Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 18, Problem 93P
To determine
The temperature at the center of the cylinder.
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Stress, ksi
160
72
150-
140
80
70
༄ ྃ ༈ ཎྜ རྦ ༅ ཎྜ ྣཧྨ ➢
130
120
110
100
90
2.0
2.8
3.6
4.4
5
Wire diameter, mm
6.0
6.8
2
7.6
8.4
Compression and extension springs.
ASTM A227 Class II
Light service
Average service
0.020
0.060
0.100
0.140
0.180
0.220
0.260
0.300
0.340
0.380
0.420
0.460
0.500
Wire diameter, in
Torsional stress due to initial tension, ksi
10
४
20
Preferred
range
100
Stress, MPa
9.2
10.0
10.8
11.6
12.4
1100
1035
965
895
825
760
Severe service
690
620
550
50
150
3456789 10 11 12 13 14 15 16
Spring index, C = DJD
FIGURE 18-21 Recommended torsional shear stress in an extension spring due to initial tension (Data from Associated
Spring, Barnes Group, Inc.)
50
200
485
Stress, MPa
Bolted Joint Design
Bolted Frames
Total Force due to door weight: P =
240
lb
Number of Bolts: N =
Distance to Bolt C/L: a =
4
N/A
Bolt Material -
Allowable shear stress of bolt material: T₂ =
x Distance from Bolt centroid to bolt: x =
y Distance from Bolt centroid to bolt: y =
Degrees per Radian-
Results
y-Load on each bolt: F, =
Moment resisted by bolt pattern: M =
Radial distance from Bolt centroid to bolt: r =
Sum squares of all radial distances: Σr²
Force on each bolt to resist moment: F, -
Angle for force composition: e=
X-Force on each bolt to resist moment: F-
y-Force on each bolt to resist moment: Fly
Total y-Force on each bolt: Fy =
Resultant force on bolt 1: R₁ =
Required shear stress area for a bolt: A₂ =
ASTM Grade
A307 Steel
10,000
0
psi
from Table 20-1
3.0
57.296
in
degrees
lb per bolt
lb-in
Formula
FS-P/N
M-Px XB
r = (x² + y²)0.5
in²
Σ
4r²
Mr
F₁ =
Στ
lb
degrees
lb
lb
lb
Minimum Bolt Diameter: Din =
Rounded up Bolt Diameter: D =
55
P.
1.5 in
2 in (3x)
1 in
This bracket…
University of Babylon
Collage of Engineering/
Al-Musayab
Department of Automobiles
Final Examination/ Stage: 3rd
Notes:
Answer 4 questions only
2023-2202
Subject: Theory of vehicles
Date: 2023\06\10-Saturday
Time: Three Hours
Course 2nd Attempt 1st
Q1: A Hooke's coupling connects two shafts whose axes are inclined at 30°. The
of the driven shaft? Find the maximum value of retardation or acceleration and
driving shaft rotates uniformly at 600 rpm. What are the extreme angular velocities
state the angle where both will occur.
(12.5 Marks)
Q2: Four masses, A, B, C, and D), revolve at equal radii and are equally spaced
along a shaft. The mass B is 7 kg, and the radius of C and D make angles of 90°
and 240°, respectively, with the radius of B. Find the magnitude of the masses A,
C, and D and the angular position of A so that the system may be completely
balanced.
(12.5 Marks)
Q3: A cam has straight worked faces that are tangential to a base circle of diameter
90 mm. The follower is a roller…
Chapter 18 Solutions
Fundamentals Of Thermal-fluid Sciences In Si Units
Ch. 18 - Prob. 1PCh. 18 - Prob. 2PCh. 18 - In what medium is the lumped system analysis more...Ch. 18 - Prob. 4PCh. 18 - Prob. 5PCh. 18 - Prob. 6PCh. 18 - Prob. 7PCh. 18 - Consider a hot baked potato on a plate. The...Ch. 18 - Prob. 9PCh. 18 - Prob. 10P
Ch. 18 - Prob. 11PCh. 18 - Prob. 12PCh. 18 - Prob. 13PCh. 18 - Prob. 14PCh. 18 - Consider a 1000-W iron whose base plate is made of...Ch. 18 - Metal plates (k = 180 W/m·K, ρ = 2800 kg/m3, and...Ch. 18 - A 5-mm-thick stainless steel strip (k = 21 W/m·K,...Ch. 18 - A long copper rod of diameter 2.0 cm is initially...Ch. 18 - Prob. 21PCh. 18 - Steel rods (ρ = 7832 kg/m3, cp = 434 J/kg·K, and k...Ch. 18 - Prob. 23PCh. 18 - The temperature of a gas stream is to be measured...Ch. 18 - Prob. 25PCh. 18 - A thermocouple, with a spherical junction diameter...Ch. 18 - Prob. 27PCh. 18 - Prob. 28PCh. 18 - Carbon steel balls (ρ = 7833 kg/m3, k = 54 W/m·K,...Ch. 18 - Prob. 31PCh. 18 - Prob. 32PCh. 18 - Prob. 33PCh. 18 - Prob. 34PCh. 18 - Prob. 35PCh. 18 - Prob. 36PCh. 18 - Prob. 37PCh. 18 - Prob. 38PCh. 18 - Prob. 39PCh. 18 - A body at an initial temperature of Ti is brought...Ch. 18 - In a meat processing plant, 2-cm-thick steaks (k =...Ch. 18 - Prob. 42PCh. 18 - Prob. 43PCh. 18 - Prob. 45PCh. 18 - Prob. 46PCh. 18 - Prob. 47PCh. 18 - Prob. 48PCh. 18 - Prob. 49PCh. 18 - A long iron rod (ρ = 7870 kg/m3, cp = 447 J/kg·K,...Ch. 18 - Prob. 51PCh. 18 - A long 35-cm-diameter cylindrical shaft made of...Ch. 18 - Prob. 54PCh. 18 - Prob. 55PCh. 18 - Prob. 56PCh. 18 - A father and son conducted the following simple...Ch. 18 - Prob. 58PCh. 18 - Prob. 59PCh. 18 - Citrus fruits are very susceptible to cold...Ch. 18 - Prob. 61PCh. 18 - Prob. 63PCh. 18 - Prob. 64PCh. 18 - Prob. 65PCh. 18 - White potatoes (k = 0.50 W/m·K and α = 0.13 × 10−6...Ch. 18 - Prob. 67PCh. 18 - Prob. 68PCh. 18 - Prob. 69PCh. 18 - Consider a hot semi-infinite solid at an initial...Ch. 18 - Prob. 71PCh. 18 - Prob. 72PCh. 18 - Prob. 73PCh. 18 - Prob. 74PCh. 18 - Prob. 75PCh. 18 - Prob. 76PCh. 18 - Prob. 77PCh. 18 - Prob. 78PCh. 18 - Prob. 79PCh. 18 - Prob. 81PCh. 18 - Prob. 82PCh. 18 - Prob. 83PCh. 18 - Prob. 84PCh. 18 - Prob. 85PCh. 18 - Prob. 86PCh. 18 - Prob. 88PCh. 18 - Prob. 89PCh. 18 - A 2-cm-high cylindrical ice block (k = 2.22 W/m·K...Ch. 18 - Prob. 91PCh. 18 - Prob. 93PCh. 18 - Prob. 94RQCh. 18 - Large steel plates 1.0-cm in thickness are...Ch. 18 - Prob. 96RQCh. 18 - Prob. 97RQCh. 18 - Prob. 98RQCh. 18 - Prob. 99RQCh. 18 - Prob. 100RQCh. 18 - Prob. 101RQCh. 18 - Prob. 102RQCh. 18 - The water main in the cities must be placed at...Ch. 18 - Prob. 104RQCh. 18 - Prob. 105RQCh. 18 - Prob. 106RQCh. 18 - Prob. 107RQ
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- Problem 18-26 Added Extension Springs Spring Material ASTM A227 Modulus of Elasticity of the Material in Shear: G 1.150E+07 psi Average Service Max Operating Load: F₁ = 100 lb Max Length between attachment points: L₁ = 60.00 in 20.00 lb 26.00 1.400 Min Operating Load: F₁ = Min Length between attachment points: L₁ = Maximum Outside Diameter = in in Results Note: you select a wire diameter from the "US steel wire gage" column in table 18-2 Formula k = AF/AL k = (F0-F1)/(Lo - L₁) Spring Rate: k = lb/in Assumed Trial Outside Diameter: OD = Assumed Trial Mean: D ma Assumed Design Stress in Spring: Tda in 1.070 in 102,000 psi Assumed Wahl Factor: K = 1.2 Calculated Wire Diameter: Dwa Actual Wire Diameter: Dw Actual outer diameter: OD = Actual inner diameter: ID= Spring Index: C = See Figure 18-8 Dw= [8KF Dm πTd 1/3 in 5' 5' 5' 5' This corresponds to US Steel 9 wire gage ID = Dm - Dw C = Dm/Dw 4C - 1 0.615 K = + 4C - с Wahl Factor: K = 8KFDm 8KFC T = TD πD Stress in Spring at F = Fo: To psi…arrow_forwardCHAIN DRIVE DESIGN Initial Input Data: Application: Garage Door Opener Drive type: AC Motor Driven machine Chain and Sprocket to pull the door up Degrees per Radian: 57.2958 degrees Sprocket Diameter: D = 1.690 in Number of strands: Chain number: 1 40 Service factor: 1.3 Table 7-10 No. of teeth Computed Data: Actual Motor Power Input: 0.000 hp Sprocket Speed (for sprocket attached to gear shaft) Design power: 0.00 rpm 0 hp 11 12 0.06 0.15 0.29 0.56 0.99 1.09 1.61 2.64 TABLE 7-7 Horsepower Ratings-Single Strand Roller Chain No. 40 0.500 inch pitch 10 25 50 100 180 200 300 500 700 900 1000 1: 0.06 0.14 0.27 0.52 0.91 1.00 1.48 2.42 3.34 4.25 4.70 ! 3.64 4.64 5.13 13 0.07 0.16 0.31 0.61 1.07 1.19 1.75 2.86 3.95 5.02 5.56 Design Decisions-Chain Type and Teeth Numbers: 14 Chain number: Use Table 7-7 Chain pitch: p = in 15 Number of Teeth: N = Per Table 7-7 16 0.08 0.20 0.39 0.75 1.32 1.46 2.15 3.52 0.07 0.17 0.34 0.66 1.15 1.28 1.88 3.08 0.08 0.19 0.36 0.70 1.24 1.37 2.02 3.30 4.55 5.80…arrow_forwardInput Data: Torque needed to overcome rolling friction in rollers, slides and other moving parts, except for Motor and Worm Gear the worm gear T₁ = Length of travel of door: Time for door to open or close: LD = 50 lb-in. 90 in t= 12.5 seconds Pitch diameter for chain sprocket: DPC 1.690 in Weight of Door: P = No. of worm threads: Nw = Worm Pitch diameter: Dw Diametral pitch: Pd Normal pressure angle: Degrees per Radian: Number of gear teeth: Calculated Data: Linear velocity of door and chain (in/sec): Linear velocity of door and chain (ft/min): Output Speed of Gear and Sprocket: Upward Force due to Weight of Door: Фо = = NG= 240 lb 2 1.250 in 12 14.5 degrees 57.2958 degrees 28 Vα= in/sec VC= ft/min NG = rpm FD lb Net Upward Force on Door: Fou lb Torque on gear ignoring rolling friction: TG = lb-in. Formula = FDU FD-2 x Fo (note: Fo is the Max Operating load of the extension springs). This is also the initial tension in the chain. TG = FDU X DPC/2 This is the also the torque on the…arrow_forward
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