Widener - CE - 342

Measuring Forces in Truss Members Using Strain GagesCE 206 (C.E. Lab #4) Conducted: February 25th, 2010 Submitted to: Prof. Brown, March 11th, 2010Tom Wienckowski Lab Group C Members: Andrew Dougan Kyle Fischer Matt SmithTable of ContentsAbstract Intr

Widener - CE - 342

Abstract3/5 Introduction/Background(Objectives,contestrules,7.5lbscement)3/5 Methods&Procedures Designapproach(okinbackground)5/5 BeamGeometry(okinresults)3/5 MixDesign(okinresults)5/5 BuildingFormwork5/5 PouringConcrete5/5 Testingcylinders&beams5/5 Predi

Widener - CE - 342

BeamType ModulusofRupture(psi) 2x4onendHemFir 470.8 2x6onendHemFir 419.11 2x4onendDouglasFir 933.77 2x4FlatHemFir 440.81 2x6FlatHemFir 507.11 BeamType MaximumShearStress(psi) 2x4onendHemFir 525 2x6onendHemFir 825 2x4onendDouglasFir 630 2x4FlatHemFir 525 2

Widener - CE - 342

Midspan FreeendNoLoad0.8 0.9Load16.994lbs 0.64 0.44 Load16.994lbs 0.73 0.44 EE MEDeflections 0.466" 0.160" Deflections 0.466" 0.160" EE MEMidspan FreeendNoLoad0.8 0.6 MM EMFringe1 Fringe2 Fringe3exp3558.33 7116.67 10674.99Stressesvs.Distance

Widener - CE - 342

Measuring Tensile Properties of Metal SpecimensCE 206 (C.E. Lab #3) Conducted: January 28th, 2010 Submitted to: Prof. Brown, February 4th, 2010Tom Wienckowski Lab Group 3 Members: Andrew Dougan Kyle FischerTable of ContentsAbstract Introduction Method

Widener - CE - 342

Methods and ProceduresClick to edit Master subtitle style6/9/10Concrete Mix Designed to Fill:Two 40 x 5.5 x 2.5 inch beams Three four inch by 8 inch cylinders One slump test with a volume of a 6 inch by 12 inch cylinder6/9/10Batch Weights for 1 Cubi

Widener - CE - 342

Widener - CE - 342

Group A B C DAverage Compressive Strength psi 1235 1750 900 1330Range lb1000 8760 830 2450CompressiveStrength W/CRatio 1235 0.57 1750 0.41 900 0.63 1330 0.48900 1235 1330 17500.63 0.57 0.48 0.41Water/CementRatiovs.CompressiveStrength0.65 0.6 0.55

Widener - CE - 342

Sheet1 Strain (in/in),Stress (psi),Force (lbf), 0.0000227,106.4428259,20.9,177.7777778 0.00001045,152.2794496,29.9, 0.00001715,170.1048032,33.4, 0.0000219,186.4022693,36.6, 0.00001945,205.2462146,40.3, 0.0000398,259.2315713,50.9, 0.0000293,323.4028444,63.

Widener - CE - 342

Force (lbf) Position (in) 5.42 0 8.13 0 8.15 0 7.95 0 8.34 0 8.64 0 9.34 0 10.01 0 10.73 0.01 11.63 0.01 12.54 0.01 13.62 0.01 14.81 0.01 15.78 0.01 17.09 0.01 18.11 0.01 19.4 0.01 20.7 0.01 22 0.02 23.2 0.02 24.5 0.02 26.1 0.02 27.2 0.02 28.7 0.02 30 0.0

Widener - CE - 342

Methods and Procedures A concrete mix had to be designed and prepared to be able to fill two concrete beams, 40 inches by 5.5 inches by 2.5 inches, three 4 inch by 8 inch cylinders, and one slump test with a volume of a 6 inch by 12 inch cylinder. The con

Widener - CE - 342

Concrete Mix Designs and Compression TestsCE 206 (C.E. Lab #2) Conducted: January 21st, 2010, and February 16th, 2010 Submitted to: Prof. Brown, February 25th, 2010Tom Wienckowski Lab Group C Members: Andrew Dougan Kyle Fischer Matt SmithTable of Conte

Widener - CE - 342

Widener - CE - 342

Tom Wienckowski CE 206 Civil Engineering Lab Assignment #1 Technical Writing: Review a Technical Journal Article Select a technical article in an ASCE journal (Materials or Structures) that involves measurements taken by the authors. Review the article wi

Widener - CE - 342

Abstract 4 /5Introduction 3/5Background on calculations to find safe load on truss (yield, pin stress, buckling) 2/5Methods & Procedures- Description of trusses with figures showing geometry, joint labels (A,B,C) & strain gages 0/5- Experimental proc

Widener - CE - 342

Lab work 10/10Abstract 5/6Introduction 4/6Methods & Procedures 10/12Results & Discussion 12/15Conclusion 4/6Grammar & Mechanics 9/10Presentation of Data & Results 7/10Calculations 22/25Total = 83Hot Mix Asphalt Superpave Volumetric Design andCo

Widener - CE - 342

CE 206 Project Presentations Evaluation EVALUATOR: Prof. Brown PROJECT: MKAT 13.5 PTS = B-PRESENTER: Tom WienckowskiBeginning = 1 Developing = 2 Competent = 3 Accomplished = 4 Exemplary = 5Content = 3 Organization = 3 Manner of Delivery = 2 Visual Aids

Widener - CE - 342

Beam analysis Using Photoelastic MethodsCE 206 (C.E. Lab #7) Conducted: April 15th, 2010 Submitted to: Prof. Brown, April 22nd, 2010Tom WienckowskiTable of ContentsAbstract Introduction Methods and Procedures Results and Discussion Conclusions Referen

Widener - CE - 342

Tom Wienckowski CE 206 Civil Engineering Lab Assignment #1 Technical Writing: Review a Technical Journal Article Select a technical article in an ASCE journal (Materials or Structures) that involves measurements taken by the authors. Review the article wi

Widener - CE - 342

Wooden Beam TestsCE 206 (C.E. Lab #6) Conducted: April 1st, 2010 Submitted to: Prof. Brown, April 8th, 2010Tom WienckowskiTable of ContentsAbstract Introduction Methods and Procedures Results and Discussion Conclusions References AppendixPage3 4 5 6

Widener - CJ - 305

Tom Wienckowski Prof. Pisani CJ 305E A Exam #1 1.) Define and exam the concept of Evidence. You are also to make sure you define each type and give examples of each. Your answers should also include defining the different levels of proof as discussed in c

UAA - PHYS - 112

38. We label these wires 1 through 5, left to right, and use Eq. 29-13. Then, (a) The magnetic force on wire 1 is7 1 1 1 250i 2l 25 ( 410 T m A ) ( 3.00A ) (10.0m) + j= j F1 = + j= 2 d 2d 3d 4d 24d 24 ( 8.00 102 m ) 0i 2 l 12= (4.69 104 N) j. (b) Simi

UAA - PHYS - 112

1. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must be directed due south. Since B = 0i 2 r , i=2 rB0=2 0.080 m 39 10 6 T 4 10 T m A7bgch = 16 A.(b) The current must be from west to east to produce a field whi

UAA - PHY112 - 1356

4. Eq. 29-1 is maximized (with respect to angle) by setting = 90 ( = /2 rad). Its value in this case is i ds dBmax = 0 2 . 4 R From Fig. 29-36(b), we have Bmax = 60 1012 T. We can relate this Bmax to our dBmax by setting ds equal to 1 106 m and R = 0.025

UAA - PHY112 - 1356

5. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with P do not contribute to the field at that point. Using Eq. 29-9 (with = ) and the right-hand rule, we find that the cu

UAA - PHY112 - 1356

6. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in segments AH and JD do not contribute to the field at point C. Using Eq. 29-9 (with = ) and the right-hand rule, we find that the current in the semicircul

UAA - PHY112 - 1356

7. (a) The currents must be opposite or antiparallel, so that the resulting fields are in the same direction in the region between the wires. If the currents are parallel, then the two fields are in opposite directions in the region between the wires. Sin

UAA - PHY112 - 1356

8. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with C do not contribute to the field at that point. Eq. 29-9 (with = ) indicates that the current in the semicircular arc

UAA - PHY112 - 1356

9. (a) BP1 = 0i1 / 2 r1 where i1 = 6.5 A and r1 = d1 + d2 = 0.75 cm + 1.5 cm = 2.25 cm, and BP2 = 0i2 / 2 r2 where r2 = d2 = 1.5 cm. From BP1 = BP2 we getr 1.5 cm i2 = i1 2 = ( 6.5 A ) = 4.3A. 2.25 cm r1 (b) Using the right-hand rule, we see that the cu

UAA - PHY112 - 1356

10. (a) Since they carry current in the same direction, then (by the right-hand rule) the only region in which their fields might cancel is between them. Thus, if the point at which we are evaluating their field is r away from the wire carrying current i

UAA - PHY112 - 1356

11. (a) We find the field by superposing the results of two semi-infinite wires (Eq. 29-7) and a semicircular arc (Eq. 29-9 with = rad). The direction of B is out of the page, as can be checked by referring to Fig. 29-6(c). The magnitude of B at point a i

UAA - PHY112 - 1356

12. With the usual x and y coordinates used in Fig. 29-43, then the vector r pointing i from a current element to P is r = s + R . Since ds = ds , then | ds r | = Rds. i j Therefore, with r = s 2 + R 2 , Eq. 29-3 gives dB =0 iR ds . 2 4 ( s + R 2 )3/ 2(

UAA - PHY112 - 1356

13. We assume the current flows in the +x direction and the particle is at some distance d in the +y direction (away from the wire). Then, the magnetic field at the location of a proton with charge q is B = ( 0i / 2 d ) k. Thus, F = qv B = 0iq2 dev kj.

UAA - PHY112 - 1356

14. The fact that By = 0 at x = 10 cm implies the currents are in opposite directions. ThusBy =0i2 0i2 4 1 = 2 ( L + x) 2 x 2 L + x x 0i1using Eq. 29-4 and the fact that i1 = 4i2 . To get the maximum, we take the derivative with respect to x and set e

UAA - PHY112 - 1356

15. Each of the semi-infinite straight wires contributes 0i 4 R (Eq. 29-7) to the field at the center of the circle (both contributions pointing out of the page). The current in the arc contributes a term given by Eq. 29-9 pointing into the page, and this

UAA - PHY112 - 1356

16. Initially, we have Bnet,y = 0, and Bnet,x = B2 + B4 = 2(o i /2d) using Eq. 29-4, where d = 0.15 m . To obtain the 30 condition described in the problem, we must have Bnet, y = Bnet, x tan(30) i B1 B3 = 2 0 tan(30) 2 d where B3 = o i /2d and B1 = 0i /

UAA - PHY112 - 1356

17. Our x axis is along the wire with the origin at the midpoint. The current flows in the positive x direction. All segments of the wire produce magnetic fields at P1 that are out of the page. According to the Biot-Savart law, the magnitude of the field

UAA - PHY112 - 1356

18. We consider Eq. 29-6 but with a finite upper limit (L/2 instead of ). This leads toB=0iL/22R ( L / 2) 2 + R 2.In terms of this expression, the problem asks us to see how large L must be (compared with R) such that the infinite wire expression B

UAA - PHY112 - 1356

19. Each wire produces a field with magnitude given by B = 0i/2r, where r is the distance from the corner of the square to the center. According to the Pythagorean theorem, the diagonal of the square has length 2a , so r = a 2 and B = 0i 2 a . The fields

UAA - PHY112 - 1356

20. Using the law of cosines and the requirement that B = 100 nT, we have2 B12 + B2 B 2 = cos = 144 , 2 B1 B2 1where Eq. 29-10 has been used to determine B1 (168 nT) and B2 (151 nT).

UAA - PHY112 - 1356

21. Our x axis is along the wire with the origin at the right endpoint, and the current is in the positive x direction. All segments of the wire produce magnetic fields at P2 that are out of the page. According to the Biot-Savart law, the magnitude of the

UAA - PHY112 - 1356

22. Using the Pythagorean theorem, we have i i B = B + B = 0 1 + 0 2 4 R 2 R 2 2 2 1 2 2 2which, when thought of as the equation for a line in a B2 versus i22 graph, allows us to identify the first term as the y-intercept (1 1010) and the part of the se

UAA - PHY112 - 1356

23. (a) As illustrated in Sample Problem 29-1, the radial segments do not contribute to BP and the arc-segments contribute according to Eq. 29-9 (with angle in radians). If k designates the direction out of the page then B=0 ( 0.40 A )( rad ) 0 ( 0.80 A

UAA - PHY112 - 1356

24. In the one case we have Bsmall + Bbig = 47.25 T, and the other case gives Bsmall Bbig = 15.75 T (cautionary note about our notation: Bsmall refers to the field at the center of the small-radius arc, which is actually a bigger field than Bbig!). Dividi

UAA - PHY112 - 1356

25. We use Eq. 29-4 to relate the magnitudes of the magnetic fields B1 and B2 to the currents (i1 and i2, respectively) in the two long wires. The angle of their net field is = tan1(B2 /B1) = tan1(i2 /i1) = 53.13.The accomplish the net field rotation de

UAA - PHY112 - 1356

26. Letting out of the page in Fig. 29-55(a) be the positive direction, the net field is B=0i1 0i2 4 R 2 ( R / 2)from Eqs. 29-9 and 29-4. Referring to Fig. 29-55, we see that B = 0 when i2 = 0.5 A, so (solving the above expression with B set equal to ze

UAA - PHY112 - 1356

27. The contribution to Bnet from the first wire is (using Eq. 29-4) B1 =0i1 (4 107 T m/A)(30 A) k= k = (3.0 106 T)k. 2 r1 2 (2.0 m)The distance from the second wire to the point where we are evaluating Bnet is r2 = 4 m 2 m = 2 m. Thus, B2 =0i2 (4 107

UAA - PHY112 - 1356

28. (a) The contribution to BC from the (infinite) straight segment of the wire is BC1 = 0i2 R.The contribution from the circular loop is BC 2 = 0i2R. Thus,7 3 1 ( 4 10 T m A ) ( 5.78 10 A ) 1 7 BC = BC1 + BC 2 = 1 + = 1 + = 2.5310 T. 2R 2 ( 0.018

UAA - PHY112 - 1356

29. Using the right-hand rule (and symmetry), we see that B net points along what we will refer to as the y axis (passing through P), consisting of two equal magnetic field ycomponents. Using Eq. 29-17, i | Bnet | = 2 0 sin 2 r where i = 4.00 A, r = r = d

UAA - PHY112 - 1356

30. Initially we have Bi =0i 0i + 4 R 4 rusing Eq. 29-9. In the final situation we use Pythagorean theorem and write i i 2 B 2 = Bz2 + By = 0 + 0 . f 4 R 4 r 2 2If we square Bi and divide by Bf , we obtain Bi B f [(1/ R) + (1/ r )]2 . = (1/ R) 2 + (1

UAA - PHY112 - 1356

31. Consider a section of the ribbon of thickness dx located a distance x away from point P. The current it carries is di = i dx/w, and its contribution to BP isdBP = Thus, BP = dBP = 2w d 0di2 x= 0idx2 xw.0id +wdx 0i w (4107 T m A)(4.61106 A) 0

UAA - PHY112 - 1356

32. By the right-hand rule (which is built-into Eq. 29-3) the field caused by wire 1s current, evaluated at the coordinate origin, is along the +y axis. Its magnitude B1 is given by Eq. 29-4. The field caused by wire 2s current will generally have both an

UAA - PHY112 - 1356

33. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with P do not contribute to the field at that point. We use the result of Problem 29-21 to evaluate the contributions to

UAA - PHY112 - 1356

34. We note that when there is no y-component of magnetic field from wire 1 (which, by the right-hand rule, relates to when wire 1 is at 90 = /2 rad), the total y-component of magnetic field is zero (see Fig. 29-62(c). This means wire #2 is either at +/2

UAA - PHY112 - 1356

35. Eq. 29-13 gives the magnitude of the force between the wires, and finding the xcomponent of it amounts to multiplying that magnitude by cos = the x-component of the force per unit length isFx 0i1i2 d 2 (4107 T m/A)(4.00 103 A)(6.80 103 A)(0.050 m) =

UAA - PHY112 - 1356

36. Using Eq. 29-13, the force on, say, wire 1 (the wire at the upper left of the figure) is along the diagonal (pointing towards wire 3 which is at the lower right). Only the forces (or their components) along the diagonal direction contribute. With = 45

UAA - PHY112 - 1356

37. Using a magnifying glass, we see that all but i2 are directed into the page. Wire 3 is therefore attracted to all but wire 2. Letting d = 0.500 m, we find the net force (per meter length) using Eq. 29-13, with positive indicated a rightward force:| F

LSU - ENGL - 1005

Summer Syllabus: ENGL 1005Textbook: Writing in the Disciplines: A Reader and Rhetoric for Academic Writers, 6th ed. by M. L. Kennedy and W. J. Kennedy. This syllabus is a tentative overview of the summer term. Print it out and keep it with your course ma

University of Phoenix - BUS - 210

Finagle A Bagel Finagle A Bagels marketing mix. Product Variable: Goods - Finagle A Bagels main product is bagel. We do bagels is their motto and they focus on the brand, which differentiates them from competitors. Any items in the menu always serve with

University of Texas - ECON - 304L

Statistics for Business and Economics6th EditionChapter 1 Why Study Statistics?Statistics for Business and Economics, 6e 2007 Pearson Education, Inc.Chap 1-1Chapter GoalsAfter completing this chapter, you should be able to:Explain how decisions are

University of Texas - ECO 304L - 33530

Statistics for Business and Economics6th EditionChapter 2 Describing Data: GraphicalStatistics for Business and Economics, 6e 2007 Pearson Education, Inc.Chap 2-1Chapter GoalsAfter completing this chapter, you should be able to: Identify types o