Integrated Marketing Communications Essay Example for Free

Integrated Marketing Communications Essay Answer1: The Vocalpoint program launched by PG was a well sought out mean to market the products by using the word of mouth marketing strategy. Inclusion of the mothers from around the country was a cunning decision made by the managers at PG the reason for this being was that mothers are the ones who usually purchase the products for the household or domestic use so convincing them and then make them work in marketing the new offering so an immaculate move by the management of PG. Another benefit which PG gained by devising this Vocalpoint program was that, housewives are hard to target as they do not use technology that often, rather they spend a big part of their leisure time in gossiping with their friends hence the best possible way to market the offerings to them was by this word of mouth publicity. PG earned pretty much short term and long term benefits by applying this strategy; the most outstanding one was that launching this program helped the firm in increasing the sales by a considerable extent which obviously is the sole aim of launching any marketing campaign. In addition to this external clients of PG also earned huge profits by using this program and so as a result PG got its share in the profits, this amount was as higher as 1 million USD. The job of the RD department got less hectic as it was able to get the desired feedback on the latest products; this was helpful for it in making changes to the existing products and also to the ones which are undergoing the process of development. One of the long term benefits which came to PG’s way as a result of this program was that it got a solid loyal customer base on which it can capitalize in the forthcoming years. The ‘connect and develop† program of PG also got immense support as a result of this program and the marketing intelligence of the company overall increased. Better customer relationship management is also one aspect of the Vocalpoint program which was advantageous for the firm because it helped the managers in comprehending the consumer’s mindset and thus forecasting the consumers’ behavior became easily. The customers on the other hand became increasingly brand loyal to the company as they were getting tailor made products for themselves because PG was able to understand their latent and covert needs in an adept manner like no one else was doing. The overall impact of this project on the position of the firm was immaculate as it was a win-win situation for the company; the managers were getting the ins and outs of the target market in no time whatsoever so the overall time consumed in the processes of product development got reduced. This decrease in the overall product development time acted as a competitive advantage for the managers as they were able to deliver the market with the products it wanted in much lesser time than the competitors. Another addition was that the managers at the firm were getting customer generated feedback on the latest products and that too in negligible time so this enabled them to alter their strategies if needed be; the consumer on the other hand were liking this as they were getting tailor-made products for their usage (Sirgy 1998). Answer2: Analyzing from the business perspective one has to say that the programs such as Vocalpoint and Tremor must go on as they are beneficial for the businesses in the longer run and same is the case with PG. Having said that one has to admit that these projects and programs are certainly not the ones, which are based on the ethical principles and norms laid by the society. In my opinion there are various ethical issues which are present in such kind of WoM programs. The first and the foremost being is that the person to whom the offering is being marketed is unaware of the reality he is perceiving the interaction as a social one but in general the other person is acting as a sales person for a specific firm. My objection regarding this is that if a person is being marketed something then he must have prior knowledge of this. For me there is a thin line between marketing and deceiving and this difference has to be understood by the firms and the marketing managers; if there are ample ethical ways of marketing then what is the point in using the deceptive means? Another issue in my opinion is that by hiring or deploying WoM marketers the managers are making the society increasingly materialistic even the sacred relationships like friendship etc are being used just as an asset and even that without the other person being aware of this all. Objectifying the society is one thing which to me is intolerable and as this eventually ruins the whole basic structure of the civilization which really is denting in terms of the future perspectives. Considering the Tremor program I can firmly state that this was one project on which I have serious reservations; the reason for this being is that PG was using minors (children under 18 years ) to market its products; this is one ridiculous way of marketing. The reason for this is that this is the age where the person learns the most, from its society and surroundings and if the society is indulged in inducing the materialistic thoughts in kids then this will be injurious for the entire social setup; this is the stage at which the children must be taught how to honor and value the relationships but such deceptive marketing projects destroys the whole brought up of a child. One legal issue which too arises as a result of the inclusion of minors in the marketing programs is that these kids are underage legally so they cannot be made a part of any marketing project without the prior consent of their guardians (Malachowski 2001); this is one thing which was completely neglected by the managers at PG and hence they violated the rules and regulations laid by the government. In the Tremor programs the minors were not allowed to tell the reality to the person to which they are interacting this is something which is immoral and to me for this the firm has to be penalized; can we teach our younger generations to indulge into immoral activities just for the sake of money? This is a million dollar question which has to be answered by the managers at PG. The Tremor program targeted the minors too which is also an unethical way of making business the reason for this is that the underage people are immature hence they cannot make decisions for themselves (Malachowski 2001); influencing a buyer’s decision by using rational marketing ploys is a justifiable act but making the kids addict by using deceptive means and backdoor channels has to be abandoned as this certainly is not favorable for the social composition. In sum commercialization of human relationships is one thing which must be stopped by the marketing managers if someone is indulging in WoM marketing of a firm then he should disclose his affiliation as this is beneficial for the society; involving money in the activities of daily life will hurt the society in the longer run. In the Vocalpoint program the managers at PG were looking to select the mothers which are more social; the main reason behind this move was that, as mother is the most sacred and trustworthy relationship in this universe so mothers enjoy a unique position in the society; using their social stature for the marketing purposes was an unreasonable attempt. Answer 3:  The value of WoM is immense in the present global marketing scenario, the primary reason for this is that due to the increase in the globalization the nature of the competition has increased exponentially and so to do business and earn profits a firm has to market it’s offering to the target buyers. With the increase in the intensity of competition and technology the media of marketing have too increased and thus in such an overall marketing environment WoM publicity is something which is of great significance (Bothma 2003). Direct marketing and interactive marketing have always been profitable ways of interacting and influencing the consumers the reason behind this is that first of all these means are less expensive than the other marketing tactics (Tuckwell 2004). In addition to this the marketers get to know more about the ins and outs of the buyer’s black box; as known that these are the two methods which are an integral part of the IMC mix hence applying them together makes it easier for the marketers to timely comprehend the latent and covert needs of the buyers and thus they can make the necessary changes to the offerings and strategies. WoM advocacy is beneficial in the present scenario because it enables the marketers to interact with the customers and hence they can obtain the customer generated feedback regarding their products in addition to this, as there is no such middle channel involved in between the marketer and the buyer (as the WoM marketer is a part of the company) so it becomes easier to get the message of the consumer. The firms once get the timely feedback can make suitable adjustments and alterations in the product designs, strategies etc and in this way eventually WoM publicity acts as a source of competitive advantage for the company in this era of fierce competition. After understanding the consumer psyche if need be, then suitable sales promotions can be launched which are an essential component of the overall IMC mix (Blakeman 2009) The holistic marketing concept has to be applied as per the book because consumer centricity is something plays a vital role in the overall growth of the firm. Establishing long term business relations with the buyers is only possible if the market is provided with what is demanded by it. The nature of the relationship between the firm and its related publics should be exceptional, as this affects the performance of the company in the longer run WoM publicity is one thing which helps in building stronger relationships between the publics and the company (Baker 2001).

Lessons Learned from Nathaniel Hawthornes The Scarlet Letter Essay

Lessons Learned from The Scarlet Letter      Ã‚  Ã‚   The Scarlet Letter by Nathaniel Hawthorne is considered by many to be a classic novel. In The Scarlet Letter, Hawthorne provides his audience with a real sense of the consequences of unconfessed sin, isolation from society, and the presence of evil everywhere. Through his portrayal of the main characters, his choice of setting, and his implied moral lessons, he teaches lessons that must be learned for humans to continue living in harmony with one another.    The setting of The Scarlet Letter provides a powerful connection between fact and fiction. Events such as the Salem witch trials, which occurred not long after the events of The Scarlet Letter, establish credibility for Hawthorne in that recounting historical details such as fear of witches makes him seem like he actually knows what he is talking about. Hawthorne's writing style has made him "one of the most widely read nineteenth century authors" (Jacobson 4). He upholds Puritan values and concepts while employing the classic allegorical characters of romanticism. One such Puritan value is that the devil resides in the forest. Anytime characters in The Scarlet Letter enter the forest, it is certain that something terrible just happened, is happening, or will happen soon. One such example is when the governor's sister, who is suspected of witchcraft, enters the forest and invites Hester to go with her. It is this woman who represents the idea that evil exists everywhere, especially among those who refuse to acknowledge its existence. Salem is a particularly intriguing setting because of its witch infamy. The occurrence of the witch trials in Salem creates an atmosphere where the evil and the right... ... Hawthorne's allegorical approach at real life situations provides his readers with a sense of accomplishment: a sense that if they learn lessons from others, then they will not have to learn from first hand experience. Although on the surface it may seem like another tale of Puritanistic virtue, The Scarlet Letter is the embodiment of life itself. After reading this novel, one may find that many events in real life today can relate directly to events in The Scarlet Letter.    Works Cited and Consulted:    Chase, Richard (1996). "The Lessons of the Scarlet Letter." Readings on Nathaniel Hawthorne (pp. 145-152). San Diego: Greenhaven.      Hawthorne, Nathaniel. The Scarlet Letter. New York: St. Martins, 1991.      Jacobson, Gary. The Critical Response to Nathaniel Hawthorne's The Scarlet Letter. New York: Greenwood, 1992.   

Soil Mechanics by Jerry Vandevelde

SOIL MECHANICS (version Fall 2008) Presented by: Jerry Vandevelde, P. E. Chief Engineer GEM Engineering, Inc. 1762 Watterson Trail Louisville, Kentucky (502) 493-7100 1 National Council of Examiners for Engineering and Surveying http://www. ncees. org/ 2 STUDY REFERENCES †¢ Foundation Engineering; Peck Hanson & Thornburn †¢Introductory Soil Mechanics and Foundations; Sowers †¢NAVFAC Design Manuals DM-7. 1 & 7. 2 †¢Foundation Analysis and Design; Bowles †¢Practical Foundation Engineering Handbook; Brown 3 Soil Classification Systems * Unified Soil Classification System * AASHTO Need: Particle Sizes and Atterberg Limits 4Particle Sizes (Sieve Analysis) (Well Graded) (Poorly Graded) 0. 1 5 Atterberg Limits Liquid, Plastic & Shrinkage Limits Plasticity Index (PI) PI = Liquid Limit – Plastic Limit (range of moisture content over which soil is plastic or malleable) 6 UNIFIED SOIL CLASSIFICATION SYSTEM ASTM D-2487 7 8 Ref: Peck Hanson & Thornburn 2nd Ed. Effe ctive Size = D10 10 percent of the sample is finer than this size D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm 0. 1 0. 1 9 Uniformity Coefficient (Cu) = D60/D10 Coefficient of Curvature (Cz) = (D30)2/(D10xD60) D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm 0. 1 10 Well Graded – Requirements 50% coarser than No. 00 sieve Uniformity Coefficient (Cu) D60/D10 >4 for Gravel > 6 for Sand Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 1 to 3 11 Is the better graded material a gravel? 81% Passing No. 4 18% Finer No. 200 0. 1 0. 1 12 Gravel if > 50 Percent Coarse Fraction retained on No. 4 sieve % Retained on No. 200 = 82% 1/2 = 41% 19% (100-81) retained on No. 4 sieve (gravel) 19< 41 half of coarse fraction 81% Passing No. 4 18% Finer No. 200 ? sand 0. 1 (â€Å"S†) 13 Well Graded Sand? Uniformity Coefficient (Cu) > 6 = D60/D10 Coefficient of Curvature (Cz) = 1 to 3 = (D30)2/(D10xD60) 14 D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 3mm 0. 1 Well Graded Sand? Uniformity Coefficient (Cu) D60/D10 = 1. 6/. 03 = 53 > 6 D60 = 1. 6mm D30 = 0. 2mm D10 = 0. 03mm Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 0. 22/(. 03Ãâ€"1. 6) = 0. 83 12% Passing No. 200 sieve: GM, GC, SM, SC 0. 1 >12% passing No. 200 sieve Since = â€Å"S† ? SC or SM 16 What Unified Classification if LL= 45 & PI = 25? From sieve data SC or SM 0. 1 A) â€Å"SC† B) â€Å"SM† C) â€Å"CL† or D) â€Å"SC & SM† 17 Unified Classification Answer is â€Å"A† ? SC 18 AASHTO (American Association of State Highway and Transportation Officials) 19 What is the AASHTO Classification? 65% Passing No. 10 40% Passing No. 0 18% Finer No. 200 1) 18 % passing No. 200 sieve 2) 65% passing No. 10 sieve 3) 40% passing No. 40 sieve 4) assume LL = 45 & PI = 25 20 18 percent passing No. 200 sieve; 65 percent passing No. 10 sieve 40 percent passing No. 40 sieve; assume LL = 45 & PI = 25 21 AASHTO Classification 1 2 3 4 4 1) 18 % passing No. 200 sieve 2) 65% passing No. 10 sieve 3) 40% passing No. 40 sieve 4) assume LL = 45 & PI = 25 22 AASHTO Group Index 23 Mass-Volume (Phase Diagram) †¢ Unit volume of soil contains: Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil – Air (gases) – Water (fluid) – Solid Particles 24 Moisture Content = ? eight of water/ weight of dry soil ? = Ww/Wd water loss/(moist soil weight – water loss) ? = Ww/(Wm-Ww) and ? =(Wm-Wd)/Wd 25 Mass – Volume Relationships Density or Unit Weight = Moist Unit Weight = ? m ? ?m = Wm/Vt = ? d + ? ?d ? = (? m – ? d )/ ? d ? ?d + ? d = ? m ? m= (1+ ? ) ? d ? d = ?m/(1+ ? ) b 26 Total Volume = ? Volume (solid + water + air) = Vs+Vw+Va ? Va = Vt – Vs- Vw Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 27 Relationship Between Mass & Volume Volume = Mass/(Specific Gravity x Unit Weight of Water) = Ws/(SGxWw) Va Total Volume Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 28Specific Gravity = weight of material/ weight of same vol ume of water Soil Specific Gravity Typical Range 2. 65 to 2. 70 Specific Gravity of Water = 1 29 Saturation = S expressed as percent S = volume of water/ volume of voids x 100 Total Volume Va Air Total S = Vw/Vv x 100 Ww Ws Weight Vt Vv Vw Vs Water Wt Soil Always ? 100 30 Porosity n = volume of voids/ total volume n = Vv/Vt Void Ratio e = volume of voids/ volume of solids e = Vv/Vs Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 31 What is the degree of saturation for a soil with: SG = 2. 68, ? m = 127. 2 pcf & ? = 18. 6 percent A) 88. 4 Total Volume VaAir Total Vt Vv Vw Vs Water Ww Ws Weight B) 100. 0 Wt Soil C) 89. 1 32 What are the porosity and degree of saturation for a soil with: SG = 2. 68, ? m = 127. 2 pcf & ? = 18. 6 percent = 107. 3pcf ?d = ? m/(1+ ? ) = 127. 2/(1. 186) Total Volume Va Air Total Vt Vv Vw Vs Water Soil Ww Weight Wt Ws Ww = ? m- ? d = 19. 9 pcf Vw = Ww/62. 4 = 0. 319 cf Vs = ? d /(SGx62. 4) = 0. 642 cf Va = Vt – Vw – Vs = 1- 0. 319 – 0. 642 = 0. 039 cf Vv = Vw + Va = 0. 358 cf 33 What are the porosity and degree of saturation for a soil with: SG = 2. 68, ? m = 127. 2 pcf & ? = 18. 6 percent Vw = 0. 319 cf, Vs = 0. 642 cf, Vv = 0. 358 cf Total VolumeVa Air Total Degree of Saturation = Vw/Vv x 100 Ww Weight Wt Ws Vt Vv Vw Vs Water = 0. 319/0. 358 x 100 = 89. 1% Soil Answer is â€Å"C† 34 Ref: NAVFAC DM-7 35 Borrow Fill Adjustments Borrow Material Properties: ?m = 110 pcf & ? = 10% Placed Fill Properties: ? d = 105 pcf & ? = 20% How much borrow is needed to produce 30,000 cy of fill? How much water must be added or removed from each cf of fill? Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 36 Borrow Fill Adjustments Borrow Material Properties: ?m = 110 pcf & ? = 10% ?d = ? m /(1+? ) = 110/(1. 10) =100 pcf; Ww = 110-100=10 lbs Placed Fill Properties: ? = 105 pcf & ? = 20% Ww = ? x ? d = 0. 2x 105 = 21 lbs Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 37 Borro w Fill Adjustments Borrow Properties: ? m = 110 pcf, ? d =100 & ? = 10% Placed Fill Properties: ? d = 105 pcf & ? = 20% Since borrow ? d =100pcf & fill ? d =105pcf, 105/100 =1. 05 It takes 1. 05 cf of borrow to make 1. 0 cf of fill For 30,000 cy, 30,000 x 1. 05 = 31,500 cy of borrow Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 38 Borrow Fill Adjustments Borrow Material Properties: Ww = 10 lbs Placed Fill Properties: Ww = 21 lbs Water supplied from borrow in each cf of fill = 10 x 1. 5 = 10. 5 lbs; 21 lbs – 10. 5 = 10. 5 lbs short/1. 05 cf 10. 5lbs/1. 05 cy = 10 lbs of water to be added per cf borrow Total Volume Va Air Total Vt Vv Vw Vs Water Ww Ws Weight Wt Soil 39 Proctor: Moisture Density Relationships Establishes the unique relationship of moisture to dry density for each specific soil at a specified compaction energy MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 D ry D ensity (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) 40 Proctor: Moisture Density Relationships †¢ 4† mold 25 blows †¢ 6† mold 56 blows Standard – 5. 5 lb hammer – dropped 12 in – 3 layers Standard: ASTM D-698 AASHTO T-99 Modified: ASTM D-1557 AASHTO T-150 †¢ Modified – 10 lb hammer – dropped 18 in – 5 layers 41 PROCTOR COMPACTION TEST Maximum Dry Density – Highest density for that degree of compactive effort Optimum Moisture Content – Moisture content at which maximum dry density is achieved for 42 that compactive effort Proctor: Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%)What density is required for 95% Compaction? What range of moisture would facilitate achieving 95% compaction? 43 Proctor: M oisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) 104 x . 95 = 98. 8 pcf A 95% B Range of moisture is within the curve A to B (14 to 24 %) 44 Proctor: Zero Air Voids Line Relationship of density to moisture at saturation for constant specific gravity (SG) Can’t achieve fill in zone right of zero air voids line ZMOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) 45 Proctor: Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) If SG = 2. 65 & moisture content is 24% What dry density achieves 100% saturation? A) 100. 0 pcf B) 101. 1 pcf 46 Proctor: Moisture Density RelationshipsMOISTURE-DENSITY RELATIONSHIP 108. 0 106. 0 104. 0 Dry Density (pcf) 102. 0 100. 0 98. 0 96. 0 94. 0 92. 0 90. 0 88. 0 8. 0 10. 0 12. 0 14. 0 16. 0 18. 0 20. 0 22. 0 24. 0 26. 0 28. 0 Moisture Content (%) X ?d=SG62. 4/(1+? SG/100) ? d=2. 65Ãâ€"62. 4/(1+24Ãâ€"2. 65/100) ? d=101. 1 pcf Answer is â€Å"B† 47 Ref: Peck Hanson & Thornburn Static Head 48 Calculate effective stress at point x Ref: Peck Hanson & Thornburn Saturated Unit Weight ? sat 5’ ? sat = 125 pcf Moist Unit Weight ? M Dry Unit Weight ? Dry 7’ Submerged (buoyant) Unit Weight = ? sat – 62. 4 x 49 Calculate effective stress at point x Ref: Peck Hanson & ThornburnTotal Stress at X 5’ ? sat = 125 pcf = 5 x 62. 4+ 7x 125= 1187psf Pore Pressure at X 7’ = 12 x 62. 4 = 749 psf Effective Stress at X = 1187-749= 438 psf x or (125-62. 4) x 7=438 psf 50 Ref: Peck Hanson & Thor nburn Downward Flow Gradient 51 Downward Flow Gradient 3’ Total Stress at X = 5 x 62. 4+ 7x 125= 1187psf Pore Pressure at X ? sat = 125 pcf 7’ = (12-3) x 62. 4 = 562 psf Effective Stress at X = 1187-562 = 625 psf 5’ x or 438 + 3 x 62. 4 = 625psf see previous problem 52 Upward Flow Gradient Ref: Peck Hanson & Thornburn 53 One Dimensional Consolidation ?e/pn 54 Primary Phase Settlement (e log p) ? H = (H x ? )/(1+eo) eo ? H H 55 Consolidation Test Pre-consolidation Pressure Cc = slope of e log p virgin curve est. Cc = 0. 009(LL-10%) Skempton Rebound or recompression curves 56 56 e- l o g p Calculate Compression Index; Cc 1. 50 1. 40 1. 30 Void Ratio (e) 1. 20 1. 10 ksf 0. 1 1 4 8 16 32 (e) 1. 404 1. 404 1. 375 1. 227 1. 08 0. 932 1. 00 0. 90 A) 0. 21 B) 0. 49 57 0. 80 0. 1 1 10 100 Pr essur e ( ksf ) Cc is the slope of the virgin e-log p e- l o g p Cc = -(e1-e2)/log (p1/p2) 1. 50 Cc=-(1. 375-1. 227)/log(4/8) Cc = 0. 49 Answer is â€Å"B† ksf 0. 1 1 4 8 16 3 2 (e) 1. 404 1. 404 1. 375 1. 227 1. 08 0. 932 1. 40 Cc Void Ratio (e) . 30 1. 20 1. 10 1. 00 0. 90 0. 80 0. 1 1 10 100 Pr essur e ( ksf ) 58 Permeability Constant Head Conditions †¢ Q=kiAt †¢ Q= k (h/L)At †¢ k=QL/(Ath) 59 If Q =15cc & t = 30 sec what is the permeability k=QL/(Ath) 10cm 5cm A) 0. 01 cm/sec B) 0. 01Ãâ€"10-2 cm/sec 25cm2 C) 0. 1 cm/sec 60 Constant Head Permeability Calculate k Q =15cc & t = 30 sec †¢ k=QL/(Ath) †¢ k= 15(5)/(25(30)10) †¢ k= 0. 01 cm/sec Answer is â€Å"A† 10cm 5cm 25cm2 61 Falling Head Permeability †¢ k=QL/(Ath) (but h varies) †¢ k=2. 3aL/(At) log (h1/h2) †¢ where a = pipette area †¢ h1 = initial head †¢ h2 = final head 62 If t = 30 sec; h1= 30 cm; h2 = 15 cm L= 5 cm; a= 0. cm2; A= 30 cm2; calculate k A) 2. 3Ãâ€"10-3 cm/sec B) 8. 1Ãâ€"10-6 cm/sec C) 7. 7Ãâ€"10-4 cm/sec 63 Falling Head Permeability k=2. 3aL/(At) log (h1/h2) k= 2. 3 (0. 2) 5 /(30Ãâ€"30) log (30/15) k= 7. 7Ãâ€"10-4 cm/sec Answer is â€Å"C† 64 †¢Flow lines & head drop lines must intersect at right angles †¢All areas must be square †¢Draw minimum number of lines †¢Results depend on ratio of Nf/Nd Flow Nets 6ft 2ft 65 Q=kia=kHNf /Nd wt (units = volume/time) w= unit width of section t=time Flow Nets 6ft 66 What flow/day? assume k= 1Ãâ€"10-5 cm/sec =0. 0283 ft/day Q= kH (Nf /Nd) wt Q= 0. 0283x8x(4. 4/8)x1x1 Q= 0. 12 cf/day 2ft Flow Nets ft 67 Check for â€Å"quick conditions† pc =2(120)= 240 psf (total stress) Flow Nets Below water level use saturated unit weight for total stress ?= 2(62. 4) = 124. 8 (static pressure) = 1/8(8)(62. 4)= 62. 4 (flow gradient) = 240-(124. 8+62. 4) 2ft 2ft 6ft p’c = pc -(? + ) p’c = 52. 8 psf >0, soil is not quick ?sat=120 pcf 68 Stress Change Influence (1H:2V) For square footing z=Q/(B+z)2 69 If Q= 20 kips, Calculate the vertical stress increase at 7 feet below the footing bottom 5’ 8’ 7’ 70 If Q= 20 k ips, Calculate the vertical stress increase at 7 feet below the footing bottom 5’ 8’ z = 0000 (8+7)(5+7) 7’ z = 111 psf 71 Westergaard (layered elastic & inelastic material) If B= 6. 3’ in a square footing with 20 kips load, what is the vertical stress increase at 7’ below the footing bottom? 72 Westergaard Q = 20 kips B = 6. 3’ Z = 7’ z = ? 73 Westergaard 7’/6. 3’ = 1. 1B z = 0. 18 x 20000/6. 32 = 90. 7 psf 74 Boussinesq (homogeneous elastic) Q = 20 kips B = 6. 3’ Z = 7’ z = ? 75 Boussinesq Z/B = 1. 1 z = 0. 3 x 20000/6. 32 = 151 psf 76 Thanks for participating in the PE review course on Soil Mechanics! More questions or comments? You can email me at: [email  protected] com 77

What Is the 5 Number Summary

There are a variety of descriptive statistics. Numbers such as the mean, median, mode, skewness, kurtosis, standard deviation, first quartile and third quartile, to name a few, each tell us something about our data. Rather than looking at these descriptive statistics individually, sometimes combining them helps to give us a complete picture. With this end in mind, the five-number summary is a convenient way to combine five descriptive statistics. Which Five Numbers? It is clear that there are to be five numbers in our summary, but which five? The numbers chosen are to help us know the center of our data, as well as how spread out the data points are. With this in mind, the five-number summary consists of the following: The minimum – this is the smallest value in our data set.The first quartile – this number is denoted Q1 and 25% of our data falls below the first quartile.The median – this is the midway point of the data. 50% of all data falls below the median.The third quartile – this number is denoted Q3 and 75% of our data falls below the third quartile.The maximum – this is the largest value in our data set. The mean and standard deviation can also be used together to convey the center and the spread of a set of data. However, both of these statistics are susceptible to outliers. The median, first quartile, and third quartile are not as heavily influenced by outliers. An Example Given the following set of data, we will report the five number summary: 1, 2, 2, 3, 4, 6, 6, 7, 7, 7, 8, 11, 12, 15, 15, 15, 17, 17, 18, 20 There are a total of twenty points in the dataset. The median is thus the average of the tenth and eleventh data values or: (7 8)/2 7.5. The median of the bottom half of the data is the first quartile. The bottom half is: 1, 2, 2, 3, 4, 6, 6, 7, 7, 7 Thus we calculateQ1 (4 6)/2 5. The median of the top half of the original data set is the third quartile. We need to find the median of: 8, 11, 12, 15, 15, 15, 17, 17, 18, 20 Thus we calculateQ3 (15 15)/2 15. We assemble all of the above results together and report that the five number summary for the above set of data is 1, 5, 7.5, 12, 20. Graphical Representation Five number summaries can be compared to one another. We will find that two sets with the similar means and standard deviations may have very different five number summaries. To easily compare two five number summaries at a glance, we can use a boxplot, or box and whiskers graph.