions at least once a month as it will slowly revert to the orthophosphate form causing a decrease in dispersive action. 

Water - either distilled or demineralized water. It should be kept at the same temperature as the test is to be run. The basic temperature as the test is to be run. The basic temperature for the hydrometer test is 20°C.



PROCEDURE 

Test Procedure 



Step 1- Taking the weight of the beaker

Step 1- Taking 50 g of oven dry soil in a beaker
Step 2- Making a 4% solution of dispersing agent
Step 2- Mixing the solution thoroughly
Step 3-Take 125 cc of mixture prepared in step 2

Step 3- ... and add it to the soil taken in step 1

Step 3- This should be allowed to soak for about 8 to 12 hours
Step 4- Take a 1000cc graduated cylinder and add 875cc of distilled water plus 125 CC of dispersing mixture from step 3
Step 5- Record the temperature of the cylinder of step 4
Step 6- Put the hydrometer in the cylinder of step 5 and record the two corrections Fm and Fz
Step 7- mix the soil prepared in step 3 using a spatula

Step7- ... Pour it into the mixer cup, wash the soils that stick to the sides of the beaker using distilled water

Step 8- add distilled water to the soil in mixer cup to make it about two-thirds full.
Step 8- Mix it for about 2 minutes using the mixer

Step 9- Pour the mix into the second graduated 1000-cc cylinder. Make sure that all of the soil solids are washed out of the mixer cup. Fill the graduated cylinder with distilled water to bring the water level to 1000 cc mark
Step 10- at the end of step 9, you should have these two cylinders
Step 11- After each reading the hydrometer is put into the transparent cylinder (R)


Sieve Analysis of 71 mm Material 

After the final hydrometer reading, was the suspension through a 71 mm sieve with tap water. Dry the retained material in an oven at 110o ± 5oC. 

Sieve the material on the following sieves: 71 mm, 160 mm, 250 mm, 400 mm, and 900 mm. 

Enter the cumulative weights to the left side of the column marked "WEIGHT PASSING" on Form MR-8 (Figure 205-10-2). 
Composite Correlation for Hydrometer 

Hydrometer reading corrections are required to compensate for temperature and density changed in the dispersing fluid. Corrections are determined by doing a "blank" test without soil, as described below. 

At the same time as the test is being run on the soil, prepare a second sedimentation cylinder with 125 ml of the dispersing solution but no soil. Fill the cylinder to the 100 ml mark with water and mix well by shaking. 

Place the cylinder in the same area or in a water bath with the soil test cylinder. 
Take hydrometer readings periodically in the same manner as for the soil test. Because the readings do not vary much, take only 3 or 4 readings over the sedimentation period. 

Conditions. Enter the corrected readings in the column marked "CORRECTED FOR CALGON & TEMPERATURE."


RESULTS AND CALCULATIONS 

Calculations 

Use the formulas given to calculate each of the following quantities: 

Hydroscopic Moist. = (Air Dry Wt. - Oven Dry Wt)/Oven Dry Wt.  x 100 

% Pass. 2.00 mm Sieve (Total Sample Wt. - Wt. Ret. 2.00)/Total Sample Wt.  x 100 


Oven dry wt. of dispersed sample 

Wd = ( Air Dry Wt. x 100 )/100 + % Hydroscopic Moist. 

Wt. of Dispersed Sample Corrected to total sample basis = w 

W = (Oven Dry Wt. x 100 )/% Pass. 2.00 mm 



Calculate grain diameter from each hydrometer reading using the formula following, and enter the data in the correct column on the form. 

Grain Diameter = K L/T 

All the values required for the calculations are taken from the data previously recorded in the various columns. 

Calculated percentage (P) of total test sample in suspension for each hydrometer reading and enter the data in the final column on the form. Use the following formula to calculate P: 

P = R/W x 100

Where R = corrected by hydrometer reading 
W = dispersed weight of sample corrected to total sample

Calculate the sieve analysis portion of the test and enter the values in the column on the form marked "PERCENTAGE PASSING." 

For the fraction retained on the 2.00 mm sieve, base calculations on the total air dry weight of sample. 

For the fraction passing the 2.00 mm sieve, adjust the cumulative weights passing each sieve to the “weight dispersed basis” or Wd as previously calculated. First obtain the weight accumulated on the balance for the material from the 71mm sieve to the 2.00mm sieve. Subtract this weight from Wd and use this difference as an adjustment to each sieve size as shown in the following example

Assume weight dispersed or Wd = 49.5 g 
Weight after washing over 71 mm and drying = 15 
Adjustment = 34.5 
Sieve Cumulated Wt. Adjusted Wt. 
2.00 mm 15.0 g 49.5 g 
400 mm 20.5 g 45.0 g 
71 mm 1.0 g 35.5 g 

Calculate percent passing each sieve using the formula: 
% Pass. = (Adjusted Wt/ W). x 100 
                     



Conclusion

We did the test for soil which was taken from the project “STOR UBAT JKN WILAYAH, KL” and the result which we obtain from the soil was 34% of clay, 40% of silt, 24% of sand and 2% of gravel. The classification of soil at different place is not same. We may not get the same result from another place.



ASTM D422-63 Standard is followed 




Report on how to design high way on a hilly area


Introduction

Romans was the people who started building long length roads, roads that connected to Europe, who saw the ability to move quickly as essential for both military and civil reasons. The Roman approach to road design is essentially the same as that in current use. The roads were constructed of several different layers, increasing in strength from the bottom. The lowest layer was normally rubble, intermediate layers were made of lime bound concrete and the upper layer was a flag or lime grouted stone slabs. The thickness of the layers was varied according to the local ground conditions.

            Highway designs vary widely and can range from a two-lane road without margins to a multi-lane, grade separated freeway. In law the word highway is often used as a legal term to denote any public road, ranging from freeways to dirt tracks. An interconnected set of highways can be variously referred to as a "highway system", a "highway network" or a "highway transportation system".

            This report consists of a highway which connects two highways A to B, through a declining hilly area, and this area needs to be cut and fill. Consist of a sag curve and crest curve. The minimum gradient through the road is 7%. And the first 50m is 0% and straight.



Speed parameter and route selection

 Route selection

Human beings are natural effort minimizes, notably when it involves moving around. When given the opportunity, they will always try to choose the shortest path to go from one place to another. Transportation, as an economic activity, replicates this process of minimization, notably by trying to minimize the friction of distance between locations. Shorter times and lower costs are looked upon by individuals as well as by multinational corporations. For an individual, it is often only a matter of convenience, but for a corporation it is of strategic importance as a direct monetary cost is involved.

The Road we have designed is 864 meters in length.

 Speed parameter


Speed or velocity, it measure how much of distance a particle or vehicle moves in a period of time. In human there are three thing which influence driver choice of speed; Personal; which means how tired he she is or how occupies he/she is,  Vehicle; the type of vehicle he/she is driving, External; this is things outside the vehicle, like how the road is, how the weather is.
The design speed of this road is 80 Km/H from A to B of 864 meters.


 Design calculations

 Horizontal curvature and alignment

Design speed of road, V =  80 Km/h,                         e = 0.07 Max,              f=0.26
Rmin     =                 V2              .           
                   127 ( emax + fmax)              
Rmin     =                 802              .           
                   127 ( 0.07 + 0.26)                       
Rmin     =       157.5 M



Rate of rotation
   LSD    =       (n.e)     x    VD


                     0.025         3.6
  LSD    =       (0.03 x 0.06)     x    80


                          0.025                  3.6
       LSD    =       80 M




Plan transition length
       LT      =          2/3 x LSD


       LT      =          2/3 x 80


       LT      =          53 M




Chainage at TP leading
       ChTP1             =          130+81 = 160 Tan 52


            =          132.97 M
       ChTP1             =          132.97+145  = 277.97


            =          278 M

Vertical Alignment

Crest curve

A         =          0 – (-7)
            =          7
K = 24,            V =  80 Km/h
Min crest =      7 x 24 = 168 m
Ha       =          912 + 0                        =912
Hb       =          912 – 0.07x 168/2       = 906.12
Hc       =          (912 +906.12)/2          = 909.06
Hd       =          (909.6 + 912)/2           = 910.53
Cha      =          140 – 168/2                 =56 m
Chb     =          140+168/2                   =224 m



   Hp     =          Ha +[(p/100) X] + [ (-A/200L) X2]


          =          912 + 0 - 0.0033


           =          911.99


Table 1 ; Crest curve in every 20 m intervals
Chainge
x
ha
(p/100)
x
A/200L
x2
Hr
56
4
912
0
4
-0.00021
16
912.00
60
24
912
0
24
-0.00021
576
911.88
80
44
912
0
44
-0.00021
1936
911.60
100
64
912
0
64
-0.00021
4096
911.15
120
84
912
0
84
-0.00021
7056
910.53
ok
140
104
912
0
104
-0.00021
10816
909.75
160
124
912
0
124
-0.00021
15376
908.80
180
144
912
0
144
-0.00021
20736
907.68
200
164
912
0
164
-0.00021
26896
906.40
ok

sag curve

A         =          -7 – (3)
            =          -10
sag       =          200 m adopt
Ha       =          865.7  + 0.07x100       = 872.7
Hb       =          865.7  + 0.03x 100      = 868.7
Hc       =          (872.7  +868.7)/2        = 869.92
Hd       =          (869.92+ 865.7 )/2      = 867.92
Cha      =          768 – 200/2                 =668 m
Chb     =          768+200/2                   =868 m
 Chd   =          768



Table 2 ; Crest curve in every 20 m intervals
Chainge
x
ha
(p/100)
x
A/200L
x2
Hr
668
20
872.7
0.07
20
-0.00025
400
871.40
688
40
872.7
0.07
40
-0.00025
1600
870.30
708
60
872.7
0.07
60
-0.00025
3600
869.40
728
80
872.7
0.07
80
-0.00025
6400
868.70
748
100
872.7
0.07
100
-0.00025
10000
868.20
768
120
872.7
0.07
120
-0.00025
14400
867.90
ok
788
140
872.7
0.07
140
-0.00025
19600
867.80
808
160
872.7
0.07
160
-0.00025
25600
867.90
828
180
872.7
0.07
180
-0.00025
32400
868.20
848
200
872.7
0.07
200
-0.00025
40000
868.70
ok
Earth work volume calculation 


From the graph there is around 74 fill in section and 3 cut section.
Cut and fill ratio         =     1 : 24.6

Volume of fill is;
=          74 – 3  = 71 nos
=          71 x 12 x 20
=          17,040 M3 of soil for fill

Volume of cut is;
=          3nos
=          3 x 12 x 20
=          720 Mof soil to be cut

 Reference

[1] Traffic and Highway engineering, Nicholas J garber, lester A Hoel, 3rd edition.
[2] Highway Engineering, Martin Rogers
[3] http://people.hofstra.edu/geotrans/eng/ch2en/meth2en/ch2m2en_2ed.html

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