ch05a

Engineering Engineering Mechanics:Mechanics:

STATICSSTATICS

Anthony Bedford and Wallace Fowler

SI Edition

Teaching SlidesChapter 5:

Objects in Equilibrium

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Chapter OutlineChapter Outline

The Equilibrium Equations 2-Dimensional Applications

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5.1 5.1 The Equilibrium EquationsThe 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:

Σ F = 0 (5.1)2. The sum of the moments about any point is

zero:

Σ Many point = 0 (5.2)

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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.

5.1 5.1 The Scalar Equilibrium EquationsThe Scalar Equilibrium Equations

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5.1 5.1 The Equilibrium EquationsThe 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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-Dimensional Applications

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 arrow

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

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5.2 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 (c) Slider on a shaft

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5.2 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-Dimensional Applications

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

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5.2 5.2 2-Dimensional Applications2-Dimensional Applications

Table 5.1

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5.2 5.2 2-Dimensional Applications2-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°

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5.2 5.2 2-Dimensional Applications2-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

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5.2 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 5.2 2-Dimensional Applications2-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 Example 5.1 Reactions at Pin & Roller SupportsSupports

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 Example 5.1 Reactions at Pin & Roller SupportsSupports

StrategyStrategy 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 Example 5.1 Reactions at Pin & Roller SupportsSupports

SolutionSolutionDraw 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

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Example 5.1 Reactions at Pin & Roller Example 5.1 Reactions at Pin & Roller SupportsSupportsSolutionSolutionApply 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 Example 5.1 Reactions at Pin & Roller SupportsSupports

SolutionSolutionConfirm 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 ThinkingCritical Thinking In drawing free-body diagrams, you should try to

choose the correct directions of the reactions because it helps to develop your physical intuition

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Example 5.1 Reactions at Pin & Roller Example 5.1 Reactions at Pin & Roller SupportsSupportsCritical ThinkingCritical 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 Example 5.1 Reactions at Pin & Roller SupportsSupports

Critical ThinkingCritical 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

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Example 5.2 Reactions at a Fixed SupportExample 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

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Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support

StrategyStrategy 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 SupportExample 5.2 Reactions at a Fixed Support

SolutionSolution 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.

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Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support

SolutionSolutionDraw the Free-Body Diagram:

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Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support

SolutionSolutionApply 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.

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Example 5.2 Reactions at a Fixed SupportExample 5.2 Reactions at a Fixed Support

Critical ThinkingCritical 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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ExExerciseercise 5. 5.66 Diving BoardDiving 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.

AnswersAx=0; Ay=-1.85 KN;By=2.74 KN

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ExExerciseercise 5. 5.77 Ironing BoardIroning 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

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ExExerciseercise 5. 5.1616 A Person Doing Push-upsA 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

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ExExerciseercise 5. 5.19 Beam with cable passing 19 Beam with cable passing through pulleythrough 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

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Exercise 5.26 WheelbarrowExercise 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 NF= 106.1 N

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Exercise 5.34Exercise 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= -153.5 N

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Exercise 5.36 TrussExercise 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

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Exercise 5.61Exercise 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.

AnswersAX= 3045 NAY= -185 N F= 1845 N