Statics and Strength of Materials for Architecture and
Transcrição
Statics and Strength of Materials for Architecture and
Statics and Strength of Materials Onouye Kane Fourth Edition ISBN 978-1-29202-707-4 9 781292 027074 Statics and Strength of Materials for Architecture and Building Construction Barry S. Onouye Kevin Kane Fourth Edition Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. ISBN 10: 1-292-02707-X ISBN 10: 1-269-37450-8 ISBN 13: 978-1-292-02707-4 ISBN 13: 978-1-269-37450-7 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America Analysis of Selected Deter minate Str uctur al Systems FBD (c) FBD (d) Using FBD (c), C g MC = 0 D - 2.4 k112¿ 2 - 0.8 k14¿ 2 + 5 k118¿2 +3.44 k124¿ 2 - Ax 128¿ 2 = 0 28Ax = 140.56 ‹ Ax = 5.02 k1;2 For the internal pin forces at C, C g Fx = 0 D + 5 k - 5.02 k - 0.8 k + Cx = 0 ‹ Cx = + 0.82 k C g Fy = 0 D - 3.44 k + 2.4 k + Cy = 0 ‹ Cy = + 1.04 k Going back to FBD (b) of the entire frame, solve for Bx: C g Fx = 0 D + 5 k - 5.02 k - 0.8 k + 0.8 k - Bx = 0 ‹ Bx = - 0.02 k The negative sign in the r esult for Bx indicates that the original direction assumed in the FBD was incorrect. ‹ Bx = 0.02 k1:2 174 Analysis of Selected Deter minate Str uctur al Systems Problems Determine all support and pin for ces for the multifor ce member diagrams listed below. 28 29 30 175 Analysis of Selected Deter minate Str uctur al Systems 31 32 33 176 Analysis of Selected Deter minate Str uctur al Systems 6 RETAINING WALLS As the name implies, retaining walls are used to hold back (retain) solid or other granular material to maintain a difference in ground elevation. A dam is a retaining wall used to resist the lateral pressure of water or other fluids. There are three general types of r etaining walls: (a) the gravity wall (Figur e 76), (b) the r einforced concrete cantilever retaining wall (Figure 77), and (c) the reinforced concrete cantilever r etaining wall with counterforts (Figure 78). Gravity retaining walls are generally built of plain concrete or masonry. Height h is generally less than four feet (1.3 m). A gravity wall depends on its mass to give it stability against the horizontal for ces from the soil. Sliding resistance (friction) is developed between the concrete and soil at the base. Some major dams are constructed as gravity wall systems, but understandably, the base dimensions are immense. Figure 76 Gravity retaining wall. Figure 77 wall. Reinforced cantilever retaining Figure 78 Counterfort wall. Reinforced concrete cantilever retaining walls are the most frequently used type of retaining wall, with an effectiveness up to a height (h) of about 20 to 25 feet (6 to 7.6 m). Stability of this wall type is achieved by the weight of the structure and the weight of the soil on the heel of the slab base. Sometimes a shear key is included at the bottom of the slab base to incr ease the wall’s r esistance to sliding. Retaining walls should have their foundations well below the frost line, and adequate drainage (weep) holes near the bottom of the wall should be provided to permit the water accumulation behind the wall to escape. As the height of a r etaining wall increases, the bending moment in the cantilever wall incr eases, requiring more thickness. The addition of counterforts (vertical triangular-shaped cross-walls) provides the additional depth at the base to absorb the large bending stresses. Counterfort walls behave like one-way slabs that span horizontally between the counterforts. Counterforts are called buttresses when this same configuration is used for the r etained earth that is on the flat side of the wall. Saturated loose sand or gravel, granular soil, or mud cause pressures against retaining walls in a manner similar to true fluids (liquids) by exerting a horizontal pr essure. In true liquids, like water , the horizontal pr essure and the vertical pr essure are the same at a given depth. However, in soil, the horizontal pressure is less than the vertical pressure, with the ratio dependent on the physical 177 Analysis of Selected Deter minate Str uctur al Systems properties of the soil. Soil pr essure, as with liquids, increases proportionately with its depth below grade (Figure 79). Lateral pressure increases linearly from zero at the top to a maximum at the bottom of the footing. p = ω¿ * h where p = the magnitude of the earth pressure in psf or 1kN>m2 2 Figure 79 FBD of a gravity retaining wall. ω¿ = the “equivalent” fluid weight (density) of the soil in pounds per cubic feet. V alues range from a minimum of 30 pcf (for well-graded, clean gravelsand mixes) to 60 pcf (for clayey sands). SI values are 4.7 to 9.4 kN>m3. h = soil depth in feet 1m2. Therefore, P = 1 * 1pmax * h2 * 1 ft. or 1 m 2 where P = the lateral force (pounds, kips, N, or kN) based on the area of the pressure distribution acting on a 1-footwide (1-m-wide) strip of wall. pmax = the maximum pressure at depth h (psf or kN/m2) Equivalent fluid pressure against a retaining wall may create conditions of instability. Retaining walls are susceptible to three failure modes: (a) sliding—when the friction at the footing base is insufficient to resist sliding; (b) overturning about the toe—when the lateral for ce produces an overturning moment greater than the stabilizing moment from the wall’s weight, slab base weight, and the soil mass above the heel; and (c) excessive bearing pr essure at the toe—when the combination of the vertical downward force and the compression at the toe caused by the horizontal force exceeds the allowable bearing pressure of the soil. The pressure distribution under the base (Figur e 80) depends upon the location and magnitude of the r esultant (vertical and horizontal forces) force as it passes through the footing base. Figure 80 footing. 178 Bearing pressure under the wall Analysis of a cantilever r etaining wall requires that the equilibrium summation of moments about the toe is stable; that is, the weight of the wall plus the backfill on the heel exceeds the overturning moment of the active soil pressure by a factor of at least 1.5 (a safety factor imposed by building codes). Once a stable configuration is achieved, the soil pressure distribution on the footing must be calculated to ensur e that the bearing pr essures are within allowable limits for the soil on site.