ERRATA for PE Mechanical Engineering: Machine Design · PDF fileERRATA for PE Mechanical...

ERRATA for PE Mechanical Engineering: Machine Design and Materials Practice Exam

ISBN: 978-1-932613-77-3 Copyright 2016

Errata posted 12/18/2017

of 7

Revisions are shown in red. Question 108, p. 13:

Critical speed = 2gEI215

AL ρ

where:

L = length of lead screw

E = modulus of elasticity

I = area moment of inertia

ρ = density

A = cross-sectional area

g = acceleration of gravity

Question 121, p. 23:

In the figure, M = 27,000 in.-lb Question 510, p. 46:

Option A should read as follows:

(A) 0.19


ISBN: 978-1-932613-77-3 Copyright 2016


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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:


The first sentence should read as follows: The hydraulic cylinder shown in the figure has a 2.75-in.-diameter piston (Area, A = 5.94 in2) and is subjected to a maximum load from a maximum pressure of 3,000 psi. Pins have been added to the figure as shown:


2.75 in.

EFFECTIVE LENGTH = 50 in.

dcollar

F2— F

2—

T

dmean

dcollar = 1.75 in.fcollar = 0.05 dmean = 1.50 in.fscrew = 0.08θ = 3.033º4 tpi

θ


ISBN: 978-1-932613-77-3 Copyright 2016


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Solution 121, p. 79:

2 2

uf (3g f )J

6+

=

( )

( )( )

2 2u

2 2

2 2

2 2

2 2

2 2

2 2

2

0.707J 0.707 h J h f (3g f )6

f gr

2

6 f gmrJ 2 0.707 h f (3g f )

6 f gh

1.414 f (3g f )

6 1.5 60.357 in.

1.414 (12,000) 1.5 in. (3 6

M

M

22.25)

3/

7,

8 n.

00

i

0

= = +

+=

+′′τ = =+

+=

′′τ +

+= =

× +

=

THE CORRECT ANSWER IS: (B)

g = 6 Mg = 6 in.f = 1.5 in.M = 27,000 in.-lb


ISBN: 978-1-932613-77-3 Copyright 2016


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Force on ceiling = mg = 400 lb Tension in chains: 2T cos 30° = 400 lb T = 231.2 lb

Components TH = T sin 30° = 115.6 lb

TV = T cos 30° = 200.2 lb

Force on 400 lb block

∑Fy = 0 2FF =2μN = 400 lb

or μN = 200 lb

Free-body diagram of 1/2 device

30° 30°T T

400 lb

FFFF

N N

mg = 400 lb

44 in.

TH

TV

13 in.

0y

0x0

13 in.

N

FF

10 in.

+


ISBN: 978-1-932613-77-3 Copyright 2016


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Solution 510, p. 94 (continued):

ΣM0 = 0

–13TV – 44TH – 13 FF + 10N = 0

–(13)(200.2) – (44)(115.6) – (13)(200) + 10N = 0

N = 1,028.9 lb

μN = 200

μ(1,028.9) = 200

200μ 0.191,028.9

= =

THE CORRECT ANSWER IS: (A)

+


ISBN: 978-1-932613-77-3 Copyright 2016


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Check maximum compression load

σ = P/A

Allowable force, P = σA = ( )( )( )6 2432 × 10 N/m 0 100 m 0 25 m. .

Pcompression = 10,800,000 N This is the maximum compressive load that can be sustained.

( )2

2eπ E=L/k

σ 2

cr 2eff

π El

IP C= C = 1 for pinned

( ) ( )33 0.250 0.100bhI = =12 12

( ) ( )

( )

2 9 5

cr 2

π 70 10 2 083 10P =

2

−× ×.

5 4I = 2.083 ×10 m− crP = 3,594,670 N

effl = 1 m

( ) ( )2 9 5

cr 2

π 70 10 2 083 10P =

1. −× ×

crP = 14,376,282 N Exceeds compressive capacity so use Pcompression.

Failure is increased by compression

cr

P failure in compressionP fai

10 800 000 3 03 594 6lure in buckli 0ng 7

, , ., ,

== =



ISBN: 978-1-932613-77-3 Copyright 2016


of 7


Point (5) is outside the area of acceptable design (fatigue failure). Points on the Goodman Line have a limited fatigue life.



crcr

rod

Pσ =A

Using safety factor of 2 to Sy.

cr

rod

P 36,500 18, 250 psiA 2

= =

Pcr = P Apiston = 3,000 × 5.94

= 17,820 lbf

crrod

2rod rod

rod rod

P 17,820A 0.976418,250 18,250πd 4AA d 1.11

4 π

= = =

= = =

THE CORRECT ANSWER IS: (A)

Previously posted errata continued on next page


ISBN: 978-1-932613-77-3 Copyright 2016


Page 1 of 1

5sin12

Revisions are shown in red.


Line 1 of the solution should read as follows:

Previously posted errata continued on next page


ISBN: 978-1-932613-77-3 Copyright 2016


Revisions are shown in red.


Line 2 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) 2

(D) 3


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:


The first paragraph should read as follows:

If Task C is delayed by 3 days, but Task C has 2 days of slack from initial critical path,

then ABCE becomes the new critical path, and the total completion delay will be 1 day.


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