Mechanics of Materials
11th Edition
ISBN: 9780137605460
Author: Russell C. Hibbeler
Publisher: Pearson Education (US)
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Textbook Question
Chapter 1.7, Problem 19FP
If the eyebolt is made of a material having a yield stress of σY = 250 MPa, determine the minimum required diameter d of its shank. Apply a factor of safety F. S. = 1.5 against yielding.
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Stress, ksi
160
72
150-
140
80
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༄ ྃ ༈ ཎྜ རྦ ༅ ཎྜ ྣཧྨ ➢
130
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100
90
2.0
2.8
3.6
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5
Wire diameter, mm
6.0
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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
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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 1 Solutions
Mechanics of Materials
Ch. 1.2 - Determine the resultant internal normal force,...Ch. 1.2 - Determine the resultant internal normal force,...Ch. 1.2 - Determine the resultant internal normal force,...Ch. 1.2 - Determine the resultant internal normal force,...Ch. 1.2 - Determine the resultant internal normal force,...Ch. 1.2 - Determine the resultant internal normal force,...Ch. 1.2 - The shaft is supported by a smooth thrust bearing...Ch. 1.2 - Determine the resultant internal normal and shear...Ch. 1.2 - Determine the resultant internal torque acting on...Ch. 1.2 - Determine the resultant internal loadings in the...
Ch. 1.2 - The shaft is supported by a smooth thrust bearing...Ch. 1.2 - Determine the resultant internal loading on the...Ch. 1.2 - Determine the resultant internal loading on the...Ch. 1.2 - The 800-lb load is being hoisted at a constant...Ch. 1.2 - Determine resultant internal loadings acting on...Ch. 1.2 - Determine the resultant internal normal force...Ch. 1.2 - Determine the resultant internal loadings on the...Ch. 1.2 - Determine the resultant internal loadings on the...Ch. 1.2 - The blade of the hacksaw is subjected to a...Ch. 1.2 - The blade of the hacksaw is subjected to a...Ch. 1.2 - Determine the resultant internal loadings on the...Ch. 1.2 - Determine the resultant internal loadings on the...Ch. 1.2 - The sky hook is used to support the cable of a...Ch. 1.2 - Determine the resultant internal torque acting on...Ch. 1.2 - Determine the resultant internal loadings acting...Ch. 1.2 - Determine the resultant internal loadings on the...Ch. 1.2 - Determine the resultant internal loadings on the...Ch. 1.2 - The metal stud punch is subjected to a force of...Ch. 1.2 - The metal stud punch is subjected to a force of...Ch. 1.2 - Determine the resultant internal loadings acting...Ch. 1.2 - A force of 80 N is supported by the bracket....Ch. 1.2 - The curved rod has a radius r and is fixed to the...Ch. 1.2 - The pipe assembly is subjected to a force of 600 N...Ch. 1.2 - If the drill bit jams when the handle of the hand...Ch. 1.2 - The curved rod AD of radius r has a weight per...Ch. 1.2 - A differential element taken from a curved bar is...Ch. 1.5 - The uniform beam is supported by two rods AB and...Ch. 1.5 - Determine the average normal stress on the cross...Ch. 1.5 - Determine the average normal stress on the cross...Ch. 1.5 - If the 600-kN force acts through the centroid of...Ch. 1.5 - Determine the average normal stress at points A,...Ch. 1.5 - Determine the average normal stress in rod AB if...Ch. 1.5 - A 175-lb woman stands on a vinyl floor wearing...Ch. 1.5 - Determine the largest intensity w of the uniform...Ch. 1.5 - The specimen failed in a tension test at an angle...Ch. 1.5 - The built-up shaft consists of a pipe AB and solid...Ch. 1.5 - If the material fails when the average normal...Ch. 1.5 - If the block is subjected to a centrally applied...Ch. 1.5 - The plate has a width of 0.5 m. If the stress...Ch. 1.5 - The member is subjected to a tensile force of 200...Ch. 1.5 - The boom has a uniform weight of 600 lb and is...Ch. 1.5 - Determine the average normal stress in each of the...Ch. 1.5 - If the average normal stress in each of the...Ch. 1.5 - Determine the maximum average shear stress in pin...Ch. 1.5 - The 150-kg bucket is suspended from end E of the...Ch. 1.5 - The 150-kg bucket is suspended from end E of the...Ch. 1.5 - If the pedestal is subjected to a compressive...Ch. 1.5 - The beam is supported by two rods AB and CD that...Ch. 1.5 - The beam is supported by two rods AB and CD that...Ch. 1.5 - The beam is supported by a pin at B and a short...Ch. 1.5 - The railcar docklight is supported by the...Ch. 1.5 - The plastic block is subjected to an axial...Ch. 1.5 - During a tension test, the wooden specimen is...Ch. 1.5 - The bar has a cross-sectional area of 400(106) m2....Ch. 1.5 - The bar has a cross-sectional area of 400(106) m2....Ch. 1.5 - Prob. 54PCh. 1.5 - The 2-Mg concrete pipe has a center of mass at...Ch. 1.5 - The 2-Mg concrete pipe has a center of mass at...Ch. 1.5 - The pier is made of material having a specific...Ch. 1.5 - Prob. 58PCh. 1.5 - The uniform bar, having a cross-sectional area of...Ch. 1.5 - Prob. 60PCh. 1.5 - Prob. 61PCh. 1.5 - The triangular blocks are glued along each side of...Ch. 1.5 - The triangular blocks are glued along each side of...Ch. 1.5 - Prob. 64PCh. 1.5 - Determine the maximum magnitude P of the load the...Ch. 1.5 - Prob. 66PCh. 1.5 - Prob. 67PCh. 1.7 - Rods AC and BC are used to suspend the 200-kg...Ch. 1.7 - If it is subjected to double shear, determine the...Ch. 1.7 - Determine the maximum average shear stress...Ch. 1.7 - If each of the three nails has a diameter of 4 mm...Ch. 1.7 - The strut is glued to the horizontal member at...Ch. 1.7 - Determine the maximum average shear stress...Ch. 1.7 - If the eyebolt is made of a material having a...Ch. 1.7 - If the bar assembly is made of a material having a...Ch. 1.7 - Determine the maximum force P that can be applied...Ch. 1.7 - The pin is made of a material having a failure...Ch. 1.7 - If the bolt head and the supporting bracket are...Ch. 1.7 - Six nails are used to hold the hanger at A against...Ch. 1.7 - If A and B are both made of wood and are 38 in....Ch. 1.7 - Prob. 70PCh. 1.7 - The connection is made using a bolt and nut and...Ch. 1.7 - Determine the required cross-sectional area of...Ch. 1.7 - Prob. 73PCh. 1.7 - The spring mechanism is used as a shock absorber...Ch. 1.7 - Prob. 75PCh. 1.7 - The hangers support the joist in such a way that...Ch. 1.7 - Prob. 77PCh. 1.7 - Prob. 78PCh. 1.7 - The two aluminum rods AB and BC have diameters of...Ch. 1.7 - The cotter is used to hold the two rods together....Ch. 1.7 - Prob. 81PCh. 1.7 - The 60mm60mm oak post is supported on the pine...Ch. 1.7 - Prob. 83PCh. 1.7 - Prob. 84PCh. 1.7 - The assembly consists of three disks A, B, and C...Ch. 1.7 - Prob. 86PCh. 1.7 - Prob. 87PCh. 1.7 - Prob. 88PCh. 1.7 - Prob. 89PCh. 1.7 - Prob. 90PCh. 1.7 - Prob. 91PCh. 1.7 - Prob. 92PCh. 1.7 - Prob. 93PCh. 1.7 - The aluminum bracket A is used to support the...Ch. 1.7 - If the allowable tensile stress for the bar is...Ch. 1.7 - The bar is connected to the support using a pin...Ch. 1 - The beam AB is pin supported at A and supported by...Ch. 1 - The long bolt passes through the 30-mm-thick...Ch. 1 - Determine the required thickness of member BC to...Ch. 1 - The circular punch B exerts a force of 2 kN on the...Ch. 1 - Determine the average punching shear stress the...Ch. 1 - The 150 mm by 150 mm block of aluminum supports a...Ch. 1 - The yoke-and-rod connection is subjected to a...Ch. 1 - The cable has a specific weight (weight/volume)...
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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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