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Cont1nuous BeamsGeneral Load1ng
General Solut1on of the ThreeMember Equat1ons
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a.m.inst.ce AEFB assumed b.sc bending moments calculated cantilever cent centre of gravity CKLD continuous beam deflection deformation at point deformation due denominator determinate system direction elastic equations elastic weights equal and opposite equation of elasticity equations of equilibrium Example.—Find external loads fitl force H forces or moments frame fixed hinge horizontal force horizontal thrust illus illustrations indeterminate structures influence line internal forces KH(i live load load wk M-diagram m.inst.ce method midspan Mohr's rule moments due obtained occurs once-indeterminate opposite hand ordinates post 9s previous example principle of virtual Published 1939 Reinforced Concrete resultant roller bearing rth span shear shown in Fig side Similarly solution statically indeterminate step support moments support reaction supporting forces symmetrical system of forces three equations three times indeterminate three-member equations total load two-hinged frame unknown Y1 vertical deflection Xu X2 zer0 zero