Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 9.6, Problem 9.161P
To determine
Complete the derivation of equation 9.47 such that express the parallel axis theorem for mass products of inertia.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
A hollow metal sphere of outer radius or = 2 cm is heated internally with a variable output electricheater. The sphere loses heat from its surface by convection and radiation. The heat transfercoefficient is 22 W/ m2 - o C and surface emissivity is 0.92. The ambient fluid temperature is 20 o C andthe surroundings temperature is 14 oC. Construct a graph of the surface temperature corresponding toheating rates ranging from zero to 100 watts. Assume steady state. Use a simplified model forradiation exchange based on a small gray surface enclosed by a much larger surface at 14 o C.
2. A program to make the part depicted in Figure 26.A has been created, presented in figure 26.B,
but some information still needs to be filled in. Compute the tool locations, depths, and other
missing information to present a completed program. (Hint: You may have to look up geometry
for the center drill and standard 0.5000 in twist drill to know the required depth to drill).
Dashed line indicates -
corner of original stock
Intended toolpath-tangent -
arc entry and exit sized to
programmer's judgment
026022 (Slot and Drill Part)
(Setup Instructions.
(UNITS: Inches
(WORKPIECE MAT'L: SAE 1020 STEEL
(Workpiece: 3.25 x 2.00 x0.75 in. Plate
(PRZ Location G54:
( XY 0.0 Upper Left of Fixture
( TOP OF PART 2-0
(Tool List:
)
(
T04
T02 0.500 IN 4 FLUTE FLAT END MILL)
#4 CENTER DRILL
'
T02
0.500 TWIST DRILL
N010 GOO G90 G17 G20 G49 G40 G80 G54
N020 M06 T02 (0.5 IN 4-FLUTE END MILL)
R0.750
N030 S760 M03 G00 x
N040 043 H02 2
Y
(P1)
(RAPID DOWN -TLO)
P4
NO50 MOB (COOLANT ON)
N060 G01 X
R1.000
N070…
6–95. The reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown. If the wood has an allowable bending stress of σallow=1.5 ksi, determine the required minimum thickness t of the rectangular cross section of the tie to the nearest 18 in. Please include all steps. Also if you can, please explain how you found Mmax using an equation rather than using just the moment diagram. Thank you!
Chapter 9 Solutions
Vector Mechanics for Engineers: Statics and Dynamics
Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - Prob. 9.8PCh. 9.1 - 9.9 through 9.11 Determine by direct integration...Ch. 9.1 - 9.9 through 9.11 Determine by direct integration...
Ch. 9.1 - Prob. 9.11PCh. 9.1 - Prob. 9.12PCh. 9.1 - 9.12 through 9.14 Determine by direct integration...Ch. 9.1 - 9.12 through 9.14 Determine by direct integration...Ch. 9.1 - Prob. 9.15PCh. 9.1 - Prob. 9.16PCh. 9.1 - Prob. 9.17PCh. 9.1 - Prob. 9.18PCh. 9.1 - Determine the moment of inertia and the radius of...Ch. 9.1 - Prob. 9.20PCh. 9.1 - Determine the polar moment of inertia and the...Ch. 9.1 - Prob. 9.22PCh. 9.1 - Prob. 9.23PCh. 9.1 - 9.23 and 9.24 Determine the polar moment of...Ch. 9.1 - Prob. 9.25PCh. 9.1 - Prob. 9.26PCh. 9.1 - Prob. 9.27PCh. 9.1 - Prob. 9.28PCh. 9.1 - Prob. 9.29PCh. 9.1 - Prove that the centroidal polar moment of inertia...Ch. 9.2 - 9.31 and 9.32 Determine the moment of inertia and...Ch. 9.2 - 9.31 and 9.32 Determine the moment of inertia and...Ch. 9.2 - 9.33 and 9.34 Determine the moment of inertia and...Ch. 9.2 - 9.33 and 9.34 Determine the moment of inertia and...Ch. 9.2 - Determine the moments of inertia of the shaded...Ch. 9.2 - Determine the moments of inertia of the shaded...Ch. 9.2 - Prob. 9.37PCh. 9.2 - Prob. 9.38PCh. 9.2 - Prob. 9.39PCh. 9.2 - Prob. 9.40PCh. 9.2 - 9.41 through 9.44 Determine the moments of inertia...Ch. 9.2 - 9.41 through 9.44 Determine the moments of inertia...Ch. 9.2 - Prob. 9.43PCh. 9.2 - Prob. 9.44PCh. 9.2 - 9.45 and 9.46 Determine the polar moment of...Ch. 9.2 - Prob. 9.46PCh. 9.2 - 9.47 and 9.48 Determine the polar moment of...Ch. 9.2 - 9.47 and 9.48 Determine the polar moment of...Ch. 9.2 - To form a reinforced box section, two rolled W...Ch. 9.2 - Two channels are welded to a d 12-in. steel plate...Ch. 9.2 - Prob. 9.51PCh. 9.2 - Two 20-mm steel plates are welded to a rolled S...Ch. 9.2 - A channel and a plate are welded together as shown...Ch. 9.2 - Prob. 9.54PCh. 9.2 - Two L76 76 6.4-mm angles are welded to a C250 ...Ch. 9.2 - Prob. 9.56PCh. 9.2 - Prob. 9.57PCh. 9.2 - 9.57 and 9.58 The panel shown forms the end of a...Ch. 9.2 - Prob. 9.59PCh. 9.2 - Prob. 9.60PCh. 9.2 - Prob. 9.61PCh. 9.2 - Prob. 9.62PCh. 9.2 - Prob. 9.63PCh. 9.2 - Prob. 9.64PCh. 9.2 - Prob. 9.65PCh. 9.2 - Prob. 9.66PCh. 9.3 - 9.67 through 9.70 Determine by direct integration...Ch. 9.3 - 9.67 through 9.70 Determine by direct integration...Ch. 9.3 - 9.67 through 9.70 Determine by direct integration...Ch. 9.3 - Prob. 9.70PCh. 9.3 - Prob. 9.71PCh. 9.3 - Prob. 9.72PCh. 9.3 - Prob. 9.73PCh. 9.3 - 9.71 through 9.74 Using the parallel-axis theorem,...Ch. 9.3 - Prob. 9.75PCh. 9.3 - 9.75 through 9.78 Using the parallel-axis theorem,...Ch. 9.3 - Prob. 9.77PCh. 9.3 - Prob. 9.78PCh. 9.3 - Determine for the quarter ellipse of Prob. 9.67...Ch. 9.3 - Determine the moments of inertia and the product...Ch. 9.3 - Determine the moments of inertia and the product...Ch. 9.3 - 9.75 through 9.78 Using the parallel-axis theorem,...Ch. 9.3 - Determine the moments of inertia and the product...Ch. 9.3 - Determine the moments of inertia and the product...Ch. 9.3 - Prob. 9.85PCh. 9.3 - 9.86 through 9.88 For the area indicated,...Ch. 9.3 - Prob. 9.87PCh. 9.3 - Prob. 9.88PCh. 9.3 - Prob. 9.89PCh. 9.3 - 9.89 and 9.90 For the angle cross section...Ch. 9.4 - Using Mohrs circle, determine for the quarter...Ch. 9.4 - Using Mohrs circle, determine the moments of...Ch. 9.4 - Prob. 9.93PCh. 9.4 - Using Mohrs circle, determine the moments of...Ch. 9.4 - Using Mohrs circle, determine the moments of...Ch. 9.4 - Using Mohrs circle, determine the moments of...Ch. 9.4 - For the quarter ellipse of Prob. 9.67, use Mohrs...Ch. 9.4 - 9.98 though 9.102 Using Mohrs circle, determine...Ch. 9.4 - Prob. 9.99PCh. 9.4 - 9.98 though 9.102 Using Mohrs circle, determine...Ch. 9.4 - Prob. 9.101PCh. 9.4 - Prob. 9.102PCh. 9.4 - Prob. 9.103PCh. 9.4 - 9.104 and 9.105 Using Mohrs circle, determine the...Ch. 9.4 - 9.104 and 9.105 Using Mohrs circle, determine the...Ch. 9.4 - For a given area, the moments of inertia with...Ch. 9.4 - it is known that for a given area Iy = 48 106 mm4...Ch. 9.4 - Prob. 9.108PCh. 9.4 - Prob. 9.109PCh. 9.4 - Prob. 9.110PCh. 9.5 - A thin plate with a mass m is cut in the shape of...Ch. 9.5 - A ring with a mass m is cut from a thin uniform...Ch. 9.5 - A thin elliptical plate has a mass m. Determine...Ch. 9.5 - The parabolic spandrel shown was cut from a thin,...Ch. 9.5 - Prob. 9.115PCh. 9.5 - Fig. P9.115 and P9.116 9.116 A piece of thin,...Ch. 9.5 - A thin plate of mass m is cut in the shape of an...Ch. 9.5 - Fig. P9.117 and P9.118 9.118 A thin plate of mass...Ch. 9.5 - Determine by direct integration the mass moment of...Ch. 9.5 - The area shown is revolved about the x axis to...Ch. 9.5 - The area shown is revolved about the x axis to...Ch. 9.5 - Determine by direct integration the mass moment of...Ch. 9.5 - Fig. P9.122 and P9.123 9.123 Determine by direct...Ch. 9.5 - Prob. 9.124PCh. 9.5 - Prob. 9.125PCh. 9.5 - Prob. 9.126PCh. 9.5 - Prob. 9.127PCh. 9.5 - Prob. 9.128PCh. 9.5 - Prob. 9.129PCh. 9.5 - Knowing that the thin cylindrical shell shown has...Ch. 9.5 - A circular hole of radius r is to be drilled...Ch. 9.5 - The cups and the arms of an anemometer are...Ch. 9.5 - Prob. 9.133PCh. 9.5 - Determine the mass moment of inertia of the 0.9-lb...Ch. 9.5 - Prob. 9.135PCh. 9.5 - Prob. 9.136PCh. 9.5 - A 2-mm thick piece of sheet steel is cut and bent...Ch. 9.5 - A section of sheet steel 0.03 in. thick is cut and...Ch. 9.5 - A corner reflector for tracking by radar has two...Ch. 9.5 - A farmer constructs a trough by welding a...Ch. 9.5 - The machine element shown is fabricated from...Ch. 9.5 - Determine the mass moments of inertia and the...Ch. 9.5 - Determine the mass moment of inertia of the steel...Ch. 9.5 - Prob. 9.144PCh. 9.5 - Determine the mass moment of inertia of the steel...Ch. 9.5 - Aluminum wire with a weight per unit length of...Ch. 9.5 - The figure shown is formed of 18-in.-diameter...Ch. 9.5 - A homogeneous wire with a mass per unit length of...Ch. 9.6 - Determine the mass products of inertia Ixy, Iyz,...Ch. 9.6 - Determine the mass products of inertia Ixy, Iyz,...Ch. 9.6 - Determine the mass products of inertia Ixy, Iyz,...Ch. 9.6 - Determine the mass products of inertia Ixy, Iyz,...Ch. 9.6 - 9.153 through 9.156 A section of sheet steel 2 mm...Ch. 9.6 - Prob. 9.154PCh. 9.6 - Prob. 9.155PCh. 9.6 - 9.153 through 9.156 A section of sheet steel 2 mm...Ch. 9.6 - Prob. 9.157PCh. 9.6 - Prob. 9.158PCh. 9.6 - Prob. 9.159PCh. 9.6 - Prob. 9.160PCh. 9.6 - Prob. 9.161PCh. 9.6 - For the homogeneous tetrahedron of mass m shown,...Ch. 9.6 - Prob. 9.163PCh. 9.6 - Prob. 9.164PCh. 9.6 - Prob. 9.165PCh. 9.6 - Determine the mass moment of inertia of the steel...Ch. 9.6 - Prob. 9.167PCh. 9.6 - Prob. 9.168PCh. 9.6 - Prob. 9.169PCh. 9.6 - 9.170 through 9.172 For the wire figure of the...Ch. 9.6 - Prob. 9.171PCh. 9.6 - Prob. 9.172PCh. 9.6 - Prob. 9.173PCh. 9.6 - Prob. 9.174PCh. 9.6 - Prob. 9.175PCh. 9.6 - Prob. 9.176PCh. 9.6 - Prob. 9.177PCh. 9.6 - Prob. 9.178PCh. 9.6 - Prob. 9.179PCh. 9.6 - Prob. 9.180PCh. 9.6 - Prob. 9.181PCh. 9.6 - Prob. 9.182PCh. 9.6 - Prob. 9.183PCh. 9.6 - Prob. 9.184PCh. 9 - Determine by direct integration the moments of...Ch. 9 - Determine the moment of inertia and the radius of...Ch. 9 - Prob. 9.187RPCh. 9 - Prob. 9.188RPCh. 9 - Prob. 9.189RPCh. 9 - Two L4 4 12-in. angles are welded to a steel...Ch. 9 - Prob. 9.191RPCh. 9 - Prob. 9.192RPCh. 9 - Prob. 9.193RPCh. 9 - Prob. 9.194RPCh. 9 - Prob. 9.195RPCh. 9 - Determine the mass moment of inertia of the steel...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- 6–53. If the moment acting on the cross section is M=600 N⋅m, determine the resultant force the bending stress produces on the top board. Please explain each step. Please explain how you got the numbers and where you plugged them in to solve the problem. Thank you!arrow_forwardSolving coplanar forcesarrow_forwardComplete the following problems. Show your work/calculations, save as.pdf and upload to the assignment in Blackboard. 1. What are the x and y dimensions for the center position of holes 1,2, and 3 in the part shown in Figure 26.2 (below)? 6.0000 7118 Zero reference point 1.0005 1.0000 1.252 Bore C' bore 1.250 6.0000 .7118 0.2180 deep (3 holes) 2.6563 1.9445 3.000 diam. slot 0.3000 deep. 0.3000 wide 2.6563 1.9445arrow_forward
- Complete the following problems. Show your work/calculations, save as.pdf and upload to the assignment in Blackboard. missing information to present a completed program. (Hint: You may have to look up geometry for the center drill and standard 0.5000 in twist drill to know the required depth to drill). 1. What are the x and y dimensions for the center position of holes 1,2, and 3 in the part shown in Figure 26.2 (below)? 6.0000 Zero reference point 7118 1.0005 1.0000 1.252 Bore 6.0000 .7118 Cbore 0.2180 deep (3 holes) 2.6563 1.9445 Figure 26.2 026022 (8lot and Drill Part) (Setup Instructions--- (UNITS: Inches (WORKPIECE NAT'L SAE 1020 STEEL (Workpiece: 3.25 x 2.00 x0.75 in. Plate (PRZ Location 054: ' XY 0.0 - Upper Left of Fixture TOP OF PART 2-0 (Tool List ( T02 0.500 IN 4 FLUTE FLAT END MILL #4 CENTER DRILL Dashed line indicates- corner of original stock ( T04 T02 3.000 diam. slot 0.3000 deep. 0.3000 wide Intended toolpath-tangent- arc entry and exit sized to programmer's judgment…arrow_forwardA program to make the part depicted in Figure 26.A has been created, presented in figure 26.B, but some information still needs to be filled in. Compute the tool locations, depths, and other missing information to present a completed program. (Hint: You may have to look up geometry for the center drill and standard 0.5000 in twist drill to know the required depth to drill).arrow_forwardWe consider a laminar flow induced by an impulsively started infinite flat plate. The y-axis is normal to the plate. The x- and z-axes form a plane parallel to the plate. The plate is defined by y = 0. For time t <0, the plate and the flow are at rest. For t≥0, the velocity of the plate is parallel to the 2-coordinate; its value is constant and equal to uw. At infinity, the flow is at rest. The flow induced by the motion of the plate is independent of z. (a) From the continuity equation, show that v=0 everywhere in the flow and the resulting momentum equation is მu Ət Note that this equation has the form of a diffusion equation (the same form as the heat equation). (b) We introduce the new variables T, Y and U such that T=kt, Y=k/2y, U = u where k is an arbitrary constant. In the new system of variables, the solution is U(Y,T). The solution U(Y,T) is expressed by a function of Y and T and the solution u(y, t) is expressed by a function of y and t. Show that the functions are identical.…arrow_forward
- Part A: Suppose you wanted to drill a 1.5 in diameter hole through a piece of 1020 cold-rolled steel that is 2 in thick, using an HSS twist drill. What values if feed and cutting speed will you specify, along with an appropriate allowance? Part B: How much time will be required to drill the hole in the previous problem using the HSS drill?arrow_forward1.1 m 1.3 m B 60-mm diameter Brass 40-mm diameter Aluminum PROBLEM 2.52 - A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of brass (E₁ = 105 GPa, α = 20.9×10°/°C) and portion BC is made of aluminum (Ę₁ =72 GPa, α = 23.9×10/°C). Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 42°C, (b) the corresponding deflection of point B.arrow_forward30 mm D = 40 MPa -30 mm B C 80 MPa PROBLEM 2.69 A 30-mm square was scribed on the side of a large steel pressure vessel. After pressurization, the biaxial stress condition at the square is as shown. For E = 200 GPa and v=0.30, determine the change in length of (a) side AB, (b) side BC, (c) diagnonal AC.arrow_forward
- Please solve in detail this problem thank youarrow_forward0,5 mm 450 mm 350 mm Bronze A = 1500 mm² E = 105 GPa प 21.6 × 10-PC Aluminum A = 1800 mm² £ = 73 GPa = a 23.2 × 10-PC PROBLEM 2.58 Knowing that a 0.5-mm gap exists when the temperature is 24°C, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to -75 MPa, (b) the corresponding exact length of the aluminum bar.arrow_forward0.5 mm 450 mm -350 mm Bronze Aluminum A 1500 mm² A 1800 mm² E 105 GPa E 73 GPa K = 21.6 X 10 G < = 23.2 × 10-G PROBLEM 2.59 Determine (a) the compressive force in the bars shown after a temperature rise of 82°C, (b) the corresponding change in length of the bronze bar.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
International Edition---engineering Mechanics: St...Mechanical EngineeringISBN:9781305501607Author:Andrew Pytel And Jaan KiusalaasPublisher:CENGAGE L
International Edition---engineering Mechanics: St...
Mechanical Engineering
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:CENGAGE L
moment of inertia; Author: NCERT OFFICIAL;https://www.youtube.com/watch?v=A4KhJYrt4-s;License: Standard YouTube License, CC-BY