Question 9.7: Figure 9.47 shows a 1 mm thick sheet of steel, one edge of w...
Figure 9.47 shows a 1 mm thick sheet of steel, one edge of which is fully restrained whilst the opposite edge is subjected to a uniformly distributed tension of total value 40 kN. For the material Young’s modulus, E = 200 GN/m² and Poisson’s ratio, ν = 0.3, and plane stress condition can be assumed
(a) Taking advantage of any symmetry, using two triangular membrane elements and hence
the assembled stiffness matrix derived for the previous Example, 9.6, determine the nodal
displacements in global coordinates.
(b) Determine the corresponding element principal stresses and their directions and illustrate
these on a sketch of the continuum
(a) Advantage can be taken of the single symmetry by modelling only half of the continuum. Figure 9.48 shows suitable node and dof. labelling, and division of the upper half of the continuum into two triangular membrane elements. Reference to the previous Example, 9.6, will reveal that the assembled stiffness matrix derived in answering this question can, conveniently, be utilised in solving the current example.
To simulate the clamped edge, dofs. 5 to 8 need to be suppressed, i.e. u_{3} = υ_{3} = u_{4} = υ_{4} = 0. Additionally, whilst node number 2 should be unrestrained in the x-direction, freedom in the y-direction needs to be suppressed to simulate the symmetry condition, i.e. υ_{3}= 0.
Applying these boundary conditions and hence partitioning the structural stiffness matrix result from Example 9.6, gives the reduced equations as
\begin{bmatrix} X_{1} \\ Y_{1} \\ X_{2} \end{bmatrix}=10^{7} [N/m]\begin{bmatrix} 14.835 & 7.143 & -3.846\\ 7.143 & 14.835 & -3.846 \\-3.846 & -3.846 & 14.835 \end{bmatrix} \begin{bmatrix} u_{1} \\ \upsilon _{1} \\ u_{2} \end{bmatrix} i.e \left\{P_{\alpha }\right\} =[K_{\alpha \alpha }]\left\{p_{\alpha }\right\}
Inverting [K_{\alpha \alpha }] to enable a solution for the displacements from \left\{P_{\alpha }\right\} =[K_{\alpha \alpha }]^{-1}\left\{p_{\alpha }\right\}
where adj [K_{\alpha \alpha }]=10^{14} \begin{bmatrix} 205.286 & -91.175 & 29.583 \\-91.175 & 205.286 & 29.583\\ 29.583 & 29.583 & 169.055 \end{bmatrix}
and det [K_{\alpha \alpha }] = 10^{21}[14.835(205.286) – 7.143(91.175) – 3.846(29.583)] = 2280.4 × 10^{21}
Then [K_{\alpha \alpha }]^{-1}=10^{-10} \begin{bmatrix} 90.03& -39.98 &12.97 \\-39.98& 90.03& 12.97\\ 12.97& 12.97& 74.13\end{bmatrix}
With reference to §9.4.7, the nodal load column matrix corresponding to a uniformly distributed load of 10 kN/m, will be given by
\left\{P_{\alpha }\right\} =\begin{bmatrix} X_{1} \\ Y_{1} \\ X_{2} \end{bmatrix}=10^{3}\begin{bmatrix} 10\\ 0 \\ 10 \end{bmatrix}_{[N]}
Hence, the nodal displacements are found from
\left\{P_{\alpha }\right\} =[K_{\alpha \alpha }]^{-1}\left\{P_{\alpha }\right\}
Substituting
\begin{bmatrix} u_{1} \\ \upsilon _{1} \\ u_{2} \end{bmatrix}=10^{-10}\begin{bmatrix} 90.03 & -39.98 & 12.97 \\ -39.98& 90.03 & 12.97 \\12.97 & 12.97 & 74.13 \end{bmatrix}10^{3} \begin{bmatrix} 10\\ 0 \\ 10 \end{bmatrix}=10^{-6} \begin{bmatrix} 103\\ -27 \\ 87\end{bmatrix}_{m}=\begin{bmatrix} 0.103\\ -0.027\\ 0.087\end{bmatrix}_{mm}The required nodal displacements are therefore u_{1} = 0.103 \ mm, \upsilon _{1} = -0.027 \ mm and u_{2} = 0.087 \ mm.
(b) With reference to §9.9, element direct and shearing stresses are found from
where, from Example 9.5,
[B]=\frac{E}{2a(1-\upsilon ^{2})} \begin{bmatrix} y_{23} & y_{31} & y_{12}& \upsilon x_{32} & \upsilon x_{13} & \upsilon x_{21 }\\ \upsilon y_{23} & \upsilon y_{31} & \upsilon y_{12}& x_{32} & x_{13} & x_{21 }\\\frac{1-\upsilon}{2} x_{32} & \frac{1-\upsilon}{2} x_{13} & \frac{l-\upsilon}{2} x_{21}& \frac{l-\upsilon}{2} y_{23} & \frac{l-\upsilon}{2} y_{31} & \frac{l-\upsilon}{2} y_{12 } \end{bmatrix}Evaluating the stresses for each element:
Element a
\begin{bmatrix} \sigma _{xx} \\\sigma _{yy} \\ \sigma _{xy} \end{bmatrix}=\frac{200\times 10^{9}}{2\times 2(1-0.3^{2})}\begin{bmatrix} -2 & 2 & 0& -0.6 & 0 & 0.6 \\-0.6 & 0.6 & 0& -2& 0 & 2 \\ -0.7& 0 & 0.7& -0.7 & 0.7 & 0\end{bmatrix}\begin{bmatrix} 0 \\87×10^{-6} \\ 0 \\0\\0\\0 \end{bmatrix} =\begin{bmatrix} 9.56×10^{ 6}\\2.87 ×10^{6} \\ 0 \end{bmatrix}_{N/m^{2}}The required principal stresses for element a are therefore σ_{1}=9.56 \ MN/m^{2}(T) and σ_{2}=2.87 \ MN/m^{2}(T) , and illustrated in Fig 9.49
Element b
\begin{bmatrix} \sigma _{xx} \\\sigma _{yy} \\ \sigma _{xy} \end{bmatrix}=\frac{200\times 10^{9}}{2\times 2(1-0.3^{2})}\begin{bmatrix} 0 & 2 & -2& -0.6 & 0.6 & 0 \\0 & 0.6 & -0.6& -2& 2 & 0 \\ -0.7& 0.7 & 0& 0 & 0.7 & -0.7\end{bmatrix}\begin{bmatrix} 87×10^{-6}0 \\103 ×10^{-6} \\ 0 \\0\\-27×10^{-6}\\0 \end{bmatrix} =\begin{bmatrix} 10.43×10^{ 6}\\0.43×10^{6} \\ 2.92× 10^{6} \end{bmatrix}_{N/m^{2}}The principal stresses are found from
\sigma _{1},\sigma _{2}=\frac{1}{2}(\sigma _{xx}+\sigma _{yy})\pm \frac{1}{2} \sqrt{[(\sigma _{xx}-\sigma _{yy})^{2}} +4 \sigma _{xy}^{2}]Substituting gives
\sigma _{1},\sigma _{2}= \left\{\frac{1}{2}(10.43+0.43)\pm \frac{1}{2} \sqrt{[(10.43-0.43)^{2} +4\times 2.92^{2}]} \right\} 10^{6} \ N/m^{2} =(5.43 \pm 5.79)10^{6} \ N/m^{2}giving \sigma _{1} = 11.22 \ MN/m^{2} (T) and \sigma _{2} = 0.36 \ MN/m^{2} \ (C)
The directions are found from
substituting gives
θ=\frac{1}{2} \tan ^{-1}[2×2.92/(10.43 -0.43)]=15.14°The required principal stresses for element b are therefore (\sigma _{1} = 11.22 \ MN/m^{2} (T) and \sigma _{2} = 0.36 \ MN/m^{2} \ (C) and are illustrated in Fig. 9.49.
