Topic#4 Equilibrium of a Rigid Body PDF

Helwan National University Faculty of Engineering GEN 0004: Mechanics Part One Static Ch4 :Equilibrium of a Rigid Body Dr. Mohamed Salah  Equilibrium of a Rigid Body (2D) Condition of Equilibrium If the resultant force and couple moment are both equal to zero, th...

Helwan National University Faculty of Engineering GEN 0004: Mechanics Part One Static Ch4 :Equilibrium of a Rigid Body Dr. Mohamed Salah  Equilibrium of a Rigid Body (2D) Condition of Equilibrium If the resultant force and couple moment are both equal to zero, then the body is said to be in equilibrium: σ 𝒇𝒙 = 𝟎 𝑹 = σ𝑭 = 𝟎 σ 𝒇𝒚 = 𝟎 𝑴𝑹 𝑶 = σ 𝑴𝑶 = 𝟎 Where O is any point on a rigid body Free body diagram 2  Support Reactions We consider the various types of reactions that occur at supports and points of contact between bodies subjected to coplanar force systems. As a general rule: If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction. If rotation is prevented, a couple moment is exerted on the body. 3  Support Reactions Roller, Rocker or Smooth Surface One unknown two Pin or Hinge unknown Fixed Support three unknown 4  Equilibrium of a Rigid Body (2D) Procedure for solving problems 1- Draw the Free Body Diagram a-Weight: 𝒘 b-Applied force: Any force on the body, T , 𝑓𝑠 , f, … c-Support Reactions: 5  Equilibrium of a Rigid Body (2D) 2- Force resolution 3-Apply the equilibrium equations σ 𝒇𝒙 = 𝟎----------------------------1 𝐑 = σ 𝐅Ԧ = 𝟎 σ 𝒇𝒚 = 𝟎----------------------------2 σ 𝑴𝑶 = 𝟎------------------------------------------------------3 4- Solve the resulting equations 6  Example 1 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 Figure. Neglect the weight of the beam. Solution: Rocker pin 𝑩𝒙 𝑩𝒚 𝑨𝒚 7  Example 1 𝟔𝟎𝟎 𝒔𝒊𝒏 𝟒𝟓 600 N 200 N 𝟔𝟎𝟎 𝒄𝒐𝒔 𝟒𝟓 45° 𝑩𝒙 𝑨𝒚 𝑩𝒚 100 N ↶ + σ 𝑀𝐵 =0 100 × 2 + 600 sin 45 × 5 − 600 cos 45 × 0.2 −(𝐴𝑦 × 7) = 0 100 ×2 + 600 sin 45×5 − 600 cos 45×0.2 𝐴𝑦 = = 319.5 𝑁 7 8  Example 1 → + ෍ 𝑓𝑥 = 0 600 cos 45 −𝐵𝑥 = 0 𝐵𝑥 = 600 cos 45 𝐵𝑥 = 424.3 𝑁 ↑ + ෍ 𝑓𝑦 = 0 𝐴𝑦 = 319.5 𝑁 𝐵𝑦 +𝐴𝑦 −600 𝑠𝑖𝑛45 −100 −200 = 0 𝐵𝑦 = 405 𝑁 9  Example 2 The member shown in Figure is pin-connected at A and rests against a smooth support at B. Determine the horizontal and vertical components of reaction at the pin A. Smooth 𝑩 Solution: pin 𝑨𝒙 𝑨𝒚 10  Example 2 𝑩 𝒄𝒐𝒔 𝟑𝟎 𝑩 𝑩 𝒔𝒊𝒏 𝟑𝟎 𝑨𝒙 𝑨𝒚 60 N ↶ + σ 𝑀𝐴 =0 ⟹ 𝐵 × 0.75 − 60 × 1 −90 = 0 ∴ 𝐵 = 200 𝑁 → + σ 𝑓𝑥 = 0 ⟹ 𝐴𝑥 −𝐵 sin 30 = 0 ∴ 𝐴𝑥 = 100 𝑁 ↑ + σ 𝑓𝑦 = 0 ⟹ 𝐴𝑦 − 𝐵 cos 30 −60 = 0 ∴ 𝐴𝑦 = 233.2 𝑁 11  Equilibrium of a Rigid Body 3D Condition of Equilibrium The conditions for equilibrium of a rigid body subjected to a three-dimensional force system require that both the resultant force and resultant couple moment acting on the body be equal to zero. σ 𝑴𝒙 = 𝟎 σ 𝑭𝒙 = 𝟎 σ 𝑴𝑶 = 𝟎 σ 𝑴𝒚 = 𝟎 σ𝑭 = 𝟎 σ 𝑭𝒚 = 𝟎 σ 𝑴𝒛 = 𝟎 σ 𝑭𝒛 = 𝟎 12  Equilibrium of a Rigid Body 3D Procedure for solving problems 1- Draw the Free Body Diagram a-Weight: 𝒘 b-Tension force in cable: T A A T B c-Spring force: 𝒇𝒔 = 𝒌 ∗ 𝒔 Where 𝒔 = 𝒍 − 𝒍𝟎 13  Equilibrium of a Rigid Body 3D d-Support Reactions: Roller or Smooth Surface One unknown Ball and socket three unknown Fixed Support six unknown 14  Equilibrium of a Rigid Body 3D 2- Force representation 3-Apply the equilibrium equation σ 𝑴𝑶 = 𝟎 4- Equating coefficients 𝒊Ԧ σ 𝑴𝒙 = 𝟎 -----------------1 𝒋Ԧ σ 𝑴𝒚 = 𝟎-----------------2 𝒌 σ 𝑴𝒛 = 𝟎 -----------------3 5- Solve the resulting equations 15  Equilibrium of a Rigid Body 3D 6-Apply the equilibrium equation 𝑭 = 𝟎 7- Equating coefficients 𝒊Ԧ σ 𝑭𝒙 = 𝟎 -----------------4 𝒋Ԧ σ 𝑭𝒚 = 𝟎-----------------5 𝒌 σ 𝑭𝒛 = 𝟎 -----------------6 8- Solve the resulting equations 16  Example 3 Draw the free-body diagram of the 5 x 8 ft plate weighing 270 lb and is supported by a ball and socket joint at A and by two cables. Solution: 17  Example 4 The assembly shown in Figure is used to support the load has a weight of 200 N. Determine the reaction at the ball-and-socket joint A and the tension in each of the wires. Solution: 18  Example 4 The free body diagram of rigid body ABCE is shown in Figure 19  Example 4 Express the tensions and forces in Cartesian vectors form. The coordinates of points A, B, G, C, D, F, and E are: A=(0, 0, 0), B=(4, 0, 0), G=(4, 2, 0), C=(4, 4, 0), D=(4, 0, 3), F=(2, 4, 3), and E=(2,4, 0). Thus, 𝐴Ԧ = 𝐴𝑥 𝑖Ԧ + 𝐴𝑦 𝑗Ԧ + 𝐴𝑧 𝑘 𝑇𝐵𝐷 = 𝑇𝐵𝐷 𝑘 𝑊 = −200𝑘 Ԧ 3𝑘 −4𝑗+ −4 3 𝑇𝐶𝐷 = 𝑇𝐶𝐷 = 𝑇𝐶𝐷 ( 𝑗Ԧ + 𝑘) 5 5 5 𝑇𝐸𝐹 = 𝑇𝐸𝐹 𝑘 20  Example 4 Applying the equilibrium equation. First summing moments about point A: σ 𝑀𝐴 = 0 𝑟Ԧ𝐴𝐵 × 𝑇𝐵𝐷 + (𝑟Ԧ𝐴𝐶 × 𝑇𝐶𝐷 ) + (𝑟Ԧ𝐴𝐸 × 𝑇𝐸𝐹 ) + (𝑟Ԧ𝐴𝐺 × W) = 0 𝑖Ԧ 𝑗Ԧ 𝑘 𝑖Ԧ 𝑗Ԧ 𝑘 𝑖Ԧ 𝑗Ԧ 𝑘 𝑖Ԧ 𝑗Ԧ 𝑘 4 0 0 + 4 4 0 + 2 4 0 + 4 2 0 =0 0 0 𝑇𝐵𝐷 0 −0.8𝑇𝐶𝐷 0.6𝑇𝐶𝐷 0 0 𝑇𝐸𝐹 0 0 −200 Equating the respective 𝑖 , 𝑗Ԧ, 𝑘 components to zero, we have: 𝒊Ԧ 2.4𝑇𝐶𝐷 + 4𝑇𝐸𝐹 − 400 = 0----------------------------------(1) 𝒋Ԧ −4𝑇𝐵𝐷 − 2.4𝑇𝐶𝐷 − 2𝑇𝐸𝐹 + 800 = 0---------------------(2) 𝒌 −3.2𝑇𝐶𝐷 = 0 ∴ 𝑇𝐶𝐷 = 0 21  Example 4 From equations 1&2 𝑇𝐸𝐹 = 100 𝑁 and 𝑇𝐵𝐷 = 150 𝑁 Second summing forces equal to zero: σ 𝐹Ԧ = 0 𝐴Ԧ + TBD + 𝑊 + TCD + TEF = 0 Equating the respective 𝑖 , 𝑗Ԧ, 𝑘 components to zero, we have: 𝒊Ԧ 𝐴𝑥 = 0 4 𝒋Ԧ 𝐴𝑦 − 𝑇𝐶𝐷 = 0 𝐴𝑦 = 0 5 3 𝒌 𝐴𝑧 + TBD − 200 + 𝑇𝐶𝐷 + TEF = 0 𝐴𝑧 = −50 𝑁 5 22  Example 5 The pipe assembly supports the vertical loads shown. Determine the components of reaction at the ball and socket joint A, and the tension in the cables BC and BD. Solution: 23  Example 5 The free body diagram of rigid body AB is shown right figure. 24  Example 5 Express the tensions and forces in Cartesian vectors form. The coordinates of points A, B, E, and H are: A=(0, 0, 0), B=(0, 1.5, 1), C=(3, 0, 2), D=(-3, 0, 2), E=(0, 4.5, 2), and H=(0,6, 2). Thus, 𝐴Ԧ = 𝐴𝑥 𝑖Ԧ + 𝐴𝑦 𝑗Ԧ + 𝐴𝑧 𝑘 3Ԧ𝑖 − 1.5Ԧ𝑗 + 1𝑘 3 1.5 1 𝑇𝐵𝐶 = 𝑇𝐵𝐶 = 𝑇𝐵𝐶 ( 𝑖Ԧ − 𝑗Ԧ + 𝑘) 3.5 3.5 3.5 3.5 −3Ԧ𝑖 − 1.5Ԧ𝑗 + 1𝑘 −3 1.5 1 𝑇𝐵𝐷 = 𝑇𝐵𝐷 = 𝑇𝐵𝐷 ( 𝑖Ԧ − 𝑗Ԧ + 𝑘) 3.5 3.5 3.5 3.5 𝐹Ԧ1 = −4𝑘 𝐹Ԧ2 = −3𝑘 25  Example 5 Applying the equilibrium equation. First summing moments about point A: σ 𝑀𝐴 = 0 𝑟Ԧ𝐴𝐵 × 𝑇𝐵𝐶 + (𝑟Ԧ𝐴𝐵 × 𝑇𝐵𝐷 ) + (𝑟Ԧ𝐴𝐻 × 𝐹Ԧ1 ) + (𝑟Ԧ𝐴𝐸 × 𝐹Ԧ2 ) = 0 𝑟Ԧ𝐴𝐵 = 1.5 𝑗Ԧ + 𝑘 𝑟Ԧ𝐴𝐻 = 6 𝑗Ԧ + 2𝑘 𝑟Ԧ𝐴𝐸 = 4.5 𝑖Ԧ + 2Ԧ𝑗 Ԧi Ԧj k Ԧi Ԧj k 0 1.5 1 + 0 1.5 1 3 −1.5 1 −3 −1.5 1 𝑇 𝑇 𝑇 𝑇 𝑇 𝑇 3.5 𝐵𝐶 3.5 𝐵𝐶 3.5 𝐵𝐶 3.5 𝐵𝐷 3.5 𝐵𝐷 3.5 𝐵𝐷 Ԧi Ԧj k Ԧi Ԧj k + 0 6 2 + 0 4.5 2 = 0 0 0 −4 0 0 −3 26  Example 5 Equating the respective 𝑖 , 𝑗Ԧ, 𝑘 components to zero, we have: 3 3 𝒊Ԧ 𝑇 + 𝑇 − 37.5 = 0----------------------------------(1) 3.5 𝐵𝐶 3.5 𝐵𝐷 3 3 𝒋Ԧ 𝑇𝐵𝐶 − 𝑇𝐵𝐷 = 0------------------------------------------(2) 3.5 3.5 4.5 4.5 𝒌 − 𝑇 + 𝑇 = 0-----------------------------------------(3) 3.5 𝐵𝐶 3.5 𝐵𝐷 Solving equations (1) through (3), 𝑇𝐵𝐶 = 𝑇𝐵𝐷 = 21.87𝐾𝑁 27  Example 5 Second summing forces equal to zero: σ 𝐹Ԧ = 0 𝐴Ԧ + 𝑇𝐵𝐶 + 𝑇𝐵𝐷 + 𝐹Ԧ1 + 𝐹Ԧ2 = 0 Equating the respective 𝑖 , 𝑗Ԧ, 𝑘 components to zero, we have: 3 3 𝒊Ԧ 𝐴𝑥 + 𝑇𝐵𝐶 − 𝑇𝐵𝐷 = 0 𝐴𝑥 = 0 3.5 3.5 1.5 1.5 𝒋Ԧ 𝐴𝑦 − 𝑇𝐵𝐶 − 𝑇𝐵𝐷 = 0 𝐴𝑦 = 18.74 𝑘𝑁 3.5 3.5 1 1 𝒌 𝐴𝑧 + 𝑇𝐵𝐶 + 𝑇𝐵𝐷 − 7 = 0 𝐴𝑧 = −5.5 𝑘𝑁 3.5 3.5 28