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    Civ-306 Estructuras Hiperestaticas Ii Ejemplo Portico: Problema 1

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    Jose Ramirez
    This document provides an example problem for calculating the stiffness matrix (K1) of a portal frame structure. It includes: 1) The geometry, properties, and node locations of two beams that make up the frame. 2) The equations used to calculate the individual element stiffness matrices for each beam based on their cross sectional properties. 3) The assembled global stiffness matrix (K1) for the entire two-member frame that relates applied forces and displacements at each node.

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    Civ-306 Estructuras Hiperestaticas Ii Ejemplo Portico: Problema 1

    Uploaded by

    Jose Ramirez
    37 views5 pages
    This document provides an example problem for calculating the stiffness matrix (K1) of a portal frame structure. It includes: 1) The geometry, properties, and node locations of two beams that make up the frame. 2) The equations used to calculate the individual element stiffness matrices for each beam based on their cross sectional properties. 3) The assembled global stiffness matrix (K1) for the entire two-member frame that relates applied forces and displacements at each node.

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    Original Title

    EXAMEN MESA

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    This document provides an example problem for calculating the stiffness matrix (K1) of a portal frame structure. It includes: 1) The geometry, properties, and node locations of two beams that make up the frame. 2) The equations used to calculate the individual element stiffness matrices for each beam based on their cross sectional properties. 3) The assembled global stiffness matrix (K1) for the entire two-member frame that relates applied forces and displacements at each node.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    37 views5 pages

    Civ-306 Estructuras Hiperestaticas Ii Ejemplo Portico: Problema 1

    Uploaded by

    Jose Ramirez
    This document provides an example problem for calculating the stiffness matrix (K1) of a portal frame structure. It includes: 1) The geometry, properties, and node locations of two beams that make up the frame. 2) The equations used to calculate the individual element stiffness matrices for each beam based on their cross sectional properties. 3) The assembled global stiffness matrix (K1) for the entire two-member frame that relates applied forces and displacements at each node.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 5

    Univ.

    José Freddy Quispe Bustinza

    CIV-306 ESTRUCTURAS HIPERESTATICAS II

    Problema 1.
    EJEMPLO PORTICO

    E 25400000

     Matriz de rigidez de cada barra


    Barra 1 1 2

    xi 0 yi 0
    A 0.3 0.4 = 0.12
    xf 3 yf 4
    3
    0.3 0.4
    I = 0.0016
    12
    2 2
    L xf xi + yf yi =5
    xf xi l m 0 0 0 0 0.6 0.8 0 0 0 0
    l = 0.6
    L m l 0 0 0 0 0.8 0.6 0 0 0 0
    yf yi 0 0 1 0 0 0 0 0 1 0 0 0
    m = 0.8 T1 =
    L 0 0 0 l m 0 0 0 0 0.6 0.8 0
    0 0 0 m l 0 0 0 0 0.8 0.6 0
    0 0 0 0 0 1 0 0 0 0 0 1

    A E 2 12 E I 2 A E 12 E I 6 E I A E 2 12 E I 2 A E 12 E I 6 E I
    l + m l m m l + m l m m
    L 3 L 3 2 L 3 L 3 2
    L L L L L L
    A E 12 E I A E 2 12 E I 2 6 E I A E 12 E I A E 2 12 E I 2 6 E I
    l m m + l l l m m + l l
    L 3 L 3 2 L 3 L 3 2
    L L L L L L
    6 E I 6 E I 4 E I 6 E I 6 E I 2 E I
    m l m l
    2 2 L 2 2 L
    L L L L
    K1
    A E 2 12 E I 2 A E 12 E I 6 E I A E 2 12 E I 2 A E 12 E I 6 E I
    l + m l m m l + m l m m
    3 3 2 3 3 2
    L L L L L L L L L L
    A E 12 E I A E 2 12 E I 2 6 E I A E 12 E I A E 2 12 E I 2 6 E I
    l m m + l l l m m + l l
    3 3 2 3 3 2
    L L L L L L L L L L
    6 E I 6 E I 2 E I 6 E I 6 E I 4 E I
    m l m l
    2 2 2 2
    L L L L L L

    221952.9216 290735.3088 7802.88 221952.9216 290735.3088 7802.88


    290735.3088 391548.5184 5852.16 290735.3088 391548.5184 5852.16
    7802.88 5852.16 32512 7802.88 5852.16 16256
    K1 =
    221952.9216 290735.3088 7802.88 221952.9216 290735.3088 7802.88
    290735.3088 391548.5184 5852.16 290735.3088 391548.5184 5852.16
    7802.88 5852.16 16256 7802.88 5852.16 32512

    FEP de la Barra 1

    You created this PDF from an application that is not licensed to print to novaPDF printer (http://www.novapdf.com)
    Univ. José Freddy Quispe Bustinza

    2
    P 1.5 2.5
    P 8 MB = 2.8125
    2
    4
    2
    P 2.5 2.5
    RB 3 2 = 5.46875
    2 4
    4

    2
    2
    P 1.5 1.5
    P 1.5 2.5 RA 3 2 = 2.53125
    2 4
    MA = 4.6875 4
    2
    4

    RA 2.53125
    0 0
    MA 4.6875
    FEP1 =
    RB 5.46875
    0 0
    MB 2.8125

    Barra 2 2 3
    xi 3 yi 4
    A 0.3 0.3 = 0.09
    xf 6 yf 4
    3
    0.3 0.3
    I = 0.000675
    12
    2 2
    L xf xi + yf yi =3
    xf xi l m 0 0 0 0 1 0 0 0 0 0
    l =1
    L m l 0 0 0 0 0 1 0 0 0 0
    yf yi 0 0 1 0 0 0 0 0 1 0 0 0
    m =0 T2 =
    L 0 0 0 l m 0 0 0 0 1 0 0
    0 0 0 m l 0 0 0 0 0 1 0
    0 0 0 0 0 1 0 0 0 0 0 1

    A E 2 12 E I 2 A E 12 E I 6 E I A E 2 12 E I 2 A E 12 E I 6 E I
    l + m l m m l + m l m m
    L 3 L 3 2 L 3 L 3 2
    L L L L L L
    A E 12 E I A E 2 12 E I 2 6 E I A E 12 E I A E 2 12 E I 2 6 E I
    l m m + l l l m m + l l
    L 3 L 3 2 L 3 L 3 2
    L L L L L L
    6 E I 6 E I 4 E I 6 E I 6 E I 2 E I
    m l m l
    2 2 L 2 2 L
    L L L L
    K2
    A E 2 12 E I 2 A E 12 E I 6 E I A E 2 12 E I 2 A E 12 E I 6 E I
    l + m l m m l + m l m m
    3 3 2 3 3 2
    L L L L L L L L L L
    A E 12 E I A E 2 12 E I 2 6 E I A E 12 E I A E 2 12 E I 2 6 E I
    l m m + l l l m m + l l
    3 3 2 3 3 2
    L L L L L L L L L L
    6 E I 6 E I 2 E I 6 E I 6 E I 4 E I
    m l m l
    2 2 2 2
    L L L L L L

    762000 0 0 762000 0 0
    0 7620 11430 0 7620 11430
    0 11430 22860 0 11430 11430
    K2 =
    762000 0 0 762000 0 0
    0 7620 11430 0 7620 11430
    0 11430 11430 0 11430 22860

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    Univ. José Freddy Quispe Bustinza

    FEP de la Barra 2 q 5

    2 2 3
    q 2 2 2 q 2 2
    MA 3 + 10 = 2.666667 MB 5 3 = 0.666667
    60 2 L 60 L L
    L

    L=3
    2 3
    q 2 2 2 q 2 2 2
    RA 10 5 2 = 4.185185 RB 5 = 0.814815
    20 2 L 2 L
    L 20 L

    0 0
    RA 4.185185
    MA 2.666667
    FEP2 =
    0 0
    RB 0.814815
    MB 0.666667

     Vector de fuerzas externas


    0
    0
    0
    0
    F 0
    0
    0
    0
    0

     Vector de fuerzas y momentos de empotramiento perfecto

    1 0 0 0 0 0 2.53125
    0 1 0 0 0 0 0
    0 0 1 0 0 0 4.6875
    0 0 0 1 0 0 5.46875
    Tr1 0 0 0 0 1 0 EFEP1 Tr1 FEP1 = 0
    0 0 0 0 0 1 2.8125
    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 0 0
    1 0 0 0 0 0 0
    Tr2 0 1 0 0 0 0 EFEP2 Tr2 FEP2 = 4.185185
    0 0 1 0 0 0 2.666667
    0 0 0 1 0 0 0
    0 0 0 0 1 0 0.814815
    0 0 0 0 0 1 0.666667

    2.53125
    0
    4.6875
    5.46875
    FEP EFEP1 + EFEP2 = 4.185185
    0.145833
    0
    0.814815
    0.666667

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    Univ. José Freddy Quispe Bustinza

     Matriz de Rigidez Ensamblada

    1 0 0 0 0 0 221952.922 290735.309 7802.88 221952.922 290735.309 7802.88 0 0 0


    0 1 0 0 0 0 290735.309 391548.518 5852.16 290735.309 391548.518 5852.16 0 0 0
    0 0 1 0 0 0 7802.88 5852.16 32512 7802.88 5852.16 16256 0 0 0
    0 0 0 1 0 0 221952.922 290735.309 7802.88 221952.922 290735.309 7802.88 0 0 0
    T
    Tr1 = 0 0 0 0 1 0 EK1 Tr1 K1 Tr1 = 290735.309 391548.518 5852.16 290735.309 391548.518 5852.16 0 0 0
    0 0 0 0 0 1 7802.88 5852.16 16256 7802.88 5852.16 32512 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 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 0 0 0 0 0 0 0 0 0 0 0 0 0
    1 0 0 0 0 0 0 0 0 762000 0 0 762000 0 0
    T
    Tr2 = 0 1 0 0 0 0 EK2 Tr2 K2 Tr2 = 0 0 0 0 7620 11430 0 7620 11430
    0 0 1 0 0 0 0 0 0 0 11430 22860 0 11430 11430
    0 0 0 1 0 0 0 0 0 762000 0 0 762000 0 0
    0 0 0 0 1 0 0 0 0 0 7620 11430 0 7620 11430
    0 0 0 0 0 1 0 0 0 0 11430 11430 0 11430 22860

    221952.92 290735.31 7802.88 221952.92 290735.31 7802.88 0 0 0


    290735.31 391548.52 5852.16 290735.31 391548.52 5852.16 0 0 0
    7802.88 5852.16 32512 7802.88 5852.16 16256 0 0 0
    221952.92 290735.31 7802.88 983952.92 290735.31 7802.88 762000 0 0
    K EK1 + EK2 = 290735.31 391548.52 5852.16 290735.31 399168.52 5577.84 0 7620 11430
    7802.88 5852.16 16256 7802.88 5577.84 55372 0 11430 11430
    0 0 0 762000 0 0 762000 0 0
    0 0 0 0 7620 11430 0 7620 11430
    0 0 0 0 11430 11430 0 11430 22860

     Matriz de fuerzas de desplazamientos conocidos


    0 8722.059264 8724.5905
    0.03 11746.455552 11746.4556
    0 175.5648 170.8773
    0 8722.059264 8716.5905
    d 0 fo K d= 11746.455552 FFEP F FEP fo = 11750.6407
    0 175.5648 175.7106
    0 0 0
    0 0 0.8148
    0 0 0.6667

     Matriz de Rigidez Reducida


    0 0 0 0
    0 0 0 0
    1 0 0 0
    32512 7802.88 5852.16 16256
    0 1 0 0
    T 7802.88 983952.92 290735.31 7802.88
    Tr 0 0 1 0 Kr Tr K Tr =
    5852.16 290735.31 399168.52 5577.84
    0 0 0 1
    16256 7802.88 5577.84 55372
    0 0 0 0
    0 0 0 0
    0 0 0 0
    170.8773
    T 8716.590514
    FFEPr Tr FFEP =
    11750.640737
    175.710633

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    Univ. José Freddy Quispe Bustinza

     Desplazamientos en los nudos


    0.003617
    1 0.0001831
    Dr Kr FFEPr =
    0.0294585
    0.0072284
    0 0
    0 0.03
    0.003617 0.003617
    0.0001831 0.000183
    D Tr Dr = 0.0294585 dD d+D= 0.029458
    0.0072284 0.007228
    0 0
    0 0
    0 0

     Reacciones

    147.5139 147.5139
    137.6677 137.6677
    0 0
    0 0
    R (K dD)) + FEP = 0 R R F= 0
    0 0
    139.5139 139.5139
    142.6677 142.6677
    254.7564 254.7564

     Esfuerzos en cada barra

    0 147.513878 198.642503
    0.03 137.66772 35.41047
    T 0.003617 0 0
    D1 Tr1 dD = F1 K1 D1 + FEP1 = T1 F1 =
    0.000183 139.513878 193.842503
    0.029458 137.66772 29.01047
    0.007228 168.802351 168.802351

    0.000183 139.513878 139.513878


    0.029458 137.66772 137.66772
    T 0.007228 168.802351 168.802351
    D2 Tr2 dD = F2 K2 D2 + FEP2 = T2 F2 =
    0 139.513878 139.513878
    0 142.66772 142.66772
    0 254.756366 254.756366

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