E-Book of Statics

CHAPTER 1-BASICS-A- Problem Solving & Significant Digits

STATICS - CASE STUDY

Introduction

Problem DescriptionClick to view movie (98k)

Susan, a new engineering student, is being very careful to write down each calculation step as she proceeds through a complex problem. Her friend Sam, on the other hand, wants to round all numbers to only two digits to get done faster.

What is known:

The traffic light weighs 146.7 lb. Cable A is 17.49° from the horizontal. Cable B is 11.46° from the horizontal.

Questions

What is Statics?Click to view movie (99k)

What is the tension in both cables using four significant digits and two significant digits? Does rounding numbers to only two digits still provide an accurate answer?

Approach

Solve the problem using the basic steps in engineering problem solving: 1) read 2) draw diagrams 3) write equations 4) solve and 5) check.

For step 3, use the summation of forces in both the x and y directions:

ΣFx = 0 ΣFy = 0 Solve twice, once with four significant digits

and once with two significant digits.

Force Diagram

STATICS - THEORY

Introduction

Example of beam in equilibrium when forces are acting in multiple directions

Welcome to Statics. For most engineering students this will be your first serious engineering class. Like most engineering classes, Statics relies on previous courses (math and physics) and will be used in future courses that you will take.

Statics has all the basics of a regular engineering course, but is not too complex. The course revolves around just two equations,

ΣF = 0 and ΣM = 0

Another way to think about these two main equations is that all forces (F) and moments (M) acting on a object must be in equilibrium. If this was not the case, then the object would move. In Statics it is assumed that nothing moves, but in Dynamics (the following course) this restriction is removed.

Before solving actual statics problems, units, solution methods and significant digits need to be discussed.

General Solution Procedure

Example problem solved with solution steps

Click to view movie (114k)

The study of Statics involves only a few basic concepts and equations. The difficulty in statics is learning how to apply the equations to complex and different situations.

In engineering, following a set procedure in solving problems saves time and decreases the number of errors. Generally, the following five steps will help solve problems in statics.

1. Read and understand the question(s).a) Identify all given data.b) Identify exactly what is being asked.

2. Make sketches and draw free-body diagrams. 3. Set up the problem and apply all relevant

equations. 4. Perform the calculations. 5. Check and review the answer.

Fundamental Concepts

As the name implies, Statics refers to the static

equilibrium of all forces and moments acting on a body. This condition can be written in mathematical form as

ΣF = 0 Forces

ΣM = 0 Moments

While not all problems have both forces and moments in all three directions, both equations need to be satisfied in all three directions for the object to be in static equilibrium. Even though these two equations are simply, applying them to a particular structure can be difficult. Where are the forces, what are their directions, what is known, etc. become the main issues in the most Static problems.

Significant Digits

In most engineering problems, there will be numerous intermediate steps and results. When writing down these intermediate results, it is important to use enough digits to minimize rounding errors. However, if too many digits are used, it takes a long time to write out the equations, and errors are easily made in keeping track of all numerical values.

In general, numerical values can be rounded to four significant digits without loss of accuracy. Even using three digits is usually safe for most problems. However, round all numbers the same. If even one number is rounded to 2 significant digits and all the rest have four significant digits, the problem can still be off by a large amount (over 2%).

Please note, significant digits is not quite the same as the just rounding decimal places. Significant digits are applicable to numbers both large and small whether nor not they have decimals. The table below shows a number of examples that illustrate that all numbers can be rounded to a given number of significant digits.

Every number in the calculator should not be round just for the sake of rounding. Let the calculator keep all digits.

ActualNumber

Significant Digits

1 2 3 4

5234.4567 5000 5200 5230 5234

0.003452 0.003 0.00345 0.00345 0.003452

98.2384 100 98 98.2 98.24

23000 20000 23000 23000 23000

729.5 700 730 730 729.5

0.0239 0.02 0.024 0.0239 0.0239

STATICS - CASE STUDY SOLUTION

Using the basic force-equilibrium equations in the x and y directions, the forces FA and FB can be determined:

ΣFx = 0 ΣFy = 0

Sum the forces in both directions gives:

-FA cosα + FB cosθ = 0 FA sinα + FB sinθ - W = 0

Four Significant Digits

Substitute numerical values up to four significant digits into the previous two equations.

-FA cos17.49 + FB cos11.46 = 0 FA sin17.49 + FB sin11.46 -146.7 = 0

Solve the two equations for the unknown values of FA and FB:

FA = 488.1 - 0.6434 FA

FA = 297.0 lb FB = 289.0 lb

Two Significant Digits

Round the given numerical values to two significant digits and substitute into previous equations.

-FA cos17 + FB cos11 = 0 FA sin17 + FB sin11 - 150 = 0

Solve the two equations for the unknown values of FA and FB:

The numerical values in each step are also rounded

to two significant digits:

FA = 520 - 0.64 FA

FA = 320 lb FB = 320 lb

Notice the value of FA is 23 lb, or 7.7% higher, than when four significant digits were used. The percentage of error will vary from problem to problem, but when only two significant digits are used, the answer will generally be off by more than 1 percent. Therefore, rounding to only two significant digits should be avoided.

CHAPTER 1- B-UNITS:STATICS - CASE STUDY

Introduction

Toy Cars on KitchenClick to view movie (440k)

Jimmy has a choice between two remote-control cars, a 1967 Ford Mustang and a 1976 Ford Pinto. He would like to have the faster one, but the models are manufactured in different countries and thus have a different maximum speed printed in the specifications.

What is known:

The maximum speed of the remote-control 1967 Mustang model is 29 ft/s.

The maximum speed of the remote-control 1976 Pinto model is 10 m/s.

Question

What is the maximum speed of the 1967 Mustang and the 1976 Pinto remote-control model in both mi/hr (mph) and km/hr? Which car is faster?

Approach

Convert the speed of both cars to a common set of units, mph and km/hr, and then compare the speeds.

CHAPTER-2-A- Scalars, Vectors and Operations


Introduction


NASA is planning an unmanned mission to Mars. Mission planners need the precise position of the planet relative to Earth on the launch date in order to determine the proper velocity for the spacecraft. If the distance or angle is incorrect, the spacecraft could be lost.

What is known:

The radius of Earth's orbit re is 150 x 106 km. The radius of Mars's orbit rm is 207 x 106 km. The angle α between Earth and Mars relative

to the Sun is 30°. Assume both Earth's and Mars' orbits around

the Sun is circular.

Question

Problem Diagram

What is the position of Mars relative to Earth on the launch day?

Approach

Draw a diagram and label the key points and known position vectors.

Determine the relationship between the unknown position vector and those that are given.

Using geometry, solve for the length and angle of the unknown vector.

STATICS - THEORY

Scalars vs. Vectors

Position VectorsClick to view movie (55k)

Force VectorsClick to view movie (27k)

Any quantity that is represented by a positive or negative number is called a scalar. Mass, volume, and distance are examples of scalars. Scalars are generally represented by letters in plain type such as A = 5 kg. Mathematical operations on scalars follow the laws of elementary mathematics and algebra.

A vector represents a quantity that has both magnitude and direction. Examples of vectors include position, force, moment (torque) and velocity. Mathematical operations involving vectors follow the rules of vector algebra, which is discussed below.

Vectors are generally represented graphically by an arrow. The length of the arrow represents the magnitude, and the direction is defined by the angle between the vector and some reference axis.

Vectors are represented symbolically by a boldface letter such as B = 10 km north. The magnitude of a vector is a positive quantity and is written as a scalar such as B = |B|. In the next section, 2-D Vectors, vector notation will be presented which makes working with vectors easier.

Multiplication of a Vector and a Scalar

Vectors multiplied by ScalarsClick to view movie (62k)

If the vector B is multiplied by the scalar a, the result is the vector aB. If the scalar a is negative, then the vector is in the opposite direction.

Division of the vector B by a scalar a can be described as a special case of multiplication.

Graphical Vector Addition

Parallelogram LawClick to view movie (32k)

Two vectors, A and B, can be added graphically by using the parallelogram law. The vectors are joined at the tails and parallel lines are drawn from the head of each vector. The intersection point of the parallel lines is the head of the vector A + B.

The vectors A and B can also be added by joining them in a head-to-tail fashion. When adding B to A, the tail of B is placed at the head of A. The result R extends from the tail of A to the head of B. We can also find the same result by adding A to B. This means that vector addition is commutative,

A + B = B + A

Graphical Vector Subtraction

Vector subtraction can be written as a special case of

Vector AdditionClick to view movie (25k)

vector addition,

A - B = B + (-A)

Vector SubtractionClick to view movie (15k)

Multiple VectorsClick to view movie (24k)


Algebraic Solution

Problem Diagram


The position vector from the Sun to Earth is

re = re 0o (counter-clockwise from vertical)

= 150 × 106 km 0o

The position vector from the Sun to Mars:

rm = rm α

= 207 × 106 km 30o

The difference between re and rm is the vector from Earth to Mars. Using the head-to-tail method gives,

rm = re + rem

This equation can be rearranged to give the unknown position vector rem,

rem = rm - re = 207× 106 km 30o - 150 × 106 km 180o

The vector calculator on the simulation page can be used to add the vectors, which gives

rem = 107.56 × 106 km 74.21o

Geometric Solution


The problem can also be solved using pure geometry. First, define the position vector from Earth to Mars using the equation

rm = rem β

Next, form a triangle with the three vectors and use the Law of the Cosine,

rem = (re2 + rm

2 - 2re rm cosα )0.5

= 107.56 × 106 km (scalar only, not vector)

Then the internal angle θ, can be determined using the Law of the Sine,

rem / sinα = rem / sinθ

θ = sin-1 (rem sinα /rem ) = 105.79o

Finally, the desired angle β is

β = 180o - θ = 74.21o

The position vector from Earth to Mars is

rem = 107.56 × 106 km 74.21o

Comments

This problem could also be solved using standard vector notation with unit vectors, i and j. Vector notation is introduced in the next several sections and is used extensively throughout Statics.

CHAPTER-2-B- 2-D VectorsSTATICS - CASE STUDY

Introduction

Pulley Failure in the FactoryClick to view movie (427k)

A pulley assembly is used to move heavy objects around a factory floor. As the assembly moves, the total force on the main pulley changes. At a certain angle, the main pulley fails from the tension in the cable.

What is known:

The forces on the pulley at failure are as follows:

(1) F1 = 500 lb downward(2) F2 = 500 lb, 30° from downward

Question

Pulley Force Diagram

What is the total force on the pulley and in what direction does it act?

Approach

Solve for the rectangular components of the two forces on the pulley.

Combine the two forces using vector addition to find the total force exerted on the pulley.

Solve for the magnitude and direction of the total force.

STATICS - THEORY

Vector Components

Vector ComponentsClick to view movie (58k)

In the previous section, vectors were described using its magnitude and direction. This makes mathematical operations difficult (cannot simply add angles). To help solve this problem, vectors are usually split into components. For example, if two vectors, F1 and F2, give the vector F when added together, then F1 and F2

are said to be components of the vector F:

F = F1 + F2

If the vectors F1 and F2 are perpendicular to each other, they are called rectangular, or Cartesian, components. If the x-y axis is oriented along these components, they are labeled Fx and Fy respectively:

X-Y Axis OrientationClick to view movie (26k)

Vector Components Diagram

F = Fx + Fy

If two vectors, i and j, have a magnitude of one and are in the x and y direction respectively, then F can be written as

F = Fxi + Fyj

The vectors i and j are called Cartesian unit vectors, and Fx and Fy are the scalar components of the vector F. This configuration is the most common to describe a vector and allows vectors to be added and subtracted quickly and easily.

If the angle θ is measured from the x axis counterclockwise, then the scalar components are

Fx = F cosθ Fy = F sinθ

General Unit Vectors

General Unit Vector uF

Unit vectors in the x and y directions (i and j) where used the above paragraphs, but unit vectors can also be used in any direction. Unit vectors give direction where magnitude gives the length of the vector. Unit vectors are defined as

uF = F/F

and have a magnitude of 1.

Vector Addition and Subtraction

Addition of Multiple VectorClick to view movie (28k)

When a set of vectors are described using Cartesian unit vectors (i and j), then the resultant vector is just the addition (or subtraction) each components, giving

FR = F1 + F2 +... = ΣFxi + ΣFyj

where ΣFx = F1x + F2x +... ΣFy = F1y + F2y +...

Note, Fn-x and Fn-y are the components of each individual vector where n is the numbering of each

Vector Angle Relationship

vector being added.

The magnitude and direction of the resultant vector, FR, can be determined just like any individual vector by,

FR = (FRx + FRy)0.5

θ = tan-1 (FRy/FRx)


Cartesian Notation

Cartesian Notation Diagram

Choose an axis system as shown and write the two vectors in Cartesian notation. It is important that the correct sign is used.

F1 = 0i - 500j lb

F2 = 500 sin30i - 500 cos30j lb

= 250i - 433j lb

Addition of Forces

Addition of Forces Diagram

Now that the two vectors are described using the unit vectors, i and j, they can been easily added. Adding the two forces will give the total force in Cartesian notation,

FR = F1 + F2 = 250i - (500 + 433)j

= 250i - 933j lb

The magnitude of the total force can be determined by squaring the i and j magnitudes, and then taking the square root,

FR = (2502 + 9332)0.5 = 965.9 lb

The angle at which the total force acts is,

θ = tan-1 (250/933) = 15o

CHAPTER-2-C- 3-D VectorsSTATICS - CASE STUDY

Introduction

Power BoatClick to view movie (185k)

Boat Angle Diagram

An insurance company is investigating an accident in which a speedboat driver ran over some debris, lost control, and crashed. The debris bent the driveshaft of one of the propellers and redirected the thrust (force) it exerts on the boat.

What is known:

The thrust F of each propeller is 1000 lb.

The thrust directions are

described by the angles α = 10° β = 5° γ = 20°

Question

What is the total thrust (force) on the boat in Cartesian components after the driveshaft was bent?

Approach

3D rotation movieClick and Rotate Boat

Draw a diagram labeling all known angles.

Using geometry, solve for the Cartesian components of the two propeller forces.

Combine the components to get the total thrust (force) on the boat.

STATICS - THEORY

3-D Vector Components

3-D Vector ComponentsClick to view movie (38k)

Cartesian Components

Recall, from the previous section, a vector in the x-y plane in can be written in Cartesian notation as,

F = Fxi + Fyj

Here i and j are unit vectors in the x and y directions. Likewise, a vector can be written in the three dimensional x-y-z space in Cartesian form,

F = Fxi + Fyj + Fzk

The new component, k is the unit vector in the z direction. The components Fx, Fy, and Fz can be determined from the magnitude and direction of F,

Fx = F cosθx Fy = F cosθy Fz = F cosθz

The angle θi is the angle that F makes with the i axis as shown in the diagram at the top left.

Right Hand Coordinate SystemClick to view movie (226k)

Right-handed Coordinate System

The order of the x-y-z coordinates does matter. If the three coordinates are aligned so that the thumb is in the x-direction when the first finger is in the y-direction and the second finger is in the z-direction, then it is called a right-handed coordinate system. Otherwise, it would be a left-handed coordinate system. Engineering always uses right-handed coordinate system. There is nothing wrong with a left-handed system, but all books, equations and technical papers use a right-handed system for convention.

Vector Addition

Direction CosinesClick to view movie (55k)

Addition of 3-D vectors is the same as for 2-D except that three components must be added instead of two. If two vectors F1 and F2 exist, then the addition of F1 and F2 is,

FR = F1+ F2

FR = (F1x + F2x)i + (F1y + F2y)j + (F1z + F2z)k

The magnitude of the resultant vector is determined by applying the Pythagorean theorem,

FR = ( FRx2 + FRy

2 + FRz2 )0.5

The direction cosines are useful in some problems to identify each direction component. They are given as,

cosθx = FRx /FR cosθy = FRy /FR cosθz = FRz /FR

3D Unit Vector

3-D Unit Vector

General unit vectors can be determined in 3D using the definition of unit vectors, as

uF = F/F

Substituting vector notation and magnitude for F, gives

This relationship is useful when the direction of the vector is needed.

Position Vectors

Position Vector

In statics, force and moment vectors are the most commonly but many times a position vector is needed to help determine the direction of the force or moment vector. Position vectors are the same as all vectors, but they describe direction and distance,

r = xi + yj + zk

where x, y, and z are distances (scalars). Generally, position vectors are determined by its two end points, giving

r = (xB - xA)i + (yB - yA)j + (zB - zA)k

The directional unit vector for a position vector is

ur = r/|r

Spherical Angles Diagram

Spherical Coordinate AnglesClick to view movie (66k)

Spherical Coordinate Angles

If the direction of the vector is defined by the two angles θ and φ, (shown at the left) as is commonly done in spherical coordinate systems, then the components are given by

Fx = F sinφ cosθ Fy = F sinφ sinθ Fz = F cosφ

The spherical angles θ and φ are given by,

φ = cos-1 FRz /FR

θ = sin-1 FRy /(FR sinφ) = cos-1 FRx /(FR sinφ)


Normal Propeller (F1)

With the x-y-z axes oriented as shown, the direction of the normal propeller is defined by the spherical angles θ and φ:

θ1 = 175o

φ1 = 90o

The magnitude is F1 = 1000 lbs. The cartesian components can be calculated by,

Normal Propeller Force Components Click to view movie (54k) F1x = F sinφ1 cosθ1

= 1000 sin90 cos175 = -996.2 lb F1y = F sinφ1 sinθ1 = 1000 sin90 sin175 = 87.2 lb F1z = F cosφ1 = 1000 cos90 = 0 lb

Vector F1 can be written in Cartesian notation as

F1 =-996.2i + 87.2j + 0k lb

Bent Propeller Force Components Click to view movie (111k)

Summing Components

Total Force Components Click to view movie (63k)

Bent Propeller (F2)

For the bent propeller, the spherical angles are more difficult to find but can be found from geometry. The animation at the left shows how to find these angles. They are,

θ2 = 200o

φ2 = 80.59o

The magnitude F2 is given as 1000 lb. The component magnitudes are

F2x = F sinφ2 cosθ2 = -927.0 lb F2y = F sinφ2 sinθ2 = -337.4 lb F2z = F cosφ2 = 163.5 lb

Vector F2 can be written in Cartesian notation as

F2 =-927.0i - 337.4j + 163.5k lb

Next, add F1 and F2 to get the total force on the boat.

FT = F1 + F2

= -1923.2i - 250.2j + 163.5k

Angles for F2

CHAPTER-2-D- Dot ProductsSTATICS - CASE STUDY

Introduction

Construction Trolley SystemClick view movie (243k)

Displacement Vectors

Some workers are refurbishing the Eiffel Tower. The construction crew utilizes a trolley system to transport equipment from point A on the ground to point B on the platform they are working on. The trolley is pulled by two cables at the base of the tower.

What is known:

With an axis system oriented as shown, the position vectors of points A and B arerA = 175i + 0j + 0k mrB = 39i + 70j + 29k m

When the trolley is halfway between points A and B, the forces exerted on the trolley by the cables areF1 = -943.7i - 221.7j + 245.4k mF2 = -919.4i - 216.0j - 328.6k m

Question

What is the component of the force exerted on the trolley that is parallel to the line AB?

Approach

Find the unit vector in the direction of the line AB.

Find the total force exerted on the trolley by the two cables.

Solve for the component of the total force that is parallel to the unit vector for line AB.

STATICS - THEORY

Vector Multiplication

In the previous two sections, addition and subtraction of 2-D and 3-D vectors were illustrated. When "multiplying" two vectors, a special types of multiplication must be used, called the "Dot Product" and the "Cross Product". This section deals with only the dot product. The cross product is presented in a later section.

Dot Product

Dot Product AngleClick to view movie (45k)

The dot product of two vectors A and B is defined as the product of the magnitudes A and B and the cosine of the angle θ between them:

A • B = |A| |B| cosθ

The dot product can also be calculated by

A • B = Ax Bx + Ay By + Az Bz

where vectors A and B are given as

A = Ax i + Ay j + Az k B = Bx i + By j + Bz k

Applications of the Dot Product

Component Parallel to a LineClick to view movie (74k)

Component Perpendicular to a LineClick to view movie (65k)

Using the dot product, the angle between two known vectors A and B, can be determined as

If the direction of a line is defined by the unit vector u, then the scalar component of the vector A parallel to that line is given by

A|| = A • u

The vector component parallel to that line is given by

A|| = (A • u) u

Using the properties of vector addition, or the Pythagorean theorem, we can also determine the scalar and vector components of A perpendicular to the line:


Unit Vector Parallel to AB

Adding rA and rB Click to view movie (75k)

The direction of line AB is defined by the position vector rAB, which can be found from vector addition,

rA + rAB = rB

rAB = rB - rA

= (39 - 175)i + (70 - 0)j + (29 - 0)k

= -136i + 70j + 29k

The unit vector in the direction of rAB can be determined by dividing rAB by its magnitude rAB,

rAB = (-1,362 + 702 + 292)0.5 = 155.7 m

uAB = rAB/rAB

= -0.8736i + 0.4496j + 0.1863k Total Force on Trolley

Add the two forces applied to the trolley to find the total force,

FT = F1 + F2

= -1,863i - 437.7j - 83.2k N

Component Parallel to uAB

Force Component to Parallel to uAB Click to view movie (41k)

The scalar component of the total force parallel to the direction uAB can be found by using the dot product,

FT|| = FT • uAB

= -1,863 (-0.8736) - 437.7 (0.4496) - 83.2 (0.1863)

= 1,415 N

This is just the magnitude of the total force acting in the direction of the cable AB. This can also be represented as a vector by multiplying this scalar component by the unit vector uAB, giving,

FT|| = FT|| uAB

= -1,236i + 636.3j + 263.7k N

The magnitude or FT|| has not changed but is now represented as a vector.

CHAPTER-3-A-Equilibrium & Free Body Diagrams


Introduction

Tram Car Stalled over Canyon

A new cable transport system has been built for a local national park. With it, tourists can travel cheaply to the other side of the canyon. Unfortunately, the engineers who built it have not worked out all the kinks. The cable machinery has locked-up, and the passengers cannot get down.

What is known:

The cable car is symmetric about its own vertical axis, and the supports connecting it to the cable are separated by 60°.

The cable to the right of the joint is angled at 20°, and the cable to the left is angled at 5°, with respect to the horizon.

Angle Diagram

Questions

Using a free-body diagram, how would we represent the forces acting on the point at which the cable car supports are connected to the cable?

Approach

Isolate the point at which the forces are acting and draw a simple sketch of the object.

Then draw vector approximations of the forces acting on the point.

STATICS - THEORY

Drawing FBD's

Free Body Diagram ExampleClick to view movie (106k)

A Free-body diagram (FBD) is an essential tool when the forces on an object need to be determined using equilibrium equations. They help focus attention on the object of interest in order to determine the forces acting on it. Creating FBD's is a straightforward process:

1. Identify the object that will be isolated. 2. Make an approximate sketch of the object

removed from the surroundings and include its relative dimensions and angles.

3. Draw vector approximations of the external forces and body forces, and label them.

4. Choose an appropriate coordinate system.

Coordinate System

The coordinate system is arbitrary when constructing a free-body diagrams. However, it is best to choose a coordinate system that will simplify future calculations. This may be done before or after making the diagram.

Forces Due to Rope/CablesClick to view movie (95k)

Equilibrium

An object is in equilibrium when each point inside the object has the same constant velocity (generally, velocity is zero for static's problems). Therefore, an object in equilibrium is not experiencing any accelerations, so Newton's 2nd Law can be written as

ΣF = ma = 0

The vector sum of the external forces acting on an object in equilibrium is zero.

Representation of Forces

"Line of Action" of a Force Vector

A force acts through a line of action, which is a vector representation of a force's magnitude and direction. On a rigid body, the force can be applied to any point on the object along the line of action of the force.

Forces may be either internal (such as the force of a piston in an engine) or external (such as the force of a baseball bat against the ball). They may also act across the object's surface or at any point in its body.

The force of gravity is an important factor when significant masses are involved. The weight of an object in a gravitational field can be expressed

|W| = mg

Forces Due to PulleyClick to view movie (62k)

where m is the object's mass and g is the gravitational constant (g = 9.81 m/s2 = 32.2 ft/s2 on Earth's surface).

Ropes and cables exert tension forces that only pull on an object and are therefore always represented as forces that are directed away from the object. Unless stated otherwise, we will always assume that ropes and cables are straight and their force is constant along the length.

A pulley serves to change the direction of the force of a rope/cable, but the tension will be the same on both sides if we assume a mass-less pulley (not true in dynamic problems).

Planar Surface Contact ForcesClick to view movie (245k)

Contact forces result from contact between the surfaces of two objects. A contact force can usually be split into two components: a normal force perpendicular to the plane of contact and a frictional force parallel to the plane of contact. Frictional forces, however, can be neglected if the surface is smooth.

Springs exert contact forces in a mechanical device. The force can be represented as

|F| = k |L - Lo|

where, k is the spring constant, L is the length after deflection, and Lo is the length before deflection. The use of springs when modeling complicated real-world structures is a useful tool.

Curved Surface Contact ForcesClick to view movie (280k)

Spring ForcesClick to view movie (107k)


Free Body Diagram SolutionClick to view movie (108 kB)

It is best to choose a coordinate system that will simplify future calculations. In this case, there is no preferable system to choose, so we will keep the original coordinate system with the x axis parallel to the horizon, and the y axis vertical.

Next, isolate the object or point through which the forces act. This occurs on the cable where the car is connected.

To complete the Free-Body Diagram (FBD) all the forces need to be drawn. The tension forces in the cable are labeled T1 and T2 because they are not in the same direction nor are they equal. In addition, the cable car is symmetric about the y axis, and therefore the support forces will be equal in magnitude. These forces are labeled FS.

Now the angles need to be specified. But since the axis has not been altered, they will be exactly the same as before.

Alternate SolutionClick view movie (98 kB)

Alternate Free-Body Diagram

Another free-body diagram, of the passenger car, could also be created. This diagram shows the weight of the car downward, and the force of the supports angled vertically. Details are given in the animation at the left.

STATICS - EXAMPLE

Example

The pulley system for an elevator is shown in the figure on the left. This mechanical system consists of 5 pulleys and 2 electric motors. The maximum allowable weight for the elevator is 5 kip.

a. What is the minimum required force for each motor?

b. If one of the electro-motors is removed and instead the cable is directly attached to the wall, what will be the minimum required force for the remaining single motor?

Solution

a. As with all statics problem, a free-body diagram will assist in solving the problem. In this example, all forces acting on the elevator cabin is first analyzed. The 5000 lb weight is divided evenly between the cables due to symmetry. Consequently the force of each cable will be

P = 5000 / 2 = 2,500 lb

Next, assume the cable force in each motor is F. The diagram on the left shows the pulley system with the external forces and with the elevator forces.

The free-body diagram for pulley number 2 is shown on the left. Note that the free-body diagram for the pulley number 4 would be similar. Summing the forces in the vertical direction gives,

ΣFy = 0 0 = 2 T2 - 2,500 lb T2 = 1,250 lb

Since the cable is continuous,

T2 = F = 1,250 lb

b. If one of the electro-motors is removed, then the cable is directly attached to the wall. The free-body diagram and the solution procedure would be the same. In this case the reaction force at the wall is equal to 1,250 lb. Therefore removing one of the motors will not affect the final answer. Also, by removing one of the electro-motors, both cost and energy will be saved for the consumer and manufacturer.

CHAPTER-3-B-2-D Forces (point)STATICS - CASE STUDY

Introduction

Bull too Heavy for ScalesClick to view movie (588k)

On Bull Gates Ranch, a prized bull needs to be weighed before being transported to the county fair. However, the bull's weight is too high to register on the current scale (1000 lb maximum). Consequently, a jury-rigged system of pulleys has been built in order to determine the bull's true weight.

What is known:

When the pulley system is used and a 75 lb counterweight is attached to the end of the rope, the scale registers the bull's apparent weight as 980 lb.

Questions

a) What is the bull's actual weight? b) If the bull's actual weight is 1368 lb, what magnitude counterweight will cause the scale to read 1000 lb?

Approach

Start with the right end of the rope and work backward to determine the tension in each segment using free-body diagrams for each pulley.

Apply equations of equilibrium, ΣFy = 0, for each pulley.

Consider all forces acting on the bull, including the scale.

STATICS - THEORY

2D Equilibrium

Example Problem Solution StepsClick to view movie (428k)

Forces acting on a object can be approximated as vectors with a magnitude and direction. When there are unknowns in the system, the equations of equilibrium can be used to find those unknowns.

Recall in the previous section, the equation for equilibrium is

ΣF = 0

This equation can be expanded for each direction (only x- and y-direction is examined in this section, the next section includes the z-direction)

ΣF = ΣFxi + ΣFyj = 0

For the expanded equation, each direction must be in equilibrium, or

ΣFx = 0 ΣFy = 0

This means that when an object is in equilibrium, the components of forces acting along the arbitrary coordinate system will cancel each other out. These relationships can be used to determine the unknown forces using simple algebra.

Rope Force Remains Constant in a Pulley System

Pulleys

A pulley redirects the force of ropes and cables. If no mass is given for the pulley (mass-less pulley), then the tension forces in a rope will be the same on either side of the pulley.


Part a) - Bull's Actual Weight

Problem Layout with Given Information

Free-Body Diagram of Bull

A good place to start for all statics problems is a free-body diagram (FBD). However, in this problem, because there are numerous objects, it can be confusing which object (or objects) to use in a FBD. Since the bull seems to be central to the problem, it would be a good place to start (note, there is no "wrong" place to start). The FBD of the bull is shown at the left. Summing forces in the y-direction gives,

ΣFy = Fscale + TCH + TEG - Wbull = 0

Wbull = 980 lb + TCH + TEG

The rope tensions TCH and TEG must be equal since it is assumed that the sling holding the bull is symmetrical. This gives,

Wbull = 980 lb + 2TCH (1)

However, you will notice that there are still two unknowns, Tch and Wbull. More information is needed.

Pulley E and D Diagrams

The next logical object to examine is pulley C since it is connected to the bull. Its FBD is shown at the left along with the rope connected to the counterweight. Summing forces on pulley C gives,

ΣFy = TBC + TDC - TCH = 0

The rope must have the same tension since the pulleys are assumed mass-less,

TBC = TDC = F = 75 lb

The rope tension TCH can be calculated as,

TCH = 150 lb

Substitute TEG and TCH into Eq. 1 to find the weight of the bull,

Wbull = 1280 lb

Part b) - Counterweight

In part a), the bull's weight was found. Now, the problem is reversed, the bull weight is known, 1,368 lb,

and the counterweight needs to be determined to make the scale register 1000 lb.

FBD of Cow for Part b)

This time it is easier to start with the bull and work though the pulleys to find the counterweight. The FBD for the bull is shown at the left.

Summing forces in the y-direction gives the tension in the ropes TEG and TCH,

ΣFy = Fscale + TCH + TEG - Wbull = 0

TCH + TEG = 368 lb

It can be assumed (as done in part a) that the bull's weight is equally distributed between the two ropes holding the cow. Thus TCH equals TEG giving,

TCH = TEG = 368/2 = 184 lb

Pulley C Free-Body Diagram

The next step is to analyze the pulley C since the tension is rope CH is known. The FBD of the pulley is given at the left. Summing forces gives,

ΣFy = TBC + TDC + TCH = 0

The tensions TBC and TDC are equal,

TBC = 184/2 = 92 lb

Since the rope has the same tension as it goes from pulley C to the counterweight,

F = TFA = TAB = TBC

The magnitude of the counterweight needed to just register 1000 lb on the scale is

F = 92 lb

STATICS - EXAMPLE

Example

Two boxes are hung from a ceiling and wall as shown in the diagram. Each box is supported by two cables that are interconnected through rings. What is the maximum tension in any rope other than the ropes directly connected to the boxes?

Solution

In this example, there are two different loads that interact

through a system of cables. Since all ropes are connected to at least one ring, it would seem reasonable to focus on the ring objects and make sure they are in static equilibrium.

Free-body Diagrams of Ring E

Drawing a free-body diagram of ring E, identifies two rope forces, FDE and FEC, that are not known. The direction of the loads are known since the force must act in the direction of the rope.

The rope forces can be determined by applying force equilibrium in the x and y directions.

ΣFx = 0 FDE cos30 + FEC cos55 = 0 FDE = 0.6623 FEC

ΣFy = 0 -10 + FDE sin30 + FEC sin 55 = 0 0.6623 FEC (0.5) + FEC (0.8192) = 10

Solving the two equations give, FEC = 8.693 lbFDE = 5.757 lb

Free-body Diagrams of Ring C

Now that the force in rope EC is know, ring C can be analyzed. Again, drawing a free-body diagram of ring C identifies two unknown forces, FAC and FCB. Applying the equilibrium equations gives,

ΣFx = 0 FCB cos45 - 8.693 cos55 = 0 FCB = 7.051 lb

ΣFy = 0 FAC + 7.051 sin45 - 8 - 8.693 sin 55 = 0 FAC = 10.14 lb

Maximum force occurs in rope AC, Fmax = 10.14 lb

CHAPTER-3-C- 3-D Forces (point)STATICS - CASE STUDY

Introduction

Radio Tower

3D Tower ViewClick to view 3D movie (19k)

The radio antenna for station WEML sits on a nearby hill for better transmission. The support cables have been pre-tensioned to keep the antenna stable, which causes a known force on the antenna. In addition, a storm with high winds is adding a significant lateral load on the antenna. The storm could cause problems because the cable anchors were not placed to provide maximum support.

What is known:

The locations of the cable anchors on the ground (in reference to point O) are A(x,y,z) = (66.5 m, -86.5 m, 0 m) B(x,y,z) = (-26.3 m, -86.5 m, -63.0 m) C(x,y,z) = (-68.0 m, -86.5 m, 42.8 m)

The wind exerts a 1 kN force, Fwind, on the antenna in the negative z direction, Fwind.

The antenna experiences a 5.2 kN compression force, Ftower, due to the pre-tensioning in the cables.

The cables will fail at tensions greater than 4 kN.

Force Diagram

Questions

Will any of the cables break due to the force of the wind on the antenna?

Approach

Use Cartesian coordinates to define cable forces.

Apply the equilibrium equations to solve for the cable tensions. (This will involve 3 equations and 3 unknowns.)

Remember, cables experience only tension forces.

STATICS - THEORY

Equilibrium


The theory behind solving for three-dimensional forces is closely related to that of solving for two-dimensional forces.

The equation for equilibrium is

ΣF = 0

For three dimensions, this can be expanded to

ΣF = ΣFxi + ΣFyj + ΣFzk = 0

For the expanded equation, each of the three directions must be equal to zero, giving

ΣFx = 0 ΣFy = 0 ΣFz = 0

The equilibrium equation has been separated into three components corresponding to the x, y, and z axes. Since each equation is independent of the others, the equations can be used to determine up to three unknowns.


3D Tower ViewClick to view 3D movie (18k)

A coordinate system has been provided with the origin located at point O.

The top of the tower, point O, must be in static equilibrium, thus

ΣFo = Fwind + Ftower + FA + FB + FC = 0

In order to sum all the vectors, it is easier to convert each force into Cartesian vector. Then each direction, i, j, and k, must equate to zero.

FW = -1.0k kN FT = 5.2j kN

Coordinate System Diagram

Substitute the forces into the equilibrium equation above gives,

ΣFo = -1.0k + 5.2j + 0.6095 TAi - 0.7928 TAj - 0.2387 TBi - 0.7850 TBj - 0.5717 TBk - 0.5760 TCi - 0.7327 TCj + 0.3625 TCk = 0

Sum each of its three components.

0.6095 TA - 0.2387 TB - 0.5760 TC = 0 -0.7928 TA - 0.7850 TB - 0.7327 TC = -5.2 - 0.5717 TB + 0.3625 TC = 1.0

The simultaneous solution of these three linear equations provides,

TA = 3.217 kN TB = 0.3239 kN TC = 3.270 kN

Since all tensions are positive and below 4 kN, the antenna will be safe. However, if the wind magnitude had caused tension TB to act in a negative direction, the antenna could have become unstable, or one of the other cables could have broken.

CHAPTER-4-A-Moment of Force: 2-D Scalar


Introduction

Problem Diagram

After attempting to form a workers' union with his shipmates, a seaman is forced to walk the plank for mutiny. The moment generated at the base of the plank depends on the man's weight, his location, and the angle of the boat.

What is known:

The man has a weight of W lb and is standing a distance of x ft from the base of the plank.

The boat is tilted an angle of α, measured in degrees.

Question

What is the moment generated about point A, which is located at the base of the plank?

Approach

Determine the perpendicular distance from point A to the line of action of the force as a function of x and α.

Calculate the moment about point A by multiplying the force times the perpendicular distance.

STATICS - THEORY

Moment of a Force

Moment of a ForceClick to view movie (54k)

The moment of a force about a point is a measure of the tendency of the force to rotate a body about that point. The body does not necessarily have to rotate about that point, but the moment defines how the force is trying to rotate the body. A good example of a moment is when a force applied is applied to a wrench. The wrench may not actually turn, but the force generates a turning force (i.e. moment) around the bolt.

Moments are often referred to as torques, and the terms can be used interchangeably. In scalar notation, a moment about a point O is

Mo = r F

Here F is the magnitude of the force and r is the perpendicular distance to the line of action of the force.

Effect of Force PositionClick to view movie (33k)

Effect of Force MagnitudeClick to view movie (18k)

The orientation of the moment is in the same direction as the rotation of the body if the body were allowed to rotate. A moment can be symbolized as a curved vector around the rotation point.

If a force is applied at the point O or the line of action of the force passes through point O, then the moment about point O is zero, and the force has no tendency to rotate the body about that point.

Effect of a Force DirectionClick to view movie (52k)

Moment Direction and Sign

Components of a Moment

Principle of Moments

Moment ComponentsClick to view movie (33k)

The principle of moments, which is also referred to as Varignon's theorem, states that the moment of a force about a point is equal to the sum of the moments of the force's components about the point. Therefore, if the components of a force are known, it may be simpler to determine the moment of each component and then add them.


Solution DiagramClick to view movie (271k)

The solution simply requires the total moment about the point A. The moment generated about point A is the man's weight times the perpendicular distance from point A

With the ship at an angle of α, the perpendicular distance from point A to the line of action of the force, d, is

d = x cosα ft

The moment is then just the distance D times the force, W, giving

MA = W d = Wx cosα ft-lb

STATICS - EXAMPLE

Example

Example Graphic

A library is using a new ladder to reach books above 6 feet. The normal force on the walls cannot exceed 200 lb or ladder will damage the wall or books. What is the maximum allowable weight, W, on the ladder. (The wall is friction-less and the friction on the floor is large enough to not let the ladder slip.)

Solution

Like most static problems, the first step in solving any problem is to construct a free-body diagram. From the free body diagram at the left, there are three unknown forces, W, Ax and Ay. The wall reaction is known since the maximum wall force is 200 lb. The unknown forces can be found by applying the force and moment equilibrium equations,

ΣF = 0 ΣM = 0

W can be found by writing the moment equation about the pivot point, A. In this case the moment due to the reaction force of 200 lb must cancel out the moment of W about point A.

Before the moment can be solved, the distance D needs to be found.

The moment equation will give

ΣMA = 0 200 (6) - W (3.464) = 0 W = 346.4 lb

Is the length of the upper arm important in the design calculations? How?

CHAPTER-4-B-Moment of Force: 3-D Scalar


Introduction


3D VisualizationClick to view 3D movie (13k)

After designing an elaborate plumbing system, a plumber's apprentice uses a pipe wrench to tighten the connections. Unknown to him, he is placing unnecessary moments on other parts of the structure that could bend the pipes and lead to leaks or failures.

What is known:

The plumber exerts a force of 80 N on the pipe wrench as shown.

All of the necessary dimensions are shown

STATICS - THEORY

The Moment Vector

Moment VectorClick to view movie (435k)

The moment about a point in 3-D space can be determined from the same basic scalar equation as in previous section on 2D scalar moments.

Mo = r F

Here F is the magnitude of the force, and r is the perpendicular distance to the line of action of the force.

A force acting on a body in 3-D space tries to rotate the body in a plane defined by the force's line of action and the point under consideration. Since this plane is often difficult to visualize or define, a double-headed vector is used to define the axis about which the force is tending to rotate the body. The magnitude of the vector is the magnitude of the moment generated by the force.

The moment of a force in 3-D space can also be calculated using the Principle of Moments discussed in in the previous section. However, as it was seen,

Effect of Force AngleClick to view movie (263k)

determining angles and distances in 3-D space can be very difficult. The next section will introduce a vector approach to calculating moments that greatly simplifies the process and reduces the number of steps necessary in calculating a moment vector.

Right Hand RuleClick to view movie (256k)

Moment vs Actual RotationClick to view movie (143k)


Joint AClick to view movie (311k)

Joint A

The line of action of the force is 30 cm from joint A. Therefore, the magnitude of the bending moment is

MA = F dA = (80 N) (30 cm) = 2,400 N-cm

Joint B

The distance from joint B to the line of action of F is found from the Pythagorean theorem:

dB = (302 + 452)0.5 = 54.08 cm

Substitute and solve for the magnitude of the

moment:

MB = (80 N) (54.08 cm) = 4,327 N-cm

Joint CClick to view movie (353k)

Joint C

Calculate the moment arm and the moment at joint C in the same manner as for joint B:

dC = (152 + 452)0.5 = 47.43 cm

MC = (80 N) (47.43 cm) = 3,795 N-cm

Joint DClick to view movie (339k)

Joint D and E

Calculate the moment of F at joints D and E in the same manner as for joints B and C:

dD = (152 + 152)0.5 = 21.21 cm

MD = (80 N) (21.21 cm) = 1,697 N-cm

dE = (302 + 152)0.5 = 33.54 cm

ME = (80 N) (33.54 cm) = 2,683 N-cm

Maximum Moment

The joint with the largest moment is joint B, which has a moment of 4,327 N cm, or in more common units 43.27 N-m.

STATICS - EXAMPLE

Example

Fixed Bracket with Force

A single force of 14 kN acts on a bracket that is fixed to a wall. What is the total moment in vector format at point A due to the force?

Solution

To find moment about a point, the position vector from the point of interest to any point on the force line of action is crossed with the force vector. In equation form this is

MA = r × F

Position Vector r from Point A to the Force Line of Action

For this problem, the position vector is drawn from point A to point E giving

rAE = 6i - k

Another position could be used, such from A to D.

The force vector can be determined by multiply its magnitude with its unit directional vector. The location of point D and E is (8,8,2) and (6,0,5) respectively. The force vector is

F = F uDE

= -3.191i - 12.764j + 4.786k

Substituting into the cross product gives,

This determinate can be evaluated as,

MA = -12.76i - 25.52j - 76.58k

CHAPTER-4-C-Moment of Force: 3-D Vector


Introduction


Known information

In the previous section, the moment generated by a wrench as analyzed. However, it was determined that the moment changed as the wrench was rotated around the pipe. It is possible that the magnitude of the moment will increase even though the magnitude of the force remains constant.

What is known:

The wrench rotates from α = 0° to α = 90°. The force is applied at point Q. The necessary dimensions are given.

Questions

What is the maximum moment generated at joint B as the wrench rotates from α = 0° to α = 90°?

Approach

Determine the position vector from joint B to point Q for any angle α.

Determine the Cartesian components of F as a function of the angle α.

Solve for the moment of F at joint B using the vector description of moment.

STATICS - THEORY

The Cross Product

Cross Product Direction and SignClick to view movie (431k)

The dot product was previously introduced as a way of "multiplying" vectors where the result is a scalar,

A • B = AB cosθ

The second method for "multiplying" vectors is called the cross product, and the result is a vector. The cross product of two vectors A and B gives the vector C that has a magnitude of

|C| = |A × B| = AB sinθ

Here θ is the angle between the vectors A and B.

The vector C is perpendicular to the plane defined by the vectors A and B, and the direction is determined from the right-hand rule.

Cross Product with Cartesian Vectors

If the vectors A and B are in Cartesian form, then it can be shown that the cross product of A and B is given by the determinant of the unit vectors and the Cartesian components of A and B.

It is interesting to note that this determinant could also be written as,

The results will be the same for either form.

Vectors r and F are always perpendicular to M ( = r × F)

Cross Product Direction and SignClick to view movie (258k)

Moment as a Cross Product

If the moment of a force F about the point O is represented by the vector Mo, then it can be shown that

Mo = r × F

Here r is the position vector of any point on the line of action of F with respect to point O. Expanding the determinant form of the cross product, gives

Mo = (ry Fz - rz Fy)i + (rz Fx - rx Fz)j + (rx Fy - ry Fx)k

From this equation, the Cartesian components of the moment vector Mo are readily found.

Moment of a Force About a Line or Axis

Moment About an AxisClick to view movie (315k)

In many problems, it is necessary to determine the moment of a force about a certain line or axis, such as an axle on a car. Since the moment of a force is a vector, the dot product can be used to determine its component parallel to any line a.

Ma = (Mo • ua) ua = [(r × F) • ua] ua

Here ua is the unit vector in either direction of the line a.


Position Vectors

Position VectorsClick to view movie (325k)

Coordinate systems can be attached to any point, but generally there is one or two points that are more convenient than others. For this problem, the coordiante system is placed at far right edge of the pipes. The position vector from the right edge of the pipe to joint B is

rB = rBxi + rByj + rBzk

= 45i + 0j + 15k

The position vector of point Q (location of applied force F) can be determined when the wrench is rotated at an angle of α as

rQ = rQxi + rQyj + rQzk

= 0i + 30 cosαj + (60 + 30 sinα)k

From vector addition, the position vector from joint B to point Q is

rB + rBQ = rQ

or rBQ = rQ - rB

= -45i + 30 cosαj + (45 + 30 sinα)k

Known Information

Force Vector

The system of two forces can be replaced by an equivalent system consisting of a force FR and couple MR applied at point O. The force FR is found by summing the individual forces,

F = Fxi + Fyj + Fzk

= 0i - 80 sinαj + 80 cosαk

Moment Vector

The moment of the force about joint B is given by

MB = rBQ ×> F

The cross product in determinant form is

Moment Variation with Angle α

Expanding the equation gives

MB = (rBQy Fz - rBQz Fy)i + (rBQz Fx - rBQx Fz)j + (rBQx Fy - rBQy Fx)k

Substituting the appropriate values and simplifing to determine the moment at joint B gives

MB = (2,400 + 3,600 sinα)i + (3,600 cosα)j + (3,600 sinα)k

If the magnitude of MB is plotted as a function of the angle α, then the moment is a maximum at α = 90°, with a value of 6,997 N-cm or about 70 N-m. This is approximately 62% greater than the moment at α = 0°.

CHAPTER-4-D- Couples and Equivalent Systems


Introduction

Force directions acting on the misaligned firework

Several types of fireworks use two rockets to cause the fireworks to spin rapidly and rise up in the air. If the thrust from the two rockets is not aligned properly, the fireworks will move in a certain direction as well as spin. This direction can be determined from an equivalent force-couple system or force resultant.

What is known:

Each rocket has a thrust of T = 0.5 lb. One of the engines is misaligned by α = 10°. The necessary dimensions are shown.

Questions

1. What is the equivalent force-couple system at point O?

2. If one is possible, what is the force resultant?

Approach

Use a vector approach to determine the equivalent force and moment at point O.

Solve for the position of the force resultant, if one is possible.

STATICS - THEORY

Moment of a Couple

Couple DefinitionClick to view movie (551k)

A couple is defined as a pair of forces having equal magnitude and opposite direction with different but parallel lines of action. Since the total force in any direction is zero, the net effect of a couple is a pure moment, or torque. The magnitude of this moment is the magnitude of one of the forces times the perpendicular distance between the forces,

M = F d

The moment vector is perpendicular to the plane containing the lines of action of the two forces.

Using a vector approach, the moment of a couple is defined as

M = r × F

Here r is the position vector from one force to the other, and F is the force to which r is directed.

Equivalent Systems

Two systems are equivalent if and only if the sum of forces for the two systems is equal and the sum of moments about an arbitrary point in both systems is equal:

(ΣF)1 = (ΣF)2

(ΣMp)1 = (ΣMp)2 Equivalent Force-Couple System

Equivalent Force-CoupleClick to view movie (350k)

Using the concept of equivalent systems, any system of forces and moments can be replaced by a single force and couple at an arbitrary point P. Applying the conditions for equivalent systems gives

FR = ΣF (1)

MRP = ΣMP (2)

Here each MP is given by the equation

MP = rPF × F (3)

Force Resultant of a System

Single Force ResultantClick to view movie (235k)

If the force and couple of an equivalent force-couple system are perpendicular to one another, we have the equation

FR • MRP = 0

The system can be reduced to a single force FR

located at a point Q. The moments of the two systems can be equated about the point P giving,

MRP = rPQ × FR (4)

By substituting Eqs. 1, 2, and 3 into Eq. 4; performing the cross products; and equating the i, j, and k components; the three scalar equation are,

Σ(rPFy Fz - rPFz Fy ) = rPQy ΣFz - rPQz ΣFy

Σ(rPFz Fx - rPFx Fz ) = rPQz ΣFx - rPQx ΣFz

Σ(rPFx Fy - rPFy Fx ) = rPQx ΣFy - rPQy ΣFx

Solving this system of three equations for the three unknowns rPQx, rPQy, and rPQz gives the position vector rPQ of the force FR,

rPQ = rPQxi + rPQyj + rPQzk

FR = ΣF


Direction of Forces

Although scalars can be used to solve this problem, vector notation will be used. The standard directions will be used for the coordinate system. The two rocket forces that turn the firework are

F1 = 0i - Tj + 0k

F2 = -Tsinαi + Tcosαj + 0k

where T is the trust, 0.5 lb and α is the misaligned rocket angle of 10o. The position vectors from point O to each force's point of application are

r1 = -3i + 0j + 0k

r2 = 3i + 0j + 0k

Solution of a)

Equivalent Force and Moment

Calculating Equivalent ForceClick to view movie (27k)

The system of two forces can be replaced by an equivalent system consisting of a force FR and couple MR applied at point O. The force FR is found by summing the individual forces.

FR = ΣF = F1 + F2 = -Tsinαi + T(cosα - 1)j + 0k

Substituting the known values, α = 10o and T = 0.5 lb, gives

FR = -0.08682i - 0.007596j + 0k lb

The moment MR is the sum of the individual moments.

MRo = ΣMo = r1× F1 + r2× F2

The cross products can be written as determinants, giving,

= 0i - 0j + 3T(1 + cosα)k

Substituting the known values gives

MRo = 0i - 0j + 2.977k in-lb

Solution of b)

Single Force ResultantClick to view movie (26k)

Since all the forces act in the same plane, there is a single force resultant for this problem. The resultant force is the same FR as before but it is located at a different point Q, which must satisfy the equation

MRo = rOQ × FR

Writing this equation in determinant form and substituting known vectors gives,

2.977k = [0 - rOQz (-0.007596)]i - [0 - rOQz (-0.08682)]j + + [(rOQx (-0.007596) - rOQy(-0.08682)]k

From the i and j terms, rOQz must be zero. However, the k term gives

2.977 = -0.007596 rOQx + 0.08682 rOQy

From inspecting the diagram, it is noted that point Q was chosen to be directly vertical from O. Thus, the rOQx must be zero. This gives,

2.977 = 0.08682 rOQy

Therefore

rOQ = 0i + 34.29j + 0k in

Applying the single force FR at the point Q has the same effect as the two original forces F1 and F2.

CHAPTER-4-E- Introduction to Distributed Loads


Introduction

Problem Diagram

Many tall buildings sway back and forth due to the force of wind currents acting on them. When designing tall buildings, engineers must consider these "wind loads" so that the building will be safe and stable in high winds.

What is known:

The wind load can be modeled as a linear distributed load.

The wind load has a magnitude of 40 lb/ft at the base of the building.

The wind load has a magnitude of 120 lb/ft at the top of the building.

The building is 500 ft tall.

Question

What is the force resultant of the wind load and where is it located?

Approach

Break the distributed load into its simpler composite parts.

Determine the single force resultants for each of the composite parts.

Combine the two resultants into a single force resultant.

STATICS - THEORY

General Distributed Load with Load Intensity of f(x)

(units force/distance)

In many static problems, applied loads are given as distributed force loads. This is similar to stacking sand bags on a beam so that the load is distributed across the beam instead of at one location (point load). To help make the problem easier to solve, it is convenient to convert the distributed load into equivalent point loads.

Discrete Distributed Force

Forces Parallel to the Y axisClick to view movie (268k)

Forces Parallel to the X-Y PlaneClick to view movie (253k)

If the loading on the object is a set of parallel discrete forces, the resultant force is simply the sum of all the forces, or

FR = ΣFi

In the previous section, three scalar equations were derived that determine the position r of a force FR that represents the force resultant to a system of discrete forces. If all forces are parallel to the y direction then the three scalar equations simplify to

Σ(zF) = z' ΣF

Σ(xF) = x' ΣF

Here x and z are the Cartesian coordinates of the individual forces, and x' and z' are the coordinates of the force resultant FR.

Rearranging these two equations gives

If the forces are further restricted so that they all lie in the x-y plane, then z' = 0 and only the second equation applies.

Distributed ForcesClick to view movie (275k)

Continuous Distributed Force

If instead of a system of point loads, consider a continuous distributed force f(x) that acts in the x-y plane and is parallel to the y axis, then through calculus the second equation (x') above becomes

The force resultant is simply the force magnitude FR given by

The force magnitude FR is located a distance x' from the origin.

Uniform and Triangular Line Loads

Resultant Force and Location forTwo Distributed Loading Types

In the case of a uniform line load as shown, it is unnecessary to perform the integrations because the force resultant is always the value of the distributed load multiplied by the distance over which it acts. The location of the force resultant is always the center point (centroid) of the distributed load.

For a triangular line load, it can be shown that the force resultant is one half of the peak value of the distributed load multiplied by the distance over which it acts. The location of the force resultant is two-thirds of the distance from the vertex to the peak value of the load.

Composite Line LoadClick to view movie (170k)

Distributed Load Simplification

Composite Loading

A load of the type shown is said to be a composite load since it can be generated with a combination of simpler loads such as a uniform and a triangular line load.

Loads of this nature can be converted to force resultants by splitting the load into its composite parts, solving for the force resultant of each part, and then combining the forces into a force resultant for the entire load.

As an example, the diagram at the left can be splitting to a triangular and rectangular distributed load. The two separate loadings can be combined as,

FR = F1 + F2

xR FR = x1 F1 + x2 F2


Integration Method

Integration Method Click to view movie (167k)

Load Diagram

In order to integrate the distributed wind load, determine a function f(y) that describes the wind load.

f(y) = 40 + 80/500 y lb/ft

The force resultant is given by

= 40(500) + 0.08(500)2 = 40,000 lb

Its location is given by

= 20(500)2/40,000 + 0.05333(500)3/40,000

= 125.0 + 166.7

= 291.7 ft

Composite Parts Method

Wind Distributed Load Splitinto Two Parts

For the composite parts method, first break the wind load into two parts: a uniform load of 40 lb/ft and a triangular load of 0 lb/ft at the base and 80 lb/ft at the top of the building.

The force resultant of the uniform load and its location is

F1 = 40 (500) = 20,000 lb

y1 = 250 ft

The force resultant of the triangular load is

F2 = 0.5 (80) (500) = 20,000 lb

Its location is two-thirds of the distance from the vertex to the peak value

Composite Parts MethodClick to view movie (196k)

y2 = 2/3 (500) = 333.3 ft

Nest, combine these two forces into a single force FR located at the position y'.

FR = ΣF = F1 + F2 = 40,000 lb

Its location is given by

Substituting F1, y1, F2, and y2 and simplifying the equation gives

y' = 1.166 x 107 / 40,000

= 291.7 ft

STATICS - EXAMPLE

Example

New Tri-Axle Dump TruckSand Load

A new 10 kN tri-axle truck is designed to carry sand but the design engineer notes that if it is loaded incorrectly the truck can tip. Thus, it is necessary for the truck to be stable even with both of the rear wheels on 3rd axle go flat.

The center of gravity (cg) of just the truck is located at 0.5 m from the front edge of the sand. The distributed load of the sand can be modeled as a third order equation, -0.17 x3 + 0.38 x2 + 0.77 x + h kN/m. The center of gravity (cg) of just the sand is not known. What will be the maximum allowable h before the load will be unsafe when the back rear wheels are flat?

Solution

When both rear wheels become flat, there will be no vertical force acting on them and the truck may become unstable. Additionally, when the truck starts tipping, the force on the front wheel will go to zero because at that time the moment about the middle wheels wants to rotate the truck.

The model can be simplified as a beam with a pinned support at the location of the 2nd axle and two forces (weight of the truck and the weight of the sand pile). Notice the back rear wheels and front wheels do not exert a reaction force.

Previously, it was determined that the resultant force FS for a distributed load over a line is,

It was also shown that the location of the line of action of FS is given by

The sand pile is distributed from 0 to 3 meters. The resulting point force is,

Also the location of this force can be found by

The moment of this force about the rear wheel should cancel with the moment of the weight

ΣM = 0

The equation can be simplified to

1.5 h = 2.079

This gives a maximum value of h as 1.386 m.

CHAPTER-5-A- 2-D and 3-D SupportsSTATICS - CASE STUDY

Introduction

Problem Introduction AnimationClick to view movie (209k)

Louis the Lamp is tired of his lonely existence up in the attic and attempts to escape through a trap door in the floor. He must determine which of the supports he should remove so that the door will open.

What is known:

The trap door is supported by a hinge and a bearing with a circular shaft on one side.

The other side is supported by a sliding bolt, which is equivalent to a bearing with a square shaft.

Questions

Hinge Diagram

What type of reaction forces and moments does each of the supports exert on the trap door?

Approach

Analyze the conventional methods by which rigid bodies are held stationary, or are held together.

Determine the type of forces and moments that these supports exert on a rigid body in reaction to externally applied loads.

There is not enough data to determine the exact magnitude of each reaction. Only determine the type of reaction.

STATICS - THEORY

Basic 2D Boundary Conditions

Before the equilibrium of rigid bodies can be investigated, the supports that hold them in place, or hold them to other objects, must be first analyzed.

Supports that are commonly found in statics can be represented by stylized models called support conventions. An actual support may be a close approximation of a model.

The forces and moments exerted on a rigid body by its supports are called reactions. These forces and moments are reacting to external loads that are applied to the rigid body.

In general, if a support prevents translation in a given direction, then the support exerts a force in that direction. If rotation is prevented, then the support exerts a couple, or moment, in the direction of the rotation.

Supports can be broken down into two categories: 2-D supports and 3-D supports. The table to the left shows common 2-D support conventions.

To better understand the relationship between support conventions and support reactions, detailed explanation of three of the more commonly used support conventions are presented below. They are the pin support, the roller support, and the fixed support.

Pin Support

Pin Support

With a pin support, a bracket and an object are connected by means of a smooth pin passing through the object and connected to the bracket.

The pin prevents translation but permits rotation. Thus, the pin support exerts forces in any direction, but it cannot exert a moment about the axis of the pin. In 2-D, the reaction force of a pin can be broken down into two component forces parallel to the x and y axes.

Roller Support

Roller Support

The roller support is similar to a pin support but mounted on wheels. It cannot exert a couple about any axis, nor can it exert a force parallel to the surface on which the roller support rests. It can, however, exert a force perpendicular to the surface on which it rests.

The roller permits rotation about any axis. It also permits translation in any direction parallel to the surface. It prohibits translation towards the surface. In 2-D, the reaction force of the roller support can be represented by one force perpendicular to the surface.

Fixed Support

Fixed Support

The fixed support prevents both translation and rotation about any axis. Thus, the fixed support prevents translation and rotation in any direction. In 2-D, the fixed support can be represented by component forces parallel to the x and y axes, and a couple that is perpendicular to the x-y plane.

The table below includes a more comprehensive presentation of both 2D and 3D support conventions and their reactions. Click on any graphic to view a detailed animation of the support mechanism.

2D 3D

Smooth Surface(movie 36k)

One force acting normal to the surface.

Smooth Surface (movie 474k)


Rope/Cable (movie 48k)

One collinear force acting along the axis of the rope or cable.

Rope/Cable (movie 260k)

One collinear force acting along the axis of the rope or

cable.

Link (movie 48k)

One collinear force acting along the axis

of the link.

Roller (movie 504k)


Roller (movie 41k)

One force acting normal to the surface

on which the roller rests.

Bearing: Circular Shaft (movie 331k)

Two force components acting parallel to the coordinate axes,

and two moments perpendicular to the axis of the

bearing.

Fixed Collar (movie 33k)

One force acting perpendicular to the

sleeve, and one moment.

Hinge (movie 344k)

Three force components acting parallel to the coordinate axes,

and two moments perpendicular to the axis of the

bearing.

Pin (movie 28k)

Two force components acting

parallel to the coordinate axes.

Fixed (movie 116k)

Three force components acting parallel to the coordinate axes,

and three moments perpendicular to the axis of the

bearing.

Fixed (movie 25k)

Two force components acting

parallel to the coordinate axes, and

one moment.



To simplify the analysis, determine the reactions that each support exerts on the trap door along each axis of a Cartesian coordinate system. Each reaction is assumed to be acting in the positive direction.

Support 1

Support 1: Sliding HingeClick to view movie (342k)

The first support appears to be a standard hinge at a quick glance. Upon further examination, it becomes clear that this support is actually a bearing with a circular shaft that can slide along the z-axis (standard hinges cannot). This type of support has two unknown reaction forces and two unknown reaction moments along the coordinate axes that are perpendicular to the axis of the bearing. There are, however, no reaction forces or moments along the axis of the bearing. Thus, the bearing can both translate and rotate along this axis.

Reaction at Support 1

Support 2

Support 2: Standard HingeClick to view movie (229k)

The second support is a standard hinge. A hinge has two unknown reaction forces and two unknown reaction moments along the coordinate axes that are perpendicular to the axis of the hinge. There is also a reaction force along the axis of the hinge (i.e. does not slide along the z-axis). There is, however, no reaction moment along the axis of the hinge. Thus, the hinge may only rotate about this axis.


Support 3

Support 3: BoltClick to view movie (178k)

The third support is the bolt. This support does not permit any rotation around any axis. It has two unknown reaction forces and two unknown reaction moments along the coordinate axes that are perpendicular to the axis of the bearing. There is also a reaction moment along the axis of the bolt. Thus, the bolt only permits translation along the axis of the bolt.


CHAPTER-5-B-Equilibrium in 2-DSTATICS - CASE STUDY

Introduction

Problem Description AnimationClick to view movie (422k)

A utility truck uses an extendible arm to raise and lower supplies and equipment. If the arm is raised and then rotated perpendicular to the length of the truck, it will affect the equilibrium of the truck.

What is known:

The initial length of the arm is 8 m. The arm is raised 30° above the horizontal. The mass of the truck is 2,000 kg. The arm is hoisting a 400 kg load of bricks. The mass of the arm and the length of the

truck can be neglected.

Question

Dimension Diagram

When the arm reaches a stable position perpendicular to the length of the truck and 30° up from the horizontal, what are the normal forces exerted on the right and left wheels by the ground?

Approach

Assume that the normal forces acting on the wheels, the center of mass of the truck, and the center of mass of the bricks all lie in the same plane.

Make a free-body diagram. Sum the forces and moments acting on the

truck. Solve for the normal forces acting on the

wheels.

STATICS - THEORY

ForcesClick to view movie (257k)

For a rigid body in equilibrium, the sum of all the forces and the sum of all the moments must be zero,

ΣF = 0 ΣM = 0

Using rectangular coordinates, these equations as be expressed by the vector equations,

ΣF = ΣFxi + ΣFyj + ΣFzk = 0

ΣM = ΣMxi + ΣMyj + ΣMzk = 0

For the expanded equation, each coefficient of the i, j, and k unit vectors must equal zero for static equilibrium.

MomentsClick to view movie (250k)

ΣFx = 0 ΣFy = 0 ΣFz = 0

ΣMx = 0 ΣMy = 0 ΣMz = 0

The equilibrium equation has been separated into three components corresponding to the x, y, and z axes for both the forces and moments. Since each equation is independent of the others, the equations can be used to determine up to six unknowns for a full 3D problem.

Equilibrium Equations

"Two-dimensional" is used to describe problems in which the forces reside in a particular plane (x-y for example), and the axes of all moments are perpendicular to the plane.

Because there is no force in the z direction, and no moments about the x- or y-axis, three of the six independent scalar equations are automatically satisfied. The remaining equations for a rigid body in 2-D equilibrium are

ΣFx = 0 ΣFy = 0 ΣFz = 0

Since there is a maximum of three independent equations for a rigid body in two-dimensional equilibrium, only three unknown forces or couples can be solved.


Use the support conventions presented in

previous section to make a free-body diagram of the truck and bricks. Assume that the normal forces acting on the wheels, the center of mass of the truck, and the center of mass of the bricks all lie in the x-y plane.

Summing the forces in both the x and y direction, and the moments about point the center of gravity of the truck, point A, gives,

ΣFx = 0 (1) ΣFy = N1 + N2 - m1g - m2g = 0 (2)


Before Brickes are RaisedClick to view movie (40k)

After Bricks are RaisedClick to view movie (50k)

ΣMA = N2 L1 - N1 L1 - m2g L2 cosθ = 0 (3)

The first equation shows that there is no force exerted on the wheels by the ground in the horizontal direction. Using the third equation, solve for N2,

N2 = N1 + L2/L1 m2g cosθ

and then substitute into the second equation,

N1 + N1 + L2/L1 m2g cosθ = g(m1 + m2)

N1 = g/2 (m1 + m2) - L2/L1 m2g/2 cosθ

Solving

N1 = 9.81/2 (2,000 + 400) - (8/1.4) (400)(9.81)/2 cos30

N1 = 2,063 N

Substituting back into the second equation gives N2 as,

N2 = g(m1 + m2) + N1

N2 = 9.81 (2,000 + 400) - 2,063

N2 = 21,480 N

Notice that indeed

N1 + N2 = 23,540 N

= g(m1 + m2)

STATICS - EXAMPLE

Example

Beam with Distributed Load

Calculate the reaction forces acting at A and B on the beam.

Solution

First convert the distributed loads into point loads and find their locations.

For a rectangular uniform distributed load, the magnitude of the resultant force will be the product of the length of the rectangle and value of the distributed load. The location of the force resultant is always the center point (centroid) of the distributed load. The magnitude of the resultant force will be

(200 N/m) (4 m) = 800 N

and it will be at 2 m away from support A.

For a triangular line load, the force resultant is one half of the peak value of the distributed load multiplied by the distance over which it acts. The location of the force resultant is two-thirds of the distance from the vertex to the peak value of the load. The magnitude of the resultant force will be

(1/2) (2) (500 - 200) = 300 N

and it is at 1.33 m from left vertex.

The actual reactions can be found by applying equilibrium equations

There will be three possible reactions at A and B namely Ax, Ay and By. which can be found by applying the equilibrium equations.

ΣMA = 0 By (6) - 800 (2) - 300 (3.33) - 150 (5.8) = 0 By = 578.2 N

Equating forces give,

ΣFy = 0 Ay + By - 800 - 300 - 150 = 0 Ay = 671.8 N

Since there is no horizontal applied load, Ax will be zero.

CHAPTER-5-C-Equilibrium in 3-DSTATICS - CASE STUDY

Introduction

Problem Description AnimationClick to view movie (165k)

As Louis the Lamp is making his great escape from the lonely attic, he must jump onto the trap door that will set him free. Before he leaps, however, he must know the tension he will place on the rope that supports the door.

What is known:

Louis has a mass of 0.2 slugs and is located in the center of the door.

The trap door is supported by a ball and socket and a bearing with a circular shaft on one side. The other side is supported by a rope that is attached to the wall.

The trap door and the rope have the dimensions shown.

Assume that the bearing does not exert any moments on the door, and neglect the mass of the door.

Problem Description Diagram

Questions

What are the reaction components at each of the supports? What is the tension in the rope?

Approach

Draw a free-body diagram. Force diagrams

for each joint is shown below. Apply the equilibrium equations for a rigid

body in equilibrium under a three-dimensional system of forces.

Solve the six equations for the six unknowns.

Rope Force Diagram

Ball and Socket Force Diagram

Hinge Force Diagram (all moment reactions are assumed zero.)

STATICS - THEORY

In the 2D equilibrium section, it was shown that the sum of all forces and the sum of all moments acting on a rigid body in equilibrium must be equal to zero,

ΣF = 0 ΣM = 0

Using rectangular coordinates, these vector equations can be expanded into

ΣF = ΣFxi + ΣFyj + ΣFzk

ΣM = ΣMxi + ΣMyj + ΣMzk

These also represent six independent scalar equations,

ΣFx = 0 ΣFy = 0 ΣFz = 0

ΣMx = 0 ΣMy = 0 ΣMz = 0

Note that the sum of the moments can be evaluated about any point. Further equations can be obtained by summing the moments about other points, but they will not be independent of the first moment equations.

Because there are not more than six independent scalar equations, it is not possible to solve for more than six unknowns.


Force Diagram

Position Vectors Diagram

The process for analyzing a rigid body in equilibrium subject to a 3-D system of forces is similar to that of a 2-D system of forces. Begin by drawing a free-body diagram.

To use the equilibrium equations, the tension in the rope must be expressed in terms of its rectangular components. Using the coordinates of points D and E, the position vector from D to E is

rDE = rAE - rAD

= (1i + 1j + 0k) - (2i + 0j + 1k)

rDE = -1i + 1j - 1k

Divide rDE by its magnitude to get a unit vector that is parallel to the axis of the rope.

uDE = -0.5774i + 0.5774j - 0.5774k

Use this unit vector to express the tension in the rope in terms of rectangular components:

T = T uDE

= T ( -0.5774i + 0.5774j - 0.5774k)

Using the equilibrium equations, sum the forces,

ΣFx = FAx + FBx - 0.5774 T = 0 (1) ΣFy = FAy + FBy + 0.5774 T + (0.20 slug)(-32.2 ft/s2) = 0 (2) ΣFz = FAz - 0.5774 T = 0 (3)

Next, sum the moments about point A. This will will give a vector equation representing moments around all three axis that go through A, x-axis, y-axis, and z-axis.

ΣMA = rAB × FB + rAD × T + rAC x (0.20 slug)(-32.2 ft/s2)j = 0

ΣMA = (-2FBy - 0.5774 T + 6.44)i + (2FBx + 0.5774 T)j + (1.1548T - 6.44)k

ΣMx = -2FBy - 0.5774 T + 6.44 = 0 (4) ΣMy = 2FBx + 0.5774 T = 0 (5) ΣMz = 1.1548T - 6.44 = 0 (6)

Solving Eqs. 1-6 simultaneously gives the components of reaction in each support, and the tension in the rope.

FAx = 4.830 lb FAy = 1.610 lb FAz = 3.220 lb FBx = -1.610 lb FBy = 1.610 lb T = 5.577 lb

CHAPTER-5-D- Indeterminate ObjectsSTATICS - CASE STUDY

Problem Description

Introduction

In the lesser-known tale of the Trojan Cow, the would-be invaders of a castle attempt to cross the moat while hidden in the belly of a giant, wooden gift-cow. To cross the moat, an engineer wedges a wooden bridge between the walls of the moat.

What is known:

The cow exerts a resultant vertical force at a distance d from the left edge of the bridge.


There are three ways the bridge can be wedged against the walls of the moat:1) The bridge is in contact with the rough surface of the walls on each end.2) The bridge has smooth surface on both ends which allows slipping (similar to a roller).3) The bridge is in contact with the rough surface of the walls on one end, and has a smooth surface on the other.

2) Smooth Surface Contact (slip)

3) Smooth and Rough Surface Contact

Question

Which method of laying the bridge will enable the engineer to determine the reactions at the supports?

Approach

When the bridge is in contact with the rough surface of the moat wall, a support force is exerted with components parallel and perpendicular to the surface of the wall.

When the bridge has a smooth surface (similar to a roller) on the end, a support force is exerted only perpendicular to the wall.

Evaluate the problem in 2-D.

STATICS - THEORY

With a statically indeterminate object, the reactions exerted on the object by its supports cannot be found using only the equilibrium equations. There are two types of static indeterminacy.

Type 1: Redundant Supports

Redundant Support on Beam

If the supports of a rigid body exert more unknown reactions than the maximum number of equilibrium equations available, then the system is statically indeterminate, and additional equations will be required to determine the reactions.

This occurs when a rigid body has more restraints than the minimum number necessary to maintain equilibrium; the supports are said to be redundant. The difference between the number of reactions and the number of equilibrium equations is called the degree of redundancy.

Free Body Diagram

As an example of a rigid body under the action of a 2-D system of redundant supports, consider a horizontal beam that is fixed at one end and supported by a roller at the other end as illustrated in the diagrams to the left.

There are four unknown reactions,

FAx FAy MA FBy

However, there are only three independent equilibrium equations,

ΣFx = FAx = 0

ΣFy = FAy + FBy - F = 0

ΣMA = MA - 1/2 LF + LFBy = 0

The horizontal beam in the previous diagram has a degree of redundancy of one. With only three equilibrium equations, all four unknowns cannot be determined.

In some cases, however, some of the reactions can be determined by using the equilibrium equations. In this case, for example, the first equation indicates that FAx = 0.

Systems such as this one can be solved by supplementing the equilibrium equations with additional equations that relate the reactions to the deformation of the rigid body. This is the subject of Mechanics of Materials.

Type 2: Improper Supports

If the supports of a rigid body are unable to maintain equilibrium under the applied loads, then the rigid body will move, and the system is improperly

supported or unstable. There are two common scenarios that can lead to improper supports.

Parallel Reaction ForcesClick to view movie (22k)

Parallel Reaction Forces

If the supports exert only parallel reaction forces, then the rigid body is free to move in directions perpendicular to the reactions. Consider a horizontal beam supported by two rollers, If an external load is applied that has a horizontal component, there is no reaction force to prevent the beam from moving in the horizontal direction, and the beam has improper supports.

Concurrent Reaction Forces

Concurrent Reaction ForcesClick to view movie (24k)

If the supports exert only concurrent forces-that is, the lines of action of all the support forces intersect at one point-and the external loads produce a moment about the point of intersection, then the rigid body is not in equilibrium. Consider a horizontal beam supported by two rollers, each on an incline. The lines of action of the reaction forces of the rollers intersect. If an external load is applied that creates a moment about the point of intersection, then the beam will not be in equilibrium.

Note that if the line of action of the external load intersects the same point as the reaction forces, there will be no moment about the point of intersection, and the beam will be in equilibrium.


Need the cow?Click to view movie (11k)

In each of the three methods for laying the bridge, the bridge is subjected to a 2-D system of forces. Each method will be analyzed separately to determine which allows the engineer to find all the reactions at the supports.

Method 1

Begin with a free-body diagram. With Method 1, the bridge is (in essence) supported at each end by pin supports. Thus, each support exerts two components of force; one parallel and one perpendicular to the moat wall. There are four unknowns,

FAx FAy FBx FBy

There are only three independent equilibrium equations for a 2-D system. Since the number of unknowns is one greater than the total number of

equilibrium equations, this method of laying the bridge leads to a statically indeterminate system with a degree of redundancy of one.

The engineer is unable to solve for the reaction forces using only the equilibrium equations.

Method 2

Method 2

Begin with a free-body diagram. With Method 2, each support exerts a force that is perpendicular to the moat wall. These reactions are concurrent, as the lines of action of each reaction force intersect at point P.

The weight of the cow creates an unbalanced moment about point P. Thus, the bridge is unstable due to improper supports and will not remain in equilibrium under the weight of the cow. There are only two unknowns,

FAy FBx

Even thought there are three independent equilibrium equations, there are infinite solutions for the two unknowns. The engineer is unable to determine a unique solution for the reactions using the equilibrium equations.

Method 3

Method 3

Begin with a free-body diagram. With Method 3, there are three unknowns,

FAy FBx FBy

There are three independent equilibrium equations. The reaction forces are neither parallel nor concurrent, and this system is statically determinate. The engineer can solve for the reactions using the equilibrium equations.

CHAPTER-5-E- Two- and Three-Force Members


Introduction

House StructureClick to view movie (804k)

Problem Diagram

An architect has designed a greenhouse with a flat glass roof and a curved glass wall. Each section of the frame for this structure is two pieces. Where should the air conditioning unit be placed so that its weight is evenly distributed?

What is known:

The entire weight of the air conditioner is resting on one section of the frame.

The section of frame consists of a quarter-circular arch with a radius of 12 ft, and a beam with a length of 24 ft.

The arch is connected to the beam with a pin.

The frame is supported at points A and C with pin supports.

The air conditioning unit weighs 200 lb, and is resting on the beam.

Question

Where should the air conditioner be placed so that pins A, B, and C all have the same force magnitude acting at their respective joints?

Approach

Knowing that the frame is composed of a two-force and a three-force member, determine the geometry of the support reactions at A, B, and C.

Use equilibrium equations to find equations for the reaction forces at B and C.

Use geometry and the equations for the reaction forces at B and C to determine the position of the air conditioner which produces equal reaction force magnitudes at A, B, and C.

STATICS - THEORY

The analysis of some equilibrium problems can be facilitated if one or more of the members is subjected to only two or three forces. Knowing what structural members are two forces generally makes solving the problem easier.

Member BD is a Two-Force Member

Both Forces on Two-Force Members Must Be Equal and Collinear

Two-Force MemberClick to view movie (41k)

Two-Force Members

A two-force member is a rigid body with no force couples, acted upon by a system of forces composed of, or reducible to, two forces at different locations.

The most common example of the a two force member is a structural brace where each end is pinned to other members as shown at the left. In the diagram, notice that member BD is pinned at only two locations and thus only two forces will be acting on the member (not considering components, just the total force at the pinned joint).

Two-force members are special since the two forces must be co-linear and equal. This can be proven by taking a two force member with forces at arbitrary angles as shown at the left. If moments are summed at point B then force FD cannot not have any horizontal component. This requires FD to be vertical. Then the forces are summed in both directions, it shows FB must also be vertical. Furthermore, the two forces must be equal.

There are three criteria for a two-force member:

1. The forces are directed along a line that intersects their points of application.

2. The forces are equal in magnitude. 3. The forces are opposite in direction.

Three-Force Members

Three-Force MemberClick to view movie (36k)

A three-force member is a rigid body with no force couples, acted upon by a system of forces composed of, or reducible to, three forces at three different locations. Because all three forces act at different locations on the member, their direction and magnitude are not known.

There are two criteria for a three-force member:

1. The three forces must be coplanar. 2. The forces must be either concurrent or

parallel.


Two Force Member AB

Three Force Member BC

Begin with free-body diagrams isolating each piece of the frame as shown in the diagrams at the left.

The free-body diagram of the quarter-circular arch shows that it is a two-force member. Thus, the reaction forces at pins A and B must be equal, opposite, and collinear. From geometry, it is known that FA and FB act at a 45° incline and

FA = - FB

The free-body diagram of the beam shows that it is a three-force member. Thus, the forces acting on it must be concurrent at point P. Note that the force acting on the beam at point B is equal and opposite the force acting on point B of the arch.

Using geometry, you know that the angle θ of the reaction force at C is a function of the distance x:

θ = tan-1 (x/(24-x))

Equilibrium Equations

Applying the equilibrium equations to the beam BC gives

ΣFx = FB cos45 - FC cosθ = 0

ΣFy = FB sin45 - 200 + FC sinθ = 0

Solving the equations simultaneously for FB and FC,

FB = 200 cosθ/(cos45 (cosθ + sinθ))

FC = 200/(cosθ + sinθ)

Setting the expression for FB equal to the expression for FC, solve to obtain the value of θ for which FB is equal to FC,

FB = FC ==> cosθ/cos45 = 1

θ = 45o

Substitute θ = 45° into the equation for θ to determine the value of x for which FB is equal to FC,

45o = tan-1 (x/(24-x))

x = 12 ft

Substitute θ = 45° into FB or FC to obtain the value of FB and FC,

FB = FC = 200/(cos45 + sin45)

= 141.4 lb

Since the arch is a two-force member,

FA = FB = 141.4 lb

CHAPTER-6-A-2-D Trusses: Method of Joints


Introduction

Problem AnimationClick to view movie (368k)

Trusses are commonly used to span distances between two walls. There are many truss designs, but one of the most common a Howe truss due to its simplicity and efficiency. Trusses usually need to carry both roof loads and ceiling loads. Both types of loads are modeled can be modeled as point loads at the truss joints.

What is known:

The width of the truss is 10 ft. The roof incline angle, θ, is 40°. There are loads at each joint as shown in

the diagram at the left Loads are shown in the diagram.

Problem Diagram

Questions

Which member will have the largest tension force and the largest compression force? What is the magnitude of those member forces?

Approach

Use the method of joints to analyze each pin joint using the basic statics equations,

Howe Truss

ΣFx = 0ΣFy = 0

Each joint must be in equilibrium. Only two unknown member forces can be determined per joint. Each member is a two-force member.

STATICS - THEORY

Various Types of Trusses

Trusses

Trusses are composed of thin structural members that are in compression or tension. Since there are only two joints in a given member, only two forces can act on the member. This means all members are two-force members and thus the member load acts in the direction of the two pins. If the member is straight (no curvature) then there is no bending moment in the member.

When modeling trusses, it is assumed that

1. all loads are applied at joints2. all joints are pins, and support no moment

Load in Two-Force Member Acts in the Direction of the Joints

Method of Joints

Method of JointsClick to view movie (56k)

Each Joint Must be in Equilibrium

One of the basic methods to determine loads in individual truss members is called the Method of Joints. Like the name states, the analysis is based on joints. Each joint is treated as a separate object and a free-body diagram is constructed for the joint.

Because each and every joint must be in equilibrium, the basic force equations can be applied to each joint,

ΣFx = 0 ΣFy = 0

Each joint will only have two equations to solve for member forces since there is no moment at the joint. The means only two unknown member forces can be solved at a single joint and the order in which the joints are solved is important.

Care must be taken in drawing force vectors. A compression member will 'push' the joint, but a tension member will 'pull' the joint.

Method of SectionsClick to view movie (92k)

Method of Sections

A second method to solve complex trusses is called the Method of Sections. This method analyzes whole sections of a truss, instead of joints. This method is described in detail in the following section, 2-D Trusses: Sections.


Support

Free Body Diagram (FBD) of Truss

When solving any truss, the reaction forces at each support should be determined before trying to calculate individual member loads. Starting with a FBD of the entire truss, the basic moment equilibrium equation gives,

ΣM1 = 0 2.5(4/3 + 4) + 5(4/3 + 4) + 7.5(4/3 + 4) - 10R = 0 R = 8.0 kips

Due to symmetry, the reaction forces at both the left and right wall are equal,

RL = RR = R = 8 kips

Joint 1

Joint 1

With the reactions known, it is now possible to analyze joint 1. Note that both F31 and F21 are drawn away from the joint, representing tension.

The two unknowns forces, F31 and F21, can be found using the two force equilibrium equations,

ΣFx = 0 - F21 - F31 cos40 = 0

ΣFy = 0 F31 sin40 + R = 0

Substituting known values and solving gives,

F31 = -8/sin40 = -12.45 kips F21 = -F31 cos40 = 9.534 kips

Joint 2

Joint 2

The value of F21, 9.534 kips, was determined at joint 1 and can now be used at joint 2.

The force in member F32 is easy to determine since all members and loads are perpendicular. Summing the forces in the vertical gives,

ΣFy = 0 F32 = 1.333 kips

Since there are only two horizontal member forces, they must be equal,

ΣFx = 0 F21 = F42 = 9.534 kips

Joint 3

Joint 3

The value of F31, -12.45 kips, and F32, 1.333 kips, were determined from joint 1 and joint 2, respectively. It is now possible to determine forces, F53 and F43, at joint 3,

ΣFx = 0 = cos40 [-F53 - F43 + (-12.45)] F53 = -F43 - 12.45

ΣFy = 0 = sin40 [F53 - F43 - (-12.45)] - 4 - 1.333

Eliminating F53 gives,

0 = -2F43 - 4/sin40 - 1.333/sin40 F43 = -4.148 kips

and

F53 = 4.148 - 12.45 = -8.302 kips

Joint 4

Joint 4

Now that F42, 9.534 kips, and F43, -4.148 kips, are known, joint 4 can be analyzed. Due to symmetry, the left member forces are the same as the right member forces. There is only one unknown force, F54. UsingΣFy = 0 gives,

F54 + 2(-4.148) sin40 - 1.333 = 0 F54 = 6.666 kips

Summary

Final Loads (kips) in Truss Members

As shown by the force diagram, the largest tensile force is 9.534 kips and the largest compressive force is -12.45 kips.

The best truss angle, θ, that minimizes the member loads can be determined by using the truss simulationin this section. The best angle is the largest angle possible in the simulator, 65 degrees. Thus a steep truss reduces the load. However, the cost is high and buckling problems will also need to be considered.

STATICS - EXAMPLE

Example

Truss Supporting its own Weight

Determine the force in each member due to the weight of the truss. Assume the weight of the members are uniform along their length and have a mass density ρ of 9 kg/m. (g = 9.81 m/s2)

Solution

Equivalent Loading for Truss Member

The only loading on the truss is the weight of each member. Since all loads on a truss must act at the joints, the distributed weight of each member must be split between the two joints. Assume the weight of each member is a vertical force, half of which is placed at each end of the member as shown in the diagram on the left.

The weight of each member is

WAB = ρgLAB = 9 (9.81) 2 = 176.6 N

WBD = WED = WAB = 176.6 N

WEB = ρgLEB

= 9 (9.81)(22 + 22)0.5

= 9 (9.81)(2.828) = 249.7 N

WBC = ρgLBC

= 9 (9.81)(12 + 22)0.5

= 9 (9.81)(2.236) = 197.4 N

WDC = ρgLDC

= 9 (9.81) (1) = 88.29 N

The weight of each member is considered as an external force on the end points. The external force at each joint is half of the weight of the bar. The table below summarizes how the weight of each member is distributed to its own joints. The table also gives the total force at each joint.

MemberJoint

A B C D EAB 88.29 88.29 0 0 0BC 0 98.70 98.70 0 0DB 0 88.29 0 88.29 0BE 0 124.8 0 0 124.8CD 0 0 44.14 44.14 0ED 0 0 0 88.29 88.29

total88.29 400.1 142.8 220.7 213.1

Joint C

Now that the total load for all the joints are know, each joint can be analyzed. Joint C is a good point to start with, since there are only two unknown member forces, FBC and FCD.

The summation of forces in y direction for the joint C gives,

ΣFy = 0 (WBC+ WDC) / 2 - FBC (LDB / LBC) = 0 (197.4 + 88.29) / 2 - FBC (2 / 2.236) = 0 FBC = 159.7 N

The summation of the forces in x direction gives,

ΣFx = 0 FBC (LCD / LBC) + FCD = 0 159.7 (1 / 2.236) + FCD = 0 FCD = -71.41 N

Joint D

Joint D

ΣFx = 0 FDE - FCD = 0 FDE = - 71.41 N

ΣFy=0 FBD - 220.7 = 0 FBD = 220.7 N

Joint D

Joint B,

ΣFy= 0 400.1 + FBD+ FBE (LDB / LEB) + FBC (LDB / LBC) = 0 400.12 + 220.73 + FBE (2 / 2.828) + 159.7 (2 / 2.236) = 0 FBE = -1,080 N

ΣFx=0 FAB + FBE (LDE / LEB) - FBC (LDC / LBC) = 0 FAB - 1,080 (2 / 2.828) - 159.7 (1 / 2.236) = 0 FAB = 835.2 N

Notice, the boundary reactions at A and E were not needed. This is unusual and is due to the structure being cantilevered. Generally, reactions need to be determined first, then member forces.

CHAPTER-6-B- 2-D Trusses: Method of Sections


Introduction

Car Driving over Bridge Click to view movie (403k)

Dimension Diagram

At the local highway department, the bridge designer is required to calculate the force in the top horizontal member of a new bridge truss. The member force is need to help determine the correct type of material and the size of the member.

What is known:

The car mass is 1,500 kg. The car is located at joint 2. The bridge length is 16 m (4 bays). The bridge height is 4 m. Joint 1 is a pin support, and joint 5 is a

roller support.

Question

When the car is over joint 2, what is the force in member 6-7?

Approach

First, solve for the support reactions. Next, section off part of the truss and apply

the three equilibrium equations:

ΣFx = 0

ΣFy = 0

ΣM = 0

STATICS - THEORY

Method of Sections

Recall, for all plane truss structures, the following three conditions are assumed:

1. All members are in the same plane.2. All loading is at the joints.3. All joints are pinned (free rotation).

These conditions require all member forces to act in the direction of the member, and there are no moment loads in the truss members (assuming straight

Making Section CutsClick to view movie (60k)

members).

In the previous Method of Joints, each joint was independently analyzed which resulted in numerous joint calculations. If only one member force is desired it does not make sense to solve for many joints. The Method of Sections is an alternative method to solve for interior member forces that is simpler than the Method of Joints.

In the Method of Sections, the truss is cut into two sections. The removed section is replace with unknown member forces acting in the direction of the cut member. They can be pointing in either direction, but generally, they are drawn away from the section to represent tension.

The unknown member forces at the section cut can be solved using the equilibrium equations, ΣFx = 0, ΣFy = 0, and ΣM = 0.

Since there are only three equilibrium equations, the truss section cut should be located where there are only three unknown member forces.

Mid-Span Loads

Mid-Span LoadsClick to view movie (70k)

It was previously stated that all external loads on a truss must be at the joints. If there is load at a location that is not a joint, the load needs to be split and applied to the nearest joints.

Distributing the mid-span load to the adjacent joints is only an approximate solution so that the structure can be analyzed as a truss. If there is large mid-span loads, the structure should be analyzed as a frame with possible bending moments in the members. Frame analysis is beyond this course and is not addressed in statics courses.


Solve for Truss Reactions

Force Diagram

The force in the top member 6-7 needs to be determined when the 1,500 kg car is at joint 2.

First, solve for reactions R1 and R5 by applying both the moment and force equilibrium equations for the whole truss.

ΣM1 = 0 -4 (14.72) + 16 R5 = 0 R5 = 3.68 kN

ΣFy = 0 R1 + R5 - 14.72 = 0 R1 = 11.04 kN

Cut Truss at Required Member

Cut 1 Diagram

Since, the member 6-7 needs to be determined, that member must be included in the cut. However, there is is usually more than one location to make a cut.

It is easiest to make a cut where only three or less members are cut. Thus, a vertical cut through members 6-7, 6-3 and 2-3 was done as shown at the left. Notice that there are only three unknowns, which can be solved for with the three equilibrium equations.

ΣFy = 0 11.04 - 14.72 - F63 cos45 = 0 F63 = -5.204 kN

ΣM1 = 0 -4 (14.72) - 4cos45 F63 - 4sin45 F63 - 4 F67 = 0 -58.88 - 2 [(4) (0.7071) ( -5.204)] = 4 F67 F67 = -7.360 kN compression

For this cut location, you need to use at least two of the three equilibrium equations. The number of equations may be reduced by other cut locations.

Alternate Cut

Cut 2 Diagram

By carefully choosing where the cut is made, the number of calculations can be reduced.

For example, if a cut is made through members 6-7, 3-7, 3-8, and 3-4 it is possible to solve for the member force F67 with one equilibrium equation even though four members are cut.

Since all unknowns, except F67, go through

joint 3, the moment about joint 3 has only one unknown.

ΣM3 = 0 -8 (11.04) - 4 F67 + 4 (14.72) = 0 F67 = -7.360 kN compression

The solution is identical to the solution for the previous cut.

STATICS - EXAMPLE

Example

Complex Truss with Loading

Determine the force in members BE, BJ, DJ, and IJ. (ADJ and HDB are not continues, each consist of two elements.)

Solution

Reaction Forces at End Supports

The first step is to find the support reaction forces. Vertical reaction force at F can be found by taking the moment about L.

ΣML= 0 = - 6Fy + 5Gy + 4Hy + 3Iy + 2Jy

= - 6Fy + 5(2) + 4(3) + 3(5) + 2(2) = - 6Fy + 41 = 0

Fy = 6.833 kN

Summation of forces in the y direction should be zero.

ΣF= 0 Fy + Ly - (Gy + Hy + Iy + Jy) = 0 6.833 + Ly - (2 + 3 + 5 + 2) = 0 Ly = 5.167 kN

Another method to find Ly is to take the moment about F. It is a good way to check to see if the previous value for Ly was found correctly.

ΣMF = 0 = 6Ly - ( Gy + 2Hy + 3Iy + 4Jy )

= 6Ly - ( 2 + 2(3) + 3(5) + 4(2) ) = 6Ly - 31 = 0

Ly = 5.167 kN

Method of Section (Cut 1)to Determine FBE

To find the value of the forces it is easier to use method of sections. Since the problem asks for four member forces, BE, BJ, DJ, and IJ, the problem needs to be solved taking two cuts. The first cut , shown at the left, will be used to solve for member force in member BE. Then another cut will be used to find the other three member forces.

Moments about J can be taken to find the value of the FBE (notation EJ represents the length of the member EJ).

ΣMJ = 0 = 2L + FBE(EJ)

JBE is a right triangle and BE is perpendicular to EJ. The length of KE and JK is 1. Based on Pythagorean theorem, EJ is the square root of 2. (EJ = 1.414).

2L + FBE(1.414) = 0 2(5.167) + FBE(1.414) = 0 FEB = -7.308 kN

Method of Section (Cut 2)to Determine FJB, FJD, and FIJ

Now that FBE is known, another cut can be made to find FJB, FJD, and FIJ. To find FJD and FJB, moment about L can be used with the sum of the vertical forces (notation JL mean length of JL)

ΣML = 0 = JL (2kN) - FJD(BD) - FJB(JL) = 2(2) - FJD(1.414) - FJB(2)

ΣFy = 0 = FJD + FJB + 5.167 - 2 -7.308(sin45)

FJD and FJB can be found by solving the above system of equations to give

FJD = -0.8327(10-3) kN (very small)

FJB = 2.000 kN

FJD is found to be 0.8 newton which is negligible and is because of the rounding errors.

Summation of forces in x direction can be written in

order to find the value of FIJ

ΣFx = 0 = FEB(cos45) + FJD(cos45) + FIJ

= -7.308 (0.707) + 0 (0.707) + FIJ

FIJ = 5.167 kN

FIJ can also be found by taking the moment about joint B

ΣMB = 0 = L(JL) - FJD(BD) - FIJ(BJ) = 5.167(2) + 0 (1.414) - FIJ(2)

FIJ = 5.167 kN

CHAPTER-6-C-3-D Trusses/Space Trusses


Introduction


The lead engineer on the new Starfinder rocket must determine the force in each member of the inter-stage truss structure. If the load is too high, the first-stage thrust will buckle the truss members.

What is known:

The base ring radius is 1 m. The top ring radius is 0.75 m. The base and top ring are 1 m apart. The total thrust of the first-stage main engine

is 2,400 kN (300 kN per joint).

Delta 2 Rocket (US)Click to view movie (798k)

Mu-3SII Rocket (Japan) Click to view movie (84k)

Dimension Diagram

Question

Space Shuttle (US) Click to view movie (122k)

What is the force in each of the truss members of the truss structure between the second- and third-stage rockets when the first-stage main engine is firing?

Approach

Sum all forces at a joint (method of joints).

ΣF = 0

Find the 3D unit vectors for all forces at the joint. Find the unknown magnitudes of the member forces.

STATICS - THEORY

Similar to 2-D trusses, 3-D trusses (space trusses) require all members to have a pin joint at each end. Also, all loads must act at the joints.

The above two conditions insure that member forces act in the direction of the members. Without knowing the direction of the load, truss problems could

Method of Joints

Method of JointsClick to view movie (451k)

Space truss problems can be solved by choosing a joint with only three unknown member forces and one or more known load forces. The forces at the joint are summed in all three directions which will produce three equations and three unknowns.

ΣFx = 0 ΣFy = 0 ΣFz = 0

The method of joints is usually the easiest method since the moment equilibrium equations are difficult to apply in 3D.

Vector notation should be used with 3-D problems because of the complexity of 3-D geometry.

Method of Sections

Method of SectionsClick to view movie (446k)

If only a single-member load is needed, then the method of sections can be used, just like with 2D planar trusses. However, with 3D trusses, the section that is cut will have three moment equations in addition to the three force equations to find the unknowns.

ΣFx = 0 ΣFy = 0 ΣFz = 0

ΣMx = 0 ΣMy = 0 ΣMz = 0


Loading at Joint BClick to view movie (604k)

Free Body Diagram of Joint BClick to view 3D movie (156k)

Top View at Point B

Since all members are symmetrical around the inter-stage structure, all members will have the same load. Using the method of joints, the forces at joint B on the bottom ring is analyzed by summing the forces,

ΣFB = 0

The free-body diagram includes the external thrust load and all member forces acting at joint B. Since Joint B is in static equilibrium, the summation of all force vectors must equal zero. Since this is a 3D problem, all forces will be represented in the vector i, j, k format.

The total thrust load of 2,400 kN is evenly distributed around the inter-stage structure so each joint will have to withstand a vertical load of 300 kN.

FT = 300 k

Truss Diagram

Use the location of points B and E to define the unit directional vector of FBE.

Ex = 0.75 sin22.5 = 0.2870 m Ey = -0.75 cos22.5 = -0.6929 m

FBD of Joint B

Radius Effect on LoadClick to view movie (56k)

Example of Space Truss

Ez = 1.0 m

Bx = 0 m By = -1 m Bz = 0 m

The length of member BE is

BE = ((0.2870 - 0)2 + (-0.6929 - (-1))2 + (1 - 0)2 )0.5 = 1.085 m

The vector FBE in i, j, k format is

The member force FBD is similar to FBE except the x component is reversed. The magnitudes will be the same:

FBD = FBE (-0.2645i + 0.2830j + 0.9217k)

If forces are summed in the z direction, ΣFz = 0, only one unknown remains, FBE. Solving for FBE gives

2 FBE 0.9217 + 300 = 0

FBE = -162.7 kN compression

STATICS- EXAMPLE

Example

Determine the force in the member CD, BD and AD of the space truss shown in the figure. Note that points A, B and C are in the x-z plane and member CD is parallel to y-axis.

Solution

First, the co-ordinates of joints A, B, C and D need to be found.

From the figure it can be seen that ΔABC is an equilateral triangle with each side of 5 feet. The height of the equilateral triangle will be equal to 5 cos30o, or 4.330 feet. Thus, the coordinates each joint will be,

A = [0, 0, 0] B = [-5, 0, 0]

C = [-2.5, 0, 4.33] D = [-2.5, 5, 4.33]

Since, there are only three unknown forces at joint D, the force analysis of the truss will begin at joint D. Recall, with 3D trusses, three equations can be used at each joint.

Let the force in member AD, BD and CD be labeled as FAD, FBD and FCD respectively. Expressing each force acting at joint D in the vector notation gives,

W = -250 k FAD = FAD (-2.5i + 5 j + 4.33k) / 7.071 FBD = FBD (2.5i + 5j + 4.33k) / 7.071 FCD = FCD 5j / 5

All forces have to be in equilibrium, giving

ΣF = 0 W + FAD + FBD + FCD = 0

Equating the i, j and k components gives,

ΣFx = 0 0.3536 FBD - 0.3536 FAD = 0 FBD = FAD

ΣFz = 0 0.6124 FAD + 0.6124 FBD - 250 = 0 By substituting FAD = FBD

0.6124 FAD + 0.6124 FAD - 250 = 0

FAD = 204.1 lb = FBD

ΣFy = 0 0.7071 FAD + 0.7071 FBD + FCD = 0 FCD = -288.6 lb

Members AD and BD are in tension and member CD is in compression.

CHAPTER-6-D-Frames and MachinesSTATICS - CASE STUDY

Introduction

Work Bench ToolsClick to view movie (855k)

An owner has forgotten the combination to an old lock. In order to remove the lock, he will need to use a pair of bolt-cutters.

What is known:

A force of 45 lb is applied on each handle. The dimensions of the bolt-cutters are:

Bolt-Cutter Dimensions


Question

If 45 lb of pressure on each handle is required to cut the lock, what is the total force exerted at point Q?

Approach

Use symmetry to simplify the problem. Disassemble the tool and label the known

and unknown forces acting on each segment.

Identify two-force members Identify joints common to multiple members. Solve using the equations of equilibrium.

STATICS - THEORY

Frames and machines are structures similar to trusses except that they have at least one member that is a multi-force member. A multi-force member is an independent piece of the system that has three or more forces acting on it.

Frames are designed to support loads and are generally stationary. Machines contain moving parts that transmit or alter applied forces.

Equilibrium

Frame ExampleClick to view movie (91k)

Frame SolutionClick to view movie (102k)

Begin by isolating each segment of the frame and labeling the known and unknown forces. This is usually done by choosing free-body diagrams that provide the simplest path to a solution. These diagrams may consist of single members, sections, or entire structures.

Next, identify two-force members and represent the forces with two equal and opposite vectors that act along the line of action.

Determine which forces are shared by more than one member. It is important that the forces are represented correctly on each member. This means that the forces represented on one member will be opposite in direction on another.

Verify that the number of unknowns does not exceed the equilibrium equations. There are three equilibrium equations that can be used per member,

ΣFx = 0 ΣFy = 0 ΣFz = 0

Finally, use the equations of equilibrium to solve the problem.


Bolt Cutter Solution Click to view movie (344k)

FBD for Cutter Section

FBD for Handle Section

Equivalent Pliers Force Click to view movie (276k)

This problem requires a number of steps to solve for the force at Q.

First, separate the components of the bolt-cutters and label the forces acting on each part. Since the bolt-cutters are symmetric, only the components of the top half will be considered.

From the free-body diagram of the cutter section, one can observe that there is only one force in the x direction. This means that

ΣFx = 0 = Bx

If force Bx is equal to zero, the diagram for the handle section shows that

ΣFx = Ax - Bx = 0

Solving for Bx gives

Bx = Ax = 0

Now, the y component of force B needs to be determined. To do this, take the moment about point A for the handle,

ΣMA = By (1.25") - 45 lb (7") = 0

Solving for By gives

By = 252 LB

Now, find the force at Q by summing the moments about point C for the cutter section,

ΣMC = By (2.5") - Qtop (0.75") = 0

Solving for Q then yields the amount of force required by the top components to cut the lock,

Qtop = 840 LB

Therefore, the total amount of force required to cut the lock is the sum of the top and bottom resultants at Q. This yields

Qtotal = 1,680 LB

If the same force is applied to machines such as pliers or metal snips, the resultant force will vary due to the different configurations. The approximate

Equivalent Metal Snips ForceClick to view movie (275k)

magnitude for the metal snips is comparable to the bolt-cutters, but the pliers' magnitude is much smaller. Thus each tool's design plays a major role in its function.

In this problem, only the total force exerted at point Q was needed. In reality, it is the stress applied to the lock that will cause it to break. Using bolt-cutters and metal snips, the force is applied over a small area due to the cutting edges. However, with pliers, the force is applied over a larger area and therefore produces less stress. In this respect, many different factors are important in tool design.

CHAPTER-7-A- Centroid: Line, Area and Volume


Introduction

Geometry Diagram

In order to determine a submarine's buoyancy characteristics, designers must accurately determine the centroid, center of mass, and center of gravity for various parts of the sub.

What is known:

The radius of the nose cone is given by r(x) = B - CxD.

The length of the nose cone is 100 ft. The base of the nose cone has a radius 20 ft. The shape exponent D is 4, but as a first

approximation, the designers use D = 1 in order to simplify the mathematics.

The density of the nose cone varies linearly from 1 slug/ft3 at the base to 2 slug/ft3 at the tip.

Question

As a first approximation, assume the nose code is a simple cone shape (D = 1 in the above equation). Where is the centroid, center of mass, and center of gravity?

Approach

Use symmetry to determine the y and z locations of the centroid, center of mass, and center of gravity.

Solve for the x location by integrating the appropriate quantities from the base to the tip of the nose cone.

STATICS - THEORY

Recall, the moment of a force about a point is given by the magnitude of the force times the perpendicular distance from the point to the force.

Using the same definition, the moment of an area about a point is the magnitude of the area times the perpendicular distance to the point. When the moment of an area about a point is zero, that point is called the centroid of the area.

The same method can be used to determine the centroid of a line or the centroid of a volume by taking the moment of the line or the moment of the volume.

Finding the location of a centroid is very similar to finding the location of the force resultant of a distributed force as covered in the moment chapter.


Centroid of a Line

The coordinates for the centroid of a line can be determined by using three scalar equations,

Centroid of an Area

Centroid of an Area Click to view movie (220k)

The centroid of an area can be determined by using three similar equations:

Centroid of a Volume

Centroid of a VolumeClick to view movie (163k)

Similarly, centroid of a volume can be determined by

Use of Symmetry

Centroid and Planes of SymmetryClick to view movie (348k)

Finding the centroid of a body is greatly simplified when the body has planes of symmetry. If a body has a single plane of symmetry, then the centroid is located somewhere on that plane. If a body has more than one plane of symmetry, then the centroid is located at the intersection of the planes.

Center of Mass

Three Planes of SymmetryClick to view movie (190k)

The centroid of a volume defines the point at which the total moment of volume is zero. Similarly, the center of mass of a body is the point at which the total moment of the body's mass about that point is zero. The location of a body's center of mass can be determined by using the following equations,

Here ρ is the density of the body. If ρ is constant throughout the body, then the center of mass is exactly the same as the centroid.

Center of Gravity

The center of gravity of a body is the point at which the total moment of the force of gravity is zero. The coordinates for the center of gravity of an object can be determined with

Here g is the acceleration of gravity (9.81m/s2 or 32.2 ft/s2). If g is constant throughout the body, then the center of gravity is exactly the same as the center of mass.


Centroid

CentroidClick to view movie (216k)

For the first approximation, the radius of the nose cone is given by the equation (recall, D is assumed to be 1)

r(x) = B - Cx ft

Since the radius is known to be 20 ft at x = 0, the constant B can be determined as

r(0) = B - C(0) => B = 20

At the other end, the nose cone tapers to zero, so at x = 100 ft, the radius is zero. Using this condition, the constant C can be determined,

r(100) = 20 - 100 C = 0

C = 0.2

Based on symmetry, the centroid must lie somewhere along the x axis.

Integrating the VolumeClick to view movie (123k)

The x coordinate of the centroid using the equation,

To perform the integrations, a slice of the nose cone that is perpendicular to the x axis can be used. For any slice, the area is

A(x) = π r(x)2

The volume will be the area times the slice thickness, dx, giving

V(x) = π r(x)2 dx

Combining equations gives

Here L is the length of the nose cone. Expand the numerator and denominator and integrate gives

The final result is

Center of Mass

Centroid and Center of MassClick to view movie (186k)

Since the density varies from 1 slug/ft3 at the base to 2 slug/ft3 at the tip, the density is given by the equation

ρ(x) = 1 + 0.01x slug/ft3

The density does not vary in the y or z direction, therefore

Substitute the density and area equations into the center of mass equation gives

Expand and integrate gives

The final result is

Notice the distance is slightly larger than the centroid, which corresponds to the increasing density toward the front.

Center of Gravity

Since the nose cone is relatively small, variation of gravity over its volume can be neglected. Thus, the center of gravity and the center of mass are the same point.

CHAPTER-7-B- Centroid: Composite Parts


Introduction


A luggage transport system for a new airport is having some problems. The railway track bends when the luggage carrier passes over it and will probably break. To analyze this bending, engineers must determine the neutral axis of the beam, which for a homogeneous beam passes through the centroid of the cross section. (The area moment of inertia of the cross section will be found in a latter section).

What is known:

The dimensions of the cross section are as shown.

Question

Where is the centroid of the cross section?

Approach

Break the beam cross section into simpler, component parts.

Determine the centroid and area of each

component part. Use the method of composite parts to

calculate the x and y locations of the centroid of the cross section.

STATICS - THEORY

Equilibrium

In the previous section, Centroid: Line, Area and Volume , it was shown that the centroid of an object involves evaluating integrals of the form

where Q is a line, area, or volume, depending on the centroid that is required. The same equation can also be used for the other two directions by just substituting y or z for x. If there are several objects, then the centroid of the entire system is given by

where n is the number of objects in the system. This equation can be simplify as

where is the centroid location of the ith object, and Qi is the length, area, or volume of the ith object, depending on the type of centroid.

Centroid of a System of Lines

For a system of lines, the coordinates of the centroid are

Here and are the centroid coordinates of the ith line, and Li is the length of the ith line.

Area Centroid of a System of Objects

For a system of objects, the coordinates of the area centroid are

Here and are the area centroid coordinates of the ith object, and Ai is the area of the ith object.

Volume Centroid of a System of Objects

Centroid of Multiple ObjectsClick to view movie (61k)

For a system of objects, the coordinates of the volume centroid are

Here and are the volume centroid coordinates of the ith volume, and Vi is the volume of the ith object.

Subtraction of Material (Holes)

If an object or system of objects has a cutout or hole, then the centroid of the system can be found by considering the cutout or hole as a negative area, volume, or line length.

For example, if a system consists of a cube with a centroid at x1, y1, z1 and a volume of V1; and a hole within the cube that has a centroid at x2, y2, z2 and a volume of V2; then the coordinates of the centroid for the system are


Rail SubpartsClick to view movie (48k)

Rail Dimensions

The beam cross section can be broken into 4 composite parts consisting of 3 rectangles and a triangle as shown. Orient an axis system as shown, so that the cross section lies entirely in the x-y plane. Since the thickness is constant for all the parts, the centroid can be found using the area equations,

Composite Part 1

Part 1

Part 1 is a rectangle with an area of

A1 = (2)(3) = 6 cm2

The centroid of a rectangle lies at half its width and half its height, so for part 1

Composite Part 2

Part 2


A2 = (16)(2) = 32 cm2

Its centroid is located at

Composite Part 3

Part 3


A3 = (2)(10) = 20 cm2

Its centroid is located at

Composite Part 4

Part 4

Part 4 is a triangle with an area of

A4 = (0.5)(4)(4) = 8 cm2

The centroid of a right triangle is located two-thirds of the distance from the vertex to the other end; therefore, for part 4

Centroid of Total System

Total System

With these results, find the total area of the system is

Substituting the areas and centroid locations for each of the individual parts into the first two equations gives

CHAPTER-7-C-Distributed Loads


Introduction

Problem DescriptionClick to view movie (339\k)

An underwater archway is being added to a large aquarium so that visitors can view the fish up close. The water in the aquarium will exert a tremendous force downward onto the transparent arch. This force needs to be calculated accurately so that a safe design can be developed.

What is known:

The tunnel radius, a, of the glass arch is 9 ft. The water depth, b, is 20 ft. The length of the arch, L, is 30 ft. The water density ρ is 1.94 slug/ft3.

Question

Dimension Diagram

What is the resultant force exerted on the glass arch?

Approach

Determine the pressure exerted by the water on the surface of the arch.

Integrate the x and y components of the pressure over the surface area of the transparent arch to get the resultant force in the x and the y directions respectively.

STATICS - THEORY

Parallel Distributed Forces

Load Distributed over a Line Click to view movie (91k)

Previously, it was determined that the resultant force FR for a distributed load over a line is,

It was also shown that the location of the line of action of FR is given by

Similarly, if the load is distributed over an area as shown, then the resultant is found by

Load Distributed over a Area Click to view movie (281k)

The line of action of FR passes through the point defined by

Forces on a Submerged Surface

Pressure on a Submerged SurfaceClick to view movie (295k)

When a body is submerged in a fluid, either a liquid or a gas, then the fluid exerts a pressure normal or perpendicular to the surface. This pressure, called hydrostatic pressure, depends on the depth and density of the fluid, the acceleration of gravity, and the pressure at the surface of the liquid. The pressure at any point is

p = po + ρgd

Here po is the pressure at the surface of the liquid, ρ is the density of the fluid, g is the acceleration of gravity, and d is the depth at that point.

Generally, devices that measure pressure actually measure the amount above some reference pressure; this is called gage pressure. If the reference pressure is the pressure at the surface of the liquid, then the gage pressure at a point is simply

p = ρgd


Pressure Distribution

Solving for a Single Slice Click to view movie (217k)

Water Pressure Resultant Force Click to view movie (74k)

The pressure at a point on the submerged arch is perpendicular to the surface and depends on the water depth at that point. If an axis system is oriented at the center of the arch, the pressure at any height y can be determined as

p = ρg(b - y)

where ρ is the water density, g is the acceleration of gravity and b is the total depth of the water to the floor. If the angle θ is as shown, then

y = a sinθ

and

p = ρg(b - a sinθ)

The pressure at any angle θ can be broken into x and y components as

px = -p cosθ = ρg(a sinθ - b) cosθ

py = -p sinθ = ρg(a sinθ - b) sinθ

Resultant Force

Integrate Over Surface Area S

The resultant force is found by integrating the pressure over the arch surface, S, which means integrating θfrom 0 to π and multiplying by the length of the arch.

For the x component:

Integrate or use integral tables to find

Rx = 270 ρg [a/2 sin2θ - b sinθ]0π

Evaluating the integral for Rx gives

Rx = 0 lb

The y component is

Integrating and evaluating for Ry

Ry = 270ρg [a/2 (-cosθ sinθ + θ) b cosθ]0π

Ry = 270ρg (aπ/2 - 2b) lb

Finally, substitute for the known density, water depth, and arch radius to give

Ry = -436,200 lb = -436 kips

Therefore, in vector form, the resultant force is

R = 0i - 436j + 0k kips

CHAPTER-7-D-Area Moment of Inertia


Introduction

Luggage Railway SystemClick to view movie (351k)

Once again, the engineers need to study the luggage transport system that was examined previously. This time, they need to understand the effects of the bending on the rail itself by determining the moments of inertia of the cross sectio with respect to an axis through the centroid of the rail.

What is known:

The centroid is located at x = 10.62 cm and y = 4.218 cm (calculated previously).

The dimensions of the cross section are as shown.

Question

What are the moments of inertia for the entire cross section with respect to the centroidal axis?

Centroid and Dimension Diagram

Approach

Determine the moments of inertia for each individual part with respect to its centroid.

Use the parallel axis theorem and the method of composite parts to determine the moments of inertia for the centroidal axis of the cross section.

STATICS - THEORY

Moment of InertiaClick to view movie (93k)

Moment of Inertia for a Plane AreaClick to view movie (59k)

Integrals of the form

are called first moments of the area. Accordingly, integrals of the form

are called second moments of the area, or area moments of inertia.

Moments of Inertia for a Plane Area

Polar Moment of InertiaClick to view movie (101k)

For an area A that lies in the x-y plane as shown, the area moments of inertia about the x and y axes are

Moments of inertia that are calculated about the centroid of the area are denoted

Ix' Iy'

The moments of inertia for basic shapes are tabulated in Sections Appendix.

Polar Moment of Inertia for a Plane Area

Polar Moment of Inertia for a Plane AreaClick to view movie (56k)

The moment of inertia for an area that lies in the x-y plane can also be calculated about the z axis, which is known as the polar moment of inertia. The polar moment of inertia of the area A is calculated as

If the polar moment of inertia is calculated at the centroid of the area, it is denoted

Jx' = Ix' + Iy'

The polar moment of inertia is commonly used when calculating the torsion of shafts.

Parallel Axis Theorem for an Area

Parallel Axis TheoremClick to view movie (23k)

If the area moments of inertia about the centroid are known, then the moments of inertia about any other parallel axis can be found as

Ix = Ix' + Ady2

Iy = Iy' + Adx2

Iz = Iz' + Ad2

Parallel Axis Theorem (Polar Moment of Inertia)Click to view movie (24k)

Parallel Axis Theorem from the Centroid

Moments of Inertia for Composite Areas

Moment of Inertia - Composite PartsClick to view movie (51k)

Just as the centroid of an area can be calculated by breaking the area into simpler composite parts, the moments of inertia of a complicated area can be calculated by breaking the area into simpler composite parts.

For an arbitrary axis, the moments of inertia for an area made of composite parts are given by

This technique also works for polar moment of inertia.

Subtraction of Material (Holes)

Previously, in the Centroid: Composite Parts section, it was shown that when the centroid of line, area, or volume are calculated, holes or cutouts can be accounted for by considering them a negative line length, area, or volume respectively. When calculating moments of inertia, we can deal with cutouts and holes in the same manner. Both the moments of inertia about the centroid of the hole as well as the area of the hole are considered negative when we are summing the composite parts.


Composite Parts - Moments of Inertia

Dimension Diagram

Object Split into Sections Click to view movie (146k)

After breaking the cross section into its composite parts, determine the area, the location of the centroid, and the centroidal moments of inertia for each part.

The centroid and the area of each part were found in the previous section, Centroid: Composite Parts. The moments of inertia for each part can be found from the tables in the Sections Appendix.

For the axis system as shown, the properties for part 1 are,

x1 = 1 cm y1 = 3.5 cm A1 = 6 cm2

Ix'1 = 1/12 2(3)3 = 54/12 cm4

Iy'1 = 1/12 3(2)3 = 24/12 cm4

The properties for part 2 are,

x2 = 8 cm y2 = 1 cm A2 = 32 cm2

Ix'2 = 1/12 16(2)3 = 128/12 cm4

Iy'2 = 1/12 2(16)3 = 8192/12 cm4

And the properties for part 3 are,

x3 = 15 cm y3 = 7 cm A3 = 20 cm2

Ix'3 = 1/12 2(10)3 = 2000/12 cm4

Iy'3 = 1/12 10(2)3 = 80/12 cm4

Part 4 is a triangle and its properties are,

x4 = 17.33 cm y4 = 10.67 cm A4 = 8cm2

Ix'4 = 1/36 4(4)3 = 256/36 cm4

Iy'4 = 1/36 4(4)3 = 256/36 cm4

Distances from Global Centroid

These can be used with the following equations to find the moments of inertia of the entire cross section with respect to the centroid of the cross section.

Here dxi and dyi are the distances from the centroid of the cross section (global centroid) to the centroid of any part i. These distances are given by the equations:

Part x y

1 1 3.5

2 8 1

3 15 7

4 17.33 10.67

combined 10.616 4.218Centroids for Each Part and Total(from Centroid: Composite Parts)

Substituting x, y and the results of the above equations give,

dx1 = 10.616 - 1 = 9.616 cm dy1 = 4.218 - 3.5 = 0.718 cm

dx2 = 10.616 - 1 = 2.616 cm dy2 = 4.218 - 1 = 3.218 cm

dx3 = 10.616 - 15 = -4.384 cm dy3 = 4.218 - 7 = -2.782 cm

dx4 = 10.616 - 17.33 = -6.714 cm dy4 = 4.218 - 10.67 = -6.452 cm

Substituting each dxi, dyi, Ix'i, and Iy'i into the basic summation equations give

Ix = 1,011 cm4

Iy = 2,217 cm4

In the theory page, the polar moment of inertia was show to be equal to

Jz = Ix + Iy

Substituting for Ix and Iy gives the polar moment,

Jz = 3,228 cm4

Since this is not a symmetrical cross section, the product of inertia, Ixy is not zero. The produce of inertia

is used for un-symmetrical bending which is not covered in this statics eBook.

STATICS - EXAMPLE

Area Between Curve and x and y-axis

Example 1

Find moment of inertia of the shaded area abouta) x axisb) y axis

Solution (a)

Recall, the moment of inertia is the second moment of the area about a given axis or line.

For part a) of this problem, the moment of inertia is about the x-axis. The differential element, dA, is usually broken into two parts, dx and dy (dA = dx dy), which makes integration easier. This also requires the integral be split into integration along the x direction (dx) and along the y direction (dy). The order of integration, dx or dy, is optional, but usually there is an easy way, and a more difficult way.

Cross-section Area

For this problem, the integration will be done first along the y direction, and then along the x direction. This order is easier since the curve function is given as y is equal to a function of x. The diagram at the left shows the dy going from 0 to the curve, or just y. Thus the limits of integration is 0 to y. The next integration along the x direction goes from 0 to 4. The final integration from is

Expanding the bracket by using the formula, (a-b)3 = a3 - 3 a2 b + 3 a b2 - b3

Solution (b)

Similar to the previous solution is part a), the moment of inertia is the second moment of the area about a given axis or line. But in this case, it is about the y-axis, or

Cross-section Area

The integral is still split into integration along the x direction (dx) and along the y direction (dy). Again, the integration will be done first along the y direction, and then along the x direction. The diagram at the left shows the dy going from 0 to the curve, or just y. Thus the limits of integration is 0 to y. The next integration along the x direction goes from 0 to 4. The final integration from is

Comment

The area is more closely distributed about the y-axis than x-axis. Thus, the moment of inertia of the shaded region is less about the y-axis as compared to x-axis.

Example 2

Determine the moment of inertia of y = 2 - 2x2 about the x axis. Calculate the moment of inertia in two different ways. First, (a) by taking a differential element, having a thickness dx and second, (b) by using a horizontal element with a thickness, dy.

Solution

a) The area of the differential element parallel to y axis is dA = ydx. The distance from x axis to the center of the element is named y.

y = y/2

Using the parallel axis theorem, the moment of inertia of this element about x axis is

For a rectangular shape, I is bh3/12. Substituting Ix, dA, and y gives,

Performing the integration, gives,

(b) First, the function should be rewritten in terms of y as

the independent variable. Due to the x2 term, there is a positive and negative form and it can be expressed as two similar functions mirrored about y axis. The function on the right side of the axis can be expressed as

The area of the differential element parallel to x axis is

Performing the integration gives,

Performing a numerical integration on calculator or by taking t = 2(2 - y) the above integration can be found as,

As expected, both methods (a) and (b) provide the same answer.

CHAPTER-8-A- Internal Forces and Moments


Introduction

A structural engineer is designing a tire swing for some new playground equipment. For the design to be successful, the material used for the swing beam must be capable of withstanding the internal loads generated by the swing and its occupant.

What is known:

The beam length L is 8 ft. The tire is connected at the mid-span of the

beam.

Playground Swing Structure The design load, F, for the swing is 300 lb.

Question

Force Diagram

What are the internal loads present in the beam at a location just to the left of the swing when the design load is applied? Assume the beam to be a pinned-roller configuration.

Approach

Solve for the external reactions. Cut the beam at the point of interest to reveal

the internal loads as equivalent external loads.

Form a free-body diagram with the appropriate sign convention for the two pieces.

Solve for the internal loads on either diagram.

STATICS - THEORY

3D Internal Forces and MomentsClick to view movie (118k)

3D Force ComponentsClick to view movie (148k)

Internal loads in a structural member are the result of externally applied loads. The external loads are transmitted to different parts of the structure through these internal loads.

The internal loads can be determined by the method of sections. A member is cut at the point of interest, and the internal loads are revealed as equivalent external loads.

Since the member was in equilibrium before being cut, these equivalent external loads must keep the member in equilibrium.

In general 3D, internal loads have three force components (one normal and two shear) and three moment components (one twisting and two bending).

For the common 2D case, there are two forces, an axial force and a shear force, plus one moment. These forces and moments are shown in the diagram below.

3D Moment ComponentsClick to view movie (123k)

Unknown Internal Loads, V, A, and M Replace Deleted 2D Structure Section to Maintain Equilibrium

Sign Convention

Standard Positive Sign Convention for Internal Shear, Moment and Axial Force

After the member is cut, a free-body diagram can be drawn for either section, but unknown equivalent external loads must be included to maintain equilibrium. A correct and consistent sign convention for the internal loads must be used when we draw the free-body diagram. The sign convention shown at the left describes what is meant by a positive force or moment. This convention is the most common sign convention for forces and moments.

Forces and moments acting in the direction shown are considered to be positive.

Solving for Internal Forces

Cut Location

Before solving for internal loads, the external reaction generally need to be determined. This is done with the normal equilibrium equations applied to the completed structure.

The first step in determining internal loads, is to make a cut at the desired location. Each location can have different internal loads.

After the cut is done, a free-body diagram is drawn for one of the two sections (generally a right and left section. Remember to replace the removed section with unknown the internal loads. The internal loads can now be determined by letting the sum of forces and sum of moments equal zero. If there are axial loads, then also sum forces along the direction of the member.

Internal loads can be calculated on either face of a

cut section. The values calculated for the two faces must be identical since they are actually the same location.

In the following example, the internal forces at the cut location are found by cutting the beam and solving for either the piece to the left of the cut or the piece to the right of the cut.

Left Side of Cut

Left Section

If the left section is used, then a free-body diagram should be drawn for just that section, as shown at the left. Notice that the internal forces are required to insure the left section stays in equilibrium without the right section. The direction of the moment and shear are according to the proper sign convention.

The axial internal force is not drawn since there are no axial external loads. The axial internal load is zero for all location in this example.

If the the forces in the vertical direction are summed, then the shear force V1 is

ΣFy = 0 Ay - V1 = 0 ==> V1 = Ay

Next, summing the moments about the cut, the bending moment, M1, can be determined,

ΣMcut = 0 M1 - a Ay = 0 ==> M1 = a Ay

Recall, the reaction force Ay is determined before the cut is made and is considered a known quantity.

Right Side of Cut

Right Section

If the section to the right side of the cut is used, the free-body diagram will be of only the right section of the beam as shown at the left. Again notice that the internal forces are drawn according to the sign convention.

Summing the forces in the vertical direction, gives

ΣFy = 0 By - F + V2 = 0 ==> V2 = F - By

Summing the moments about the cut gives

ΣMcut = 0 c By - b F - M2 = 0 M2 = c By - b F

Since both methods calculate the shear and moment at cut, the results must be the same.


Free Body Diagram

Form a free-body diagram and solve for the external reactions giving,

ΣMA = 0 L By - F L/2 = 0 By = F/2 = 300/2 = 150 lb

ΣFy = 0 Ay + By - F = 0 Ay = 300 - 150 = 150 lb

Solution of Internal Forces

Left Section

If the beam is cut just to the left of the applied load, then the proper free-body diagram for the section to the left of the cut is as shown.

Sum the forces and moments to solve for the unknown internal loads:

ΣFy = 0 Ay - V = 0 V = Ay = 150 lb

ΣMA = 0 M - V L/2 = 0 M = LV/2 = 8 (150) / 2 = 600 lb-ft

Right Section

The problem can also be solved by using the right section after making the cut. The free-body diagram for the right side section is as shown at the left. Summing the forces and moments give,

ΣFy = 0 V + By - F = 0 V = 300 - 150 = 150 lb

ΣMB = 0 -V L/2 + F L/2 - M = 0

M = -150 (8/2) + 300 (8/2) = 600 lb-ft

As expected, both methods obtain the same results, a shear force of 150 lb and a bending moment of 600 lb-ft.

CHAPTER-8-B- Shear and Moment Diagrams I


Introduction

Problem Graphic

A painter is working from a ladder propped against an overhang. The unusual position places dangerous loads on the ladder and could cause it to fail.

What is known:

The ladder length L is 20 ft. The ladder angle α is 70°. The overhang height h is 12 ft. The painter's weight W is 200 lb. The painter is 3 ft from the top of the ladder.

Questions

Problem DImensions

1. If the ladder were to fail due to the shear force, where would it fail?

2. If it were to fail due to the bending moment, where would it fail? Assume the ladder is pinned at its base.

Approach

Solve for the external reactions. Cut the ladder at an arbitrary point. Form a free-body diagram with the

appropriate sign convention. Solve for the internal loads as a function of

position along the length of the ladder. Repeat steps 2-4 for each section of the

ladder to get a complete shear force diagram and bending moment diagram.

STATICS - THEORY

Free Body Diagram before Making Section Cuts

From a design point of view, it is often necessary to know the maximum and minimum internal loads in a structure and where they are located.

If the shear force and bending moment are calculated and graphed, then the maximum and minimum of each are easily identified and located.

For the simply-supported beam with only one point force load, the reactions can be found as

Ay = F/3 By = 2F/3

The section of the beam to the left of the applied load will have an expression for the shear force and bending moment that will differ from the section to the right of the applied load. Therefore, the sections must be evaluated separately. In other words, there is not a single function that will model shear or moment from A to B.

Section 1 Cut and Analysis

If right side of the cut is used, then summing the forces give,

By + V1 - F = 0 V1 = F - By = F/3

Summing the moments about the left edge gives the bending moment as

-F(2/3L - x) + By(L - x) - M1 = 0 M1 = By(L - x) - F(2/3L - x) = xF/3

As expected, both methods give the same results.

Section 1

Cutting the beam at an arbitrary location x in section 1 and forming free-body diagrams for the piece to the left and to the right of the cut will result in the diagrams shown.

Either side of the cut beam can be used to solve for the internal shear force and internal bending moment. Generally, the simpler side is used. In this case, if the left side is used, internal shear load is

ΣFy = 0 Ay - V1 = 0 V1 = Ay = F/3

Summing the moments about the right edge give,

ΣMcut = 0 M1 - x (F/3) = 0 M1 = xF/3

For any location x between x = 0 and x = 2/3 L, the shear and moment are given by

V1 = F/3 0 ≤ x ≤ 2/3 L (1)

M1 = 2F/3 0 ≤ x ≤ 2/3 L (2)

Section 2 Cut and Analysis

If the piece to the left of the cut is used, then the vertical forces are

Ay - F - V2 = 0 V2 = F/3 - F = -2F/3

Summing the moments about the right edge gives the bending moment as

-(F/3) x + F(x - 2L/3) + M2 = 0 M2 = 2F(L - x)/3

Again, both methods produce the same results.

Section 2

This process can be repeated for for the section to the right of the load. The free-body diagrams for the two pieces are shown at the left. For this section, it is easier to solve for the piece to the right of the cut.

ΣFy = 0 2F/3 + V2 = 0 V2 = -2F/3

The moments about the left edge can be used to determine the bending moment.

ΣMcut = 0 (2F/3) (L - x) - M2 = 0 M2 = 2F(L - x)/3

Thus, for any location x between x = 2/3 L and x = L, the shear and moment are given by

V2 = -2F/3 2/3 L ≤ x ≤ L (3)

M2 = 2F(L - x)/3 2/3 L ≤ x ≤ L (4)

Shear and Moment Diagrams


If the shear equations (Eqs. 1 and 3) are graphed on one single axis and the moment equations (Eqs. 2 and 4) on another single axis, the shear and moment diagrams are obtained as shown on the left.

The location for maximum and minimum shear force and bending moment are easily found and evaluated.

Number of Sections

Number of Sections

The number of sections required for complete shear and moment diagrams depends on the beam configuration and loading. The beam must be sectioned at each point that loading or support conditions change. A cut must be made in each section of the beam so that the shear and moment can be evaluated over that section.

Multiple sections are required since shear and moment is not a continuous function over changing supports and loads.


Free Body Diagram DevelopmentClick to view movie (44k)

Ladder Free-Body DiagramRotated to Horizontal Position

A free-body diagram of the ladder reveals three unknown reaction forces, Ax, Ay and By. To simplify calculations, the x-y coordinate system is aligned with the ladder. The painter's weight is split into its x and y component, giving

Fx = F sinα = 200 sin70 = 187.9 lb

Fy = F cosα = 200 cos70 = 68.40 lb

The ladder can be considered as a beam, and oriented in the horizontal position. This will make it easier to plot the shear and moment as a function of x. The free-body diagram for the ladder in the horizontal position is shown at the left.

As with most static problem, the reaction forces, Ax, Ay, and By, should be determined first. To find By, sum the moments about the left end to solve for the overhang reaction.

ΣMA = -17 (68.40) + By 12.77 = 0 By = 91.06 lb

Summing the forces in the x and y directions gives

the ground reactions as,

ΣFx = Ax - 187.9 = 0 Ax = 187.9 lb

ΣFy = Ay + By - 68.40 = 0 = Ay + 91.06 - 68.40 = 0 Ay = -22.66 lb

Immediately, there is reason for concern since the reaction at A is negative, which indicates the ladder will lift off the ground. However, assume that the ladder is anchored to the ground and proceed with the problem.

Since the ladder has forces acting at three locations, the shear and moment needs to be analyzed in three different sections. The shear and moment may not be continuous across a load or support.

Section 1

Section 1

Each section needs to be cut and analyzed for shear and moment as a function of x from the left edge. The results can then be plotted and the maximum shear and moment can be easily identified.

First, cut the beam at the arbitrary point x and then a draw a free-body diagram the left side as shown. Either the left or the right side can be used but generally, the simpler section is used. Next, sum the forces to give,

ΣFx = 0 A1 = -187.9 lb

ΣFy = 0 V1 = -22.66 lb

Sum the moments about the cut edge,

ΣMcut = 0 M1 - (-22.66)x = 0 M1 = -22.66 x lb-ft

It should be noted that these three results are only good for section 1 (0 ≤ x ≤ 12.77 ft).

Section 2

Section 2

Section 2 is between the weight of the person and the corner of the roof overhand, point B. Again, cut the ladder at an arbitrary point in section 2 and draw a free-body diagram for the left side of the ladder, as shown. Sum the forces and moments to find repeat the process using the piece to the right of the cut:

ΣFx = 0 A2 = -187.9 lb

ΣFy = 0 -V2 - 22.66 + 91.06 = 0 V2 = 68.40 lb

ΣMcut = 0 M2 - (-22.66)x - 91.06 (x - 12.77) = 0 M2 = 68.4 x - 1,163 lb-ft

These three results are only good for section 1 (12.77 ≤ x ≤ 17 ft).

Section 3

Section 3

The final cut is at the far right section of the ladder. For this cut, it is easier to analyzes the right side of the cut. Since there are no loads on this part of the ladder, all forces and moments are zero.

A3 = 0

V3 = 0

M3 = 0

These three results are only good for section 1 (17≤ x ≤ 20 ft).


Now that the shear and moment is known for each section of the ladder, the results can be plotted. The resulting graphics are called the shear diagram and moment diagram.

Since the same x was used for all three sections, the each equation for each section can be easily plotted as shown at the left.

Failure in Shear

If the ladder were to fail in shear, one would expect it to fail somewhere in section 2 because the shear force, with a magnitude of 68.4 lb, is a maximum in

that section.

Failure in Bending

If the ladder were to fail in bending, the failure should occur at the point of contact with the wall, where the bending moment magnitude is a maximum of 290 lb-ft.

> Statics- Example

Beam Loading and Supports

Example

Draw the shear and moment diagram for the beam shown in figure.

Solution

Beam Free-body Diagram

Before the shear and moment can be determined at any internal location, the boundary conditions need to be calculated. There will be three possible reactions at A and B namely Ax, Ay and By. To calculate them, the distributed load is converted into an equivalent point load of (3 kN/m) (3 m) = 9 kN which is located 3 m from support A as shown in the diagram.

The unknown reactions can be found by applying the three standard equilibrium equations. Taking moments about support A and assuming counter clockwise (CCW) moment as positive gives,

ΣMA = 0 By (6) - (5) (3) - (9) (3) = 0 By = 7 kN

Equating vertical forces gives,

ΣFy = 0 Ay + By - 5 - 9 = 0 Ay = 7 kN

Since there is no horizontal applied load, Ax will be zero. (Reaction forces can also be calculated by dividing sum of forces acting on the beam by 2, since the beam and loading are symmetrical.)

Beam Sections

Since the beam has forces acting at four locations, the shear and moment needs to be analyzed in four different sections. The shear and moment may not be continuous across a load or support.

Each section needs to be cut and analyzed for shear and moment as a function of x from the left edge. The results can then be plotted and the maximum shear and moment can be easily identified.

Section 1

Section 1

To find the shear and moment in section 1, cut the beam at an arbitrary point x in section 1 and then draw a free-body diagram as shown on the left. Next, sum the forces to give,

ΣFy = 0 7 - V1 = 0 V1 = 7 kN

Summing the moments about the cut edge gives,

ΣMcut = 0 M1 - 7x = 0 M1 = 7x kN-m

It should be noted that these results are only good for section 1 (0 ≤ x ≤ 1.5 m).

Section 2

Section 2

Section 2 is between the start of distributed load and the point load. Again, cut the beam at an arbitrary point x in section 2 and then draw a free-body diagram as shown on the left. Sum the forces to give,

ΣFy = 0 7 - 3 (x - 1.5) - V2 = 0 V2 = 11.5 - 3x kN


ΣMcut = 0 M2 + 3(x - 1.5)(x - 1.5) / 2 - 7x = 0 M2 = 7x - 1.5(x - 1.5)2 kN-m

These results are only good for section 2 (1.5 ≤ x ≤

3m).

To assist in graphing the shear and moment equations, the exact values can be calculated at each end of the beam section. When x = 1.5, the shear force will be, V2 = 11.5 - (3)1.5 = 7 kNand the moment will be, M3 = (7)1.5 - 1.5(1.5 - 1.5)2 = 10.5 kN-m

When x = 3, the shear force will be, V2 = 11.5 - (3)3 = 2.5 kNand the moment will be, M3 = (7)3 - 1.5(3 - 1.5)2 = 17.625 kN-m

Section 3

Section 3

Section 3 is between point load of 5 kN and end of universal distributed load. The free-body diagram for the beam will be as shown on the left. Sum of the forces gives,

ΣFy = 0 7 - 5 - 3(x - 1.5) - V3 = 0 V3 = 6.5 - (3)x kN


ΣMcut = 0 M3 - (7)x + 5(x - 3) + 3(x - 1.5)(x - 1.5) / 2 = 0 M3 = 2x + 15 - 1.5(x - 1.5)2 kN-m

These results are only good for section 3 (3 ≤ x ≤ 4.5m).

When x = 3, the shear force will be, V3 = 6.5 - (3)3 = -2.5 kNthe moment will be, M3 = (2)3 + 15 - 1.5(3 - 1.5)2 = 17.625 kN-m

When x = 4.5, the shear force will be, V3 = 6.5 - (3)4.5 = -7 kNthe moment will be, M3 = (2)4.5 + 15 - 1.5(4.5 - 1.5)2 = 10.5 kN-m

Section 4

Section 4

The final cut is at the far right section of the beam. The free-body diagram for the beam will be as shown on the left. Sum of the forces gives,

ΣFy = 0 7 - 5 - 3(3) - V4 = 0 V4 = -7 kN


ΣMcut = 0 M4 - (7)x + 5(x - 3) + 3(3)(x - 3) = 0 M4 = (7)x - 14(x-3) kN-m

These results are only good for section 3 (4.5 ≤ x < 6m).

When x = 4.5, the shear force will be, V4 = -7 kNthe moment will be, M4 = (7)4.5 - 14(4.5 - 3 ) = 10.5 kN-m

When x = 6, the shear force will be, V4 = -7 kNthe moment will be, M4 = (7)6 - 14(6 - 3) = 0 kN-m


Now that the shear and moment is known for each section of the beam, the results can be plotted. The resulting graphics are called the shear diagram and moment diagram.

Since the same x was used for all three sections, the each equation for each section can be easily plotted as shown at the left.

Since beam and loading are symmetrical, shear force diagram and bending moment diagram can also be drawn by solving first 2 sections only. For shear moment diagram other half will be anti-mirror image of first 2 sections and for bending moment diagram other half will be mirror image of first 2 sections which can be seen in the figure shown on the left.

CHAPTER-8-C-Shear and Moment Diagrams II


Introduction


Engineers need to test the wing of a new aircraft before actual flight testing. To simulate the lift force during flight, a hydraulic jack applies a force of 4,500 lb at on the wing.

What is known:

The wing weight is 165 lb/ft. The wing length is 54 ft. The hydraulic point load is 4,500 lb, and is

located 49 ft from the wing root.

Force Diagram

Question

What is the maximum shear and bending moment for this loading condition?

Approach

Replace the distributed wing weight load with an equivalent point load.

Determine the reactions for the equivalent system.

Develop the shear and moment diagrams with the distributed load in place.

STATICS - THEORY

Distributed Load Diagram

Common Distributed Loads

To find reactions, a distributed force can be replaced with an equivalent point load.

For a general distributed load f(x) as shown, the equivalent point load Fr is found by integrating the distributed load over its length.

For the point load, FT, to be equivalent, the moment generated about any point must also be equivalent. Thus,

This gives the distance xr to the equivalent point load as,

Note that xr is the distance to the centroid of the distributed load.


Wing Modeled as a Cantilever Beam - Force Diagram

The wing can be modeled as a simple cantilevered beam with a distributed load of 165 lb/ft and point force load of 4,500 lb.

The first step in solving this problem is to determine the reaction force and moment at the fixed end, A. This can be done more efficiently by replacing thedistributed load with a point load of magnitude Wr at location xr.

Wing Model as Cantilever BeamClick to view movie (31k)

= (165 lb/ft)(54 ft) = 8,910 lb

The location of the point load will be

= (54 ft)/2 = 27 ft

Wing Free Body Diagram

With a free-body diagram (shown at left) the reaction force and moment can be determined.

ΣFy = 0 Ay - 8,910 + 4,500 = 0 Ay = 4,410 lb

ΣMA = 0 Mr - 8,910 (27) + 4,500 (49) = 0 Mr = 20,070 lb-ft

There will be two sets of shear and moment equations, one to the left of F and one to the right.

Section 1 (0 ≤ x ≤ 49)

Section 1 FBD

After making the section cut, and developing a free-body diagram for the left side of the wing, sum the forces to get the shear equation.

ΣFy = 0 4,410 - 165x - V1 = 0 V1 = 4,410 - 165x

Summing the moments give the moment equation,

ΣMcut = 0 M1 + (165x) (x/2) + 20,070 - 4,410x = 0 M1 = -82.5x2 + 4,410x - 20,070

Section 2 (49≤ x ≤ 54)

Section 2 FBD

Once again, first make a cut in section 2 (left of point B). Next, construct a free-body diagram and then sum forces to get the shear equation.

ΣFy = 0 4,410 - 165x + 4,500 - V2 = 0 V2 = 8,910 - 165x

Summing the moments give the moment equation,

ΣMcut = 0 M2 + (165x) (x/2) + 20,070 - 4,410x - 4,500(x - 49) = 0 M2 = -82.5x2 + 8,910x - 240,570


Maximum Internal Shear

The shear equations for both sections can now be plotted to help identify the maximum and their locations. Both shear equations are linear lines, and shown at the left.

The maximum shear occurs at the far left where the wing is fixed to the aircraft body. The magnitude is 4,410 lb.

Maximum Internal Moment

Like the shear, the two moment equations can be plotted to help identify the maximum internal moment. However, since the equations are quadratic curves, they are more difficult to plot than the linear lines for the shear.

There are a few techniques that can assist in plotting curves. If the second derivative is positive, then the curve is facing upward. For both equations, their second derivatives are negative, so the they face downward.

The maximum can be determined by equating the first derivative to zero. For the first section equation, that becomes,

d(M1)/dx = -165x + 4,410 = 0

x = 26.73 ft

The maximum value at 26.73 ft is

M26.73 = -82.5(26.73)2 + 4,410(26.73) - 20,070

Mmax = 38,860 lb-ft

CHAPTER-8-D- Shear, Moment and Load Relation


Introduction

Problem Graphic

The wing of a P-38 is designed to withstand a 5g maneuver without damage. In order to design the structure, engineers needed to know the internal loads generated by the aerodynamic lift, and the mass and torque of the engine.

What is known:

The total airplane mass mt is 9,050 kg. The engine mass me is 735 kg. The engine torque Te is 3,000 Nm. The distance d from the centerline to the

engine is 2.4 m. The span b of one wing is 7.92 m. The dynamic load factor G is 5. The acceleration due to gravity is 9.8 m/s2. The lift distribution is given by:

L(x) = (mt gG/b)(x/b+1) N/m

Force Diagram

Question

What are the internal loads in the wing structure of the airplane during a 5g maneuver? Due to symmetry, only one wing needs to be calculated.

Approach

Develop shear and moment diagrams for the aerodynamic lift, engine mass, and engine torque separately using the relationships between load, shear, and moment.

Using the principle of superposition, combine the three diagrams to get the shear and moment diagrams for the complete structure.

STATICS - THEORY

In the previous two sections, methods where

presented to construct shear and moment diagrams. However, there are some important relationships between loading, shear and moment which can aid in the construction of the diagrams.

Beam Element with Distributed Load

Example of Loading Function "q" equalingthe Slope of the Shear Diagram

Example of Change in the Shear Force

Relationship between Shear Force and Distributed Loads

For a distributed load, q, the forces can be summed in the y direction to give,

ΣFy = V - (V + ΔV) - q Δx = 0

This can be rearrange to give,

ΔV/Δx = -q

If the limit of both sides is taken as Δx goes to zero, then the relationship between shear force and distributed loads becomes

dV/dx = -q

This can also be written in integral form as

Graphical examples of both these equations are shown at the left.

Beam Element with Distributed Load

Relationship between Bending Moment and Shear Force

Similar to the shear force in the previous paragraphs, the moments can be summed for a beam element with a distributed load of "q", giving,

ΣM = -M + (M+ ΔM) - (V + ΔV)Δx - q Δx (Δx/2) = 0

This equation can be rearranged to give,

Example of Shear Function "V" equalingthe Slope of the Moment Diagram

Example of Change in the Bending Moment

ΔM/Δx = V + ΔV + q (Δx/2)

Using the previous relationship for ΔV gives

ΔM/Δx = V - q Δx + q (Δx/2) = V - q (Δx/2)

Now, take the limit of both sides as Δx goes to zero. Last term vanishes and a direct relationship be between bending moment and shear force is identified as,

dM/dx = V

This can also be written as an integral relationship,

Two examples using the above equations are shown at the left.

Beam Element with Point Load

Effect of a Point Load

For a beam element subjected to a point load, the forces can be summed in the vertical to give,

ΣFy = V - (V + ΔV) - F = 0

ΔV = -F

This represents that point loading will cause a jump in the shear diagram equal to the point load. The moment will not jump with a point load.

Effect of a Point Load

Beam Element with Point Moment

Effect of a Point Moment

Effect of a Point Moment

For a beam element subjected to a point moment as shown, the shear force at that point is unaffected; but if the moments are summed about the left edge, then

ΣM = ΔM - (V + ΔV)Δx - T = 0

If the limit as Δx goes to zero is taken, then it becomes

ΔM = -T

Principle of Superposition

For beams with several different loads, it is often easier to solve for the shear and moment diagrams for the individual loads and then combine them to find the total shear and moment diagrams. This is known as the method, or principle, of superposition.

For the cantilevered beam shown, shear and moment diagrams can be constructed for the two point loads and for the distributed load, separately; they can then

Principle of Superposition

be combined to form the full shear and moment diagrams.


Loading on P38 Wing

The total solution can be found by determining the shear force distribution and bending moment distribution of the three separate types of loading (distributed lift, engine mass, and engine torque) and then combining them through the principle of superposition.

It is important to use the same x-coordinate system for each of the diagrams so that they can be superimposed. The lift distribution is given as an equation where x is positive from the the center of the aircraft, as shown in the diagram at the left.

1. Distributed Lift

The more difficult loading is the linear distribution loading due to the lift on the wing. The shear force distribution on the wing can be calculated by integrating the loading function, the lift distribution, from the left free end to any point, x, on the wing,

Distributed Lift LoadClick to view movie (30k)

The integration constant, C1, can be determined by using the boundary condition that the shear force on the wing tip (x = -b) must be zero. Thus,

Similarly, the bending moment can be found by integrating the shear force distribution, as

The bending moment force on the wing tip must be zero which is used to determine the integration constant. The moment function becomes,

2. Engine Mass

Engine Mass as Point LoadClick to view movie (43k)

Next, consider the shear and moment diagrams for the engine point force load. The shear force is zero up to the engine location, so

V2A(x) = 0 -b ≤ x ≤ -d

At the engine location, there is a jump in shear force due to the weight of the engine, giving

V2B(x) = -Fe = -me g G -d ≤ x ≤ 0

The bending moment distribution is found by integrating the shear force distribution in both sections, giving:

M2A(x) = 0 -b ≤ x ≤ -d

M2A(x) = -me g G x + C3

The integration constant can be determined by noting that the moment at x = -d must be zero (engine locations). This gives the final moment equation as

M2A(x) = -me g G (x + d) -b ≤ x ≤ -d

3. Engine Torque

Engine Torque LoadClick to view movie (29k)

The engine torque has no effect on the shear force so the shear equation is simply zero.

V3(x) = 0

The bending moment is zero up to the engine location:

M3A(x) = 0

At the engine location, there is a jump in bending moment due to the engine torque, giving

M3B(x) = -T

Total Solution

Total Solution from all Three LoadedClick to view movie (38k)

The complete shear and moment diagrams can be constructed by adding the individual shear and moment equations.

From x = -b to x = -d, this gives

From x = -d to x = 0, the equations are

The critical points for the shear force and bending moment can now easily be located and evaluated.

Expanded Diagrams for Loads

CHAPTER-9-A- Friction ISTATICS - CASE STUDY

Introduction


Crazy Al has just acquired a 1967 Ford Mustang and wants to do a thorough checkup on it before he gives it a price. Since the hydraulic jack is broken, he has decided that in order to get a good look at the underside of the car, he must drive it up a jury-rigged ramp that rests against a large wooden crate.

What is known:

The tires create a 1,000 N force perpendicular to the ramp. Assume that the box cannot move.

The length of the ramp is 3 m. The distance from the start of the ramp, point A, to the front wheels is 2 m.

The ramp is elevated at a 25° angle. The coefficient of static friction at point

B is 0.2.

Question

Known Information

What is the minimum value for the coefficient of static friction at point A that will prevent the ramp from moving?

Approach

Find the normal force at point B. Use the normal force to find the friction

force at B. Use the equilibrium equations to solve for

the reaction forces at point A, and then find the coefficient of friction.

STATICS - THEORY

Friction on a Block in EquilibriumClick to view movie (48k)

In previous sections, it was assumed that objects were frictionless. Consequently, the only reaction forces were normal to the surface. This ideal approximation always involves relative errors because no surface is perfectly frictionless.

There are several types of frictional forces used in mechanics. However, dry friction will be the primary form considered in this section.

Dry Friction

Mechanics of FrictionClick to view movie (120k)

Dry friction, also known as Coulomb friction, occurs between two rough, solid surfaces.

Frictional forces always oppose the tendency towards motion and act tangential to the surfaces in contact.

Static friction describes surfaces that are in contact but not moving. The maximum static friction is represented by

fmax = μs N

Here, μs is a proportionality constant called the coefficient of static friction. This equation only gives the force of friction when motion is impending. If motion is not impending, then the friction force, fs, will be less than the maximum, or

fs ≤ μs N

Friction vs ForceClick to view movie (62k)

Thus, only when motion is just starting to take place will the friction force be a maximum.

On the other hand, if motion has started, then kinetic friction is used, or

fk = μk N

Here, μk is the coefficient of kinetic friction. This equation only applies when an object is in motion. It is generally true that μk is less than μs and may vary slightly with velocity. In general, friction simply written as

f = μ N

It is also helpful to consider the resultant of the frictional and normal forces. The angle α from the normal force to the resultant can be calculated as

tanα = f/N = μ

As the static friction approaches fmax, α reaches a maximum angle φs.

tanφs = μs

Likewise, a similar equation for kinetic friction can be written as

tanφk = fk/N = μk

The angles φs and φk define the limiting positions of the reactions between surfaces. There are typically three types of friction problems.

Type 1: Impending Motion

Impending motion is known to exist because the system is on the verge of slipping. The equilibrium equations and basic equation Fmax = μs N hold true.

Type 2: Status Unknown

Neither impending motion nor motion is known. Using equilibrium equations, the friction force direction on a surface is assumed and solved. Comparing the results give

a. f < (fmax = μs N) System is still in equilibrium.

b. f = (fmax = μs N) Motion is impending.c. f > (fmax = μs N) This condition would violate

equilibrium and therefore is impossible. Instead, it indicates that the system is in motion and maximum friction is μkN.

Type 3: Relative Motion

Relative motion exists between the surfaces. Therefore, the properties of kinetic friction would be in effect.


Development of the Free Body DiagramClick to view movie (107k)

Free Body Diagram

In this problem, the ramp AB may slide if the friction force at A is too low. To find the coefficient of friction that will prevent the ramp from sliding, the conditions under which the ramp will be in motion needs to be determined.

First, draw a free-body diagram of the ramp with the friction acting to the left since the ramp will want to slide to the right.

Using equilibrium equations, the moment about point A can be found as,

ΣMA = NB (3 m) - FC (2 m) = 0

NB = (100 N) (2 m) / (3 m) = 666.7 N

Since motion is impending at each point and the normal force at point B is known, the friction at point B is

fB = μB NB = (0.2)(666.7 N) = 133.3 N

The equilibrium equations can now be used to find the normal and frictional forces at point A.

ΣFx = 0

fA = FC sinα - NB sinα - fB cosα

Impending MotionClick to view movie (137k)

= (1000 - 666.7) sin25 - 133.3 cos25 = 20.0 N

ΣFy = 0

NA = FC cosα - NB cosα - fB sinα = (1000 - 666.7) cos25 - 133.3 sin25 = 358.4 N

Now that the normal and frictional forces are known, the coefficient of friction can be determined as

μA = fA/NA = 0.056

This is the lowest value of μA needed to prevent motion. For lower values of μA, the ramp is in motion, and for higher values, the ramp is in equilibrium.

STATICS - EXAMPLE

Example Graphic

Example

It is observed that when a flat bed of the dump truck is raised to an angle of 22.5o, a box that was loaded on the bed begins to slide. Determine the static coefficient of friction between box and the surface of bed of dump truck. Also check if the box will tip before it slides. The box is 6 feet high and 4 feet wide and its center of gravity is at its center.

Simplified System Diagram

Solution

The problem can be simplified as a box slide down a hill as shown in the figure on the left. The center of gravity is 3 feet from the bottom.

There are three possible forces acting on the box, its weight, reaction force due to the ground and the friction force. The location and direction of these forces are shown on the free-body diagram of the box on the left. Let the resultant normal force, N be acting at point A. There are two unknowns, N and Fs.

Note that the co-ordinate system is rotated by 22.5o in clockwise direction to simplify the calculations. For equilibrium,

ΣFy = 0 N - w cos 22.5 = 0

N = w cos 22.5

ΣFx = 0 w sin 22.5 - Fs = 0

Fs = μ N = μ w cos 22.5

Putting back Fs in ΣFx equation gives,

w sin 22.5 = μ w cos 22.5 μ = tan 22.5 = 0.4142

Notice from the calculation that static coefficient of friction, μ, is independent of the weight of a box. Also, knowing the inclination angle of the bed is a convenient way of measuring the coefficient of static friction.

Free-body Diagram of the System

To check whether box will tip or slide first, let the resultant normal force, N be acting at point B which is at distance x as shown in the figure on the left.

Taking moments about point 'B' and assuming counter clockwise (CCW) moment as positive gives,

ΣMB = 0 w (cos 22.5) x - w (sin 22.5) 3 = 0 w (cos 22.5) x = w (sin 22.5) 3 x = (tan 22.5) 3 x = 1.243

'x' is less than 2 ft which means the box will slide before it can tip. Also notice from the calculation that distance 'x' is independent of the weight of a box but it does depend on the inclination angle and the distance between the center of gravity and the base.

CHAPTER-9-B-Friction IISTATICS - CASE STUDY

Introduction


Wedge AnimationClick to view movie (140k)

King Ignoramus Maximus has established a lowest bid construction policy for the building of new temples. As a result, a column erected in the temple of Apollo was several inches too short. In order to raise the column, engineers will use two small marble wedges.

What is known:

The weight of the column is 1000 lb. The weight of each wedge is 100 lb. The coefficient of friction is 0.3. The angle of the wedge is 15°. The column and top wedge are constrained

against movement to the left.

Questions

Force Diagram

1. What is the smallest horizontally applied force that will start the column moving upward?

2. If the applied force and lateral constraints are removed, will the system remain in equilibrium?

Approach

Determine the forces acting on each wedge.

Use any special properties of friction that could be helpful to simplify the problem.

STATICS - THEORY

Forces Acting on WedgeClick to view movie (810k)

Semi-Graphical ApproachClick to view movie (368k)

A wedge is a block with two flat faces that make a small angle with each other.

Depending on the angle and the coefficients of friction acting on the wedge, the weight lifted can be many times the applied force P.

Wedge problems can be often solved by using a semi-graphical approach that involves drawing the known vectors and using the laws of sines and cosines to find the unknown forces.

Wedges are usually constrained against rotation. Therefore, only force equilibrium needs to be considered.

Transferring forces by using wedges can often be considered in terms of mechanical advantage (M.A.) as

M.A. = direct force / wedge force

In this formula, the numerator is the amount of direct force needed to accomplish the task normally, and the denominator is the force applied to the wedge to do the same task.

A well-designed wedge would have a mechanical advantage greater than one. However, wedges are often designed to remain in place after an applied load is removed. This type of wedge is called self-locking. Self-locking wedges often have a mechanical advantage less than one but are still viable designs.


Solution of a)

Solution a)Click to view movie (159k)

The solution requires finding the smallest magnitude of P required to move the column. First, consider the forces acting on wedge A. Force equilibrium provides two equations,

ΣFx = 0 N1 - f2 cos15 - N2 sin15 = 0

ΣFy = 0 N2 cos15 - f1 - Wcol - WA - f2 sin15 = 0 N2 cos15 - f1 - 1,000 - 100 - f2 sin15 = 0

Since motion is impending, the friction forces are

Free Body Diagram for Wedge A

f1 = μ N1 = 0.3 N1 f2 = μ N2 = 0.3 N2

If these two friction equations are used, the four equations can be reduced to two equations and two unknowns, N1 and N2.

N1 - 0.3 N2 cos15 - N2 sin15 = 0 N2 cos15 - 0.3 N1 - 0.3 N2 sin15 = 1,100

N1 = 0.5486 N2

0.8883 N2 - 0.3 N1 = 1,100

Solving gives

N2 = 1,520 lb and N1 = 833.3 lb

and the friction forces are

f2 = 456.0 lb and f1 = 250.0 lb

Free Body Diagram for Wedge B

Now that the forces on the bottom surface of Wedge A are known, wedge B can be analyzed. First, sum the forces in the vertical direction, to give,

ΣFy = 0 f2 sin15 + N3 - WB - N2 cos15 = 0 (456) sin15 + N3 - 100 - (1,520) cos15 = 0

N3 = 1,450 lb and f3 = 435 lb

Finally, P can be determined by summing forces on wedge B in the horizontal,

ΣFx = 0 f3 + f2 cos15 + N2 sin15 - P = 0 (435) + (456) cos15 + (1,520) sin15 - P = 0

Solving for P gives,

P = 1,269 lb

Solution of b)

Solution b)Click to view movie (453k)

Wedge A and B Interaction

The applied force and lateral constraints have been removed. As a result, the direction of potential motion has changed, and the frictional forces will reverse direction.

The equations of equilibrium on wedge A will produce

ΣFx = 0 f2 cos15 - N2 sin15 = 0 f2 = 0.2679 N2

ΣFy = 0 N2 cos15 + f2 sin15 - Wcol - WA = 0 N2 cos15 + (0.2679 N2) sin15 - 1,000 - 100 = 0

Since the status of the wedge is being analyzed, it cannot be assumed that f2 = μN2. The resulting forces are

N2 = 1,063 lb and f2 = 285 lb

The friction due to impending motion is determined as

f2 max = μ N2 = 318.9 lb

Since this value is greater than the friction due to equilibrium, wedge A must be in equilibrium.

The minimum coefficient of friction value required to maintain equilibrium is 0.268.

Since the top wedge is not moving, it is impossible for the bottom wedge to move. This can be verified with a free-body diagram of wedge B.

Also, if the column force were to increase due to the weight from the ceiling, the wedge would still not move. Feel free to experiment with this and other factors in the simulation.

STATICS- Example

Example

The vertical position of a machine (block A) is adjusted by moving wedge B. The coefficient of static friction between all surfaces is 0.3. Determine the horizontal force, P, acting on wedge B, that is required to

a) raise the block A (acting on the right side), andb) lower the block A (acting on the left side).

Solution (a) - Load P to Raise Block A

Raising the Machine

The first part of this problem asks for the the smallest value of the force, P, to raise the machine. This force will act on the right side of the wide of block A, as shown, in order to push block A upward. As with all static problems, a free-body diagram will help identify all forces acting on an object.

Since there is a known force of 1,500 lb acting on block A, this object will be analyzed first. All forces acting on the block 'A' are shown in the free-body diagram on the left. The frictional forces are,

Free-body Diagram of Block A

f1 = μ N1 = 0.3 N1

f2 = μ N2 = 0.3 N2

Applying the equilibrium equations gives,

ΣFx = 0 N1 - f2 cos 12 - N2 sin 12 = 0 N1 - (0.3) (0.9781) N2 - 0.2079 N2 = 0 N1 = 0.5013 N2

ΣFy = 0 N2 cos 12 - f1 - 1,500 - f 2 sin 12 = 0 0.9781 N2 - 0.3 N1 - 1,500 - (0.3)(0.2079) N2 = 0 0.9158 N2 - 0.3 N1 = 1,500

Solving above two equations gives,

N2 = 1,959.8 lb N1 = 982.4 lb

and the frictional forces are,

f2 = 587.9 lb f1 = 294.7 lb

Free-body Diagram of Block B

Now that the forces on the bottom surface of block A are known, wedge B can be analyzed. First, sum the forces in the vertical direction, to give,

ΣFy = 0 N3 + f2 sin 12 - N2 cos 12 = 0 N3 + (587.9) (0.2079) - (1,959.8) (0.9781) = 0 N3 = 1,794.7 lb and f3 = 538.5 lb


ΣFx = 0 N2 sin 12 + f3 + f2 cos 12 - P = 0 (1,959.8) (0.2079) + 538.5 + (587.9) (0.9781) = P

P = 1,521 lb

Lowering the Machine

Solution (b) - Load P to Lower Block A

The second part of this problem asks for the the smallest value of the force, P, to lower the machine. This force will act on the left side of the wide of block A, as shown, in order to lower block A downward.

Free-body Diagram of Block A

Since force P is acting in the opposite direction than in the previous solution, the forces acting on block A will be in opposite directions. The new forces acting on the block A are shown in the free-body diagram on the left. Frictional forces are given by,

f1 = μ N1 = 0.3 N1

f2 = μ N2 = 0.3 N2

Applying the equilibrium equations gives,

ΣFx = 0 f2 cos 12 - N2 sin 12 - N1= 0 (0.3) (0.9781) N2 - 0.2079 N2 - N1 = 0 N1 = 0.0855 N2

ΣFy = 0 N2 cos 12 - f1 - 1500 + f2 sin 12 = 0 (0.9781) N2 - 0.3N 1 - 1500 - (0.3)(0.2079) N2 = 0 0.9157 N2 - 0.3N1 = 1500

Solving above two equations gives,

N2 = 1,685.3 lb N1 = 144.1 lb

and the frictional forces are,

f2 = 505.6 lb f1 = 43.23 lb

Free-body Diagram of Block B

Now that the forces on the bottom surface of Wedge A are known, wedge B can be analyzed. First, sum the forces in the vertical direction, to give,

ΣFy = 0 N3 - f2 sin 12 - N2 cos 12 = 0 N3 - (505.6) (0.2079) - (1,685.3) (0.9781) = 0

N3 = 1,753.5 lb and f3 = 526.1 lb


ΣFx = 0

N2 sin 12 - f3 - f2 cos 12 + P = 0 (1,685.3) (0.2079) - 526.1 - (505.6) (0.9781) - P = 0

P = - 670.2 lb = 670.2 lb ←

Comments

To keep the machine in equilibrium one should apply a force of magnitude greater then 670.2 lb in ← direction but less than 1521 lb in the same direction. If the applied force is more than 1521 lb, the machine will be raised and if applied force is less than 670.2 lb then the machine will be lowered due to it's own weight.

CHAPTER-10-A- Virtual WorkSTATICS - CASE STUDY

Introduction


A car jack is constructed with four pinned members and a threaded shaft. The shaft must be designed to withstand the tension that results when hoisting a heavy automobile.

What is known:

The members of the jack are pinned together at points A, B, C, and D.

The length of each member is 0.2 m. The weight of the automobile exerts a force

of 1,960 N on the jack. The threaded shaft is pinned to the jack at

points C and D. Under normal conditions, the jack is

designed to support weight when the bottom members are at an angle between θ = 30° and θ = 60°.

Questions

Force Diagram

At what angle θ is the tension in the shaft greatest under normal conditions (30o ≤ θ ≤ 60o)? At what angleθ is the tension in the shaft least?

Approach

Determine the virtual work performed by the weight of the automobile and the tension in the shaft.

Knowing that the total virtual work must be equal to zero, derive an equation for the tension T as a function of the angle θ.

Use graphic analysis to determine the maximum and minimum values of T under normal operating conditions.

STATICS - THEORY

Work of a Force

Work of a ForceClick to view movie (32k)

A force does work when it undergoes a displacement in the direction of the force. If a force moves a small distance dr, the work done by the force is given by

dU = F • dr

If the angle between the force and the vector dr is θ, then the scalar equation is

dU = F ds cosθ

Here ds is the magnitude of the vector dr.

Work of a Couple

Work of a CoupleClick to view movie (24k)

A couple, or moment, does work when it undergoes a rotation in the plane of the couple. If the couple rotates a small angle dθ, then the work done by the couple is given by

dU = M = dθ

If the rotation takes place in the plane of the couple, then the scalar equation is

dU = M dθ

Virtual Work

Virtual Work of a ForceClick to view movie (47k)

The virtual work of a force or couple is defined as the work done by an imaginary or virtual displacement. In scalar notation, the virtual work of a force is given by

δU = F δs cosθ

In x and y components,

δU = Fx δx + Fy δy

Here Fx and Fy are the forces in the x and y directions, and δx and δy are the virtual displacements in the x and y directions. The virtual work of a couple, or moment, is

δU = M δθ

Principle of Virtual Work

Virtual Work of a CoupleClick to view movie (18k)

The principle of virtual work states that a system of rigid bodies is in equilibrium if the total virtual work done by all external forces and couples acting on the system is zero for each independent virtual displacement of the system. Mathematically, this means that equilibrium for a single-degree-of-freedom system is

δU = 0

From this equation, the equilibrium position or equilibrium forces can be determined.


Free-body Diagram Before Displacement

Free-body Diagram After Displacement

To solve for the tension in the shaft, the principle of virtual work can be used. Begin with a free-body diagram of the jack as shown at the left.

Hold point A fixed and let point B undergo a virtual displacement δy in the positive y direction. The deflection is assumed to be small even though it is large in the diagram.

Because point A is fixed, the reaction forces at A does no work.

Rod Tension and Virtual Displacement

Point B Virtual DisplacementClick to view movie (159k)

Right Side of Jack After Displacement

The tension in the shaft can be broken down into two equal and opposite components, T1 and T2. Because these components remain horizontal, they do no work in the vertical direction.

The total virtual work performed during the virtual displacement is the sum of the virtual work performed by the two components of the tension, and the virtual work performed by the weight of the automobile,

δU = δUT1 + δUT2 + δUF = T1 δx + (-T1)(-δx) + (-F) δy

The total virtual work performed during the virtual displacement is the sum of the virtual work performed by the two components of the tension, and the virtual work performed by the weight of the automobile,

δU = 2T δx - F δy

Notice that the equation for the virtual work is expressed in terms of the virtual displacements. To solve, express the virtual displacements in terms of a common variable. The vertical displacement of point B can be expressed in terms of the angle θ,

y = 2L sinθ δy = 2L cosθ

The horizontal displacement of points C and D can

also be expressed in terms of the angle θ:

x = 2L cosθ δx = -L sinθ

Substitute the expressions for δx and δy into δU:

δU = 2T(-L sinθ) - F 2L cosθ = 0

T = -F/tanθ

The tension, T, vs. the angle, θ, is graphed in the diagram at the left. Observe that the tension increases as θ approaches 0. Thus, under normal operating conditions (30o≤ θ ≤ 60o), the angles at which the tension is the greatest and the least are

θT max = 30o

θT min = 60o

STATICS- Example

Example

Slider-crank Mechanism

A common mechanism that converts rotary motion into reciprocating motion or vice-versa is sometimes referred as a "Slider-crank mechanism", which is shown at the left. It has wide range of applications like pumps, engines, and compressors.

Generally, the mechanism consists of two rigid links and a piston that are connected by frictionless joints and constrained to move in a single plane. The link AC is called a "crank-shaft" and the link CB is called a "connecting rod". The slider-crank mechanism has only one degree of freedom since the location of both links can be specified by the single independent coordinate θ

In this case, the crank-shaft, AC, weighs 20 N and connecting rod, CB, weighs 35 N. The weight of the links pushes the piston towards right. To keep this system in equilibrium, a force, F, of 70 N pushes in the opposite direction. Determine angle θ for the equilibrium of the slider-crank mechanism.

Solution

To solve for the equilibrium angle θ, the principle of virtual work can be used.

Free-body Diagram of the Slider-crank Mechanism

Begin with a free-body diagram of the slider-crank mechanism as shown at the left. If the origin is established at the fixed support A, the location of F and center of gravity for each link can be specified by the position coordinates xB, yW1, and yW2, respectively.

Let point B undergoes a virtual displacement δxB in the negative x direction. The deflection is assumed to be small even though it is shown large in the diagram.

Because point A is fixed, the reaction forces at A perform no work and By does not move in the direction of the force so no work is performed by By.

Expressing the position coordinates of point B and

center of gravity of links in terms of the independent coordinate θ and taking the derivatives to find virtual displacements yields,

xB = 0.5 cosθ + 0.5 cosθ = 1 cosθ

δxB = -sinθ δθ

yW1 = 0.25 sinθ δyW1 = 0.25 cosθ δθ

yW2 = 0.25 sinθ δyW2 = 0.25 cosθ δθ

As shown in the free-body diagram, an increase in θ (i.e. δθ) causes a decrease in xB and an increase in yW1 and yW2

The total virtual work performed during the virtual displacement is the sum of the virtual work performed by the force F, and the virtual work performed by the weight of the connecting rod and crank-shaft,

δU = 0 -20 δyW1 - 35 δyW2 - F δxB = 0

Putting results obtained earlier in the above equation gives,

20 (0.25) cosθ δθ + 35 (0.25) cosθ δθ + 70 (-sinθ δθ = 0

(13.75 cosθ - 70 sinθ) δθ = 0

Noting that δθ can not be zero,

θ = tan-1(13.75 / 70) = 11.11o

Equilibrium of crank-shaft mechanism is obtained when crank-shaft is rotated by 11.11o in the anti-clockwise direction.

CHAPTER-10-B- Potential EnergySTATICS - CASE STUDY

Introduction


The defending soldiers of a castle are launching a watermelon over the castle wall at an advancing army. To do this, they have designed a catapult with a heavy spring.

What is known:

The catapult and spring have the dimensions shown.

The spring constant k is 160 lb/ft, and the spring is in a relaxed position when the catapult is vertical (i.e. the relaxed length is 5 ft).

The collar of the spring is attached to a frictionless bearing.

The center of mass of the catapult (without the watermelon) is located at 2L/3.

Questions

Dimensions

After the watermelon has been launched, at what angles θ can the catapult come to a rest? Are these equilibrium positions stable?

Approach

If the system is conservative, we can find an equation for the potential energy of the catapult in terms of θ.

The values of θ that yield zero for the derivative of the potential energy indicate the equilibrium positions for the catapult.

Use the second derivative of the potential energy to determine which positions are stable, and which are unstable.

STATICS - THEORY

Conservative Forces

Conservative ForcesClick to view movie (47k)

When a force is displaced over a path with a finite length s, the work done can be found by integrating the differential work over the path as

If the total work of a force is independent of the path s, then the force is said to be a conservative force. Gravity and spring forces are examples of conservative forces, but friction is nonconservative

Gravitational Potential Energy

Gravitational Potential EnergyClick to view movie (44k)

The gravitational potential energy of a body is defined as the weight of the body times the distance y above a horizontal reference line,

Vg = Wy = mgy

If the body is below the reference line, it will have a negative potential energy.

Elastic Potential Energy

Spring Elastic PotentialClick to view movie (42k)

The elastic potential energy of a spring is given by

Ve = 1/2 ks2

Here k is the spring constant and s is the distance the spring is stretched or compressed from its unstretched length.

Potential Energy and Equilibrium

If a body is subject to only gravitational and elastic forces, the total potential energy, also called the potential function V, is given by

V = Vg + Ve

For a single-degree-of-freedom system at a point of equilibrium, the rate-of-change of total potential energy with respect to displacement goes to zero. For a small displacement q (q can be either distance or rotation), this can be stated mathematically as

dV/dq = 0

From this equation, the equilibrium position can be determined.

Potential Energy and Stability

Stable EquilibriumClick to view movie (24k)

A body in equilibrium is said to be in one of three states: stable, neutrally stable, or unstable. After we determine the equilibrium condition, we can determine the stability of that condition from the second derivative of the potential function.

If the second derivative of the potential function V is greater than zero, the body is in a stable equilibrium:

Unstable EquilibriumClick to view movie (20k)

If the second derivative is less than zero, the body is in an unstable equilibrium:

Neutrally Stable EquilibriumClick to view movie (21k)

If the second derivative is zero, higher-order derivatives must be used to determine the stability. Only if all derivatives of V are zero is the body neutrally stable:


The only forces acting on the catapult are the spring force and the force of gravity. Since both of these are conservative forces, the system is conservative, and the potential energy can be used to determine equilibrium positions and stability.

Begin with a free-body diagram with the origin of the x-y coordinate system located at the pivot point of the catapult.

Dimensions

The potential energy associated with the force of gravity can be determined as

Vg = (2/3 L cosθ) mg

Next, the potential energy associated with the spring can be calculated as

VL = 1/2 k s2 = 1/2 k (1/2 L sinθ)2

Thus, the total potential energy of the system is

V = Vg + VL = 2/3 mgL cosθ + 1/8 L2 k sin2θ

Equilibrium

Stable EquilibriumClick to view movie (112k)

Unstable EquilibriumClick to view movie (142k)

The equilibrium positions for the catapult can be found by taking the first derivative of the potential energy,

dV/dθ = -2/3 mgL sinθ + 1/4 L2 k sinθ cosθ = L sinθ (1/4 Lk cosθ - 2/3 mg)

When the catapult is in equilibrium, the derivative of the potential energy is zero.

dV/dθ = 0

There are three values of θ for which the catapult is in equilibrium. The first is relatively easy to determine as

L sinθ = 0 θ = 0o

The second and third values can be found by setting the expression in parenthesis equal to zero.

1/4 Lk cosθ - 2/3 mg = 0 θ = cos-1 (8 mg/(3Lk) ) = cos-1 ( 8(100)/(3 (5)(160) ) θ = ± 70.5o

Stability

To determine the stability of the three equilibrium positions, take the second derivative of the potential energy.

d2V/dθ2 = L cosθ (1/4 Lk cosθ - 2/3 mg) + 1/4 L2 k sin2θ

Next, substitute θ = 0° into the equation for the second derivative of the potential energy which gives,

d2V/dθ2 = L (1/4 Lk - 2/3 mg) > 0

Because the second derivative of the potential energy is greater than zero, θ = 0 is a stable equilibrium position.

Now, substitute θ = ±70.5 into the equation for the second derivative of the potential energy.

d2V/dθ2 = L cos70.5 (1/4 Lk cos70.5 - 2/3 mg) + 1/4 L2 k sin270.5

d2V/dθ2 < 0

Because the second derivative of the potential energy is less than zero, θ = ±70.5° are unstable equilibrium positions.

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