Notebook - Secrets of Engineeringsecretsofengineering.net/Statics/PDF_files/Statics_Notebook.pdfWhen forces are concurrent (have a common point of application) their resultant will

Introduction to Statics.PDF Edition – Version 0.95

Notebook

Helen Margaret Lester PlantsLate Professor Emerita

Wallace Starr VenableEmeritus Associate Professor

West Virginia University, Morgantown, West Virginia

© Copyright 2010 by Wallace Venable

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Contents: Notebook

0 Instructions to the Student

1 Forces as Vectors Definition of Forces. Characteristics of Forces. Principle of Transmissibility. Resultant of two Forces.

2 Vector Notations: Graphic Vectors Definition of Vectors. Definition of Scalars.

3 Vectors: Algebraic Representation Vector Algebra. Notation. Cartesian Unit Vectors. Coordinate Systems. Vectors as Sums. Vectors as Products.

4 Vector Addition: Resultant Forces Graphical Vector Addition. Algebraic Vector Addition. Resultants of Force Systems.

5 Components of Forces Definition of Components. Orthogonal Components Non-Orthogonal Components.

6 Equilibrium of a Particle: Concurrent Force Systems Newton's First Law. Definition of a Particle.

7 Vector Products Product of a Scalar and a Vector. Dot Product. Cross Product. Properties of Vector Products.

8 Moment about a Point Definition of Moment . Moments by Vector Products.

9 * Moment about a Line

10 Moments of a Force System: Resultant of a Coplanar Force System 24 Moment of a Force. Resultants of Non-Concurrent Force System.

11 First Moments First Moment of Area. Forming Integrals. First Moment of Volume about a Plane. First Moment of Volume about a Line.

12 Centroids Definition. Centroids by Symmetry. Centroids of Composite Areas

13 Center of Mass and Center of Pressure Center of Gravity. Center of Mass. Center of Pressure. Resultant of Distributed Load

14 Free Body Diagrams of Single-Body Systems

15 Equilibrium of Bodies Conditions for Equilibrium. Problem Solving Procedure. Statically Indeterminate Bodies.

16 Free Body Diagrams of Multi-Body Systems Diagrams of Members. Diagrams of Systems. Necessary and Sufficient Diagrams.

17 Equilibrium of Frames

18 *Trusses: Method of Joints Simplifying Assumptions. Method of Joints. Special Joints.

19 * Trusses: Method of Sections

20 Equilibrium of Non-Coplanar Force Systems Three Dimensional Equilibrium.

21 Couples and Resultants with Couples Definition of Couple. Moment of a Couple. Characteristics of a Couple. Equivalent Couples.. Force and a Couple.

22 * Equivalent Force Systems Resultants of Parallel Force Systems. Resultants of Non-Coplanar Force Systems. Summary of Resultants.

23 Equilibrium with Couples Reactions on Members Subjected to Couples. Cantilevers. Machine Elements.

24 Introduction to Friction Observation. Limiting Friction. Coefficient of Friction. Equilibrium.

25 Friction Determination of Impending Motion. Determination of Coefficient of Friction. Systems with Friction.

26 Friction of Machine Elements 27 Belt Friction

28 Moments of Inertia of Geometric Areas Definition. Forming Integrals. Polar Moment of Inertia. Radius of Gyration. Properties of Areas Table.

29 Moments of Inertia of Composite Areas Units. Moment about a Common Axis. Parallel Axis Theorem. Moment about a Centroidal Axis. Choice of Reference Axis.

30 Moments of Inertia of Mass Definition. Moment of Inertia of Common Geometric Shapes. Composite Bodies. Radius of Gyration.

Index

Page 0-1

Unit 0Instructions to the Student

Using the Notebook

The notebook has been designed to provide the student with concise recapitulation of the major points in the units at the same time that it provides an annotated index for ready reference and review.

It is an integral part of the programs so that you are directed to work first in the programs, then to summarize your work in the notebook and to return to the programs for the next topic.

Every item called for in the notebook is referred to a frame in the program where the item is discussed. For example, a question followed by the notation (8-23) means that the answer will be found in Frame 8-23. This enables you to check your reply and to build up a completely correct notebook for future reference, at the same time that it provides you with immediate access to the programmed treatment of the topic should you wish more detail.

Since the entire notebook forms an annotated index to the material, it seems reasonable that the alphabetical index to topics should be included in the notebook. Thus, you may look up a given topic in the alphabetical index. There, you will be referred to the appropriate page of the notebook for a concise treatment of the topic. Should that not be sufficient the parenthetical numbers will direct you to the portion of the unit which gives a detailed treatment of the topic.

The authors hope that the notebook will help you in your night-before-the-examination reviews of the programs and that the reference system will make it possible for you to locate quickly material which you may recall and wish to find at a later date.

$ Return to Frame 1-20

Page 1-1

Unit 1 Forces as VectorsForce may be defined as __________________________________________________

___________________________________________________________________________ (1-5)


Mass, Weight, and Force of Gravity

Sir Isaac Newton postulated a law of gravitational attraction, primarily to use for explaining planetary motion.

For our earthly work, his equation is

where F is the force of gravity on a body, m is the mass of the body, M is the mass of the earth, r is the radius of the earth, and G is the universal constant of gravitation. According to experimental evidence, G = 66.73 x 10-12 m3/(kg • s2)

We define the local gravitational constant as g, and we get the simple equation

F = m g

Note that the earth is not a perfect sphere, so the precise value of r, and therefor g, is different at different locations.

According to some measurements made by the U.S. Coast & Geodetic Survey, at sea level g varies from 32.088 feet per second squared (fps2) [9.780 meters per second squared (mps2)] at the equator to 32.258 [9.832 ] at the North Pole. It also decreases by 0.003 fps2

for each increase of 1000 feet in altitude [ 0.003 mps2 per 1000 m].

In this book we will use values of g = 32.2 fps2 (a good value for Washington, DC or New York) and g = 9.81 mps2 (a good value for Paris or London) for our calculations.


Page 1-2 The characteristics of a force are

1. _____________________________________________________________________

2. _____________________________________________________________________ 3. _____________________________________________________________________ (1-14)

The direction of a force is defined by

1. _____________________________________________________________________

2. _____________________________________________________________________ (1-18)


Principle of Transmissibility

The principle of transmissibility states that a force may be applied at any point along its line of action without changing its external effect on a rigid body. (The internal effect will

usually be changed.)

For example, force A has the same external effect on the body as force B or force C, since all three forces have the same line of action. Thus we could say that A, B, and C are "equivalent" forces.


Page 1-3Resultant of Two Forces

Experimental evidence shows us that forces do not obey the laws of arithmeticaddition. Instead they add according to the "parallelogram law."

This means that forces A and B acting together on a particle have exactly the same effect on the particle as force C which is proportional to the diagonal of the parallelogram constructed on forces A and B. Force C is called the resultant of A and B. This is an observed law. It cannot be proven mathematically.


Page 2-1

Unit 2 Vector Notation: Graphic Vectors

Vectors are those quantities which have ____________________ and ___________________

and add by the ______________________________________________________________ (2-9)

Scalars are quantities which have only ______________________ and add by

____________________________________________________________________________ (2-8)

(There are a few quantities which are neither vector nor scalar but they are rare.)


Vector quantities may be represented by directed line segments. In this case the length

of the line indicates _________________ , the slope of the line indicates _______________ ,

and the position of the arrowhead indicates ________________________ (2-14, 2-15, 2-16)


A position vector shows the position of one point with respect to another.

The sketch below shows the position of _____ with respect to ______ .

In this case _______ is called the reference point. (2-22)

Page 2-2 Problem 2-1

A buried treasure is located 6 miles west and 8 miles south of your home. Sketch its position vector with respect to your home. Determine its distance and bearing from your home.

In the above example a variation of the parallelogram law was used to add two components. Draw the components and add them by the parallelogram law.


Page 3-1

Unit 3 Vectors: Algebraic Representation

Vector Algebra

Notation R = 7e indicates a vector whose magnitude is ______ and whose direction is

parallel to unit vector ____________ . (3-4)

M = Mm is a vector of magnitude _____ directed parallel to ________ (3-6)

Cartesian Unit Vectors

The symbol i denotes a vector of ____________ magnitude parallel to the

__________________ , j is a similar vector parallel to the ___________ , and k is a similar

vector parallel to the ____________ (3-9)

Coordinate Systems

The discussion will be limited to coordinate systems which are _____________________

and __________________________ . Sketch such a system below.

(3-16) A right hand system is one in which _______________________________________________

_________________________________________________________________________________

_________________________________________________________________________________

__________________________________________________________________________ . (3-14)


Page 3-2 Vectors as Sums The vector, R, lying in the yz plane may be represented by the sum of its rectangular components thus:

R = ________________________

The vector Q = 7i + 4j - 5k may be shown as follows:


Vectors as Products

The vector Q = 7i + 4j - 5k may also be written as the product of a magnitude and a unit vector as follows:

Q = _________________________________ The vector R shown in the figure may be written as the product of its magnitude (35) and a unit vector as follows:

R = _________________________________


Page 4-1

Unit 4Vector Addition and Resultant Forces

One graphical method of vector addition may be stated as follows:

_________________________________________________________________________________

_________________________________________________________________________________

_________________________________________________________________________________

___________________________________________________________________________ . (4-5)

Problem 4-1

Find the vector sum of the following vectors by using graphical addition.

A = 10i + 12j B - = -8i + 6j C - = 3i-6j

Problem 4-2

Use graphical addition to add the vectors shown.


Page 4-2Problem 4-3

Write the vectors in Problem 4-2 as sums of components and find A + B algebraically.

A = ______ i + ______ j B = ___________________________

A + B = ___________________________________

Problem 4-4

A = 12i + 3j + 6kB = -10i + 8j + 7kFind: A + B = ________________________________________ A - B = ________________________________________ B - A = ________________________________________

$ Return to Frame 4-12Problem 4-5

Find the resultant of the forces shown.


Page 4-3

Resultants of Force Systems

When forces are concurrent (have a common point of application) their resultant will also have the same point of application. The resultant of such a force system can be obtained by adding the forces vectorially and showing the resultant acting through the intersection of the lines of action of the original forces. Thus C shown below is the resultant of forces A and B. Stated algebraically

When forces are not concurrent the magnitude and direction of the resultant can be found by simple addition but its point of application must be found by the use of moments. (You will learn about problems of this nature in Unit 10.)

Page 5-1

Unit 5Components of Forces

If a vector is replaced by two or more vectors, which, when taken together, have the same effect as the original vector, the replacing vectors are called components of the original vector. A resultant vector is the sum of its components.


Problem 5-1

Break the force shown into three components, each of which is parallel to a coordinate axis.

Problem 5-2

Divide the 100 lb force into rectangular components, one of which lies along AB. Evaluate the components and show them on the sketch.


Page 5-2The steps in solving a component problem by means of a figure are as follows:

1. ______________________________________________________________________________

2. ______________________________________________________________________________

3. ______________________________________________________________________ (5-12)

Can a component be larger than the resultant? _________________ (5-19)

Problem 5-3

A is one component of R. Find the component which, when added to A, will produce R by a trigonometric solution. Then check it by drawing the figure to scale and measuring.


Page 5-3Problem 5-4

Break the 500 lb force shown into two components, one horizontal and one parallel to ab. Solve the problem trigonometrically and check your answer graphically.

When directions of both components are given the most direct solution involves the law of

______________________ .

When the magnitude and direction of one component are given, the most direct solution

involves the law of _________________________ . (5-29)


Page 5-4 Problem 5-5

The force F = 25i + 16j + 30k has one component, C1 = 25i + 25j. Find theother component which, added to C1 , will be equal to F.


Problem 5-6

Find the components in the indicated directions.


Page 6-1

Unit 6Equilibrium of a Particle: Concurrent Force Systems

In the last part of the seventeenth century Sir Isaac Newton, by the observation of natural phenomena, formulated three basic laws upon which all Newtonian mechanics is based. It should be emphasized that these laws cannot be proven mathematically but may be verified by physical measurement.

Statics is concerned primarily with the first of these laws which may be stated as follows:

If the resultant of all forces acting on a particle is zero, the particle will remain at rest (if originally at rest) or will travel at a constant speed along a straight path (if originally in motion).

When the resultant of all forces on a particle is zero, the particle is said to be in equilibrium.


A particle is a body whose equilibrium is ________________________________________

______________________________________________________________________________ .

A body may be considered a particle if the forces acting on it form a ________________force system.

Draw an example of a particle acted upon by an appropriate force system.


Page 6-2Problem 6-1

The figures show a detail of a joint in a truss and its free body diagram. Find the two unknown forces.


Page 6-3Problem 6-2

A balloon is moored by three cables as shown. The upward lift on the balloon is 500 lb. Find the tension in each cable.

In solving problems involving equilibrium of a particle, the steps in the solution are as follows:

1. _________________________________________________________________________

2. _________________________________________________________________________

3. _________________________________________________________________________

4. _________________________________________________________________________ (6-33)


Page 7-1

Unit 7Vector Products

Product of a Scalar and a Vector

The product of a scalar and a vector is a ______________ directed _______________ to the original vector. (7-3)

Example: R = 4i + 3j and Q = 3R Q = ______________________________ (7-4)

Dot Product

The dot product of two vectors is a scalar and is given by the following expression:

A % B = ________________________ (7-7)Example:

Vector A has a magnitude of 20 units. Vector B has a magnitude of 5 units.

A % B = _____________________ (7-8)

Dot Products of Unit Vectors

R = 3i -4j - 2kS = 9i + 6j 3k–

R % S = _______________________________$ Return to Frame 7-14

Page 7-2Cross Product

The cross product of any two vectors may be obtained from the following expression

A } B = ______________________ (7-17)

The magnitude of A } B is _______________________________________________ (7-16)

The direction of A } B is _________________________________________________ (7-17)

The sense of A } B is ____________________________________________________ (7-18)

Evaluate and sketch B } C


The magnitude of the cross product of parallel unit vectors is ______________________

The magnitude of the cross product of perpendicular unit vectors is ________________

___________________________________________________________________________ (7-30)


Page 7-3 While it is quite possible to find the cross product of the two vectors

by multiplying it out term by term, most people find it easier to write it as a determinant.

We can evaluate this determinant in several ways. One way that is used by many is to rewrite the first two columns and take ordinary products along the diagonals subtracting the south-western diagonals from the south-eastern diagonals thus

Another way that amounts to the same thing, but avoids rewriting is to follow first the plus, then the minus, path from each unit vector, taking products as you go.

Collecting terms yields the same form as before. There are several other ways of achieving the same result.


Page 7-4Example:

A = 7i + 4j + 6kB = 2i + 3j + 4k

A } B = _______________________A % B = _______________________B } A = _______________________B % A = _______________________

Properties of Products

In this unit we have observed (not proved) the following laws governing multiplication involving vectors.

Multiplication of a vector by a scalar is distributive.

Example: a(x + y) = ax + ayThe dot product of two vectors is distributive.

Example: a % (b + c) = a % b + a % cThe dot product of two vectors is commutative.

Example: u % v = v % uThe cross product of two vectors is distributive.

Example: A } (B + C) = A } B + A } CThe cross product of two vectors is not commutative.

Example: A } B ≠ B } AThe cross product is not associative.

Example: (P } Q ) } R ≠ P } ( Q } R)Proof of all these statements may be found in any text on vector analysis.


Page 8-1

Unit 8Moment About a Point

Moment of a Vector with Respect to a Point

Definition (in your own words)

Moment of a vector with respect to a point is ____________________________________

________________________________________________________________________ (8-10)

Problem 8-1

Find the moment of the vector

which acts through the point (0,4,0) with respect to the origin.


Page 8-2Moments by Vector Products

It is frequently advantageous to use vector algebra to compute moments. This is done

with the ___________________ or _________________ product.

M = ________________________The vector a denotes a vector which runs from __________________________ to

_____________________________________________________ (8-18)

Problem 8-2

The vector V = -3i + 4j + 5k passes through the point B which has the coordinates (0,0,4). Find its moment about the point (0,2,2).


Page 9-1

Unit 9Moment About a Line

Moment of a Vector About a Line

The moment of a vector about a line is a rectangular component of the moment of the vector about a point on the line. The component is parallel to the direction of the line.

In the figure above, e is a unit vector along line L-L, P is a point on the line, and a is an arm from P to the vector V.

The moment of the vector V about point P is given by the equation M = a } V

Since M % e = M cos θ , the expression (a } V) % e represents the magnitude of the rectangular component of the moment along L-L. This magnitude is then multiplied by the unit vector along the line in order to form a vector component.


Problem 9-1

A force F = 5i - 3j + 2k acts through the point P with coordinates (-3,1,1). Find the moments (1.) about the x axis, (2.) about the y axis, (3.) about a line through the origin and the point Q with coordinates (-8,-4,1).

Mx = ______________________________My = ______________________________M0-Q = ______________________________


Page 10-1

Unit 10Moments of a Force System: Resultant of a Coplanar Force System

Moment of a Force System

State in your own words the definition of the resultant moment of a force system.

_______________________________________________________________________________

_______________________________________________________________________________

________________________________________________________________________ (10-7)

Resultant of a Non-concurrent Force System

The resultant of a two dimensional force system is a force which:

(a) has a magnitude and direction determined by (in words) ________________________

___________ or the equation R = ______________ (10-16)

(b) has a line of action which must be located such that (in words) ___________________

________________________________________________________________________________

________________________________________________________________________ (10-17)

Page 10-2Problem 10-1

Given the system of forces:

(a) Find its moment about the origin 0.

M0 = _____________________________(b) Find its moment about P, the point (4,4).

MP = _____________________________(c) Find the resultant of the force system and locate its line of action on the drawing above.

R = ______________________________


Page 11-1

Unit 11First Moments

First Moment of Area

The rule for successful selection of an element for determination of first moment

by integration is: ___________________________________

___________________________________________________

_____________________________________________ (11-8)

Draw an appropriate element on the figure for finding the first moment of the shaded area with respect to the x-axis.

The general expression for the first moment of an area with respect to an axis is:

Qx = _____________________

The steps in the determination of a first moment are as follows:

1. Sketch the problem.2. Select an element. Dimension it in terms of x and y and known constants.

3. Write the integral Qy = ∫ xdA or Qx = ∫ ydA as required.4. From your dimensioned sketch write dA in terms of x, y, and known constants.5. From the sketch determine your limits.6. Reduce your expression for xdA or I ydA to an expression in one variable by

substituting x in terms of y or vice versa.7. Integrate and evaluate your expression.


Page 11-2

Problem 11-1

Find the first moment of the quarter-circle shown with respect to the x-axis.


Page 11-3First Moment of a Line

The method for finding the first moment of a line about an axis is very similar to that used for first moments of areas. The difference is that a differential line element is used, rather than an area.

Problem 11-2

Find the first moment of a quarter of a circular arc of radius a about the x- axis.


First Moments of Volume about a Plane

The first moment of a volume about a plane can be found by extending the two dimensional methods into three dimensions. Work the problem on the next page.


Set up the integral for the first moment of a hemisphere about its base. Work it out if you choose.


First Moment of Volume about a Line

This topic, while similar to those preceding is more complex and requires the use of multiple integration. An example problem is worked out in some detail in Frames 11-28 through 11-30. Should you wish practice in this area it is suggested that you find Qy for the same volume.


Page 12-1

Unit 12Centroids

The centroid of an area is defined as the point at which ____________________________

________________________________________________________________________________

_________________________________________________________________________ (12-2)

The distance from the centroid of a given area to a specified axis may be found by

________________________________________________________________________________

_________________________________________________________________________ (12-5)

For the area shown A = 6 in.2, Qy = 18 in.3 and Qx = 22 in.3

G denotes ________________________________

yG = ______________________________________

xG = _________________________________ (12-5)

Fill in the locations of the centroids of the figures below. You will find it convenient to know these.


Page 12-2Centroids by Symmetry

The centroid of any symmetrical area will fall ______________________________ (12-16)

By symmetry, find the centroids of the areas below.

(12-18)Problem 12-1

Find the centroid of the area shown and locate it on the sketch.


Page 13-1

Unit 13Center of Mass and Center of Pressure

Center of Gravity

Every body can be considered to be made up of a large number of small particles, each having a mass, m. The earth exerts an attractive force on each particle which we call the weight of the particle. If we call each of the individual forces Wi , we might draw a picture of a body acted upon by its own weight that would look like this:

The resultant of all the weights would be

where W is the total weight of the body. The resultant would have a line of action such that its moment about 0 would equal the sum of the moments of all the little weights.

where q is a vector from the origin to some point on the line of action of W. The simplest spot would be where it pierces the x-z plane, where

XG and ZG are two of the three coordinates of the center of gravity of the body. If you expand equation (2.) you can write

If you work out the cross products and break down the results into coefficient equations you will get

The third coordinate of the center of gravity can be found by rotating the body and the coordinate system repeating the process we just completed giving

The center of gravity is that point in a body at which its entire weight may be concentrated as a resultant force without changing the external effects on the body. If we shrink our small elements of mass to the “infinitesimals” of calculus, these equations may be re-written as


Page 13-2Center of Mass

The coordinates of the center of mass, or mass-center, are given by the expressions

Problem 13-1

Locate the mass center of a homogeneous hemisphere of density ρ.


The "ice-cream cone" shown is to be cast in a material weighing 25 lb/cu ft for an advertising display. Find its mass center.


Page 13-3Center of Pressure

If a distributed force system is considered as a system of parallel forces acting on a plane, the total force exerted is the resultant of the system and the center of pressure is the point at which the resultant acts on the plane.

Total Force

$ Return to Frame 13-26Another way of saying the same thing is

(13-27)

This shows us that the total force is proportional to the _____________________________

__________________________________________________________________________ (13-29)

Location of Center of Pressure


Page 13-4The resultant force F must be applied so as to produce a moment about 0 equal to M0.

This means that the resultant force must pass through the _________________________

_________________________________________________________________________ (13-34)

Problem 13-3

Find and locate the resultant force for the distributed load shown.


Find and locate the distributed load shown.


Page 14-1

Unit 14Free Body Diagrams of Single-Body Systems

Free Body Diagrams

The process of drawing a free body diagram (FBD) consists of two steps:

(1.) ______________________________________________________________________________

(2.) _______________________________________________________________________ (14-3)

The weight of the body is one of the forces which must generally be included.

It acts through ___________________________________________ (14-9) which often

coincides with __________________________________________________ (14-10)

Problem 14-1

Use what you have learned so far and draw the free body diagram for this beam.



Draw a FBD of this system.


Page 15-1

Unit 15Equilibrium of Bodies

Equilibrium of Rigid Bodies

A rigid body is in equilibrium when the external forces acting on it form a system of forces equivalent to zero. This means that the necessary and sufficient conditions for the equilibrium of a rigid body are:

#F = 0 and #M0 = 0The moments may be taken about any point whatsoever and must always equal zero.


Problem 15-1

Use the information above to solve for T2 in terms of T1 assuming the pulley shown to be perfectly frictionless. Also find the reaction in terms of T1.



Find the reactions at A and B for the beam shown, and check your work.


Find the tension in the cable at A and the reactions at C.

(Continue on to the next page when you have done the problem.)

Page 15-3

The steps in solving a problem involving the equilibrium of a rigid body are as follows:

(1.) ____________________________________________________________________________

(2.) ____________________________________________________________________________

(3.) ____________________________________________________________________________

(4.) ____________________________________________________________________________

(5.) ____________________________________________________________________ (15-19)


The maximum number of unknowns that can be found by the laws of statics for a given force system is as follows:

(1.) Concurrent coplanar _____________

(2.) Concurrent parallel coplanar _____________

(3.) Parallel coplanar _____________

(4.) General coplanar _____________ (15-25)

If the number of unknowns exceeds the number that can be found by statics, the system is statically indeterminate. This does not mean it cannot be solved. It often can be solved by including additional information obtained from the deformation of bodies. It is, however, insoluble by analysis as a completely rigid body.


Page 16-1

Unit 16Free Body Diagrams of Multi-Body Systems

Problem 16-1

Bars AB and BC are of equal length and weight. A rests on a frictionless surface while joints B and C are pinned. DE is a weightless cord.

Draw free body diagrams of each of the two bars.


Draw a FBD of the system consisting of both bars and the cord in the problem above.



Draw a free body diagram of the system below.

Draw free body diagrams of each of the members of the system above.


Page 16-3Free Body Diagrams of Systems

Reactions

In statics, the word "reactions" is used to refer to __________________________________

__________________________________________________________________________ (16-4)

Draw a FBD of the system below which shows the reactions.

Necessary and Sufficient FBDs

If it is necessary to find all the unknown forces acting on a system it is necessary to draw

as many free body diagrams as ___________________________________________ (16-30)

When only selected forces are to be found it is possible to find them from a few of the possible FBDs for the system.

Draw the FBD which will be needed to find the tension in cable AB and the forces at C.


Page 17-1

Unit 17Equilibrium of Frames

Problem 17-1

A piece of metal is bent to make a pipe rack as shown. Two similar lengths of pipe, each weighing 20 lb, are in the rack. Find the force exerted by the wall on the lower pipe.



Find all the forces acting on the horizontal member.

(Continue on to the next page.)


Find the forces on member AB at A and B.


Page 18-1

Unit 18Trusses: Method of Joints

Trusses are rigid bodies made up of a number of members fastened at their ends. In their analysis three simplifying assumptions are made.

1. ______________________________________________________________________________

2. _____________________________________________________________________________

3. ______________________________________________________________________ (18-3)

This results in a series of two force members, so that the line of action of the force on any

member is ______________________________________________________ __ (18-7)

Thus, the line of action of such a force is apparent by inspection.

Method of Joints

In the method of joints the forces on members are determined by drawing free body diagrams of joints and solving #F = 0 for each joint.

Special Joints

Case I. When a joint is acted upon by four forces arranged in two collinear pairs opposite forces must be equal. Complete the free body diagrams below.

Page 18-2Case II. When a joint connects three members, two of which form a straight line, the forces in the collinear members are equal and the force in the third member is zero.

Example

Typical Joints

Draw the FBD and write the equation for joint E.

Special Joints

Draw the FBDs and write the equations for joints D and B.

Joint K is not a special joint because members KL and KJ are ________________________ .



A Pratt roof truss is loaded as shown. Determine the force on each of the members of the truss by the method of joints.


Page 19-1

Unit 19Trusses: Method of Sections

Method of Sections

In the method of sections, the first step is to draw a line on the truss which _________

________________________________________________________________________________

Next, a FBD is made of _________________________________________________________

The solution for the forces on the members is then made by using the equations #F = 0 and #M = 0. (19-6)

Problem 19-1

Find the forces in AB, BD, DE, and CE by using a combination of the method of joints and the method of sections.


Page 20-1

Unit 20Equilibrium of Non-Coplanar Force Systems

Problem 20-1

A 7 ft boom is held in place by a ball and socket at A and two cables, BC and BD. Find the magnitude of P if the tensions in the cables are known to be equal. Also find the force acting along AB.



The 600 lb homogeneous plate is supported in a horizontal position by two wires at A and B and a roller at C. Find all reactions.



A slender homogeneous rod weighing 50 kilograms is supported by a ball and socket at A, and a smooth wall at B, and a cable BC. Find all unknown forces acting on the bar.


Page 21-1

Unit 21Couples and Resultants with Couples

Couples

A couple is defined as ___________________________________________________________

__________________________________________________________________________ (21-5)

Moment of Couple

The coplanar forces F1 and F2 make up a couple and the coordinate axes are chosen so that the forces lie in the x-y plane.

The moments of F1 and F2 about the general point (x,y,z) are

(1.) M1 = ____________________________ M2 = ____________________________hence the total moment

MT = (r1 } F1) + (r2 } F2)Making use of the fact that F1 = -F2 and rearranging the equation we find

(2.) MT = _______________________Thus demonstrating that: "The moment of a couple is the moment of one of the forces about any point on the line of action of the other force, and it has the same moment about any other point in space."


Page 21-2Characteristics of a Couple

The characteristics of a couple are magnitude and direction.

The magnitude of a couple is given by _____________________________________________

_________________________________________________________________________ (21-16)

The direction of a couple is given by _______________________________________________

__________________________________________________________ (21-16, 21-17, or 21-25)

Couples are equivalent if ________________________________________________________

_________________________________________________________________________ (21-21)

Represent a couple of 60k ft-lb as:

(1.) two 20 lb forces in the x-y plane (2.) a curved arrow

(3.) a vector



Resolve the force at A into an equivalent force system consisting of a force through B and a couple.

Problem 21-2

Replace the 20 kip eccentric load on the column with an equivalent force system consisting of a force acting through G, the centroid of the cross-section and a couple.


Page 22-1

Unit 22Equivalent Force Systems

Resultants

Problem 22-1

Find the resultant of the coplanar system shown and the point where its line of action intersects line AB. Show your results on a sketch.


Problem 22-2

What is the resultant couple acting on the valve stem shown?



Find the resultant of the force system shown.



A square concrete foundation slab supports three columns as shown. The columns carry the loads shown. Determine the resultant load and locate the point at which its line of action passes through the foundation. Show your answer on the sketch.



The figure represents a brace and bit. At the instant shown it is in the horizontal xz plane and the x-axis coincides with the bit. Find an equivalent force system consisting of a force through the "business end" of the bit and a couple.


Page 22-5Summary of Resultants

(1.) The resultant of a concurrent force system is a _________________________ , acting

through the _________________________________________________________ .

(2.) The resultant of a system of couples is a _______________________________ .

(3.) The resultant of a system of parallel forces is a _______________________, acting

parallel to the forces of the system or a ________________________________ .

(4.) The resultant of a coplanar force system is a _______________________,

or a _________________________ .

If, in the latter two cases #F = 0 and #M0 ≠ 0, the resultant is a _________________ .

If, in the latter two cases #F ≠ 0, the resultant is a ____________________, and its

location is given by the equation ______________________________

For every general force system there is a unique equivalent system consisting of a single force and a single couple which causes rotation about an axis parallel to the force. This

system is called a _____________________________ . (22-35 and 22-36)

In any cases in which #F = 0 and #M0 = 0, the resultant is __________________, and the

body is in ______________________________ .


Page 23-1

Unit 23Equilibrium with Couples

Problem 23-1

Compute the reactions on the member which is subjected to a couple and a force.



The cantilever beam shown carries a load of 1500 kilograms. Find the reactions at the wall.



A rope is fixed to pulley B at one end and passes over pulley A to a weight. If the torque on the shaft to which A is fastened is 30 ft-lb, what couple must be applied to B?What force does the pulley A exert on its shaft?



The torsion bar suspension on the original Volkswagen Beetle was somewhat like the drawing above. A 2000 Newton wheel load is applied to the axle G. Joints A, B and H are frictionless. What couple C must be applied to link EH by the torsion bar?



The possible snow loading on the roof of a bus-stop shelter is 50 lb/ft of support. What will the reaction at the base be?


Page 23-6Shear and Moment on a Section

The steps in finding the shear and moment on a section are:

1. ______________________________________________________

2. ______________________________________________________

3. _____________________________________________________ (23-41)

Problem 23-6

Find the shear and moment on sections a-a and b-b of the beam below.


Page 24-1

Unit 24Introduction to Friction

Friction

Observations on Friction

Friction is the name of the force that opposes the motion of one surface relative to another. Its magnitude depends primarily on the nature of the two surfaces involved. Its direction is opposite to the motion, and is tangent to the contact surfaces. It is presumed to be caused by a tendency of the surfaces to mesh together.

Friction is either fluid friction or dry friction. The former applies not only to moving fluids but also to problems involving lubricated surfaces. The latter applies to problems involving rigid bodies in contact along non-lubricated surfaces. Since fluid friction is best considered in connection with fluid mechanics we will limit our present study to dry friction.


There are two sorts of dry friction, static friction and kinetic friction. The first opposes impending motion; the second opposes motion actually in progress.

This course will deal only with problems involving static friction but you will encounter kinetic friction in dynamics.

Limiting friction is defined as ___________________________________________________

________________________________________________________________________ (24-11)

and implies ____________________________________________________________ (24-12)

The coefficient of friction is found by _____________________________________________

_______________________________________________________________________________

and is given by the equation ____________________________________________ (24-15)

Page 24-2

Coefficient of Static Friction

The coefficient of friction* must be determined experimentally. It is the result of the roughness or smoothness of the materials in contact. The following table shows some of the published results for common materials.

Approximate Coefficients of Static Friction for dry surfaces

Steel on steel 0.4 to 0.8

Wood on wood 0.2 to 0.5

Wood on metal 0.2 to 0.6

Metal on stone 0.3 to 0.7

Metal on leather 0.3 to 0.6

Wood on leather 0.2 to 0.5

Earth on earth 0.2 to 1.0

Cast iron on cast iron 0.3 to 0.4

Rubber on concrete 0.6 to 0.8

Rubber on ice 0.05 to 0.2

If, in practice, one needs the coefficient of friction between less common materials or under specific circumstances, he must determine it experimentally.

The coefficient of kinetic friction is also determined experimentally. For a given pair of materials it will always be than the coefficient of static friction.

* The term "coefficient of friction" will always refer to static friction throughout these units.


f < f' implies _____________________________________________________________

f = f' implies _____________________________________________________________

f > f' implies _____________________________________________________ (24-32)

Page 24-3

List the steps in solving problems involving static friction.

1. _____________________________________________________________________________

2. _____________________________________________________________________________

3. _____________________________________________________________________________

4. _____________________________________________________________________ (24-32)

Problem 24-1

The coefficient of friction between the 300 kilogram block and the plane is 0.30. Is the 100 kilogram weight sufficient to keep the block from sliding down the plane? Prove your answer.


Page 25-1

Unit 25Friction

Problem 25-1

The coefficient of friction between a homogeneous 300 lb block and the floor is 0.40.

Determine the maximum height above the floor at which a force sufficient to cause motion of the block will cause it to slip and not tip.


Page 25-2Experimental Determination of Coefficient of Friction

A popular method for determining the coefficient of friction between two materials is to place a block of one material on a surface of the other and then to slowly increase the inclination of the surface until the block just begins to slide. The maximum angle at which the block remains at rest is called the "angle of repose", and this angle is carefully measured.

Free Body Diagram

#F = __________________________________

Using the equilibrium equation and designating the angle of repose as _________ , we can find the coefficient of friction from the relationship

µ = _________________________________

It is relatively easy to make precise measurements of the coefficient of friction with the method you have just studied because the measurement is a geometric one. It is very difficult to construct systems which measure the forces on the block at the instant it begins to move on a horizontal surface.



What is the minimum force necessary to move the block B?



What force is needed to pull the 25 lb wedge from under the 50 lb block?


Page 26-1

Unit 26Friction of Machine Elements

Problem 26-1What is the minimum value of P necessary to maintain equilibrium in the system shown?

$ Return to Frame 26-8Problem 26-2 What force must the girl exert at A on the trash can in order to move it to the right?



Find the minimum value of P necessary to keep the drum from turning as a result of the application of the 300 N-m couple.



Find the minimum value of P needed to move the wedge to the right. The coefficient of friction at all contact surfaces is 0.25, the cylinder weighs 1000 lb and the wedge weighs 200 lb.

(Continued on next page.)

Page 26-4Problem 26-4 (continued)


Page 27-1

Unit 27Belt Friction

Friction on Flat Belts

Consider the case of a flexible flat belt which is wrapped around a cylinder. The angle through which the belt remains in contact with the cylinder is β and the tensions in the end portions of the belt are TL and TS as shown. Assume that TL is larger than TS , and that the belt is about to slip on the drum.

A diagram of an element of the belt which subtends an angle of dθ can be drawn as shown on the right.

$ Return to Frame 27-3 but keep your notebook open.

Summation of forces on the element gives the equations _______________________ = 0

in the x direction, and _____________________________ = 0 in the y direction.

Taking advantage of the fact that for small angles

is very much smaller than the other terms the equations may be reduced to

df = ______________ and dN = _______________

Since motion impends df = µ dN

Therefore dT = µ dT dθ


Page 27-2The technique of separation of variables may be applied to this equation and the following definite integral established.

Evaluation of the integral yields the result

Form I. ______________ = _________________

which may also be expressed as

Form II. ______________ = _________________

Form II is the more useful and more used.


A rope is wrapped 3/4 of a turn around a fixed round bar as shown. How much force must be applied to the end of the rope to lift the 30 kilogram weight?



Determine the force P needed to cause the 450 lb block to begin to move to the left.


Page 28-1

Unit 28Moments of Inertia of Geometric Areas

Moment of Inertia

Moment of inertia, or second moment, is the name for the commonly encountered mathematical quantity

where s is the distance to the element dA from a required axis. For example

I y = ___________ (28-10)

The element dA is always selected ____________________ to the required axis. (28-11)

The procedural steps in finding I y are:

1. _____________________________________________________________________________

2. _____________________________________________________________________________

3. _____________________________________________________________________________

4. _____________________________________________________________________________

5. _____________________________________________________________________ (28-16)

Problem 28-1

Find I y for the triangle shown.



Find I x for the triangle shown. The x-axis passes through the centroid.


Page 28-3Polar Moment of Inertia

The polar moment of inertia, J, about a point 0 is defined as

where ρ is the distance from the point to the element of area. Evaluation by means of direct integration is most efficiently accomplished by the use of polar coordinates. Fortunately there is an indirect way which is much easier.

Looking at the sketch and remembering Pythagoras we see that

Stated in words, the polar moment of inertia is equal to the sum of the moments of inertia about any two perpendicular axes which intersect at 0.

$ Return to Frame 28-33Radius of Gyration

The radius of gyration of an area with respect to an axis or point is defined physically as the distance from the axis or point at which the entire area could be concentrated without changing the moment of inertia about the axis or the point.

The mathematical definition is more to the point.


Page 28-4

Page 29-1

Unit 29Moments of Inertia of Composite Areas

Since moment of inertia is obtained from the summation by integration of the product of a distance squared and an area, its dimensions are length2 x length x length = length4

Moments of inertia of areas are used primarily in “strength of materials” problems. Since in American engineering mechanics problems the areas involved are of a size most conveniently measured, and shown on drawings, in inches, and stresses are expressed in Pounds per square inch, the moment of inertia of an area is most commonly and usefully expressed in inches4. We will use in4 as the proper form for answers in these units in ACU.

In SI, the situation is a bit more complex. Although are often given on drawings in centimeters, in calculations meters are the correct unit. We will use m4 as the proper form for answers in these units in SI.

Moment of Inertia of Composite Areas

Moment of inertia for a composite area may be found by direct addition or subtraction of the moments of inertia of the parts as long as the moments of inertia of the parts are all known about the desired axis.

Problem 29-1

Use subtraction to find I x and I y for the scalene triangle shown.


Page 29-2Parallel Axis Theorem

The parallel axis theorem enables us to find the moment of inertia of an area about any axis in terms of a parallel axis passing through the centroid of the area. It is most conveniently expressed in terms of the transfer equation

I xa = ________________________

In this equation I xa is ______________________________________ ,

I xG is __________________ , and d is _________________________________________ .

Problem 29-2

The circle is 4 centimeters in diameter. Find I y

The transfer equation may also be used in reverse to find I xG if the moment of inertia is known about any x-axis.

Problem 29-3

Find I xG for the semi-circle by use of the transfer equation.

Complete the table on page 28-4 and return to Frame 29-24.

Page 29-3Choice of Reference Axis

There is no right or wrong reference axis. There are, however, axes that result in easier work than others.

Easy work usually results from an axis chosen according to one or more of the following criteria:

(1.) _____________________________________________________________________ (29-34)

(2.) ______________________________________________________________________ (29-36)

(3.) ______________________________________________________________________ (29-36)

(4.) ______________________________________________________________________ (29-37)

(5.) ______________________________________________________________________ (29-38)

Problem 29-4

Find the moment of inertia about a horizontal axis through the centroid (I xG) for the area shown. I'll leave it to you to practice drawing the table this time.


Page 30-1

Unit 30Moments of Inertia of Mass

The moment of inertia of a mass, of density δ , is given by the integral

I = ________________________________________________________________ (30-3)

In selecting the element it is necessary to exercise care that the entire element be a constant distance from the axis in question. This generally results in a multiple integration, for which the best reference is a calculus book.

As you work through this unit you will record the results of evaluating the integrals for several common engineering shapes on the following page.

Page 30-2Moment of Inertia of Geometric Shapes

Cylinder about its geometric axis.

I z =

Thin disk about a diametric-axis through the mass center.

I y =

Cylinder about a diametric-axis through the mass center.

I x =

Rectangular parallelepiped about an axis parallel to a longitudinal face and passing through the mass center.

I x =

Slender rod of any cross-section about an axis perpendicular to the longitudinal axis and passing through the mass center.

I x =

Sphere about any axis through its mass center.



The mallet shown has a head weighing 8 lb and a handle weighing 2 lb.

1. Determine I A-A for this body in slug-ft2.

2. Determine the moment of inertia of the mallet about an axis parallel to the A-A axis but passing through its mass center.


Page 30-4

Radius of Gyration

The radius of gyration of a mass about a given axis is defined as _____________________

_________________________________________________________________________________

It is calculated from the equation k = __________________________ (30-41)

Problem 30-2

Find the radius of gyration of the body in Problem 30-1 about an axis through the mass center parallel to A-A.

k G = _______________

Find the radius of gyration of the body about A-A.

k A-A = ______________


The roly-poly shown has a mass of 1 kilogram and a radius of gyration of 8 centimeters about its mass center. What is its moment of inertia about an axis through its mass center?


Documents

Notebook - Secrets of Engineeringsecretsofengineering.net/Statics/PDF_files/Statics_Notebook.pdfWhen forces are concurrent (have a common point of application) their resultant will