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Lecture 2 The Concept of Strain Problem 21 A thin walled steel pipe of length 60 cm, diameter 6 cm, and wall thickness 012 cm is stretched 001 cm axially, expanded 0001 cm in diameter, and twisted through 1o Determine the strain components of the pipe«»¬¼ 1 1 2 3 n n n n 3 Principal Strain Tensor •Since, the principal stress tensor contains only normal stresses, the principal strain tensor will contain only normal strains •Our procedure for finding the principal strains follows 1 2 3 00 00 00 H H HLow Benzene Reformate 27
F E a ¬ ¼ Âp4fL 5 ÚÓ àß & âß c Y 5 à Y âð ͵ z Í µÙB ᦠF E a ¬ ¼ý µ w à K Âp4 VDT Ò 3, ýx F × , * U 4ß w à Ê E a ¬ ¼ Âp4fL 5 ÚÓ àß & âß c Y 5 àÙ z K U 7v K 5 & âµ U z M ¶Ù ð < Oð y B ᦠF N Ê w âð y B ᦠF bÁ Âp4 VDT Ò 3YA Ò , ýx F × , * U 4ß F 2 K 5 w âð y B(Remember a column times a row is an outer product and results in a matrix) The matrix T I ce L kk 2 Letªº¬¼ Examples Find the derivative 1 ye 5x 2 ye x2 3 y xex 4 y x exx 5 2 eexx y 6 yeln x Examples Find the equation of the tangent line to the graph of the function at the given point 1 ye 2xx, 2,12 2 f x e x3 ln , 1,0 Example Use implicit differentiation to find dy/dx given
Ce is the signªº¬¼ ªº¬¼ cos 2 sin 2 sin(2 ) 1 ( ) cos 2 2 Result from AM modulation with tone freq uency 1 ( ) cos 2 2 T cos 2 ( ) cos 2 ( ) cos 2 ( ) cos 2 ( ) he differen C C C m NB FM C C C AM C C m C m CC C m C m A f t f t f t t A f t A f f t f f t t A f t A f f tfft?E i j V ªºV V «» H ¬¼ H H 0 1 31ÖÖ P 2 E j V ªº§«»¨¸¨¸ H «»¬¼©¹ 8 Answer (1) Hint The given setup is equivalent to 4 capacitors Sol 12 34 4 PQ CC CC C C §¨¸¨¸ ©¹ ©¹ 0 0 0 0 00 2 2 2 2 2 2 PQ A C d AA dd H uu HH 0 eq HA K d K eq = 6 9 Answer (3) Hint u
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¬¼ 1 2 2 e 2 2 2 I kI MM 6 2 2 2 quantum e 2 e Ik kIk V V V V Absorption Phase Dark Field X Noise of detector output X X Fringe visibility Total exposure Total number of phase steps I M H 6 Impacts of Electronic Noise Rejection Experimental Studies«»¬¼ 0 0 0 1 1 1 x x y y z z x sx j y sy j z sz j V ZH V ZH V ZH c c c 3 0 3 0 3 0 2 2 2 x x y y z z x x tL y y tL z z tL H V H V H V §c ¨¸ ' ©¹ §c ¨¸ ' ¨¸ ©¹ §c ¨¸ ' ©¹ L?J J J J J P J D ªº «» ¬¼ ' D 175/r 0 ' E E E nn 1 72 Macroscale simulation of plasma assisted combustion Challenges • Multiphysics problems photons, electrons, electronic and vibrational excitations • Electromagnetic field, acoustic waves, shockwaves, ignition and
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Ù 5 s j ;Chapter 8 1 For a multiple state model where there are two states i State 0 is a person is alive ii State 1 is a person is dead Further you are given that a person can transition from State 0 to State 1 but not back again«»¬¼¬¼ p p p X X p X X J J J J AB JJ =T J J J J BD TT ) ªº' «» ¬¼' p X X f Normal equation T ªºªº' ªº
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¬¼ is (a) Resistance (b) Resistivity (c) Electrical conductivity (d) Electromotive force 2 The orbital velocity of a satellite close to the surface of the earth is v if this satellite starts orbiting at an altitude of half the earth's radius, the orbital velocity will be (a) 2 3 v (b) 2 5 v (c) 2 7 v (d) 2 9 v 3B A C 3750 veh k km ªº «»¬¼ km u hr ªº «»¬¼ I II kkjj(II)=170(I) 0 uf0 uf0 D E max (I) (I)080 (I) 4000 44 ukfj veh q hr ªº «» ¬¼)0 100 22 j opt k veh k km ªº «» I ¬¼« » « »¬ ¼ ¬ ¼ Rigorous coupledwave analysis (RCWA) is an extremely efficient algorithm for modeling us scattering through alldielectric structures It is most efficient for devices with low to moderate dielectric contrast THE CONVOLUTION MATRICES r ªº «» «» «» «» «» «» «» «» «» ¬¼ 0 2 2 2 ii xx iii yy d dz ª
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ªº¬¼= ip and () () i l P q t q t P ªº¬¼= iq This definition is broad enough to apply to simple GNSS signals such as the GPS C/A code, for which there is only an inphase component with all symbols taking on the same, rectangular shape, or to more complicated GNSS signals such as the GPS L1 civil signal (L1C) that include in andG¬¼¬ ¼ ¯¿32 By observation maximum shear force and bending moment occurs at r=b (closest point to the center line) WWD Vb ªº2 max L b 2 b L cD 2gg¬¼ WLDªº§·22§L·W§L L«»¨¸LL ¨¸ ¨ ¸D 29gg«»¬¼©¹©9¹ ©9 10¹ wL2D V max 364 g WD½11 Mbªº max ®¾32Lb 32 rª b Lb º b L c b L c b g¯¿32¬¼¬ ¼Buhlmann – Straub Credibility 1 Individual losses on a policy are distributed as a Pareto with D 6 and T The parameter is uniformly distributed between 4000 and 6000
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¬ ¼ ¬ ¼'cb4pb bn ½ ° ¼b1&» !J j Y j j ªº «» «» «» «» ¬¼ The nodal equation describing this power system is Y V =I bus Fault current calculations using the impedance matrix With all other voltage sources set to zero, the voltage at bus 2 is –V f, and the current entering the bus 2
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H J mol ' ' ' ' ªº¬¼ ' u u ' 8 A particle of unit mass is moving on a plane Its trajectory, in polar coordinates, is given by r t t t t( ) , ( ) ,2 T where t is time The kinetic energy of the particle at time t=2 is A 4 B 12 C 16 D 24 Answer C Solution 2 2 2 2 2 2sec 1¬¼ 2 04 8 5 32 0 order = 2 x 3, 2 = number of rows, 3 = number of columns Special Type of Matrices 1 Square Matrix A m x n matrix is called as a square matrix if m = n ie number of rows = number of columns The elements ij a when i = j 2 are called diagonal elements Example ªº «» ¬¼ 12 45 2 Diagonal Matrix¬¼ ªº¬¼ Examples Use the quotient rule to find the derivative 1 2 4 53 t gt t 2 1 x hx x 3 3 sin x fx x 4 3 2 4 8 26 5 sin x x x y xx Examples Find the derivatives using
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J O O u «»¬¼ u «»¬¼ u «»¬¼ u « ¼ ¬ » ¦ ¦ ¦ ¦ 2 2 2 y y y q W ªº q W * * 1 ** 11 ,,,,, ki W i jk k J W i jk kj X X D OD u ªº «»¬¼ u «»¬¼ ¦ ¦¦ 2 2 2 y y y The second derivative of loglikelihood with respect to each item parameter can be obtained as ** ** 1 ** 1 2 1 * 1, 1, 0, 2 1 j j j j j j q k j k q Gas Dehydration Chapter 11 Based on presentation by Prof Art Kidnay Updated Copyright © 19 John Jechura (jjechura@minesedu)J¸vwhulohq v¾uhohuh pdndohohul rnxunhq \d gd gl\dorjoduó yh vhv nd\ówoduóqó glqohunhq kdufdgóáóqó dpdq g¤klo gháloglu 2nxpd yh glqohph nóvópoduóqó nhqgl kóóqód j¸uh \dspdoóvóqó 2nxpd yh\d glqohph l©lq lkwl\dfóqó rodq v¾uh qh ndgduvd r v¾uh\l nxoodqóq %x ©doó !pd v¾uhvl khvdsodpdvóqó
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¬ ¼ ¬ ¼ ³³³ ³³³ A A, A, ¦ ¦ ¦ A i i i j j j k i inlet j outlet k rxns d dV F F R dV dt ZU A Z U Z U ¦ ¦ ¦ > @ A, A A i i i j k i inlet j outlet k rxns dV F F R dV dt or in terms of moles (and molar concentration, though mole fractions could also be used) ªº ¦ ¦ ¦,,¬¼ ³³³ A A A i A j k i inlet j outlet k«»¬¼ 1,, 0 0 kk nk c l l thus 0's are in the first k positions Let T >0 0 1 0 0 @ e k, a row vector with all zeros except for a 1 in position k What is the matrix T I ce k?F(x, y, z) = z i y j x k across the unit sphere x2 y2 z2 = 1 •Using the parametric representation r(Φ, θ) = sin Φ cos θ i sin Φ sin θ j cos Φ k F(r(Φ, θ)) = cos Φ i sin Φ sin θ j sin Φ cos θ k Example 4
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I i j ee j i j f k u u f k u u eeªº ()e i ee j kku f kku f ½°° «»® ¾ ® ¾ «»¬¼ ¯¿°°¯¿ ^ ` ^ ` () ( ) ( ) ( ) e e i e j e e e u f u f ½°° ªº¬¼ ¯¿°° k k q f k q f 8 SPRING ELEMENT cont • Stiffness matrix –It is square as it relates to the same number of forces as the displacements –It is symmetric7hvw $b3dshu b &rgh % 0rfn whvw iru (( 0dlq $gydqfhg $dndvk (gxfdwlrqdo 6huylfhv /lplwhg ± 5hjg 2iilfh $dndvk 7rzhu 3xvd 5rdg 1hz 'hokl >3djh @'c ° ¼b1)d g0 !Çbu bm Îb p !c ºb2aî bbdb) éb b a b &ïb bab b"aï ò ã Ä "5!
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« » « » «»¬¼ ¬ ¼ ¬ ¼ 1 1 J Linear Dependence and Rank of a Matrix •Linear Dependence When a linear function of the columns (rows) of a matrix produces a zero vector (one or more columns (rows) can be written as linear function of the other columns (rows))Radiationenergy (transferredor absorbed) J They can also be defined as mass kg ªº «» ¬¼ J radiation fieldquantity massinteraction coefficient kg ªº u «» ¬¼ 23 DOSIMETRIC QUANTITIES FUNDAMENTALS 231 General IntroductionAlgebraic Geometry A Personal View CSE 590B James F Blinn cse590b@cswashingtonedu Mailing List Subscribe at https//mailmancswashingtonedu/
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¬ ¼ Please note that this implies that if E is the standard ordered basis for \ n , then the change‐of‐basis matrix to J is simply J 1¬¼¬ ¼ ¬ ¼ The basic equation for powerflow analysis is derived from the nodal analysis equations for the power system For example, for a 4bus system, where Y ij are the elements of the bus admittance matrix, V i are the bus voltages, and I i are the currents injected at each node The node equation at bus i can be written as ¦ n j I iJ j jj jj B j j j ªº «» ¬¼ Y Y Solution * 2 2 * 22 2 1, 2 * 2 (0) 2 ( ) ( ) 22 1S 1 1 05 ( 5 15)(10 0) 5 1470 Guess 10 0 (this is known as a flat sta rt) 0 1000 0000 3 1 4 n ik k k k i vv V Y V Y V j Vj j V V
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P K R tR= SOªº¬¼(3) where pº¼ J 3 1 JR K p p p= ªº¬¼ 1 2 3 111 T 1 2 3 22 33 ªº ªº¬¼ «» «» «»¬¼ K p p p U V d d d J d 11 R UV T SVD cleanup 1 4 11 = Kp t d Translation and scale recoveryProblem 25, Serway, R A, and Jewett, J W Jr, Principles of Physics, 4th ed (Belmont, CA Brooks/Cole – Thomson Learning, 06) A rod of length L lies along the x axis with its left end at the origin It has a nonuniform charge density OD x, where α is a positive constantM r d r f l i ;
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