Ch05a

Engineering Mechanics: STATICS Anthony Bedford and Wallace Fowler SI Edition Teaching Slides Chapter 5: Objects in Equilibrium

Chapter Outline  

The Equilibrium Equations 2-Dimensional Applications

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5.1 The Equilibrium Equations 

When an object acted upon by a system of forces & moments is in equilibrium, the following conditions are satisfied: 1. The sum of the forces is zero: ΣF=0 (5.1) 2. The sum of the moments about any point is zero: Σ Many point = 0 (5.2)

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5.1 The Scalar Equilibrium Equations 



When the loads and reactions on an object in equilibrium form a two-dimensional system of forces and moments, they are related by three scalar equilibrium equations: Σ Fx = 0 (5.3) Σ Fy = 0 (5.4) Σ Many point = 0 (5.5) More than three independent equilibrium equations cannot be obtained from a two-dimensional free-body diagram, which means we can solve for at most three unknown forces or couples. (C) 2005 Pearson Education South Asia Pte Ltd

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5.1 The Equilibrium Equations  Eqs.

(5.1) & (5.2) imply that the system of forces & moments acting on an object in equilibrium is equivalent to a system consisting no forces & no couples  If the sum of the forces on an object is zero & the sum of the moments about 1 point is zero, then the sum of the moments about every point is zero

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5.2 2-Dimensional Applications 

Supports:  Forces

& couples exerted on an object by its supports are called reactions, expressing the fact that the supports “react” to the other forces & couples or loads acting on the object 20 KN-m

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5.2 2-Dimensional Applications 

Supports:  Some

very common kinds of supports are represented by stylized models called support conventions if the actual supports exert the same reactions as the models

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5.2 2-Dimensional Applications 

The Pin Support:  Figure

a: a pin support  a bracket to which an object (such as a beam) is attached by a smooth pin that passes through the bracket & the object  Figure b: side view

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5.2 2-Dimensional Applications  To

understand the reactions that a pin support can exert:  Imagine holding the bar attached to the pin support  If you try to move the bar without rotating it (i.e. translate the bar), the support exerts a reactive force that prevents this movement  However, you can rotate the bar about the axis of the pin  The support cannot exert a couple about the pin axis to prevent rotation

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5.2 2-Dimensional Applications  Thus,

a pin support can’t exert a couple about the pin axis but it can exert a force on the object in any direction, which is usually expressed by representing the force in terms of components  The arrows indicate the directions of the reactions if Ax & Ay are positive  If

you determine Ax or Ay to be negative, the reaction is in the direction opposite to that of the (C) 2005 Pearsonarrow Education South Asia Pte Ltd

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5.2 2-Dimensional Applications  The

pin support is used to represent any real support capable of exerting a force in any direction but not exerting a couple  Used

in many common devices, particularly those designed to allow connected parts to rotate relative to each other

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5.2 2-Dimensional Applications 

The Roller Support: A

pin support mounted on wheels  Like a pin support, it cannot exert a couple about the axis of the pin  Since it can move freely in the direction parallel to the surface on which it rolls, it can’t exert a force parallel to the surface but can exert a force normal (perpendicular) to this surface

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5.2 2-Dimensional Applications  Other

commonly used conventions equivalent to the roller support:

 The

wheels of vehicles & wheels supporting parts of machines are roller supports if the friction forces exerted on them are negligible in comparison to the normal forces

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5.2 2-Dimensional Applications A

plane smooth surface can also be modeled by a roller support:

 Beams

& bridges are sometimes supported in this way so that they will be free to undergo thermal expansion & contraction

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5.2 2-Dimensional Applications  These

supports are similar to the roller support in that they cannot exert a couple & can only exert a force normal to a particular direction (friction is neglected)

(a) Pin in a slot

(b) Slider in a slot

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(c) Slider on a shaft

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5.2 2-Dimensional Applications  In

these supports, the supported object is attached to a pin or slider that can move freely in 1 direction but is constrained in the perpendicular direction  Unlike the roller support, these supports can exert a normal force in either direction

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5.2 2-Dimensional Applications 

The Fixed Support:  The

fixed support shows the supported object literally built into a wall (built-in)

 To

understand the reactions:  Imagine holding a bar attached to the fixed support

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5.2 2-Dimensional Applications  If

you try to translate the bar, the support exerts a reactive force that prevents translation  If you try to rotate the bar, the support exerts a reactive couple that prevents rotation  A fixed support can exert 2 components of force & a couple

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5.2 2-Dimensional Applications  The

term MA is the couple exerted by the support & the curved arrow indicates its direction  Fence posts have fixed supports  The attachments of parts connected so that they cannot move or rotate relative to each other, such as the head of a hammer & its handle, can be modeled as fixed supports

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5.2 2-Dimensional Applications 

Table 5.1 summarizes the support conventions commonly used in 2-D applications:

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5.2 2-Dimensional Applications Table 5.1

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5.2 2-Dimensional Applications 

Free-Body Diagrams:  By

using the support conventions, we can model more elaborate objects & construct their free-body diagrams in a systematic way  Example:  a beam with a pin support at the left end & a roller support on at the right end & is loaded with a force F  The roller support rests on a surface inclined at 30° (C) 2005 Pearson Education South Asia Pte Ltd

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5.2 2-Dimensional Applications  To

obtain a free-body diagram of the beam, isolate it from its supports  Complete the free-body diagram by showing the reactions that may be exerted on the beam by the supports  Notice that the reaction at B exerted by the roller support is normal to the surface on which the support rests (C) 2005 Pearson Education South Asia Pte Ltd

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5.2 2-Dimensional Applications  Example:  The

object in this figure has a fixed support at the left end  A cable passing over a pulley is attached to the object at 2 points

 Isolate

it from its supports & complete the free-body by showing the reactions at the fixed support & the forces exerted by the cable

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5.2 2-Dimensional Applications

Don’t forget the couple at the fixed support  Since we assume the tension in the cable is the same on both sides of the pulley, the 2 forces exerted by the cable have the magnitude T 

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5.2 2-Dimensional Applications  Once

you have obtained the free-body diagram of an object in equilibrium to identify the loads & reactions acting on it, you can apply the equilibrium equations

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Example 5.1 Reactions at Pin & Roller Supports The beam in Fig. 5.1 has a pin at A & roller supports at B & is subjected to a 2-kN force. What are the reactions at the supports?

Fig. 5.1

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Example 5.1 Reactions at Pin & Roller Supports Strategy To determine the reactions exerted on the beam by its supports, draw a free-body diagram of the beam isolated from the supports. The free-body diagram must show all external forces & couples acting on the beam, including the reactions exerted by the supports. Then determine the unknown reactions by applying equilibrium equations

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Example 5.1 Reactions at Pin & Roller Supports Solution Draw the Free-Body Diagram: Isolate the beam from its supports & show the loads & the reactions that may be exerted by the pin & roller supports. There are 3 unknown reactions: 2 components of force Ax & Ay at the pin support & a force B at the roller support (C) 2005 Pearson Education South Asia Pte Ltd

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Example 5.1 Reactions at Pin & Roller Supports Solution Apply the Equilibrium Equations: Summing the moments about point A: Σ Fx = Ax − Bsin 30° = 0 Σ Fy = Ay + Bcos 30° − 2 kN = 0 Σ Mpoint A = (5 m)(Bcos 30°) − (3 m)(2 kN) = 0

Solving these equations, the reactions are: Ax = 0.69 kN, Ay = 0.80 kN & B = 1.39 kN

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Example 5.1 Reactions at Pin & Roller Supports Solution Confirm that the equilibrium equations are satisfied: Σ Fx = 0.69 kN − (1.39 kN)sin 30° = 0 Σ Fy = 0.80 kN + (1.39 kN)cos 30° − 2 kN = 0 Σ Mpoint A = (5 m)(1.39 kN)cos 30° − (3 m)(2 kN) = 0

Critical Thinking 

In drawing free-body diagrams, you should try to choose the correct directions of the reactions 31

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Example 5.1 Reactions at Pin & Roller Supports Critical Thinking 

However, if you choose an incorrect direction for a reaction in drawing the free-body diagram of a single object, the value you obtain from the equilibrium equations for that reaction will be negative, which indicates that its actual direction is opposite to the direction you chose  E.g. if we draw the free-body diagram of the beam with the component Ay pointed downward:

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Example 5.1 Reactions at Pin & Roller Supports Critical Thinking  Equilibrium

equations: Σ Fx = Ax − Bsin 30° = 0 Σ Fy = −Ay + Bcos 30° − 2 kN = 0 Σ Mpoint A = (5 m)(Bcos 30°) − (3 m)(2 kN) = 0

 Solving,

we obtain: Ax = 0.69 kN, Ay = −0.80 kN & B = 1.39 kN

 The

negative value of Ay indicates that the vertical force exerted on the beam by the pin support at A is in the direction opposite to the arrow, i.e. the force is 0.80 kN upward 33 (C) 2005 Pearson Education South Asia Pte Ltd

Example 5.2 Reactions at a Fixed Support The object in Fig. 5.2 has a fixed support at A & is subjected to 2 forces & a couple. What are the reactions at the support?

Fig. 5.2 (C) 2005 Pearson Education South Asia Pte Ltd

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Example 5.2 Reactions at a Fixed Support Strategy Obtain a free-body diagram by isolating the object from the fixed support at A & showing the reactions exerted at A, including the couple that may be exerted by a fixed support. Then determine the unknown reactions by applying the equilibrium equations.

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Example 5.2 Reactions at a Fixed Support Solution Draw the Free-Body Diagram: Isolate the object from its support & show the reactions at the fixed support. There are 3 unknown reactions: 2 force components Ax & Ay & a couple MA. (Remember that we can choose the directions of these arrows arbitrarily) Also resolve the 100-N force into its components. (C) 2005 Pearson Education South Asia Pte Ltd

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Example 5.2 Reactions at a Fixed Support Solution Draw the Free-Body Diagram:

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Example 5.2 Reactions at a Fixed Support Solution Apply the Equilibrium Equation: Summing the moments about point A: Σ Fx = Ax + 100cos 30° N = 0 Σ Fy = Ay − 200 N + 100sin 30° N = 0 Σ Mpoint A = MA + 300 N-m − (2 m)(200 N) − (2 m)(100cos 30° N) + (4 m)(100sin 30° N)= 0 Solving these equations, Ax = −8.86 N, Ay = 150.0 N & MA = 73.2 N-m. (C) 2005 Pearson Education South Asia Pte Ltd

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Example 5.2 Reactions at a Fixed Support Critical Thinking 

Why don’t the 300 N-m couple & the couple MA exerted by the fixed support appear in the first 2 equilibrium equations?  A couple exerts no net force  Also, because the moment due to a couple is the same about any point, the moment about A due to the 300 N-m counterclockwise couple is 300 N-m counterclockwise

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Exercise 5.6 Diving Board 

The masses of the person and the diving board are 54 kg and 36 kg, respectively. Assume that they are in equilibrium.

(a) Draw the free-body diagram of the diving board. (b) Determine the reactions at the supports A and B. Answers (C) 2005 Pearson Education South Asia Pte KN Ltd Ax=0; Ay=-1.85 KN;By=2.74

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Exercise 5.7 Ironing Board The ironing board has supports at A and B that can be modeled as roller supports. (a) Draw the free-body diagram of the ironing board. (b) Determine the reactions at A and B. 

Answers

Ay=79.2 N By=144.2 N (C) 2005 Pearson Education South Asia Pte Ltd

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Exercise 5.16 A Person Doing Push-ups 

A person doing push-ups pauses in the position shown. His mass is 80 kg. Assume that his weight W acts at the point shown. The dimensions shown are a = 250 mm, b = 740 mm, and c = 300 mm. Determine the normal force exerted by the floor (a) on each hand, (b) on each foot.

Answers Force on each hand=FH=293.3 N Force on each feet=FF=99.1 N (C) 2005 Pearson Education South Asia Pte Ltd

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Exercise 5.19 Beam with cable passing through pulley (a) Draw the free-body diagram of the beam. (b) Determine the tension in the cable and the reactions at A.

Answers AX=554 N AY=-160 N T=640 N (C) 2005 Pearson Education South Asia Pte Ltd

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Exercise 5.26 Wheelbarrow The total weight of the wheelbarrow and its load is W = 100 lb. (a) If F = 0, what are the vertical reactions at A and B? (b) What force F is necessary to lift the support at A off the ground? Answers AX= 0 N AY= 269.2 N FB= 230.8 N F= 106.1 N (C) 2005 Pearson Education South Asia Pte Ltd

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Exercise 5.34 The forklift is stationary. The sign’s weight WS = 160 N acts at the point shown. The 50-N weight of the bar AD acts at the midpoint of the bar. Determine the tension in the cable AE and the reactions at D. Answers TAE= 165.2 N, DX= -155.2 N DY(C)=2005 -153.5 N Pearson Education South Asia Pte Ltd

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Exercise 5.36 Truss This structure, called a truss, has a pin support at A and a roller support at B and is loaded by two forces. Determine the reactions at the supports. Answers AX= -1.828 KN AY= 2.10 KN BY= 2.46 KN (C) 2005 Pearson Education South Asia Pte Ltd

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Exercise 5.61 The dimensions a = 2 m and b = 1 m. The couple M = 2400 N-m. The spring constant is k = 6000 N/m, and the spring would be unstretched if h = 0.The system is in equilibrium when h = 2 m and the beam is horizontal. Determine the force F and the reactions at A.This structure, called a truss, has a pin support at A and a roller support at B and is loaded by two forces. Determine the reactions at the supports. Answers AX= 3045 N AY= -185 N F= 1845 N (C) 2005 Pearson Education South Asia Pte Ltd

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