8.1 Illustrative problems

Example 8.1   Problem Statement. The frame depicted in Fig. 8.1 is of height $h$ $= 4{\,}\mathrm{m}$ and of span length $l = 8{\,}\mathrm{m}$. The beam 2-4 is loaded with a vertical concentrated load $F_{1}$ $= 1.0\lambda {\,}\mathrm{kN}$ (here, $\lambda$ is the load factor, see Eq. (7.1)). A horizontal load $F_{2}$ $= 0.5 \lambda {\,}\mathrm{kN}$ acts at joint 2. For the beam, the full plastic moment $M_{pl}$ $= 30.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$, and for the columns, $M_{pl}$ $= 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$.

Figure 8.1: The frame ESTsn3            

Let us assume that the flexural rigidity of a column $EI_{p} = 2\cdot 10^{4} {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}^{2}$ and that of the beam $EI_{r} = 3.0EI_{p}$; the axial rigidity of a column $EA_{p} = 4.6\cdot 10^{6}{\,}\mathrm{kN}$ and that of the beam $EA_{r} = 8.8\cdot 10^{6}{\,}\mathrm{kN}$; the shear rigidity of a column $GA_{rp} = 0.4EA_{p}$ and that of the beam $GA_{rr} = 0.4EA_{r}$.

We wish to compute the collapse load factor $\lambda_{C}$ at which the frame will actually fail.

Problem Solving. We use the EST method to solve the problem of an n times statically indeterminate frame. The steps of the load incremental method are listed below.

1. The load of an n=3 times statically indeterminate frame is increased $\lambda$ times ($F_{1} =$ $1.0\cdot \lambda ,{\,} F_{2}=0.5\cdot \lambda$, see Fig. 8.1) to the appearance of the full plastic moment $M_{pl}$. For that
   • compose a structural system for an n=3 times statically indeterminate frame (Fig. 8.2(a)) with a load factor $\lambda = 1.0$;
   • find the node where a 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, where $M_{4} = 1.021{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (see Fig. 8.2(a));
   • find the load factor $\lambda = \lambda_{0}$ for a moment $M_{pl} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ to appear at node 4. We find that for a full plastic moment to develop at node 4, the load factor is to be $\lambda_{0}=14.685$ as shown in Fig. 8.2(b);
   • find the next node where a 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 at this node to achieve $M_{pl\,r} = 30.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$, $\Delta M_{5}$ is needed to add, and that $\Delta M_{5} = 15 - 12.2727 = 2.7273{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ as can be seen in Figs. 8.2(b) and 8.3(a).
2. In an n=2 times statically indeterminate frame, load is increased ( $\lambda=\lambda_{0} +\lambda_{1}$) times to the appearance of the second full plastic moment $M_{pl}$. For that
   • compose a structural system for an n=2 times statically indeterminate frame (Fig. 8.3(a)) with a load factor $\lambda = 1.0$;
   • find the load factor $\lambda_{1}$ for a moment $M_{pl} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ to appear at node 5. We find that for the appearance of a plastic moment at node 5, the load factor is to be $\lambda_{1}=\Delta M_{5}/M_{5}= 2.7273/0.8525=3.1993$ as shown in Fig. 8.3(a);
   • add the boundary moments that are equal to the full plastic moment $M_{pl} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (related to $\lambda_{0}$) to the ends of elements 3 and 4 at hinge 4 (see Fig. 8.3(b));
     
     
   • find the displacements and internal forces of the frame shown in Fig. 8.3(b). Here, the load factor $\lambda=\lambda_{0} +\lambda_{1} = 17.884$, the corresponding loads $F_{1}=17.884{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ and $F_{2}=8.9422{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$;
   • check if the dissipation D at plastic hinge 4 satisfies the condition (7.19). We find that ${D_{4}}=-\left( M^{r}_{pl}\varphi_{r}+M^{l}_{pl}\varphi_{l}\right) = -\left(-7.5000\cdot 10^{-3} + 8.5190\cdot 10^{-18}\right) = 0.0075 > 0$;
   • find the next node at which a full plastic moment $M_{pl\,r} = 30.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ or $M_{pl\,l} =$ will form. We find that this is node 3 (see Fig. 8.3(b)). We also find that at this node, $\Delta M_{3}$ is needed to add to achieve $M_{pl} = 30.0{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$, and that $\Delta M_{3} = 30 - 27.6923 = 2.3077{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ as shown in Figs. 8.3(b) and 8.4(a).
3. In an n=1 times statically indeterminate frame, load is increased ( $\lambda=\lambda_{0} +\lambda_{1}+\lambda_{2})$ times to the appearance of the third full plastic moment $M_{pl}$. For that
   • compose a structural system for an n=1 times statically indeterminate frame (Fig. 8.4(a)) with a load factor $\lambda = 1.0$;
   • find the load factor $\lambda_{2}$ for a moment $M_{pl} = 30.0{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ to appear at node 3. We find that for the appearance of a plastic moment at node 3, 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);
   • add the boundary moments that are equal to the full plastic moment $M_{pl} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (related to $\lambda_{0}$) to the ends of elements 3 and 4 at hinge 4 (see Fig. 8.3(b));
     
     
   • add the boundary moments that are equal to the full plastic moment $M_{pl} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (related to $\lambda_{1}$) to the end of element 4 at plastic hinge 5 (see Fig. 8.4(b));
     
     
   • find the displacements and internal forces of the frame shown in Fig. 8.4(b). Here, the load factor is $\lambda=\lambda_{0} +\lambda_{1} +\lambda_{2} = 18.900$, the corresponding loads are $F_{1}=18.900{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ and $F_{2}=9.450{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$;
   • check if the dissipation D at plastic hinges 4 and 5 satisfies the condition (7.19). We find that ${D_{4}}=-\left( M^{r}_{pl}\varphi_{r}+M^{l}_{pl}\varphi_{l}\right) = -\left(-8.7000\cdot 10^{-3} - 1.2000\cdot 10^{-3}\right) =$ $0.0099 > 0$ and ${D_{5}}=-\left( M^{r}_{pl}\varphi_{r}\right) = -\left(-1.2000\cdot 10^{-3}\right) = 0.0012 > 0$;
   • find the next node at which a 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 1 (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 $\Delta M_{1}$ $= 15 - 8.400 = 6.600{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ as shown in Figs. 8.4(b) and 8.5(a).

Figure 8.2: Bending moment diagrams of an n = 3 times indeterminate frame

[$F_{1} = 1{\,}\mathrm{kN}, F_{2} = 0.5{\,}\mathrm{kN}$]

[$F_{1} = 14.685{\,}\mathrm{kN}, F_{2} = 7.3427{\,}\mathrm{kN}$]

Figure 8.3: Bending moment diagrams of an n = 2 times indeterminate frame

[ $F_{1} = 1{\,}\mathrm{kN}, F_{2} = 0.5{\,}\mathrm{kN}$]

[ $F_{1} = 17.884{\,}\mathrm{kN}, F_{2} = 8.9422{\,}\mathrm{kN}$]

Figure 8.4: Bending moment diagrams of an n = 1 times indeterminate frame

[ $F_{1} = 1{\,}\mathrm{kN}, F_{2} = 0.5{\,}\mathrm{kN}$]

[$F_{1} = 18.900{\,}\mathrm{kN}, F_{2} = 9.450{\,}\mathrm{kN}$]

Figure 8.5: Bending moment diagrams of a statically determined frame (n = 0)

[$F_{1} = 1{\,}\mathrm{kN}, F_{2} = 0.5{\,}\mathrm{kN}$]

[$F_{1} = 20.00{\,}\mathrm{kN}, F_{2} = 10.00{\,}\mathrm{kN}$]

1. As the last step of the procedure, load is increased in a statically determined frame (n=0) to the appearance of the full plastic moment $M_{pl}^{\left( n\right)}$. A plastic hinge forms and the frame becomes a mechanism. The load computed in which the moment exceeds $M_{pl}^{\left( n\right)}$ is equal to the true collapse load. In the n=0 times statically indeterminate (statically determinate) frame, load is increased $\lambda$ ( $\lambda=\lambda_{0} +\lambda_{1}+\lambda_{2}+\lambda_{3})$ times to the appearance of the fourth full plastic moment $M_{pl}$. For that
   • compose a structural system for an n=0 times statically indeterminate frame with a load factor $\lambda = 1.0$ (see Fig. 8.5(a));
   • find the load factor $\lambda_{3}$ for a moment $M_{pl} = 15.0{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ to develop at node 1. We find that for the appearance of the plastic moment at node 1, the load factor is to be $\lambda_{3}=\Delta M_{1}/M_{1} = 6.600/6.000 = 1.1000$ as shown in Fig. 8.5(a);
   • add the boundary moments that are equal to the full plastic moment $M_{pl} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (related to $\lambda_{0}$) to the ends of elements 3 and 4 at hinge 4 (see Fig. 8.5(b));
   • add the boundary moments that are equal to the full plastic moment $M_{pl} = 15.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (related to $\lambda_{1}$) to the end of element 4 at plastic hinge 5 (see Fig. 8.5(b));
   • add the boundary moments that are equal to the full plastic moment $M_{pl} = 30.0 {\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (related to $\lambda_{2}$) to the end of elements 2 and 3 at plastic hinge 3 (see Fig. 8.5(b));
   • find the displacements and internal forces of the frame shown in Fig. 8.5(b). Here, the load factor is $\lambda=\lambda_{0} +\lambda_{1} +\lambda_{2} +\lambda_{3} = 20.000$, the corresponding loads are $F_{1}=20.000{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ and $F_{2}=10.000{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$;
   • check if the dissipation D at plastic hinges 4 and 5 satisfies the condition (7.19). We find that
     ${D_{4}}=-\left( M^{r}_{pl}\varphi_{r}+M^{l}_{pl}\varphi_{l}\right) = -\left(-2.66667\cdot 10^{-2} - 1.0000\cdot 10^{-2}\right) = 0.03667 > 0$ ,
     ${D_{5}}=-\left( M^{r}_{pl}\varphi_{r}\right) = -\left(-1.00000\cdot 10^{-2}\right) = 0.01000 > 0$ and
     ${D_{3}}=-\left( M^{r}_{pl}\varphi_{r}+M^{l}_{pl}\varphi_{l}\right) = -\left(-3.8333\cdot 10^{-2} -3.8333\cdot 10^{-2}\right) = 0.0767 > 0$.

The four plastic hinges (nodes 1, 3, 4, and 5) produce a mechanism leading to collapse at the loads $F_{1}=20.000{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ and $F_{2}=10.000{\,}\mathrm{kN}\hspace*{-2pt}\cdot\hspace*{-2pt}\mathrm{m}$ (related to the load factor $\lambda = 20.000$).

An idealized relation between the load $F_{2}$ and deflection at node 3 is displayed in Fig. 8.6. The first yield moment occurs at node 4 shown in Fig. 8.5(b) and is marked with the letter A in Fig. 8.6. The following yield moments occurring at nodes 5, 3, and 1 shown in Fig. 8.5(b) are marked with the letters B, C, D in Fig. 8.6.

Figure 8.6: Load-deflection relation for the frame ESTsn3