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Study Guide

Statics/Strength ofConstructionMaterials

INSTRUCTIONS TO STUDENTS 1

LESSON ASSIGNMENTS 5

LESSON 1: INTRODUCTION TO STATICS 7

LESSON 2: STRESS AND STRAIN 13

LESSON 3: PROPERTIES OF LOADS AND CROSS SECTIONS 19

LESSON 4: BEAMS, COLUMNS, AND LOAD SYSTEMS 27

ANSWERS 35

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INTRODUCTION

This course introduces the science of structural analysis. Itexplains how to analyze simple structural components andsystems, and how to evaluate the results of more complexanalysis. When you’ve completed this course you’ll have amuch better idea of why this type of analysis is a critical partof the design process. You’ll also understand when to involveothers in decisions that may impact a structural design.

This part of your program is based on the textbook, A

Structures Primer. It’s divided into nine assignments; eachassignment covers a specific skill critical to understandingstructural analysis and performing basic analysis yourself.The study material for your program consists of

1. Your textbook, A Structures Primer. It contains theassigned readings and supplementary exercises at theend of each chapter. These are assigned as self-checkexercises.

2. This study guide, including an introduction to your pro-gram and presents a summary of the material you’ll coverin each lesson. This guide also contains the followingfeatures:

n An assignments page that lists all of the readingassignments for your textbook

n Introductions to your lessons

n Self-check exercises you should complete as part ofeach assignment

n Answers to the self-check exercises

As you now know, your textbook is covered by the readingassignments in this guide. Your textbook, A Structures Primer,

is the heart of this course. It’s very important that you readthe material in the text and study it until you’re completelyfamiliar with it. This is the material on which your examina-tions will be based. Each chapter begins with a list of learningobjectives. Read through the learning objectives so that you’llknow what to expect when you read through the chapter.

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After you complete each chapter, you can use the list oflearning objectives to review the important points of thechapter.

A STUDY PLAN

We’ve divided the contents of your textbook, A Structures

Primer, into four lessons for you to study. For each lesson,you’ll read part of the textbook. Then, you’ll complete anexamination on the material you read for the lesson. Thisstudy guide contains a list of your lesson assignments. Thelesson examinations are found online. Be sure to read all the

material in both the textbook and this study guide before you

attempt to complete your examinations.

To get the most benefit out of each of your lessons, we sug-gest that you follow these steps:

Step 1: In this study guide, read the introduction to Assign-ment 1. This is the first reading assignment ofLesson 1. Pay attention to the new ideas and con-cepts that are introduced, and carefully note thepages in your textbook where the reading assign-ment begins and ends.

Step 2: Skim the assigned pages in your textbook to get ageneral idea of their contents.

Step 3: Now, read the assigned pages in the textbook. Tryto see the “big picture” of the material during thisfirst reading.

Step 4: Next, go back and study the assigned pages in yourtextbook carefully. Pay careful attention to all details,including the illustrations, charts, and diagramsincluded in the textbook. Take notes on the impor-tant points and terms in a notebook, if you wish.

Instructions to Students2

Step 5: At the end of the reading assignment, review whatyou’ve learned by completing the SupplementaryExercises in the textbook. Write out the answers ona separate piece of paper, if you wish. Try to answerthe questions on your own without looking them upin the textbook. Don’t worry about making a mis-take. The purpose of answering these exercises isto review the material and to help you recognize the areas that you may need to study again. Afteryou’ve done the Supplementary Exercises, checkyour answers with the answers in the back of thisstudy guide to confirm that you did them correctly.If you answered any questions incorrectly, reviewthe material for that topic until you’re sure thatyou understand it. Note that these questions areprovided only for you to review your learning. Youwon’t be graded on them in any way. Do not sendyour Supplementary Exercise answers to the school.

Step 6: Repeat Steps 1 through 5 for each of the remainingreading assignments in the lesson.

Step 7: When you’ve finished reading all the assigned text-book pages for the lesson and you’re sure that you’recomfortable with the material, complete the exami-nation for that lesson. Each examination contains a number of multiple-choice questions. Take yourtime as you complete the examination. You may go back to your textbook to review material at anytime when you’re working on the examination.Remember, your examinations are taken online.Complete your lesson exam as soon as you’reready. Don’t wait to finish another lesson beforecompleting the examination.

Instructions to Students 3

Step 8: Repeat these steps until all four lessons have beencompleted.

Remember, you may ask your instructor for help whenever youneed it. Your instructor can answer your questions, provideadditional information, and provide further explanation ofyour study materials. E-mail your questions to your instruc-tor, and he or she will see to it that you receive the neededinformation.

Now, look over the lesson assignments. Then, begin your studyof A Structures Primer with Lesson 1.

Good luck with your course!

Instructions to Students4

Lesson 1: Introduction to Statics

For: Read in the Read in study guide: the textbook:

Assignment 1 Pages 7–10 Pages 1–25


Examination 498806 Material in Lesson 1

Lesson 2: Stress and Strain





Lesson 3: Properties of Loads and Cross Sections





Lesson 4: Beams, Columns, and Load Systems






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ntsNote: To access and complete any of the examinations for this study

guide, click on the appropriate Take Exam icon on your “My Courses”

page. You shouldn’t have to enter the examination numbers. These

numbers are for reference only if you have reason to contact Student

Services.

NOTES

Lesson Assignments6

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Introduction to StaticsThe first lesson in your course explores how forces act onstructures. Real-world structures like bridges and buildingssupport many different loads applied at many different points.This lesson explains how to evaluate, combine, and determinethe effect of these forces on a simple structure.

Structures are made up of several members. Various types ofjoints connect these members to each other and to the surfaceon which the structure rests. This lesson introduces commonstructural joints and describes the types of loads each of themcan support.

OBJECTIVES

When you complete this lesson, you’ll be able to

n Draw free body diagrams indicating the magnitude anddirection of forces

n Determine the resultant force of concurrent force systems

n Identify types of structural connections and the types ofloads they carry

n Analyze beam reactions

ASSIGNMENT 1Read this introduction to Assignment 1. Then study Chapter 1,“Forces, Moments and Loads” on pages 1–25 in this textbook.

This assignment introduces the forces and moments that acton a structure. You’ll learn several methods that help visual-ize how the structure carries these loads and how to simplifythe representation of multiple forces or moments to more easilycalculate the results. You’ll also learn several graphical meth-ods using drafting tools to determine the resultant forces. Whileengineers and designers may rely on the mathematical meth-ods found in this book, the graphical methods described hereare especially useful for drafters, builders, and construction

Statics/Strength of Construction Materials8

managers who want to quickly check a design. These graphicalmethods allow you to approximate the loading on a particularstructure and to determine whether the design is appropriatefor the desired purpose. Although the initial design is completedby engineers or architects using codes and standards, veryoften changes are requested that require additional decisionsto be made in the field or on the job site. It’s important thatthe responsible people on the job site understand how loads,forces, and moments affect the structure, and have a generalunderstanding of whether the potential changes are reason-able and should be brought to an engineer for further study.

Throughout this course, it’s important to consider both themagnitude (or amount) of a force as well as the direction inwhich it acts. As you calculate resultant forces (the result ofadding several forces together) you’ll need to know whetherthe force is acting in tension (pulling on the structure), or incompression (pressing on the structure). Obviously, a load thatpushes on a structure can’t be “pushed back” by a rope, cable,or chain. The same is true when a rotational force (known asa moment) acts on a structure. The moment can’t be opposedby a joint that uses hinges, for instance, as a hinge won’t stopthe moment from rotating the joint. As this assignment isintended to help you understand how the forces and momentsact, it’ll also help you generally understand why particularstructural components are present at various locations through-out a design.

The methods shown in your textbook determine the resultantforce using ratios (or proportions), trigonometry, and graphicalapproaches. When the forces are all acting on a single point(known as a concurrent force system), graphical methods canbe used to sum the forces. The parallelogram method is usedto sum two forces, and the graphic addition method is usedfor three or more concurrent forces. These methods can beaccomplished with drafting tools to lay out and measurelengths and angles. When laying out a problem using agraphical method, use the largest practical scale so that your answers will be more accurate. For instance, a scale of 1� = 100 lbs would allow you to represent a force equal to725 lbs with a line that was 7.25 inches long. If your scalewas 1� = 10 lbs, you would need to draw a line that was

Lesson 1 9

72.5� long, and if your scale was 1� = 1000 lbs, the 725 lbforce would be represented with a line only .725� long. Thiswould make it more difficult to produce very precise results.

To break forces into their components (forces acting alongperpendicular axes), you’ll use proportions or trigonometry.Trigonometry simply uses relationships between a triangle’sangles, lengths, and the ratio of one side’s length to another.These ratios can be easily calculated using a calculator, providedyou understand a few simple trigonometric relationships likethe ones described in Figure 1.

Don’t think of trigonometry as a theoretical advanced form of math that’s to be feared. Many builders use trigonometrywithout even knowing it. A 4-inch-pitch roof is a roof that

a

Right Angle

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a2+ b 2 = c 2

Trigonometric Relationships

sin =ac

cos =bc

tan =ab

ß = 90 –

b = c 2 – a 2

c = a2 b 2+

= sin -1 ac

= cos -1 bc

= tan -1 ab

a = c 2 – b 2

=

=

=

=

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Angles in a Right Triangle

Length of a Right Triangle’s Sides

FIGURE 1—This trigonome-try chart is a useful toolthat shows how to deter-mine unknown values fromvarious combinations ofknown values.


rises 4 inches for every 12 inches or onefoot of run. Stairs are laid out the sameway. Surveyors use angular degrees to finda location and rise or fall over a given dis-tance to measure the slope of land. All ofthese practices can be expressed as trigo-nometric functions. This chapter of yourtextbook applies these methods through-out, and it’s good for you to become familiarwith each.

When solving the problems, note that sometimes the “horizon” is mentioned, orcompass bearings such as north, south,east, and west. Horizon refers to a horizon-tal line with east at the right hand end of the line, or the three o’clock position.Angles are then listed as decimal degreesand either above or below the horizon,along with the compass bearing (northeast,southeast, southwest, or northwest asshown in Figure 2).

After you’ve read pages 1–25 in the textbook and com-pleted all of the Sample Problems on pages 25–35 and theSupplementary Exercises found on pages 35–38, checkyour answers with those in the back of this study guide.When you understand the material in Assignment 1, moveon to Assignment 2.

ASSIGNMENT 2Read this introduction to Assignment 2. Then study Chapter 2,“Static Equilibrium and Reactions” on pages 39–56 in the textbook.

The previous assignment introduces external forces and how they’re combined into an equivalent resultant force. Yourtextbook now examines the effect of applied loads on a struc-ture. The opposing forces within the structure are referred toas reactions, which in the cases presented in your textbookact equally and opposite in direction to the applied loads.

N

NW

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Horizon

FIGURE 2—Compass bearings and the horizonare sometimes used to indicate angular direc-tions. Similar to the use of o’clock positions toidentify a point of interest, this is a commonway to indicate angles that isn’t dependent onactual compass directions.

Lesson 1 11

Structures from bridges and towers to buildings and tele-phone poles most often exist in a state of static equilibrium.

This means that all of the forces balance out, and that thestructure isn’t moving. Static equilibrium allows us to studya building with static loads such as snow on a roof, peopleon a floor, and the weight of the building itself. The reactionscaused by dancers moving on a dance floor, or wind causinga bridge to sway can’t be determined using the methods ofstatic equilibrium. Also, we’ll ignore the effects of expansionand contraction due to temperature change.

To simplify the solution to a static-equilibrium structureproblem, we’ll need to determine how the ends of the compo-nents in the structure are joined or mounted. The three typesof joints are pinned, roller, and fixed. All three of these arefound in the common structures with which you should befamiliar. A pinned joint, the most common, is similar to abolted or riveted joint that allows the components to pivotaround the fastener. A hinge on a door is an example of apinned joint. This type of joint can only support a force act-ing through the center point or axis of the pin. A pinned jointcan’t resist a moment, just as an open door won’t providemuch resistance to a force that swings the door further openor closed.

A roller joint is similar to someone standing on roller skates.This type of joint only can support a force that’s perpendicu-lar to the surface on which the roller sits. You would expect a skater at the top of a ramp to roll down. Like the pinnedjoint, the roller joint can’t maintain its position on the rampby resisting a moment (which in this case is due to the forceof gravity). A common application of a roller joint is a bridgespan and for mounting long structures on a foundation. Oftenone end of the bridge or structure is fixed to prevent movement,and the other end is placed on a roller to allow the structureto expand and contract with changes in temperature.

The final joint type is a fixed joint. Common types of fixed jointsinclude welded connections in a structure, a column in a con-crete base, or a lamppost or signpost with the bottom boltedrigidly to a concrete foundation. This type of joint can supportforces and moments in all directions.


When you know the type of joint, analyzing the loads on thestructure can be simplified by setting unsupported forces andmoments equal to zero. In some cases this simplificationreduces the number of unknowns, allowing for easier calcula-tion of reactions and moments.

When analyzing reactions in beams, you must also considerthe type of load. Common types include forces acting on asingle point, such as a weight hanging from a rope or chain,and distributed loads acting along a length of the beam, orover an area. For instance, a snow load on a roof is quantifiedas so many pounds per square foot. When analyzing loads onan actual structure, the load over an area is converted into adistributed load along the length of a component such as aroof rafter or floor joist. To convert the load from pounds persquare foot to pounds per foot distributed along a structure,the portion of the load supported by an individual componentmust be calculated. This is done using the tributary width,

which for many structures will be the spacing of the floorjoists or roof rafters. Be careful when making these calcula-tions not to mix units, since sometimes the distance betweencenters of the structure is given in inches, and the length ofthe beam is given in feet. Note that 18� O.C. refers to compo-nents spaced 18 inches apart as measured from the center ofone component to the next. On center is a common construc-tion term used throughout the problems in your textbook.

After you’ve read pages 39–56 and completed the SampleProblems on pages 56–65 and the Supplementary Exercisesfound on pages 65–67, check your answers with those inthe back of this study guide. When you understand thematerial in this lesson, complete the Lesson 1 Examination.

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Stress and StrainStructural analysis relies, in part, on the process of simplifi-cation where a designer represents actual components withsimilar, more ideal ones. One such simplified component isthe link. This lesson focuses on loads carried by links andtrusses, the more complex and commonly occurring struc-tural body made up of several connected links. This lessonalso introduces stress and strain, the basic quantities thatdefine the effect of loads on structural components.

OBJECTIVES


n Evaluate how loads act on links

n Identify different types of trusses based on the structuralfunction they serve

n Analyze the loads on trusses

n Characterize types of stress including tension, compres-sion, shear, and bending

n Describe the relationship between stress, strain, andmodulus of elasticity

ASSIGNMENT 3Read this introduction to Assignment 3. Then study Chapter 3,“Trusses.” on pages 68–83 in the textbook.

Trusses are a special type of structure made up of links pinnedto each other. The ends of the assembly can be fixed, pinned,or placed on rollers to fasten the entire structure to a founda-tion. Remember that when analyzing the effect of loads on astructure, knowing the type of joint allows the unsupportedforces and moments to be set to zero.

Trusses are commonly found in some types of bridges, towers,structures holding signs across a highway, and roof supports.Since the internal joints of a truss are pinned, trusses support


no moments except at an end fixed to a foundation. Also, linksin a truss only support compression or tension forces that actin the direction of their length. This means the direction of theforces in each link is easily determined. Using simple trigono-metric relationships, you’ll use the lengths and angles of thelinks to determine the component forces. This allows you todetermine component forces in the x and y directions. Sincethe truss is assumed to be in static equilibrium, the sum ofthe forces in the x direction will equal zero, and the sum of theforces in the y direction will equal zero. These relationshipshold true for individual links, joints, or the entire structure.

You’ll use one of two methods to solve truss problems—themethod of joints or the method of sections. Both methods willwork for any problem, but you’ll determine which method touse to provide the easiest solution.

When using the method of joints, isolate one joint and deter-mine the individual forces acting on the joint by summingcomponents to zero. The link-force direction is known since itacts along the length of the link. Just as when solving a puz-zle, you’ll want to select a joint that has the fewest unknowns.Then once those forces are calculated, you can move to theother end of the link and start to solve the forces on that joint,continuing this process until you determine all of the linkforces.

The method of sections involves cutting through a structure’slinks with an imaginary line, then only considering a portionof the structure at a time. When using this method, rememberto include both the reaction forces and loads on the portionyou’re analyzing. Also, place link forces on the ends of the cutlinks. Make sure that these link forces are shown parallel to thecut link. Then, sum forces in the x and y directions, as well asmoments about a point to start calculating the unknown forces.If you cut the section at the proper point, forces that act directlythrough the point about which you’re summing moments canbe ignored. To simplify the solution process, it’s important thatyou study the structure before you cut the section. Once youhave some of the link forces solved, move on to others. Youmay even solve portions of the problem using a combinationof both methods.

Lesson 2 15

As you’ve just learned, each link supports no moments andcarries only those forces that act along the direction of thelink. That means a horizontal link can’t support a verticalforce component nor can a vertical link support a horizontalforce. The force in the link can be compression, tension, orzero. If the force is zero, that means that under these specificconditions, that link could be removed without affecting thestructure. Also, if a link supports a tension force, a cable canreplace it. Once you start to visualize how the forces act, you’llbetter understand the design of the structure, recognizing whysome components are sized to withstand higher loads whileother links simply add stability.

After you’ve read pages 68–83 in the textbook and com-pleted the Sample Problems on pages 84–100 and theSupplementary Exercises found on pages 101–103, checkyour answers with those in the back of this study guide.When you understand the material in Assignment 3, moveon to Assignment 4.

ASSIGNMENT 4Read this introduction to Assignment 4. Then study Chapter 4,“Stresses and Strain” on pages 104–114 in the textbook.

In your last assignment, you learned that applied loads causedifferent reactions at various points on the structure andinternal or link forces that vary from link to link and joint tojoint. There are two other values that are critical to fully under-standing the actual strength of a link or joint and the amountthe link deforms under load. These values are referred to asstress and strain.

The stress in a link depends on both the link force and thecross-sectional area of the link. It’s important that a designerselect a structural member that’s strong enough to withstandthe designed loads without selecting one much heavier thanrequired. There are two reasons for this. One is simply an eco-nomic reason—the stronger or larger the member, the more itusually costs. The other reason is that by selecting membersmuch larger and heavier than required, the designer is puttingadditional loads on the structure due to the weight of the over-sized components. This requires other components be designedstronger to handle the additional weight.


This assignment introduces three basic types of stress. Thebasic stresses are compression, tension, and shear. Compressionand tension are known as normal stresses since they act per-pendicular to (in this case) the link’s cross-sectional area.Compression stress and tension stress can be easily definedas the link force in compression or tension divided by the cross-sectional area of the link.

Compression stress would be commonly occur in a columnsupporting a parking deck. The column would be subject to acompression stress due to the weight of the deck (Figure 3). Asthe cross-sectional area of the column increases (larger-dia-meter pipe) while the weight resting on it remains unchanged,

FIGURE 3—This columnsupports the weight of theparking deck above. Theweight of the deck actsalong the center of thecolumn, loading it in compression. The metalcorrugated sheet is usedas a form to pour the con-crete floor above, and isleft in place. Some meth-ods of construction woulduse plywood and bracing,which would be removedafter the concrete cured.

Lesson 2 17

the stress decreases. Since the column is encased in a con-crete footer in the ground, the base of the column could alsobe subject to a moment. This will induce an additional stressdiscussed later in your course.

A cable or guy wire supporting a telephone pole experiencestension stress equal to the force in the cable divided by itscross-sectional area. If a smaller-diameter cable were used,the lower cross-sectional area would result in a higher ten-sion stress. Remember that the cable can support only atension load, acting in the direction of its length. The cablecan’t support a moment or compressive force.

Shear stress is the stress that acts parallel to the link’s cross-sectional area, or perpendicular to its centerline. Shear stresscan be thought of as the resistance of two surfaces of a compo-nent to slide across each other. A link pin or a bolt supportinga weight like the one shown in Figure 4 is subject to shearstress. A failure due to the shear stress causes the componentto shear or tear apart, with the faces of the failed cross sectionsliding across each other.

FIGURE 4—The bolt and U-bracket in this assembly are subject to shear forces. If they faildue to an excessive load, the bolt or bracket will shear—tearing the component apart. Thebending moment, about which you’ll soon learn more, acting on this “S” link is starting tobend it open.


Bending stress, a combination of both tension and compres-sion stresses, results from the bending of a link. This usuallyoccurs in structures when a beam is loaded so that it deflects.When loaded from the top, the lower side of a horizontal beamelongates as a result of tension stress, while the upper part ofthe beam contracts due to compressive stress. This conditionis well illustrated by Figure 5-1 in your textbook.

Throughout this assignment your textbook assumes thatstructures are rigid and don’t change shape (except whendiscussing bending stresses). In real life, compression stresscauses a component to shrink or shorten while tension stresscauses a component to elongate. The amount the componentshrinks or elongates is measured as a change in length. Thischange in length divided by the original length is referred toas strain. Strain is a dimensionless value, meaning that boththe change in length and the component’s length must bemeasured in the same units. A change in length of .03 inchesover 12 feet must be calculated as .03 inches / (12 feet � 12inches/foot) = .03 inches / 144 inches = 2.08 � 10–4 inches.

Modulus of elasticity (E) is a material property that’s criticalto the analysis of structures. This value, which measures thematerial’s stiffness, is calculated by dividing the stress in amember by the strain. This value is constant over a range ofstress and strain that defines the material’s elastic range. Whenloaded in its elastic range, a structural component returns toits original shape after the load is removed. At loads abovethe elastic range, the component remains deformed after loadremoval. Loading of this type is within the plastic range, andresults in permanent damage being done to the component.When loaded in the plastic range the component may not failcompletely; however, damage is present and additional loadswill cause more damage. Later in your studies, you’ll rely ona material’s modulus of elasticity when determining the sizeand shape of a structural component required to handle agiven load.

After you have read pages 104–114 in the textbook andcompleted the Sample Problems on pages 114–117 andthe Supplementary Exercises found on page 117, checkyour answers with those in the back of this study guide.When you understand the material in this lesson, com-plete the Lesson 2 Examination.

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Properties of Loads andCross SectionsThis lesson introduces three tools used throughout all kindsof structural analysis work. These tools include diagrams thatrepresent in a clear and visual manner the loads acting on astructural component like a beam. You’ll also learn to repre-sent various shapes by identifying their centroid, the singlepoint that’s treated as the “center” of a two-dimensional shape,and moment of inertia. As you’ll learn, a shape’s moment ofinertia represents, among other things, how well the shaperesists the bending that results from some types of loads.

OBJECTIVES


n Represent loads, including shear and moments, on aload diagram

n Define the terms centroid and moment of inertia andexplain why they’re useful

n Locate the centroid of various two-dimensional objects

n Calculate the moment of inertia for various cross sec-tional shapes

ASSIGNMENT 5Read this introduction to Assignment 5. Then study Chapter 5,“Load, Shear, and Moment Diagrams” on pages 118–135 in thetextbook.

You’ve already learned how graphical methods help visualizeand solve force problems. Similar techniques allow you todetermine variations in load, shear, and moment along thelength of a beam. First, consider all of the external loads acting on the beam and calculate the reaction forces andmoments. To accomplish this you’ll start by summing theforces in each direction (x and y) and setting them equal tozero. Also, sum the moments about a point and set them


equal to zero. Use the type of end conditions (roller, pin, orfixed) to reduce the number of unknowns, then solve theproblem.

By studying the sample problems in your textbook, you shoulddevelop an ability to anticipate how the diagrams will look foreach type of load and end condition. For instance, notice howthe shear diagram includes a rectangular section when a loadis placed at a single point on the beam. This rectangular sheardiagram translates to a triangular section when the same por-tion of the beam is represented on the moment diagram. Thepeak or height of the triangle is equal to the length of the cor-responding rectangular section of the shear diagram multipliedby the height (or load value) of the rectangular section. Referto Figure 5-16 in your textbook for a graphical illustration ofthis condition. In this figure, the 200-pound shear at point Aoccurs along the 3-foot length from point A to the center ofthe beam. To find the peak height of the moment diagram’striangle:

200 lbs � 3 ft = 600 ft-lbs

Note that since the beam in Figure 5-16 is simply supported(placed on rollers at both points A and B), the beam ends can’tsupport a moment. The moment in the beam is introduced bythe deflection of the beam, causing it to bow down in the cen-ter. Remember, from your study of stress and strain, that thiscauses bending stress in the horizontal beam, which results incompression loading of the beam’s upper half and tension load-ing of the lower half. The bending stress in this case will be ata maximum value in the center of the beam’s length.

It’s helpful to draw the load, shear, and moment diagrams ina consistent order. By consistently placing the load diagramon top, the shear diagram in the middle, and the momentdiagram on the bottom, for instance, you’ll be better able touse the same method of visualizing each problem. It will alsohelp others who might work on or evaluate your analysis.

So far, all of the beam examples in your textbook were symmet-rically loaded about the center of the beam and supported atboth ends. However, this isn’t always the case. You’ll solvecases that aren’t loaded or supported symmetrically using thesame methods that you used for solving the earlier problems.Start with the load diagram (solve for the reaction forces and

Lesson 3 21

moments), and then create the diagrams for each section ofthe load or beam, working from one end to another. If a beamis supported with a portion of it overhanging, the moment iszero at some point along the beam’s length. This point isreferred to as the inflection point and can be approximatedgraphically by plotting known points on the moment diagramand then sketching the curve.

After you’ve read pages 118–135 in the textbook and com-pleted the Sample Problems on pages 135–150, and theSupplementary Exercises found on pages 150–152, checkyour answers with those in the back of this study guide.When you understand the material in Assignment 5, moveon to Assignment 6.

ASSIGNMENT 6Read this introduction to Assignment 6. Then study Chapter 6,“Properties of a Cross Section” on pages 153–167 in the textbook.

Earlier in this course, you learned that a beam or structure’sperformance depended on the loads acting on it, the propertiesof the material from which it’s made, and its cross-sectionalarea. Remember, a beam’s cross-sectional area is the area ofthe surface created by slicing across the beam. A pipe has anannular cross section (two concentric circles), an I-beam hasa shape like the letter “I,” and a piece of angle iron has an L-shaped cross section. Each of these shapes reacts differentlyto loads. In this chapter, you’ll study several properties ofvarious beam cross sections.

While you may not yet be familiar with centroids and momentsof inertia, you’ll soon calculate these values for simple shapes.Be sure to refer to Tables 6-1 and 6-4 and Appendix A. Thesecontain formulas for common shapes and calculated values forvarious rectangular sizes.

The centroid or center of area is a term that describes thesingle point at which all of the area of a cross section can berepresented. From this location we can also calculate two othervalues, x and y. x is the distance to the centroid from the yaxis you choose as a reference, and y is the distance to thecentroid from the reference x axis. Refer to Table 6-1 and Figure 6-1 in your textbook. This concept is especially useful


when combing several simple shapes to locate the centroid ofa more complex shape. To locate the centroid of a complexshape, break it into simple shapes whose areas and centroidsare given in Table 6-1. Then use the following formulas

Where

x and y are the centroid of the composite section, at which thecomplex section composed of two or more simple sections

ΣA is the sum of the areas of all of the simple sections

ΣAx is the sum of each individual area multiplied by the dis-tance (in the x direction) from the simple section’s centroid tothe reference y axis

and ΣAy is the sum of each individual area multiplied by thedistance (in the y direction) from the simple section’s centroidto the reference x axis

With experience, you’ll learn how to quickly break more complex shapes into simpler ones (such as rectangles, cir-cles, and triangles). Also, you’ll learn to sometimes subtractan area out of the composite. For instance if the compositeshape includes a hole, you can calculate the values of theentire composite, and then subtract the area of the hole.Often, several methods can be used to solve the same prob-lem. Different solutions depend on the shapes you chooseand whether you add or subtract areas. Choose the shapesand methods that make it easier for you. For instance, theexample shown in Figures 6-3 and 6-4 was broken into threepieces, the center rectangle, and two triangles. You mightinstead choose to create a rectangle equal to the outsidedimensions of the shape (24 � 24) and then subtract a trian-gular void from each side.

A cross section’s moment of inertia measures its stiffness andstrength. This quantity varies based on how the area is dis-tributed about the neutral axis. The cross section’s neutralaxis passes through its centroid and is the axis about whichthe part bends. As shown in Figure 6-10 of your textbook,

yAy

A= Σ

Σ

xAx

A= Σ

Σ

Lesson 3 23

the moment of inertia differs depending on whether the beamsits on its short or long face. This makes sense to someonewho’s walked across a wooden plank that deflected or bowed.The plank is much stiffer and better resists deflection whenplaced so that the load acts on its short face. This is whyfloor joists, roof rafters, and trusses are built so that thebeams stand up, rather than lie flat. This configuration sup-ports the load better and provides a stiffer structure. Figure 5shows floor joists placed along a beam. Note that both thefloor joists and beam are placed to take advantage of the highermoment of inertia.

FIGURE 5—Floor joists and beams are oriented so the load is placed on the short face. Theresulting higher moment of inertia allows the structure to be stiffer than if the load wasapplied to the rectangular cross section’s longer face. Cross braces have been added to provide lateral support.


As you study the formula in your textbook, notice that a rectangular section’s moment of inertia includes the heightcubed. This means that if you double the height, the momentof inertia will increase by a factor of eight (two cubed). For acircular section (such as a rod or pipe), the formula includesthe radius raised to the fourth power. This means that if youdouble the radius, the moment of inertia will increase by 16(two to the fourth power). Small changes in size or orientationresult in large differences in the moment of inertia.

When adding the moments of inertia for simple shapes to calculate the moment of inertia of a more complex one, youcan’t always simply add or subtract the moments of inertiafor each individual shape. To calculate the overall moment ofinertia of a complex shape, calculate the moments of inertiaof individual simple shapes, and then add to each the valueof that section’s area times the distance from its centroid tothe neutral axis squared. First, the neutral axis is locatedusing the method described earlier. For example, to calculatethe distance between the neutral axis and the reference xaxis use

Locate the neutral axis and calculate the distances betweenthe centroids of the simple shapes and this axis. Then findthe individual cross-sectional area, and the moment of inertiafor each simple shape. Finally, for each simple shape calculatea value equal to the moment of inertia plus the cross-sectionalarea multiplied by the distance from its centroid to the neutralaxis squared (I + Ay2). The overall moment of inertia will bethe sum of these values. To help keep all of your calculationsin order, sketch the composite shape, identify the individualshapes, locate and place the neutral axis, and then fill outtables like those in your textbook. One important note—if youcalculate cross section and moment of inertia values for com-mon structural members such as I-beams, angles, rectangulartubing, don’t expect to get the exact same answer as thosefound in a table or chart for structural members. This isbecause you’ll simplify the geometry by assuming sharp cor-ners, while the table or chart will take into account the actualshape of the member, including radii and fillets.

yAy

A= Σ

Σ

Lesson 3 25

After you’ve read pages 153–167 in the textbook andcompleted the Sample Problems on pages 167–175 andthe Supplementary Exercises found on pages 175–178,check your answers with those in the back of this studyguide. When you understand the material in this lesson,complete the Lesson 3 Examination.


NOTES

Beams, Columns, andLoad SystemsThe final lesson in your course continues to examine thebehavior of beams under load. In this case you’ll determinehow much a beam will deflect when subjected to a definedload and end conditions. The lesson also introduces a finalsimplified structural component, the column. Columns arevertical components loaded in compression. Their reaction to loads depends on both their cross sectional area and length.

The lesson concludes by introducing the analysis of structuralsystems. These systems come much closer to resemblingreal-world structures in that they’re made up of two or moresimplified components.

OBJECTIVES


n Evaluate beam loading and cross sections to determineshear and bending stress

n Determine beam deflection based on loading and cross-sectional area

n Describe and calculate the lateral deflection in a beam

n Analyze loads and resulting stresses in columns

n Analyze loads and resulting stresses in structural systems

ASSIGNMENT 7Read this introduction to Assignment 7. Then study Chapter 7,“Beam Analysis and Design” on pages 179–196 in the textbook.

This chapter ties together all of the previous chapters, usingwhat you’ve learned to analyze and design beams. Remember,beams are load-carrying members that lie horizontally. A beam’send conditions determine how it reacts to loads. For our pur-poses, we’ll assume that all of the reactions and deflections actin the plane represented by the figure drawn on the paper,

27

Le

ss

on

4L

es

so

n 4


and that there are no lateral deflections attempting to flex thebeam into or out of the paper. In buildings the floor decking andother surface materials help prevent lateral deflection, whilecross braces help stabilize the beams. Refer back to Figure 5to view the cross braces in a representative floor system.

Remember that in this assignment you’ll ultimately calculatestress, which represents a force acting on an area. The firststep in analyzing stresses in a beam involves summing forcesand moments. At this step, any distributed loads can bereplaced with a force acting at a single point. You’ll next create the load, shear, and moment diagrams. Make sure tomark load values, shear values, locations where the curveschange direction or cross zero, and peaks.

This assignment introduces an important new term, section

modulus. In beams with rectangular cross sections, the sec-tion modulus is

where b = base of the rectangle, and h = height

The section modulus and the moment diagram’s maximummoment determine the beam’s bending stress. This stress isgreatest at the outer faces of the beam, with the maximumtension bending stress typically found on the bottom face andthe maximum compression stress is its top face.

where M is the maximum moment from the moment diagram.

If the areas above and below the cross section’s neutral axisaren’t equal (the beam’s cross section isn’t symmetric acrossthe neutral axis), then the tension and compression stressesand forces at the beam’s outer surfaces won’t be equal. Theresulting compression and tension forces that create thesestresses form a couple, which creates a moment. If the sec-tion isn’t symmetric about the neutral axis (which is knownas asymmetrical), then you’ll calculate two values of the

f bendingstressM

Sb = =

Sbh=

2

6

Lesson 4 29

distance from the neutral axis to the extreme outside faces—one for tension and one for compression. You’ll then use aslightly different formula for bending stress:

where fb = the bending stress (tension or compression)

M = maximum moment calculated from the moment diagram

c = distance from the neutral axis to the extreme outer face(tension or compression)

I = moment of inertia about the same axis the beam bendsaround

With an asymmetric beam, M and I will be the same for boththe tension and compression bending stress calculation whilethe value of c varies depending on where the neutral axis posi-tion in relation to each face.

Like bending stress, shear stresses are also distributed overthe face of the cross section. The calculation for shear stress(fv ) is

where V = shear found from the shear diagram

I = the moment of inertia

b = the width of the shear plane being considered

Q = the moment about the neutral axis of the area above theshear. Q = A � y.

Figure 7-16 in your textbook shows how shear stress varies atdifferent planes, and shows how to calculate the values. Theshaded areas (A) rest above the shear plane, and y representsthe distance from the neutral axis to the centroid of area A.

This assignment ends with a discussion of beam deflection,which represents the distance the beam moves or flexes fromits unloaded or free state. Structural designs often depend ona limited amount of deflection. In these cases it’s important

fM c

Ib =

fQV

I bv =


that a beam doesn’t deflect more than intended. Deflectioncalculations make use of many of the values you’ve alreadystudied, as well as “rules of thumb,” including:

Roofs: Total load deflection should be less than the span of the roof divided by 180 while live load (snow) deflectionshould be less than the span of the roof divided by 240.

Floors: Total load deflection should be less than the span ofthe floor divided by 240 while live load deflection should beless than the span divided by 360.

Deflection guidelines are found in Table 7-1 of your textbook.Be very careful when using tables because they’re valid onlyfor the specific load type and end conditions shown. Also notethat equations in Table 7-1 include the constant 1,728, whichconverts from the standard inches and feet normally used forcross sections or beam sizes and lengths. When working withthese equations, use the standard inch dimensions for cross-sectional area and beam sizes and feet for the length values.The mathematical conversion of these units is already part ofthe formula. To use this table with the SI (metric) units of mil-limeters and meters, replace the value 1,728 with 103. Payspecial attention to the approximation methods shown onpages 193 to 196 of your textbook, and the special caseswhere these methods apply.

After you’ve read pages 179–196 in the textbook andcompleted the Sample Problems on pages 197–204 andthe Supplementary Exercises found on pages 204–206,check your answers with those in the back of this studyguide. When you understand the material in Assignment7, go on to Assignment 8.

ASSIGNMENT 8Read this introduction to Assignment 8. Then study Chapter 8,“Columns” on pages 207–213 in the textbook.

As you’ve learned, beams lie horizontally and support loadsthat cause deflection. This assignment introduces anothercommon structural member known as a post or column.Columns end conditions are either pinned, roller, or fixed.However, columns are loaded parallel to their length and tend

Lesson 4 31

to compress. Column length determines how it fails when over-loaded. Columns short enough to fail by crushing are referredto as short columns. A long column is one that fails by buckling,or lateral deflection. This deflection isn’t due to a lateral load,but is the result of a load that flexes the long column to theside. There’s no particular length that defines a short or longcolumn. The length is determined by a combination of factorsincluding material properties and the column’s cross section.

If the load isn’t too great, buckling in a long column will beelastic. This means that when the load is released, the col-umn returns to its original shape. However, if the load is toogreat the column remains deformed after the load is removed.It’s critical to understand that as the column starts to buckle,the geometry of the column changes to include a bow. In thiscase the compressive force no longer acts through the centerof the column, and instead causes both compression and amoment. This quickly magnifies the effect of the load on thecolumn, making it dramatic failure possible, even without anyfurther loading.

Other terms that are used for long columns include the radiusof gyration (r) and the slenderness ratio. The radius of gyration

measures the resistance to buckling and is calculated as thesquare root of the moment of inertia about a given axis dividedby the cross-sectional area. Note that columns made from cer-tain shapes such as rectangular posts have a different radiusof gyration in the x and y axis. One of these axes will be moresusceptible to buckling. Lateral bracing can be used on thesesides to prevent the column from buckling as quickly. Addingbraces to a long column makes the column behave more liketwo shorter columns.

A column’s slenderness ratio compares its length to its radiusof gyration. The higher the slenderness ratio, the lower thecritical load at which it buckles. A column’s effective length,

which is used in the buckling equation, depends on the typeof end supports. The equations were developed using a col-umn pinned at both ends. In practice this isn’t always thecase. If the column is pinned at one end and fixed at theother, the column’s effective length is only 70% of the actuallength. If the column is fixed at both ends, then its effectivelength drops to 50% of its actual length. This should makesense, since a column that’s fixed at one or both ends would


be restricted from buckling as easily as one that was free to pivot at the ends. It’s important to remember that a smallchange in length can quickly decrease the load that a columncan support without buckling. Also, remember to use the col-umn’s weaker axis when calculating critical loads and buckling.

After you have read pages 207–213 in the textbook andcompleted the Sample Problems on pages 213–216 andthe Supplementary Exercises found on pages 216–217,check your answers with those in the back of this studyguide. When you understand the material in Assignment8, go on to Assignment 9.

ASSIGNMENT 9Read this introduction to Assignment 9. Then study Chapter 9,“Systems and Loads” on pages 218–227 in the textbook.

The final assignment considers beams, columns, and otherstructural components as a load-carrying system. You’ll firstreview the types of loads placed on the structures. Most loadson roofs and floors are distributed, such as snow on a roof ormany people standing on a floor. In fact, the weight of thestructure is a distributed load. Concentrated loads couldinclude air-conditioning units on the roof and pieces ofequipment placed on a floor.

Always-present loads (like the weight of the structure) arereferred to as dead loads. Live loads are those not alwayspresent such as snow in the winter, or a room that’s occu-pied only part of a day. When you evaluate a design, makesure it accommodates the total load that’s equal to the sumof both the live and dead loads.

Often loads are stated as pounds per square foot. You’ll usually simplify that as a distributed load on one particularcomponent such as a floor joist. To do this, convert the loadper area to a distributed load along the length of the joist.This was previously accomplished by dividing the load per

Lesson 4 33

area by the tributary width, or the width between components.A typical tributary width would be two feet when floor joistsare spaced two feet on center. Remember that if the tributarywidth is given in another set of units (such as 24� O.C.), thenyou’ll need to convert that value to the same units you’re usingfor the rest of the equation.

Once you’ve distributed the load onto a joist, calculate thereactions and moments. You can then check the size of theselected material to see if it meets the strength requirementsof the design. Be careful that the final load isn’t too close tothe maximum allowed. Factors of safety are always used toreduce the allowed loads to a point well below the maximumload the structure can carry. This technique accommodatesvariations in materials, and covers assumptions made to sim-plify the calculations. For instance, a standard 2 � 10 pieceof lumber used for floor joists is actually undersized and willhave a lower cross-sectional area than the assumed 20 squareinches. With the calculations completed, you can now trans-fer the reaction forces to the support beams and analyze theirdesign. Finally, transfer the reactions from the beam to thecolumns and analyze their design.

Your assignment ends with some valuable terms found inFigure 9-6. Almost all structural components analyzed in thiscourse were either horizontal or vertical. However, roofs andother building components are often pitched like the slopedroof in Figure 9-6. Note that the dead load, or weight of theroof system, is calculated along the entire length of the roofrafters. From our earlier discussions of triangles, you’ll needto calculate the length of the roof rafter from the pitch of theroof and the total width of the house. On the other hand, liveloads like snow loads are calculated on only the horizontalprojection of the rafters. In the example shown, the length ofthe rafter is 7.81 feet, while the horizontal projection is only 6 feet. Figure 6 shows a typical roof frame system. Note thatthe truss members in this system help support the load.


After you’ve read pages 218–227 in the textbook andcompleted the Sample Problems on pages 227–233 andthe Supplementary Exercises found on pages 233–234,check your answers with those in the back of this studyguide. When you understand the material from this les-son, complete the Lesson 4 Examination.

FIGURE 6—A typical roof system is shown with the rafters, roof, and truss members.

Chapter 1

E1-1. Determine the components (Fx and Fy ) of the forcesshown below.

c a b

F F F

FF

y x

y

= + = 7 + 12 = 13.89

13.89 =

7 =

12

= (7) 13.

2 2 2 2

889 =

(7) 2.4 kN13.89

= 1.21 kN

=13.89

= (12) 2.4 (12)

Fx

F kkN13.89

= 2.07 kN

F

F

x

y

= (cos 30 )= 100 lbs cos 30 = 86.60 lbs

= (sin 30 )

o

o

o

F

F

( )

== 100 lbs sin 30 = 50.0 lbso( )

35

Answers

Answers

E1-2. Using the parallelogram method, determine the magnitude, direction, andsense of the resultant of the concurrent force system shown below.

The forces F1 and F2 are drawn at their designated direction and magnitudeusing a scale of 1 inch = 40 lbs

Answers36

E1-3. Using the parallelogram method, determine the magnitude, direction, andsense of the resultant of the concurrent force system shown below.

Use a scale of 1 inch = 20 kN

Answers 37

E1-4. Determine the magnitude, direction, and sense of the resultant of the concurrent force system shown below using the graphic method.

Use a scale of 1 inch = 50 lbs

Answers38

E1-5. Determine the magnitude, direction, and sense of the resultant of the concurrent force system shown below using the graphic method.

Use a scale of 1 inch = 20 kN

Answers 39

E1-6. Verify your answer to Problem 1.2 using the component method of determin-ing the resultant of a concurrent force system.

c a b = + = 4 + 12 =

4 =

12 =

12.65 =

253

2 2 2 2 12.65F F F1y 1x 1 llbs

12.65 = 20 lbs

= 4 20 lbs = 80 lbs

= 12 20 lbs = 24

F

F

1y

1x

( )( ) 00 lbs

Σ − −ΣF F F

F F Fy 1y 2y

x 1x 2x

= + = 80 lbs 100 lbs = 20 lbs= + = 240 lbs + 100 lbs = 340 lbs

F

F

2y

2x

= 141.4 lbs sin 45 = 100 lbs

= 141.4 lbs cos 45 = 100

o

o

( )( ) lbs

R = + = 20 lbs + 340 lbs = 340.6 lbs

= tan2

2 2 2 2

1

Σ Σ −

−

F Fy x ( ) ( )θ

00 lbs340 lbs

= 3.4 from the horizon, Southeasto⎛⎝⎜

⎞⎠⎟

Answers40

E1-7. Using the component method, verify your answer to Problem 1-4.

R = + = 31.17 lbs + 287 lbs = 288.82 lbs

=

2 2 2 2Σ Σ − −F Fy x ( ) ( )

θ ttan31.17 lbs287.13 lbs

= 6.20 below the horiz1 o− ΣΣ

F

F

y

x

= oon, Southwest

F

F

2y

2x

= 86.6 lbs sin 30 = 43.3 lbs

= 86.6 lbs cos 30 = 75

o

o

( )( )

−llbs

F

F

1y

1x

= 300 lbs sin 45 = 212.13 lbs

= 300 lbs cos 45 = 212

o

o

( )( ) − ..13 lbs

Σ − − −Σ −F

Fy

x

= 212.13 lbs 43.3 lbs 200 lbs = 31.17 lbs= 212.13 lbs + 75 lbs 150 lbs = 287.13 lbs− −

c a b = + = 4 + 3 =

4 =

3 =

5 =

250 lbs5

= 50

2 2 2 2 5F F F3y 3x 3 llbs

= 4 50 lbs = 200 lbs

= 3 50 lbs = 150 lbs

F

F

3y

3x

( )( )

−−

Answers 41

E1-8. Using the component method, verify your answer to Problem 1-5.

E1.9 Determine the resultant of the parallel force system shown in the figurebelow. Note that since all forces point down, identifying this as the positivedirection simplifies the calculations.

R = 2 kN + 1 kN + 3 kN = 6 kN= 2 kN(1 m) + 1 kN(3.5 m) + 3 ∑ M A kkN(4.5 m) = 19.0 kN-m

=R(x) MA∑

F F

F F

1y 1

1x 1

= (sin 60 ) = 50 kN (sin60 ) = 43.30 kN

= (cos

o o

660 ) = 50 kN (cos 60 ) = 25 kNo o −

∑∑

F

Fy

x

= 43.30 kN 75 kN = 31.70 kN= 25 kN +100 kN = 75 kN

− −−

R F F

F

y x = = ( 31.70 kN) 75 kN) = 81.42 kN

= tan

2 2 2 2

1

∑ + ∑ +∑

−

−

(

θ yy

xF∑=

31.70 kN75 kN

= 22.91 from the horizon, Southeasto

Answers42

E1-10. Determine the moments of the forces about the points A in the figure below.

E1-11. Determine the moments of the forces about the points B in the figure below.

∑ −MB = 12 kip-ft + 2 kips(18 ft) = 24 kip-ft

F

My

A

= 12(20 lbs) = 240 lbs= 240 lbs(8 ft) 160 lbs(16 ft) ∑ − − == 4,480.00 ft-lbs−

6 kN( ) = 19 kN-m

= 19 kN-m

6 kN = 3.17 m

x

x

c

F Fy

= + = 12 + 5 = 13

12 =

13 =

260 lbs13

= 20 lbs

2 2 2 2a b

Answers 43

Answers44

Chapter 2

Determine the components of the reactions for the beams in Problems 2.1through 2.10 shown below.

E2.1.

E2.2.

Σ − −↑

F R

R

R

y Ay

Ay

Ax

= 0 = 600 lbs 300 lbs + 550 lbs +

= 350 lbs = 0

Σ − −M R

R

A B

B

= 0 = 600 lbs (9 ft) 300 lbs (15 ft) + (18 ft)

=5,4000 ft-lbs + 4,500 ft-lbs

18 ft= 550 lbs ↑

c b

F F F

F

y x

y

= + = 4 +3 = 5

4 =

3 =

5 =

50 kN5

= 10 kN

= 4

2 2 2 2a

(10 kN) = 40 kN= 3 (10 kN) = 30 kNFx

E2.3.

E2.4.

E2.5.

Σ −↑

F R

R

y A

A

= 0 = 4,400 lbs + 2,200 lbs +

= 2,200 lbs

W L

ML

A

= = 220 lbs/ft (20 ft) = 4,400 lbs = 0 = 4,400 lbω

Σ − ss (10 ft) + (20 ft)

=44,000 ft-lbs

20 ft = 2,200 lbs

R

R

By

By

= 0

↑

RB x

Σ − − −F R

R

y B

B

= 0 = 4 kips 8 kips 8 kips + 13.33 kips +

= 6.667 kips ↑

ΣMB = = 4 kips (20 ft) + 8 kips (10 ft) + 8 kips (5 ft) 0 −− (15 ft)

=80 kip-ft + 80 kip-ft + 40 kip-ft

15 ft= 13.3

R

R

A

A 33 kips ↑

Σ − −M R

R

A B

B

= 0 = 40 kN(1 m) 60 kN(4 m) + (8 m)

=40 kN-m + 2440 kN-m

8 m = 35 kN

= 0 = 40 kN 60 kN + 35 kN +

↑

Σ − −F Ry AAy

R

F R

R

Ay

x Ax

Ax

= 65 kN = 0 = 30 kN += 30 kN

↑Σ −

→

Answers 45

E2.6.

Σ − −↑

F R

R

R

y Ay

Ay

Ax

= 0 = 60 kips + 36 kips

= 24 kips = 0

W L

ML

A

= = 2.5 kips/ft (24 ft) = 60 kips= 0 = 60 kips (1ω

Σ − 22 ft) + (20 ft)

= 720 kip-ft

20 ft = 36 kips

R

R

B

B ↑

W LL

M A

= = 6 kN/m (4 m) = 24 kN

24= 0 = 36 kN (1.5 m) ω

Σ − − kN (5 m) + (7 m)

= 54 kN-m + 120 kN-m

7 m = 24.86 k

R

R

B

B NN ↑

Σ − −↑

F R

R

R

y Ay

Ay

A

= 0 = 36 kN 24kN + 24.86kN +

= 35.14 kN

xx = 0

Answers46

E2.7

E2.8.

W L

ML

A

= = 2 kip/ft (18 ft) = 36 kips= 0 = 36 kips (9 fω

Σ − tt) 4 kips (30 ft) + (24ft)

= 324 kip-ft + 120 kip-f

− R

R

B

B

tt24 ft

= 18.5 kips ↑

Σ − −F R

R

y Ay

Ay

= 0 = 36 kips 4 kips + 18.5 kips +

= 21.5 kips = 0

↑RAx

WL

R R

L

Ay B

= 2

= 26 kip/ft (18 ft)

2 = 234 kips

= = 234 kips

ω

22 = 117 kips

= 0

↑

RAx

Answers 47

E2.9

E2.10.

WL

W L

1L

2 L

= 2

= 600 lbs/ft (9 ft)

2 = 2,700 lbs

= = 600 lb

ω

ω ss (18 ft) = 10,800 lbs

WL

M

L

A

= 2

= 36 kN/m (9 m)

2 = 162 kN

= 0 = 162 kN (6 m) +

ω

Σ − RR

R

By

By

(9 m)

=972 kN-m

9 m = 108 kN ↑

R

F R

R

Bx

y A

A

= 0= 0 = 162 kN + 108 kN +

= 54 kN

Σ −↑

Answers48

Chapter 3

E3-1. Using the method of joints, determine the link force in all of the membersshown in the truss below. Indicate whether the members are in compressionor tension.

Σ − −M RA By= 0 = 2,700 lbs (6 ft) 10,800 lbs (18 ft) + (27 ftt)

=16,000 ft-lbs + 194,000 ft-lbs

27 ft = 7,800 lbs RBy ↑

R

F R

R

Bx

y A

A

= 0= 0 = 2,700 lbs 10,800 lbs + 7,800 lbs +

=

Σ − − 5,700 lbs ↑

Answers 49

E3-2. Determine the link forces in all the members of the truss shown below usingthe method of joints. Indicate whether the members are in compression ortension.

NOTE: Since the configuration of the members and the loads are symmetri-cal, only the left half of the truss is shown above. The right half is amirror image of the left half.

Answers50

E3-3. Using the method of joints, determine the link force in all of the membersshown in the truss elow. Indicate whether the members are in compressionor tension.

LEFT HALF

Answers 51

RIGHT HALF

Answers52

E3-4. Using the method of joints, determine the link force in all of the membersshown in the truss below. Indicate whether the members are in compressionor tension.

LEFT HALF

Answers 53

RIGHT HALF

Answers54

E3.5. Verify your answers to Problem 3-4 by determining the stress in CD, IH, andDI by the method of sections.

• Cut a section through the members CD, IH, and DI.

• Move the link force CD along its line of action to the Joint C.

• Convert CD into components.

• Move the link force DI along its line of action to the Joint I.

• Convert DI into component forces

• The line of action of CDy, IH, DIy, and DIx extend through the point I leavingCDx as the only component link force to consider when taking moments.

• Take moments about the point I.

Answers 55

Since the only other member that can support a vertical component is mem-ber DI, we can determine the link force by summing forces in the y direction.

The one remaining link force is IH, which can be obtained by summingforces in the x direction.

∑ −M CD

CD

I x

x

= 0 = ( 4 kips + 1 kip) (9 ft) + (9 ft)

= 27 kiip-ft

9 ft = 3 kips

CD CD

CD

y x

y

3 =

9 =

3 kips9

= 0.33 kips

= (3) (0.33 kips) = 1 kkip

= (3 kips) + (1 kip) = 3.16 kips C2 2CD

Σ − −F DI

DIy y

y

= 0 = 1 kip 2 kips + 4 kips + 1 kips + = 2 kip

DI DI

DI

x y

x

9 =

6 =

2 kips6

= 0.33 kips

= 9 (0.33 kip) = 3 kipps

= (2 kips) + (3 kip) = 3.16 kips C2 2DI

Σ − −F IH

IHx = 0 = 3 kips 3 kips + = 6 kips T

Answers56

E3-6. Using the Method of Sections, determine the stresses in members AB, AI,

and IJ. Using a second section, calculate the stresses in CD, CG, and GH.

Σ −F R

R

y Ay

Ay

= 0 = 212.1 kN +

= 212.1 kN ↑

RJ y = 212.1 kN-m + 424.2 kN-m + 636.3 kN-m + 848.4 kN-m

10 m = 212 kN ↓

RJx = 0 = 0 = 21.21 kN (10 m) + 21.21 kN (20 m) + 21.21 ∑ M A kkN (30 m)

+ 21.21 kN (40 m) 10 m)− RJ y (

Answers 57

Cut a section through AB, AI, and JI.

IJ

F AI

AIx x

x

= 212.1 kips T = 0 = 84.84 kN = 84.84 k

∑ −NN = AIy

AI AI AIy x = + = (84.84 kN) + (84.84 kN) = 120 kN C2 2 2 2

∑ FF AB

ABy = 0 = 212.1 kN + 212.1 kN + 212.1 kN 84.84 kN − − − = 127.26 kN C

Answers58

Cut a section through CD, CG, and GH.

Σ − −M GH

G

C = 0 = 21.21 kN (20 m) 21.21 kN (20 m) + (10 m)

HH =424.20 kN-m + 212.10 kN-m

10 m = 63.63 kN T

Σ − −F CG

CG CG

CG

x x

x y

= 0 = 21.21 kN 21.21 kN + = 42.42 kN =

= (42.42 kN) + (42.42 kN) = 60.0 kN C2 2

Σ − −F CD

CDy = 0 = 63.63 kN 42.42 kN = 21.21 kN C

Answers 59

E3-7 Shown below is a structure composed of two and three force members.Determine the reactions A and E and at all the pinned connections.

R

F R

R

Ay

x Ax

Ax

= 30 kips = 0 = 25 kips = 25 kips

↓

←Σ −

Σ −M RE

RE

A= 0 = 25 kips (12 ft) + (10 ft)

= 300 kip-ft

10 ft== 30 kips

= 0 = 30 kips

↑

∑ −F Ry Ay

Answers60

Free Body Diagram 1

C

F C

C

x

y y

y

= 8.33 kips = 0 = 30 kips

= 30 kips

←

↓∑ −

Using Free Body Diagram 1

M DC x∑ − = 0 = +30 kips (5 ft) (9 ft)

= 150 kip-ft

9 = 16.67 kips

= 0 = 16.67 kips +

D

F

x

x

←

∑ − 25 kips − Cx

Answers 61

Free Body Diagram 2

Free Body Diagram 2 verifies the structure is in static equilibrium.

Chapter 4

E4-1. A 3/4� diameter bolt is subjected to a tensile force of 12 kips as shown below.What is the average shear stress in the bolt?

Answers62

E4-2. A concrete cylinder 18 inches high is subjected to a compression test. At3,000 psi the strain is .003 inch/inch.

a. What was the change in length of the cylinder at 3000 psi?

b. What is the modulus of elasticity of the concrete?

E4-3. A steel column with an area of 15.6 in2 and must supports an axial load of240 kips. If the column is 16 feet high, how much did the column shrinkwhen the load was applied? Use E = 29 � 103 ksi.

E4.4. Determine the elongation of a 1 inch diameter steel rod, 10 feet long used tosupport a 15 kip tensile force. Use E = 29 � 103 ksi.

E4.5. What load would be required to cause a steel column with a 3 in2 cross sec-tion to shrink 0.25 inches if the column is 15 feet high? Use E = 29 � 103

ksi.

P = = 29 10 kips/in (1.39 10 in/in 3 in 3 2 3E Aε × × − ) ( )== 120.8 kips

ε = =Δ × −l

l

.25 in15 ft (12 in/ft)

= 1.39 10 in/in

=

3

E = ƒε ε

P / A

A=2 2

2

4=

4 = 0.79 in

=

= 15 kips (10

1 in π π (

Δ

d

P

E A

)

l l fft) (12 in/ft)

29 10 kips/in (.79 in ) = 0.08 in

3 2 2×

A

P

A

d

v

= 4

= 75 in4

= .44 in

= = 12 kips.44 in

=

2 22

2

π π

ƒ

(. )

27.2 ksi

a. = = .003 (18 in) = 0.05 in

b. = = 3,000 lb

Δƒ

l l ε

εE c ss/in

.003 in/in = 1 10 psi

26×

Δ×

l l

= P

E A =

240 kips (16 ft) (12 in/ft)29 10 kip/in (3 2 115.6 in )

= 0.10 in2

Answers 63

Answers64

Chapter 5

Construct the load, shear and moment diagrams for Problems E5.1 throughE5-10.

E5-1.

E5.2.

Answers 65

E5-3.

Answers66

E5-4.

Answers 67

E5-5.

Answers68

E5-6.

Answers 69

E5-7.

Answers70

E5-8.

Answers 71

E5-9.

Answers72

E5-10.

Answers 73

Chapter 6

E6-1. Determine the centroid of the section shown below.

yAy

= =200 in (10 ) (15 )

200 in

2 2

2 2

50.27 in

50.27 in

∑∑

″ ″−−A

== 8.3 in

= 5 inx

A

A

1

2

= 4

= 4

= 20 10 = 200 in

22

2

850.27 in− − −

π π ″

″ × ″

d ( )=

2

Answers74

E6-2. Determine the centroid of the section shown below (SI units).

xAx

A = =

2,592 mm (18 mm) 300 mm (10.5 mm)2,592 mm

2 2

2

∑∑

−− 3300 mm

= 19 mm2

yAy

A = =

2,592 mm (36 mm) 300 mm (46 mm)2,592 mm 3

2 2

2

∑∑

−− 000 mm

= 34.7 mm2

A

A1

2

= 36 mm (72 mm) = 2,592 mm = 15 mm (20 mm) = 300 m

2

− − mm2

Answers 75


y Ay

A= =

312 in 13 48 in 23312 in 48 in

= 2 2

2 2

∑∑

″ − ″−

( ) ( )111.2 in

= 6 inx

A

A1

2

= 2 (4 ) (6 ) = 48 in = 12 (26 ) = 312 in

2

2

− ″ ″ −″ ″

Answers76


A

x Ax

A

2 = 12 (5 ) = 60 in

= = 84 in (3 ) 60 in (3.

2

2 2

− ″ ″ −∑∑

″ − 55 )84 in 60 in

= 1.75 in2 2

″−

y

A1

= 7 in = (6 ) (14 ) = 84 in2″ ″

Answers 77

E6-5. Determine the moment of inertia of the section shown in Problem 6-4.

PieceArea(in2)

Icy

(in4)d

(in)d2

(in2)Ad2

(in4)Ic + Ad2

(in4)

1 84 252 1.25 1.56 131.25 383.25

2 –60 –125 1.75 3.06 –83.75 308.75

� 24 74.5

Moment of Inertia about the axis:

= = 14 6

12

3

y

Icy1

bh ( ) (″ ″))3

12= 252 in

= = 12 (5 )

12 = 125 in

4

3 34

12I cy2

bh − ″ ″ −

I

I I I

c x2

x cx1 cy2

= = 5 (12 )

12 = 720 in

= = 12

3 34− − ″ ″ −

−

bh

11,372 in 720 in = 652 in4 4 4−

Moment of Inertia about the axis:

= =6 14

12

3

x

Icx1

bh ( ) ( )″ ″ 33

12 = 1,372 in4

Answers78

E6-6 Determine the moment of inertia of the section shown in Figure below (SI units).

PieceArea

(mm2)y

(mm)Ay

(mm3)

1 45 30.5 1,372.5

2 75 16.5 1,237.5

3 80 2 160

� 200 2,770

Answers 79

E6-7. The steel beam shown in Figure 6-7 has a moment of inertia about the xaxis of 103 in4 and an area of 4.71 in2. If a 1/4� � 6� steel plate is welded to the bottom of the beam, what is the moment of inertia of the composite section?

PieceArea

(mm2)Icx

(mm4)d

(mm)d2

(mm2)Ad2

(mm4)Ic + Ad2

(mm4)

1 45 33.75 16.65 277.22 12,475.0 12,509

2 75 3,906.25 2.65 7.02 526.69 4,433

3 80 106.67 11.85 140.42 11,233.8 11,340

� 28,282

I

I

c x1

c x2

= = 15 (3)

12 = 33.75 mm

= =3 (25)

12

12

12

3 34

3 3

bh

bh== 3,906.25 mm

= =20 (4)

12 = 106.67 mm

4

3 34

12I c x3

bh

y = = 2,770 mm200 mm

= 13.85 mm3

2

∑∑Ay

A

Answers80

PieceArea(in2)

Icx

(in4)d

(in)d2

(in2)Ad2

(in4)Ic + Ad2

(in4)

1 4.7 103 1.45 2.20 10.33 113.33

2 1.5 0.01 4.64 21.56 32.33 32.34

� 146

I

I

c x1

c x2

= 103 in

= = 6 in (.25 in)

12 = 0.01 in

4

3 34

12

bh

y = = 29.56 in6.20 in

= 4.77 in3

2

∑∑Ay

A

PieceA

(in2)y

(in)Ay

(in3)

1 4.7 6.25 29.38

2 1.5 .125 0.19

� 6.2 29.56

Answers 81

E6.8. The steel beam shown in Figure 6-8 has a moment of inertia about the xaxis of 11,493 cm4 and an Area of 94.8 cm2. Determine Ix of the compositesection if a 1 � 15 cm plate is welded to the top of the beam and a 1 � 10cm plate is welded to the bottom of the beam.

Answers82

Chapter 7

E7-1. Determine the bending stresses in the sections shown below. The beam mustsupport a moment of 8,000 ft-lbs.

PieceArea(in2)

Icx

(in4)d

(in)d2

(in2)Ad2

(in4)Ic + Ad2

(in4)

1 15 1.25 12.6 158.76 2,281.4 2,382.65

2 94.8 11,493 0.55 0.30 28.68 11,521.68

3 10 0.83 13.7 187.69 1,876.9 1,877.73

� 15,782

I

I

I

c x1

c x2

= = 15 cm (1 cm)

12 = 0.125 cm

= 11,493 cm12

3 34

4

bh

cc x3 = = 10 cm (1 cm)

12 = 0.83 cm

12

3 34bh

y = = 1 701 cm119.8 cm

= 14.2 cm3

2

∑∑Ay

A

PieceArea(cm2)

y(cm)

Ay(cm3)

1 15 26.8 402.0

2 94.8 13.65 1,294.0

3 10 .5 5.0

� 119.8 1,701.0

Answers 83

Answers84

E7-2. What is the maximum concentrated load, (P), that the beam shown belowcan support if Fb =1400 psi?

E7-3. The T-beam shown below must support a shear force of 13 kips.

a) What is the maximum shear stress (fv ) in the section?

b) What is the shear stress at the top of the web?

MPL

PM

L

=4

= 4

=4 (3,733.3 ft-lbs)

15 = 995.56 lbs

Sbh

M F Sb

= 6

= 3 (8 )

6 = 32 in

= =1400 lbs

in

(32 in

2 23

2

3

″ ″

))(12 in/ft)

= 3,733.3 ft-lbs

I x = = 6 in (10 in)

12= 500 in

= = 8,000 ft-lbs

12

3 34bh

Mc

Ibƒ ((12 in/ft) (5 in)

500 in = 960 psi

4

PieceArea(in2)

Icx

(in4)d

(in)d2

(in2)Ad2

(in4)Ic + Ad2

(in4)

1 6 0.5 2.5 6.25 37.5 38

2 5 41.67 3 9 45.0 86.67

� 124.67

PieceA

(in2)y

(in)Ay

(in3)

1 6 10.5 63

2 5 5 25

� 11 88

y

Ic x1

= = 88 in11 in

= 8 in

= = 6 1

12 =

3

2

3

12

∑∑

″ ″

Ay

A

bh ( )3

00.5 in

= = 0.5 (10 )

12 = 41.67 in

4

3 34

12I c x2

bh ″ ″

Answers 85

E7.4 A 6� � 14� beam supports the loads shown below. Determine the maximumshear stress.

a) Using the bottom half of the section:

= = 8 (0.5Q Ay ″ ″)) (3 ) = 12 in

= = 12 in (13 kips)124.67 in

3

max

3

4

″

ƒv

QV

Ib (.5in) = 3.34 ksi

b Using the top portion of the beam:

= = 6 (1 ) (2.

)

Q Ay ″ ″ 55 ) = 15 in

= = 15 in (13 kips)

124.67 in (0.5in

3

3

4

″

v

QV

Ib )) = 3.13 ksi

= 2

= 1,000 lbs + 400 lbs/ft (16 ft)

2 = 4,200 lbs

V PL+ ω

v == 3V2A

= 3 (4,200 lbs)2 (6 ) (14 )

= 75 psi″ ″

Answers86

E7-5. Calculate the total load deflection for a 2� � 10� beam, spanning 16 feet, thatsupports a uniform distributed load of 120 lbs/ft. Use E = 1.5 � 106 psi.

E7-6. Determine if the deflection of the steel wide flange beam shown below fallwithin the allowable limits for total and live load deflection. Use Ix = 1140 in4

and E = 29 � 103 ksi.

Δ ΔωωL

L

TTActual

= = 1.5 kips/ft2.4 kips/ft

(1.50 in) = 0..94 in < 1.00 in OK

Δ ωT

4 4

Actual=

5 (1728)384

=5 (2.4 kip/ft) (30 ft) (1L

EI

7728)384 (29 10 ksi) (1140 in )

= 1.32 in < 1.50 in OK

3 4×

ΔL Allowable=

360 =

30 ft (12 in/ft)360

= 1.00 inL

ΔTAllowable=

240 =

30 ft (12 in/ft)240

= 1.50 inL

I x = 166.67 in (Appendix A) 1728

384

4

T

4

Actual= Δ ω5 L

EI

( )==

5 (120 lbs) (16 ft) (1728)384 (1.5 10 psi) (166.67 in )

4

6 4× = 0.71 in

Answers 87

E7.7. Calculate the approximate total deflection for the 8� � 12� beam shownbelow. Assume the modulus of elasticity is 100,000 psi (105 psi).

I

L

x = 1,152 in (Appendix A)

= 8 M

=8 (1,312 ft-lbs)

(16

4

2ω

fft) = 322.27 lb/ft

2

Answers88

Answers 89

E7-8. Design a beam to support the loads shown below. The beam must have arectangular cross section in full inches with the following allowable stresses:Fb = 1200 psi, Fv =100 psi and E = 1.4 � 106 psi. Limit the total load deflec-tion to L/240 and the live load deflection to L/360.

Check Deflection:

= 240

= 14 ft (12 in/ft)

240 =TAllowable

Δ L 0.70 in

= 360

= 14 ft (12 in/ft)

360 = 0.47 inL Allowable

Δ L

Check Shear:

= 2

= 500 lbs/ft (14 ft)

2 = 3,500 lbsV

L

v

ω

=3V22A

=( )

( )3 3,500 lbs

2 65 in = 80.77 psi 100 psi OK

2<

Try: 5 13 ( = 140.3 in , = 65 in , = 915.42 in3 2 4″ × ″ S A Ix )

Trial

ML

Selection:

=8

= 500 lbs/ft 14 ft

8 = 12,250.0 f

2 2ω ( )tt-lbs

= 12,250.0 ft-lbs (12 in/ft)

1200 lbs/in = 12req’d 2

S 22.5 in3

ω ω ωT L T= + 400 lbs/ft + 100 lbs/ft = 500 lbs/ft=

Δ ωApprox.

4

4

= 5 (1728)

384

= 5 (322.27 lb/ft) (16 ft) (1

L

EI7728)

384 (10 psi) (1152 in ) = 4.1 in

5 4

E7-9. Design a beam to support the loads shown below. The beam must have arectangular cross section in full inches with the following allowable stresses:Fb = 1000 psi, Fv =110 psi and E = 1.3 � 106 psi. Limit the total load deflec-tion to L/240.

=800 lbs (7 ft) [(3) (21 ft) 4(7 ft) ] (1728)

24 (1.3 10

2 2−× 66 4

L

psi) (432 in )= 0.81 in < 1.05 in OK

UAllowable

< Δsse: 3 12 Beam″ × ″

Check Deflection:

= 240

= 21 ft (12 in/ft)

240 =TAllowable

Δ L 1.05 in

= (3 4 ) (1728)

24 T

2 2

ActualΔ −P a L a

EI

Check

V P

FV

Av

Shear: = = 800 lbs

= 3 2

= 3 (800 lbs)2 (366 in )

= 33.33 psi < 110 psi OK2

Trial Selection: = = 800 lbs (7 ft) = 5,600 lb-ft

req’

M Pa

S dd 23 = =

5,600 lb-ft (12 in/ft)1,000 lbs/in

= 67.2 inM

Fb

TTry: 3 12 ( = 72 in , = 36 in , = 432 in )3 2 4″ × ″ S A Ix

Δ ωT

4 4

Actual=

5 (1728)384

=5 (500 lb/ft) (14 ft) (17L

EI

228)384 (1.4 10 ksi) (915.42 in )

= 0.34 in < 0.70 in

6 4×< ΔΔ

× ″L Allowable

OK

Use: 5: 13 Beam

Answers90

7-10. Design a beam to support the loads shown below. The beam must have arectangular cross section in full inches with the following allowable stresses:Fb = 1300 psi, Fv =120 psi and E = 1.3 � 106 psi. Limit the total load deflec-tion to L/240 and the live load deflection to L/360.

Answers 91

Chapter 8

E8-1. Determine the critical buckling stress (ƒcr ) of a 4� � 4� column. The columnis 12� high, pinned at the top and bottom, and the material has a modulusof elasticity of 1.4 � 106 psi.

r

rcr

E

= 1.15 in

= = 1.4 10 lbs/in

(144 in /

2

2

2 6 2

ƒ ×π πl/

( )

( ) 11.15 in) = 888.1 psi

2

Δ Δωω

LL

TTApprox.

= =600 lbs/ft720 lbs/ft

(0.60 in)

= 0.50 iin < 0.53 in OKUse: 6 14 Beam″ × ″

Δ ωT

4 4

Approx.=

5 (1728)384

=5 (720 lb/ft) (16 ft) (1L

EI

7728)384 (1.3 10 ksi) (1372 in )

= 0.60 in < 0.80 in O

6 4×KK

Check Deflection:

= 240

= 16 ft (12 in/ft)

240 =TAllowable

Δ L 0.80 in

= 360

= 16 ft (12 in/ft)


Δ L

Try: 6 14 ( = 196 in , = 84 in , = 1,372 inChec

3 2 4″ × ″ S A Ix )kk Shear:= 6,120 lbs

3 (6,120 lbs)

2 (84 in )

max

2

V

V

Avƒ =

32

= = 109.29 psi < 120 psi OK

ω ω ωT L T = + = 600 lbs/ft +120 lbs/ft = 720 lbs/ft SeleTrial cction:

= 20,250.0 ft-lbs

= 20,250.0 ft-lbs (12

max

req’d

M

Siin/ft)

1300 lbs/in = 186.92 in

23

Answers92

E8-2. Determine the critical buckling load (Pcr) for the pipe column shown below.Use E = 29 � 103 ksi, a length of 20 feet and assume the column is pinnedtop and bottom.

rI

A

PEI

x

cr

= = 18.11 in7.07 in

= 1.60 in

= =

4

2

2π πl�2

22 3 4

2

(29 10 ksi) (18.11 in(240 in)

= 90 kips× )

I I Id d

d dx x x = =64

64

= 64

( )

= 64

[ 5

o i

o4 4

o4

i4− − π −

π

π π i

( in) (4 in) ] = 18.11 in4 4 4−

Ad d

=4

4

= 5 in4

4 in4

= 7.07 ino2

i2 2 2

2π π π π − −( ) ( )

Answers 93

E8-3. The column shown below has its base fixed in a concrete foundation. Thetop is pinned. Determine the critical buckling load using a height of 18 feet,and assuming material has a modulus of elasticity of 1.2 � 106 psi.

PEI

cr = = 1.2 10 psi 11.25 in

151.20 in = 5

2

2

2 6

2

π π ×le

( ) ( )( )

,,828.1 lbs

A

Iy

e

= 15 in = 11.25 in

7 (18 ft) (12 in/ft) = 151.20 i

2

l = . nn

Answers94

E8-4. The column shown below is two stories high and is braced on it weak axis at mid-height. Assume E = 1.3 � 106 psi and determine the critical load thecolumn can support.

E8-5. Determine the critical stress in the tube column shown below. The columnhas a height of 28 feet; it has a fixed connection at the top and a pinnedconnection at the bottom. The material has a modulus of elasticity of 29 � 103 ksi.

PEI

cr

e

=

=1.3 10 psi 72.0 in

(288 in) = 11,13

2

2

2 6 4

2

π π ×l

( ) ( )77.6 lbs

le

y

x

r

I

= 288 in1.73 in

= 166.47 CONTROLS

= 72.0 in4

←

Slenderness ratio axis: = 1.73 in = (24 ft) (12 in/f

x

rx

el tt) = 288 in

Slenderness ratio axis:

= 1.15 in= (12 ft) (12 in/ft)

y

ry

el = 144 in

= 144 in1.15 in

= 125.22le

yr

Answers 95

E8-6. Determine the transition point between a short and long column for a 3� � 3� column, pinned at the top and bottom. Assume a yield stress (Fy) of 1210 ksi and a modulus of elasticity of .9 � 103 ksi.

Chapter 9

Problems E9-1 through E9-5 refer to the partial framing plan shown below. Forthese problems we will assume the following allowable stresses:

Fb = 1200 psi

Fv = 100 psi

E = 1.2 � 106 psi.

We will also assume the following loads:

Roof: D = 12 psf (includes the weight of the structure)S = 40 psf

Floor: D = 15 psf (includes the weight of the structure)L = 100 psf.

le

cr

EI

P= =

9 10 ksi 6.75 in10,890 kips

= 72 2 6 4π π × (. ) ( )

44.2 in

P A F

I

P

cr y

x

= = (3 3 )1,210 ksi = 10,890 kips

= 6.75 in4

″ × ″

ccr

e

EI =

2

2

π

(l )

PEI

cr

e

=

= (29 10 ksi) 55.92 in

(235.20 in) = 28

2

2

2 3 4

2

π π ×l

( )99.33 kips

= 289.33 kips

(36 in 25 in )= 26.30 ksi

2 2ƒ

−cr

I x

e

= 108.0 in 50.08 in = 55.92 in= 0.7(28 ft)(12 in

4 4 4−l //ft) = 235.20 in

I I I

Ibh

I

x xo xi

xo

xi

= = 6 in (6 in)

12 = 108.0 in

12

3 34

= −

= = 5 in (5 in)

12 = 50.08 in

12

3 34bh

Answers96

E9-1. Using rectangular cross sections design the interior roof joist limiting thetotal defection to L/180 and the snow load deflection to L/240.

Check Shear:

= 2

= 104 lbs/ft (16 ft)

2 = 832.00 lbs

=

VL

v

ω

ƒ 332

= 3(832.00 lbs

2 20 in = 62.4 psi < 100 psi OK

2

V

A

)( )

Try: 2 10 joist ( = 33.33 in , = 20 in , = 166.673 2″ × ″ S A I x in )4

Trial Selection:

= 8

= 104 lbs/ft (16 ft)

8 = 3,328 ft

2 2

MLω

--lbs

= = 3,328 ft-lbs (12 in/ft)

1200 psi = 33.2S

M

Freq d

b

’ 88 in3

T D S

T

= + = 12 psf + 40 psf = 52 psf= trib. width Tω × == 52 psf (2 ft) = 104 lbs/ft= trib. width = 40 psfSω × S (2 ft) = 80 lbs/ft

Answers 97

E9-2. Using rectangular cross sections design the roof beam limiting the totaldefection to L/180 and the snow load deflection to L/240.

Check Deflection:

= 180

= 14 ft (12 in/ft)

180 T Allowable

Δ L== 0.93 in

= 240

= 14 ft (12 in/ft)

240 = 0.70 S Allowable

Δ Liin

Check Shear:

=2

= 416 lbs/ft (14 ft)

2 = 2,912.00 lbs

=

VL

v

ω

32

= 3 (2,912.00 lbs)

2 (60 in ) = 72.80 psi < 100 psi

2

V

A OK

SM

Fb

req’d = = 10,192 ft-lbs (12 in/ft)

1200 psi = 101.92 inn

Try: 5 12 Beam ( = 120 in , = 60 in , = 720 i

3

3 2″ × ″ S A I nn )4

Trial Selection:

= 8

= 416 lbs/ft (14 ft)

8 = 10,192 f

2 2

MLω

tt-lbs

T

T

= 52 psf= trib. width = 52 psf (8 ft) = 416 lbs/Tω × fft= trib. width = 40 psf (8 ft) = 320 lbs/ftSω × S

=5 (104 lbs/ft) (16 ft) (1728)

384 (1.2 10 psi) (166.67 i

4

6× nn )

= 0.77 in < 1.07 in and <.80 OKUse: 2 10 Inte

4

″ × ″ rrior roof joists

Δ ωT

4

Actual=

5 (1728)384

L

EI

Check Deflection:

= 180

= 16 ft (12 in/ft)

180 T Allowable

Δ L== 1.07 in

= 240

= 16 ft (12 in/ft)

240 = 0.80 S Allowable

Δ

Liin

Answers98

E9-3. Using rectangular cross sections design the interior floor joist limiting thetotal defection to L/240 and the live load deflection to L/360.

Δ×L

4

6Actual=

5 200 lbs/ft 16 ft 1728384 (1.2 10 psi)

( )( ) ( ) (443.67 in )

= 0.55 in > .53 in NG4

Δ ωT

4

4

Actual=

5 (1728)384

=5 230 lbs/ft 16 ft 1

L

EI( ) ( ) ( 7728

384 (1.2 10 psi) (443.67 in ) = 0.64 in < .80 in OK

6 4

)×

Check Deflection:

= 240

=16 ft (12 in/ft)

240 = T Allowable

Δ L00.80 in

= 360

= 16 ft (12 in/ft)


Δ L

Check Shear:

=2

= 230 lbs/ft (16 ft)

2 = 1,840 lbs

= 32

VL

Vv

ω

ƒAA

= 3 (1,840 lbs

2 44 in = 62.73 psi < 100 psi OK

2

)( )

SM

Fb

req’d = = 7,360.00 ft-lbs (12 in/ft)

1200 psi = 73.60 iin

Try: 4 11 joist ( = 80.67 in , = 44 in , = 44

3

3 2″ × ″ S A I 33.67 in )4

Trial Selection:

=8

=230 lbs/ft (16 ft)

8 = 7,360.00 ft

2 2

MLω

--lbs

T L D

T

= + = 100 psf + 15 psf = 115 psf= trib. widtTω × hh = 115 psf (2 ft) = 230 lbs/ft= trib. width = 100Lω × S psf (2 ft) = 200 lbs/ft

= 0.42 in. < 0.93 in. and < 0.70 in. OKUse: 5 12 ro″ × ″ oof beams

Δ ωT

4

4

Actual=

5 (1728)384

=5 (416 lbs/ft) (14 ft) (1728

L

EI))

384 (1.2 10 psi) (720.00 in )6 4×

Answers 99

E9-4. Using rectangular cross sections design the floor beam limiting the totaldefection to L/240 and the live load deflection to L/360.

= 0.36 in < 0.80 in and < 0.47 in OKUse: 8 14 floor″ × ″ beams

Δ ωT

4

4

Actual=

5 (1728)384

=5 920 lbs/ft) 14 1728

L

EIft( ( ) ( ))

384 (1.2 10 psi) (1,829.33 in )6 4×

Check Deflection:

= 240

= 14 ft (12 in/ft)

240 T Allowable

Δ L== 0.70 in

= 360

= 14 ft (12 in/ft)

360 = 0.47 L Allowable

Δ Liin

ƒv

V

A =

3 2

= 3 (6,440.00 lbs

2 112 in = 86.25 psi < 10

2

)( )

00 psi OK

Check Shear:

= 2

= 920 lbs/ft (14 ft)

2 = 6,440.00 lbsV

Lω

SM

Fb

req’d = = 22,540 ft-lbs (12 in/ft)

1200 psi = 225.40 inn

Try: 8 14 beam ( = 261.33 in , = 112 in , = 1

3

3 2″ × ″ S A Ix ,,829.33 in )4

Trial Selection:

=8

=920 lbs/ft (14 ft)

8 = 22,540.00

2 2

MLω

fft-lbs

T

T

= 115 psf= trib. width = 115 psf (8 ft) = 920 lbTω × ss/ft= trib. width = 100 psf (8 ft) = 800 lbs/ftSω × L

=5 200 lbs/ft) 16 ft 1728384 (1.2 10 psi) (.53 in)

4

6

( ( ) ( )×

== 463.70 in

Use: 4 12 floor joists ( = 96in , =

4

3″ × ″ S A 448 in , = 576 in )2 4I

IL

Ex req’d

=5 (1728)

384

4

L Allowable

ωΔ

Answers100

E9-5. Assume the column is 4� � 4� and determine if the loads on the column arewithin the limits of Euler’s critical buckling load (Pcr ).

P P Proof floor = + = 2,912 lbs + 6,440 lbs = 9,352 lbs = 9.35 kips

=

=1.2 10 )(21.33)

(11.5 ft 1

2

2

2 6

PEI

cr

e

π π ××l

(22 in/ft)

= 13,265 lbs = 13.27 kips

= 9.35 kips < =

2

P Pcr 113.27 kips

Assume a continuous 2 story column, pinned top and bottom,

with intermediate supports at 11.5 : ′

Answers 101

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