Simplified Structural Analysis and Design for Architects

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2 Simplified Structural Analysis and Design for Architects Second Edition Rima Taher, PhD, PE New Jersey Institute of Technology

3 Bassim Hamadeh, CEO and Publisher Kassie Graves, Director of Acquisitions and Sales Jamie Giganti, Senior Managing Editor Miguel Macias, Senior Graphic Designer Carrie Montoya, Manager, Revisions and Author Care Natalie Lakosil, Licensing Manager Kaela Martin, Associate Editor Christian Berk, Associate Production Editor Copyright 2018 by Cognella, Inc. All rights reserved. No part of this publication may be reprinted, reproduced, transmitted, or utilized in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information retrieval system without the written permission of Cognella, Inc. For inquiries regarding permissions, translations, foreign rights, audio rights, and any other forms of reproduction, please contact the Cognella Licensing Department at com. Cover image copyright 2011 by Depositphotos / Vladitto. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Printed in the United States of America ISBN: (pbk) / (br) / (hc)

4 Contents Preface vi Chapter 1 Forces and Force Systems 2 Chapter 2 Moments and Equilibrium of Forces 16 Chapter 3 Equilibrium of Non-Concurrent Coplanar Forces Truss Analysis 24 Chapter 4 Properties of Materials Stress and Strain 40 Chapter 5 Centroids and Moment of Inertia 50 Chapter 6 Structural Analysis of Beams Shear and Bending Moment 62 Chapter 7 Design Loads 78 Chapter 8 Steel Structures Structural Steels 92 Chapter 9 Steel Structural Systems 100 Chapter 10 Steel Design Design Methods and Load Combinations 118 Chapter 11 Simplified Design of Steel Beams 124 Chapter 12 Simplified Design of Steel Columns 158 Chapter 13 Wood Structures and Structural Systems 176

5 iv Simplified Structural Analysis and Design for Architects Chapter 14 Simplified Design of Wood Structures 190 Chapter 15 Reinforced Concrete Construction 208 Chapter 16 Reinforced Concrete Structures and Structural Systems 220 Chapter 17 Simplified Design of Reinforced Concrete Beams 236 Chapter 18 Foundation Systems 248 Chapter 19 Analysis of Indeterminate Structures 274 Chapter 20 Approximate Analysis of Concrete Indeterminate Structures 286 References 294

6 I wish to thank the entire Cognella staff who worked on the two editions of this book, especially Jamie Giganti, Managing Editor, Kaela Martin, and Christian Berk, Associate Editors, for their invaluable help. I also thank Mr. John Remington, Senior Field Acquisitions Editor, for giving me the opportunity to write this book, Natalie Lakosil, Senior Licensing Manager, and Carrie Montoya, Manager, Revisions and Author Care at Cognella for their help with the second edition work. Dani Skeen, Marketing Coordinator at Cognella, is helping with the marketing efforts and I wish to thank her for her contribution. Special thanks to Brian Fahey, Chelsey Rogers, Jennifer Levine, and Miguel Macias for their help with the first edition. I would like to address my sincere thanks to three of my teaching assistants at the College of Architecture & Design at New Jersey Institute of Technology who helped with the drafting of some of the figures in the final chapters of this book, Adebayo Oyeniya, Michael Natanzon and Banafsheh Soltani. Their help is greatly appreciated.

7 PREFACE

8 Structural analysis and design is a wide and complex field that is generally difficult to learn and to teach, especially within the time limits often dictated by academic programs. The purpose of this book is to introduce structures to architecture students in a simplified, clear and concise manner and to give the students an understanding of the essence of structural technology as it relates to architecture and construction. Having taught this subject for many years to architecture students, I have become familiar with the types of difficulties that these students experience when learning this material, and I have used my teaching experience in outlining the various topics in a simple and direct manner that is relatively easy to understand. This book covers the basics of statics and strength of materials, the various steel, wood and concrete structural systems, and the design of basic structural elements using these traditional structural materials and some simplified design methods. This edition also covers the various types of foundation systems and the design of simple footings. The final chapters introduce some analysis methods of inderterminate structures. Understanding the basic principles of structural technology is essential to all building professionals. However, complex engineering methodology and analysis procedures can become sometimes overwhelming, thus discouraging some of the most avid learners. My goal in writing vii

9 viii Simplified Structural Analysis and Design for Architects this book was mainly to simplify the process to allow students not just to learn about structures but to also enjoy learning the material at the same time. I hope that I have somewhat succeeded in this task and in making structural technology and engineering principles interesting and less intimidating than they usually are. This book should be helpful to undergraduate and graduate architecture students. It should also be useful to practicing architects and can be used as a reference by architecture graduates who are preparing for the Architect Registration Examination. Rima Taher, PhD, PE

10 CHAPTER ONE Forces and Force Systems

11 1.1 Fundamental Terms Mechanics Mechanics is a science that deals with bodies in motion or bodies at rest. It is divided into two fields: Statics: The branch of mechanics that deals with bodies at rest. Dynamics: The branch of mechanics that deals with bodies in motion Strength of Materials Strength of Materials is the science that studies the ability of components of structures to resist loads and the selection of materials for this purpose Force A force is an action applied on a body that can make it move or change its motion, size, or shape. 3

12 4 Simplified Structural Analysis and Design for Architects Rigid Body A rigid body is a term used in mechanics to designate a body whose size and shape do not change due to forces acting on it. The notion of a rigid body is a virtual notion Gravity Gravity is the force with which a body is pulled toward the center of the earth. The weight of a body is equal to this gravitational pull Mass The mass of a body represents the amount of matter in the body. The mass is a constant and is different from the weight Density There are two types of density: weight density and mass density. The weight density is the weight of a material per unit volume. The mass density represents the mass per unit volume Load The load is the force exerted by some body on a supporting element. For instance, the weight of the snow on a roof is a load applied on the roof Moment The moment of a force is related to the tendency of this force to cause rotation about a certain axis Vectors and Scalars Quantities can be represented using vectors or scalars. A vector quantity is given as an arrow where the length of the line is the magnitude of the quantity, while the arrowhead shows the direction. A scalar is just a number representing the magnitude only. A scalar has no direction. 1.2 Types of Forces There are different types of forces: External Forces External forces are applied to a body through an external source.

13 Forces and Force Systems Internal Forces Internal forces are exerted by the material of a body and generally represent a resistance to deformation Coplanar Forces Forces that are placed and act in the same plane are called coplanar forces Concurrent Forces Concurrent forces have lines of action that intersect or meet in a common point (Figure 1.1). F 3 F 2 0 F 1 Figure 1.1 Concurrent Forces Collinear Forces Collinear forces are forces that act along the same line Parallel Forces Parallel forces have parallel lines of action. 1.3 Principle of Transmissibility of a Force This principle states that a force is generally considered as acting at any point located on its line of action, provided that the direction and magnitude of the force remain the same. 1.4 The Principle of Action and Reaction This principle states that to every action, there is an equal and opposite reaction. For example, when a 20-lb. box is placed on a desk, the weight of the box (W) exerts a force on this desk. The desk replies with an equal and opposite force (N), as shown in Figure 1.2. This principle is also known as Newton s third law.

14 6 Simplified Structural Analysis and Design for Architects 1.5 Axial Forces Tension and Compression When a force is acting along the axis of a member such as a rod or bar, it is called an axial force. Axial forces are considered tension forces when their action tends to stretch the member (Figure 1.3a). They are called compression forces when they tend to compress the body or member in question (Figure 1.3b). 1.6 Resultant of a Force System The resultant of a force system is defined as a single force that can replace the system of forces and produce the same effect created by the system. W N Figure 1.2 Action and Reaction 1.6a Resultant of Collinear Forces To find the resultant of two collinear forces: 1. If the two forces have the same direction, keep this direction and add the magnitudes of the two forces (Figure 1.4a). (a) F F (b) F F Figure 1.3 (a) Tension (b) Compression 2. If the two forces have opposite directions, use the direction of the larger force and subtract the smaller magnitude from the larger one to get the magnitude of the resultant (Figure 1.4b).

15 Forces and Force Systems 7 1.6b Resultant of Two Concurrent Forces The resultant of two concurrent forces can be determined using either a graphic method or an analytical (algebraic) method. This resultant acts through the point of intersection. 1.6b.1 Graphic Method The resultant is the diagonal of the parallelogram constructed using these two concurrent forces, as shown in Figure 1.5a. The diagonal used is the one that passes through the intersection point. (a) 10 lb 40 lb R = 50 lb (b) 10 lb 40 lb R = 30 lb Figure 1.4 Resultant of Collinear Forces If the two concurrent forces are perpendicular, then the resultant is the diagonal of the rectangle built using these two forces, as illustrated in Figure 1.5b. The resultant of two concurrent forces can also be determined by constructing a triangle, instead of a parallelogram, as shown in Figure 1.5c. 1.6b.2 Algebraic Method Two trigonometric laws are used in this method to determine the magnitude and direction of the resultant. The first law is known as the Law of Cosines, and it is used to determine the magnitude of the resultant. The second law, known as the Law of Sines, is used to determine the direction of the resultant. The triangle of forces is drawn first, using the two concurrent forces in question; then, the laws of cosines and sines are applied to this triangle, as explained in the example below.

16 8 Simplified Structural Analysis and Design for Architects F 2 F 2 R F 2 F 2 R 0 F 1 0 F 1 0 F 1 0 F 1 (a) Parallelogram of Forces (b) Rectangle of Forces F 2 α R α F 2 0 F 1 F 1 0 (c) Triangle of Forces Figure 1.5 Resultant of Two Concurrent Forces Example 1.6b.1 Two concurrent forces of 75 lb. (F 1 ) and 100 lb. (F 2 ) make an angle of 55 (α), as shown in Figure 1.6a below. Determine the magnitude and direction of the resultant. Solution Construct a triangle of forces (Figure 1.6b) as follows: 1. Draw a vector parallel and equal to the first force (F 1 ). 2. Starting at the end of the first vector, draw a second vector that is parallel and equal to the second force (F 2 ). 3. The resultant of the two forces is represented by the closing line of the triangle. It is the third vector that starts at the starting point of F 1 and ends at the ending point of F 2. Next, write the Law of Cosines for this triangle. This law helps determine the length of the third side of a triangle when the other two sides and the angle between them are known. In this triangle of forces: R 2 = F F 2 2 2F 1 F 2 Cosθ

17 Forces and Force Systems 9 (a) F 2 = 100 lb (b) R β F 2 = 100 lb 0 α = 55 F 1 = 75 lb γ θ F 1 = 75 lb 0 α = 55 Figure 1.6 Example 1.6b.1 R 2 = (75) 2 + (100) 2 2(75)(100).Cos 125 R 2 = (5625) + (10,000) (15,000) ( 0.574) R 2 = 15, R 2 = 24,235 R = or R = 156 lb The magnitude of the resultant is approximately 156 lb. To determine the direction of this resultant, apply the Law of Sines. This is done by writing three equal ratios. Each ratio is the length of one side in the triangle in question divided by the sine of the angle facing it. In this triangle of forces: F1 F2 R = = sinβ sin γ sin θ 75 lb 100 lb 156 lb = = sinβ sin γ sin lb 156 lb = sin β sin lb sin 125 sin β = 156 lb sin β = β = γ = γ = 318. The resultant makes an angle of 31.8 with the force F 1 and an angle of 23.2 with the force F 2, as shown in the diagram. Finding one of these two angles is generally sufficient to determine the direction of the resultant.

18 10 Simplified Structural Analysis and Design for Architects 1.7 Equilibrium of Concurrent Forces A system of concurrent forces is said to be in equilibrium when the resultant of these forces is equal to zero. A body subjected to a system in equilibrium is a body at rest or in a state of uniform motion. If two concurrent forces are acting on a body, the equilibrant is the third force that would hold these two forces in equilibrium. The equilibrant is represented by the closing line of the triangle of forces, as illustrated in Figure 1.7. Equilibrant F 2 F 1 0 Figure 1.7 Equilibrant 1.8 Free-Body Diagram A free-body diagram is a diagram that shows the studied body and all the forces acting on it. These forces are represented by vectors. 1.9 Components of a Force A force can be replaced by two forces that produce the same effect. These forces are called components. Example Resolve a force of 90 lb. into two components making angles of 40 and 75, respectively, with the force.

19 Forces and Force Systems 11 Solution Draw the triangle of forces as shown in Figure 1.8. The third angle is 65. Next, write the Law of Sines, and solve it for the unknown components as follows: F 1 65 F F = 90 lb Figure 1.8 Example F1 F2 F sin 75 = sin 40 = sin 65 F1 90 lb = lb ( ) F1 = F = lb 1 F2 90 lb = lb ( ) F2 = F = lb Rectangular Components of a Force If the angle between the components of a force is 90, then the components are called rectangular components. Generally, a rectangular coordinate system is selected, and the forces are resolved into this system with rectangular components acting along the X- and Y-axes. Refer to Figure 1.9. The horizontal component (F x ) of the force (F) is: F x = F. cos α

20 12 Simplified Structural Analysis and Design for Architects y F y 0 α F F x x Figure 1.9 Rectangular Components The vertical component of the force (F) is: F y = F. sin α The force (F) can be calculated using its rectangular components: F 2 = F x 2 + F y 2 or 2 2 x y F= F + F Example A force (F) of 60 lb. makes an angle of 60 with the horizontal. Determine the rectangular components of this force. Solution F x = F. cos α = 60 lb cos 60 = 60 lb (0.500) = lb F y = F. sin α = 60 lb sin 60 = 60 lb (0.866) = lb

21 Forces and Force Systems Resultant of More than Two Concurrent Forces in a Plane The resultant of more than two concurrent forces in a plane can be determined using a graphical solution, which consists in drawing a polygon of forces, as shown in Figure Starting from a Point A in the plane, draw a vector that is parallel and equal to vector F 1. From the end of the first vector, draw a second vector that is parallel and equal to vector F 2 The closing line of the polygon is the resultant (R) of the system of forces. F 2 F 4 F 3 F 3 0 F 1 R F 2 F 4 A Figure 1.10 Polygon of Forces F Resultant of Concurrent Forces by Summation A system of concurrent planar forces can be replaced by rectangular components in a rectangular coordinate system. The origin of the coordinate system would be the intersection point of the forces. The components are then combined along the two axes by algebraic addition in order to determine the rectangular components of the resultant (R). The following example illustrates the Summation Method. Example Determine the magnitude and direction of the resultant of the system of concurrent forces shown in Figure

22 14 Simplified Structural Analysis and Design for Architects y F 3 = 95 lb F 2 = 60 lb 60 0 F 1 = 70 lb x Figure 1.10 Example Solution The rectangular components of the three forces are calculated and summarized in the following table: Force Angle with X-Axis (α) F x = F. cos α F y = F. sin α 70 lb 0 70 lb. cos 0 = lb 70 lb. sin 0 = 0.00 lb 60 lb lb. cos 90 = 0.00 lb 60 lb. sin 90 = lb 95 lb lb. cos 120 = lb 95 lb. sin 120 = lb Σ F x = 22.5 lb Σ F y = lb

23 Forces and Force Systems 15 R= F F x 2 2 ( y) ( ) R = ( 22. 5) + ( ) R = , R = lb Fy Tanθ = F = =. x θ = The resultant (R) has a magnitude of lb., and it makes an angle (θ) of with the positive side of the X-axis Equilibrium of Concurrent Forces When a system of concurrent forces is in equilibrium, the resultant of forces (R) is equal to zero. ( ) x y 2 2 x y 0 R= F F ( F ) + F ( ) ( ) = The sum of two positive numbers can be equal to zero only when each term is equal to zero: 2 ( Fx) = 0 Fx = 0 2 ( Fy) = 0 Fy = 0 As a conclusion, a system of concurrent forces is in equilibrium when the algebraic sums of the components along both axes are equal to zero.

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