SEPTEMBER Conquering the FE & PE exams Examples & Applications

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1 SEPTEMBER 2015 DR. Z s ORNER onquering the FE & PE exams Examples & Applications Topics covered in this month s column: FE IVIL Exam Topics & Number of Questions Four Important Things That You Need Math, Angle onversions, From DMS to Decimal and Vice Versa Math, Operations With omplex Numbers Math, Binary and Decimal onversions Math, Definite Integrals Statics and Mechanics of Materials entroids & Moments of Inertia alculations Determinate Beams and Frames Technology Usage (alculator)

2 FUNDAMENTALS OF ENGINEERING IVIL EXAM TOPIS omputer-based Test (BT) Total Number of Questions: 110 Time: 6 hours The new ivil FE omputer Based Test (BT) consists of 110 multiple choice questions (Each problem only one question) the examinee will have 6 hours to complete the test. Mathematics (Approx. 9 questions*) Probability and Statistics (5 questions) omputational Tools (5 questions) Ethics and Professional Practice (5 questions) Engineering Economics (5 questions) Statics (9 questions) Dynamics (5 questions) Mechanics of Materials (9 questions) ivil Engineering Materials (5 questions) Fluid Mechanics (5 questions) Hydraulics and Hydrologic Systems (10 questions) Structural Analysis ( 8 questions) Structural Design ( 8 questions) Geotechnical Engineering ( 12 questions) Transportation Engineering ( 10 questions) Environmental Engineering ( 8 questions) * Here the number of questions are the average values taken from the NEES Reference Handbook (Version 9.1 / omputer-based Test) ASE 111 FALL 2014

3 WHAT DO YOU NEED? NEES REFERENE MANUAL ALULATOR ONLY FOR FE & PE PROBLEMS ONLY FOR FE & PE MATERIAL D RING BINDER ASE 620 FALL 2014 ANVAS BAG

4 ANGLE ONVERSIONS FROM(DMS)TODEIMALDEGREES MANUALALULATIONS Degrees, minutes and seconds:(dº m' s") One degree is equal to 60 minutes and 3600 seconds: 1º = 60' = 3600" One minute is equal to 1/60 degrees: 1' = (1/60)º = º One second is equal to 1/3600 degrees: 1" = (1/3600)º = º For an angle with d (integer) degrees, m minutes, and s seconds: dº m' s" The formula to convert DMS to decimal degrees: Angle = d + m / 60 + s / 3600 Example onvert 5 degrees 25 minutes and 30 seconds angle to decimal degrees: Angle = 5º 25' 30" The decimal degrees dd is equal to: Angle = d + m/60 + s/3600 = 5º + 25'/ "/3600 = 5.425º ANGLE-315 FALL2015

5 Problem: ANGLE ONVERSIONS FROM (DMS) TO DEIMAL DEGREES USING ALULATOR onvert the angle given as 5 degrees, 25 minutes, and 30 seconds to decimal degrees using your calculator. Solution: Angle = 5º, 25', 30" ANGLE 315 FALL 2015 KEY SEQUENE

6 ANGLE ONVERSIONS FROM (DMS) TO DEIMAL DEGREES USING ALULATOR Important Keys Angle = 15º, 35', 45" KEY SEQUENE ANGLE FALL 2015

7 ANGLE ONVERSIONS FROM (DMS) TO DEIMAL DEGREES SUPPLEMENTAL PROBLEMS (1) (2) ANGLE 115 FALL 2015

8 ANGLE ONVERSIONS FROM DEIMAL TO DEGREES, MINUTES, SEONDS SUPPLEMENTAL PROBLEMS (1) (2) ANGLE FALL 2015

9 ANGLE ONVERSIONS FROM DEIMAL TO DEGREES, MINUTES, SEONDS USING ALULATOR KEY SEQUENE ANGLE FALL 2015

ANGLE 115 3 FALL 2015 ANGLE ONVERSIONS FROM

10 ANGLE FALL 2015 ANGLE ONVERSIONS FROM DEIMAL TO DEGREES, MINUTES, SEONDS USING ALULATOR

11 Binary Number System: NUMBER SYSTEMS BINARY & DEIMAL NEES Reference Handbook, Page: 213 In digital computers, binary number system (the base 2) is used. onversions from BINARY to DEIMAL or from DEIMAL to BINARY can easily be done using the calculator. Binary (base 2), decimal (base 10). Problem: Find the binary equivalent of decimal 25? Here, decimal is base 10. Turn on your calculator 1) Press MODE 2) Press 4 3) Enter 25 and press = 4) Make sure to see 25 under Dec on the screen 5) Press SHIFT then log 6) Answer: Problem: Find the decimal equivalent of binary 1111? Turn on your calculator 1) Press MODE 2) Press 4 3) Press SHIFT then press log key 4) Enter 1111 and then press = 5) Make sure to see 1111 under Bin on the screen 6) Press SHIFT then hit x 2 key 7) Answer: 15 MATH 126 FALL 2014

12 NUMBER SYSTEMS BINARY & DEIMAL (NEES RH / Page 213) Technology Usage From DEIMAL to BINARY Find the Binary equivalent of Decimal 7. KEY SEQUENE: DEIMAL (7) = BINARY (111) BNRY 180 FALL 2015

13 NUMBER SYSTEMS BINARY & DEIMAL (NEES RH / Page 213) Technology Usage From DEIMAL to BINARY Find the Binary equivalent of Decimal 51. KEY SEQUENE: BNRY FALL 2015 DEIMAL (51) = BINARY (110011)

14 NUMBER SYSTEMS BINARY & DEIMAL (NEES RH / Page 213) Technology Usage From BINARY to DEIMAL Find the Decimal equivalent of Binary KEY SEQUENE: BNRY FALL 2015 BINARY (1100) = DEIMAL (12)

15 NUMBER SYSTEMS BINARY & DEIMAL (NEES RH / Page 213) Technology Usage From BINARY to DEIMAL Binary is given. Find the Decimal equivalent. KEY SEQUENE: BINARY (101010) = DEIMAL (42) BNRY FALL 2015

16 NUMBER SYSTEMS BINARY & DEIMAL (NEES RH / Page 213) Supplemental Problems (1) (2) (3) (4) (5) BNRY 222 FALL 2015

17 SAMPLE (FE) PROBLEM MORNING SESSION MATHEMATIS Problem: Shaded Area =? 2 2 x I x. e dx 0 The value of the definite integral shown above (shaded area) is most nearly: (A) (B) () (D) ASE 325 FALL 2015

18 SAMPLE (FE) PROBLEMS MORNING SESSION MATHEMATIS 2 2 x I x. e dx 0 ASE FALL 2015 KEY SEQUENE

19 DEFINITE INTEGRALS AREAS UNDER URVES Using ASIO FX 115 ES PLUS Problem: f( x) 2 3 x The area under the graph shown is most nearly: (A) 5.06 (B) 4.25 () 2.15 (D) 1.37 ASE 355 FALL 2014

20 Solution using calculator: Important Keys Key Sequence: ASE FALL 2015

21 SAMPLE (FE) PROBLEMS MORNINGSESSION MATHEMATIS Problem: 3 I x.( e x ) dx The value of the definite integral shown above is most nearly: (A) 20 (B) 3986 () 1112 (D) omments: At first, this problem seems complex, but actually it is not. You can easily get the correct answer in about 15 seconds using your calculator. ASE418 FALL2015

22 DEFINITE INTEGRALS AREAS UNDER URVES Using alculator Solution: Some important keys: Key sequence: ASE FALL 2014

23 Problem: (Determinate Beam Analysis) 4 k 4 k 4 k 4 k 4 k P A B Support A : Pin Support B : Roller Point : 8 ft to the left of B A simple beam is loaded as shown in the figure. Using the given support conditions, answer the following questions: (1) If two support reactions are equal, the value of P (k) is, P (A) 4 (B) 8 () 10 (D) 12 (2) the vertical support reactions (kips) at the supports, A y B y (A) 18 (B) 14 () 12 (D) 10 (3) the magnitude of the shear force (k) at is most nearly, V (A) 8 (B) 6 () 4 (D) 2 (4) the maginute of the moment (k.ft) at is most nearly, M (A) 66 (B) 72 () 76 (D) 80 BM-37-G FALL 2015

24 FUNDAMENTALS OF ENGINEERING DETERMINATE BEAMS STATIS 2 kip / ft A B 6 ft 6 ft Support A : Pin Support B : Roller A simply supported beam is loaded as shown. Using the given support conditions, the magnitude of maximum bending moment (k-ft) is most nearly: +V +M (A) (B) () (D) M max =? MBM-10 SPRING 2015

25 ( { { ( ( { { RESULTANT OF TRAPEZOIDAL LOADS a R L b a b R = Resultant (Snake) R= Area of Trapezoid x L z Formulas: A = L 2 ( a + b x ( = L 2b + a 3 ( a + b ( z ( = L 2a + b 3 ( a + b ( Examples 1 a = 24 b = 16 L = 6 x = 2.80 ft. R = a = 40 b = 20 L = 9 m x = 4.0 m. R = a = 30 b = 20 L = 3 m x = 1.40 m. R = 75 4 a = 36 b = 24 L = 12 x = 5.6 m. R = 360 TRAP-30 SPRNG 2010

26 Problem: (antilever Beams) 24 kn/m 10 kn 6 kn FE/PE EXAM A 6 m B 2 m 2 m Support A : Fixed A cantilever beam is loaded as shown in the figure. Using the given loads and support condition, answer the following questions: (1) The vertical support reaction (kn) at the fixed support, A y (A) 88 (B) 105 () 113 (D) 128 (2) The bending moment (kn.m) at the fixed support, M A (A) 510 (B) 414 () 325 (D) 284 +M (3) The moment (kn.m) at, the mid-point of AB is most nearly, M (A) 90 (B) 110 () 122 (D) 134 (4) The shear force (kn) at the mid-point of AB is most nearly (A) 21.5 (B) 34.0 () 47.5 (D) 52.5 MBM-360 FALL 2010

27 Problem: (antilever Beams) A 30 kn/m 20 kn/m B 30 kn 4 3 FE/PE EXAM Support A : Fixed 3 m 2 m A cantilever beam is loaded as shown in the figure. Using the given loads and the support condition, answer the following questions: (1) The vertical support reaction (kn) at the fixed support, A y (A) 69 (B) 78 () 87 (D) 99 (2) The bending moment (kn.m) at the fixed support, M A (A) 125 (B) 185 () 205 (D) 225 (3) The horizontal support reaction (kn) at the fixed support, A x (A) 15.0 (B) 18.0 () 21.5 (D) 28.0 (4) The bending moment (kn.m) at the mid-point of AB is most nearly (A) 94.5 (B) () (D) MBM-357 FALL 2015

28 Problem: (antilever Beams) 24 kn/m 16 kn/m 8 kn 5 kn FE/PE EXAM A B Support A : Fixed 6 m 2 m 2 m A cantilever beam is loaded as shown in the figure. Using the given loads and support condition, answer the following questions: (1) The vertical support reaction (kn) at the fixed support, A y (A) 169 (B) 145 () 133 (D) 127 (2) The bending moment (kn.m) at the fixed support, M A (A) 525 (B) 450 () 375 (D) 265 +M (3) The moment (kn.m) at the mid-point of AB is most nearly, M (A) 140 (B) 153 () 162 (D) 28.0 (4) The shear force (kn) at the mid-point of AB is most nearly (A) 81.5 (B) 78.4 () 67.0 (D) 54.5 MBM-358 FALL 2010

29 Problem: (Determinate Frames) 10 kn 5 kn/m kn B FE/PE EXAM 3 m 30 kn 3 m A 4 m 4 m 2 m Support A : Hinge Support B : Roller The determinate frame is loaded as shown in the figure. Using the given loads and the support conditions, answer the following questions: (1) The magnitude of the horizontal support reaction (kn) at A, A x (A) 12.0 (B) 20.0 () 25.0 (D) 30.0 (2) The magnitude of the vertical support reaction (kn) at A, A y (A) (B) 9.25 () 8.50 (D) 7.60 (3) The magnitude of the vertical support reaction (kn) at B, B y (A) (B) () (D) (4) The magnitude of the bending moment (kn-m) at joint is most nearly, M (A) 120 (B) 90 () 80 (D) 75 MFR-566 FALL 2014

30 Problem: (Determinate Frames) 8 m A 6 m 20 kn/m 10 kn / m 12 m FE/PE EXAM Support A : Hinge Support B : Roller 5 m B The determinate frame is loaded as shown in the figure. Using the given loads and the support conditions, answer the following questions: (1) The magnitude of the horizontal support reaction (kn) at A, A x (A) 300 (B) 280 () 260 (D) 240 (2) The magnitude of the vertical support reaction (kn) at A, A y (A) 4.09 (B) 6.24 () 8.00 (D) 9.67 (3) The magnitude of the vertical support reaction (kn) at B, B y (A) (B) () (D) (4) The magnitude of the axial load (kn) in member B is most nearly, N A (A) (B) () (D) MFR-122 FALL 2015

31 Problem: 8, 25, 7, 5, 8, 3, 10, 12, 9 onsider the data set given above: (a) alculate the mean ( y ) (b) alculate the Standard Deviation (s y ) Solution: The mean is the sum of scores divided by n where n is the number of scores. 1. y y = ( )/9 n = the standard deviation may be calculated using the following formula: s y ( y y) i n 1 2 Deviation ( y y) i In order to calculate the values in the standard deviation formula, the following table may be used: Deviation = Score - Mean Score Mean Deviation (Deviation) = PRBL 210 FALL 2014

32 Standard Deviation( S y ) s y ( y y) i n = 6.32 S y = 6.32 Alternate method for calculating the Standard Deviation: (The Raw Score Method) onsider the raw scores 8, 25, 7, 5, 8, 3, 10, 12, First, square each of the scores. 2. Determine n, which is the number of scores. 3. ompute the sum of y i and the sum of y i 2 4. Then, calculate the standard deviation as illustrated below. 2 score(y i ) y i n = y i = y i = Standard Deviation( S y ) s y y 2 2 i y / i n = square root[(1161)-(87*87)/9)/(9-1)] n 1 = square root[(1161-(7569/9)/8)] = 6.32 PRBL 210 FALL 2014

33 entroid / Moments of Inertia y =? Solid Void entroid alculations 1 2 S y i A i A i y i in. 2 in. in A i y i y = S S A i 323 = = 3.44 in. 94 Moments of Inertia I o A i in. 4 in. 2 d i A i d i in. in S I cx = S I o + S A i. d i 2 = = in. 4 9 x x I cy = I y1 + I y2 = = = in SPRNG 2014

34 ENTROIDS / MOMENTS OF INERTIA SUPPLEMENTAL PROBLEMS 3 FE/PE EXAMS I cx =? y =? REF. LINE y = 3.0 in I cx = 204 in. 4 I cy = 135 in y =? y = 6.0 in I cx = 480 in. 4 I cy = 545 in y =? x y = 5.83 in. I cx = 978 in. 4 I cy = 334 in. 4 7 y y x y = 5.45 in. I cx = in. 4 I cy = in. 4 6 y =? L-111 FALL 2014

35 ENTROIDS / MOMENTS OF INERTIA SUPPLEMENTAL PROBLEMS y x y = 5.45 in. I cx = in. 4 I cy = in. 4 6 y =? y =? x y = 5.83 in. I cx = 978 in. 4 I cy = 334 in. 4 7 r = y =? y = 5.46 in. I cx = in. 4 I cy = in. 4 8 y 4- XL x I cx = in. 4 I cy = in L-125 FALL 2014

36 ENTROIDS / MOMENTS OF INERTIA SUPPLEMENTAL PROBLEMS y FE/PE EXAMS x y =? y = 5.0 in I cx = 1317 in. 4 I cy = 1733 in M y = 6.90 in. I cx = 2157 in. 4 7 r x y =? I cy = 1123 in. 4 y 3- M r 2 r = 3 4 x y =? y = 7.99 in. I cx = in. 4 I cy = in M-124 r = 3 r 11 x 4 y =? y = 5.20 in. I cx = in. 4 I cy = in L-120 FALL 2014

FUNDAMENTALS OF ENGINEERING CIVIL EXAM TOPICS. Computer-Based Test (CBT)

FUNDAMENTALS OF ENGINEERING CIVIL EXAM TOPICS Computer-Based Test (CBT) Total Number of Questions: 110 Time: 6 hours The new Civil FE Computer Based Test (CBT) consists of 110 multiple choice questions