Explanation Given: The width of the shaded area is 4 in . The height of the shaded area is 6 in . The radius is 1 in . Show the free body diagram of the shaded area as in Figure 1. Calculation: From the Figure 1, Compute the product of inertia for the shaded area with respect to the x axes. I x = 1 3 ( c + d ) ( a + b ) 3 − [ π r 4 4 + π r 2 a 2 ] (I) Here, the radius is r . Compute the product of inertia for the shaded area with respect to the y axes. I y = 1 3 ( a + b ) ( c + d ) 3 − [ π r 4 4 + π r 2 d 2 ] (II) Compute the product of inertia for the shaded area with respect to the x and y axes. I x y = [ a + b 2 ] [ c + d 2 ] ( a + b ) ( c + d ) − d a π r 2 (III) Compute the radius. R = [ I x − ( I x + I y ) / 2 ] 2 + ( I x y ) 2 (IV) Calculate the maximum principle moment of inertia. I max = I x + I y 2 + R (V) Calculate the minimum principle moment of inertia. I min = I x + I y 2 − R (VI) Calculate the direction of principal axes. tan 2 θ p 1 = − I x y ( I x − I y ) / 2 (VII) Calculate the direction of principal axes. θ p 2 = 90 ° + θ P 1 (VIII) Conclusion: Substitute 2 in . for c , 2 in . for d , 4 in . for a , 2 in . for b , and 1 in . for r in Equation (I). I x = 1 3 ( 2 + 2 ) ( 4 + 2 ) 3 − [ π ( 1 ) 4 4 + π ( 1 ) 2 ( 4 ) 2 ] = 288 − 51.050 = 236.95 in . 4 Substitute 2 in . for c , 2 in . for d , 4 in . for a , 2 in . for b , and 1 in . for r in Equation (II). I y = 1 3 ( 4 + 2 ) ( 2 + 2 ) 3 − [ π ( 1 ) 4 4 + π ( 1 ) 2 ( 2 ) 2 ] = 128 − 13.351 = 114.65 in . 4 Substitute 2 in . for c , 2 in . for d , 4 in . for a , 2 in . for b , and 1 in . for r in Equation (III). I x y = [ 4 + 2 2 ] [ 2 + 2 2 ] ( 4 + 2 ) ( 2 + 2 ) − ( 2 ) ( 4 ) π ( 1 ) 2 = ( 3 ) ( 2 ) ( 6 ) ( 4 ) − 25

Engineering Mechanics: Statics Plus Mastering Engineering with Pearson eText -- Access Card Package (14th Edition) (Hibbeler, The Engineering Mechanics: Statics & Dynamics Series, 14th Edition)

14th Edition

ISBN: 9780134160689

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Concept explainers

Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

Videos

Question

Chapter 10.7, Problem 82P

To determine

The directions (θp1,θp2) of the principal axes with origin located at point O, and the principal moments of inertia (Imax,Imin) for the area about these axes.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.7, Problem 81P

Chapter 10.7, Problem 83P

Students have asked these similar questions

PROBLEM 10: A sleeve in the form of a circular tube of length L is Nut placed around a bolt and fitted between washers at each end. The nut is then turned until it is just snug. Use material properties as follows: For the sleeve, as = 21 x 106/°C and Es = 100 GPa Washer Bolt ·L· Sleeve Bolt head For the bolt, αB = 10 × 10-6/°C and EB = 200 GPa. 1. Calculate the temperature rise that is required to produce a compressive stress of 25 MPa in the sleeve.

This problem illustrates that the factor of safety for a machine element depends on the particular point selected for analysis. Here you are to compute factors of safety, based upon the distortion-energy theory, for stress elements at A and B of the member shown in the figure. This bar is made of AISI 1006 cold-drawn steel and is loaded by the forces F = 1.100 kN, P = 8.00 kN, and T = 50.00 N·m. Given: Sy = 280 MPa. B -100 mm- 15-mm D. a) What is the value of the axial stress at point A? b)What is the value of the shear stress at point A? c)Determine the value of the Von Mises stress at point A. P F

The three steel wires, each of cross-sectional area 0.05 in2, support the weight W. Theirunstressed lengths are 74.98 ft, 74.99 ft, and 75.00 ft. Use E = 29 x 106 psi.1. Find the stress (psi) in the longest wire if W = 1500 lb.2. Determine the stress in the shortest wire if W = 500 lb ANSWERS: 6130 psi; 6930 psi

Chapter 10 Solutions

Engineering Mechanics: Statics Plus Mastering Engineering with Pearson eText -- Access Card Package (14th Edition) (Hibbeler, The Engineering Mechanics: Statics & Dynamics Series, 14th Edition)

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.

The directions ( θ p 1 , θ p 2 ) of the principal axes with origin located at point O , and the principal moments of inertia ( I max , I min ) for the area about these axes.

Solution Summary: The author calculates the product of inertia for the shaded area with respect to the x axes.

Concept explainers

Videos

The directions (θp1,θp2) of the principal axes with origin located at point O, and the principal moments of inertia (Imax,Imin) for the area about these axes.

Want to see the full answer?

Chapter 10 Solutions