Question 9.1: Figure 9.31 shows a planar steel support structure, all thre...
Figure 9.31 shows a planar steel support structure, all three members of which have the same axial stiffness, such that AE/L = 20 MN/m throughout. Using the displacement based finite element method and treating each member as a rod:
(a) assemble the necessary terms in the structural stiffness matrix;
(b) hence, determine, with respect to the global coordinates (i) the nodal displacements, and (ii) the reactions, showing the latter on a sketch of the structure and demonstrating that equilibrium is satisfied
(a) Figure 9.32 shows suitable node, dof. and element labelling. Lack of symmetry prevents any advantage being taken to reduce the calculations. None of the members are redundant and hence the stiffness contributions of all three members need to be included. All three elements will have the same stiffness matrix scalar,
i.e. \left(AE/L\right) ^{\left(a\right) }=\left(AE/L\right) ^{\left(b\right) }=\left(AE/L\right) ^{\left(c\right) }=AE/L
With 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 each element:
Element a
\alpha ^{(a)}=-\tan ^{-1} (4/3),\cos \alpha ^{(a)}=-0.6 , \sin \alpha ^{(a)}=0.8
[k^{(a)}]=\frac{AE}{L} \begin{bmatrix} 0.36 & -0.48 & -0.36 & 0.48 \\ -0.48 & 0.64 & 0.48 & -0.64 \\ -0.36 & 0.48 & 0.36 & -0.48 \\ 0.48 & -0.64 & -0.48 & 0.64 \end{bmatrix}
Element b
\alpha ^{(b)}=180° ,\cos \alpha ^{(b)}=-1 , \sin \alpha ^{(b)}=0
[k^{(b)}]=\frac{AE}{L} \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
Element c
\alpha ^{(c)}=90 ,\cos \alpha ^{(c)}=0 , \sin \alpha ^{(c)}=1
[k^{(c)}]=\frac{AE}{L} \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 1 \end{bmatrix}
The structural stiffness matrix can now be assembled using a dof. correspondence table, (ref. §9.10). Observation of the highest dof. number, i.e. 6, gives the order (size), of the structural stiffness matrix, i.e. 6 × 6. The 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 | c | |
| 1 | 3 | 3 | 5 |
| 2 | 4 | 4 | 6 |
| 3 | 1 | 5 | 1 |
| 4 | 2 | 6 | 2 |
(b) (i) Rearranging and partitioning, with u_{1}= u_{2}=u_{3}=\upsilon _{3}=0 (i.e. \left\{p_{β}\right\} = 0)
\begin{bmatrix} Y _{1} \\ Y_{2} \end{bmatrix}= \begin{bmatrix}-90\times 10^{3} \\ -72\times 10^{3} \end{bmatrix}=\frac{AE}{L}\begin{bmatrix} 1.64 & -0.64 \\-0.64 & 0.64 \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 using \left\{p_{α}\right\} = [K_{αα}]^{-1}\left\{p_{α}\right\}
adj [K_{\alpha \alpha }]=\begin{bmatrix} 0.64 & 0.64 \\0.64 & 1.64 \end{bmatrix} and det[K_{\alpha \alpha }]=0.64(AE/L)^{2}
then [K_{αα}]^{-1}==\frac{L}{AE} \begin{bmatrix} 1 &1 \\1 &2.5625 \end{bmatrix} Check: \frac{L}{AE} \begin{bmatrix} 1 &1 \\1 &2.5625 \end{bmatrix}\frac{AE}{L} \begin{bmatrix} 1 &1 \\1 &2.5625 \end{bmatrix} \begin{bmatrix} 1.64 & -0.64 \\-0.64 & 0.64 \end{bmatrix} =[I]
The required displacements are found from
\left\{P_{\beta }\right\} =[K_{\beta \alpha }]^{-1}\left\{P_{\alpha }\right\}Substituting,
\begin{bmatrix} \upsilon _{1} \\ \upsilon _{2} \end{bmatrix}=\frac{L}{AE} \begin{bmatrix} 1 & 1 \\ 1 &2.5625 \end{bmatrix} \begin{bmatrix} Y_{1} \\ Y_{2} \end{bmatrix} =-5\times 10^{-8}\begin{bmatrix} 1 & 1 \\ 1 &2.5625 \end{bmatrix}\begin{bmatrix} 90.10^{3} \\ 72.10^{3} \end{bmatrix}
=\begin{bmatrix} -8.10 \\ -13.73 \end{bmatrix}_{mm}
The required nodal displacements are therefore \upsilon _{1} = -8.10 mm and \upsilon _{2} = -13.73 mm.
(ii) With reference to §9.12, nodal reactions are obtained from
\left\{P_{\beta }\right\} =[K_{\beta \alpha }]\left\{P_{\alpha }\right\}Substituting gives
\begin{bmatrix} X_{1} \\ X_{2} \\ X_{3} \\ Y_{4} \end{bmatrix} =\frac{AE}{L} \begin{bmatrix} -0.48 & 0.48 \\ 0.48 & -0.48\\ 0 & 0 \\-1 &0\end{bmatrix} \begin{bmatrix} \upsilon _{1} \\ \upsilon _{2} \end{bmatrix} =2\times 10^{7} \begin{bmatrix} -0.48 & 0.48 \\ 0.48 & -0.48\\ 0 & 0 \\-1 &0\end{bmatrix}\begin{bmatrix} -8.10\times 10^{-3} \\ -13.725\times 10^{-3} \end{bmatrix}
=\begin{bmatrix} -54 \\ 54 \\ 0 \\ 162 \end{bmatrix}_{kN}
The required nodal reactions are therefore X_{1} =-54 kN , X_{2} =54 kN , X_{3} =0 and Y_{3} =162 kN
Representing these reactions together with the applied forces on a sketch of the structure, Fig. 9.33, and considering force and moment equilibrium, gives
\sum{F_{x} } =(54-54)kN=0
\sum{F_{y} } =(162-90-72)kN=0
\sum{M_{3} } =(54\times 4-73\times 3)kNm=0
Hence, equilibrium is satisfied by the system of forces
