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Lecture #7 Part a

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Lecture #7 Part a

2D Rigid Body Equilibrium

(Reference Chapter 5, sections 1, 2, 3 and 4 – 2D)

- Can we solve for the Reaction at each wheel on this truck using Translational Equilibrium?
- Is this problem Statically Indeterminate?

- Given: Mass of the truck is 310.56 Slugs.
- Force pulling the truck is 800 Lbs attached at the bumper 1.5 ft off the ground and acting parallel to the ground.
- The distance between the wheels is 15 ft
- Center of Gravity is 5 ft in front of the rear wheel
- There is an equal force on each tire toward the back of the truck to offset the pulling force

- Unknown external forces are also known as constraints
- Constraints on Space Diagrams need to be replaced with the correct reactionary forces on the Free Body Diagram
- In a real world problem, the type of constraint at each reaction point needs to be determined

The truck ramp has a weight of 400 lb.

The ramp is pinned to the body of the truck and held in the position by the cable. How can we determine the cable tension and support reactions ?

APPLICATIONS

How are the idealized model and the free body diagram used to do this?

Which diagram above is the idealized model?

Two smooth pipes, each having a mass of 300 kg, are supported by the tines of the tractor fork attachment.

How can we determine all the reactive forces ?

Again, how can we make use of an idealized model and a free body diagram to answer this question?

APPLICATIONS (continued)

In contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent and may cause rotation of the body (due to the moments created by the forces).

Forces on a particle

For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be equal to zero.

F = 0 (no translation)

and MO= 0 (no rotation)

Forces on a rigid body

CONDITIONS FOR RIGID-BODY EQUILIBRIUM

(Section 5.1)

Write scaler equations on board

Then we need to draw a free-body diagram (FBD) showing all the external (active and reactive) forces.

THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS

For analyzing an actual physical system, first we need to create an idealized model (above right).

Finally, we need to apply the equations of equilibrium to solve for any unknowns.

Free-body diagram (FBD)

- 1.Draw an outlined shape. Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape.

FREE-BODY DIAGRAMS (Section 5.2)

Idealized model

- 2.Show all the external forces and couple moments. These typically include: a) applied loads, b) support reactions, and, c) the weight of the body.

FREE-BODY DIAGRAMS (continued)

Idealized model

Free-body diagram

- 3.Label loads and dimensions on the FBD: All known forces and couple moments should be labeled with their magnitudes and directions. For the unknown forces and couple moments, use letters like Ax, Ay, MA, etc.. Indicate any necessary dimensions.

SUPPORT REACTIONS IN 2-D

A few example sets of diagrams s are shown above. Other support reactions are given in your textbook (Table 5-1).

As a general rule, if a support prevents translationof a body in a given direction, thena force is developedon the body in the opposite direction.

Similarly,if rotation is prevented, a couple momentis exerted on the body in the opposite direction.

Given:The operator applies a vertical force to the pedal so that the spring is stretched 1.5 in. and the force in the short link at B is

20 lb.

Draw:A an idealized model and free-body diagram of the foot pedal.

EXAMPLE

1. The beam and the cable (with a frictionless pulley at D) support an 80 kg load at C. In a FBD of only the beam, there are how many unknowns?

A) 2 forces and 1 couple moment

B) 3 forces and 1 couple moment

C) 3 forces

D) 4 forces

CONCEPT QUIZ

2. If the directions of the force and the couple moments are both reversed, what will happen to the beam?

A)The beam will lift from A.B)The beam will lift at B.C)The beam will be restrained.

D)The beam will break.

CONCEPT QUIZ

Draw a FBD of the crane boom, which is supported by a pin at A and cable BC. The load of 1250 lb is suspended at B and the boom weighs 650 lb.

GROUP PROBLEM SOLVING

GROUP PROBLEM SOLVING (continued)

FBD

Draw a FBD of member ABC, which is supported by a smooth collar at A, roller at B, and link CD.

GROUP PROBLEM SOLVING

GROUP PROBLEM SOLVING (continued)

FBD

2. How many unknown support reactions are there in this problem?

A)2 forces and 2 couple moments

B)1 force and 2 couple moments

C)3 forces

D)3 forces and 1 couple moment

ATTENTION QUIZ

1. Internal forces are not shown on a free-body diagram because the internal forces are_____. (Choose the most appropriate answer.)

A) Equal to zero B) Equal and opposite and they do not affect the calculations

C) Negligibly small D) Not important

Remaining slides = Section 5.3/5.4 –

2D Rigid Body Equilibrium!!! (Finally)

READING QUIZ (Section 5.3/5.4)

- 1. The three scalar equations FX = FY = MO = 0, are ____ equations of equilibrium in two dimensions.
- A) IncorrectB) The only correct
- C) The most commonly usedD) Not sufficient

2. A rigid body is subjected to forces as shown. This body can be considered as a ______ member.

A)Single-forceB) Two-force

C)Three-forceD) Six-force

APPLICATIONS

A

The uniform truck ramp has a weight of 400 lb.

The ramp is pinned at A and held in the position by the cable. How can we determine the forces acting at the pin A and the force in the cable ?

APPLICATIONS (continued)

A 850 lb of engine is supported by three chains, which are attached to the spreader bar of a hoist.

You need to check to see if the breaking strength of any of the chains is going to be exceeded. How can you determine the force acting in each of the chains?

A body is subjected to a system of forces that lie in the x-y plane. When in equilibrium, the net force and net moment acting on the body are zero (as discussed earlier in Section 5.1). This 2-D condition can be represented by the three scalar equations:

EQUATIONS OF EQUILIBRIUM (Section 5.3)

Fx = 0 Fy = 0 MO = 0

where point O is any arbitrary point.

Please note that these equations are the ones most commonly used for solving 2-D equilibrium problems. There are two other sets of equilibrium equations that are rarely used. For your reference, they are described in the textbook.

The solution to some equilibrium problems can be simplified if we recognize members that are subjected to forces at only two points (e.g., at points A and B).

TWO-FORCE MEMBERS & THREE FORCE-MEMBERS (Section 5.4)

If we apply the equations of equilibrium to such a member, we can quickly determine that the resultant forces at A and B must be equal in magnitude and act in the opposite directions along the line joining points A and B.

In the cases above, members AB can be considered as two-force members, provided that their weight is neglected.

This fact simplifies the equilibrium analysis of some rigid bodies since the directions of the resultant forces at A and B are thus known (along the line joining points A and B).

EXAMPLE OF TWO-FORCE MEMBERS

STEPS FOR SOLVING 2-D EQUILIBRIUM PROBLEMS

1. If not given, establish a suitable x - y coordinate system.

2. Draw a free body diagram (FBD) of the object under analysis.

3. Apply the three equations of equilibrium (E-of-E) to solve for the unknowns.

IMPORTANT NOTES

1. If there are more unknowns than the number of independent equations, then we have a statically indeterminate situation.We cannot solve these problems using just statics.

2. The order in which we apply equations may affect the simplicity of the solution. For example, if we have two unknown vertical forces and one unknown horizontal force,then solving FX = 0 first allows us to find the horizontal unknown quickly.

3. If the answer for an unknown comes outas negative number, then the sense (direction) of the unknown force is opposite to that assumed when starting the problem.

Given:The 4kN load at B of the beam is supported by pins at A and C .

Find:The support reactions at A and C.

Plan:

EXAMPLE 2D Equilibrium:

1. Put the x and y axes in the horizontal and vertical directions, respectively.

2. Determine if there are any two-force members.

3. Draw a complete FBD of the boom.

4. Apply the E-of-E to solve for the unknowns.

EXAMPLE (Continued)

FBD of the beam:

AY

4 kN

1.5 m

1.5 m

AX

C

A

B

45°

FC

Note: Upon recognizing CD as a two-force member, the number of unknowns at C are reduced from two to one. Now, using E-o-f E, we get,

+ MA = FC sin 45 1.5 – 4 3 = 0

Fc = 11.31 kN or 11.3 kN

+ FX = AX + 11.31 cos 45 = 0; AX = – 8.00 kN

+ FY = AY + 11.31 sin 45 – 4 = 0; AY = – 4.00 kN

Note that the negative signs means that the reactions have the opposite direction to that shown on FBD.

- 1. For this beam, how many support reactions are there and is the problem statically determinate?
- A) (2, Yes)B) (2, No)
- C) (3, Yes)D) (3, No)

F

F

F

F

Fixed support

F

A

B

CONCEPT QUIZ

2. The beam AB is loaded and supported as shown: a) how many support reactions are there on the beam, b) is this problem statically determinate, and c) is the structure stable?

A) (4, Yes, No)B) (4, No, Yes)

C) (5, Yes, No)D) (5, No, Yes)

Given:The jib crane is supported by a pin at C and rod AB. The load has a mass of 2000 kg with its center of mass located at G.

Assume x = 5 m.

Find: Support reactions at B and C.

GROUP PROBLEM SOLVING

2D Equilibrium

Plan:

a) Establish the x – y axes.

b) Draw a complete FBD of the jib crane beam.

c) Apply the E-of-E to solve for the unknowns.

FBD of the beam

GROUP PROBLEM SOLVING (Continued)

FAB

4

3

4 m

5

0.2 m

Cx

5 m

2000(9.81) N

Cy

First write a moment equation about Point C.

Why point C?

+ MC = (3 / 5) FAB 4 + (4 / 5) FAB 0.2 – 2000(9.81) 5 = 0

FAB = 38320 N = 38.3 kN

FBD of the beam

GROUP PROBLEM SOLVING (Continued)

FAB

4

3

4 m

5

0.2 m

Cx

5 m

2000(9.81) N

Cy

FAB = 38320 N = 38.3 kN

Now solve the FX and FY equations.

+ FX = Cx– (4 / 5) 38320 = 0

+ FY = – Cy+ (3 / 5) 38320 – 2000(9.81) = 0

Solving these two equations, we getCx = 30656 N or 30.7 kN and Cy = 3372 N or 33.7 kN

100 lb

AX

A

B

FB

AY

ATTENTION QUIZ

1. Which equation of equilibrium allows you to determine FB right away?

A) FX = 0 B) FY = 0

C) MA = 0 D) Any one of the above.

2.A beam is supported by a pin joint and a roller. How many support reactions are there and is the structure stable for all types of loadings?

A)(3, Yes) B) (3, No)

C)(4, Yes) D) (4, No)

- Solve this for the Unknown Reactions at Point A and B

2m

1.5m

1.5m

- Given that the table weighs 50 lbs, what is the maximum weight of the person sitting at it before it tips over?

F5-1: Find Support Reactions

Find Support Reactions

24,000 lb

5:20. The train car has a weight of 24,000 lb and a center of gravity at G. It is suspended from its front and rear track by six tires located at A, B, and C. Determine the normal reactions on these tires if the track is assumed to be a smooth surface and an equal portion of the load is supported at both the front and rear tires.

NC

NA

NB

5.21: Determine the horizontal and vertical components of reaction at pin A and the tension developed in cable BC used to support the steel frame.

5-40: The assembly has a weight of 250 lb and center of gravity at G1. If it is intended to support a maximum load of 400 lb placed at point G2, determine the smallest counterweight W that should be placed at B in order to prevent the platform from tipping over.

5.47: The motor has a weight of 850 lb. Determine the force that each of the chains exerts on the supporting hooks at A, B and C. Neglect the size of the hooks and thickness of the beam.

What’s better? Attaching the rope at A or B? FBD!!!!!

Example: 2D – 2 force and 3 force members:

- The last part of doing Rigid Body Equilibrium problems is understanding what needs solved.
- This has to be done by careful examination and engineering judgment (for lack of a better word)

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Accelerometer location Left/Rear

Accelerometer location Left/Front

Accelerometer location Right/Front

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Vertical acceleration, left front

Woofer Isolator

Face Mask Shock Isolator:

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Butyl elastomer bonded to g/f nylon

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Isolation mount

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