E0-190-2008_(7)Chapter_5(DIF)

1 f rt f eq5 ug f gives f g

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Unformatted text preview: ) . F(=P ) = f1+ f2+ ……+ fm = {1, 1, ….. ,1}{ f } ={R}T{ f } Eq.(5) : {u}=[G]{ f } gives, { f }=[G] –1 { u } (5) (11) (10) Substitute Eq.(11): { f }=[G] –1 { u } (11) into Eq.(10): F(=P ) = {R}T{ f } (10) then, we obtain Eq.(12): F(=P ) ={R}T{ f }={R}T[G]-1{ u } (12) Furthermore, the substitution of Eq.(8) : {u}={R} U0 (8) into Eq.(12) gives Eq.(13): F(=P ) = {R}T[G]-1{R} UO (13) Because Eq.(13): F(=P ) = {R}T[G]-1{R} UO (13) relates the applied force F(=P) to the representative displacement U0 of the foundation, the coeff...
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