Basic Theory of StructuresBasic Theory of Structures. |
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Page 53
... concentrated loads will be a series of horizontal steps , and the b.m. diagram will be polygonal . If a beam carries only uniform loading ( not necessarily con- tinous over the whole span ) the s.f. diagram will be a sloping straight ...
... concentrated loads will be a series of horizontal steps , and the b.m. diagram will be polygonal . If a beam carries only uniform loading ( not necessarily con- tinous over the whole span ) the s.f. diagram will be a sloping straight ...
Page 121
... load crossing a structure , it is a simple matter to determine the effect of a train of rolling loads from the unit influence line . A train of loads usually consists of a number of concentrated ( point ) loads linked together , so that ...
... load crossing a structure , it is a simple matter to determine the effect of a train of rolling loads from the unit influence line . A train of loads usually consists of a number of concentrated ( point ) loads linked together , so that ...
Page 141
... point 15 ft from the left - hand support , ( b ) the maximum bending moment under the 10 tonf load . Ans : ( a ) 226 tonf ft ; ( b ) 242-8 tonf ft ( I.C.E. Apr. 1952 ) 7. The following train of loads rolls across a girder , simply ...
... point 15 ft from the left - hand support , ( b ) the maximum bending moment under the 10 tonf load . Ans : ( a ) 226 tonf ft ; ( b ) 242-8 tonf ft ( I.C.E. Apr. 1952 ) 7. The following train of loads rolls across a girder , simply ...
Contents
PREFACE | 11 |
Shearing Force and Bending Moment | 41 |
Relationship between loading | 47 |
Copyright | |
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acting action angles applied arch assumed axial b.m. diagram bars base beam bending stress body bridge buckling Calculate cantilever carries centre component compression compressive stress concrete consider Construct cross-section deflection depth Determine developed direct stress displacement distance divided effect elastic equal equation equilibrium Example expression factor fibres Figure flange forces girder give given Hence horizontal in² influence line intensity internal joint lbf/ft² lbf/in² length line of s.f. load maximum maximum b.m. method moments neutral axis Note obtained pair parallel pinned plane portion position pressure principal produce ratio reaction representing resistance resultant shearing force shown in Fig shows simple simply supported space span steel strength structure strut subjected taking tensile tension tion tonf ft tonf/in² train truss uniform uniformly distributed load unit usually vector vertical wall weight wide