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Course: LIB 2800, Fall 2009
School: Wisconsin Milwaukee
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Coursehero >> Wisconsin >> Wisconsin Milwaukee >> LIB 2800

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929530,6753,1573,2783 ;44185,18429,2031,495 a56118,29304,1402,1791 a45748,32939,1459,1791 a56042,43785,1430,1868 about25085,54717,1430,8920 after9225,22207,1430,7319 agnes12771,50481,2117,10064 along41783,29276,2203,8577 an9492,18601,1459,3774 an9492,11389,1488,3698 and15287,29476,1430,5756 and52344,51025,1430,5909 ask35874,11246,1430,5184 asked9264,54803,1516,9073 at50971,43842,1459,3011 basket47959,32223,2203,10369 been24208,43184,2203,7510 before48607,14137,2089,9721 carrying9073,40265,2232,12962 dress49408,25040,2117,7815 ed9569,29447,2089,4422 end27639,6696,1201,4917 for46282,17800,2060,4231 had16240,43213,2232,5947 hand20968,35944,2861,8425 her14410,36029,2203,5032 her19176,54087,2117,5032 her43385,21348,2203,4994 her45519,50310,2203,5070 her34883,53973,2232,4994 home22988,39607,2117,8501 in30918,36001,2089,2935 in9569,36058,2203,2973 in51581,28675,2117,2935 in49294,21406,2117,2973 it25352,14280,2089,2592 long41440,14137,2861,6747 manner26038,32996,2060,12466 met36751,22121,1402,5909 name40792,54717,1402,9187 neatly39191,25698,2089,9645 not35531,14909,1402,4994 occurred30651,18487,1402,13648 of33053,6582,1345,3011 opportunity13457,11332,2060,18337 opportunity13114,18544,2089,18299 peggie41555,10531,2117,10941 peggie36293,6639,1230,8806 peggie20853,25069,2203,10865 proper16012,33110,1402,10331 purchasing32481,43756,2260,17193 question8920,14995,1402,13838 she30918,22121,1402,5108 she45748,36545,1459,5032 she9454,43957,1459,5108 something32443,40150,2203,16126 soon51086,18429,1402,7205 store9340,47591,1430,8272 street9187,25870,1488,9264 the22454,6667,1259,4765 the53144,10617,1946,4841 the5...

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Wisconsin Milwaukee - LIB - 2726
2011664,7401,1594,2762a33726,47599,1394,1688a36374,43984,1423,1841about34187,26162,1366,8901after52374,22490,1366,7328agnes31769,7344,1138,8095almost45429,18960,1366,10244and54983,36866,1338,5793and19875,29806,1394,5678and19530,476
Wisconsin Milwaukee - LIB - 2800
929530:6753:2783:1573;44185:18429:495:2031a56118:29304:1791:1402 45748:32939:1791:1459 56042:43785:1868:1430about25085:54717:8920:1430after9225:22207:7319:1430agnes12771:50481:10064:2117along41783:29276:8577:2203an9492:18601:3774:1459
Wisconsin Milwaukee - LIB - 2726
2011664:7401:2762:1594a33726:47599:1688:1394 36374:43984:1841:1423about34187:26162:8901:1366after52374:22490:7328:1366agnes31769:7344:8095:1138almost45429:18960:10244:1366and54983:36866:5793:1338 19875:29806:5678:1394 19530:47656:5640:14
Wisconsin Milwaukee - LIB - 3017
!23047:43565:400:1168"14229:46001:1242:497 24049:43640:1042:472 46014:26914:1082:472 14269:39614:1082:472 14269:43888:1122:472 13988:27138:1082:472 14149:35389:1082:472 14068:31288:1122:472.16193:7381:400:29819611423:6784:3447:969;44251:2713
Wisconsin Milwaukee - LIB - 1399
194156418:48779:2439:595217205:16597:719:516324062:17649:719:5163737903:50109:1078:6357839059:61226:924:416a37441:52789:796:655active39187:50347:3004:397afford26527:57037:3004:397all30045:54377:1027:377also24524:56998:1823:436amon
Wisconsin Milwaukee - LIB - 2789
127753:57536:690:574 12329:26966:690:593229749:58209:767:574 33611:26966:741:574327753:58882:767:574889822:61852:1074:613a12994:38251:511:435 56633:18274:511:435 42794:58902:920:574 13122:52546:792:653 52847:13146:511:435 13531:26986:895:574
Wisconsin Milwaukee - LIB - 2417
115532:32757:610:414242402:32797:635:434334928:33487:635:414432691:34178:686:414a28395:45327:508:394across20514:41953:3228:414activities22649:42584:4728:631added21251:48702:3253:394after25598:45327:2414:394 13473:54523:2465:394all-c
Wisconsin Milwaukee - LIB - 4512
20958361:61737:2118:731a30763:41461:663:534 4723:31550:816:672 32295:50857:919:791 24559:31649:485:435 10543:58690:510:415about45009:54596:3497:514activities32678:43854:5386:435ahy30993:55862:2450:672among56293:42470:4161:731an41281:5817
Wisconsin Milwaukee - LIB - 2615
-22141:28723:254:59112227:58137:711:572 31318:30183:279:51212810092:62161:1398:552229488:58117:762:532 56358:30045:355:532341893:58748:813:552:56942:30222:101:335 31852:30360:101:335a15608:28565:432:394 57298:53264:432:394add58061:4618
Wisconsin Milwaukee - LIB - 2187
'54675:58303:152:177144501:55446:685:532 22174:28235:710:57111210300:62283:1395:571214690:29004:735:571 58405:56096:761:571337093:29004:735:571 53584:58500:735:571417988:29772:761:571a50134:15566:482:413 60714:15467:507:433 10250:46639:4
Wisconsin Milwaukee - LIB - 2394
1188764:62630:1208:431238006:35426:654:41222543950:42826:1813:608a51153:43003:503:392 46947:49735:528:372 48383:56408:478:372 48735:48047:528:372acquainted43648:48891:5742:372affairs46443:45534:3551:372all52312:46359:1158:372also43849:
Wisconsin Milwaukee - CHEM - 100
Chapter 4Lesson 22 Chapter 4Empirical Formulas and Molecular Formulas Goals To understand empirical formulas versus molecular formulas To make to the connection between empirical and molecular formulas on the symbolic versus particulate level
Rose-Hulman - ECE - 454
* Profile: "SCHEMATIC1-Trans" [ C:\WEBSITE\ROSE_CLASSES\ECE454\Homework\Fall 2004\HW7\ece454 hw7 proble. Date/Time run: 11/07/04 15:12:13 Temperature: 27.0 (A) ece454 hw7 problem 1-SCHEMATIC1-Trans.dat (active) 5.0V0V-5.0V V(Vin) 5.0V2.5VSEL>
Rose-Hulman - ECE - 454
ECE 454 Homework #7Due 11/4/04Problem 1: a) Under what conditions is the circuit below an absolute value circuit (full-wave rectifier with no diode drop losses). b) Prove your result using PSpice and a 1 kHz sine wave input.R 4+R 1R 2R 3
Rose-Hulman - ECE - 454
54321VCCD+R4-V4 DC = 15D+10k+Vin R1 R2 R3 R50-V5 DC = 15+10kTransient Analysis+20k D2 D1N914 D1+10k+10k-VEE+V3 AMPLITUDE = 5 FREQUENCY = 1k-VEE 11CU3A LM324C0211D1N914 U4A -VEE LM324 V1
Rose-Hulman - ECE - 454
vti_encoding:SR|utf8-nl vti_timelastmodified:TR|26 Oct 2004 21:09:56 -0000 vti_extenderversion:SR|6.0.2.6353 vti_author:SR|HERNITER-2\herniter vti_modifiedby:SR|HERNITER-2\herniter vti_timecreated:TR|26 Oct 2004 21:09:56 -0000 vti_cacheddtm:TX|26 Oct
Alaska Anch - CE - 434
Panel Design Specication2008 APA THE ENGINEERED WOOD ASSOCIATION ALL RIGHTS RESERVED. ANY COPYING, MODIFICATION, DISTRIBUTION OR OTHER USE OF THIS PUBLICATION OTHER THAN AS EXPRESSLY AUTHORIZED BY APA IS PROHIBITED BY THE U.S. COPYRIGHT LAWS.W
Alaska Anch - CE - 434
CE 434 - Fall 2000 - Project Evaluation Form Design Project Phase IDesign Team: Organization: (20%) Calculations easy to follow All members considered Beams Columns Walls Diagrams/Framing plans showing results Summary of Results Possible 5 3 3 3 3 3
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th Edition, Chapter 14 Problem 14.1 Given: The connection in Figure 14.A (2) 6x16 beams 6x6 post No.1 DF-LBolt Diameter CM DL on column SL on column Sesimic tension in beams0.875 1 5 15 8in k k kWanted: a. Desi
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th Edition, Chapter 14 Problem 14.5 Given: The beam-girder connection shown in Figure 14.C Girder is 5.3/4" x 25.1/2" 16F-V3 DF GLB Beam is 4x14 No.1 DF-L Bolts are 3/4" diameter A307 Beam reaction 2000 lbs (DL + RLL)
Alaska Anch - CE - 434
CE434 - Fall 2000Exam #2 solution Seismic Load - roof Seismic Load - 2nd flr Problem #1 Seismic Load Diaphragm span Diaphragm depth Floor area Seismic Load/ft^2 seismic, W4 Controlling W4 max shear reaction max unit shear Problem #2 Note: This shear
Alaska Anch - CE - 434
CE434 - Fall 2000 Exam #1 SolutionProblem #1: Cq qs I Cq*qs*I = width Level Roof 4th flr 3rd flr 2nd flr Grnd 1.3 25.6 psf 1 33.28 psf 60 ft Elev (ft) 40 35 30 25 20 15 10 5 0 Ce p (psf) 52.58 49.92 46.26 46.26 w (plf) 3155 30.8 1.5 1.39 1.39 2995 2
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th Edition Problem 10.1 Given: Single story wood framed building in Fig. 10.A. The critical lateral wind force in long. direction = 15/32 STR I sheathing with 8d common nails. a. The unit shear in the shear panel along
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th Edition Problem 10.10 Given: The two story wood framed shear wall in Fig. 10C with wind forces as shown. 15/32 C-C STR I sheathing w/ 8d common nails. Rr = 2.1 k R2 = 2.6 kWanted/Solution:a. The required nailing
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th Edition Problem 10.12 Given: See TextWanted/Solution:a. The req'd nailing for each panel2nd Story: Total Load Length unit shear Use: 4000 lb 12 ft 333.3333 plfc. Maximum compressive chord forces at end of fir
Alaska Anch - CE - 434
Alaska Anch - CE - 434
Alaska Anch - CE - 434
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2002 Breyer - 3rd ed Problem 6.32 Given: the beam-column connection in Fig 6.G. Ct = 1Determine the maximum allowable beam reaction for the following conditions a. 4x12 DF-L, MC < 19% Load condition: DL + SL A = B = Beam
Alaska Anch - CE - 434
Timber DesignCE 434 - Fall 2002 Breyer - 4th ed Problem 6.35 Given the commercial roof framing in fig. 6.Ia.Check subpurlins using No. 1 & BTR DF-L. Are the AITC recommended deflection limits satisfied.Computation of Allowable Stresses for Saw
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2002 Breyer, 4th ed, Chapter 7 Problem 7.13 Material Data: No. 1 DF-L Tabulated CD = CM = Ct = CF = Factor Modified KcE = c = Geometry: dx dy F'c Ex Ey 1000.00 1600000 1600000 psi 1.15 -1.00 1.00 1.00 1.00 1.00 1.00 1.00 -
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2002 Breyer, 4th ed, Chapter 7 Problem 7.14 Geometry: dx dy lu = = = 5.50 in 7.50 in 11.00 ft DF-L No1 Area = NDS Table Size Category 41.25 in^2 4D P&T Lookup Tables Table 4A Fc 1500 1550 1350 1700 E 1700000 1800000 160000
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2002 Breyer, 4th ed, Chapter 7 Problem 7.15 Loads: D L Lr Ptl Material Data: 2 DF 20 90 40 150.00 k k k k=Table 5B F'c Ex Ey Tabulated 1950.00 1600000 1600000 psi CD = 1.25 -CM = 1.00 1.00 1.00 Ct = 1.00 1.00 1.00 CF =
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2002 Breyer, 4th ed, Chapter 7 Problem 7.15 Loads: D L Lr Ptl 20 90 40 150.00 k k k kTables 5A & 5B E 1.6 1.5 1.6 1.6 1.7 1.7 1.7 1.7 1.7 1.6 1.9 1.9 2 Fc4+ 1550 1550 1550 1600 1600 1550 1650 1650 1650 1950 2300 2100 240
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2002 Breyer, 4th ed, Chapter 7 Problem 7.18 Material Data: No. 1 DF-L P&T Tabulated CD = CM = Ct = CF = Factor Modified KcE = c = F'c 1000.00 1.15 1.00 1.00 1.00 1.15 1150 0.30 0.80 F'c 925.00 1.15 1.00 1.00 1.00 1.15 1063
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2002 Breyer, 4th ed, Chapter 7 Problem 7.23 Pdl Psl wind 5 k 15 k 200 plf Height 16 ftComputation of Allowable Stresses for Sawn Lumber Based on NDS Supplement, 2001 edition Nominal Actual Width 6.00 5.50 in Depth 10.00
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2002 Breyer, 4th ed, Chapter 7 Problem 7.3 Note: Use a truss spacing of 4'-0" O.C. when computing trib. widthDetermine the force in the bottom chord DL SL Rs Reduction SL (reduce Total Load Trib width Load Span Pnt Ld A
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th ed, Chapter 9 Problem 9.11 Given: Drag Strut from problem 9.10 Wdl 100 plf Wrll 160 plf Size = 4x14 No. 1 DF Larch Cm = 1 and Ct = 1 Max Force 5405 lbAnalyze for combined stresses Computation of Allowable Stresses
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th ed, Chapter 9 Problem 9.2 Given: Single story commerical building in Fig. 9.A. L1 80 ft L2 35 ft Plywood is 15/32 C-DX STR I. Seismic Strength (plf) 320 570 Wind Service (plf) 320 320Seismic SeismicwT wLServic
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th ed, Chapter 9 Problem 9.4 Given: Single story warehouse building in Fig. 9.B. L1 120 ft L2 70 ft Plywood is 3/8 C-DX STR I. Seismic Ultimate (plf) 330 570 Wind Service Service (plf) (plf) 236 Not Given 407 Not Given
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th ed, Chapter 9 Problem 9.9 Given: The one-story building in Fig. 9.C Roof DL 12 psf Chord As Wall DL 150 pcf C-C Plywood Wind 20 psf Wall thickn Seismic Coefficient 0.2 Wall Trib Height 10 ft RLL according to UBC Tab
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th Edition, Chapter 13 Problem 13.28 Given: The wood shear wall in Fig 13.I Wood is DF-L Cm = 1 In plane shear load4000 lbWanted:a. Req'd number of 3/4" A.B. in 2x sole plate to handle in-plane shear load.Yield
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th Edition, Chapter 13 Problem 13.29 Given: The ledger connection in Fig. 13.J Wood is DF-L 1/2" dia x 6" lag bolts @ 32" O.C. Patio DL 6 psf roof LL 10 psf Span 16 ft a. Is connection adequate if lumber installed dry
Alaska Anch - CE - 434
CE 434 - Timber Design Breyer, 4th Edition, Chapter 13 Problem 13.31 Given: Building in Fig. 13.K Wood is DF-L #1 5/8" dia x 6" lag bolts wT wL CM210 psf 275 psf 1Wanted:The number of lag bolts for connection shown. Determine Drag Strut force v
Alaska Anch - CE - 434
CE 434 - Assignment #1 - Fall 2000Problem 2.1 Given: The house framing depicted in Figure 2.Aa.Determine the roof DL on a horizontal plane Slope 4 :12 horiz load (psf) 2.6 1.2 1.7 2.9 2.6 11.0 2.0 13.0ItemSlope OnSlope Correction (psf) Asph
Alaska Anch - CE - 434
CE 434 - Assignment #1 - Fall 2000 Problem 2.12 Given: A column supports only loads from the second floor of an office building. The tributary areas and dead load are: Trib Area DL Solution: a. Determine the basic floor LL: Basic floor Live Load UBC
Alaska Anch - CE - 434
CE 434 - Fall 2000 Assignment #2 Problem 2.21 See text for problem statement Determine the average elevation of each element Element Roof 2nd Level Walls 1st Level Walls Min 22 11 0 Max 28.25 22 11 Ave 25.125 16.5 5.5WIND DESIGN LOADS - 1991 UNIFOR
Alaska Anch - CE - 434
CE 434 - Assignment #1 - Fall 2000 Problem 2.4 Given: The roof framing plan of the industrial building shown in Figure 2.D. Roof Slope is 1/4 in per foot. General construciton: Roofing 5 ply felt Sheathing 15/32-in plywood Subpurlin 2x4 @ 24" O.C. Pu
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 1993 Breyer, Design of Wood Structures, 3rd Ed. Prob. 3.1 The purpose of this problem is to compare the design values of shear and moment for a girder with different assumed load configurations. Given: The roof framing pla
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2000 Breyer, Design of Wood Structures, 3rd Ed. Problem 3.11 The plan and section of the building in Fig 3.A. Roof dead load is given below. The walls are 7.5" thick concrete. The building is a bearing wall system located
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 1993 Breyer, Design of Wood Structures, 3rd Ed. Problem 3.4 The plan and section of the building in Fig 3.B. The basic wind speed is 80 mph, and exposure B applies. The building is enclosed and has a standard occupancy cla
Alaska Anch - CE - 434
CE 434 - Fall 2000Breyer, Problem 4.14 Solution There are only three unique sets of tabulated values From Table 4D Fb Ft 1350 675 1200 825Fv 85 85f,h a,b,gFcp 625 625Fc E 925 1600000 WWPA - Beams & Stringers 1000 1600000 WWPA - Posts & Timb
Alaska Anch - CE - 434
Breyer, 4th ed, Problem 4.27 Given: Column Loads RDL RLL FDL FLL Wind Snow Seismic 10 2 8 10 6 0 0 k k k k k k k (D from roof) (Lr) (D from floors) (L) (W) (S) (E) Adjusted Load 20.00 24.00 24.35 28.00 15.00 10.12 19.69 18.75Solution Load Load Case
Alaska Anch - CE - 434
Breyer, 4th ed, Problem 4.30 All materials are No. 2 Hem Fir w/ bending about strong axis Part a RoofJoists are 2x10 @ 16" O.C. supporting roof DL and Snow Width Depth Controlling load type In Use Moisture Content Meets Repetative Member requi Form F
Alaska Anch - CE - 434
Breyer, 4th ed, Problem 5.11 Given: Width Depth Span 5.125 28.5 32 in in ft Material: 24F-V4 DF/DF Load Comb. = DL+SL, uniformly distributed Ignore LTBPart a: Area Sx Ix 146.0625 693.7969 9886.605 in^2 in^3 in^4Part b:Width Depth Span Controlli
Alaska Anch - CE - 434
Breyer, 3ed, Problem 5.17a. b. c. d. e. f.Dead Load Snow Wind Flr Live Load Seismic Roof Live Load0.9 1.15 1.6 1 1.6 1.25
Alaska Anch - CE - 434
CE 434 - Fall 2000 Problem 6.18 Given: The beam in Figure 6.CBreyer - 4th edLoad Shortest Duration Load Spans: L1 L2 Member Size: 4x12 b d Species & Stress Grade Unbraced Length, lu Moisture Content, MC Deflection Limits: Free End Between Support
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2000 Problem 6.25 Given:Breyer - 4th edThe roof rafters depicted in figure 6.D. Spacing 24 in. O.C. Span 14 ft Roof DL 15 psf Snow Load 50 psf Disregard deflection Lateral Stability is not a problem Cm 1 Ct 1 Lumber: N
Alaska Anch - CE - 434
Timber Design CE 434 - Fall 2000 Problem 6.26 GivenBreyer - 4th edSpan 25 ft d 13.5 in b 5.5 in Load is point load at midspan Support at ends onlyFb E1600 psi 1600000 psiCalculated CL KBE E'y RB FBE CL 0.438 1600000 14.19492 3477.987 0.9618
Alaska Anch - CE - 434
Yield Limit Analysis of a Bolt ( D >= .25")2005 NDS Input Data: Diameter of Bolt, D Reduced Diameter, Dr (lag screws) Member is a Lag Screw? Side Member Number of side members Is the side member steel? Thickness of Wood Side Member Thickness of Stee
Alaska Anch - CE - 434
Design of Wood Structures, 6th ed Problem 2.1 Givenlast modified: 9 Jan 2008 by: TBQThe house framing shown in Fig. 2.AWanted:A. Roof dead load D in psf on the horizontal plane. B. Wall D in psf of wall surface area. C. Wall D in plf of wall.