ERRATA for PE Mechanical Engineering: Machine Design and Materials Practice Exam ISBN: Copyright 2016 Errata posted 12/18/2017

Transcription

1 ISBN: Copyright 016 Errata posted 1/18/017 Revisions are shown in red. Question 108, p. 13: Critical speed 15 gei L ρa where: L length of lead sew E modulus of elasticity I area moment of inertia ρ density A oss-sectional area g acceleration of gravity Question 11, p. 3: In the figure, M 7,000 in.-lb Question 510, p. 46: Option A should read as follows: (A) 0.19 age 1 of 7

2 ISBN: Copyright 016 Errata posted 1/18/017 Question 516, p. 49: The last sentence should read as follows: If the element is supported in all directions at the location shown in the bottom figure, the failure load will: Question 533, p. 60: The first sentence should read as follows: The hydraulic cylinder shown in the figure has a.75-in.-diameter piston (Area, A 5.94 in ) and is subjected to a maximum load from a maximum pressure of 3,000 psi. ins have been added to the figure as shown:.75 in. EFFECTIVE LENGTH 50 in. Question 537, p. 63: F d mean F d collar θ d collar 1.75 in. f collar 0.05 d mean 1.50 in. f sew 0.08 θ 3.033º 4 tpi T age of 7

3 ISBN: Copyright 016 Errata posted 1/18/017 Solution 11, p. 79: Ju f(3g + f ) 6 J h Ju h f (3g + f ) 6 g 6 in. f 1.5 in. M 7,000 in.-lb g 6 M r f + g ( ) mr 6M f + g τ J h f (3g + f ) h 6M f + g τ f (3g + f ) ( 7, 000) ( ) in (1,000) 1.5 in. ( ) 3/ 8n. i THE CORRECT ANSWER IS: (B) age 3 of 7

4 ISBN: Copyright 016 Errata posted 1/18/017 Solution 510, p. 94: Force on ceiling mg 400 lb Tension in chains: T cos lb T 31. lb Components TH T sin lb TV T cos lb Force on 400 lb block T lb T N F F F F N mg 400 lb + Fy 0 FF μn 400 lb or μn 00 lb Free-body diagram of 1/ device T V 44 in. 13 in. 0 T H 0 y 0 x 13 in. 10 in. N F F age 4 of 7

5 ISBN: Copyright 016 Errata posted 1/18/017 Solution 510, p. 94 (continued): + ΣM0 0 13TV 44TH 13 FF + 10N 0 (13)(00.) (44)(115.6) (13)(00) + 10N 0 N 1,08.9 lb μn 00 μ(1,08.9) μ ,08.9 THE CORRECT ANSWER IS: (A) age 5 of 7

6 ISBN: Copyright 016 Errata posted 1/18/017 Solution 516, p. 97: Check maximum compression load σ /A 6 Allowable force, σa ( N/m )( m)( 0. 5m) compression 10,800,000 N sustained. This is the maximum compressive load that can be σ e π E L/k ( ) Cπ EI leff C 1 for pinned bh 3 ( 0.50)( 0.100) 3 I 1 1 ( )(. ) π ( ) 5 4 I m 3,594,670 N l eff 1m ( )(. ) π ,376,8 N Exceeds compressive capacity so use compression. Failure is ineased by compression failurein compression 10, 800, failure in buckling 3, 594, 670 THE CORRECT ANSWER IS: (B) age 6 of 7

7 ISBN: Copyright 016 Errata posted 1/18/017 Solution 519, p. 98: oint (5) is outside the area of acceptable design (fatigue failure). oints on the Goodman Line have a limited fatigue life. THE CORRECT ANSWER IS: (B) Solution 533, p. 104: σ A rod Using safety factor of to Sy. 36,500 18,50 psi A rod Apiston 3, ,80 lbf 17,80 18, 50 18, 50 Arod πdrod 4Arod Arod drod π THE CORRECT ANSWER IS: (A) reviously posted errata continued on next page age 7 of 7

8 ISBN: Copyright 016 Errata posted 3/31/017 Revisions are shown in red. Solution 506, p. 9: Line 1 of the solution should read as follows: sin 5 1 reviously posted errata continued on next page age 1 of 1

9 ISBN: Copyright 016 Errata posted /17/017 Revisions are shown in red. Question 105, p. 11: Line should read as follows: If the delivery of parts from Source Y is delayed by 3 days, the total completion delay (days) will be most nearly: (A) 0 (B) 1 (C) (D) 3 Question 111, p. 16: Line 1 should read as follows: A circular rod will be loaded in simple tension. The rod has a length of 10 in. and a diameter of 3/8 in. Data for available materials are as follows: Solution 105, p. 70: The first paragraph should read as follows: If Task C is delayed by 3 days, but Task C has days of slack from initial itical path, then ABCE becomes the new itical path, and the total completion delay will be 1 day. THE CORRECT ANSWER IS: (B) age 1 of 1