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Next: III. Plastic analysis of Up: 6. The EST method Previous: 6.2 Illustrative frame problem


6.3 Illustrative frame problem 2

Example 6.2   Problem Statement. In Fig. 6.7, a two-span frame of the height $h = 4{\,}\mathrm{m}$ is shown. Two spans of the frame are of the same length, $l = 6{\,}\mathrm{m}$. The beam of the first span carries a uniform load $q = 15{\,}\mathrm{kN/m}$. The rightmost column is loaded with horizontal concentrated loads $F_{1} = 90{\,}\mathrm{kN}$ and $F_{2} = 120{\,}\mathrm{kN}$. Each of the frame columns carries a vertical concentrated load $F_{3} = 750{\,}\mathrm{kN}$.
Figure 6.7: The two-span frame EST2
\includegraphics[width=110mm]{joonised/raamn2esmoe.eps}

Let us assume that the value of the column flexural rigidity $EI_{p} = 2*10^{4} {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}^{2}$ and beam flexural rigidity $EI_{r} = 1.5EI_{p}$; axial rigidity of the columns $EA_{p} = 4.6*10^{6}{\,}\mathrm{kN}$ and that of the beam $EA_{r} = 8.8*10^{6}{\,}\mathrm{kN}$; shear rigidity of the columns $GA_{p} = 0.4EA_{p}$ and that of the beam $GA_{r} = 0.4EA_{r}$.

We wish to compute the displacements, reactions, internal forces, and draw the axial force, shear force and bending moment diagrams.

Problem Solving. We use the EST method to solve the problem. The solving procedure includes the following.

  1. Data input: the number of frame nodes, elements, support reactions; element properties, element loads in local coordinates, node forces in global coordinates, nodal coordinates, topology and hinges, restrictions on support displacements.
  2. Assembling and solving the boundary problem equations (6.1) and (6.9) (prepared and solved by the program):
    1. writing the basic equations of a frame in transfer matrix form,

      $\displaystyle \mathbf{spA}{\cdot}\mathbf{Z} = \mathbf{B}$     (6.9)

    2. adding the compatibility equations of the displacements at nodes,
    3. adding the joint equilibrium equations,
    4. adding the side conditions (hinges),
    5. adding the restrictions on support displacements,
    6. solving the compiled system of sparse equations,
    7. producing an output: initial parameter vectors for element displacements and forces; support reactions.
  3. Output: element displacements and forces determined by the transfer matrix.

Figure 6.8: Displacement and force numbers of the two-span frame EST2
\includegraphics[width=110mm]{joonised/raamn3esmoe.eps}

1. Input data for the GNU Octave program yspESTframe2LaheWFI.m are shown in excerpt from the program: element and nodal loads - excerpt 6.5; nodal coordinates -excerpt 6.6; element properties, topology and hinges - excerpt 6.7.

Program excerpt 6.5 ( yspESTframe2LaheWFI.m )   
 

 epsdountil=0.0000001; # for iteration (max |S| increment)
 Number_of_frame_nodes=6 
 Number_of_elements=5
 Number_of_support_reactions=8
 spNNK=12*Number_of_elements+Number_of_support_reactions;
 Number_of_unknowns=spNNK 
 Displacements and forces are calculated on parts ''Nmitmeks'' of the element
 Nmitmeks=4
 Lp=12.0; # graphics axis 
# --- Element properties ---

  
EIp=20000 # kN/m^2 
EIr=40000 # kN/m^2 
EAp=4.6*10^6 
#EAp=4.6*10^15; 
EAr=6.8*10^6;
#EAr=6.8*10^15 
GAp=0.4*EAp;    
GAr=0.4*EAr;


 
baasi0=EIp/5.6  # scaling multiplier for the displacements
#baasi0=1.0;

#Element load  in local coordinates 
#  qz     qx     qA      qL 
# Uniformly distributed load in local coordinate z and x directions
LoadsqONelement=4;
esQkoormus=zeros(LoadsqONelement,4,ElementideArv);

esQkoormus(1,1:4,1)=[0.0 0.0 0.0 4.0]; 
esQkoormus(1,1:4,2)=[15.0 0.0 0.0 6.0]; 
esQkoormus(1,1:4,3)=[0.0 0.0 0.0 4.0]; 
esQkoormus(1,1:4,4)=[0.0 0.0 0.0 6.0]; 
esQkoormus(1,1:4,5)=[0.0 0.0 0.0 4.0];

 
#
# Point load in local coordinate z and x directions  kN
#  Fz, Fx, aF (coordinate of the point of force application)
LoadsF_on_Element=5; 
esFjoud=zeros(LoadsF_on_Element,2,ElementideArv);

esFjoud(1,1:3,1)=[0.0 0.0 4.0];
esFjoud(1,1:3,2)=[0.0 0.0 6.0];   
esFjoud(1,1:3,3)=[0.0 0.0 4.0]; 
esFjoud(1,1:3,4)=[0.0 0.0 6.0];
esFjoud(1,1:3,5)=[120.0 0.0 2.0];

 
#
# Node forces in global coordinates
# sSolmF(forces,1,nodes); forces=[Fx; Fz; My]
sSolmF = zeros(3,1,SolmedeArv);

#sSolmF(:,1,1)= 0.0
sSolmF(2,1,2)= 750.0; % Fz
#sSolmF(:,1,3)= 0.0
sSolmF(2,1,4)= 750.0; % Fz
#sSolmF(:,1,5)= 0.0
sSolmF(1,1,6)= -90.0; % Fx
sSolmF(2,1,6)= 750.0; % Fz

 
#
#s1F(1,1,1)=0.0;  # force Fz
#s1F(2,1,1)=0.0;  # force Fz
#s1F(3,1,1)=0.0;  # force My


# Support shift - tSiire#
# Support shift is multiplied by scaling multiplier
tSiire = zeros(3,1,SolmedeArv);

#tSiire(:,1,1)= 0.0
#tSiire(2,1,1)= 0.01*baasi0
#tSiire(:,1,2)= 0.0
#tSiire(:,1,3)= 0.0
#tSiire(:,1,4)= 0.0
#tSiire(:,1,5)= 0.0
#tSiire(:,1,6)= 0.0

Program excerpt 6.6 ( yspESTframe2LaheWFI.m )   
#==========
#     Nodal coordinates
#==========
krdn=[#  x       z
        0.0     0.0;    % node 1
        0.0    -4.0;    % node 2
        6.0     0.0;    % node 3
        6.0    -4.0;    % node 4
       12.0     0.0;    % node 5
       12.0    -4.0];   %% node 6
#========== 
#
#==========
#  Restrictions on support displacements (on - 1, off - 0)
# Support  No   u   w  fi  
#==========       
tsolm=[1     1 1 0;   % node 1
       3     1 1 1;   % node 3
       5     1 1 1];  % node 5       
#==========

Program excerpt 6.7 ( yspESTframe2LaheWFI.m )   
# ------------- Element properties, topology and hinges ---------
elasts=[# Element properties 
#            n2 - end of the element
#                  n1 - beginning of the element
#                       N, Q, M - hinges at the end of the element
#                                N, Q, M - hinges at the beginning of the element
# 
 EIp EAp GAp   2     1    0 0 0    0 0 1;    % element 1
 EIr EAr GAr   4     2    0 0 0    0 0 0;    % element 2
 EIp EAp GAp   4     3    0 0 0    0 0 0;    % element 3
 EIr EAr GAr   6     4    0 0 0    0 0 1;    % element 4
 EIp EAp GAp   5     6    0 0 0    0 0 0];   % element 5
# 1 - hinge 'true' (axial, shear, moment hinges)
#

2. Assembling and solving the boundary problem equations (6.5), carried out by the function Lahe2FrameDFIm(baasi0,Ntoerkts,esQkoormus,esFjoud,sSolmF,tsolm,tSiire, krdn,selem). The program has numbered the displacements and forces of the element ends of the frames as shown in Figs. 6.2 and 6.8.

The results of iteration of the axial forces are given in excerpt 6.7 from the computing diary.

Computing diary excerpt 6.7 ( yspESTframe2LaheWFI.m )   
SIvec =
Element  Linear     1         2         3         4
  1    -824.978  -828.823  -828.821  -828.821  -828.821
  2     -25.760   -21.535   -21.499   -21.499   -21.499
  3    -775.104  -773.182  -773.186  -773.186  -773.186
  4     -79.661   -77.786   -77.769   -77.769   -77.769
  5    -739.918  -737.994  -737.993  -737.993  -737.993

The unscaled initial parameter vectors of the elements are shown in excerpt 6.8 from the computing diary.

Computing diary excerpt 6.8 ( yspESTframe2LaheWFI.m )   
-------- Scaling multiplier for displacements = 1/baasi0 -------- 
============================================================================
 Unscaled initial parameter vector 
Element No    u          w          fi           S          H          M 
----------------------------------------------------------------------------
   1   -0.000e+00   0.000e+00   1.166e-02     828.821     21.499      0.000
   2   -3.056e-02   7.207e-04  -1.311e-04      21.499    -78.821    111.330
   3   -0.000e+00   0.000e+00   0.000e+00     773.186     56.270   -157.134
   4   -3.058e-02   6.723e-04  -2.423e-03      77.769    -12.007      0.000
   5    6.417e-04   3.065e-02   4.838e-03     737.993     12.231    -72.037  
----------------------------------------------------------------------------

The support reactions of the frame in global coordinates are shown in excerpt 6.9 from the computing diary.

Computing diary excerpt 6.9 ( yspESTframe2LaheWFI.m )   
Support reactions begin from X row: 61  
===========================================
 No         X        Node    Cx <=> 1 
                             Cz <=> 2
                             Cy <=> 3 
------------------------------------------- 
 61   +2.149932e+01    1         1 
 62   -8.288206e+02    1         2 
 63   +5.626986e+01    3         1 
 64   -7.731859e+02    3         2 
 65   -1.571336e+02    3         3 
 66   +1.322308e+02    5         1 
 67   -7.379935e+02    5         2 
 68   -2.395078e+02    5         3 
--------------------------------------------

3. Output: the element displacements and forces determined by the transfer matrix are given in excerpt 6.10 from the computing diary.

The horizontal displacements diagram of the frame EST2 is depicted in Fig. 6.9. The bending moment M, shear force Q and axial force N diagrams of the frame EST2 are shown in Fig. 6.10.

Figure 6.9: Horizontal displacements diagram of the frame EST2
\includegraphics[width=105mm]{joonised/hor_ii3e.eps}
               (bracketed are values of linear solution)      

Figure 6.10: Internal forces diagrams of the two-span frame EST2
\includegraphics[width=115mm]{joonised/nep_ii3e.eps}
[Axial force N diagram]

\includegraphics[width=125mm]{joonised/qep_ii3e.eps}
                         [Shear force Q diagram] 		    

\includegraphics[width=125mm]{joonised/mom_ii3e.eps}
                [Bending moment diagram]
                                                                              (bracketed are values of linear solution)


Computing diary excerpt 6.10 ( yspESTframe2LaheWFI.m )   
#=================================================================================
 Element displacements and forces determined by transfer matrix 
#================================================================================= 
Displacements and forces of element no 1 of length 4.000 m 
The element is divided into 4 parts 
 displacement u - 0.00000e+00 -1.80178e-04 -3.60357e-04 -5.40535e-04 -7.20714e-04
 displacement w - 0.00000e+00 -1.14016e-02 -2.12611e-02 -2.81000e-02 -3.05647e-02
 rotation    fi - 1.16608e-02  1.08844e-02  8.58719e-03  4.86412e-03 -1.31089e-04
 normal force N - -828.82062   -828.82062  -828.82062    -828.82062  -828.82062
 shear force  Q -  -31.16401   -30.52051    -28.61656     -25.53080   -21.39067
 moment force M -    0.00000   -30.94922    -60.62029     -87.78785  -111.32995
------------------
 axial  force S - -828.82062   -828.82062   -828.82062   -828.82062   -828.82062
 transv force H -  -21.49932    -21.49932    -21.49932    -21.49932    -21.49932
----------------------------------------------------------------------------

Displacements and forces of element no 2 of length 6.000 m 
The element is divided into 4 parts 
 displacement u - -3.05647e-02 -3.05695e-02 -3.05742e-02 -3.05789e-02 -3.05837e-02
 displacement w -  7.20714e-04  3.71930e-03  7.67193e-03  7.50247e-03  6.72336e-04
 rotation    fi - -1.31089e-04 -3.02184e-03 -1.68423e-03  2.19229e-03  6.91421e-03
 normal force N -  -21.49932   -21.49932   -21.49932   -21.49932  -21.49932
 shear force  Q -   78.82344    56.38559    33.85683    11.27349  -11.32803
 moment force M -  -111.32995    -9.90955    57.78135    91.63364   91.59272
------------------
 axial  force S -  -21.49932   -21.49932   -21.49932   -21.49932  -21.49932
 transv force H -   78.82062    56.32062    33.82062    11.32062  -11.17938
----------------------------------------------------------------------------
Displacements and forces of element no 3 of length 4.000 m 
The element is divided into 4 parts 
 displacement u - 0.00000e+00 -1.68084e-04 -3.36168e-04 -5.04252e-04 -6.72336e-04
 displacement w - 0.00000e+00 -3.44769e-03 -1.17895e-02 -2.18995e-02 -3.05837e-02
 rotation    fi - 0.00000e+00  6.40393e-03  9.75665e-03  9.92896e-03  6.91421e-03
 normal force N - -773.18589  -773.18589   -773.18589   -773.18589   -773.18589
 shear force  Q -  -56.26986   -61.22129    -63.81357    -63.94679    -61.61583
 moment force M -  157.13359    98.19802     35.47841    -28.60837    -91.59272
------------------
 axial  force S - -773.18589  -773.18589   -773.18589   -773.18589   -773.18589
 transv force H -  -56.26986   -56.26986    -56.26986    -56.26986    -56.26986
----------------------------------------------------------------------------
Displacements and forces of element no 4 of length 6.000 m 
The element is divided into 4 parts 
 displacement u - -3.05837e-02 -3.06008e-02 -3.06180e-02 -3.06351e-02 -3.06523e-02
 displacement w -  6.72336e-04  4.07754e-03  6.11281e-03  5.41621e-03  6.41733e-04
 rotation    fi - -2.42253e-03 -1.96544e-03 -5.96843e-04  1.67528e-03  4.83769e-03
 normal force N - -77.76918    -77.76918    -77.76918    -77.76918    -77.76918
 shear force  Q -  12.19491     12.15936     12.05292     11.87622     11.63028
 moment force M -   0.00000     18.27458     36.44263     54.39821     72.03667
------------------
 axial  force S -  -77.76918   -77.76918   -77.76918   -77.76918  -77.76918
 transv force H -   12.00651    12.00651    12.00651    12.00651   12.00651
----------------------------------------------------------------------------
Displacements and forces of element no 5 of length 4.000 m 
The element is divided into 4 parts 
 displacement u - 6.41733e-04 4.81300e-04 3.20867e-04 1.60433e-04  0.00000e+00
 displacement w - 3.06523e-02 2.41507e-02 1.49071e-02 4.86941e-03 -6.63532e-15
 rotation    fi - 4.83769e-03 8.02360e-03 1.03047e-02 8.60626e-03  5.15213e-15
 normal force N - -737.99349  -737.99349  -737.99349  -737.99349   -737.99349
 shear force  Q -  -15.80100   -18.15219  -139.83562  -138.58218   -132.23082
 moment force M -   72.03667    55.00768    35.95515  -103.68339   -239.50781
------------------
 axial  force S - -737.99349  -737.99349  -737.99349  -737.99349   -737.99349
 transv force H -  -12.23082   -12.23082  -132.23082  -132.23082   -132.23082
----------------------------------------------------------------------------

Testing a static equilibrium for the frame

Consider next a static equilibrium of the frame shown in Fig. 6.11. Let us project the forces onto the X-axis (see Fig. 6.11),

$\displaystyle \begin{array}{ccl} \Sigma X = 0; & & 21.49932 + 56.26986 + 132.2308 - 90 -120 = -2.0\cdot 10^{-5} \end{array} \qquad$     (6.10)

Figure 6.11: Boundary forces of the two-span frame EST2
\includegraphics[width=120mm]{joonised/raamn2esSmoe.eps}

and onto the Z-axis,

$\displaystyle \begin{array}{ccl} \Sigma Z = 0; & & - 828.8206 - 773.1859 - 737.9935 + 15\cdot6.0 + 3\cdot 750 = 0.0 \end{array}$     (6.11)

We now write equation (6.12) of the sum of the moments and the moments of the forces acting about point 1 shown in Fig. 6.11:

$\displaystyle \Sigma M_{1} = 0;\qquad -157.1336-239.5078+773.1859\cdot 6+737.9935\cdot12 +120.0\cdot2.0 \qquad$      
$\displaystyle +90.0\cdot4.0-750.0\cdot\left( 12.0-0.0306523\right) -750.0\cdot\left( 6.0-0.0305837 \right) \quad$      
$\displaystyle - 15\cdot6.0\cdot\left( 3.0-0.0305742\right) +0.0305647\cdot750.0 = -1.797\cdot 10^{-3} \:\:\:\: \quad$     (6.12)

At node 2, the horizontal displacement $u_{2{\,}4} = - 3.05647{\cdot}10^{-2}$ (see excerpt 6.10 from the computing diary). The additional moment $\Delta M$ from node (member) displacements

$\displaystyle \Delta M = 0;\qquad -750.0\cdot\left(-0.0306523\right) -750.0\cdot\left( -0.0305837 \right) \qquad \qquad \qquad$      
$\displaystyle - 15\cdot6.0\cdot\left(-0.0305742\right) +0.0305647\cdot750.0 = 71.602{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$     (6.13)

The calculations with equations (6.10), (6.11), and (6.12) have verified the static equilibrium of the frame.

Figure 6.12: Elements of the frame EST2
\includegraphics[width=0.62\textwidth]{joonised/yspESTframe2e.eps}



Figure 6.13: Sparsity pattern of matrix spA of the frame EST2
\includegraphics[width=0.72\textwidth]{joonised/yspESTframe2_spars.eps}

Table 6.1 gives the values for internal reactions 6.1 6.2 6.3 (contact forces 6.4) determined by the EST method. These can be compared with the results ($M_{def}$) 6.6 obtained with the slope-deflection method. The table describes the boundary values (internal reactions) after the second iteration, whereas those in table 16.17 of [Lah12] had been obtained in the first iteration.



Table 6.1: Comparison of the internal reactions values of the frame EST2
Element Linear theory 2nd order theory
    No $M_{def}$ $M^{I}_{est}$ $S^{I}_{est}$ $H^{I}_{est}$ $M^{II}_{est}$ $S^{II}_{est}$ $H^{II}_{est}$
$\left[\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m} \right]$ $\left[\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m} \right]$ $\left[\mathrm{kN}\right]$ $\left[\mathrm{kN}\right]$ $\left[\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m} \right]$ $\left[\mathrm{kN}\right]$ $\left[\mathrm{kN}\right]$
1 begin. 0.0000 0.000 824.978 25.760 0.000 828.820 21.499
1 end -103.1977 -103.040 -824.978 -25.760 -111.330 -828.820 -21.499
2 begin. 103.1977 103.040 25.760 -74.978 111.330 21.499 -78.820
2 end 76.9767 76.829 -25.760 -15.022 91.593 -21.499 -11.179
3 begin. -138.8954 -138.774 775.104 53.901 -157.133 773.186 56.270
3 end -76.9767 -76.829 -775.104 -53.901 -91.593 -773.186 -56.270
4 begin. 0.0000 0.000 79.661 -10.082 0.000 77.769 -12.007
4 end 60.3488 60.490 -79.661 10.082 72.037 -77.769 12.001
5 begin. -60.3488 -60.490 739.918 10.339 -72.037 737.993 12.231
5 end -220.5814 -220.868 -739.918 -130.339 -239.508 -737.993 -132.231

The boundary value problem (differential equations together with a set of boundary conditions) is well posed with the EST method.


next up previous contents index
Next: III. Plastic analysis of Up: 6. The EST method Previous: 6.2 Illustrative frame problem
andres
2014-09-09