Chapter 5 Equilibrium of a Rigid Body MCE111 PDF

Mechanics (MCE111) Chapter 5 Equilibrium of a Rigid Body 1 Simple Distributed Load Sometimes, a body may be subjected to a loading that is distributed over its surface. For example, the pressure of...

Mechanics (MCE111) Chapter 5 Equilibrium of a Rigid Body 1 Simple Distributed Load Sometimes, a body may be subjected to a loading that is distributed over its surface. For example, the pressure of the wind on the face of a sign, the pressure of water within a tank, or the weight of sand on the floor of a storage container, are all distributed loadings. The pressure exerted at each point on the surface indicates the intensity of the loading. It is measured using pascals Pa (or N/m2) in SI units or lb/ft2 in the U.S. Customary system. 2 1  Simple Distributed Load Consider the beam (or plate) in Fig. a that has a constant width and is subjected to a pressure loading that varies only along the x axis. This loading can be described by the function p = p(x) N/m2. we can also represent it as a coplanar distributed load by multiply the loading function by the width b m of the beam, so that w(x) = p(x) b N/m. The resultant force is equivalent to the area under the loading diagram and has a line of action that passes through the centroid or geometric center of this area. 3 Example 4.22: A distributed loading of p = (800x) Pa acts over the top surface of the beam shown. Determine the magnitude and location of the equivalent resultant force. 𝑃 = 800 𝑥 𝑁/𝑚2 𝑊 = 𝑃. 𝑏 𝑁/𝑚 𝑊 = 800 𝑥 0.2 = 160 (𝑥) 𝑁/𝑚 𝑎𝑡 𝑥 = 9.0 𝑚, 𝑒𝑛𝑑 𝑜𝑓 𝑏𝑒𝑎𝑚 ; 𝑊 = 160 9 = 1440 𝑁/𝑚 The magnitude of the resultant force is equivalent to the area of the triangle. 𝐹𝑅 = 0.5 ∗ 1440 ∗ 9 = 6480 𝑁 The line of action of FR passes through the centroid C of this triangle. 4 2  Distributed Load - The external effects caused by a coplanar distributed load acting on a body can be represented by a single resultant force. - This resultant force is equivalent to the area under the loading diagram and has a line of action that passes through the centroid or geometric center of this area. Uniform Distributed Load Uniform Distributed Load F1 F2 F1 W2 W t/m W1 L 2L/3 L/3 L F1 = W * L F1 = W1 * L F2 = 0.5 (W2-W1) * L 5 Example 4.23: The granular material exerts the distributed loading on the beam as shown in Fig. a. Determine the magnitude and location of the equivalent resultant of this load. 𝐹1 = 0.5 9 100 − 50 = 225 𝐼𝑏 𝐹2 = 50 9 = 450 𝐼𝑏 The lines of action of these parallel forces act through the respective centroids of their associated areas and therefore intersect the beam at x1 = 1/3 (9 ft) = 3 ft x2 = 1/2 (9 ft) = 4.5 ft The magnitude of the resultant force is, 𝐹𝑅 = 225 + 450 = 675 𝐼𝑏 location of FR with reference to point A could be estimated as follows 𝐹1 𝑥1 + 𝐹2 𝑥2 = 𝐹𝑅 𝑥ҧ → 225 3 + 450 4.5 = 675 𝑥ҧ 𝑥ҧ = 4.0 𝑓𝑡 6 3  Mechanics (MCE111) Chapter 5 Equilibrium of a Rigid Body 7 Chapter 5 Equilibrium of a Rigid Body Objectives. To develop the equations of equilibrium for a rigid body. To introduce the concept of the free-body diagram for a rigid body. To show how to solve rigid-body equilibrium problems using the equations of equilibrium. 8 4  Conditions for Rigid-Body Equilibrium. If this resultant force and couple moment are both equal to zero, then the body is said to be in equilibrium. Mathematically, the equilibrium of a body is expressed as 𝐹𝑅 = ෍ 𝐹 = 0 𝑀𝑅𝑜 = ෍ 𝑀𝑜 = 0 The first of these equations' states that the sum of the forces acting on the body is equal to zero. The second equation states that the sum of the moments of all the forces in the system about point O, added to all the couple moments, is equal to zero. 9 Support Reactions. A support prevents the translation of a body in a given direction by exerting a force on the body in the opposite direction. A support prevents the rotation of a body in a given direction by exerting a couple moment on the body in the opposite direction. Types of Supports https://www.youtube.com/watch?app=desktop&v=xLPbqH8Yk5c https://www.quora.com/How-many-support-reaction-on-the-simply-supported-beam https://shop21403.sosoutremer.org/content?c=roller+support+reactions&id=6 10 5 11 Example 5.1: Draw the free-body diagram of the uniform beam shown in Fig. a. The beam has a mass of 100 kg. To draw the free-body diagram, - Replace the existing supports by their relevant reactions - Replace the uniform load (if existing) to equivalent point load (at its centroid) The fixed support at A refer to 3 unknown 𝐴𝑥 , 𝐴𝑦 , 𝑀𝐴 Weight of the beam W = 100 * 9.81 = 981 N To get the reactions σ 𝐹𝑥 = 0, 𝐴𝑥 = 0 σ 𝐹𝑦 = 0, 𝐴𝑦 − 1200 − 981 = 0 ; 𝐴𝑦 = 2181 𝑁 σ 𝑀𝐴 = 0 −𝑀𝐴 + 1200 2 + 981 3 = 0; 𝑀𝐴 = 5343 N. m 12 6  Example 5.5: Determine the horizontal and vertical components of reaction on the beam caused by the pin at B and the rocker at A as shown in Fig. a. Neglect the weight of the beam. a- Draw the free-body diagram (replace supports by reactions, shows all loads in the x-y components) To get the reactions σ 𝐹𝑥 = 0, +→ ; 600 cos 45 − 𝐵𝑥 = 0; 𝐵𝑥 = 424.3 𝑁 ← ෍ 𝐹𝑦 = 0, +↑ 𝐴𝑦 + 𝐵𝑦 − 600 sin45 − 200 − 100 = 0 𝐴𝑦 + 𝐵𝑦 = 724.26 𝑁 σ 𝑀𝐴 = 0 600 𝑐𝑜𝑠45 0.2 + 600 𝑠𝑖𝑛45 2 + 100 5 + 200 7 − 𝐵𝑦 7 = 0 𝐵𝑦 = 404.77 N σ 𝑀𝐵 = 0 −600 𝑐𝑜𝑠45 0.2 + 600 𝑠𝑖𝑛45 5 + 100 2 − 𝐴𝑦 7 = 0 𝐴𝑦 = 319.49 N 13 Example 5.8: The box wrench in Fig. a is used to tighten the bolt at A. If the wrench does not turn when the load is applied to the handle, determine the torque or moment applied to the bolt and the force of the wrench on the bolt. a- Draw the free-body diagram (replace supports by reactions, shows all loads in the x-y components) σ 𝑀𝐴 = 0 12 𝑀𝐴 − 52 0.3 − 30 sin 60 (0.7) = 0; 𝑀𝐴 = 32.6 N. m 13 5 σ 𝐹𝑥 = 0, +→ ; 𝐴𝑥 − 52 + 30 cos 60 = 0; 𝐴𝑥 = 5 𝑁 → 13 12 σ 𝐹𝑦 = 0, +↑ ; 𝐴𝑦 − 52 − 30 sin 60 = 0; 𝐴𝑦 = 74 𝑁 ↑ 13 15 7  Example 5.9: Determine the horizontal and vertical components of reaction on the member at the pin A, and the normal reaction at the roller B in Fig. a a- Draw the free-body diagram (replace supports by reactions, shows all loads in the x-y components) σ 𝑀𝐴 = 0 𝑁𝐵 cos 30 (6) − 𝑁𝐵 sin 30 2 − 750 3 = 0; 𝑁𝐵 = 532.6 Ib σ 𝐹𝑥 = 0, +→ ; 𝐴𝑥 − 532.6 sin 30 = 0; 𝐴𝑥 = 268.1 𝐼𝑏 → σ 𝐹𝑦 = 0, +↑ ; 𝐴𝑦 − 750 + 532.6 𝑐𝑜𝑠 30 = 0; 𝐴𝑦 = 288.8 𝐼𝑏 ↑ 16 Example 5.12: Determine the support reactions on the member in Fig. a. The collar at A is fixed to the member and can slide vertically along the vertical shaft. a- Draw the free-body diagram (replace supports by reactions, shows all loads in the x-y components) σ 𝐹𝑥 = 0, +→ ; 𝐴𝑥 = 0 𝑁 σ 𝐹𝑦 = 0, +↑ ; 𝑁𝐵 − 900 = 0; 𝑁𝐵 = 900 𝑁 ↑ σ 𝑀𝐴 = 0 𝑀𝐴 + 𝑁𝐵 (3 + 1 cos 45) − 900 1.5 − 500 = 0; 𝑀𝐴 = −1486.4 𝑁. 𝑚 𝑀𝐴 = 1486.4 𝑁. 𝑚 17 8  Fundamental problems 5.3: The truss is supported by a pin at A and a roller at B. Determine the support reactions. a- Draw the free-body diagram (replace supports by reactions, shows all loads in the x-y components) σ 𝐹𝑥 = 0, +→ ; 𝐴𝑥 + 5 cos 45 = 0 ; 𝐴𝑥 = − 3.54 𝐾𝑁 → = 3.54 𝐾𝑁 (←) σ 𝑀𝐴 = 0 𝐵𝑦 (6 + 6 cos 45) − 10 2 + 6 cos 45 − 5 (4) = 0; 𝐵𝑦 = 8.047 𝐾𝑁 ↑ σ 𝐹𝑦 = 0, +↑ ; 𝐴𝑦 + 8.047 − 10 − 5 sin 45 = 0; 𝐴𝑦 = 5.488 𝐾𝑁 ↑ 𝑩𝒚 𝑨𝒙 𝑨𝒚 18 Problem 5.13: Determine the reactions at the supports (roller at A and pin at B) a- Draw the free-body diagram (replace supports by reactions, shows all loads in the x-y components) σ 𝑀𝐴 = 0 𝟒𝟓𝟎 𝟑𝟔𝟎𝟎 −450 1 − 3600 3 + 𝐵𝑦 (6) = 0; 𝑩𝒚 = 𝟏𝟖𝟕𝟓 𝐍 ↑ 𝟏𝒎 𝑩𝒙 σ 𝐹𝑦 = 0, +↑ ; 𝐴𝑦 + 1875 − 450 − 3600 = 0; 𝑵𝑨 = 𝟐𝟏𝟕𝟓 𝑵 ↑ σ 𝐹𝑥 = 0, +→ ; 𝑩𝒙 = 𝟎𝑵 𝑩𝒚 𝑨𝒚 20 9  Problem 5.14: Determine the reactions at the supports (rocker at A and pin at B) a- Draw the free-body diagram (replace supports by reactions, shows all loads in the x-y components) σ 𝑀𝐵 = 0 4000 2.5 − 𝑁𝐴 (3) = 0; 𝑵𝑨 = 𝟑𝟑𝟑𝟑. 𝟑 𝐍 ↑ 4 σ 𝐹𝑦 = 0, +↑ ; 𝐵𝑦 + 3333.3 − 4000 = 0; 𝑩𝒚 = −𝟏𝟑𝟑 𝑵 ↑ 5 𝑩𝒚 = 𝟏𝟑𝟑 𝑵 ↓ 𝑵𝑨 σ 𝐹𝑥 = 0, +→ ; −𝐵𝑥 − 4000 3 = 0; 𝐵𝑥 = −2400 𝑁 ← 𝑩𝒙 5 𝑩𝒙 = 𝟐𝟒𝟎𝟎 𝑵 → 𝑩𝒚 21 Thank You 23 10