ANSWERS September 2014

Transcription

1 NSWERS September 2014 nswers to selected questions: Sheet # (1) (2) () (4) SCE-55 D SCE-86 D SCE-88 D C MCM-21 MCM-12 D MMC-80 C C D MCM-52 D D MCM-1 C D D MCM-51 D D MCM-57 D D MCM-60 D MLS-12 C D NS 240 FLL 2014

2 CSIO / fx-115 ES PLUS MODE SHIFT SETUP (1) COMP (2) CMPLX () STT (4) SE-N (5) EQN (6) MTRIX (7) TLE (8) VECTOR (1) Mth-IO (2) Line-IO () Deg (4) Rad (5) Gra (6) Fix (7) Sci (8) Norm Frequently asked two Number Systems: 1- Decimal Number System (base 10) 2- inary Number System (base 2) In Decimal System 10 different digits are used to create any number, but in inary System only 0s and 1s are used to create any number DECIML INRY SCE 622 SPRNG 2014

3 inary Number System: NUMER SYSTEMS INRY & DECIML NCEES Reference Handbook, Page: 21 In digital computers, binary number system (the base 2) is used. Conversions from INRY to DECIML or from DECIML to INRY can easily be done using the calculator. inary (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 ) Enter 25 and press = 4) Make sure to see 25 under Dec on the screen 5) Press SHIFT then log 6) nswer: Problem: Find the decimal equivalent of binary 1111? Turn on your calculator 1) Press MODE 2) Press 4 ) Press SHIFT then press log key 4) Enter 1111 and then press = 5) Make sure to see 1111 under in on the screen 6) Press SHIFT then hit x 2 key 7) nswer: 15 MTH 126 FLL 2014

4 NUMER SYSTEMS INRY & DECIML NCEES Reference Handbook, Page: 21 Important Keys IMGE 55 FLL 2014

5 Problem: (eam Deflections) w = 2 kips / ft P = 12 kips W16 x ft 12 ft W 16 x 40 E = 29,000 ksi For the simple beam shown the beam weight is included in the uniform load. Determine the maximum deflection and the slope at (in radians). Solution: We will use NCEES-Reference Handbook, page 155 and 81. W 16 x 40 FOR DEFLECTIONS : ( 12 I = 518 in 4 FOR SLOPES : ( 12 2 ) ) Finding the maximum deflection: The maximum deflection will be at the midpoint of the span. For quick calculations when using US unit systems, architects and engineers use conversion factors like (12 ) and (12 2 ). For DEFLECTIONS this conversion factor is (12 ) and for SLOPES the conversion factor will be (12 2 ). 4 5 wl 1 PL max 84 EI 48 EI = 5 4 (2.0) (24) 84 (12 ) + (29,000)(518) 1 48 (12) (24) (29,000)(518) (12 ) = = 1.91 inches Finding the slope at support : 2 wl PL 24 EI 16 EI 1 (2.0) (24) 2 = (12 ) + 24 (29,000)(518) (12) (24) (29,000)(518) 2 (12 ) = = Radians = Radians DEF 216 FLL 2014

6 Problem: (eam Deflection) 10 kip 2.5 k/ft W 18 x 50 W 18 x ft E = 29,000 ksi cantilever beam is loaded as shown. Knowing that the beam weight is included in the uniform load, determine the maximum deflection. Solution: We will use NCEES Ref. Handbook, Page 155 and page kip 2.5 k/ft W 18 x 50 max I = 800 in 4 Maximum deflection: This is a cantilever beam with a fixed support at. The maximum deflection will be at the free end as shown. Using the formulas from NCEES Reference Handbook, page 82 and the unit conversion factor of (12 ). 1wL 1 PL max 8 EI EI 4 = (2.5) (14) (29,000)(800) (12 ) + 1 (10) (14) (29,000)(800) (12 ) = = inches DEF-228 FLL 2014

7 Problem: (Deflections) FE/PE EXM w = 1.5 k/ft 8 k.ft L = 10 ft section E = 29 x 10 6 psi cantilever beam is loaded as shown. The beam weight is included in the uniform load. Using the given cross-section and the modulus of elasticity answer the following questions: (1) the max. deflection (in.) of the beam is most nearly (d ) () () (C) (D) (2) the slope (rad.) at the free end of the beam is most nearly (q ) () () (C) (D) () the moment at the support is most nearly (M ) () 75 () 8 (C) 90 (D) 105 DEF-69 FLL 2014

8 Problem: (Friction) FE EXM P 5 4 W = 200 lb P = 60 lb m = 0.25 s m = 0.20 k m m : coefficient of static friction s : coefficient of kinetic friction k 60-lb force acts on a 200-lb crate on an inclined plane as shown in the figure. Using the listed data, answer the following questions: (1) the magnitude of the friction force (lb) acting on the crate: ( F ) () 45.5 () 50.0 (C) 55.5 (D) 60.0 (2) the magnitude of the normal force (lb) acting on the crate: (N ) () 140 () 150 (C) 160 (D) 180 () the maximum friction force (lb) acting on the crate: (F m ) () 2.9 () 40.0 (C) 5.5 (D) 62.0 FRCT-15 FLL 2014

9 Problem: (Friction) P = 60 lb FE EXM q = 25 o P = 60 lb W = 220 lb q = 25 o m = 0.28 s m s : coefficient of static friction 60-lb force acts on a 220-lb crate on an inclined plane as shown in the figure. Using the listed data answer the following questions: (1) the magnitude of the friction force (lb) acting on the crate F () 26.8 () 0.5 (C) 8.6 (D) 45.4 (2) the magnitude of the normal force (lb) acting on the crate N () 50 () 10 (C) 280 (D) 225 () the maximum friction force (lb) acting on the crate (F m ) () 62.9 () 78.4 (C) 8.5 (D) 92.8 FRCT-12 FLL 2014

10 DETERMINTE FRMES Support Reactions 6 kip (4) FR-410 Pin 2 k/ft 9 ft C 5 ft 5 ft D 1.5 k/ft 5 kip Roller 6 ft 6 ft nswers x = 1.0 k y = 15 k y = 9 k 18 kip k/ft (5) FR-74 k/ft C D k/ft 12 ft nswers x = 0 y = 26.5 k y = 18.5 k Pin 9 ft 9 ft Roller FR-6 (6) FR-7 Pin 2 k/ft 9 ft 2 k/ft C 9 ft 20 kip k/ft 4 k/ft D 12 ft Roller nswers x = 6.0 k y = k y = k LFR-48 FLL 2014

11 DETERMINTE FRMES Support Reactions (1) k/ft 6 kip nswers FR-22-C Roller 8 ft x = 24 k y = 9 kip y = 15 kip Pin 6 ft 4 ft 4 ft 12 kip 4 kip (2) FR ft 2 k/ft Roller nswers x = 2.0 k y = 9. k y = 0.67 k Pin 12 ft 6 ft 6 ft 6 ft 6 kip 12 k-ft () FR ft 1.5 k/ft Roller nswers x = 24.0 k y = 6.5 k y = 17.5 k Pin 12 ft 6 ft 6 ft 6 ft LFR-42 FLL 2014

12 Problem: (Frame with internal hinge) 2 m Joint C : Pin 8 kn / m C 1.5 m FE/PE EXM 1.5 m 1.5 m Support : Pin Support : Pin The frame shown has an internal pin at C. Using the given load and support conditions, answer the following questions: (1) the vertical support reaction (kn) at the left support, y ().26 () 4.86 (C) 5.44 (D) 6.75 (2) the horizontal support reaction (kn) at the left support, x () 5.48 () 6.15 (C) 7.4 (D) 8.86 () the vertical support reaction (kn) at the right support, y () 5.22 () 6.4 (C) 7.14 (D) 8.20 (4) the maginute of the axial load (kn) in C is most nearly, N C () 1.00 () 1.4 (C) 2.05 (D) 2.65 FRMC-0 FLL 2014