# Ch.05 Equilibrium of a Rigid Body

July 28, 2017 | Author: CK_85_3 | Category: Force, Chemical Equilibrium, Rotation, Mechanics, Mechanical Engineering

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Equilibrium of a Rigid Body...

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2/13/2013

Engineering Mechanics – Statics

5.01

Equilibrium of a Rigid Body

Engineering Mechanics – Statics

5.02

Equilibrium of a Rigid Body

Chapter 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

05. Equilibrium of a Rigid Body

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.03

Nguyen Tan Tien

Equilibrium of a Rigid Body

§1. Conditions for Rigid-Body Equilibrium - The force and couple system acting on a body can be reduced to an equivalent resultant force and resultant couple moment at an arbitrary point 𝑂

- A rigid body is in equilibrium ⟺ 𝐹𝑅 = ∑𝐹𝑖 = 0, 𝑀𝑅𝑂 = ∑𝑀𝑂 = 0 • The sum of the forces acting on the body is equal to zero • The sum of the moments of all forces in the system about point 𝑂, added to all the couple moments, is equal to zero HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.05

Nguyen Tan Tien

Equilibrium of a Rigid Body

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.04

Nguyen Tan Tien

Equilibrium of a Rigid Body

§1. Conditions for Rigid-Body Equilibrium - The two equations of equilibrium for a rigid body 𝐹𝑅 = ∑𝐹 = 0 𝑀𝑅𝑂 = ∑𝑀𝑂 = 0 where 𝑂 is an arbitrary point - Equilibrium in two dimensions 𝐹𝑥 = 0 𝐹𝑦 = 0 𝑀𝑧 = 0 - Note 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛, 𝑠𝑒𝑛𝑠𝑒 𝐹 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 known value − unknown value ? HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.06

Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram - A Free-Body Diagram: a sketch of the isolated or free body which shows all the pertinent weight forces, the externally applied loads, and the reaction from its supports and connections acting upon it by the removed elements - General rules for support reactions • 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

§2. Free-Body Diagram - Support Reactions • Roller: prevents the beam from translating in the vertical direction, the roller will only exert a force on the beam in this direction

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

1

2/13/2013

Engineering Mechanics – Statics

5.07

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Support Reactions • Roller: prevents the beam from translating in the vertical direction, the roller will only exert a force on the beam in this direction

𝐹

5.08

Equilibrium of a Rigid Body

§2. Free-Body Diagram • Pin: prevents translation of the beam in any direction

⊥ 𝑏𝑒𝑎𝑚 ?

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

Engineering Mechanics – Statics

5.09

Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram • Pin: prevents translation of the beam in any direction

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.10

Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram • Fixed Support: prevents both translation and rotation of the beam

𝐹

∥ 𝑂𝑦 ? ∥ 𝑂𝑥 = 𝐹𝑥 + 𝐹𝑦 ? ? ?

− 𝑀 ? 𝐹

∥ 𝑂𝑦 ? ∥ 𝑂𝑥 = 𝐹𝑥 + 𝐹𝑦 ? ? ?

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.11

Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Supports for Rigid Bodies Subjected to 2D Force Systems • Cable: the reaction is a tension force which acts away from the member in the direction of the cable

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.12

Equilibrium of a Rigid Body

§2. Free-Body Diagram • Roller: the reaction is a force which acts perpendicular to the surface at the point of contact

− 𝐹 ?

− 𝐹 ?

• Weightless link: the reaction is a force which acts along the axis of the link

• Roller or pin in confined smooth slot: the reaction is a force which acts perpendicular to the slot

− 𝐹 ?

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

− 𝐹 ?

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

2

2/13/2013

Engineering Mechanics – Statics

5.13

Equilibrium of a Rigid Body

§2. Free-Body Diagram

Engineering Mechanics – Statics

5.14

• Rocker: the reaction is a force which acts perpendicular to the surface at the point of contact

• Member pin connected to collar on smooth rod: the reaction is a force which acts perpendicular to the rod

− 𝐹 ?

− 𝐹 ?

• Smooth contacting surface: the reaction is a force which acts perpendicular to the surface at the point of contact

• Smooth pin or hinge: the reactions are two components of force, or the magnitude and direction of the resultant force

− 𝐹 ?

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

Equilibrium of a Rigid Body

§2. Free-Body Diagram

𝐹

Nguyen Tan Tien

5.15

Equilibrium of a Rigid Body

§2. Free-Body Diagram

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.16

? ?

Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Some typical examples of actual supports

• Member fixed connected to collar on smooth rod: the reactions are the couple moment and the force which acts perpendicular to the rod − − 𝐹 ?, 𝑀 ? • Fixed support: the reactions are the couple moment and the two force components, or the couple moment and the magnitude and direction of the resultant force 𝐹 HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.17

− ? ,𝑀 ? ? Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Internal Forces

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.18

Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Weight and the Center of Gravity

𝑊 = 𝑚𝑔 𝑚: mass, 𝑘𝑔 𝑔: gravity acceleration, 𝑚/𝑠 2 HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

3

2/13/2013

Engineering Mechanics – Statics

5.19

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Idealized Models

Engineering Mechanics – Statics

5.20

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Example 5.1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100𝑘𝑔

Solution

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.21

Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Example 5.2 Draw the free-body diagram of the foot lever. The operator applies a vertical force to the pedal so that the spring is stretched 36𝑚𝑚. and the force in the short link at 𝐵 is 90𝑁

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.22

Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Example 5.3 Two smooth pipes, each having a mass of 300𝑘𝑔, are supported by the forked tines of the tractor. Draw the free-body diagrams for each pipe and both pipes together

Solution Solution Spring force 𝐹𝑠 = 36 × 3.5 = 126𝑁

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.23

Nguyen Tan Tien

Equilibrium of a Rigid Body

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.24

Nguyen Tan Tien

Equilibrium of a Rigid Body

§2. Free-Body Diagram - Example 5.4 Draw the free-body diagram of the unloaded platform that is suspended off the edge of the oil rig. The platform has a mass of 200𝑘𝑔 Solution

Problems - Prob.5.1 Draw the free-body diagram of the 50𝑘𝑔 paper roll which has a center of mass at 𝐺 and rests on the smooth blade of the paper hauler. Explain the significance of each force acting on the diagram

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

4

2/13/2013

Engineering Mechanics – Statics

5.25

Equilibrium of a Rigid Body

Engineering Mechanics – Statics

5.26

Equilibrium of a Rigid Body

Problem - Prob.5.2 Draw the free-body diagram of member 𝐴𝐵, which is supported by a roller at 𝐴 and a pin at 𝐵. Explain the significance of each force on the diagram

Problem - Prob.5.3 Draw the free-body diagram of the dumpster 𝐷 of the truck, which has a weight of 5000𝑁 and a center of gravity at 𝐺 . It is supported by a pin at 𝐴 and a pin-connected hydraulic cylinder 𝐵𝐶 (short link). Explain the significance of each force on the diagram

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.27

Nguyen Tan Tien

Equilibrium of a Rigid Body

Engineering Mechanics – Statics

5.28

Nguyen Tan Tien

Equilibrium of a Rigid Body

Problem - Prob.5.4 Draw the free-body diagram of the beam which supports the 80𝑘𝑔 load and is supported by the pin at 𝐴 and a cable which wraps around the pulley at 𝐷 . Explain the significance of each force on the diagram

Problem - Prob.5.5 Draw the free-body diagram of the truss that is supported by the cable 𝐴𝐵 and pin 𝐶. Explain the significance of each force acting on the diagram

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.29

Nguyen Tan Tien

Equilibrium of a Rigid Body

Engineering Mechanics – Statics

5.30

Nguyen Tan Tien

Equilibrium of a Rigid Body

Problem - Prob.5.6 Draw the free-body diagram of the bar, which has a negligible thickness and smooth points of contact at 𝐴, 𝐵, and 𝐶. Explain the significance of each force on the diagram

§3. Equations of Equilibrium - The conditions for equilibrium in two dimensions ∑𝐹𝑥 = 0 ∑𝐹 = 0 ⟹ ∑𝐹𝑦 = 0 ∑𝑀 = 0 ∑𝑀𝑂 = 0 - Alternative sets of equilibrium equations • The first alternative set ∑𝐹𝑥 = 0 ∑𝑀𝐴 = 0 ∑𝑀𝐵 = 0 • The second alternative set ∑𝑀𝐴 = 0 ∑𝑀𝐵 = 0 ∑𝑀𝐶 = 0

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

5

2/13/2013

Engineering Mechanics – Statics

5.31

Equilibrium of a Rigid Body

§3. Equations of Equilibrium - Example 5.5 Determine the horizontal and vertical components of reaction on the beam caused by the pin at 𝐵 and the rocker. Neglect the weight of the beam Solution Free-body Diagram Equations of Equilibrium +→ ∑𝐹𝑥 : 600𝑐𝑜𝑠450 − 𝐵𝑥 = 0 + ↑ ∑𝐹𝑦 : 𝐴𝑦 − 600𝑠𝑖𝑛450 −100 + 𝐵𝑦 = 0

Engineering Mechanics – Statics

Engineering Mechanics – Statics

5.33

Nguyen Tan Tien

Equilibrium of a Rigid Body

Equilibrium of a Rigid Body

+↺ ∑𝑀𝐴 : 100 × 0.5 − 𝑇 × 0.5 = 0 ⟹ 𝑇 = 100𝑁, 𝐴𝑥 = 50𝑁, 𝐴𝑦 = 187𝑁

+↺ ∑𝑀𝐵 : −7𝐴𝑦 + 600𝑠𝑖𝑛450 × 5 −600𝑐𝑜𝑠450 × 0.2 +100 × 2 = 0 ⟹ 𝐴𝑦 = 319𝑁, 𝐵𝑥 = 424𝑁, 𝐵𝑦 = 405𝑁 HCM City Univ. of Technology, Faculty of Mechanical Engineering

5.32

§3. Equations of Equilibrium - Example 5.6 The cord supports a force of 100𝑁 and wraps over the frictionless pulley. Determine the tension in the cord at 𝐶 and the horizontal and vertical components of reaction at pin 𝐴 Solution Free-body Diagram Equations of Equilibrium +→ ∑𝐹𝑥 : −𝐴𝑥 + 𝑇𝑠𝑖𝑛300 = 0 + ↑ ∑𝐹𝑦 : 𝐴𝑦 − 𝑇𝑐𝑜𝑠450 − 100 = 0

Note: The tension remains constant as the cord passes over the pulley HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

Nguyen Tan Tien

5.34

Equilibrium of a Rigid Body

§3. Equations of Equilibrium - Example 5.7 The member 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 Solution Free-body Diagram Equations of Equilibrium +→ ∑𝐹𝑥 : 𝐴𝑥 − 𝑁𝐵 𝑠𝑖𝑛300 = 0 + ↑ ∑𝐹𝑦 : 𝐴𝑦 − 𝑁𝐵 𝑐𝑜𝑠300 − 60 = 0

§3. Equations of Equilibrium - Example 5.8 The box wrench 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 Solution Free-body Diagram Equations of Equilibrium

+↺ ∑𝑀𝐴 : 𝑁𝐵 × 0.75 − 60 × 1 − 90 = 0 ⟹ 𝑁𝐵 = 200𝑁 𝐴𝑥 = 100𝑁 𝐴𝑦 = 233𝑁

+ ↑ ∑𝐹𝑦 : 𝐴𝑦 − 52 13 − 30𝑠𝑖𝑛600 = 0

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.35

Nguyen Tan Tien

Equilibrium of a Rigid Body

5

+→ ∑𝐹𝑥 : 𝐴𝑥 − 52 13 + 30𝑐𝑜𝑠300 = 0 12

12

+↺ ∑𝑀𝐴 : 𝑀𝐴 − 52 13 × 0.3 −

30𝑠𝑖𝑛600 × 0.7 = 0 ⟹ 𝑀𝐴 = 32.6𝑁𝑚, 𝐴𝑥 = 5𝑁, 𝐴𝑦 = 74𝑁 HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.36

Nguyen Tan Tien

Equilibrium of a Rigid Body

§3. Equations of Equilibrium - 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 Solution Free-body Diagram Equations of Equilibrium +→ ∑𝐹𝑥 : 𝐴𝑥 − 𝑁𝐵 𝑠𝑖𝑛300 = 0 + ↑ ∑𝐹𝑦 : 𝐴𝑦 − 500 + 𝑁𝐵 𝑐𝑜𝑠300 = 0

§3. Equations of Equilibrium - Example 5.10 The uniform smooth rod is subjected to a force and couple moment. If the rod is supported at 𝐴 by a smooth wall and at 𝐵 and 𝐶 either at the top or bottom by rollers, determine the reactions at these supports. Neglect the weight of the rod Solution Free-body Diagram Equations of Equilibrium +→ ∑𝐹𝑥 : 𝐶𝑦′𝑠𝑖𝑛300 + 𝐵𝑦′𝑠𝑖𝑛300 − 𝐴𝑥 = 0

+↺ ∑𝑀𝐴 : −500 × 3 + 𝑁𝐵 𝑐𝑜𝑠300 × 6 −𝑁𝐵 𝑠𝑖𝑛300 × 2 = 0 ⟹ 𝑁𝐵 = 536𝑁 𝐴𝑥 = 268𝑁 𝐴𝑦 = 286𝑁

+ ↑ ∑𝐹𝑦 : −300 + 𝐶𝑦′𝑐𝑜𝑠300 + 𝐵𝑦′𝑐𝑜𝑠300 = 0

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

+↺ ∑𝑀𝐴 : −𝐵𝑦′ × 2 + 4000 − 𝐶𝑦′ × 6 +300𝑐𝑜𝑠300 × 8 = 0 ⟹ 𝐵𝑦′ = −1000.0𝑁, 𝐶𝑦′ = 1346.4𝑁 𝐴𝑥 = 173𝑁 HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

6

2/13/2013

Engineering Mechanics – Statics

5.37

Equilibrium of a Rigid Body

§3. Equations of Equilibrium - Example 5.11 The uniform truck ramp has a weight of 400𝑁 and is pinned to the body of the truck at each side and held in the position shown by the two side cables. Determine the tension in the cables

Engineering Mechanics – Statics

Free-body Diagram Equations of Equilibrium ∑𝑀𝐴 = −𝑇𝑠𝑖𝑛100 × 0.165 + 400 × 0.125𝑐𝑜𝑠300 = 0 ⟹ 𝑇′ = 𝑇/2 = 755.6𝑁

Engineering Mechanics – Statics

5.39

Nguyen Tan Tien

Equilibrium of a Rigid Body

§4. Two- and Three-Force Members - The solutions to some equilibrium problems can be simplified by recognizing members that are subjected to only two or three forces - Two-Force Members • Forces applied at only two points on the member

Equilibrium of a Rigid Body

+↺ ∑𝑀𝐴 : 𝑀𝐴 − 500 − 900 × 1.5 +𝑁𝐵 × (1𝑐𝑜𝑠450 + 3) = 0 ⟹ 𝐴𝑋 = 0 𝑁𝐵 = 900𝑁 𝑀𝐴 = −1.49𝑘𝑁𝑚 = 1.49𝑘𝑁 ↻

Solution

HCM City Univ. of Technology, Faculty of Mechanical Engineering

5.38

§3. Equations of Equilibrium - Example 5.12 Determine the support reactions on the member in the figure. The collar at 𝐴 is fixed to the member and can slide vertically along the vertical shaft Solution Free-body Diagram Equations of Equilibrium +→ ∑𝐹𝑥 : 𝐴𝑥 = 0 + ↑ ∑𝐹𝑦 : 𝑁𝐵 − 900 = 0

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.40

Nguyen Tan Tien

Equilibrium of a Rigid Body

§4. Two- and Three-Force Members - Three-Force Members • A member is subjected to only three forces • Moment equilibrium can be satisfied only if the three forces form a concurrent or parallel force systems • If the lines of action of 𝐹1 and 𝐹2 intersect at point 𝑂, then the line of action of 𝐹3 must also pass through point 𝑂 so that the forces satisfy: ∑𝑀𝑂 = 0

• Force equilibrium:

𝐹𝐴 = −𝐹𝐵

• Moment equilibrium: ∑𝑀𝐴 = 0 or ∑𝑀𝐵 = 0 𝐹𝐴 ↑↓ 𝐹𝐵 ⟹ |𝐹𝐴 | = |𝐹𝐵 | HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.41

• If the three forces are all parallel, the location of the point of intersection 𝑂 will approach infinity Nguyen Tan Tien

Equilibrium of a Rigid Body

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.42

Nguyen Tan Tien

Equilibrium of a Rigid Body

§4. Two- and Three-Force Members - Example 5.13 The lever 𝐴𝐵𝐶 is pin supported at 𝐴 and connected to a short link 𝐵𝐷. If the weight of the members is negligible, determine the force of the pin on the lever at 𝐴 Solution Free-body Diagram Equations of Equilibrium 𝜃 = 𝑡𝑎𝑛−1 (0.7/0.4) = 60.30 +→ ∑𝐹𝑥 : 𝐹𝐴𝑐𝑜𝑠𝜃 − 𝐹𝑐𝑜𝑠450 + 400 = 0 + ↑ ∑𝐹𝑦 : 𝐹𝐴𝑠𝑖𝑛𝜃 − 𝐹𝑠𝑖𝑛450 = 0 ⟹ 𝐹𝐴 = 1.07𝑘𝑁 𝐹 = 1.32𝑘𝑁

Fundamental Problems - F5.1 Determine the horizontal and vertical components of reaction at the supports. Neglect the thickness of the beam

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

7

2/13/2013

Engineering Mechanics – Statics

5.43

Equilibrium of a Rigid Body

Engineering Mechanics – Statics

5.44

Equilibrium of a Rigid Body

Fundamental Problems - F5.2 Determine the horizontal and vertical components of reaction at the pin 𝐴 and the reaction on the beam at 𝐶

Fundamental Problems - F5.3 The truss is supported by a pin at 𝐴 and a roller at 𝐵. Determine the support reactions

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.45

Nguyen Tan Tien

Equilibrium of a Rigid Body

Engineering Mechanics – Statics

5.46

Nguyen Tan Tien

Equilibrium of a Rigid Body

Fundamental Problems - F5.4 Determine the components of reaction at the fixed support 𝐴. Neglect the thickness of the beam

Fundamental Problems - F5.5 The 25𝑘𝑔 bar has a center of mass at 𝐺 . If it is supported by a smooth peg at 𝐶, a roller at 𝐴, and cord 𝐴𝐵, determine the reactions at these supports

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.47

Nguyen Tan Tien

Equilibrium of a Rigid Body

Fundamental Problems - F5.6 Determine the reactions at the smooth contact points 𝐴, 𝐵, and 𝐶 on the bar

Engineering Mechanics – Statics

5.48

Nguyen Tan Tien

Equilibrium of a Rigid Body

§5. Free-Body Diagrams (3D) - Supports for Rigid Bodies Subjected to 3D Force Systems • Cable: the reaction is a force which acts away from the member in the known direction of the cable − 𝐹 ? • Smooth surface support: the reaction is a force which acts perpendicular to the surface at the point of contact − 𝐹 ?

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

8

2/13/2013

Engineering Mechanics – Statics

5.49

Equilibrium of a Rigid Body

§5. Free-Body Diagrams (3D) • Roller: the reaction is a force which acts perpendicular to the surface at the point of contact − 𝐹 ? • Ball and socket: the reactions are three rectangular force components − − − 𝐹 = 𝐹𝑥 ? + 𝐹𝑦 ? + 𝐹𝑧 ?

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.51

Nguyen Tan Tien

Equilibrium of a Rigid Body

§5. Free-Body Diagrams (3D) • Single thrust bearing: the reactions are three force and two couple-moment components

Engineering Mechanics – Statics

5.50

• Single journal bearing with square shaft: the reactions are two force and three couple-moment components − − 𝐹 = 𝐹𝑥 ? + 𝐹𝑧 ? − − − 𝑀 = 𝑀𝑥 ? + 𝑀𝑦 ? + 𝑀𝑧 ? HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.52

Engineering Mechanics – Statics

5.53

Nguyen Tan Tien

Equilibrium of a Rigid Body

§5. Free-Body Diagrams (3D) - Some typical examples of actual supports

ball-and-socket joint

journal bearing

thrust bearing

Equilibrium of a Rigid Body

− − − 𝐹 = 𝐹𝑥 ? + 𝐹𝑦 ? + 𝐹𝑧 ? − − 𝑀 = 𝑀𝑥 ? + 𝑀𝑧 ? • Fixed support: the reactions are three force and three couple-moment components

− − − 𝐹 = 𝐹𝑥 ? + 𝐹𝑦 ? + 𝐹𝑧 ? − − 𝑀 = 𝑀𝑦 ? + 𝑀𝑧 ? HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

§5. Free-Body Diagrams (3D) • Single hinge: The reactions are three force and two couplemoment components

− − − 𝐹 = 𝐹𝑥 ? + 𝐹𝑦 ? + 𝐹𝑧 ? − − 𝑀 = 𝑀𝑥 ? + 𝑀𝑧 ? • Single smooth pin: the reactions are three force and two couple-moment components

Equilibrium of a Rigid Body

§5. Free-Body Diagrams (3D) • Single journal bearing: the reactions are two force and two couple-moment components which act perpendicular to the shaft − − 𝐹 = 𝐹𝑥 ? + 𝐹𝑧 ? − − 𝑀 = 𝑀𝑥 ? + 𝑀𝑧 ?

− − − 𝐹 = 𝐹𝑥 ? + 𝐹𝑦 ? + 𝐹𝑧 ? − − 𝑀 = 𝑀𝑦 ? + 𝑀𝑧 ? HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.54

Nguyen Tan Tien

Equilibrium of a Rigid Body

§5. Free-Body Diagrams (3D) - Example 5.14 Consider the two rods and plate, along with their associated free-body diagrams. The 𝑥 , 𝑦 , 𝑧 axes are established on the diagram and the unknown reaction components are indicated in the positive sense. The weight is neglected Solution pin

• Free-body Diagrams

Properly aligned journal bearings at 𝐴, 𝐵, 𝐶

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

The force reactions developed by the bearings are sufficient for equilibrium since they prevent the shaft from rotating about each of the coordinate axes

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

9

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Engineering Mechanics – Statics

5.55

Equilibrium of a Rigid Body

§5. Free-Body Diagrams (3D)

Engineering Mechanics – Statics

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Equilibrium of a Rigid Body

§6. Equations of Equilibrium (3D) - Vector Equations of Equilibrium ∑𝐹 = 0 ∑𝑀𝑂 = 0 - Scalar Equations of Equilibrium ∑𝐹 = ∑𝐹𝑥 𝑖 + ∑𝐹𝑦 𝑗 + ∑𝐹𝑧 𝑘 = 0

Pin at 𝐴 and cable 𝐵𝐶

Moment components are developed by the pin on the rod to prevent rotation about the 𝑥 and 𝑧 axes

∑𝑀𝑂 = ∑𝑀𝑥 𝑖 + ∑𝑀𝑦 𝑗 + ∑𝑀𝑧 𝑘 = 0 or ∑𝐹𝑥 = 0, ∑𝐹𝑦 = 0, ∑𝐹𝑧 = 0 ∑𝑀𝑥 = 0, ∑𝑀𝑦 = 0, ∑𝑀𝑧 = 0

Properly aligned journal bearing at 𝐴 Only force reactions are developed by the bearing and hinge on the plate to prevent rotation about each and hinge at 𝐶. Roller at 𝐵 coordinate axis. No moments at the hinge are developed HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.57

Nguyen Tan Tien

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy - To ensure the equilibrium of a rigid body, it is not only necessary to satisfy the equations of equilibrium, but the body must also be properly held or constrained by its supports - Redundant constraints: when a body has redundant supports, that is, more supports than are necessary to hold it in equilibrium, it becomes statically indeterminate - Statically indeterminate: there will be more unknown loadings on the body than equations of equilibrium available for their solutions

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.58

Nguyen Tan Tien

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy - Example: the beam is shown together with its free-body diagram The beam is statically indeterminate because of additional (or redundant) supports reactions

There are five unknown 𝑀𝐴 , 𝐴𝑥 , 𝐴𝑦 , 𝐵𝑦 , 𝐶𝑦 for which only three equilibrium equations can be written ∑𝐹𝑥 = 0, ∑𝐹𝑦 = 0, ∑𝑀𝑂 = 0

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.59

Nguyen Tan Tien

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy - Example: the pipe is also statically indeterminate because of additional (or redundant) supports reactions

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.60

Nguyen Tan Tien

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy Statically indeterminate the number of unknown reactive forces > the number of the derived static equilibrium equations How to solve ? - The additional equations needed to solve statically indeterminate problems are generally obtained from the deformation conditions at the points of supports - This is done in courses dealing with “Mechanics of Materials”

The pipe assembly has eight unknowns, for which only six equilibrium equations can be written

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

10

2/13/2013

Engineering Mechanics – Statics

5.61

Equilibrium of a Rigid Body

Engineering Mechanics – Statics

5.62

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy - Improper constraints: having the same number of unknown reactive forces as available equations of equilibrium does not always guarantee that a body will be stable when subjected to a particular loading - For example, the pin support at 𝐴 and the roller support at 𝐵 for the beam are placed in such away that the lines of action the reactive forces are concurrent at point 𝐴

§7. Constrains and Statical Determinacy - In three dimensions, a body will be improperly constrained if the lines of action of all the reactive forces intersect a common axis - For example, the reactive forces at the ball-and-socket supports at 𝐴 and 𝐵 all intersect the axis passing through 𝐴 and 𝐵

- Consequently, the applied loading 𝑃 will cause the beam to rotates lightly about 𝐴 , and so the beam is improperly constrained

- Note: Since the moments of these forces about 𝐴 and 𝐵 are all zero, then the loading 𝑃 will rotate the member about the 𝐴𝐵 axis

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

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Nguyen Tan Tien

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy - Another way in which improper constraining leads to instability occurs when the reactive forces are all parallel

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.64

Nguyen Tan Tien

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy - Example 5.15 The homogeneous plate has a mass of 100𝑘𝑔 and is subjected to a force and couple moment along its edges. If it is supported in the horizontal plane by a roller at 𝐴 , a ball-and-socket joint at 𝐵, and a cord at 𝐶, determine the components of reaction at these supports Solution Free-body Diagram

- Note: the summation of forces along the 𝑥 axis will not be equal zero HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.65

Nguyen Tan Tien

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy Equations of Equilibrium ∑𝐹𝑥 = 0: 𝐵𝑥 = 0 ∑𝐹𝑦 = 0: 𝐵𝑦 = 0 ∑𝐹𝑧 = 0: 𝐴𝑧 + 𝐵𝑧 + 𝑇𝐶 − 300 − 981 = 0 ∑𝑀𝑥 = 0: 𝑇𝐶 × 2 + 981 × 1 + 𝐵𝑧 × 2 = 0 ∑𝑀𝑦 = 0: 300 × 1.5 + 981 × 1.5 −𝐵𝑧 × 3 − 𝐴𝑧 × 3 − 200 = 0 ⟹ 𝐴𝑧 = 790𝑁 𝐵𝑥 = 0 𝐵𝑦 = 0 𝐵𝑧 = −217𝑁 𝑇𝐶 = 707𝑁 HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.66

Nguyen Tan Tien

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy - Example 5.16 Determine the components of reaction that the ball-and-socket joint at 𝐴, the smooth journal bearing at 𝐵, and the roller support at 𝐶 exert on the rod assembly Solution Free-body Diagram Equations of Equilibrium ∑𝐹𝑥 = 0: 𝐴𝑥 + 𝐵𝑥 = 0 ∑𝐹𝑦 = 0: 𝐴𝑦 = 0 ∑𝐹𝑧 = 0: 𝐴𝑧 − 900 + 𝐵𝑧 + 𝐹𝐶 = 0 ∑𝑀𝑥 = 0: −900×0.4 +𝐵𝑧 ×0.8 +𝐹𝐶 ×1.2 = 0 ∑𝑀𝑦 = 0: −900 × 0.4 + 𝐹𝐶 × 0.6 = 0 ∑𝑀𝑧 = 0: 𝐵𝑥 × 0.8 = 0 ⟹ 𝐴𝑦 = 0, 𝐴𝑥 = 0, 𝐴𝑧 = 750𝑁, 𝐵𝑥 = 0, 𝐵𝑧 = −450𝑁, 𝐹𝐶 = 600𝑁 HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

11

2/13/2013

Engineering Mechanics – Statics

5.67

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy - Example 5.17 The boom is used to support the 75𝑁 flowerpot. Determine the tension developed in wires 𝐴𝐵 and 𝐴𝐶 Solution Free-body Diagram Equations of Equilibrium

Engineering Mechanics – Statics

Engineering Mechanics – Statics

Nguyen Tan Tien

5.69

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy 18 18 𝐹 + 𝐹𝐴𝐶 − 450 𝑖 7 𝐴𝐵 7 12 12 + − 𝐹𝐴𝐵 + 𝐹𝐴𝐶 𝑘 = 0 7 7 ⟹ ∑𝑀𝑥 = 0:

18 𝐹 7 𝐴𝐵

+

18 𝐹 7 𝐴𝐶

− 450 = 0

∑𝑀𝑦 = 0: 0 = 0 ∑𝑀𝑧 = 0: −

12 𝐹 7 𝐴𝐵

+

12 𝐹 7 𝐴𝐶

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.70

Engineering Mechanics – Statics

5.71

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy Applying the force equation of equilibrium

∑𝐹 = 0:

𝐹𝐴 + 𝑇𝐸 + 𝑇𝐷 + 𝐹 = 0

∑𝑀𝐴 = 0: 𝑟𝐶 × 𝐹 + 𝑟𝐵 × 𝑇𝐸 + 𝑇𝐷 = 0 ⟹ 𝐴𝑥 + 𝑇𝐸 𝑖 + 𝐴𝑦 + 𝑇𝐷 𝑗 + 𝐴𝑧 − 200 𝑘 = 0 0.5𝑖 + 𝑗 − 𝑘 × −200𝑘 + 𝑖 + 2𝑗 − 2𝑘 × 𝑇𝐸𝑖 + 𝑇𝐷𝑗 = 0 ⟹ 𝐴𝑥 + 𝑇𝐸 𝑖 + 𝐴𝑦 + 𝑇𝐷 𝑗

Equilibrium of a Rigid Body

𝐹𝐴 = 𝐴𝑥 𝑖 + 𝐴𝑦 𝑗 + 𝐴𝑧 𝑘

=0

Nguyen Tan Tien

Nguyen Tan Tien

§7. Constrains and Statical Determinacy - Example 5.18 Rod 𝐴𝐵 is subjected to the 200𝑁 force. Determine the reactions at the ball-and-socket joint 𝐴 and the tension in the cables 𝐵𝐷 and 𝐵𝐸 Solution Free-body Diagram Equations of Equilibrium 𝑇𝐸 = 𝑇𝐸 𝑖 𝑇𝐷 = 𝑇𝐷 𝑗 𝐹 = −200𝑘

⟹ 𝐹𝐴𝐵 = 87.5𝑁 𝐹𝐴𝐶 = 87.5𝑁

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Equilibrium of a Rigid Body

∑𝑀𝑂 = 0: 𝑟𝐴 × 𝐹𝐴𝐵 + 𝐹𝐴𝐶 + 𝑊 = 0 2 6 3 ⟹ 6𝑗 × 𝐹 𝑖− 𝐹 𝑗+ 𝐹 𝑘 + 7 𝐴𝐵 7 𝐴𝐵 7 𝐴𝐵 2 6 3 − 𝐹𝐴𝐶𝑖 − 𝐹𝐴𝐶𝑗 + 𝐹𝐴𝐶𝑘 − 75𝑘 = 0 7 7 7 18 18 ⟹ 𝐹 + 𝐹 − 450 𝑖 7 𝐴𝐵 7 𝐴𝐶 12 12 + − 𝐹𝐴𝐵 + 𝐹𝐴𝐶 𝑘 = 0 7 7

𝑟𝐴𝐵 2𝑖 − 6𝑗 + 3𝑘 𝐹𝐴𝐵 = 𝐹𝐴𝐵 = 𝐹𝐴𝐵 𝑟𝐴𝐵 22 + (−6)2 +32 2 6 3 = 𝐹𝐴𝐵 𝑖 − 𝐹𝐴𝐵 𝑗 + 𝐹𝐴𝐵 𝑘 7 7 7 𝑟𝐴𝐶 −2𝑖 − 6𝑗 + 3𝑘 𝐹𝐴𝐶 = 𝐹𝐴𝐶 = 𝐹𝐴𝐶 𝑟𝐴𝐶 (−2)2 +(−6)2 +32 2 6 3 = − 𝐹𝐴𝐶 𝑖 − 𝐹𝐴𝐶 𝑗 + 𝐹𝐴𝐶 𝑘 7 7 7 HCM City Univ. of Technology, Faculty of Mechanical Engineering

5.68

§7. Constrains and Statical Determinacy 2 6 3 𝐹𝐴𝐵 = 𝐹𝐴𝐵 𝑖 − 𝐹𝐴𝐵 𝑗 + 𝐹𝐴𝐵 𝑘 7 7 7 2 6 3 𝐹𝐴𝐶 = − 𝐹𝐴𝐶 𝑖 − 𝐹𝐴𝐶 𝑗 + 𝐹𝐴𝐶 𝑘 7 7 7 𝑊 = −75𝑘

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.72

Nguyen Tan Tien

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy - Example 5.19 The bent rod is supported at 𝐴 by a journal bearing, at 𝐷 by a ball-and-socket joint, and at 𝐵 by means of cable 𝐵𝐶. Using only one equilibrium equation, obtain a direct solution for the tension in cable 𝐵𝐶. The bearing at 𝐴 is capable of exerting force components only in the 𝑧 and 𝑦 directions since it is properly aligned on the shaft Solution Free-body Diagram

+ 𝐴𝑧 − 200 𝑘 = 0 2𝑇𝐷 − 200 𝑖 + −2𝑇𝐸 + 100 𝑗 + 𝑇𝐷 − 2𝑇𝐸 𝑘 = 0 ⟹ 𝑇𝐷 = 100𝑁, 𝑇𝐸 = 50𝑁, 𝐴𝑥 = −50𝑁, 𝐴𝑦 = −100𝑁, 𝐴𝑧 = 200𝑁 HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

12

2/13/2013

Engineering Mechanics – Statics

5.73

Equilibrium of a Rigid Body

§7. Constrains and Statical Determinacy Equations of Equilibrium The cable tension may be obtained directly by summing moments about an axis that passes through points 𝐷 and 𝐴 𝑟𝐷𝐴 1 1 𝑢= =− 𝑖− 𝑗 = −0.7071(𝑖 + 𝑗) 𝑟𝐷𝐴 2 2 The sum of the moments about this axis is zero ∑𝑀𝐷𝐴 = 𝑢∑ 𝑟 × 𝐹 = 0

Engineering Mechanics – Statics

5.74

Equilibrium of a Rigid Body

Fundamental Problems - F5.7 The uniform plate has a weight of 500𝑁. Determine the tension in each of the supporting cables

⟹ 𝑢 𝑟𝐵 × 𝑇𝐵 + 𝑟𝐸 × 𝑊 = 0 −0.7071(𝑖 + 𝑗) −𝑗 × 𝑇𝐵𝑘 − 0.5𝑗 × −981𝑘 −0.7071(𝑖 + 𝑗) −𝑇𝐵 + 490.5 𝑖 = 0 −0.7071 −𝑇𝐵 + 490.5 + 0 + 0 = 0 ⟹ 𝑇𝐵 = 490.5𝑁 HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.75

=0

Nguyen Tan Tien

Equilibrium of a Rigid Body

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.76

Nguyen Tan Tien

Equilibrium of a Rigid Body

Fundamental Problems - F5.8 Determine the reactions at the roller support 𝐴, the balland-socket joint 𝐷, and the tension in cable 𝐵𝐶 for the plate

Fundamental Problems - F5.9 The rod is supported by smooth journal bearings at 𝐴, 𝐵 and 𝐶 and is subjected to the two forces. Determine the reactions at these supports

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Engineering Mechanics – Statics

5.77

Nguyen Tan Tien

Equilibrium of a Rigid Body

Engineering Mechanics – Statics

5.78

Nguyen Tan Tien

Equilibrium of a Rigid Body

Fundamental Problems - F5.10 Determine the support reactions at the smooth journal bearings 𝐴, 𝐵, and 𝐶 of the pipe assembly

Fundamental Problems - F5.11 Determine the force developed in cords 𝐵𝐷, 𝐶𝐸, and 𝐶𝐹 and the reactions of the ball-and-socket joint 𝐴 on the block

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

13

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Engineering Mechanics – Statics

5.79

Equilibrium of a Rigid Body

Fundamental Problems - F5.12 Determine the components of reaction that the thrust bearing 𝐴 and cable 𝐵𝐶 exert on the bar

HCM City Univ. of Technology, Faculty of Mechanical Engineering

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14