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Lone Star College System, Woodlands *

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MECE3360

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Mechanical Engineering

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Oct 30, 2023

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pdf

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5

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MECE 3369 Solid Mechanics Solution to Assignment #1: Equilibrium (Statics), 1D Bars, and Indicial Notations 1. Support Reactions: We will consider segment AE of the boom, so we must first determine the pin reactions at A . Since member CD is a two-force member, it acts like a cable and therefore exerts a force F CD having a known direction. The free-body diagram of the boom is shown in Fig. 1a. Applying the equations of equilibrium, Figure : 1a (Anti-clockwise is +ve) Σ M A = 0; F CD (3/5)(2 m)–(5000 N)(3 m) = 0 = > F CD = 12500 N (+ve x-direction is +ve) Σ F x = 0; A x – (12500 N)(4/5) = 0 = > A x = 10000 N (+ve y-direction is +ve) Σ F y = 0; A y + (12500 N)(3/5) - (5000 N) = 0 = > A y = 2500 N Free-Body Diagram: The free-body diagram of segment AE is shown in Fig. 1b Figure : 1b Equations of Equilibrium (+ve x-direction is +ve) Σ F x = 0; N E + (10000 N) = 0 = > N E = -10000 N = -10 kN ( Ans. ) (+ve y-direction is +ve) Σ F y = 0; V E + (2500 N) = 0 = > V E = -2500 N = -2.5 kN ( Ans. ) (Anti-clockwise is +ve) Σ M E = 0; M E +(2500 N)(1 m) = 0 = > M E = -2500 N.m = -2.5 kN.m ( Ans. ) 2 points 2 points 1 point 1 point 2 points 2 points
2. Gravity-loaded bar. The end deflection will be given by- = ! L 0 R ( x ) A ( x ) E ( x ) dx = 1 AE ! L 0 R ( x ) dx Figure : 2a To determine the internal force, equilibrium is to be applied. Using the equilibrium equation dR ( x ) dx + b = 0 and integrating it, one can get R(x) . Another approach is making section cuts and explicitly summing forces in the x -direction. Let’s make a section cut at an arbitrary location x and sum the forces on the lower section, as shown in Fig. 2a. This tells us that R(x) = (L-x)A γ . Inserting into the previous equation gives- = 1 AE ! L 0 R ( x ) dx = γ L 2 2 E 3 points 3 points 4 points
3. Apply equilibrium at sections just right of point B and just left of point B. (a) Just right of point B: Figure : 3a Σ F x = 0; (+ve x-direction is +ve) = > R P (1 a/L ) = 0 = > R = P (1 a/L ) (b) Just left of point B: Figure : 3b Σ F x = 0; (+ve x-direction is +ve) = > R + P P (1 a/L ) = 0 = > R = P ( a/L ) Correction: This should be R = P ( a/L ) With the strain field (shown in the figure), we can compute the displacement as: u ( L ) u (0) = " L 0 R ( X ) AE d X = P (1 a L ) a AE + Pa/L AE ( L a ) = 0 which indicates that under these forces the bar does not undergo any net deformation! So, the internal force diagram, axial stress diagram and axial strain diagram turn out to be as shown in the figure below. 3 points 3 points 2 points
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