Chapter 2 Stresses and Strains in Flexible Pavements 1 Layers of Flexible Pavements Surface (wearing) Course Binder Course Base Course Subbase Course Subgrade 2 Contents Single layer analysis Two-layer system Three-layer system Viscoelastic solution 3 Single Layer Analysis r t z 4 Boussinesq Theory (1885) Homogeneous elastic half-space A concentrated load is applied Stresses‚ strains‚ and deflections are calculated
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International Business School Maastricht Theme 1.1 Introduction to International Business Course outline for Finance & Accounting |Department: |International Business English stream | |Year: |2013-2014 | |Block: |1.1
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1.8 0.4 A1 A2 A3 B1 B2 D1 D2 E1 E2 E3 *Figures in percentage This round of analysis has been conducted using data from IRS 2008 (round 20)*. We have drawn a sub-sample of 39‚441 5 THE NEW SEC SYSTEM DISTRIBUTION OF HOUSEHOLDS Urban New SEC System 13.2 12.9 11.6 12.6 10.2 7.6 9.8 8.6 5.1 4.7 2.6 1.1 A1 A2 A3 B1 B2 C1 C2 D1 D2 E1 E2 E3 Current Urban SEC system The current urban SEC
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L7: Memory Basics and Timing Acknowledgements: Materials in this lecture are courtesy of the following sources and are used with permission. Nathan Ickes Rex Min Yun Wu J. Rabaey‚ A. Chandrakasan‚ B. Nikolic. Digital Integrated Circuits: A Design Perspective. Prentice Hall/Pearson‚ 2003. L7: 6.111 Spring 2006 Introductory Digital Systems Laboratory 1 Memory Classification & Metrics Read-Write Memory Random Access Non-Random Access SRAM FIFO DRAM Non-Volatile
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From the table‚ the result of dehydration of secondary alcohol is expected‚ and it undergoes E1 elimination. As supposed in the theory and mechanism‚ because elimination can occur both sides of secondary carbocation‚ so there are three products: 1-butene‚ trans-2-butene and cis-2-butene with three peaks. The main product is trans-2-butene‚ because it is the most stable compound. However‚ the result of dehydration of primary alcohol is not expected. It is supposed to form only one product‚ 1-butene
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each material we were going to be using‚ and the numbers we chose to use differed from our reference text slightly. We used 6.5 mL of our assigned butanol and only 4 mL of sulfuric acid. Going back to our reaction mechanisms‚ we were looking at SN1 and SN2 reactions‚ so we of course used the hydrobromic acid as our acid catalyst which would be protonating the hydroxyl group of our butanol‚ to make a better leaving group‚ so that the bromine radical would be able to attack the leftover carbocation
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126 (2001) MATHEMATICA BOHEMICA No. 1‚ 237–246 BLOCK DIAGONALIZATION J. J. Koliha‚ Melbourne (Received June 15‚ 1999) Abstract. We study block diagonalization of matrices induced by resolutions of the unit matrix into the sum of idempotent matrices. We show that the block diagonal matrices have disjoint spectra if and only if each idempotent matrix in the inducing resolution double commutes with the given matrix. Applications include a new characterization of an eigenprojection
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while using sulfuric acid as a catalyst. The sulfuric acid will accelerate the chemical reaction with being consumed in the process‚ it will increase the concentration of the protonated alcohol‚ which then can form an alkyl bromide by either an SN1 or SN2 reaction. The reaction for both mechanisms depends on the concentration of protonated alcohol and the catalyst should increase the rate of the reaction. This could increase the amount of alkyl bromide produced‚ but it could increase the side
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convention of Hofmann’s Rule through an E2 elimination because of the presence of a strong‚ bulky base and resulted in the major product of 1-butene. The acid-catalyzed hydration of a primary and secondary alcohol followed the convention of Zaitsev’s Rule proceeded with E1 elimination for the secondary alcohol. 1-butanol could not proceed with E1 because the unstable primary carbocation would have to undergo a 1‚2-hydride or 1‚2-methyl shift to proceed‚ and an E2 elimination is more efficient. The major
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|’P’. Since q1 is negative‚ its field E1 is directed toward q1‚ while the field E2 | | |is directed away from charge q2 since it is positive. | | The magnitudes of these fields can be immediately calculated: E1(at P) = k q1 /(r1)2 = (9 x 109)(6 x 10-9)/(1.52) = 24 N/C E2(at P) = k q2 /(r2)2 = (9 x 109)(4
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