Explanation Given: The width of the triangular area is 9 in . The height of the triangular area is 6 in . Show the free body diagram of the triangular area as in Figure 1. From the Figure 1, Compute the moment of inertia of the triangular area about the x axes using the parallel-axis theorem. I x = I ¯ x ′ + A d y 2 = 1 36 b a 3 + [ 1 2 b a ] [ 1 3 a ] 2 (I) Here, the moment of inertia for the beam’s cross-sectional area about the x axis is I x , the first integral of the moment of inertia of the area about the centroidal axis in x direction is I ¯ x ′ , half of the distance from the centroid is d y , and an area is A . Compute the moment of inertia of the triangular area about the y axes using the parallel-axis theorem. I y = I ¯ y ′ + A d x 2 = 1 36 a b 3 + [ 1 2 a b ] [ 2 3 b ] 2 (II) Here, the moment of inertia for the beam’s cross-sectional area about the y axis is I y , half of the distance from the centroid is d x , and first integral of the moment of inertia of the area about the centroidal axis in y direction is I ¯ y ′ . Calculate the moment of inertia of the triangular area about the x and y axes. I x y = b 2 a 2 8 (III) Calculate the maximum and minimum principal of moments. I max / min = I x + I y 2 ± [ ( I x − I y ) / 2 ] 2 + ( I x y ) 2 (IV) Compute the orientation of the principal axes. tan 2 θ p 2 = − 2 I x y I x − I y (V) Compute the orientation of the principal axes. θ p 1 = θ p 2 − 90 ° (VI) Conclusion: Substitute 6 in . for a and 9 in . for b in Equation (I). I x = 1 36 ( 9 ) ( 6 ) 3 + [ 1 2 ( 9 ) ( 6 ) ] [ 1 3 ( 6 ) ] 2 = 54 + 108 = 162 in . 4 Substitute 6 in . for a and 9 in . for b in Equation (II). I y = 1 36 ( 6 ) ( 9 ) 3 + [ 1 2 ( 6 ) ( 9 ) ] [ 2 3 ( 9 ) ] 2 = 121

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 72P

To determine

The direction (θp1,θp2) of the principal axes having an origin at point O, and the principal moments of inertia (Imax,Imin) for the triangular area about the axes.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.7, Problem 71P

Chapter 10.7, Problem 73P

Students have asked these similar questions

I had a theoretical question about attitude determination. In the attached images, I gave two axis and angles. The coefficient of the axes are the same and the angles are the same. The only difference is the vector basis. Lets say there is a rotation going from n hat to b hat. Then, you introduce a intermediate rotation s hat. So, I want to know if the DCM produced from both axis and angles will be the same or not. Does the vector basis affect the numerical value of the DCM? The DCM formula only cares about the coefficient of the axis and the angle. So, they should be the same right?

3-15. A small fixed tube is shaped in the form of a vertical helix of radius a and helix angle y, that is, the tube always makes an angle y with the horizontal. A particle of mass m slides down the tube under the action of gravity. If there is a coefficient of friction μ between the tube and the particle, what is the steady-state speed of the particle? Let y γ 30° and assume that µ < 1/√3.

The plate is moving at 0.6 mm/s when the force applied to the plate is 4mN. If the surface area of the plate in contact with the liquid is 0.5 m^2, deterimine the approximate viscosity of the liquid, assuming that the velocity distribution is linear.

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 direction ( θ p 1 , θ p 2 ) of the principal axes having an origin at point O , and the principal moments of inertia ( I max , I min ) for the triangular area about the axes.

Solution Summary: The author explains the moment of inertia of the triangular area about the x axes using the parallel-axis theorem.

Concept explainers

Videos

The direction (θp1,θp2) of the principal axes having an origin at point O, and the principal moments of inertia (Imax,Imin) for the triangular area about the axes.

Want to see the full answer?

Chapter 10 Solutions