Question 9.2: Figure 9.34 shows the members and idealised support conditio...
Figure 9.34 shows the members and idealised support conditions for a roof truss. All three members of which are steel and have the same cross-sectional area such that AE = 12 MN throughout. Using the displacement based finite element method, treating the truss as a pin-jointed plane frame and each member as a rod:
(a) assemble the necessary terms in the structural stiffness matrix;
(b) hence, determine the nodal displacements with respect to the global coordinates, for the condition shown in Fig. 9.34.
(c) If, under load, the left support sinks by 5 mm, determine the resulting new nodal displacements, with respect to the global coordinates.
(a) Figure 9.35 shows suitable node, dof. and element labelling. Lack of symmetry prevents any advantage being taken to reduce the calculations. However, since both ends of the horizontal member are fixed it is redundant therefore and does not need to be considered further.
All three members have the same AE, hence
\left(AE\right)^{a} =\left(AE\right)^{b}=AEWith reference to §9.7, the element stiffness matrix with respect to global coordinates is given by
[k^{(e)}]=(\frac{AE}{L} )^{(e)}\begin{bmatrix} \cos ^{2} \alpha & & & \\ \sin \alpha \cos \alpha & \sin ^{2}\alpha & symmetric & \\ -\cos ^{2} \alpha & -\sin \alpha \cos \alpha& \cos ^{2} \alpha & \\ -\sin \alpha \cos \alpha & -\sin ^{2}\alpha & \sin \alpha \cos \alpha &\sin ^{2}\alpha \end{bmatrix} ^{(e)}Evaluating the stiffness matrix for both elements:
Element a
L^{\left(a\right) } =2m , \alpha ^{\left(a\right) }=60° , \cos \alpha ^{\left(a\right)}=1/2 , \sin \alpha ^{\left(a\right)}=\sqrt{3} /2
[k^{\left(a\right) }]=\frac{AE}{8}\begin{bmatrix} 1 & \sqrt{3}&-1&-\sqrt{3} \\ 3 & 3 &-\sqrt{3} &-3 \\-1&-\sqrt{3}&1&\sqrt{3}\\-\sqrt{3}&-3& \sqrt{3}&3 \end{bmatrix}
Element b
L^{\left(b\right) } =2\sqrt{3} m , \alpha ^{\left(b\right) }=150° , \cos \alpha ^{\left(b\right)}=-\sqrt{3} /2 , \sin \alpha ^{\left(a\right)}=1 /2
[k^{\left(b\right) }]=\frac{AE}{8}\begin{bmatrix} \sqrt{3} &-1&-\sqrt{3} &1 \\ -1 & 1/\sqrt{3} &1 &-1/\sqrt{3}&\\-\sqrt{3}&1& \sqrt{3}&-1 \\1 &-1/\sqrt{3} & -1& 1/\sqrt{3} \end{bmatrix}
The structural stiffness matrix can now be multi-assembled using a dof. correspondence table, (ref. §9.10), and will be of order 6 × 6. Only the upper sub-matrices need to be completed, i.e. [K_{αα}] \text{ and } [K_{αβ}] , since the reactions are not required in this example. The necessary structural governing equations and hence the required structural stiffness matrix are therefore given as
| Row/column in [K^{(e)}] |
Row/column in [K] |
|
| a | b | |
| 1 | 3 | 5 |
| 2 | 4 | 6 |
| 3 | 1 | 1 |
| 4 | 2 | 2 |
(b) Corresponding to u_{2}= \upsilon _{2}=u_{3}=\upsilon _{3}=0 , the partitioned equations reduce to:
\begin{bmatrix} X _{1} \\ Y_{2} \end{bmatrix}= \begin{bmatrix}-20\times 10^{3} \\ -40\times 10^{3} \end{bmatrix}=\frac{AE}{8}\begin{bmatrix} 2.7321 & 0.7321 \\0.7321 & 3.5774 \end{bmatrix} \begin{bmatrix} \upsilon _{1} \\ \upsilon _{2} \end{bmatrix} i.e. \left\{p_{α}\right\} = [K_{αα}]\left\{p_{α}\right\}
Inverting [K_{αα}] to enable a solution for the displacements from \left\{p_{α}\right\} = [K_{αα}]^{-1}\left\{p_{α}\right\}
adj [K_{\alpha \alpha }]=\begin{bmatrix} 3.5774& -0.7321 \\-0.7321 & 2.7321 \end{bmatrix} and det[K_{\alpha \alpha }]=(AE/8)^{2} 9.2378
Then [K_{αα}]^{-1}
=\frac{8}{AE} \begin{bmatrix} 0.3873 &-0,079251 \\-0.07925 &0.2958 \end{bmatrix} Check: \frac{8}{AE} \begin{bmatrix} 0.3873 &-0.079251 \\-0.07925 &0.2958 \end{bmatrix}\frac{8}{AE} \begin{bmatrix} 2.7321 &0.7321 \\0.7321 &3.57741 \end{bmatrix} =[I]
Hence, the required displacements are given by
\left\{P_{α }\right\} =[K_{\alpha \alpha }]^{-1}\left\{P_{\alpha }\right\}Substituting, \begin{bmatrix} u_{1} \\ \upsilon _{1} \end{bmatrix} =\frac{8}{12×10^{6}}\begin{bmatrix} 0.3873 & -0.079251 \\ -0.07925&0.2958 \end{bmatrix}\begin{bmatrix}-20.10^{3} \\ -40.10^{3} \end{bmatrix}=\begin{bmatrix} -3.051 \\ -6.83\end{bmatrix}_{mm}
The required nodal displacements are therefore u_{1} = -3.05 \ mm and \upsilon _{1} = -6.83 \ mm.
(c) With reference to §9.12, for non-zero prescribed displacements, i.e. \left\{p_{β} \right\} ≠ \left\{0 \right\}, the full partition of the governing equation is required, namely,
\left\{P_{\alpha }\right\} =[K_{\alpha \alpha }]\left\{P_{\alpha }\right\}+[K_{\alpha \beta }]\left\{P_{\beta }\right\}
Rearranging for the unknown displacements
\left\{P_{\alpha }\right\} =[K_{\alpha \alpha }]^{-1}\left\{P_{\alpha }\right\}-[K_{\alpha \alpha }]^{-1}[K_{\alpha \beta }]\left\{P_{\beta }\right\}Evaluating
[K_{\alpha \alpha }]^{-1}[K_{\alpha \beta }]=\begin{bmatrix} 0.3873& -0.079251 \\ -0.07925&0.2958 \end{bmatrix}\begin{bmatrix} -1 & -1.7321 &-1.7321 &1\\ -1.7321 & -3 &1 & -0.5773 \end{bmatrix} = \begin{bmatrix} -0.25 & -0.4331 &-0.75 &0.4331\\ -0.4331 & -0.75&0.4331& -0.25 \end{bmatrix}and [K_{\alpha \alpha }]^{-1}[K_{\alpha \beta }]\left\{P_{\beta }\right\}=\begin{bmatrix} -0.25 & -0.4331 &-0.75 &0.4331\\ -0.4331 & -0.75&0.4331& -0.25 \end{bmatrix} \begin{bmatrix} 0 \\-5\times 10^{-3}\\0\\0 \end{bmatrix} =\begin{bmatrix} 2.1655\\3.75 \end{bmatrix} _{mm}
Recalling from part (b) that
[K_{\alpha \alpha }]^{-1}\left\{P_{\alpha }\right\}=\begin{bmatrix} -3.05\\-6.83 \end{bmatrix}_{mm}and substituting into the above rearranged governing equation,
i.e \begin{bmatrix} u_{1} \\ \upsilon _{1} \end{bmatrix}=\begin{bmatrix} -3.05\\-6.83 \end{bmatrix}-\begin{bmatrix} 2.1655\\3.75\end{bmatrix}=\begin{bmatrix} -5.22\\-10.58\end{bmatrix}_{mm}
yields the required new nodal displacements, namely, u_{1} = -5.22 \ mm and \upsilon _{1} = – 10.58 \ mm.
