Chapter 5
5-1 MSS: DE: 1 - 3 = S y /n n= Sy n= Sy 1 - 3
1/2
2 2 = A - AB + B
1/2
2 2 2 = x - x y + y + 3x y
(a) MSS:
1 = 12, 2 = 6, 3 = 0 kpsi 50 n= = 4.17 Ans. 12 = (122 - 6(12) + 62 ) 1/2 = 10.39 kpsi, 12 2
2
DE:
n=
50 = 4.81 10.39
Ans.
12 (b) A , B = 2
Chapter 4
4-1 (a)
k1 k2 k3 F y
k= so (b)
k1 k2 k3 F y
F ; y
y=
F F F + + k1 k2 k3 Ans.
k=
1 (1/k1 ) + (1/k2 ) + (1/k3 )
F = k1 y + k2 y + k3 y k = F/y = k1 + k2 + k3 Ans.
(c)
k2 k1 k3
1 1 1 + = k k1 k2 + k3
k=
1 1 + k1 k2 + k3
-1
4-2 For a torsion bar, k
Chapter 8
8-1 (a)
2.5 mm 25 mm 5 mm 2.5
Thread depth = 2.5 mm
Ans.
Width = 2.5 mm Ans. dm = 25 - 1.25 - 1.25 = 22.5 mm dr = 25 - 5 = 20 mm l = p = 5 mm Ans. Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. dm = 22.5 mm dr = 20 mm l = p = 5 mm
Chapter 9
9-1 Eq. (9-3): F = 0.707hl = 0.707(5/16)(4)(20) = 17.7 kip Ans. 9-2 Table 9-6: all = 21.0 kpsi f = 14.85h kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64(4) = 18.56 kip Ans. 9-3 Table A-20: 1018 HR: Sut = 58 kpsi, S y = 32 kpsi 1018 CR: Sut =
Chapter 10
10-1
1" 4"
1" 2
1" 4"
1" 2
10-2
A = Sd m dim( Auscu ) = dim(S) dim(d m ) = kpsi inm m dim( ASI ) = dim(S1 ) dim d1 = MPa mmm MPa mmm . ASI = m Auscu = 6.894 757(25.40) m Auscu = 6.895(25.4) m Auscu kpsi in For music wire, from Table 10-4: Auscu
Chapter 11
11-1 For the deep-groove 02-series ball bearing with R = 0.90, the design life x D , in multiples of rating life, is xD = The design radial load FD is FD = 1.2(1.898) = 2.278 kN From Eq. (11-6), C10 540 = 2.278 0.02 + 4.439[ln(1/0.9)]1/1.483 =
DESIGN OF HELICAL COMPRESSION SPRING - METHOD 2
Specify wire diameter, design stresses, and number of coils. Compute mean diameter.
1. Enter data for forces and lengths
2. Specify material, shear modulus, G, and an estimate for design stress
3. Enter tria
Chapter 6
Note to the instructor: Many of the problems in this chapter are carried over from the previous edition. The solutions have changed slightly due to some minor changes. First, the calculation of the endurance limit of a rotating-beam specimen Se
1.
2.
3.
4.
5.
6.
DESIGN OF HELICAL COMPRESSION SPRING - METHOD 1
Specify mean diameter and design stresses. Compute wire diameter, number of coils.
Enter data for forces and lengths
Specify material, shear modulus, G, and an estimate for design stress
En
4/19/16
MEC 410 Agenda 4/19/2016
Announcement and Deadlines
Exam #3 on Thursday, April 28, 2016
HW#9 is due April 21, 2016
Design Project #2 XY table designdue on Tuesday, April
26, 2016
Discussions
My travel schedule & lecture dates
AEC
conference:
4/7/16
MEC 410 Agenda 4/7/2016
Announcement and Deadlines
HW#8 will be assigned (due on April 14, 2016)
Design Project #2 XY table design
Design
of a XY table for motion control using power screws
to use the handout for the calculation of dynamics of
4/27/16
MEC 410 Agenda 4/26/2016
Announcement and Deadlines
Exam #3 on Thursday, April 28, 2016
Includes
the following chapters with handouts and supplementary
materials: Chapters 7, 17, 18, 21
Three parts: Part (I)short questions; Part (II)3 problems
4/19/16
Discussions on HW#9
and Project #2
HW#9: problems 4 and 5
Reected iner<a for belt-pulley drive
Reected iner<a of ball screw drive systems
Note the use of units and dimensional analysis to
ensure the consiste
MEC410 Handout
Theory and Application of Timing Belt
MEC410, Spring 2016
I. Kao
This handout presents basic theory of timing belt and its application. The terminology and geometric relationship are also presented.
1
Introduction
Timing belts, integrating
Chapter 1
Problems 1-1 through 1-4 are for student research. 1-5 (a) Point vehicles
v x
Q= Seek stationary point maximum
cars v 42.1v - v 2 = = hour x 0.324
dQ 42.1 - 2v =0= v* = 21.05 mph dv 0.324 Q* = (b) 42.1(21.05) - 21.052 = 1368 cars/h Ans. 0.324
v
Chapter 2
2-1 From Table A-20 Sut = 470 MPa (68 kpsi), S y = 390 MPa (57 kpsi) 2-2 From Table A-20 Sut = 620 MPa (90 kpsi), S y = 340 MPa (49.5 kpsi) Ans. 2-3 Comparison of yield strengths: Sut of G10 500 HR is Syt of SAE1020 CD is 620 = 1.32 times larger
Chapter 3
3-1
W 2 A RA 1 RB B A RA 1 1 RB W 2 B
(a)
1 RD 3 RA RB 2 D
(b)
1 RC C
A
B
W
(c)
W 1 RC RB RA RB 2 RA W
(d) (e)
A
RA
1
2 W RBx B RBx 1 RB RBy RBy
Scale of corner magnified
(f)
Chapter 3
15
3-2 (a)
2 kN 60 2 30 2 kN 60 RB RA 30 RA RB 90
R A = 2 si
Chapter 13
13-1 d P = 17/8 = 2.125 in dG = N2 1120 dP = (2.125) = 4.375 in N3 544
NG = PdG = 8(4.375) = 35 teeth Ans. C = (2.125 + 4.375)/2 = 3.25 in Ans. 13-2 n G = 1600(15/60) = 400 rev/min Ans. p = m = 3 mm Ans. C = [3(15 + 60)]/2 = 112.5 mm Ans. 13-3
Design and Analysis of Machine Elements
!
!
Power/Energy Transmission II:
!
Belt Drives
!
!
!
Power/Energy Transmissions
!
Power Transmissions
!
!
Power screws converts rotations into translations!
Many applications require transmission of rotations fro
I
Keys, Couplings, and Seals
The Big Picture
You Are the Designer
11-1
Objectives of This Chapter
11-2
Keys
11-3
Materials for Keys
11-4
Stress Analysis to Determine Key Length
11-5
Splines
11-6
Other Methods of Fastening Elements to Shafts
11-7
Couplings
Helical Gears, Bevel Gears,
and Wormgearing
The Big Picture
You Are the Designer
10-1
Objectives of This Chapter
10-2
Forces on Helical Gear Teeth
10-3
Stresses in Helical Gear Teeth
104 Pitring Resistance for Helical Gear Teeth
10-5
Design of Helical Gea
*
Shaft Design
The Big Picture
You Are the Designer
12-1
Objectives of This Chapter
12-2
Shaft Design Procedure
12-3
Forces Exerted on Shafts by Machine Elements
124
Stress Concentrations in Shafts
12-5
Design Stresses for Shafts
12-6
Shafts in Bending an
Kinematics of Gears
The Big Picture
You Are the Designer
300
8-1
Objectives of This Chapter
8-2
Spur Gear Styles
8-3
Spur Gear Geometry: Involute-Tooth Form
8-4
Spur Gear Nomenclature and Gear-Tooth Features
8-5
Interference between Mating Spur Gear Teeth
MEC410 H ANDOUT: C RITICAL ROTATING S PEED IN S HAFT D ESIGN
MEC 410, Spring 2016
I. Kao
The critical speed of a rotating shaft can be estimated using the lumped mass model with the Rayleigh equation, as presented in Appendix A. The fundamental natural fr