Explanation Given: The total height of the beam's cross-sectional area is 200 mm . The total width of the beam's cross-sectional area is 200 mm . Explanation: To calculate the moment of inertia for the beam’s cross-sectional area about the x axis, subdivide the cross-sectional area into 3 segments as follows: Part 1: Rectangle Part 2: Rectangle Part 3: Rectangle Show the subdivided beam's cross-sectional area along with their respective centroids as in Figure 1: From Figure (1), Write the height and the width of the rectangle in part 1: b 1 = 3 0 mm h 1 = 17 0 mm Here, the height of rectangle in part 1 is h 1 and the width of rectangle in part 1 is b 1 . Write the height and width of the rectangle in part 2: b 2 = 17 0 mm h 2 = 3 0 mm Here, the height of the rectangle in part 2 is h 2 and the width of the rectangle in part 2 is b 2 . Write the height and width of the rectangle in part 3: b 3 = 3 0 mm h 3 = 10 0 mm Here, the height of the rectangle in part 3 is h 3 and the width of the rectangle in part 3 is b 3 . Calculations: Calculate the distance between centroid and y axis about the x axis for rectangle in part 1: d x 1 = h 2 + h 1 2 (I) Substitute 30 mm for h 2 and 170 mm for h 1 in Equation (I). d x 1 = 30 + 170 2 = 115 mm Calculate the distance between centroid and y axis about the axis for the rectangle in part 2: d x 2 = h 2 2 (II) Substitute 30 mm for h 2 in Equation (II). d x 2 = 30 2 = 15 mm Calculate the distance between centroid and y axis about the x axis for rectangle in part 3: d x 3 = h 3 2 (III) Substitute 100 mm for h 3 in Equation (III). d x 3 = 100 2 = 50 mm Conclusion: Apply parallel axis theorem in Figure 1: I y = ∑ I ¯ y ′ + A d x 2 = ( I ¯ y ′ + A d x 2 ) 1 + ( I ¯ y ′ + A d x 2 ) 2 + ( I ¯ y ′ + A d x 2 ) 3 =

INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

14th Edition

ISBN: 9780133918922

Author: Russell C. Hibbeler

Publisher: PEARSON

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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…

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Chapter 10.4, Problem 43P

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The moment of inertia of the beam's cross-sectional area about the y axis.

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Chapter 10.4, Problem 42P

Chapter 10.4, Problem 44P

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Sign in PDF Lecture W09.pdf PDF MMB241 - Tutorial L9.pdf File C:/Users/KHULEKANI/Desktop/mmb241/MMB241%20-%20Tutorial%20L9.pdf II! Draw | I│Alla | Ask Copilot + of 4 D Topic: Kinetics of Particles: - Forces in dynamic system, Free body diagram, newton's laws of motion, and equations of motion. TQ1. The 10-kg block is subjected to the forces shown. In each case, determine its velocity when t=2s if v 0 when t=0 500 N F = (201) N 300 N (b) TQ2. The 10-kg block is subjected to the forces shown. In each case, determine its velocity at s-8 m if v = 3 m/s at s=0. Motion occurs to the right. 40 N F = (2.5 s) N 200 N 30 N (b) TQ3. Determine the initial acceleration of the 10-kg smooth collar. The spring has an unstretched length of 1 m. 1 σ Q ☆ Q 6 ا الى ☑

Sign in PDF Lecture W09.pdf PDF MMB241 - Tutorial L9.pdf File C:/Users/KHULEKANI/Desktop/mmb241/MMB241%20-%20Tutorial%20L9.pdf II! Draw | I│Alla | Ask Copilot + 4 of 4 | D TQ9. If motor M exerts a force of F (10t 2 + 100) N determine the velocity of the 25-kg crate when t kinetic friction between the crate and the plane are μs The crate is initially at rest. on the cable, where t is in seconds, 4s. The coefficients of static and 0.3 and μk = 0.25, respectively. M 3 TQ10. The spring has a stiffness k = 200 N/m and is unstretched when the 25-kg block is at A. Determine the acceleration of the block when s = 0.4 m. The contact surface between the block and the plane is smooth. 0.3 m F= 100 N F= 100 N k = 200 N/m σ Q Q ☆ ا الى 6 ☑

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INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

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