8.1.1 The n=3 times statically indeterminate frame

Example 8.2 Problem Statement. Consider the frame shown in Fig. 8.1 and described in Example 8.1. Our aim is to compute the load factor $\lambda$ (see Fig. 8.7(b)) at which the first plastic moment occurs.

**Figure 8.7:** Load cases of the n = 3 times indeterminate frame ESTsn3
$\includegraphics[width=74mm]{joonised/raamPiirk2eSn.eps}$ [ $F_{1} = 1{\,}\mathrm{kN}, F_{2} = 0.5{\,}\mathrm{kN}$ ]		$\includegraphics[width=74mm]{joonised/raamPiirk3eSn.eps}$ [ $F_{1} = 14.685{\,}\mathrm{kN}, F_{2} = 7.3427{\,}\mathrm{kN}$ ]

Problem Solving. Now we apply the two load cases shown in Fig. 8.7.

Load case 1 (Fig. 8.7(a)). We determine the node at which the first full plastic moment $M_{pl\,r} = 30.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ or $M_{pl\,l} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ will appear. We find that this is node 4: $M_{4} =$ $1.021{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (see Fig. 8.2(a)). Next we find the load factor $\lambda = \lambda_{0}$ for a full plastic moment $M_{pl} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ to appear at node 4: $\lambda_{0}=14.685$ (see Fig. 8.2(a)).
Load case 2 (Fig. 8.7(b)). We determine the node at which the second full plastic moment $M_{pl\,r} = 30.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ or $M_{pl\,l} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ will appear. We find that this is node 5 (see Fig. 8.2(b)). We also find that to achieve $M_{pl} = 15.0{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ at this node, $\Delta M_{5}$ is needed to add, and that $\Delta M_{5} = 15 - 12.2727 = 2.7273{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ as shown in Figs. 8.2(b) and 8.3(a).

We use the EST method described in Chapter 3, ``Statically indeterminate problems'', and carry out the following steps of calculations:

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, side conditions, restrictions on support displacements.

$\displaystyle \mathbf{spA}{\cdot}\mathbf{Z} = \mathbf{B} \vspace*{-6pt}$ (8.1)
Assembling and solving the boundary problem equation (8.1) (prepared and solved by the function LaheFrameSnDFIm.m):
1. writing the basic equations of a frame in transfer matrix form,
2. adding the compatibility equations of the displacements at nodes,
3. adding the joint equilibrium equations,
4. adding the side conditions (moment hinges),
5. adding the restrictions on support displacements,
6. solving the system of sparse equations,
7. producing an output: initial parameter vectors for element displacements and forces; support reactions.
Output: element displacements and forces determined by the transfer matrix.

1. Input data for the GNU Octave program spESTframeSn3LaheWFI.m are given in excerpts from the program: element and nodal loads - excerpt 8.1; nodal coordinates - excerpt 8.2; element properties, topology and hinges - excerpt 8.3.

Program excerpt 8.1 ( spESTframeSn3LaheWFI.m )

 Number_of_frame_nodes=5
 Number_of_elements=4
 Number_of_support_reactions=6
 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=10.0; % graphic axis
# ---- Load variants ----- 
load_variant=1;
#load_variant=2;
# --- 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/4 # scaling multiplier for displacements # baasi0=1.0; h=4.0; l=4.0; l1=l % l l2=l % l Mplr=30.0 # The plastic moment of beam Mplp=15.0 # The plastic moment of column # Loads F2s=0.5; F3s=1.0;
switch (koormusvariant) case{1} disp('--- ') disp(' Load variant 1 ') disp('--- ') # # Element load in local coordinates # qz qx qA qL # Uniformly distributed load in local coordinate z and x directions Loadsq on element=4; esQkoormus=zeros(LoadsqONelement,4,ElementideArv);

esQkoormus(1,1:4,1)=[0.0 0.0 0.0 4.0];
esQkoormus(1,1:4,2)=[0.0 0.0 0.0 4.0];
esQkoormus(1,1:4,3)=[0.0 0.0 0.0 4.0];
esQkoormus(1,1:4,4)=[0.0 0.0 0.0 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 4.0]; 
#esFjoud(2,1:3,2)=[0.0 0.0 4.0];
esFjoud(1,1:3,3)=[0.0 0.0 4.0]; 
esFjoud(1,1:3,4)=[0.0 0.0 4.0];

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

#sSolmF(:,1,1)= 0.0;
sSolmF(1,1,2)= F2s; # 0.5;
sSolmF(2,1,3)= F3s; # 1.0;
#sSolmF(:,1,4)= 0.0
#sSolmF(:,1,5)= 0.0

# #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

case{2} disp('--- ') disp(' Load variant 2 ') disp('--- ') load myfile30.mat plastF1F2; Flambda0=plastF1F2(1,1); F2s03=plastF1F2(1,2) F3s03=plastF1F2(1,3)
# Element load in local coordinates # qz qx qA qL # Uniformly distributed load in local coordinate z and x directions Loadsq on element=4; esQkoormus=zeros(LoadsqONelement,4,ElementideArv);

esQkoormus(1,1:4,1)=[0.0 0.0 0.0 4.0];
esQkoormus(1,1:4,2)=[0.0 0.0 0.0 4.0];
esQkoormus(1,1:4,3)=[0.0 0.0 0.0 4.0];
esQkoormus(1,1:4,4)=[0.0 0.0 0.0 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 4.0]; 
#esFjoud(2,1:3,2)=[0.0 0.0 4.0];
esFjoud(1,1:3,3)=[0.0 0.0 4.0]; 
esFjoud(1,1:3,4)=[0.0 0.0 4.0];

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

#sSolmF(:,1,1)= 0.0;
sSolmF(1,1,2)= F2s03;
sSolmF(2,1,3)= F3s03;
#sSolmF(:,1,4)= 0.0
#sSolmF(:,1,5)= 0.0

# #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

otherwise disp(' No load variant cases ') endswitch

Program excerpt 8.2 ( spESTframeSn3LaheWFI.m )
#========== # Nodal coordinates #========== krdn=[# x z 0.0 0.0; % node 1 0.0 -4.0; % node 2 4.0 -4.0; % node 3 8.0 -4.0; % node 4 8.0 0.0]; % node 5 #========== #========== # Restrictions on support displacements (on - 1, off - 0) # Support No u w fi #========== tsolm=[1 1 1 1; % node 1 5 1 1 1]; % node 5 #==========

Program excerpt 8.3 ( spESTframeSn3LaheWFI.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 # Mpl - plastic moment at the end of the element # Mpl - plastic moment at the beginning # of the element EIp EAp GAp 2 1 0 0 0 0 0 0 0.0 0.0 ; % element 1 EIr EAr GAr 3 2 0 0 0 0 0 0 0.0 0.0 ; % element 2 EIr EAr GAr 4 3 0 0 0 0 0 0 0.0 0.0 ; % element 3 EIp EAp GAp 5 4 0 0 0 0 0 0 0.0 0.0 ]; % element 4 # 1 - hinge 'true' (axial, shear, moment hinges) #

2. Assembling and solving the boundary problem equations (8.1), carried out by the function LaheFrameSnDFIm(baasi0,Ntoerkts,esQkoormus,esFjoud,sSolmF,tsolm,tSiire, krdn,selem). This function gives the unscaled initial parameter vectors of the elements and support reactions according to the load cases.

3. Output data.
Load case 1. The load factor $\lambda = \lambda_{0}$ (Flambda0), forces $F_{1}$ (F3s03) and $F_{2}$ (F2s03) are given in excerpt 8.1 from the computing diary.

Computing diary excerpt 8.1 ( spESTframeSn3LaheWFI.m )
Flambda0=15/AbsmaxminM Flambda0 = 14.685 -------------- F2s = 0.50000 F2s03=Flambda0*F2s F2s03 = 7.3427 --- F3s = 1 F3s03=Flambda0*F3s F3s03 = 14.685 ============== plastF1F2=[Flambda0 F2s03 F3s03] plastF1F2 = 14.6853 7.3427 14.6853 save myfile30.mat plastF1F2

The bending moments of the first loading case are given in excerpt 8.2 from the computing diary, and the bending moments diagram is shown in Fig. 8.2(a).

Computing diary excerpt 8.2 ( spESTframeSn3LaheWFI.m )
Moments at nodes (Sign Convention 2) ============================================================== Node Moment DaF No Fi/baasi0 DaF No Work_b Element -------------------------------------------------------------- 1 0.26429 12 0.00000e+00 9 0.00000e+00 1 2 -0.12143 6 -3.85714e-05 3 4.68367e-06 1 2 0.12143 24 -3.85714e-05 21 -4.68367e-06 2 3 1.42857 18 5.00000e-06 15 7.14286e-06 2 3 -1.42857 36 5.00000e-06 33 -7.14286e-06 3 4 -1.02143 30 1.85714e-05 27 -1.89694e-05 3 4 1.02143 48 1.85714e-05 45 1.89694e-05 4 5 0.83571 42 0.00000e+00 39 0.00000e+00 4 -------------------------------------------------------------- Note. Work_b is done by boundary forces of an element

Load case 2. The bending moments of the second loading case are given in excerpt 8.4 of the computing diary. The bending moments diagram is shown in Fig. 8.2(b). We find that the next node at which a full plastic moment $M_{pl} = 15.0{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ will appear is node 5 (DaF = 42, see computing diary excerpt 8.3, Figs. 8.8 and 8.2(b)). At this node, $\Delta M_{5} = 15 - 12.2727=$ $2.7273{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (deltaM0) is to be added as shown in excerpt 8.3 and Fig. 8.3(a).

Figure 8.8: Numeration of displacements and forces of the frame ESTsn3

$\includegraphics[width=115mm]{joonised/raamPiirk1DaFe.eps}$

Computing diary excerpt 8.3 ( spESTframeSn3LaheWFI.m )

The maximum absolute values of reduced moments AbsmaxminM = 12.273 muutujaDaF1 = 42 deltaM0=15.0-AbsmaxminM deltaM0 = 2.7273 Flambda0 = 14.685 deltaM0 = 2.7273 muutujaDaF1 = 42 save myfile31.mat Flambda0 deltaM0 muutujaDaF1

Computing diary excerpt 8.4 ( spESTframeSn3LaheWFI.m )

Moments at nodes (Sign Convention 2) ============================================================== Node Moment DaF No Fi/baasi0 DaF No Work_b Element -------------------------------------------------------------- 1 3.88112 12 0.00000e+00 9 0.00000e+00 1 2 -1.78322 6 -5.66434e-04 3 1.01007e-03 1 2 1.78322 24 -5.66434e-04 21 -1.01007e-03 2 3 20.97902 18 7.34266e-05 15 1.54042e-03 2 3 -20.97902 36 7.34266e-05 33 -1.54042e-03 3 4 -15.00000 30 2.72727e-04 27 -4.09091e-03 3 4 15.00000 48 2.72727e-04 45 4.09091e-03 4 5 12.27273 42 0.00000e+00 39 0.00000e+00 4 --------------------------------------------------------------