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Overview

Considered a standard in the field, this edition has been updated and expanded. Presents a minimum of fundamental structural mechanics and a digest of the basic issues regarding common forms of construction in major structural materials of wood, steel, concrete and masonry. Contains a new section on the topic of strength design, the latest information on model building codes and industry standards, a chapter on the basic nature and some detail of trusses. Includes a wealth of illustrative examples.
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Editorial Reviews

From the Publisher
"I recommend this as a must have book which gives me confidence that I am being diligent with my design and drawings." (Orange Bytes, 5/2006)
Booknews
The latest edition of a classic that was first published in 1938. Pragmatically focused with respect to the concerns of the designers and constructors of buildings, the volume presents a minimum of fundamental structural mechanics and a digest of the basic issues regarding common forms of construction in the major structural materials of wood, steel, concrete, and masonry. This edition adds an entire new part (Part V) devoted to the topic of strength design. It also further expands the illustrations in Part VI (old Part V), which tie together much of the rest of the work in the volume in presenting some full-blown involvements in complete building structures. Annotation c. Book News, Inc., Portland, OR (booknews.com)
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Product Details

  • ISBN-13: 9780471676072
  • Publisher: Wiley, John & Sons, Incorporated
  • Publication date: 12/2/2005
  • Series: Parker-Ambrose Series of Simplified Desi
  • Edition description: Older Edition
  • Edition number: 10
  • Pages: 656
  • Product dimensions: 5.75 (w) x 8.64 (h) x 1.53 (d)

Meet the Author

JAMES AMBROSE is Editor of the Parker/Ambrose Series of Simplified Design Guides. He practiced as an architect in California and Illinois, and as a structural engineer in Illinois. He was a professor of architecture at the University of Southern California.

PATRICK TRIPENY is Director of School of Architecture and an associate professor at the University of Utah. He teaches the architectural structures sequence in the School of Architecture and the graduate design studio. He is the recipient of several teaching awards including the ACSA/AIAS New Faculty Teaching Award in 2001 and the University of Utah's Early Career Teaching Award in 2000-2001.

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Table of Contents

Preface to the Eighth Edition
Preface to the First Edition
Introduction 1
1 Investigation of Forces 11
2 Forces Actions 44
3 Investigation of Beams and Frames 71
4 Properties of Sections 152
5 Wood Spanning Elements 187
6 Wood Columns 218
7 Fastening of Wood Structures 236
8 Steel Structural Products 259
9 Steel Beams and Framing Elements 271
10 Steel Columns 312
11 Bolted Connections 340
12 Welded Connections 367
13 Reinforced Concrete Structures 385
14 Sitecast Concrete Framing Systems 445
15 Concrete Columns 466
16 Footings 484
17 General Considerations for Masonry Structures 496
18 Design of Structural Masonry 510
19 Strength Design Methods 527
20 Strength Design of Reinforced Concrete 533
21 Strength Design of Steel Structures 542
22 General Concerns for Structures 555
23 Building One 571
24 Building Two 599
25 Building Three 612
References 649
Answers to Exercise Problems 651
Index 661
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First Chapter

Note: The figures and/or Tables mentioned in this sample chapter do not appear on the Web.

PART ONE


PRINCIPLES OF STRUCTURAL MECHANICS


This part consists of basic concepts and applications from the field of applied mechanics as they have evolved in the process of investigation of the behavior of structures. The purpose of studying this material is twofold. First, there is the general need for a comprehensive understanding of what structures do and how they do it. Second, there is the need for some factual, quantified basis for the exercise of judgment in the processes of structural design. If it is accepted that the understanding of a problem is the necessary first step in its solution, this analytical study should be seen as the cornerstone of any successful, informed design process.

Although this book makes considerable use of mathematics, it is mostly only a matter of procedural efficiency. The concepts, not the mathematical manipulations, are the critical concerns.

CHAPTER ONE
INVESTIGATION OF FORCES


Structures perform the basic task of resisting forces. Specific forces derive from various sources and must be considered in terms of their probable actions for any given structure. This chapter presents some basic concepts and procedures that are used in the analysis of the actions and effects of forces. Topics selected and procedures illustrated are limited to those that relate directly to the problem of designing ordinary building structures.

1.1 PROPERTIES OF FORCES

Force is one of the fundamental concepts of mechanics and as such does not lend itself to simple, precise definition. For our purposes at this stage of study, we may define a force as that which produces, or tends to produce, motion or a change in the motion of bodies. One type of force is the effect of gravity, by which all bodies are attracted toward the center of the earth. The magnitude of the force of gravity is the weight of a body. The amount of material in a body is its mass. In the U. S. (old English) System, the force effect of gravity is equated to the weight. In SI units, a distinction is made between weight and force, which results in the force unit of a newton. In the U. S. System, the basic unit of force is the pound, although in engineering work the kip (1000 pounds or, literally, a kilopound) is commonly used.

To identify a force, we must establish the following:

Magnitude, or the amount of the force, which is measured in weight units such as pounds or tons.

Direction of the force, which refers to the orientation of its path, called its line of action. Direction is usually described by the angle that the line of action makes with some reference, such as the horizontal.

Sense of the force, which refers to the manner in which it acts along its line of action (e. g., up or down, right or left). Sense is usually expressed algebraically in terms of the sign of the force, either plus or minus.

Forces can be represented graphically in terms of these three properties by using an arrow, as shown in Fig. 1.1a. Drawn to some scale, the length of the arrow represents the magnitude of the force. The angle of inclination of the arrow represents the direction of the force. The location of the arrowhead represents the sense of the force. This form of representation can be more than merely symbolic because actual mathematical manipulations may be performed using these force arrows. In this book, arrows are used in a symbolic way for visual reference when performing algebraic computations and in a truly representative way when performing graphical analyses.

In addition to the basic properties of magnitude, direction, and sense, some other concerns that may be significant for certain investigations are

The position of the line of action of the force with respect to the lines of action of other forces or to some object on which the force operates, as shown in Fig. 1.1b. For the beam, shifting of the location of the load (active force) affects changes in the forces at the supports (reactions).

The point of application of the force along its line of action may be of concern in analyzing for the specific effect of the force on an object, as shown in Fig. 1.1c.

When forces are not resisted, they tend to produce motion. An inherent aspect of static forces is that they exist in a state of static equilibrium, that is, with no motion occurring. In order for static equilibrium to exist, we must have a balanced system of forces. An important consideration in the analysis of static forces is the nature of the geometric arrangement of forces in a given set of forces that constitute a single system. The usual technique for classifying force systems involves consideration of whether the forces in the system are

Coplanar. All acting in a single plane, such as the plane of a vertical wall.

Parallel. All having the same direction.

Concurrent. All having their lines of action intersect at a common point.

Using these three considerations, the possible variations are given in Table 1.1 and illustrated in Fig. 1.2. Note that variation 5 in the table is really not possible because a set of coacting forces that is parallel and concurrent cannot be noncoplanar; in fact, the forces all fall on a single line of action and are called collinear.

We must qualify a set of forces in the manner just illustrated before proceeding with any analysis, whether it is to be performed algebraically or graphically.

Earlier, we defined a force as that which produces or tends to produce motion or a change of motion of bodies. Motion is a change of position with respect to some object regarded as having a fixed position. When the path of a moving point is a straight line, the point has motion of translation. When the path of a point is curved, the point has curvilinear motion or motion of rotation. When the path of a point lies in a plane, the point has plane motion. Other motions are space motions.

1.2 STATIC EQUILIBRIUM

As stated previously, an object is in equilibrium when it is

either at rest or has uniform motion. When a system of forces acting on an object produces no motion, the system of forces is said to be in static equilibrium.

A simple example of equilibrium is illustrated in Fig. 1.3a. Two equal, opposite, and parallel forces, having the same line of action, P1 and P2 , act on a body. If the two forces balance each other, the body does not move, and the system of forces is in equilibrium. These two forces are concurrent. If the lines of action of a system of forces have a point in common, the forces are concurrent.

Another example of forces in equilibrium is illustrated in Fig. 1.3b. A vertical downward force of 300 lb acts at the midpoint in the length of a beam. The two upward vertical forces of 150 lb each (the reactions) act at the ends of the beam. The system of three forces is in equilibrium. The forces are parallel and, not having a point in common, are nonconcurrent.

1.3 RESULTANT OF FORCES

The resultant of a system of forces is the simplest system (usually a single force) that has the same effect as the various forces in the system acting simultaneously. The lines of action of any system of two nonparallel forces must have a point in common, and the resultant of the two forces will pass through this common point. The resultant of two nonparallel forces may be found graphically by constructing a parallelogram of forces.

This graphical construction is based on the parallelogram law: two nonparallel forces are laid off at any scale (of so many pounds to the inch) with both forces pointing toward, or both forces pointing away from, the point of intersection of their lines of action. A parallelogram is then constructed with the two forces as adjacent sides. The diagonal of the parallelogram passing through the common point is the resultant in magnitude, direction, and line of action, the direction of the resultant being similar to that of the given forces, toward or away from the point in common. In Fig. 1.4a, P1 and P2 represent two nonparallel forces whose lines of action intersect at point O. The parallelogram is drawn, and the diagonal R is the resultant of the given system. In this illustration, note that the two forces point away from the point in common; hence, the resultant also has its direction away from point O. It is a force upward to the right. Notice that the resultant of forces P1 and P2 shown in Fig. 1.4b is R; its direction is toward the point in common.

Forces may be considered to act at any points on their lines of action. In Fig. 1.4c, the lines of action of the two forces P1 and P2 are extended until they meet at point O. At this point, the parallelogram of forces is constructed, and R, the diagonal, is the resultant of forces P1 and P 2 . In determining the magnitude of the resultant, the scale used is, of course, the same scale used in laying off the given system of forces.

Example 1. A vertical force of 50 lb and a horizontal force of 100 lb, as shown in Fig. 1.4d, have an angle of 90° between their lines of action. Determine the resultant.

Solution: The two forces are laid off from their point of intersection at a scale of 1 in. = 80 lb. The parallelogram is drawn, and the diagonal is the resultant. Its magnitude scales approximately 112 lb, its direction is upward to the right, and its line of action passes through the point of intersection of the lines of action of the two given forces. By use of a protactor, it is found that the angle between the resultant and the force of 100 lb is approximately 26.5°.

Example 2. The angle between two forces of 40 and 90 lb, as shown in Fig. 1.4e, is 60°. Determine the resultant.

Solution: The forces are laid off from their point of intersection at a scale of 1 in. = 80 lb. The parallelogram of forces is constructed, and the resultant is found to be a force of approximately 115 lb, its direction is upward to the right, and its line of action passes through the common point of the two given forces. The angle between the resultant and the force of 90 lb is approximately 17.5°.

Note that we solved these two examples graphically by constructing diagrams, even though we could have used mathematics. For many practical problems, graphical solutions give sufficiently accurate answers and frequently require far less time. Do not make diagrams too small. Remember that greater accuracy is obtained by using larger parallelograms of forces.

Problems 1.3. A- F. By constructing the parallelogram of forces, determine the resultants for the pairs of forces shown in Fig. 1.5a- f.

1.4 RESOLUTION AND COMPOSITION OF FORCES

Components of a Force

In addition to combining forces to obtain their resultant, it is often necessary to replace a single force by its components. The components of a force are the two or more forces that, acting together, have the same effect as the given force. In Fig. 1.4d, if we are given the force of 112 lb, its vertical component is 50 lb, and its horizontal component is 100 lb. That is, the 112 lb force has been resolved into its vertical and horizontal components. Any force may be considered as the resultant of its components.

Combined Resultants

The resultant of more than two nonparallel forces may be obtained by finding the resultants of pairs of forces and finally the resultant of the resultants.

Example 3. Find the resultant of the concurrent forces P1 , P2 , P3 , and P4 shown in Fig. 1.6.

Solution: By constructing a parallelogram of forces, the resultant of P1 and P2 is found to be R1 . Simarily, the resultant of P3 and P4 is R 2. Finally, the resultant of R 1 and R 2 is R, the resultant of the four given forces.

Equilibrant

The force required to maintain a system of forces in equilibrium is called the equilibrant of the system. Suppose that we are required to investigate the system of two forces, P 1 and P 2 , as shown in Fig. 1.7. We construct the parallelogram of forces and find the resultant to be R. The system is not in equilibrium. The force required to maintain equilibrium is force E, shown by the dotted line. E, the equilibriant, is the same as the resultant in magnitude and direction but is opposite in sense. The three forces, P 1 and P 2 and E, constitute a system in equilibrium. If two forces are in equilibrium, they must be equal in magnitude and opposite in sense and have the same direction and line of action. Either of the two forces may be said to be the equilibrant of the other. The resultant of a system of forces in equilibrium is zero.

Problems 1.4. A--C. Using graphical methods, find the resultants of the systems of concurrent forces shown in Figs. 1.8 a--c.

1.5 FORCE POLYGON

We can find the resultant of a system of concurrent forces by constructing a force polygon. To draw the force polygon, begin with a point and lay off, at a convenient scale, a line parallel to one of the forces, equal to it in magnitude, and having the same direction. From the termination of this line, draw similarly another line corresponding to one of the remaining forces and continue in the same manner until all the forces in the given system are accounted for. If the polygon does not close, the system of forces is not in equilibrium, and the line required to close the polygon drawn from the starting point is the resultant in magnitude and direction. If the forces in the given system are concurrent, the line of action of the resultant passes through the point they have in common.

If the force polygon for a system of concurrent forces closes, the system is in equilibrium, and the resultant is zero.

Example 4. Find the resultant of the four concurrent forces P 1 , P 2 , P 3 , and P 4 , shown in Fig. 1.9a. This diagram is called the space diagram; it shows the relative positions of the forces in a given system.

Solution: Beginning with some point such as O, shown in Fig. 1.9b, draw the upward force P 1 . At the upper extremity of the line representing P 1 , draw P 2 , continuing in a like manner with P 3 and P 4 . The polygon does not close; therefore, the system is not in equilibrium. The resultant R, shown by the dot-and-dash line, is the resultant of the given system. Note that its direction is from the starting point O, downward to the right. The line of action of the resultant of the given system shown in Fig. 1.9a has its line of action passing through the point they have in common, its magnitude and direction having been found in the force polygon.

In drawing the force polygon, the forces may be taken in any sequence. In Fig. 1.9c, a different sequence is taken, but the resultant R is found to have the same magnitude and direction as previously found in Fig. 1.9b.

Bow's Notation

Thus far, forces have been identified by the symbols P 1 , P 2 , and so on. A system of identifying forces, known as Bow's notation, affords many advantages. In this system, letters are placed in the space diagram on each side of a force, and the force is identified by two letters. The sequence in which the letters are read is important. Figure 1.10a shows the space diagram of five concurrent forces. Reading about the point in common in a clockwise manner, the forces are AB, BC, CD, DE, and EA. When a force in the force polygon is represented by a line, we place a letter at each end of the line. As an example, we read the vertical upward force in Fig. 1.10a AB; in the force polygon (Fig. 1.10b), the line from a to b represents the force AB. Use capital letters to identify the forces in the space diagrams and lowercase letters in the force polygon. From point b in the force polygon draw force bc, then cd, and continue with de and ea. Because the force polygon closes, the five concurrent forces are in equilibrium.

In all the following discussions, we read forces in a clockwise manner. It is important that this method of identifying forces be thoroughly understood. To make this method of identification clear, suppose that a force polygon is drawn for the five forces shown in Fig. 1.10a, reading the forces in sequence in a counterclockwise manner. This will produce the force polygon shown in Fig. 1.10c. We can use either method, but for consistency we use the method of reading clockwise.

Use of the Force Polygon

Two ropes are attached to a ceiling, and their ends are connected to a ring, making the angles shown in Fig. 1.11a. A weight of 100 lb is suspended from the ring. Obviously, the stress in the rope AB is 100 lb, but the magnitudes of the stresses in ropes BC and CA are unknown.

The stresses in the ropes AB, BC, and CA constitute three concurrent forces in equilibrium. The magnitude of only one of the forces is known: it is 100 lb in rope AB. Because the three concurrent forces are in equilibrium, their force polygon must close, and this fact makes it possible to find their magnitudes. Now, at a convenient scale, draw the line ab (Fig. 1.11c) representing the downward force AB, 100 lb. The line ab is one side of the force polygon. From point b, draw a line parallel to rope BC; point c will be at some location on this line. Next, draw a line through point a parallel to rope CA; point c will be at some position on this line. Because point c is also on the line though b parallel to BC, the intersection of the two lines determines point c. The force polygon for the three forces is now completed; it is abc, and the lengths of the sides of the polygon represent the magnitudes of the stresses in ropes BC and CA, 86.6 and 50 lb, respectively.

Particular attention is called to the fact that the lengths of the ropes in Fig. 1.11a are not an indication of the magnitude of the forces within the ropes; the magnitudes are determined by the lengths of the corresponding sides of the force polygon (Fig. 1.11c).

1.6 GRAPHICAL ANALYSIS OF PLANAR TRUSSES

When we use the so-called method of joints, finding the internal forces in the members of a planar truss consists of solving a series of concurrent force systems. Figure 1.12 shows a truss with the truss form, the loads, and the reactions displayed in a space diagram. Below the space diagram is a figure consisting of the free body diagrams of the individual joints of the truss. These are arranged in the same manner as they are in the truss in order to show their interrelationships. However, each joint constitutes a complete concurrent planar force system that must have its independent equilibrium. "Solving" the problem consists of determining the equilibrium conditions for all the joints. The procedures used for this solution will now be illustrated.

Figure 1.13 shows a single span, planar truss that is subjected to vertical gravity loads. This example will be used to illustrate the procedures for determining the internal forces in the truss, that is, the tension and compression forces in the individual members of the truss. The space diagram in the figure shows the truss form, the support conditions, and the loads. The letters on the space diagram identify individual forces at the truss joints, as discussed in Section 1.5. The sequence of placement of the letters is arbitrary, the only necessary consideration being to place a letter in each space between the loads and the individual truss members so that each force at a joint can be identified by a two-letter symbol.

The separated joint diagram in Fig. 1.13 provides a useful means for visualizing the complete force system at each joint as well as the interrelation of the joints through the truss members. The individual forces at each joint are designated by two-letter symbols obtained by simply reading around the joint in the space diagram in a clockwise direction. Note that the two-letter symbols are reversed at the opposite ends of each of the truss members. Thus, the top chord member at the left end of the truss is designated as BI when shown in the joint at the left support (joint 1) and is designated as IB when shown in the first interior upper chord joint (joint 2). The purpose of this procedure will be demonstrated in the following explanation of the graphical analysis.

The third diagram in Fig. 1.13 is a composite force polygon for the external and internal forces in the truss. It is called a Maxwell diagram after one of its early promoters, James Maxwell, a British engineer. The construction of this diagram constitutes a complete solution for the magnitudes and senses of the internal forces in the truss. The procedure for this construction follows.

1. Construct the force polygon for the external forces. Before this can be done, we must find the values for the reactions. There are graphic techniques for finding the reactions, but it is usually much simpler and faster to find them with an algebraic solution. In this example, although the truss is not symmetrical, the loading is, and it may simply be observed that the reactions are each equal to one half of the total load on the truss, or 5000 divided by 2 equals 2500 lb. Because the external forces in this case are all in a single direction, the force polygon for the external forces is actually a straight line. Using the two-letter symbols for the forces and starting with letter A at the left end, we read the force sequence by moving in a clock-wise direction around the outside of the truss. Thus, we read the loads as AB, BC, CD, DE, EF, and FG and the two reactions as GH and HA. Beginning at a on the Maxwell diagram, we read the force vector sequence for the external forces from a to b, b to c, c to d, and so on, ending back at a, which shows that the force polygon closes and the external forces are in the necessary state of static equilibrium. Note that we have pulled the vectors for the reactions off to the side in the diagram to indicate them more clearly. Note also that we have used lowercase letters for the vector ends in the Maxwell diagram, whereas we use uppercase letters on the space diagram. The alphabetic correlation is thus retained (A to a), preventing any possible confusion between the two diagrams. The letters on the space diagram designate points of intersection of lines.

2. Construct the force polygons for the individual joints. The graphic procedure for this construction consists of locating the points on the Maxwell diagram that correspond to the remaining letters, I through P, on the space diagram. After locating all the lettered points on the diagram, we can read the complete force polygon for each joint on the diagram. In order to locate these points, we use two relationships. The first is that the truss members can resist only forces that are parallel to the members' positioned directions. Thus we know the directions of all the internal forces. The second relationship is a simple one from plane geometry: a point may be located at the intersection of two lines. Consider the forces at joint 1, as shown in the separated joint diagram in Fig. 1.13. Note that there are four forces and that two of them are known (the load and the reaction) and two are unknown (the internal forces in the truss members). The force polygon for this joint, as shown on the Maxwell diagram, is read as ABIHA. AB represents the load; BI, the force in the upper chord member; IH, the force in the lower chord member; and HA, the reaction. Thus the location of point i on the Maxwell diagram is determined by noting that i must be in a horizontal direction from h (corresponding to the horizontal position of the lower chord) and in a direction from b that is parallel to the position of the upper chord.

The remaining points on the Maxwell diagram are found by the same process, using two known points on the diagram to project lines of known direction whose intersection will determine the location of another point. After all the points are located, we can complete the diagram and use it to find the magnitude and sense of each internal force. The process for construction of the Maxwell diagram typically consists of moving from joint to joint along the truss. After one of the letters for an internal space is determined on the Maxwell diagram, we can use it as a known point for finding the letter for an adjacent space on the space diagram. The only limitation of the process is that it is not possible to find more than one unknown point on the Maxwell diagram for any single joint. Consider joint 7 on the separated joint diagram in Fig. 1.13. To solve this joint first, knowing only the locations of letters a through h on the Maxwell diagram, we must locate four unknown points: l, m, n, and o. This is three more unknowns than can be determined in a single step, so three of the unknowns must be found by using other joints.

Solving for a single unknown point on the Maxwell diagram corresponds to finding two unknown forces at a joint because each letter on the space diagram is used twice in the force identification for the internal forces. Thus for joint 1 in the previous example, the letter I is part of the identity of forces BI and IH, as shown on the separated joint diagram. The graphic determination of single points on the Maxwell diagram, therefore, is analogous to finding two unknown quantities in an algebraic solution. As discussed previously, two unknowns are the maximum that we can solve for in equilibrium of a coplanar, concurrent force system, which is the condition of the individual joints in the truss.

When the Maxwell diagram is completed, we can read the internal forces from the diagram as follows:

1. Determine the magnitude by measuring the length of the line in the diagram, using the scale that was used to plot the vectors for the external forces.

2. Determine the sense of individual forces by reading the forces in clockwise sequence around a single joint in the space diagram and tracing the same letter sequences on the Maxwell diagram.

Figure 1.14a shows the force system at joint 1 and the force polygon for these forces as taken from the Maxwell diagram. The forces known initially are shown as solid lines on the force polygon, and the unknown forces are shown as dashed lines. Starting with letter A on the force system, we read the forces in a clockwise sequence as AB, BI, IH, and HA. Note that on the Maxwell diagram moving from a to b is moving in the order of the sense of the force, that is from tail to end of the force vector that represents the external load on the joint. Using this sequence on the Maxwell diagram, this force sense flow will be a continuous one. Thus reading from b to i on the Maxwell diagram is reading from tail to head of the force vector, which indicates that force BI has its head at the left end. Transferring this sense indication from the Maxwell diagram to the joint diagram indicates that force BI is in compression; that is, it is pushing, rather than pulling, on the joint. Reading from i to h on the Maxwell diagram shows that the arrowhead for this vector is on the right, which translates to a tension effect on the joint diagram.

Having solved for the forces at joint 1 as described, we can use the fact that the forces in truss members BI and IH are known to consider the adjacent joints, 2 and 3. However, note that the sense reverses at the opposite ends of the members in the joint diagrams. Referring to the separated joint diagram in Fig. 1.13, if the upper chord member shown as force BI in joint 1 is in compression, its arrowhead is at the lower left end in the diagram for joint 1, as shown in Fig. 1.14a. However, when the same force is shown as IB at joint 2, its pushing effect on the joint will be indicated by having the arrowhead at the upper right end in the diagram for joint 2. Similarly, the tension effect of the lower chord is shown in joint 1 by placing the arrowhead on the right end of the force IH, but the same tension force will be indicated in joint 3 by placing the arrowhead on the left end of the vector for force HI.

If we choose the solution sequence of solving joint 1 and then joint 2, it is now possible to transfer the known force in the upper chord to joint 2. Thus, the solution for the five forces at joint 2 is reduced to finding three unknowns because the load BC and the chord force IB are now known. However, we still cannot solve joint 2 because there are two unknown points on the Maxwell diagram (k and j) corresponding to the three unknown forces. An option, therefore, is to proceed from joint 1 to joint 3, at which there are presently only two unknown forces. On the Maxwell diagram, we can find the single unknown point j by projecting vector IJ vertically from i and projecting vector JH horizontally from point h. Because point i is also located horizontally from point h so that i and j are on a horizontal line from h in the Maxwell diagram, the vector IJ has zero magnitude. This indicates that there is actually no stress in this truss member for this loading condition and that points i and h are coincident on the Maxwell diagram. The joint force diagram and the force polygon for joint 3 are as shown in Fig. 1.14b. In the joint force diagram, place a zero, rather than an arrowhead, on the vector line for IJ to indicate the zero stress condition. In the force polygon in Fig. 1.14b, the two force vectors are slightly separated for clarity, although they are actually coincident on the same line.

Having solved for the forces at joint 3, proceed to joint 2, because there now remain only two unknown forces at this joint. The forces at the joint and the force polygon for joint 2 are shown in Fig. 1.14c. As explained for joint 1, read the force polygon in a sequence determined by reading in a clockwise direction around the joint: BCKJIB. Following the continuous direction of the force arrows on the force polygon in this sequence, it is possible to establish the sense for the two forces CK and KJ.

We can proceed from one end and work continuously across the truss from joint to joint to construct the Maxwell diagram in this example. The sequence in terms of locating points on the Maxwell diagram would be i-j- k-l-m-n-o-p, which would be accomplished by solving the joints in the following sequence: 1, 3, 2, 5, 4, 6, 7, 9, 8. However, it is advisable to minimize the error in graphic construction by working from both ends of the truss. Thus a better procedure would be to find points i-j-k-l-m, working from the left end of the truss, and then to find points p-o-n-m, working from the right end. This would result in finding two locations for the point m, whose separation constitutes the error in drafting accuracy.

Problems 1.6. A and B. Using a Maxwell diagram, find the internal forces in the truss in Fig. 1.15.

1.7 ALGEBRAIC ANALYSIS OF PLANAR TRUSSES

Graphical solution for the internal forces in a truss using the Maxwell diagram corresponds essentially to an algebraic solution by the method of joints. This method consists of solving the concentric force systems at the individual joints using simple force equilibrium equations. The process will be illustrated using the truss in Figure 1.13.

As with the graphic solution, first determine the external forces, consisting of the loads and the reactions. Then proceed to consider the equilibrium of the individual joints, following a sequence as in the graphic solution. The limitation of this sequence, corresponding to the limit of finding only one unknown point in the Maxwell diagram, is that only two unknown forces at any single joint can be found in a single step. (Two conditions of equilibrium produce two equations.) Referring to Fig. 1.16, the solution for joint 1 follows.

We draw the force system for the joint with the sense and magnitude of the known forces shown; however, the unknown internal forces are represented by lines without arrowheads because their senses and magnitudes initially are unknown. For forces that are not vertical or horizontal, replace the forces with their horizontal and vertical components. Then, consider the two conditions necessary for the equilibrium of the system: the sum of the vertical forces is zero, and the sum of the horizontal forces is zero.

If the algebraic solution is performed carefully, the sense of the forces will be determined automatically. However, it is recommended that whenever possible the sense be predetermined by simply observing the joint conditions, as will be illustrated in the solutions.

The problem to be solved at joint 1 is as shown in Fig. 1.16a. In Fig. 1.16b, the system is shown with all forces expressed as vertical and horizontal components. Note that although this now increases the number of unknowns to three (IH, BI v , and Bi h ), there is a numeric relationship between the two components of BI. When this condition is added to the two algebraic conditions for equilibrium, the number of usable relationships totals three so that the necessary conditions to solve for the three unknowns are present.

The condition for vertical equilibrium is shown in Fig. 1.16c. Because the horizontal forces do not affect the vertical equilibrium, the balance is between the load, the reaction, and the vertical component of the force in the upper chord. Simple observation of the forces and the known magnitudes makes it obvious that force BI v must act downward, indicating that BI is a compression force. Thus the sense of BI is established by simple visual inspection of the joint, and the algebraic equation for vertical equilibrium (with upward force considered positive)

From this equation, BI v is determined to have a magnitude of 2000 lb. Using the known relationships between BI, BI v , and BI h , we can determine the values of these three quantities if any one of them is known.

The results of the analysis to this point are shown in Fig. 1.16d, from which it may be observed that the conditions for equilibrium of the horizontal forces can be expressed. Stated algebraically (with force sense toward the right considered positive), from which it is established that the force in IH is 3000 lb.

The final solution for the joint is then as shown in Figure 1.16e. On this diagram, the internal forces are identified as to sense by using C to indicate compression and T to indicate tension.

As with the graphic solution, proceed to consider the forces at joint 3. The initial condition at this joint is as shown in Fig. 1.17a, with the single known force in member HI and the two unknown forces in IJ and JH. Because the forces at this joint are all vertical and horizontal, there is no need to use components. Consideration of vertical equilibrium makes it obvious that it is not possible to have a force in member IJ. Stated algebraically, the condition for vertical equilibrium is

It is equally obvious that the force in JH must be equal and opposite to that in HI because they are the only two horizontal forces. That is, stated algebraically

The final answer for the forces at joint 3 is as shown in Fig. 1.17b. Note the convention for indicating a truss member with no internal force.

Now proceed to consider joint 2; the initial condition is as shown in Fig. 1.18a. Of the five forces at the joint, only two remain unknown. Following the procedure for joint 1, first resolve the forces into their vertical and horizontal components, as shown in Fig. 1.18b.

Because the sense of forces CK and KJ is unknown, consider them to be positive until proven otherwise. That is, if they are entered into the algebraic equations with an assumed sense, and the solution produces a negative answer, then the assumption was wrong. However, be careful to be consistent with the sense of the force vectors, as the following solution will illustrate.

Arbitrarily assume that force CK is in compression and force KJ is in tension. If this is so, the forces and their components will be as shown in Fig. 1.18c. Then, consider the conditions for vertical equilibrium; the forces involved will be those shown in Fig. 1.18d, and the equation for vertical equilibrium will be

or

Now consider the conditions for horizontal equilibrium; the forces will be

as shown in Fig. 1.18e, and the equation will be

Note the consistency of the algebraic signs and the sense of the force vectors, with positive forces considered as upward and toward the right.

Now solve these two equations simultaneously for the two unknown forces as follows:

Because the algebraic solution produces a negative quantity for KJ, the assumed sense for KJ is wrong, and the member is actually in compression.

The final answers for the forces at joint 2 are, thus, as shown in Fig. 1.18g. In order to verify that equilibrium exists, however, the forces are shown in the form of their vertical and horizontal components in Fig. 1.18f.

When all the internal forces have been determined for the truss, we can record or display the results in a number of ways. The most direct way is to display them on a scaled diagram of the truss, as shown in Fig. 1.19a. The force magnitudes are recorded next to each member with the sense shown as T for tension or C for compression. Zero stress members are indicated by the conventional symbol consisting of a zero placed directly on the member.

When solving by the algebraic method of joints, the results may be recorded on a separated joint diagram as shown in Fig. 1.19b. If the values for the vertical and horizontal components of force in sloping members are shown, it is a simple matter to verify the equilibrium of the individual joints.

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