Explanation Given: The width of the rectangular area is 120 mm . The height of the rectangular area is 30 mm . The angle is 30 ° . Show the beam’s rectangular area as in Figure (1). From Figure 1, As the rectangular area is symmetrical about the x and y axes, write the product of inertia for the rectangular area with respect to the x and y axes as 0 . I x y = 0 Compute the moment of inertia for the rectangular area with respect to the x -axis. I x = 1 12 a b 3 (I) Compute the moment of inertia for the rectangular area with respect to the y -axis. I y = 1 12 b a 3 (II) Compute the moment of inertia for the rectangular area with respect to the u -axis. I u = [ I x + I y 2 ] + [ I x − I y 2 ] cos 2 θ − I x y sin 2 θ (III) Compute the moment of inertia for the rectangular area with respect to the v -axis. I v = [ I x + I y 2 ] − [ I x − I y 2 ] cos 2 θ + I x y sin 2 θ (IV) Compute the moment of inertia for the rectangular area with respect to the u and v axes. I u v = [ I x − I y 2 ] sin 2 θ + I x y cos 2 θ (V) Conclusion: Substitute 120 mm for a and 30 mm for b in Equation (I). I x = 1 12 ( 120 ) ( 30 ) 3 = 324 ( 10 4 ) 12 = 0.27 ( 10 6 ) mm 4 Substitute 120 mm for a and 30 mm for b in Equation (II). I y = 1 12 ( 30 ) ( 120 ) 3 = 5184 ( 10 4 ) 12 = 4.32 ( 10 6 ) mm 4 Substitute 0.27 ( 10 6 ) mm 4 for I x , 4.32 ( 10 6 ) mm 4 for I y , 0 for I x y , and 30 ° for θ in Equation (III). I u = { [ 0.270 + 4.32 2 ] + [ 0.270 − 4.32 2 ] cos 2 ( 30 ° ) − ( 0 ) sin 2 ( 30 ° ) } ( 10 6 ) = { [ 0.270 + 4.32 2 ] + [ 0.270 − 4.32 2 ] ( cos 60 ° ) − 0 } ( 10 6 ) = [ 2

14th Edition

ISBN: 9780136522409

Author: HIBBELER

Publisher: PEARSON

See similar textbooks

Concept explainers

Center of Gravity and Centroid

The term center of gravity indicates the position or location of a point about which the whole weight of a body or system is supposed to be concentrated, which means the whole weight of the body or system works through this point. It is closely related t…

Videos

Question

Chapter 10.7, Problem 70P

To determine

The moments of inertia Iu, Iv and the product of inertia Iuv for the rectangular area.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.7, Problem 69P

Chapter 10.7, Problem 71P

Students have asked these similar questions

The design of the gear-and-shaft system shown requires that steel shafts of the same diameter be used for both AB and CD. It is further required that the angle D through which end D of shaft CD rotates not exceed 1.5°. Knowing that G = 77.2 GPa, determine the required diameter of the shafts. 40 mm 400 mm 100 mm 600 mm T-1000 N-m D

Assume a Space Launch System (Figure 1(a)) that is approximated as a cantilever undamped single degree of freedom (SDOF) system with a mass at its free end (Figure 1(b)). The cantilever is assumed to be massless. Assume a wind load that is approximated with a concentrated harmonic forcing function p(t) = posin(ωt) acting on the mass. The known properties of the SDOF and the applied forcing function are given below. • Mass of SDOF: m =120 kip/g • Acceleration of gravity: g = 386 in/sec2 • Bending sectional stiffness of SDOF: EI = 1015 lbf×in2 • Height of SDOF: h = 2000 inches • Amplitude of forcing function: po = 6 kip • Forcing frequency: f = 8 Hz

13.44 The end of a cylindrical liquid cryogenic propellant tank in free space is to be protected from external (solar) radiation by placing a thin metallic shield in front of the tank. Assume the view factor Fts between the tank and the shield is unity; all surfaces are diffuse and gray, and the surroundings are at 0 K. Tank T₁ Shield, T T₁ = 100 K E1 Solar irradiation Gs ε₁ = ε₂ = 0.05 ε₁ = 0.10 Gs = 1250 W/m² E2 Find the temperature of the shield T, and the heat flux (W/m²) to the end of the tank.

Chapter 10 Solutions

ENGINEERING MECHANICS

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 moments of inertia I u , I v and the product of inertia I u v for the rectangular area.

Solution Summary: The author explains the moments of inertia I_u,

Concept explainers

Videos

The moments of inertia Iu, Iv and the product of inertia Iuv for the rectangular area.

Want to see the full answer?

Chapter 10 Solutions