9 781292 027074

ISBN 978-1-29202-707-4

Statics and Strength of Materials forArchitecture and Building Construction

Barry S. Onouye Kevin KaneFourth Edition

Statics and Strength of Materials Onouye Kane Fourth Edition

Pearson Education LimitedEdinburgh GateHarlowEssex CM20 2JEEngland 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 affi liation with or endorsement of this book by such owners.

ISBN 10: 1-269-37450-8ISBN 13: 978-1-269-37450-7

British Library Cataloguing-in-Publication DataA catalogue record for this book is available from the British Library

Printed in the United States of America

Copyright_Pg_7_24.indd 1 7/29/13 11:28 AM

ISBN 10: 1-292-02707-XISBN 13: 978-1-292-02707-4

ISBN 10: 1-292-02707-XISBN 13: 978-1-292-02707-4

Analys i s o f Se lected Determinate Str uctur a l Systems

Using FBD (c),

For the internal pin forces at C,

Going back to FBD (b) of the entire frame, solve for Bx:

The negative sign in the r esult for indicates that theoriginal direction assumed in the FBD was incorrect.

‹Bx = 0.02 k1:2

Bx

‹Bx = -0.02 kCgFx = 0 D + 5 k - 5.02 k - 0.8 k + 0.8 k - Bx = 0

‹ Cy = +1.04 kCgFy = 0 D - 3.44 k + 2.4 k + Cy = 0

‹Cx = +0.82 kCgFx = 0 D + 5 k - 5.02 k - 0.8 k + Cx = 0

‹ Ax = 5.02 k1;228Ax = 140.56

+3.44 k124¿2 - Ax128¿2 = 0CgMC = 0 D - 2.4 k112¿2 - 0.8 k14¿2 + 5 k118¿2

FBD (c) FBD (d)

174

Analys i s o f Se lected Determinate Str uctur a l Systems

Problems

Determine all support and pin for ces for the multifor cemember diagrams listed below.28

29

30

175

Analys i s o f Se lected Determinate Str uctur a l Systems

31

32

33

176

Analys i s o f Se lected Determinate Str uctur a l Systems

6 RETAINING WALLSAs the name implies, retaining walls are used to hold back(retain) solid or other granular material to maintain a dif-ference in ground elevation. A dam is a retaining wall usedto resist the lateral pressure of water or other fluids.There are three general types of r etaining walls: (a) thegravity wall (Figur e 76), (b) the r einforced concretecantilever retaining wall (Figure 77), and (c) the reinforcedconcrete cantilever r etaining wall with counterforts(Figure 78).Gravity retaining walls are generally built of plain con-crete or masonry. Height h is generally less than four feet(1.3 m). A gravity wall depends on its mass to give it sta-bility against the horizontal forces from the soil. Slidingresistance (friction) is developed between the concrete andsoil at the base. Some major dams are constructed as grav-ity wall systems, but understandably, the base dimensionsare immense.Reinforced concrete cantilever retaining walls are the mostfrequently used type of retaining wall, with an effective-ness 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 thestructure and the weight of the soil on the heel of the slabbase. Sometimes a shear key is included at the bottom ofthe slab base to incr ease the wall’s resistance to sliding.Retaining walls should have their foundations well belowthe frost line, and adequate drainage (weep) holes near thebottom of the wall should be provided to permit the wateraccumulation behind the wall to escape.As the height of a r etaining wall increases, the bendingmoment in the cantilever wall incr eases, requiring morethickness. The addition of counterforts (vertical triangu-lar-shaped cross-walls) provides the additional depth atthe base to absorb the large bending stresses. Counterfortwalls behave like one-way slabs that span horizontallybetween the counterforts. Counterforts are called buttresseswhen this same configuration is used for the r etainedearth that is on the flat side of the wall.Saturated loose sand or gravel, granular soil, or mudcause pressures against retaining walls in a manner simi-lar to true fluids (liquids) by exerting a horizontal pr es-sure. In true liquids, like water , the horizontal pr essureand the vertical pressure are the same at a given depth.However, in soil, the horizontal pressure is less than thevertical pressure, with the ratio dependent on the physical

Figure 76 Gravity retaining wall.

Figure 78 Counterfort wall.

Figure 77 Reinforced cantilever retainingwall.

177

Analys i s o f Se lected Determinate Str uctur a l 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 toa maximum at the bottom of the footing.

wherethe magnitude of the earth pressure in psf or

“equivalent” fluid weight (density) of thesoil in pounds per cubic feet. V alues range from aminimum of 30 pcf (for well-graded, clean gravel-sand mixes) to 60 pcf (for clayey sands). SI values are4.7 to .

Therefore,

where the lateral force (pounds, kips, N, or kN) basedon the area of the pressure distribution acting on a 1-foot-wide (1-m-wide) strip of wall.

the maximum pressure at depth h (psf or kN/m2)Equivalent fluid pressure against a retaining wall may cre-ate conditions of instability. Retaining walls are susceptibleto three failure modes: (a) sliding—when the friction at thefooting base is insufficient to resist sliding; (b) overturningabout the toe—when the lateral for ce produces an over-turning moment greater than the stabilizing moment fromthe wall’s weight, slab base weight, and the soil massabove the heel; and (c) excessive bearing pr essure at thetoe—when the combination of the vertical downward forceand the compression at the toe caused by the horizontalforce exceeds the allowable bearing pressure of the soil.The pressure distribution under the base (Figure 80) de-pends upon the location and magnitude of the r esultant(vertical and horizontal forces) force as it passes throughthe footing base.Analysis of a cantilever r etaining wall requires that theequilibrium summation of moments about the toe is sta-ble; that is, the weight of the wall plus the backfill on theheel exceeds the overturning moment of the active soilpressure by a factor of at least 1.5 (a safety factor imposedby building codes). Once a stable configuration is achieved,the soil pressure distribution on the footing must be cal-culated to ensure that the bearing pr essures are withinallowable limits for the soil on site.

pmax =

P =

P =

12

* 1pmax * h2 * 1 ft. or 1 m

h = soil depth in feet 1m2.9.4 kN>m3

ω¿ = the1kN>m22p =

p = ω¿ * h

Figure 80 Bearing pressure under the wallfooting.

Figure 79 FBD of a gravity retaining wall.

178