Vector Mechanics for Engineers: Statics and Dynamics
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
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 9.5, Problem 9.135P
To determine

Find the mass moment of inertia of the component with respect to x,y,z coordinate axes.

Expert Solution & Answer
Check Mark

Answer to Problem 9.135P

The mass moment of inertia of the component with respect to x coordinate axes is 7.11×103kgm2_

The mass moment of inertia of the component with respect to y coordinate axes is 16.96×103kgm2_

The mass moment of inertia of the component with respect to z coordinate axes is 15.27×103kgm2_.

Explanation of Solution

Given information:

The thickness (t) of sheet steel is 2mm.

The density ρ of steel is 7,850kg/m3.

Calculation:

Sketch the sheet steel as shown in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 9.5, Problem 9.135P

Find the mass of rectangular section 1 as shown in below:

m1=ρV1 (1)

Here, V1 is volume of rectangular section 1.

Express the volume of rectangular section 1 as follows:

V1=lwt

Substitute 120mm for l, 200mm for w, 7,850kg/m3 for ρ, and 2mm for t in Equation (1).

m1=7,850×[120mm(1m103m)×200mm(1m103m)×2mm(1m103m)]=7,850[0.120×0.200×0.002]=7,850[0.000048]=0.3768kg

Find the mass of rectangular section 2 as shown in below:

m2=ρV2 (2)

Here, V2 is volume of rectangular section 2.

Express the volume of rectangular section 2 as follows:

V2=lwt

Substitute 120mm for l, 100mm for w, 7,850kg/m3 for ρ, and 2mm for t in Equation (2).

m2=7,850×[120mm(1m103m)×100mm(1m103m)×2mm(1m103m)]=7,850[0.120×0.100×0.002]=7,850[0.000024]=0.1884kg

Panel 1:

Find the moment of inertia about x axis for panel 1 as shown below:

Ix=I¯x+md2=112(m1h2)+m1(d2) (3)

Substitute 0.3768kg for m1, 120mm for h, and (10+60)mm for d in Equation (3).

Ix=112(0.3768)(120mm×1m103mm)2+(0.3768)((100mm×1m103mm)2+(60mm×1m103mm)2)=112(0.3768)(0.12)2+(0.3768)(0.12+0.062)=0.00045216+0.00512448=5.5766×103

=5.577×103kgm2

Find the moment of inertia about y axis for panel 1 as shown below:

Iy=I¯y+md2=112(m1h2)+m1(d2) (4)

Substitute 0.3768kg for m1, 200mm for h, and (100+100)mm for d in Equation (4).

Iy=112(0.3768)(200mm×1m103mm)2+(0.3768)(100mm×1m103mm+100mm×1m103mm)=112(0.3768)(0.2)2+(0.3768)(0.12+0.12)=0.001256+0.007536=8.792×103kgm2

Find the moment of inertia about z axis for panel 1 as shown below:

Iz=I¯z+md2 (5)

Substitute 0.3768kg for m1, 112(0.3768[0.22+0.122]) for I¯z, and (100+60)mm for d in Equation (5).

Iz=112(0.3768[0.22+0.122])+0.3768[(100mm(1m103mm)2+60mm(1m103mm)2)]=0.00170816+0.005124=6.833×103kgm2

Panel 2:

Find the moment of inertia about x axis for panel 2 as shown below:

Ix=I¯x+md2 (6)

Substitute 0.1884kg for m2, 112(0.1884)[0.122+0.12] for I¯x, and (50+60)mm for d in Equation (6).

Ix=112(0.1884)[0.122+0.12]+(0.1884)((50mm×1m103mm)2+(60mm×1m103mm)2)=112(0.1884)[0.122+0.12]+(0.1884)(0.052+0.062)=3.8308×104+0.00114924=1.532×103kgm2

Find the moment of inertia about y axis for panel 2 as shown below:

Iy=I¯y+md2=112(m2h2)+m2(d2) (7)

Substitute 0.1884kg for m2, 100mm for h, and (5+200)mm for d in Equation (7).

Iy=112(0.1884)(100mm×1m103mm)2+(0.1884)5(50mm×1m103mm+200mm×1m103mm)=112(0.1884)(0.1)2+(0.1884)(0.052+0.22)=0.000157+0.00800=8.164×103kgm2

Find the moment of inertia about z axis for panel 2 as shown below:

Iz=I¯z+md2=112(m2h2)+m2(d2) (8)

Substitute 0.1884kg for m2, 120mm for h, and (6+200)mm for d in Equation (8).

Iz=112(0.1884)(120mm×1m103mm)2+(0.1884)5(60mm×1m103mm+200mm×1m103mm)=112(0.1884)(0.12)2+(0.1884)(0.062+0.22)=0.00022608+0.008214=8.44×103kgm2

Find the total mass of inertia (Ix) about x axis as shown below:

Ix=(Ix)1+(Ix)2 (9)

Here, (Ix)1 is moment of inertia about x panel 1 and (Ix)2 is moment of inertia about x panel 2.

Substitute 5.577×103kgm2 for (Ix)1 and 1.532×103kgm2 for (Ix)2 in Equation (9).

Ix=5.577×103+1.532×103=7.11×103kgm2

Thus, the mass moment of inertia of the component with respect to x coordinate axes is 7.11×103kgm2_

Find the total mass of inertia (Iy) about y axis as shown below:

Iy=(Iy)1+(Iy)2 (10)

Here, (Iy)1 is moment of inertia about x panel 1 and (Iy)2 is moment of inertia about x panel 2.

Substitute 8.792×103kgm2 for  (Iy)1 and 8.164×103kgm2 for (Iy)2 in Equation (10).

Iy=8.792×103+8.164×103=16.96×103kgm2

Thus, the mass moment of inertia of the component with respect to y coordinate axes is 16.96×103kgm2_

Find the total mass of inertia (Iz) about z axis as shown below:

Iz=(Iz)1+(Iz)2 (11)

Here, (Iz)1 is moment of inertia about x panel 2 and (Iz)2 is moment of inertia about x panel 2.

Substitute 6.833×103kgm2 for (Iz)1 and 8.440×103kgm2 for (Iz)2 in Equation (11).

Iy=6.833×103+8.440×103=15.27×103kgm2

Thus, the mass moment of inertia of the component with respect to z coordinate axes is 15.27×103kgm2_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Continuity equation A y x dx D T معادلة الاستمرارية Ly X Q/Prove that ди хе + ♥+ ㅇ? he me ze ོ༞“༠ ?
Q Derive (continuity equation)? I want to derive clear mathematics.
motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Figure. Gear B transfers 125 W of power to operating machinery in the factory, and the remaining power in the shaft is mansferred by gear D. Shafts (1) and (2) are solid aluminum (G = 28 GPa) shafts that have the same diameter and an allowable shear stress of t= 40 MPa. Shaft (3) is a solid steel (G = 80 GPa) shaft with an allowable shear stress of t = 55 MPa. Determine: a) the minimum permissible diameter for aluminum shafts (1) and (2) b) the minimum permissible diameter for steel shaft (3). c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b).

Chapter 9 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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
Background pattern image
Mechanical Engineering
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
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
moment of inertia; Author: NCERT OFFICIAL;https://www.youtube.com/watch?v=A4KhJYrt4-s;License: Standard YouTube License, CC-BY