8.1.3 The n=1 times statically indeterminate frame

Example 8.4 Problem Statement. Let us consider the frame shown in Fig. 8.1 and described in Example 8.1. We wish to compute the load factor $\lambda$ (see Fig. 8.10(b)) at which the third plastic moment occurs.

**Figure 8.10:** Load cases of the n = 1 times indeterminate frame ESTsn1
$\includegraphics[width=65mm]{joonised/raamPiirk6eSn.eps}$ [ $F_{1} = 1{\,}\mathrm{kN}, F_{2} = 0.5{\,}\mathrm{kN}$ ]		$\includegraphics[width=74mm]{joonised/raamPiirk7eSn.eps}$ [ $F_{1} = 18.900{\,}\mathrm{kN}, F_{2} = 9.450{\,}\mathrm{kN}$

Problem Solving. The two load cases shown in Fig. 8.10 are applied:

Load case 1 (Fig. 8.10(a)). We determine the load factor $\lambda_{2}$ for a moment $M_{pl} =$ $30.0 {\,}\mathrm{kN\hspace*{-2pt}\cdot\hspace*{-2pt}m}$ to appear at node 3. We find that for a plastic moment to form at node 5, the load factor is to be $\lambda_{2}=\Delta M_{3}/M_{3}= 2.3077/2.2727 = 1.0154$ as shown in Fig. 8.4(a) ( $\Delta M_{3}$ (deltaM0) was found on page ).
Load case 2 (Fig. 8.10(b)). The boundary moments that are equal to the full plastic moment $M_{pl} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ are added to the ends of elements 3 and 4 at plastic hinge 4 and 5 (see excerpt 8.9). The corresponding load factor $\lambda =\lambda_{0} +\lambda_{1} +\lambda_{2}$ is shown in Fig. 8.4(b). We find that at node 1, a fourth full plastic moment, $M_{pl\,r} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ , will appear (see Fig. 8.4(b)). We also find that at this node $\Delta M_{1}$ is needed to add to achieve $M_{pl} = 15.0{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ and that $\Delta M_{1} = 15 - 8.4000 = 6.6000{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ as shown in Figs. 8.4(b) and 8.5(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.

$\displaystyle \mathbf{spA}{\cdot}\mathbf{Z} = \mathbf{B} \vspace*{-6pt}$ (8.3)
Assembling and solving the boundary problem equation (8.3) (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 spESTframeSn1LaheWFI.m are given in excerpts from the program: element and nodal loads - excerpt 8.7; nodal coordinates - excerpt 8.8; element properties, topology and hinges - excerpt 8.9.

Program excerpt 8.7 ( spESTframeSn1LaheWFI.m )

 Number_of_frame_nodes=5
 Number_of_elements=4
 Number_of_support_reactions=5
 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('--- ') # disp('deltaM0=15.0-AbsmaxminM ') # disp(' load myfile21.mat Flambda Flambda0 Flambda1 deltaM0 muutujaDaF2 ') disp('load myfile21.mat ') load myfile21.mat # Flambda Flambda0 Flambda1 deltaM0 muutujaDaF2 # # 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('--- ') disp(' load myfile10.mat ') load myfile10.mat disp(' plastF1F2; ')% #plastF1F2=[Flambda Flambda0 Flambda1 Flambda2 F2s01 F3s01] plastF1F2; Flambda=plastF1F2(1,1) Flambda0=plastF1F2(1,2) Flambda1=plastF1F2(1,3) Flambda2=plastF1F2(1,4) F2s01=plastF1F2(1,5) F3s01=plastF1F2(1,6)
# 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)= F2s01;
sSolmF(2,1,3)= F3s01;
#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 8.8 ( spESTframeSn1LaheWFI.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 0]; % node 5 #==========

Program excerpt 8.9 ( spESTframeSn1LaheWFI.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 1 0 0 0 -Mplp 0.0 ; % element 3 EIp EAp GAp 5 4 0 0 1 0 0 1 Mplp Mplp ]; % element 4 # 1 - hinge 'true' (axial, shear, moment hinges) #

2. Assembling and solving the boundary problem equations (8.3), 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 $\lambda = \lambda_{0}+\lambda_{1}+\lambda_{2}$ (Flambda), forces $F_{1}$ (F3s01) and $F_{2}$ (F2s01) are presented in excerpt 8.9 from the computing diary.

Computing diary excerpt 8.9 ( spESTframeSn1LaheWFI.m )
deltaM0 = 2.3077 -------------- absX = 2.2727 Flambda2=deltaM0/absX Flambda2 = 1.0154 Flambda=Flambda0+Flambda1+Flambda2 Flambda = 18.900 -------------- F2s01=Flambda*F2s F2s01 = 9.4500 F3s01=Flambda*F3s F3s01 = 18.900 ============== plastF1F2=[Flambda Flambda0 Flambda1 Flambda2 F2s01 F3s01] plastF1F2 = 18.9000 14.6853 3.1993 1.0154 9.4500 18.9000 save myfile10.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. 8.3(a).

Computing diary excerpt 8.10 ( spESTframeSn1LaheWFI.m )

Moments at nodes (Sign Convention 2) ============================================================== Node Moment DaF No Fi/baasi0 DaF No Work_b Element -------------------------------------------------------------- 1 1.45455 12 0.00000e+00 9 0.00000e+00 1 2 0.54545 6 -9.09091e-05 3 -4.95868e-05 1 2 -0.54545 24 -9.09091e-05 21 4.95868e-05 2 3 2.27273 18 3.03030e-06 15 6.88705e-06 2 3 -2.27273 36 3.03030e-06 33 -6.88705e-06 3 4 0.00000 30 7.87879e-05 27 0.00000e+00 3 4 0.00000 48 -7.87879e-05 45 -0.00000e+00 4 5 0.00000 42 -7.87879e-05 39 -0.00000e+00 4 --------------------------------------------------------------

Load case 2. The bending moments of the second loading case are given in excerpt 8.12 from the computing diary, and the bending moments diagram is shown in Fig. 8.4(b). We find that the next node for a full plastic moment $M_{pl} = 15.0{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ to form at is node 1 (DaF = 12, see excerpt 8.11 from the computing diary, Figs. 8.8 and 8.4(b)). At this node, $\Delta M_{1} = 15 - 8.4000 = 6.6000{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (deltaM0) is needed to add as shown in excerpt 8.11 and Fig. 8.4(a).

Computing diary excerpt 8.11 ( spESTframeSn1LaheWFI.m )

The maximum absolute values of reduced moments AbsmaxminM = 8.4000 muutujaDaF3 = 12 AbsmaxminM = 8.4000 Flambda = 18.900 Flambda0 = 14.685 Flambda1 = 3.1993 Flambda2 = 1.0154 deltaM0 = 6.6000 muutujaDaF3 = 12 save myfile11.mat Flambda Flambda0 Flambda1 Flambda2 deltaM0 muutujaDaF3

Computing diary excerpt 8.12 ( spESTframeSn1LaheWFI.m )

Moments at nodes (Sign Convention 2) ============================================================== Node Moment DaF No Fi/baasi0 DaF No Work_b Element -------------------------------------------------------------- 1 8.40000 12 0.00000e+00 9 0.00000e+00 1 2 -0.60000 6 -9.00000e-04 3 5.40000e-04 1 2 0.60000 24 -9.00000e-04 21 -5.40000e-04 2 3 30.00000 18 8.00000e-05 15 2.40000e-03 2 3 -30.00000 36 8.00000e-05 33 -2.40000e-03 3 4 -15.00000 30 5.80000e-04 27 -8.70000e-03 3 4 15.00000 48 -8.00000e-05 45 -1.20000e-03 4 5 15.00000 42 -8.00000e-05 39 -1.20000e-03 4 -------------------------------------------------------------- At node 4, the dissipation D = -(-8.70000e-03 -1.20000e-03) = 0.00990 > 0 At node 5, the dissipation D = -(-1.20000e-03) = 0.00120 > 0

The dissipation D at plastic hinges 4 and 5 satisfies the condition (7.19):
${D_{4}}=-\left( M^{r}_{pl}\varphi_{r}+M^{l}_{pl}\varphi_{l}\right) = -(-8.70000... ...0e-03) = 0.00990{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m} > 0$ ,
${D_{5}}=-\left( M^{r}_{pl}\varphi_{r}+M^{l}_{pl}\varphi_{l}\right) = -(-1.20000e-03) = 0.00120{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m} > 0$ .