Examen306
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1. This document appears to be an analysis of the structural behavior of a system under various loads and conditions. It includes information on beam properties, applied forces, stiffness matrices, and displacement calculations. 2. Methods for calculating the rotation matrix, stiffness of individual beams, and assembling the overall structural stiffness matrix are presented. Applied loads are also broken down into different components. 3. Equations are shown for calculating the displacements of the system based on the applied forces and the overall structural stiffness matrix. Errors are reported in inverting the stiffness matrix, suggesting an instability in the system.
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Examen306
- 1. EXAMEN REVALIDA SEM II /16 ANALISIS ESTRUCTURAL AVANZADO – CIV 3306 “A” OR/ / / DOCENTE: Ing. Hugo Mercado AUXILIAR: Univ. Alfredo Tapia Velasquez NOMBRE: Univ. Michael Alex Velasquez Sandoval CI 4020016 Or. = 1/(1/r1 + 1/r2) = t/m ESTRUCTURA HIPERESTATICA DATOS B/R b h A I E cm cm m² Kg/cm² T 1 T 2 T 3 r 1 t/m ∆H cm r 2 t/m ∆T ° r 3 tm/rad GIRO - α cos α sen α R rad ° GL N GENERAL GIRO X’ Y’ X Y 1 2 3 4 5 6 B O E T re re m4
- 2. 1. CARACTERISTICAS.- C = cos α S = sen α BARRA YΔ L C S A E I EA/L EI/L m m m² m² t/m² t/m t*m t t/m MATRIZ DE ROTACIÓN- BARRA BARRA R = R = G L G L BARRA BARRA R = R = G L G L BARRA BARRA R = R = G L G L ΔX EI/L2 EI/L3 m4
- 3. MATRIZ DE RIGIDEZ DE LAS BARRAS.- BARRA G G K = K = G G K = K = BARRA G G K = K = G G K = K = BARRA G G K = K = G G K = K = BARRA G G K = K = G G K = K = BARRA G G K = K = G G K = K = BARRA G G K = K = G G K = K =
- 4. NUDO 1 G K = = 1 1 NUDO 2 G K = = 2 2 NUDO 3 G K = = 3 3 NUDO 4 G K = = 4 4 NUDO 5 G K = = 5 5
- 5. 3. VECTOR DE CARGAS.- Φ D CARGAS DIRECTAS.- Φ = ∑ P Las cargas se obtienen en coord. Globales i k I CARGAS INDIRECTAS.- Φ Las cargas se obtienen en coord. locales i j
- 6. F I T ∆ TABLA DE CARGAS.- Φ = Φ + Φ + Φ i j i j i j i j BARRA TIPO DE CARGA LOCAL ∑ δ MATRIZ DE ROTACION GLOBAL I T ∆ F δ T F Φ Φ Φ Φ Φ R Φ i j i j i j i j i j G L i j u x v y w z u x v y w z u x v y w z u x v y w z u x v y w z u x v y w z u x v y w z u x v y w z u x v y w z u x v y w z i DIRECTA INDIRECTA ∑ i DIRECTA INDIRECTA ∑ i DIRECTA INDIRECTA ∑ D F 1 D F 1 D F 1 Φ Φ Φ Φ Φ Φ Φ Φ Φ i i i i i i i i i ### x ### x ### x 0,000 0,00 1 y 3 y 5 y 0,000 0,00 0 z 0 z 0 z 0,000 0,00 ### x ### x 2 y 4 y 0 z 0 z T I P O N U D O E J E E J E N U D O E J E R E S T R N U D O E J E R E S T R N U D O E J E R E S T R
- 7. F = K ∆ - Φ F x1 y1 z1 x2 y2 z2 x3 y3 z3 x4 y4 z4 x5 y5 z5 ∆ Φ x1 y1 z1 x2 y2 z2 x3 y3 z3 x4 y4 z4 x5 y5 z5 F K ∆ Φ x2 y2 z2 x3 y3 z3 x1 y1 z1 x4 y4 z4 x5 y5 z5 x2 y2 z2 x3 y3 z3 x1 y1 z1 x4 y4 z4 x5 y5 z5
- 8. Φ K ∆
- 9. K K-1 0 0 0 0 0 0 0 0 0 0 Err:502 Err:502 Err:502 Err:502 Err:502 0 0 0 0 0 0 0 0 0 0 Err:502 Err:502 Err:502 Err:502 Err:502 0 0 0 0 0 0 0 0 0 0 Err:502 Err:502 Err:502 Err:502 Err:502 0 0 0 0 0 0 0 0 0 0 Err:502 Err:502 Err:502 Err:502 Err:502 0 0 0 0 0 0 0 0 0 0 Err:502 Err:502 Err:502 Err:502 Err:502 0 0 0 0 0 0 0 0 0 0 Err:502 Err:502 Err:502 Err:502 Err:502 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0