tag:blogger.com,1999:blog-61705197919364711472024-02-19T09:05:40.531-08:00Geotechnical Engineering and Underground Structures..Apps/Articles/Solutions for Civil Engineering Problems. Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.comBlogger80125tag:blogger.com,1999:blog-6170519791936471147.post-88215006121491412992017-11-28T17:03:00.001-08:002017-11-28T17:03:18.863-08:00Solved using formula - drainage time comparison in case of single drainage to double drainage<div dir="ltr" style="text-align: left;" trbidi="on">
<b>Problem:</b><br />
<br />
compare, using equations, how much longer it would take a clay to settle in the case of single drainage, with that in the double drainage.<br />
<br />
<b>Solution:</b><br />
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhTnp3xyYFy8iv9c1Y4-hXJvimBLuEsXa6ZkSCkejD6Km6yZ4rfZ29-v4pfFcu1aqu-tHJiAGBZY3MxJEgZn7yuozWwirxXjG8TgCjUzV48Q1ID1XyFt8C21b3JnNeRug0oOwPD94WoWr0/s1600/consolidation+time.jpg" imageanchor="1"><img border="0" height="368" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhTnp3xyYFy8iv9c1Y4-hXJvimBLuEsXa6ZkSCkejD6Km6yZ4rfZ29-v4pfFcu1aqu-tHJiAGBZY3MxJEgZn7yuozWwirxXjG8TgCjUzV48Q1ID1XyFt8C21b3JnNeRug0oOwPD94WoWr0/s640/consolidation+time.jpg" width="640" /></a><br />
<br />
I think this is the correct solution, if you find any problem with the solution, please suggest in the comment box.<br />
<br />
Thanks!</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-33777338241938358002017-04-29T20:59:00.000-07:002017-04-29T20:59:01.384-07:00solved- Find Saturated Density and effective stress from specific gravity(G), water content(w) and void ratio(w)<div dir="ltr" style="text-align: left;" trbidi="on">
Hi,<br />
<br />
Given problem is to find the effective stress at a depth of 2 m from the top of clay bed in offshore, where the water depth is 40 m. Given for clay is G =2.65, water content is 70%.<br />
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQPXts_151RQhXamNeiYkTbTsR4A0I51inO3M7ay6axCiog-OBjJuWfhXOo7l8CzHkQgBfKPxeMt7ZIBLuzJQwwB2RucWf7Dj4WKzHgTgV1Bsj6vNYv81nwM3ShZ3VElREWymNq9eXaNo/s1600/effective+stress+in+clay+offshore.jpg" imageanchor="1"><img border="0" height="376" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQPXts_151RQhXamNeiYkTbTsR4A0I51inO3M7ay6axCiog-OBjJuWfhXOo7l8CzHkQgBfKPxeMt7ZIBLuzJQwwB2RucWf7Dj4WKzHgTgV1Bsj6vNYv81nwM3ShZ3VElREWymNq9eXaNo/s640/effective+stress+in+clay+offshore.jpg" width="640" /></a><br />
<br />
Thanks for kind visit!</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-39112163380762106542016-12-22T21:15:00.000-08:002016-12-22T21:15:10.202-08:00Consolidation Settlement of Clay - Time Factor and coefficient of Consolidation- Single and Double Drainage<div dir="ltr" style="text-align: left;" trbidi="on">
Hi,<br />
One of the main factor of rate of consolidation and the time of the consolidation is the single or the double drainage. When there is a double drainage, i.e. drainage from both of the opposite faces, the drainage path becomes half of the total thickness of the clay layer. In such cases the time of settlement decreases significantly. The following example illustrates the same.<br />
<br />
<b>Example: </b>(a) A 20 mm clay layer with double drainage reaches the 90% of total consolidation in 30 minutes. Find out the coefficient of consolidation.<br />
If the same clay has a thickness of 10mm, find out the time it would take to reach 90% consolidation with (b) single drainage (c) Double drainage. Draw the settlement curve/consolidation curve in each case.<br />
<br />
<b>Solution: </b>For solution please zoom into the picture below.<br />
<b><br /></b>
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuoGtCb5Cp1mRtCOfcApro5zuws8cuaq8dVRkT7FEpATeeLJODLA1jeunzGkj1p8cSoOUVqwFDg2-7jX8_DYRUn2zNjl0E4RsgMDy-scMOr1iNXNdSW1dE6bnXYRSsWOMh3LOMNJ7umgY/s1600/consolidation+curve+time+factor++single+and+double+drainage.jpg" imageanchor="1"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuoGtCb5Cp1mRtCOfcApro5zuws8cuaq8dVRkT7FEpATeeLJODLA1jeunzGkj1p8cSoOUVqwFDg2-7jX8_DYRUn2zNjl0E4RsgMDy-scMOr1iNXNdSW1dE6bnXYRSsWOMh3LOMNJ7umgY/s640/consolidation+curve+time+factor++single+and+double+drainage.jpg" width="438" /></a><br />
<br />
Thank You!</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com1tag:blogger.com,1999:blog-6170519791936471147.post-84933188448861829462016-10-18T04:30:00.002-07:002016-10-18T04:31:55.049-07:00Expression for permeability of an Unconfined Aquifer.<div dir="ltr" style="text-align: left;" trbidi="on">
<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="font-size: large;">Hi,</span><br />
<span style="font-size: large;"><br /></span>
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1ssmM40FQdAAbYEKAr4fgqtipO4l-V6YV1tj1ljEpw3eOEmeXfkPVdldpvDdZnBYsfDXggnvg8JdCR-codkq7Sv6A3nZ7AiX4-Iud_V1Pkmn5eR8a9SBbNzsP4k5FkdEYIf3qHr0hpKI/s1600/unconfined+aquifer.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><span style="font-size: large;"><img border="0" height="256" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1ssmM40FQdAAbYEKAr4fgqtipO4l-V6YV1tj1ljEpw3eOEmeXfkPVdldpvDdZnBYsfDXggnvg8JdCR-codkq7Sv6A3nZ7AiX4-Iud_V1Pkmn5eR8a9SBbNzsP4k5FkdEYIf3qHr0hpKI/s320/unconfined+aquifer.jpg" width="320" /></span></a><span style="font-size: large;">In pumping out test, the water is pumped out through a tube well, until a steady state is reached. The water table around the well turns into a cone of depression. The maximum depression along the centre line of the well is known as the drawdown(d).</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Following assumptions are made:</span><br />
<span style="font-size: large;"><br /></span>
<br />
<ol style="text-align: left;">
<li><span style="font-size: large;">Flow follows the Darcy's law.</span></li>
<li><span style="font-size: large;">Flow is laminar.</span></li>
<li><span style="font-size: large;">Soil mass is isotropic and homogeneous.</span></li>
<li><span style="font-size: large;">Well penetrates the aquifer to reach the impervious bed.</span></li>
<li><span style="font-size: large;">Steady flow along the soil.</span></li>
<li><span style="font-size: large;">Co-efficient of permeability is constant throughout.</span></li>
<li><span style="font-size: large;">Natural water ground regime doesn't change.</span></li>
</ol>
<div>
<span style="font-size: large;">Also it is assumed that the slope of the hydraulic line is small and so it can be taken as the tangent of the angle, in place of sine of the angle. </span></div>
<div>
<span style="font-size: large;"> </span></div>
<div>
<span style="font-size: large;">or i = dz/dr</span></div>
<div>
<span style="font-size: large;"><br /></span></div>
<div>
<span style="font-size: large;">Consider flow through a cylindrical surface at a distance of 'r' and of depth 'z'. Further equation are given in the image below.</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfmoxyz2IZuc-PGcfo10cw3qtOGsA4TLgUXNtVfZOM0tt_DGPqEqMOUjrb-_dJRzTqSnjnw4Z-ruWyGcZ9AfpysapC1U4Z4TqfWBjuGC9bQ6aRGPvrita4-EQGmdXyQ6KphMxUsCtv-L4/s1600/unconfined+aquifer1.jpg" imageanchor="1"><img border="0" height="314" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfmoxyz2IZuc-PGcfo10cw3qtOGsA4TLgUXNtVfZOM0tt_DGPqEqMOUjrb-_dJRzTqSnjnw4Z-ruWyGcZ9AfpysapC1U4Z4TqfWBjuGC9bQ6aRGPvrita4-EQGmdXyQ6KphMxUsCtv-L4/s640/unconfined+aquifer1.jpg" width="640" /></a></span></div>
<div>
<span style="font-size: large;"><br /></span></div>
<div>
<span style="font-size: large;">Flow can be estimated to a rough value using the influence radius of the well. Influence radius is the radius of the circle along which the effect of the pumping are observed. It varies from 150m to 300m. According the Sichardt, it can be found as follows. </span></div>
<div>
<span style="font-size: large;"> R = 3000*d*(k)^0.5 , here d = Drawdown, k = permeability.</span></div>
<div>
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Further, permeability can be calculated using the influence radius as given in the formula in the image below.</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8CTggeHDf5N8TeUbwFS0W-089HMqHJgU3QZfNDkf_dHy-MxgkiCT8YVaQQMMQ9ZG4J6uhTHkxCgltxGYaHb1aQgw4iFodYjRU6pyZh7VMMhxJeHLJvMe5cx552fo2SSf2zpI-Y4HhRdM/s1600/Influence+radius+of+well-+pumping+out+test..jpg" imageanchor="1"><img border="0" height="176" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8CTggeHDf5N8TeUbwFS0W-089HMqHJgU3QZfNDkf_dHy-MxgkiCT8YVaQQMMQ9ZG4J6uhTHkxCgltxGYaHb1aQgw4iFodYjRU6pyZh7VMMhxJeHLJvMe5cx552fo2SSf2zpI-Y4HhRdM/s640/Influence+radius+of+well-+pumping+out+test..jpg" width="640" /></a></span></div>
<div>
<span style="font-size: large;"><br /></span></div>
<div>
<span style="font-size: large;">Thanks!</span></div>
<div>
<span style="font-size: large;"><br /></span></div>
<div>
<span style="font-size: large;">Reference : Buy the book online from the link given below. Book Title: "Soil Mechanics and Foundation Engineering" by Dr. K R Arora.</span></div>
</div>
<iframe frameborder="0" marginheight="0" marginwidth="0" scrolling="no" src="//ws-in.amazon-adsystem.com/widgets/q?ServiceVersion=20070822&OneJS=1&Operation=GetAdHtml&MarketPlace=IN&source=ac&ref=tf_til&ad_type=product_link&tracking_id=httpreviewboo-21&marketplace=amazon&region=IN&placement=8180141128&asins=8180141128&linkId=b98e4321de00fc76c3cab646a7fefbc4&show_border=false&link_opens_in_new_window=false&price_color=333333&title_color=0066C0&bg_color=FFFFFF" style="height: 240px; width: 120px;">
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Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-74466193791451809922016-10-07T02:26:00.002-07:002016-10-07T02:26:59.412-07:00Lateral Thrust on a Wall at rest against a Soil Backfill<div dir="ltr" style="text-align: left;" trbidi="on">
Hi,<br />
<br />
A lateral thrust is applied by the soil backfilled behind a wall at rest, this thrust is known as the lateral thrust at rest. Walls which are restricted against any movements or moment come under this category. Example can be a basement wall, which is topped with a RCC slab, and therefore restricts its movement. The amount of thrust on the wall depends upon the properties of the soil.<br />
<br />
Density of the soil, angle of friction, degree of the consolidation and the level of the water table are the major factors which govern the magnitude of the lateral thrust.<br />
<br />
<b>Example: </b>A 10 m high basement wall, behind which is filled a soil with density 19.7 kN/m3, c=0,<br />
OCR= 2, π = 35Β°, πΎπ = 19.7 kN/m3. Find the lateral thrust and its location from base.<br />
<b><br /></b>
<b>Solution. </b><br />
<br />
<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgr82Is8KxK4Ws4-PQr_dFmwHFafAVNX8t9D5C_Fs6AcXjHFzU-S9-lFNSXiOCfX1-m-Lyc1ms_lcs3lD-t6KTjHvDe_2R3sLivk2O4NtxUcZHSV5L4n0vM4Qy3zq7b1lEU4WNFis-zrmE/s1600/Lateral+thrust+at+rest.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" height="295" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgr82Is8KxK4Ws4-PQr_dFmwHFafAVNX8t9D5C_Fs6AcXjHFzU-S9-lFNSXiOCfX1-m-Lyc1ms_lcs3lD-t6KTjHvDe_2R3sLivk2O4NtxUcZHSV5L4n0vM4Qy3zq7b1lEU4WNFis-zrmE/s320/Lateral+thrust+at+rest.jpg" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Fig. Lateral Thrust diagram for wall at rest. </td></tr>
</tbody></table>
πΎπ = 1 - sinπ = 1 - sin35Β° = 0.426<br />
<br />
πΎπ(overconsolidated) = πΎπ(normally consolidated )βOCR<br />
πΎπ(overconsolidated) = 0.426β2 = 0.602<br />
<br />
At any depth z<br />
πβ²π£ = z.πΎπ, [pore water pressure u= 0)<br />
πβ²β = πΎπ*πβ²π£<br />
Depth, z= 0<br />
πβ²π£ = 0, πβ²β = 0<br />
Depth, z= H=10 m<br />
πβ²π£ = 10*19.7 = 197 kN/m2, <br />
πβ²β = 0.602*197 = 118.594 kN/m2<br />
<br />
Lateral Thrust = Area of the lateral stress diagram. = 1/2 (10*118.59) = 592.97 kN<br />
<br />
Lateral thrust acts at the centroid of the lateral stress diagram, i.e. at a height of H/3 from base.<br />
H/3 = 10/3 = 3.333 m from base.<br />
<br />
Thanks!</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-43459535301561582412016-09-02T19:46:00.010-07:002016-09-02T19:46:58.801-07:00When and why the mat or raft foundations are used?<div dir="ltr" style="text-align: left;" trbidi="on">
Q. When and why the mat or raft foundations are used?<br />
<br />
A.<br />
Mat or raft foundations are type of the shallow foundations, which covers the entire area beneath a structure and provides support to all the columns and walls. It is also a type of the combined footing where all the footings are combined in the following cases:<br />
<br />
<ol style="text-align: left;">
<li>When the columns are so close such that their footings nearly touch each other.</li>
<li>When the soil is weak and having the low bearing capacity.</li>
<li>Where chances of differential settlement exist due to either the existence of the different soils or variation of the moisture content.</li>
<li>or where there is a large variation in the loading in the adjacent columns.</li>
</ol>
<div>
<br /></div>
<div>
<br /></div>
<div>
reference: 'Analysis and design of substructures' by Swami Saran.</div>
<div>
<br /></div>
<div>
thanks!</div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-54050565115499461832015-05-02T11:51:00.000-07:002015-05-02T11:51:55.868-07:00Failure Criteria for Rockmasses<div dir="ltr" style="text-align: left;" trbidi="on">
Hi there!<br />
<br />
As we know, rock-masses are the complex structures. Failure of the rock-mass occur by the development of the fractures or slip surfaces, when the stresses get increased from the strength of the rock mass.<br />
The failure process in rock-mass is complex, and mathematically difficult to quantify. Four failure criteria are:<br />
<br />
<ol style="text-align: left;">
<li>Coulomb's Criteria</li>
<li>Mohr's Criteria</li>
<li>Hoek and Bray Criteria</li>
<li>Griffith's Criteria.</li>
</ol>
<div>
First three uses empirical approach, and have considerable value for the design of excavation, while the last one has value for understanding the fracture initiation in the rock.</div>
<div>
<br /></div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-9502050847412871162015-04-17T08:33:00.001-07:002015-04-17T08:33:14.086-07:00Tunneling in weak Rocks by Singh and Goel<div dir="ltr" style="text-align: left;" trbidi="on">
Hi,<br />
This book by Bhawani Singh and Rajnish K. Goel is a very easy to read and useful book for the students of the UG and PG courses of Geo-technical Engineering. I got this book from my course teacher, and it is very helpful.<br />
<br />
In this post, I want to share with you the quotes those are written in the book, at the bottom of every title line at the start of the new chapter.<br />
I loved these quotes, and they somehow motivated me to read this book.<br />
<br />
<br />
1. (Chapter No.1 INTRODUCTION)<br />
"College is where you learn how to learn" - <i>Socrates (470 - 399 B.C.)</i><br />
<br />
2. (Chapter No.2 Applications of Geophysics in tunneling and site survey activities)<br />
"And so Geology once considered mostly a descriptive and historical science has in recent years taken on the aspect of an applied science. Instead of being largely speculative as perhaps it used to be, geology has become factual, quantitative, and immensely practical. It became so first in mining as an aid in the search for metals; then in the recovery of fuels and the search for the oil; and now in engineering in the search for the more perfect adjustment of man's structure to nature's limitations and for greater safety in public works." - <i>Charles P. Berkey, Pioneer Engineering Geologist. </i><br />
<br />
3. (Chapter No.3 Terzaghi's Rock Load Theory)<br />
"The Geological engineer should apply theory experimentation but temper them by putting them into the context of the uncertainty of nature. Judgement enters through engineering." - <i>Karl Terzaghi</i><br />
<br />
4. (Chapter No.4 Rock Mass Rating (RMR))<br />
"Effectiveness of knowledge through research(E) is E = mc^2; where m is the mass of knowledge and c is communication of knowledge by publications." - <i>Z.T. Bieniawski</i><br />
<i><br /></i>
5. (Chapter No.5 Rock Mass Quality Q)<br />
"Genius is 99 percent perspiration and 1 percent inspiration." - <i>Bernard Shaw</i><br />
<i><br /></i>
<i>6. </i>(Chapter No.6 Rock Mass Number)<br />
" My attention is now entirely concentrated on rock mechanics, where my experience in applied soil mechanics can render useful services. I am more and more amazed about the blind optimism with which the younger generation invades this field, without paying any attention to the inevitable uncertainties in the data on which their theoretical reasoning is based and without making serious attempts to evaluate the resulting errors." - <i>Annual summary in Terzaghi's Diary</i><br />
<i><br /></i>
7. (Chapter No.7 Strength of Discontinuities)<br />
"Failure is success if we learn from it." - <i>Malcom S. Forbes</i><br />
<i><br /></i>
<i>8.</i>(Chapter No.8 Strength enhancement of rock mass in tunnels)<br />
"The behaviour of microscopic systems is generally described by non linear laws. (The non-linear laws may explain irreversible phenomena like instabilities, dualism, unevolving societies, cycles of growth and decay of societies. The linear laws are only linear approximation of the non-linear laws at a point in time and space.)" - <i>Ilya Prigogine, Nobel Laureate</i><br />
<i><br /></i>
<i><br /></i>
<i><br /></i>
<i>In </i><i>this book, t</i><i>here are 29 chapters in total, I think it would be a length post to write all of the quotes in one post, let me know if you want the others. If you want to buy the book, go to the Amazon link I have given above. </i><br />
<br />
Thanks.</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-79957445075376908962015-02-09T07:03:00.001-08:002015-02-09T07:03:07.705-08:00Rock Quality Designation (RQD)<div dir="ltr" style="text-align: left;" trbidi="on">
Rock Quality Designation was introduced by D.U. Deere (1964) (<i>Practical Approach to Civil Engineering by Singh and Goel) </i>as an index of assessing rock quality quantitatively. RQD is a modified per cent core recovery which incorporated only sound rock core pieces that are 100 mm or greater in length along the core axis.<br />
<br />
RQD = (Sum of core pieces >= 10 cm)/ total drill length *100<br />
<br />
Following are the methods of obtaining RQD :<br />
<br />
<b>a. Direct Method</b><br />
<b><br /></b>
International Society for Rock Mechanics(ISRM) recommends a core size of at least NX (size 54.7 mm) drilled with double-tube core barrel using a diamond bit. All the artificial fractures should be ignored while calculating the core length for RQD.<br />
The relationship between RQD and engineering quality of the rock mass as proposed by Deere in 1968 is given as below:<br />
<br />
<b>S.No. RQD (%) Rock Quality</b><br />
1. <25 Very Poor<br />
2. 25-50 Poor<br />
3. 50-75 Fair<br />
4. 75- 90 Good<br />
5. 90- 100 Excellent<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5x-mEg2PYROB2GHobQXNg0xigvzJwc0Uw39dyCq3EFVbJo_hhQZ-fJOLVT1NtKS8seasub0fnmow34Kqmh2H_Dv-vJLdsbA3VOWDzgyOMVTujj1jXGFAi80ojEuE6Ccj8lhOWGpwWWr8/s1600/RQD+1.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5x-mEg2PYROB2GHobQXNg0xigvzJwc0Uw39dyCq3EFVbJo_hhQZ-fJOLVT1NtKS8seasub0fnmow34Kqmh2H_Dv-vJLdsbA3VOWDzgyOMVTujj1jXGFAi80ojEuE6Ccj8lhOWGpwWWr8/s1600/RQD+1.jpg" height="400" width="368" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">RQD determination - Direct Method ( Photo credit- A Practical Approach to Civil Engineering by Singh and Goel)</td></tr>
</tbody></table>
<b>b. Indirect Method:</b><br />
<b>1. </b><i style="font-weight: bold;">Seismic Method : </i>The seismic survey method makes use of the elastic properties of strata that affect the velocity of the seismic waves travelling through them. This method is cheap, rapid and relatively easy to apply. We can find out the following information from this method:<br />
a) Location and configuration of bed rock and geological structures in the subsurface.<br />
b) The effect of discontinuities in rock mass may be estimated by comparing the in-situ compressional wave velocity with laboratory sonic wave velocity of intact drill core obtained from the same rock mass.<br />
RQD(5) = (Vi/Vl)^2 * 100<br />
Where, Vi is the in-situ compressional wave velocity and Vl is compressional wave velocity in the intact rock core.<br />
<br />
<b>2. <i>Volumetric Joint Count Method:</i></b><br />
The RQD can be determined by counting the number of joints(discontinuities) per unit volume Jv. A simple relationship(given by Palmstrom, 1982), <b>RQD = 115 - 3.3*Jv</b> , can be used to convert Jv into RQD for clay free rock masses.<br />
<br />
Here Jv is the number of joints per cubic meter of the rock mass.<br />
There are few other methods, which I shall cover up in the upcoming posts.<br />
<br />
Thank You!</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-35390104214634756242015-01-18T08:14:00.001-08:002015-01-18T08:14:24.637-08:00Helical Auger boring and determination of water content and specific gravity<div dir="ltr" style="text-align: left;" trbidi="on">
<b>Theory:</b><br />
Helical Augers are used to get the disturbed soil samples from the surface or near to the soil surface. The size of augers varies from 75mm to 200mm. These samples can be used to determine the nearest ground water level, or to determine the water content at various depths. Generally hand augers are used up to the depths of 6m but they can be used up to 30 m depths. Other properties such as particle size distribution and specific gravity can also be determined using these disturbed samples.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgmvzrjQVnY3WGrEZKYqoCAW8RZbyu51iRfB5SyUTINStMj3juA0vEH1bprgimmQ-SNL-Cfw5-9GOV1UK-t-AqueetC8V4b6zsfRwB-RcmPwFd6EJ2xGJZfEYDOxOMBueKgusndjiMYnKM/s1600/Hand-Augers.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgmvzrjQVnY3WGrEZKYqoCAW8RZbyu51iRfB5SyUTINStMj3juA0vEH1bprgimmQ-SNL-Cfw5-9GOV1UK-t-AqueetC8V4b6zsfRwB-RcmPwFd6EJ2xGJZfEYDOxOMBueKgusndjiMYnKM/s1600/Hand-Augers.jpg" height="320" width="291" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Hand Augers (Image source: www.bestengineeringprojects.com)</td></tr>
</tbody></table>
<br />
<b>Apparatus/Equipment</b>:<br />
<br />
<ul style="text-align: left;">
<li>Helical Auger bore with the handle</li>
<li>Scale and sample collectors.</li>
<li>Other apparatus for the water content determination and specific gravity determination, such as the oven, soil cans and pycnometer.</li>
</ul>
<div>
<b>Procedure:</b></div>
<div>
<ol style="text-align: left;">
<li>Fix the T-handle correctly to the Auger.</li>
<li>Press the auger into the ground and rotate the auger clockwise to get it sink into the ground to the required depth in the area under investigation.</li>
<li>When the annular space between the blades is filled with the soil, it is withdrawn and cleaned.</li>
<li>The cleaned auger is again inserted and the process is repeated.</li>
<li>Extension rods are attached when bore hole progresses downwards.</li>
<li>Take the samples from the various depths and then various tests are performed to check their properties.</li>
</ol>
<div>
<b>Precautions:</b></div>
</div>
<div>
<ol style="text-align: left;">
<li>The auger is rotated and pressed at uniform rate.</li>
<li>Clean the blades of the auger after the experiment is over.</li>
<li>Joining of the extension rods should be done carefully.</li>
<li>The sample should be collected at every 30cm boring and sample should be extracted whenever the strata is changed.</li>
<li>Straight driving should be done.</li>
</ol>
<div>
<b>Limitations:</b></div>
</div>
<div>
<ol style="text-align: left;">
<li>Helical Auger does not work in rocky strata.</li>
<li>Sample cannot be obtained under water.</li>
</ol>
</div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-63480315125675178722014-11-24T22:43:00.001-08:002014-11-24T22:43:07.714-08:00Introduction to Dynamic Soil Properties<div dir="ltr" style="text-align: left;" trbidi="on">
Hi, How have you been?<br />
<br />
Soil dynamics is the study of the behavior of the soil under the dynamic loads. When the loads are larger such as the earthquake loads, Blasts or the nuclear explosions, the strain rate of the soil is from 0.01% to 0.1%, while the strain rate in the soil is comparably low of order 0.0001% to 0.001 % in case of the small loading such as the loading on the foundation of the reciprocating machines.<br />
<br />
In order to analyse and design the structures subjected dynamic loading, we have to study the following dynamic properties of the soil:<br />
<br />
<ol style="text-align: left;">
<li>Dynamic moduli, such as Young's modulus of Elasticity 'E', Shear modulus 'G', and bulk modulus 'K'.</li>
<li>Poisson's ratio</li>
<li>Dynamic Elastic constants, such as the co-efficient of uniform compression 'Cu'; Co-efficient of uniform Shear 'Ct', co-efficient of non uniform compression and Co-efficient of non-uniform shear. </li>
<li>Damping Ratio</li>
<li>Liquefaction parameters, such as the cyclic stress ratio, cyclic deformation and pore water pressure response.</li>
</ol>
<div>
Dynamic properties of the soil are dependent upon the strain, and several laboratory and field techniques been developed to measure these properties over a wide range of strain amplitude.</div>
<div>
<br /></div>
<div>
Resonant column test, Ultrasonic Pulse test, Cyclic simple shear test, Cyclic torsional simple shear test, and cyclic tri-axial compression tests are the type of the laboratory tests used to determine the dynamic soil properties.</div>
<div>
<br /></div>
<div>
Thank you :)</div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-33824182096200637712014-10-26T04:58:00.003-07:002014-10-26T04:58:29.874-07:00Vertical Piles of a Group subjected to Eccentric Loading.<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
<br />
In order to determine the load distribution between the vertical piles of a group subjected to eccentric loading, we have to consider the two cases of loading:<br />
<h3 style="text-align: left;">
(a) When eccentricity is about one axis only</h3>
<div>
Vpi = V/n +- (V.e.Xi.A)/Ig</div>
<div>
<br /></div>
<div>
Where, Vpi = Load on the ith pile</div>
<div>
V = Total vertical load acting on the pile group.<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGW7Zl9OkMlwOEX98Q9woGYO0ByLiy7PD4Ji_9d2dpci5aig5MypcITlFNYd9HR6hwPx85KOXqN5P44FCq4dMv-g8YNw35rnBDe6FtPCUnpVMFmepmQbLHEMXYhcFMCbN9BAA-7ykeDz4/s1600/Eccentric+loading.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGW7Zl9OkMlwOEX98Q9woGYO0ByLiy7PD4Ji_9d2dpci5aig5MypcITlFNYd9HR6hwPx85KOXqN5P44FCq4dMv-g8YNw35rnBDe6FtPCUnpVMFmepmQbLHEMXYhcFMCbN9BAA-7ykeDz4/s1600/Eccentric+loading.jpg" height="223" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Accentric loading on pile group</td></tr>
</tbody></table>
</div>
<div>
n = Total no. of piles.</div>
<div>
e= amount of eccentricity w.r.t. the centre of the pile group. </div>
<div>
Xi = Distance of the centre of the ith pile from the centre of the pile group, measured parallel to e. </div>
<div>
Ig = Moment of Inertia of the piles about the axis normal to the direction of eccentricity. </div>
<div>
= A.X1^2 + A.X2^2 +.... + A.Xn^2</div>
<div>
A = Area of the pile cross-section and </div>
<div>
X1, X2, ...... Xn = Distance from centre of gravity of pile group to the line of each pile, measured parallel to e. </div>
<div>
Since, all the piles in the group are assumed to be identical, above equation can be written as:</div>
<div>
<br /></div>
<div>
Vpi = V/n +- (V.e.Xi.)/Sum(Xi.^2).</div>
<div>
<br /></div>
<div>
<h3>
(b) When eccentricity is about two axes:</h3>
</div>
<div>
When there is eccentricity about both the axes, individual pile loads may be determined by the method of superposition:</div>
<div>
<br /></div>
<div>
Vpi = V/n +- (V.ey.Yi.A)/Ix +- (V.ex.Xi.A)/Iy</div>
<div>
<br /></div>
<div>
where, V, n and A have the same meaning. </div>
<div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjA2FEmXB9u78LglFsz69xr8q9rZWAe2wexr7qNTmlJnmX2NmkkuYeaerpTHyGz5uEXroijYiHyf_p_5JnYD8XluEnld9oIsCaCf2NBNEXssyrQZPnpJRPJTsm7jmGRZfxPV3X1V1zJ4js/s1600/Eccentric+loading2.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjA2FEmXB9u78LglFsz69xr8q9rZWAe2wexr7qNTmlJnmX2NmkkuYeaerpTHyGz5uEXroijYiHyf_p_5JnYD8XluEnld9oIsCaCf2NBNEXssyrQZPnpJRPJTsm7jmGRZfxPV3X1V1zJ4js/s1600/Eccentric+loading2.jpg" height="202" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Eccentric loading on pile group</td></tr>
</tbody></table>
</div>
<div>
ex = Amount of eccentricity w.r.t. centre of pile group measured along x-axis.</div>
<div>
ey = Amount of eccentricity w.r.t. centre of pile group measured along y-axis.</div>
<div>
Ix = Moment of inertia of the piles about x-axis. == A.Y1^2 + A.Y2^2 +.... + A.Yn^2</div>
<div>
Iy = Moment of inertia of the piles about the y-axis = = A.X1^2 + A.X2^2 +.... + A.Xn^2</div>
<div>
Xi = distance from the centre of gravity of the pile group to the line of each pile, measured parallel to the x-axis, and </div>
<div>
Yi = distance from the centre of gravity of the pile group to the line of each pile, measured parallel to the y-axis.</div>
<div>
<br /></div>
<div>
<br /></div>
<div>
Reference: <i>Analysis and Design of Sub-Structures </i>by Swami Saran</div>
<div>
<br /></div>
<div>
Thank you :)</div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-33725283013908039812014-09-29T20:29:00.002-07:002014-09-29T20:29:42.286-07:00Underground Excavation problems<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
This post contains the brief lecture notes on the topic "Underground Excavation problems".<br />
<br />
There are three types of underground excavation problems, which we come across in general:<br />
<br />
<ol style="text-align: left;">
<li>Stability of excavation</li>
<li>Dewatering</li>
<li>Effect on the adjoining structures</li>
</ol>
<div style="text-align: left;">
In my previous post, I discussed the first part of this topic, i.e. <a href="http://soilmechanicsandfoundationengineering.blogspot.in/2014/09/construction-problems-with-underground.html">Stability of Excavation</a>. In this post, I will take up the second part, i.e. Dewatering.</div>
<h3 style="text-align: left;">
<b>Dewatering:</b></h3>
<div>
Dewatering is required for deep construction excavations, such as power houses, pumping stations, bridge foundations and so on. Practically any foundation below water table will require dewatering, however, if the soil has got low permeability i.e. less the 10^(-6) cm/sec, much problem due to seepage of water is not posed, because the discharge is small. </div>
<div>
However, if the discharge is more, that is case of permeable soils having permeability more than 10^(-5) cm/sec, dewatering is essentially required. Extent of dewatering will depend upon characteristics of soil, presence of water bearing stratas, aquifer parameters and the field permeability.</div>
<div>
The dewatering job is to estimate the rate seepage of water, for this a central circular well is constructed, and the rate of discharge is estimated for a fully penetrating well. </div>
<div>
The rate of discharge will depend upon the geometry and extent of excavation and the desired level of lowered ground water table. </div>
<div>
A no. of wells are placed around the periphery of excavation and the water is pumped out simultaneously out of the wells.</div>
<div>
<br /></div>
<div>
There are three methods of dewatering:</div>
<div>
<ol style="text-align: left;">
<li>Sump Pumping</li>
<li>Deep well pumping</li>
<li>well point dewatering.</li>
</ol>
<div>
<b>Sump Pumping: </b>If the soil is cohesive, having low permeability, shallow trenches are dug along outer edges of the excavation to collect water and to pump it out through the pumps. It is particularly suitable in clayey and silty clay, where the rate of seepage is low.</div>
</div>
<div>
This method is inexpensive and easy to operate. The greatest depth upto which this method can be adopted is 5 to 6 m below the pump level.</div>
<div>
<br /></div>
<div>
<b>Deep Well Pumping: </b>Deeps wells are installed along the sides of excavation and water is continuously pumped out. This method is suitable for highly permeable soils such as sand and silty sand. A group of wells is installed, then the effect of pumping is observed, if sufficient control is not achieved, additional wells or pumps may be added. </div>
<div>
Deep wells must be constructed in such a way that they remove the large quantity of water, without allowing the silt to enter into the well casing.</div>
<div>
<br /></div>
<div>
<b>Well Point Pumping: </b>It consist of installation of nos. of well points around usually excavation. The vertical raiser pipes are connected to the header main on ground. The water is drawn into the riser pipe and discharged by the header pipe. The distance between the well points is 1 to 1.5 m. </div>
<div>
<br /></div>
<div>
Thanks!</div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-14827038686369368792014-09-28T05:36:00.001-07:002014-09-28T05:36:36.335-07:00Consolidation Test (IS 2720- part 15 - 1986 )<div dir="ltr" style="text-align: left;" trbidi="on">
<b>AIM:</b> This test is performed to determine the consolidation properties of a given soil as listed below:<br />
<br />
<ol style="text-align: left;">
<li>Co-efficient of compressibility</li>
<li>Co-efficient of volume change</li>
<li>Co-efficient of consolidation Cv</li>
<li>Compression Index</li>
<li>Pre-consolidation Pressure, Cc</li>
<li>Co-efficient of permeability, K</li>
</ol>
<div>
<b>STANDARD REFERENCE: </b>IS: 2720(part 15) - 1986 - Determination of consolidation properties.</div>
<div>
<br /></div>
<div>
<b>THEORY: </b>Consolidation is the process of removal of pore water present in the soil gradually due to the application of sustained static load. Because of consolidation, there will be decrease in volume of soil mass.</div>
<div>
<br /></div>
<div>
<b>SIGNIFICANCE: </b>The consolidation data of soil is used to predict the rate and amount of settlement of structure founded on clay primarily due to volume change. In addition, the following information can be obtained for foundations resting on clay using consolidation data.</div>
<div>
<ol style="text-align: left;">
<li>Total settlement of foundation under any given load.</li>
<li>Time required for total settlement due to primary consolidation.</li>
<li>Settlement for any given time and load.</li>
<li>Time required for any percent of total settlement or consolidation.</li>
<li>Pressure due to which soil already has been consolidation or compressed.</li>
</ol>
<div>
<b>APPARATUS REQUIRED: </b></div>
</div>
<div>
<ol style="text-align: left;">
<li>Consolidometer</li>
<li>Loading device (Jack or lever system)</li>
<li>Ring of non-corrosive material</li>
<li>Porous stone</li>
<li>Water reservoir</li>
<li>Soil trimming tool like wire saw, knife or spatula etc.</li>
<li>Balance (0.01 gm sensitivity)</li>
<li>Dial gauge (0.01 mm accuracy)</li>
<li>Oven</li>
<li>Desiccators</li>
<li>moisture content cans</li>
<li>Stopwatch.</li>
<li>Scale ( 0.5 mm least count)</li>
</ol>
<div>
<b>PROCEDURE: </b></div>
</div>
<div>
<ol style="text-align: left;">
<li>Clean the ring and weigh it empty.</li>
<li>Measure the height and diameter of the consolidation ring.</li>
<li>For undisturbed soil specimen, insert the ring in the soil mass by pressing with hand and remove the material around the ring. The soil specimen so cut should project about one centimeter on either side of the ring.</li>
<li>Trim the specimen to flush it with the top and bottom of the ring.</li>
<li>Remove any soil sticking outside of the ring and weight the ring with the soil specimen.</li>
<li>For remoulded specimen, compact the soil in the ring to the desired density and weigh it.</li>
<li>Determine the moisture content of the extra soil removed from outside the ring.</li>
<li>Assemble the consolidometer with the ring having the soil specimen and saturated porous stones on top and bottom of the specimen. Place the filter paper between the soil specimen and the porous stone.</li>
<li>Mount the assembly on the loading frame and the dial gauge is set in position in such a way that the dial is at the beginning of its release run.</li>
<li>Connect the system to a reservoir with the same level as that of specimen. Allow the water to flow into the sample till its saturation is achieved.</li>
<li>After saturation, note initial reading of the dial gauge.</li>
<li>Apply the normal load to give the desired pressure intensity of 2.5 N/cm^2 on the soil specimen. </li>
<li>Not the dial gauge reading at elapsed times of o, 0.25, 1.0, 2.25, 4, 9, 16, 25, 36, 49, 64, 81, 100, 169, 256, 361, 600, 1440 - minutes from the instant of application of load. The dial gauge readings are taken until 90% consolidation is reached or atleast for 24 hours.</li>
<li>Increase the normal load to give the doubled pressure intensity of the previous pressure or 5 kN/cm^2.</li>
<li>On successive days, apply the loads to give the pressure of 10, 20, 40 and 80 N/cm^2 for the desired pressure intensity.</li>
<li>After the last load is applied, decrease the load 1/4 th the value of the last laod i.e. 20 N/cm^2 and allow to stand for 24 hours.</li>
<li>Note the dial gauge reading after 24 hours.</li>
<li>Further reduce the load in steps of 1/4th the previous load and repeat the observations.</li>
<li>If data for repeated loading is required, the load intensity is increased and observations are repeated.</li>
<li>Finally reduce the load to the initial setting load, keep for 24 hours and final dial gauge reading is recorded.</li>
<li>Dismantle the consolidation ring and weigh it after gently removing any surface water present.</li>
<li>Dry the specimen in the oven for 24 hours and weigh the dry soil specimen.</li>
<li>Draw graphs between:</li>
</ol>
<div>
(a) Dial gauge reading versus under root of time on normal graph sheet and determine t90 for all the pressure by square root of time fitting method.</div>
</div>
<div>
(b) Dial gauge reading versus root of time on semi log graph sheet and determine t50 for all pressure by logarithm of time fitting method.</div>
<div>
(c) Void ratio 'e' versus pressure curve.</div>
<div>
(d) Void ratio 'e' versus load of pressure on semi-log graph sheet.</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdJV_VY37q4IPtGobVREW9hGokNDHbTA_YqG4b9AQ82WkNAhNml7kmOViLbHUe0lVgW_c4gTPw79GMdM1YF4FrTZY5Ojt00tFzeRuePvvBKqvydj_02RlPhno1vYmTcSxcgcKamd-8IC8/s1600/The+square+root+of+time+fitting+method.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdJV_VY37q4IPtGobVREW9hGokNDHbTA_YqG4b9AQ82WkNAhNml7kmOViLbHUe0lVgW_c4gTPw79GMdM1YF4FrTZY5Ojt00tFzeRuePvvBKqvydj_02RlPhno1vYmTcSxcgcKamd-8IC8/s1600/The+square+root+of+time+fitting+method.jpg" height="320" width="249" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Square root time fitting method</td></tr>
</tbody></table>
<div>
<br /></div>
<div>
<b>RESULTS AND COMMENTS: </b></div>
<div>
Average value of the co-efficient of compressibility =</div>
<div>
Average value of co-efficient of consolidation, Cv = </div>
<div>
Average value of co-efficient of volume compressibility =</div>
<div>
Average value of co-efficient of permeability, K = </div>
<div>
Compression Index, Cc = </div>
<div>
<br /></div>
<div>
<br /></div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-48125945086869623052014-09-13T23:06:00.001-07:002014-09-13T23:06:47.635-07:00Forced vibrations of Single Degree Freedom System.<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
In previous article we discussed about the <a href="http://soilmechanicsandfoundationengineering.blogspot.com/2014/09/free-vibrations-with-damping-of-sdof.html">Free vibrations with the viscous damping</a>, in which we discussed three different cases when the system is overdamped, critically damped and underdamped.<br />
<br />
In today's article we will discuss about the forced vibrations of single degree freedom system. There are cases when vibrations caused by the rotating parts of machines cause steady state periodic exciting forces in the system. This exciting force can be written as follows:<br />
<br />
F(t) = F*.sin(w.t)<br />
<br />
The equation of motion for a SDOF having a damping of 'C' subjected to such an exciting force can be written as follows:<br />
m.z'' + C.z' + K.z = F(t)<br />
<br />
The equation is a linear, 2nd order differential, nonhomogeneous equation. The solution of this equation consists of two parts, namely: (i) complementary function, and (ii) particular integral. Complementary function is obtained by considering no forcing function. <br />
Therefore its equation can be written as follows:<br />
m.z'' + C.z' + K.z = 0<br />
<br />
the solution for the above equation has already been discussed in the previous article. Now the particular integral can be found by assuming an exciting force F(t), under action here.<br />
m.z'' + C.z' + K.z = F(t)<br />
<br />
This can be shown for the particular integral, that the system will vibrate harmonically with the same frequency as that of the exciting force.<br />
<br />
The complete solution is obtained by adding the complementary function and the particular integral. The complementary function is an exponentially decaying function and will die out soon, and the motion will be described by only the particular integral. There the system as a whole will vibrate harmonically with the same frequency as the forcing.<br />
<br />
Let, F` be the amplitude of the exciting force;<br />
K is the spring constant, and<br />
Az is the amplitude of the displacement, then it can be shown that<br />
<br />
Az = Magnification factor * (F`/K)<br />
F`/K describes the static state amplitude, while magnification factor will depend upon the frequency ratio and the damping ratio of the system.<br />
<br />
<br /></div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-22643164048933456872014-09-13T21:33:00.001-07:002014-09-13T21:33:43.020-07:00Free Vibrations with Damping of SDOF system<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
In previous article we discussed about the <a href="http://soilmechanicsandfoundationengineering.blogspot.com/2014/09/undamped-free-vibrations.html">undamped free vibrating system</a>. In this article we are going to talk about the free vibrations with damping for a SDOF(Single Degree of Freedom) system.<br />
For such a system, the excitation force F(t) is kept zero, while the viscous damping is available.<br />
<br />
The differential equation of motion for such a system can be written as follows:<br />
<br />
m.z'' + C.z' + K.z = 0 [ F(t) = 0]<br />
<br />
Here, 'C' is the damping constant or force per unit velocity. The solution of the above equation gives us a value of 'z' with which we can interpret the following results:<br />
<br />
<ol style="text-align: left;">
<li><b>Case 1: (C/2m) > K/m</b></li>
</ol>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNaPJyO_nYMNhGjCHfJcaaiMzOYnfumFhSO5PEMc_iDIFptocyJX7dp0Fl1ckCJuxhYbqVLNzMip-ENceVUHI-AX2dEcEG8aeDTjiD6odhkZLXbs9iScj1ahv0gDUopWVX2D6dvC_4fiI/s1600/Overdamped+SDOF+system.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNaPJyO_nYMNhGjCHfJcaaiMzOYnfumFhSO5PEMc_iDIFptocyJX7dp0Fl1ckCJuxhYbqVLNzMip-ENceVUHI-AX2dEcEG8aeDTjiD6odhkZLXbs9iScj1ahv0gDUopWVX2D6dvC_4fiI/s1600/Overdamped+SDOF+system.png" height="237" width="400" /></a></div>
<div>
<b><br /></b></div>
In this case the motion of the system is not exponential but is an exponential subsidence. Because of the relatively large damping, so much energy is dissipated by the damping force that there is not sufficient kinetic energy left to carry the mass and pass the equilibrium position. Physically this means a relatively large damping and the system is said to be over damped.<br />
<br />
<br />
<ul style="text-align: left;">
<li><b>Case 2: (C/2m)^2 = K/m</b></li>
</ul>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3OK6Xhx0bDZsF64sHIAYjfFzm-X3aZbpDNbQweyhExa1vJ8mqst9mBN9fdSj3mylgvjAHOQPhmyFOmS1BL0Hrc1TQER2I4O0EjZRm0VMhv6Uc_t9av6WHDd7rNxnE7QazkiCWEwUaqMo/s1600/Critically+damped+system.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3OK6Xhx0bDZsF64sHIAYjfFzm-X3aZbpDNbQweyhExa1vJ8mqst9mBN9fdSj3mylgvjAHOQPhmyFOmS1BL0Hrc1TQER2I4O0EjZRm0VMhv6Uc_t9av6WHDd7rNxnE7QazkiCWEwUaqMo/s1600/Critically+damped+system.png" height="242" width="400" /></a>In this case the system is called critically damped system. It is similar to the previous case, but the difference is that the amplitude may become negative as shown in the figure. </div>
<div class="separator" style="clear: both; text-align: left;">
The damping constant obtained from this case is known as critical damping constant and the ratio of the actual damping to the critical damping constant is known as the damping ratio.</div>
<div class="separator" style="clear: both; text-align: left;">
</div>
<ul style="text-align: left;">
<li><b>Case 3: </b><b>(C/2m)^2 < K/m</b></li>
</ul>
<div>
In this case the solution for displacement 'z' can be written in a sinusoidal form, and the frequency of therefore damped system is known as damped frequency.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgPT4tyztjdyy0l_XaFTHDYwR2_VnTDZw35wuGTQAotdgvukAwxQ4_mrzBpgkJ03Eic4_EP6f2s4G0sC7csfU2BrlybtBYYU_216c8cq_LNssJSSr_Qt-T7H50zKmBAn4lpPr9LtzgAPEE/s1600/Free+Vibration+of+an+Underdamped+SDOF.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgPT4tyztjdyy0l_XaFTHDYwR2_VnTDZw35wuGTQAotdgvukAwxQ4_mrzBpgkJ03Eic4_EP6f2s4G0sC7csfU2BrlybtBYYU_216c8cq_LNssJSSr_Qt-T7H50zKmBAn4lpPr9LtzgAPEE/s1600/Free+Vibration+of+an+Underdamped+SDOF.png" height="230" width="400" /></a></div>
<div>
From above discussions we can say the a freely vibrating SDOF with viscous damping is over damped if damping ratio is greater than one, critically damped if equal to one and underdamped when less than one.</div>
<div>
<br /></div>
<div>
Thank you!</div>
<br />
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-9437384199623440652014-09-13T01:35:00.000-07:002014-09-13T01:35:11.997-07:00Undamped Free Vibrations<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
In previous article we had an introduction with the system with the<a href="http://soilmechanicsandfoundationengineering.blogspot.com/2014/09/fundamentals-of-vibrations-sdof-systems.html"> single degree of freedom</a>, in this article we are going to study behavior of the system with the 1 degree of freedom which is undamped and is free to vibrate, without any external exciting force.<br />
<br />
For undamped free vibrations, the damping force and the exciting force are equal to zero. Therefore the equation of motion of the system becomes<br />
<br />
m.z'' + K.z = 0<br />
<br />
or z'' + (K/m).z = 0 ---------eqn.(1)<br />
<br />
The solution of the equation can be obtained by substituting<br />
<br />
z= A1. cos.Wn.t + A2. sin.Wn.t<br />
<br />
Where A1 and A2 are both constants and Wn is the undamped natural frequency.<br />
<br />
when we substitute this value into the above eqn. (1), we get<br />
<br />
Wn = (K/m)^(1/2)<br />
<br />
It can be shown the final equation for the displacement will be again a sinusoidal wave with a phase angle at the start of the curve, i.e. at t=0.<br />
<br />
<br /></div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-13075625621075739112014-09-13T01:05:00.000-07:002014-09-13T01:05:16.805-07:00Fundamentals of Vibrations: SDOF systems<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
<br />
In real scenarios, structural foundations have to bear many kind of dynamic loadings, such as the loading due to the earthquakes, blastings, wind loads, pile driving, machine works, quarrying and fast moving moving traffic.<br />
The characteristics of the dynamic load is that, they varies with time. Purely dynamic loads do not occur in nature. Loads are the combination of the static and dynamic load. Static loads are caused by the dead weight of the building itself, while the dynamic loads may occur due to the causes mentioned above.<br />
The pattern of variation of a dynamic load with respect to time may either be periodic or transient. The periodical motions can be resolved into sinusoidally varying components e.g. vibrations in the case of reciprocating machine foundations. Transient vibrations may have very complicated non-periodic time history e.g. vibrations due to earthquake and quarry blasts.<br />
<br />
The structures subjected to dynamic loads may vibrate along its extension, shearing, bending and may be along torsional deformation of the structure. The form of vibration mainly depends of the mass, stiffness distribution and end conditions of system.<br />
To study the response of a vibratory system, in many cases it is satisfactory to reduce it to the idealized system of lumped parameters. The simplest model consists of mass, spring and dashpot.<br />
<br />
First of all we must know, what a Degree of freedom means, well!, the number of independent coordinates which are required to define the position of a system during vibration, is called the degrees of freedom of the system. In order to understand the motion of the systems with the multi degrees of freedom, we should first analyse the single motion with the single degree of freedom.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhRC28L3rYRaNXvAArnPmTxWrf8yGlvpf_FkEC2RoS8XchC6fANbmPBw242OuoDnFjKTRUSelphjoxn_sif7MEcdlr2p-0Ak4jvE9gTVAymLQshhbASWVzr3PZ250VzJraDzBStlVXUa1U/s1600/SDOF.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhRC28L3rYRaNXvAArnPmTxWrf8yGlvpf_FkEC2RoS8XchC6fANbmPBw242OuoDnFjKTRUSelphjoxn_sif7MEcdlr2p-0Ak4jvE9gTVAymLQshhbASWVzr3PZ250VzJraDzBStlVXUa1U/s1600/SDOF.png" height="333" width="570" /></a></div>
A simple harmonic motion can be described by a simple sinusoidal function, having a time period of T. In this article we are going to discuss the vibrations of a single degree of freedom system shown in fig.3 above, consisting of a rigid mass m supported by a spring and dashpot. <br />
Damping in this system is represented by the dashpot, and the resulting damping force is proportional to the velocity. The system is subjected to an external time dependent force F(t).<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJJBdjg_aiIiQzjeJOZmr2zBCq9JJD5FBVNO2jqlZMdonJ9v915EzdLD5rgvuroCu84jcY_REtdWzWOF09v_0xT4ljbGIe0rxlMAg4PVof2ILHKa-0HRr3HnAc2FxRGTKEHsFlGIIy9Ns/s1600/SDOF1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJJBdjg_aiIiQzjeJOZmr2zBCq9JJD5FBVNO2jqlZMdonJ9v915EzdLD5rgvuroCu84jcY_REtdWzWOF09v_0xT4ljbGIe0rxlMAg4PVof2ILHKa-0HRr3HnAc2FxRGTKEHsFlGIIy9Ns/s1600/SDOF1.jpg" height="247" width="400" /></a></div>
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</div>
<div class="separator" style="clear: both; text-align: center;">
</div>
Figure above, shows the free body diagram of the mass m at any instant during the course of vibrations. The forces acting on the mass m are:<br />
<br />
<ol style="text-align: left;">
<li>Exciting force, F(t): It is the externally applied force that causes the motion of the system. </li>
<li>Restoring force, Fr: It is the force exerted by the spring on the mass and tends to restore the mass to its original position. For a linear system, restoring force is equal to K.z, where K is the spring constant and indicates the stiffness. This force always acts towards the equilibrium position of the system.</li>
<li>Damping force, Fd: The damping force is considered directly proportional to the velocity and is given by C.z' where C is called the co-efficient of viscous damping; this force always opposes the motion.</li>
<li>Inertia force, Fi: It is due to the acceleration of the mass and is given by m.z''. This force can be justified by the De-Alembert's principle.</li>
</ol>
<div>
The equation of equilibrium for the mass m is given as:</div>
<div>
m.z'' + C.z' + K.z = F(t)</div>
<div>
<br /></div>
<div>
Thank you!</div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-87278757879450368372014-09-11T20:44:00.001-07:002014-09-11T20:44:20.997-07:00Construction Problems with the Underground Structures<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
Here are Lecture notes on the construction problems related to the construction of the underground structures.<br />
The major problems which we come across while underground excavation and construction of underground structures are classified into following 3 groups:<br />
<br />
<ol style="text-align: left;">
<li>Stability of Excavation</li>
<li>Dewatering</li>
<li>Effect on adjoining structures</li>
</ol>
In this post, I am going to discuss only the introduction of the first part, i.e. stability of excavation. In the subsequent posts, I shall discuss it in details and other parts as well.<br /><ol style="text-align: left;">
</ol>
<div>
<b>1) Stability of Excavation: </b>Excavation is done manually or by chemical means depending upon the quantity of earth work involved. In case of manual excavation, the size of foundation is limited. The mechanical excavators and haulers can be used for gentle side slopes such as in road construction, thus the area involved will be larger.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjIdXccM6L80PUKaaN-3x8AxkHZWUF4e5joFR-ZeHM_3f09Wz8Jwt-I1gsrEigriaGeGUQyNgmXX1FgsGaGSvTOpcxa9-qXInyRXkMFFnFBvNU1rh1GSpsZXEYxnmlDzi8jYAKK4XVUGek/s1600/Sloped+excavation.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjIdXccM6L80PUKaaN-3x8AxkHZWUF4e5joFR-ZeHM_3f09Wz8Jwt-I1gsrEigriaGeGUQyNgmXX1FgsGaGSvTOpcxa9-qXInyRXkMFFnFBvNU1rh1GSpsZXEYxnmlDzi8jYAKK4XVUGek/s1600/Sloped+excavation.png" height="230" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Sloped excavation - for underground construction</td></tr>
</tbody></table>
</div>
<div>
Depending upon the space availability at the construction site, the excavation with a side slope as shown in figure, or a braced cut can be adopted, however the sloped excavation can be adopted only for a stable slope & free space in vicinity of the structure.</div>
<div>
If, however there is an existing building in close vicinity, then excavation with the side slopes will not be feasible and economic. </div>
<div>
There in built up & crowded areas, braced cut is a viable proposition. Braced cuts consist of making the vertical walls in the soil and suitably propping them by driving steel struts as the excavation is proceeded.<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiBmN4jVawmi1jJg6nK7eyIUyQoX4TUt0w4SbcImjDuzSDZWT44BtjOe9lZ5oXb09x0BfrNSma4oB9yc0DPindQNobvVBat3tVVR-Mcgvj1YO4MI3EIMlals5wnjC4WIquSfJQST_7jk7g/s1600/braced+cut.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiBmN4jVawmi1jJg6nK7eyIUyQoX4TUt0w4SbcImjDuzSDZWT44BtjOe9lZ5oXb09x0BfrNSma4oB9yc0DPindQNobvVBat3tVVR-Mcgvj1YO4MI3EIMlals5wnjC4WIquSfJQST_7jk7g/s1600/braced+cut.png" height="293" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Excavation with Braced Cut</td></tr>
</tbody></table>
<div class="separator" style="clear: both; text-align: center;">
</div>
</div>
<div>
At the final excavation level, the foundation is cast and the soil is backfilled to restore the original ground level, as the struts are successively pulled up. </div>
<div>
<br /></div>
<div>
For shallow footings and raft foundations, the depth seldom exceeds 2 m and hence, no extensive supporting system is required, because earth pressure is not much. If however, heterogeneous soil exist near ground surface, side protection can be applied using timber planks or small struts may be used. </div>
<div>
Deep excavations exceeding 2 basements i.e. excavation depth greater than 6 m, will require thick, adequate lateral support by using diaphragm walls or continuous bored piles and struts. </div>
<div>
In case of large excavation widths, it is not possible to have horizontal struts, because they may bend under their own weight. </div>
<div>
In such cases, inclined props or struts may be provided in between the diaphragm wall and base. Also, a series of H-pile(Heavy I -sections) may be driven throughout the excavation at close intervals. The inclined props support the diaphragm wall in the wide cuts.</div>
<div>
<br /></div>
<div>
To be continued to next post.... </div>
<div>
<br /></div>
<div>
Thank you!</div>
<div>
<br /></div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-557220274342119022014-09-08T20:11:00.000-07:002014-09-08T20:12:36.362-07:00The Foundations in Expansive Soils<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings! Here are the lecture notes for the next lecture of Advanced Foundation Engineering.<br />
<br />
The following methods can be used for the construction of foundations on expansive soils:<br />
<br />
<ul style="text-align: left;">
<li><b>A)</b> Designing the structure as a solid structure to withstand the effects of swelling & shrinkage of strata.</li>
<li><b>B)</b> Eliminating the swelling by:</li>
<li>i) stabilizing the moisture.</li>
<li>ii) Loading the soil more than the swelling pressure.</li>
<li>iii) Treating the soil, so that its behavior changes.</li>
<li>iv) Replacing the soil by non-swelling soil.</li>
<li><b>C</b>) Isolating the structure by taking down the foundation to a stratum, which is not affected by swelling.</li>
</ul>
<div>
<b>A) S</b>trengthening the structure beyond a certain limit is not economic, however, reinforcement bands can be provided at different levels, such as plinth, lintel and beams.</div>
<div>
Heavily reinforced rafts can also be provided, but the flexible structures are not practical.</div>
<div>
<br /></div>
<div>
<b>B) </b>Taking down the foundation to a level where no volume changes or water content change occur, pre-wetting techniques may also be required. For isolated structures it may not be possible.</div>
<div>
<ul style="text-align: left;">
<li>Loading the foundation with more than swelling pressure has shown better performance, but quantification of suitable load is not practical, as the factors affecting are many.</li>
<li>Stabilization of an expansive soil by lime is well known but mixing of soil and compacting it in layers is problematic and sometimes un-economical too. Usually 2 to 8% lime decreases the liquid limit(LL) and Swelling; increases the OMC and strength of expansive soil. This method is more practical for roads. For deeper treatments lime slurry may be injected, but it is not a common practice.</li>
</ul>
<div>
Thank you!</div>
</div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-49985466752782593612014-09-04T19:30:00.002-07:002014-09-04T19:30:27.289-07:00Under Reamed Piles<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
<br />
The under reamed piles are the most common and appropriate type of foundation used in expansive soils. In this case the superstructure of the building is supported over beams, which are clear off the ground surface, and span over the piles anchored at the depth where the soil has nearly constant water content. This depth is usually found to be 3.5 m in India.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjKOTCZ6vxyZN1uEWqROFEfDa5h7eyzkkAagBnMOLcspRYXb5wGtgDg2AkXS5TAj81qyfjH2C_9FGwaOcRb2R6eLLNlZvUHGBa0-hynG7zW4Cq9UIS-mH6R0KmNDpLZNxSMyqPT2wKmwI8/s1600/Under+reamed+piles.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjKOTCZ6vxyZN1uEWqROFEfDa5h7eyzkkAagBnMOLcspRYXb5wGtgDg2AkXS5TAj81qyfjH2C_9FGwaOcRb2R6eLLNlZvUHGBa0-hynG7zW4Cq9UIS-mH6R0KmNDpLZNxSMyqPT2wKmwI8/s1600/Under+reamed+piles.jpg" height="600" width="570" /></a></div>
<h3 style="text-align: left;">
Length of Under reamed Pile:</h3>
<div>
The length of pile varies from 3.5 m to 4.0 in deep deposits of black cotton soils, and pile is carried to non expansive soil layer up to a depth of at least 0.6 m. </div>
<h3 style="text-align: left;">
Spacing of Piles:</h3>
<div>
Depending upon bearing capacity, spacing of piles varies from 1.5 m to 3.0 m. A pile is to be provided under every wall junction or where a point load is acting on the plinth beam. </div>
<h3 style="text-align: left;">
Diameter of pile shaft:</h3>
<div>
It varies from 200 to 500 mm, and the ratio of the diameter of the bulb to shaft i.e. Du/D varies from 2 to 3. In case of multi-reamed piles, the first bulb should be at minimum depth of 2.Du, below the ground level. </div>
<div>
Center to center distance between 2 bulbs may vary from 1.25 to 1.5.Du. The pile is reinforced throughout its length to take care of tensile stresses. Steel reinforcement should be sufficient to take care of shrinkage effects when the pile acts as the column, so 0.8 to 6% (IS: 456-2000) of reinforcement, which we provide for the columns should be provided at least to make it as a column.</div>
<div>
<br /></div>
<div>
As per IS:2911-1980(part 3), the ultimate bearing capacity of the soil can be calculated and a factor of Safety(FOS) of 2 to 2.3 is applied to calculate the safe load/allowable load.</div>
<div>
<br /></div>
<div>
Thanks for your kind visit!</div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com1tag:blogger.com,1999:blog-6170519791936471147.post-67240989966498917972014-08-31T10:29:00.000-07:002014-08-31T10:29:04.243-07:00Identification of Expansive Soils<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
<br />
<ul style="text-align: left;">
<li><b>Identification of Expansive Soils:</b></li>
</ul>
Expansive soils can be identified by studying their mineralogy, which can be identified by the following tests:<br />
<br />
<ol style="text-align: left;">
<li>X-Ray Diffraction test.</li>
<li>Microscopic examination.</li>
<li>Differential Thermal Analysis(DTA).</li>
</ol>
<div>
There are certain simpler tests to determine the expansive characteristic from engineering point of view.</div>
<div>
<ol style="text-align: left;">
<li>Free swell test </li>
<li>Differential free swell test.<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhocynmKl4-evkixzkTeXBOXzEbIQu3G7FeOy-6LKhlY8hvcPKA3K-390p5H9HPl_LQm1fCSEslgrdKkwgfpvdn35Wa1No7Ix7DPwqXClJZ0OeJDUVJlxcPnDmjLMkDmoCV10AvgAcLeCc/s1600/Swell+Index.png" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhocynmKl4-evkixzkTeXBOXzEbIQu3G7FeOy-6LKhlY8hvcPKA3K-390p5H9HPl_LQm1fCSEslgrdKkwgfpvdn35Wa1No7Ix7DPwqXClJZ0OeJDUVJlxcPnDmjLMkDmoCV10AvgAcLeCc/s1600/Swell+Index.png" height="320" width="250" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Swelling Index determination</td></tr>
</tbody></table>
</li>
</ol>
<div>
In <b>free swell</b> test 10 gm of soil passing through 425 mic. IS-sieve if poured in a cylinder containing 100 ml of distilled water, and left undisturbed for 24 hours. The volume of the soil is measured and the free swell Sf is given as (Vs-Vi)/Vi *100.</div>
</div>
<div>
Where,</div>
<div>
Vi= Initial Volume</div>
<div>
Vs= Volume of the swelled soil.</div>
<div>
The Bentonite may swell from 1200 to 2000%; Kaolinite - about 80 % ; Illite - 30 to 80%.</div>
<div>
<br /></div>
<div>
In <b>differential free swell</b> test, two samples of 10 gm dry soil passing through 425 mic. IS - sieve are poured in 50 ml graduated cylinders, one containing non-polarizing liquid(Kerosene) and another water. Both the jars are kept undisturbed for 24 hours and differential free swell (DFS) is given as,</div>
<div>
</div>
<div>
<b>Differential Free Swell <i>(DFS) = [ Vs(w) - Vs(k)]/ Vs(k) *100</i> </b></div>
<div>
<b><br /></b></div>
<div>
Where, Vs(w) = Volume of soil in water.</div>
<div>
Vs(k) = Volume of soil in kerosene.</div>
<div>
<br /></div>
<div>
A relationship between differential free swell (DFS) and degree of expansion is given in the table below:</div>
<div>
<b> Differential Free Swell (DFS)[%]</b> <b>Degree of expansion</b></div>
<div>
<ol style="text-align: left;">
<li><b> </b><20 <b> </b>Low</li>
<li> 20-35 Moderate</li>
<li> 35-50 High</li>
<li> > 50 Very High</li>
</ol>
<div>
<b>Holtz and Gibbs(1956), </b>gave the following table to classify the expansive soils into low, moderate, high and very high expansive categories based on volume change(%), Colloidal content(%), Plasticity Index(PI) and Shrinkage Limit(SL).</div>
</div>
<div>
</div>
<div>
<b>Property Property Ranges for</b></div>
<div>
<b> Low moderate High Very High</b></div>
<div>
<ol style="text-align: left;">
<li>Volume Change(%) 0-10 10-20 20-30 >30</li>
<li>Colloidal Content(%) 0-15 10-25 20-35 >25</li>
<li>Plasticity Index(PI)(%) 0-15 10-35 20-45 >30</li>
<li>Shrinkage Limit(SL)(%) 12 8-18 6-12 10</li>
</ol>
<div>
<b><br /></b></div>
<div>
<b>R.B. Peck, W.E. Hanson & T.H. Thornburn (1974), </b>gave the following relation between the Plasticity Index(PI) and Swelling Potential:</div>
</div>
<div>
<b> Swelling Potential Plasticity Index(PI)</b></div>
<div>
<ol style="text-align: left;">
<li> Low 0-15</li>
<li> Medium 10-35</li>
<li> High 20-55</li>
<li> Very High >=35</li>
</ol>
<div>
<b>Gromko, G.J. (1974) </b>has related shrinkage limit(SL) to the linear shrinkage(LS) and degree of expansion:</div>
</div>
<div>
<b> SL(%) LS(%) Degree of expansion</b></div>
<div>
<ol style="text-align: left;">
<li><b> <10 </b>>8 Critical</li>
<li> 10-12 5-8 marginal</li>
<li> >12 0-5 non-critical</li>
</ol>
<div>
<b>Sridharan et al. </b>gave another relationship between the swell potential and free swell index.</div>
</div>
<div>
<br /></div>
<div>
Thanks for your kind visit!</div>
<div>
<br /></div>
<div>
<br /></div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-36507703271684233542014-08-28T10:48:00.001-07:002014-08-28T10:48:45.885-07:00Standard one dimensional Consolidation Test<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
<br />
Standard one dimensional consolidation test is usually carried out on saturated specimens of about 25.4 mm thick and 63.5 mm in diameter. The soil specimen is kept inside a metal ring, with a porous stone on the top and another at the bottom.<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhzqKEcycrVUbNo_oSeAdB-fSWtaGL2m4A3avIhQwuw0B1UsbHVS7L_ZuHMmvm5MucKmAfNtjN37svtwUhMIDVcHTVO9IDwRqwWy_94h3p3isE5xaBoygh7IHd-uWeLo1L27TWM0m33UhE/s1600/Consolidometer.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhzqKEcycrVUbNo_oSeAdB-fSWtaGL2m4A3avIhQwuw0B1UsbHVS7L_ZuHMmvm5MucKmAfNtjN37svtwUhMIDVcHTVO9IDwRqwWy_94h3p3isE5xaBoygh7IHd-uWeLo1L27TWM0m33UhE/s1600/Consolidometer.jpg" height="357" width="400" /></a></div>
The load P is applied on to the specimen using a lever arm, and compression of the specimen is measured with the help of a micrometer dial gauge. The load is usually doubled every 24 hours. The specimen is kept under water throughout the test.<br />
<br />
For each load increment, the specimen deformation and the corresponding time t are plotted on a semilogarithmic graph paper. The graph consists of three distinct parts:<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8TFXf7to2ENnhRx-IiEb_SLqyCMDT2ejlIUEHFtIkfBZKu5wIaauUeyHvgech2AJxTmZVB9KqssiO95_uyC9leeoq2G3Wsm52mLqKBvUAZJDrW7hIQLxbLN4nY6uP0IE26htcu-BPTAo/s1600/Consolidation+Stages.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg8TFXf7to2ENnhRx-IiEb_SLqyCMDT2ejlIUEHFtIkfBZKu5wIaauUeyHvgech2AJxTmZVB9KqssiO95_uyC9leeoq2G3Wsm52mLqKBvUAZJDrW7hIQLxbLN4nY6uP0IE26htcu-BPTAo/s1600/Consolidation+Stages.jpg" height="400" width="295" /></a></div>
<br />
<ol style="text-align: left;">
<li>Upper curved portion(Stage I). It is mainly due to the result of pre-compression.</li>
<li>A Straight line portion (Stage II), is referred to as the primary consolidation. At the end of primary consolidation, a major extent of the excess pore water generated by the increased loading is dissipated.</li>
<li>A lower straight line portion (Stage III), is called the secondary consolidation. During this stage, specimen undergoes small deformation with time.<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg6O1d3omulEhlzglKW2xEMbTcLDtS520jUOIuKJk3fNA59eLDgBLqFMvUqn-wicU7QyRKzo3bCRhHg-jW7leYZt3IXEwPrFWosUC9XGTXUkMkJSZghbSZ5GIasL42cLg1PAXKok86g4Jc/s1600/Consolidation+curve.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg6O1d3omulEhlzglKW2xEMbTcLDtS520jUOIuKJk3fNA59eLDgBLqFMvUqn-wicU7QyRKzo3bCRhHg-jW7leYZt3IXEwPrFWosUC9XGTXUkMkJSZghbSZ5GIasL42cLg1PAXKok86g4Jc/s1600/Consolidation+curve.jpg" height="297" width="400" /></a></div>
</li>
</ol>
<div>
Note that at the end of of the test for each loading, the stress on the specimen is the effective stress. Once the specific gravity of the soil solids, the initial dimensions of the specimen, and the specimen deformation at the end of each load have been determined, the corresponding void ratio can be calculated.</div>
<div>
<br /></div>
<div>
Thanks for your kind visit!</div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-35080242215497407212014-08-24T06:00:00.000-07:002014-08-24T06:00:11.345-07:00Vertical piles subjected to Eccentric Loading<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<br />
<br />
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhMwlpMlbOxxi0mOE5CfKg1afkhSQ8EbahAdXo-8fj-__VTPDjdB9n85V5BcC4JaVQKAnxMl8efkbAcqc2cfMDJY8hluC4Fd-5bTycBG7Te6tkvloSYtnHDG-OpqYi0Xhi-TZr7X1SbvsM/s1600/eccentric+loadin+on+pile+group.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhMwlpMlbOxxi0mOE5CfKg1afkhSQ8EbahAdXo-8fj-__VTPDjdB9n85V5BcC4JaVQKAnxMl8efkbAcqc2cfMDJY8hluC4Fd-5bTycBG7Te6tkvloSYtnHDG-OpqYi0Xhi-TZr7X1SbvsM/s1600/eccentric+loadin+on+pile+group.jpg" height="234" width="320" /></a>Let's have two cases of eccentric loading on a pile group.<br />
<br />
<ul style="text-align: left;">
<li><b>e is about one axis.</b></li>
</ul>
In this case load is given as, <b>Vp = V/n +- (V.e.xj.A)/Ig</b><br />
<div style="display: inline !important;">
<br /></div>
<b></b><b>V = </b>Total vertical load on pile group<br />
n = No. of the piles<br />
e = Eccentricity w.r.t. C.G.(Center of Gravity) of pile group.<br />
xj = Distance of the center of the pile group from center of pile, measured parallel to e.<br />
Ig = Moment of Inertia of pile group about axis normal to axis of eccentricity.<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMUdQotDhytezw9Rz8bWV5CaedMLXxUFoXyMNodj6-D0CpZUTAgROhfGhyphenhyphenyzfA7BGYM277ZZ6LK6C-zzPo62dbtG0jXzBewOLTAC5_WWWkjT9Py3RV4pdZqPfW9sERazRY22a5zVlAXIg/s1600/eccentric+loadin+on+pile+group.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMUdQotDhytezw9Rz8bWV5CaedMLXxUFoXyMNodj6-D0CpZUTAgROhfGhyphenhyphenyzfA7BGYM277ZZ6LK6C-zzPo62dbtG0jXzBewOLTAC5_WWWkjT9Py3RV4pdZqPfW9sERazRY22a5zVlAXIg/s1600/eccentric+loadin+on+pile+group.jpg" height="234" width="320" /></a></div>
<br />
<br />
<ul style="text-align: left;">
<li><b>e is about both the axis</b></li>
</ul>
<b>Vp = V/n +- (V.ex.xj.A)/Ix +- (V.ey.yj.A)/Iy</b><br />
<b><br /></b>
Where ex and ey are the eccentricities w.r.t. center of pile group, measured along x and y axis respectively.<br />
<br />
<br />
Thanks for your kind visit!</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0tag:blogger.com,1999:blog-6170519791936471147.post-40815173415943377412014-08-22T21:04:00.000-07:002014-08-22T21:04:14.899-07:00Uplift capacity of the Footing<div dir="ltr" style="text-align: left;" trbidi="on">
Greetings!<div>
<br /></div>
<div>
Footings are sometimes subjected to uplift or tension forces, few examples are:</div>
<div>
<ol style="text-align: left;">
<li>Legs of the elevated water tank</li>
<li>Anchorages to the anchor of the footings.</li>
<li>Footings of the transmission line towers.</li>
</ol>
<div>
Bala(1961) developed the expressions along with the following procedure for their use:</div>
</div>
<div>
<ul style="text-align: left;">
<li>A footing may be considered as shallow or deep if D<span style="font-size: x-small;">f</span>/B < H/B<span style="font-size: x-small;">limiting. </span>Which depends upon angle of internal friction. </li>
</ul>
<b> Angle of Internal friction (Degrees) H/B<span style="font-size: x-small;">limiting</span></b></div>
<div>
20 2.5</div>
<div>
25 3.0</div>
<div>
30 4.0</div>
<div>
35 5.0</div>
<div>
40 7.0</div>
<div>
45 8.0</div>
<div>
<ul style="text-align: left;">
<li><b>For shallow footing, i.e. D<span style="font-size: x-small;">f</span>/B < H/B<span style="font-size: x-small;">limiting</span></b></li>
</ul>
<ol style="text-align: left;">
<li><span style="font-size: x-small;"> </span>For Circular Footings <div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0IQUp0252jo3O1tpJbU1yiZbCvGShQdG3FqazsCbjIf-RiF-u6uA-vmuZZcvZWjMwtgvR7buEGwMQhnNDyVavcGmkwW3WCS_ikLeac1BdSzjO7CgPePQUOFy5CgMdVr8KJ2K3e0SGjPw/s1600/Uplift.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0IQUp0252jo3O1tpJbU1yiZbCvGShQdG3FqazsCbjIf-RiF-u6uA-vmuZZcvZWjMwtgvR7buEGwMQhnNDyVavcGmkwW3WCS_ikLeac1BdSzjO7CgPePQUOFy5CgMdVr8KJ2K3e0SGjPw/s1600/Uplift.jpg" height="96" width="320" /></a></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
</li>
<li>For Rectangular Footings</li>
</ol>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgJetsHiXHcMJe7ICKgX1a4rWCjo3pgtj73MmFP-mkC15-Hm99na7fqUteXhN5slBentud4rMDtfrD_2owEFXP6WrH5xi2zjgIKwLvVFjIB-7CjBT_oIOtY8tvzSdOIyevw6zO4X8DeTCY/s1600/Rectangular.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgJetsHiXHcMJe7ICKgX1a4rWCjo3pgtj73MmFP-mkC15-Hm99na7fqUteXhN5slBentud4rMDtfrD_2owEFXP6WrH5xi2zjgIKwLvVFjIB-7CjBT_oIOtY8tvzSdOIyevw6zO4X8DeTCY/s1600/Rectangular.jpg" height="116" width="400" /></a></div>
<ul style="text-align: left;">
<li><b>For Deep Footing D<span style="font-size: x-small;">f</span>/B > H/B<span style="font-size: x-small;">limiting</span></b></li>
</ul>
<ol style="text-align: left;">
<li> For Circular Footings <div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhlWF2xpAELBIVei9JMCwzmXS2bd5lt0DNeKLKXeaAyuflziaYYUEyjfBi3y1OxpWYkYwlCxvrwpbg4slBdrsKvpKjJxStmCeoTYNUA-BvZ9t-oSqvh-XIjzCX9FHsiYp-xgOCKNmTEoAo/s1600/Deep_C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhlWF2xpAELBIVei9JMCwzmXS2bd5lt0DNeKLKXeaAyuflziaYYUEyjfBi3y1OxpWYkYwlCxvrwpbg4slBdrsKvpKjJxStmCeoTYNUA-BvZ9t-oSqvh-XIjzCX9FHsiYp-xgOCKNmTEoAo/s1600/Deep_C.jpg" height="98" width="400" /></a></div>
<div class="separator" style="clear: both; text-align: center;">
</div>
</li>
<li>For Rectangular Footings<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgX_LTQr4i5H_BVJFG-7bqvGws84UmvCCOo4kqbZo6NQV_Fz0-FIqQ3VSfcKhx18-xIhOMZLvbdZEOUUVfwyHXIMISchBaViojX_PTPCjjYr02B3YTSNsKrIFhDf69O_al5JJXzJzTbElc/s1600/Deep_R.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgX_LTQr4i5H_BVJFG-7bqvGws84UmvCCOo4kqbZo6NQV_Fz0-FIqQ3VSfcKhx18-xIhOMZLvbdZEOUUVfwyHXIMISchBaViojX_PTPCjjYr02B3YTSNsKrIFhDf69O_al5JJXzJzTbElc/s1600/Deep_R.jpg" height="82" width="400" /></a></div>
</li>
</ol>
Where,</div>
<div>
Sf= Shape Factor = 1+m.Df/B - For shallow</div>
<div>
= 1+ m.H/B - For Deep</div>
<div>
K = Co-efficient of earth pressure at rest.</div>
<div>
<br /></div>
<div>
The value of m: </div>
<div>
<b>Angle of friction </b> <b> H/B<span style="font-size: x-small;">limiting </span> m</b></div>
<div>
<b> </b> 20 2.5 0.05</div>
<div>
25 3.0 0.10</div>
<div>
30 4.0 0.15</div>
<div>
<br /></div>
<div>
A factor of safety of 2.0 to 2.5 may be adopted to find the allowable bearing capacity.</div>
<div>
<br /></div>
<div>
If soil is very poor, Pu may be taken equal to W. In this case FOS of 1.5 may be taken to determine the pull capacity Pa.</div>
<div>
<br /></div>
<div>
Thanks for your kind visit!</div>
<div>
<br /><div style="text-align: center;">
<br /></div>
</div>
</div>
<div style="text-align: center;">
<br /></div>
</div>
Sanjay Sharmahttp://www.blogger.com/profile/13728855310168117244noreply@blogger.com0