Analysis and Behavior of StructuresOffering students a presentation of classical structural analysis, this text emphasizes the limitations required in creating mathematical models for analysis, including these used in computer programs. Students are encouraged to use hand methods of analysis to develop a feel for the behaviour of structures. |
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Page 142
... kN / m R 80 kN - m + 2md Reaction computation • • - EMA : 167 3.5+ 100 · 580 - Rc · 7 = 0 Rc = 116 kN गीत · 5 m RA Rc 80 kN - m 100 kN Loading diagram RA = 96 kN 96 KN 116 KN ΣFR + 116 - 100 - A - - 16 · 7 = 0 16kN 100 kN MBr 96 kN 1 ...
... kN / m R 80 kN - m + 2md Reaction computation • • - EMA : 167 3.5+ 100 · 580 - Rc · 7 = 0 Rc = 116 kN गीत · 5 m RA Rc 80 kN - m 100 kN Loading diagram RA = 96 kN 96 KN 116 KN ΣFR + 116 - 100 - A - - 16 · 7 = 0 16kN 100 kN MBr 96 kN 1 ...
Page 483
... kN / m B C 21 TH I गीता 4 m 4 m 4 m A B D MAB MBA MBC MCB MCD MDC ( 1 ) 21/4 2/3 1/4 1/2 1/4 ( 2 ) 90 60 -60 ( 3 ) 20 - 1/2 40 20 1/2 - 10 ( 4 ) -2.5 + 1/2 -5 -5 1/2 → -2.5 ( 5 ) 0.8 1/2 1.7 0.8 1/2 0.4 ( 6 ) ≈0 - 1 ...
... kN / m B C 21 TH I गीता 4 m 4 m 4 m A B D MAB MBA MBC MCB MCD MDC ( 1 ) 21/4 2/3 1/4 1/2 1/4 ( 2 ) 90 60 -60 ( 3 ) 20 - 1/2 40 20 1/2 - 10 ( 4 ) -2.5 + 1/2 -5 -5 1/2 → -2.5 ( 5 ) 0.8 1/2 1.7 0.8 1/2 0.4 ( 6 ) ≈0 - 1 ...
Page 497
... M MBC = MCB = -60 kN - m . = AB = MB = 120 kN - m and ΒΑ STEP 4 Do moment distribution using moments computed in step 3 . A MAB 120 -20 100 - 1/2 B MBA MBC 2/3 1/3 120 -60 -40 -20 1/2 80 -80 STEP 5 Compute reaction force at B. This ...
... M MBC = MCB = -60 kN - m . = AB = MB = 120 kN - m and ΒΑ STEP 4 Do moment distribution using moments computed in step 3 . A MAB 120 -20 100 - 1/2 B MBA MBC 2/3 1/3 120 -60 -40 -20 1/2 80 -80 STEP 5 Compute reaction force at B. This ...
Common terms and phrases
action analysis antisymmetric applied loads assumption axial loads calculation centroidal column complementary virtual Compute concentrated load conjugate beam constant cross section curvature diagram defined deformation system direct integration displacements and rotations distributed load Draw the final end moments equations of equilibrium equilibrium equations Example Figure final moment diagram forces and moments free body hinge horizontal indeterminate structure influence line integration joint kips kN/m left end linear linear elastic loading diagram magnitude mathematical model maximum member A-B member forces ment moment distribution moment of inertia Neglect axial deformations nonlinear materials nonprismatic numerical integration panel points positive reaction components shown in Fig sign convention simply supported beam slope spreadsheet statically determinate structures STEP strain energy stress stress-strain relation struc superposition tion truss U₁ uniform load unit load vertical deflection vertical displacement virtual force system virtual work principle zero ΕΙ