Question 9.4: The vehicle engine mounting bracket shown in Fig. 9.39 is ma...
The vehicle engine mounting bracket shown in Fig. 9.39 is made from uniform steel channel section for which Young’s modulus, E = 200 GN/m². It can be assumed for both.channels that the relevant second moment of area, I = 2 × 10^{-8} m^{4} and cross-sectional area, A = 4× 10^{-4}m^{2}. The bracket can be idealisedas two beams, the common junction of which can be assumed to be infinitely stiff and the other ends to be fully restrained. Using the displacement based finite element method, and representing the constituent members as
simple beam elements:
(a) assemble the necessary terms in the structural stiffness matrix;
(b) hence, determine for the condition shown in Fig. 9.39 (i) the nodal displacements with respect to the global coordinates, and (ii) the combined axial and bending extreme fibre stresses at the built-in ends and at the common junction..
a) Figure 9.40 shows suitable node, dof. and element labelling. The structure does not have symmetry or redundant members. The least number of beam elements will be used to minimise the hand calculations which, in this example, is two. Both elements will have the same A, E and I,
i.e. (A, E, I)^{(a)} = (A, E, I) ^{(b)}=A,E,I,
but will have different lengths, i.e. L^{(a)} and L^{(b)}.
With reference to §9.8, the element stiffness matrix inclusive of axial terms and in global coordinates is appropriate, namely:
\begin{matrix} {A\cos ^{2}\alpha +(12I\sin ^{2}\alpha )/L^{2}} & {symmetric} & {} \\ {(A-12I/L^{2})\cos \alpha \sin \alpha ,} & {A\sin ^{2}\alpha +(12I\cos ^{2}\alpha )/L^{2}} & {} \\ {(6I\sin \alpha )/L,} & {-(6I\cos \alpha )/L,} & {4I} \end{matrix}
Evaluating, for both elements, only those stiffness terms essential for the analysis:
Element a
L^{(a)}=0.1 m, α^{(a)}=180°, cos α^{(a)}=-1, sin α^{(a)}=0
No need to complete these rows and columns, for these examples
Element b
L^{(b)}=0.08 m, α^{(b)}=270°, cos α^{(b)}=0, sin α^{(b)}=-1
The structural stiffness matrix can now be assembled. Whilst the structure has a total of 9 dof., only 3 are active, the remaining 6 dof. are suppressed corresponding to the statement in the question regarding the ends being fully restrained. The node numbering adopted in Fig. 9.40 simplifies the stiffness assembly, whereby the first 3 × 3 submatrix terms for both elements are assembled in the first 3 × 3 locations of the structural stiffness matrix; these being the only terms associated with the active dofs. It follows that rearrangement is unnecessary, prior to partitioning. The necessary structural governing equations and hence the required structural stiffness matrix are therefore given as
These submatrices are not required, for this example.
(b) (i) Corresponding to u_{2} = υ_{2} = θ_{2} = u_{3} = υ_{3} = θ_{3} = 0, the partitioned equations reduce to
i.e {P_{α}}=[K_{αα}] {p_{α}}
Inverting [K_{αα}] to enable a solution for the displacements from {P_{α}}=[K_{αα}]^{1} {p_{α}}
where adj [K_{αα}]=10^{14}\begin{bmatrix} 3.7152 & -0.09&- 39.3 \\- 0.09 & 3.0769& 21.45\\-39.3& 21.45 & 9366.5 \end{bmatrix}
and det [K_{αα}]=10^{21}{89.375[104.8×0.036-(-0.24)(-0.24)]-0+0.375(0-0.375×104.8)}
=317.3085×10^{21}Then [K_{αα}]^{-1}=10^{-10}\begin{bmatrix} 11.7085 & -0.2836&- 123.8542 \\- 0.2836 &9.6969& 67.5998\\-123.8542&67.5998 & 29518.59 \end{bmatrix}
The required displacements are found from
{P_{α}}=[K_{αα}]^{1} {p_{α}}
Substituting , \begin{bmatrix} u_{1}\\ \upsilon _{1} \\ \theta _{1} \end{bmatrix}=10^{-10}\begin{bmatrix} 11.7085 & -0.2836&- 123.8542 \\- 0.2836 &9.6969& 67.5998\\-123.8542&67.5998 & 29518.59 \end{bmatrix}10^{3}\begin{bmatrix} 1\\ -2.5 \\ -0.5 \end{bmatrix}
=\begin{bmatrix}7.434 ×10^{-6} m\\ -5.833×10^{-6} m \\ -1.505 ×10^{-3} rad \end{bmatrix}The required nodal displacements are therefore u_{1}=7.434 ×10^{-6} m , \upsilon _{1}=-5.833×10^{-6} m and \theta _{1}=-1.505 ×10^{-3} rad
(b) (ii) With reference to §9.8, the element stress matrix in global coordinates is given as
[H^{(e)}]=\frac{E}{L} \begin{bmatrix} -cos \alpha -6t sin(\alpha )/L & -sin \alpha +6t cos(\alpha )/L & 4t & cos \alpha +6t sin (\alpha )/L &sin \alpha -6t cos (\alpha )/L & 2t \\-cos \alpha +6b sin(\alpha )/L & -sin \alpha -6b cos(\alpha )/L & – 4b & cos \alpha -6b sin (\alpha )/L &sin \alpha +6b cos (\alpha )/L & -2t \\-cos \alpha +6t sin(\alpha )/L & -sin \alpha -6t cos (\alpha )/L & -2t & cos \alpha -6t sin (\alpha )/L & sin \alpha +6t cos (\alpha )/L & -4t \\-cos \alpha -6b sin (\alpha )/L & -sin \alpha +6b cos (\alpha )/L & 2b & cos \alpha +6b sin (\alpha )/L & sin \alpha -6b cos (\alpha )/L & 4b \end{bmatrix}Evaluating, for both elements, only those terms essential for the analysis:
Element a
t^{(a)} = 14 ×10^{-3} m, b^{(a)} = 6 × 10^{-3} m, and recalling from part (a) L^{(a)} = 0.1 m, α^{(a)} = 180°, cos α^{(a)} = – 1, sin α^{(a)} = 0
With reference to §9.7, the element stresses are obtained from
{σ^{(e)}}=[H^{(e)}]{S^{(e)}}
where, for element a, the displacement column matrix is
{S^{(a)}}={u_{1 } υ_{1 } θ_{1 } u_{2 } υ_{2} θ_{2 }} in which u_{2 } = υ_{2} = θ_{2 } =0 in this example
Substituting for element a and letting superscript i denote extreme inner fibres and superscript o denote extreme outer fibres, gives
\begin{bmatrix} \sigma ^{i}_{1} \\\sigma ^{o}_{1} \\\sigma ^{i}_{2} \\ \sigma ^{o}_{2} \end{bmatrix} =2\times 10^{12}\begin{bmatrix}1 | & -0.84 | & 56\times 10^{-3} &| &| &| & \\ 1 | & 0.36 | & -24\times 10^{-3} &| &| &| & \\1 | & 0.84 | & -28\times 10^{-3} &| &| &| & \\1 | & -0.36 |& 12\times 10^{-3} &| &| &| & \end{bmatrix} \begin{bmatrix} 7.434\times 10^{-6} \\ -5.833\times 10^{-6} \\ 1.505\times 10^{-3} \\0\\0\\0 \end{bmatrix}= \begin{bmatrix} -143.89\times 10^{6} \\ 82.91\times 10^{6} \\ 89.35\times 10^{6} \\ -17.05\times 10^{6}\end{bmatrix}The required element stresses are therefore \sigma ^{i}_{1} = 143.89 MN/m^{2} (C), \sigma ^{o}_{1} = 82.91 MN/m^{2}(T), \sigma ^{i}_{2} = 89.35 MN/m^{2}(T) and \sigma ^{o}_{1} = 17.05 MN/m^{2} (C).
Element b
t^{(b)} = 6 ×10^{-3} m, b^{(b)} = 14 × 10^{-3} m, and recalling from part (a) L^{(b)} = 0.08 m, α^{(b)} = 270°, cos α^{(b)} = 0, sin α^{(b)} = -1
[H^{(b)}]=\frac{200\times 10^{9}}{0.08} \begin{bmatrix} 0.45 | & 1| & 24\times 10^{-3}&| &| &| \\-1.05 | & 1| & -56\times 10^{-3}&| &| &| \\-0.45 | &1| &-12\times 10^{-3}&| &| &| \\ 1.05 |& 1 | & 28 \times 10^{-3} &| &| &| \end{bmatrix}Again ,the element stresses are obtained from
{σ^{(e)}}=[H^{(e)}]{S^{(e)}}
where, for element b, the displacement column matrix is
{S^{(b)}}={u_{1 } υ_{1 } θ_{1 } u_{3 } υ_{3} θ_{3 }} in which u_{3 }= υ_{3}= θ_{3 } =0 in this example
Substituting for element b gives
\begin{bmatrix} \sigma ^{0}_{1} \\\sigma ^{i}_{1} \\\sigma ^{0}_{3} \\ \sigma ^{i}_{3} \end{bmatrix} =2.5\times 10^{12}\begin{bmatrix} 0.45 | & 1| & 24\times 10^{-3}&| &| &| \\ -1.05 | & 1| & -56\times 10^{-3}&| &| &| \\-0.45 |&1|&-12\times 10^{-3}&| &| &| \\ 1.05 |& 1| & 28 \times 10^{-3} &| &| &| \end{bmatrix} \begin{bmatrix} 7.434\times 10^{-6} \\ -5.833\times 10^{-6} \\ 1.505\times 10^{-3} \\0\\0\\0 \end{bmatrix}= \begin{bmatrix} -96.52\times 10^{6} \\ 176.60\times 10^{6} \\ 22.20\times 10^{6} \\ -100.42\times 10^{6}\end{bmatrix}_{N/m^{2}}The required element stresses are therefore \sigma ^{i}_{1} = 176.60 MN/m^{2} (T), \sigma ^{o}_{1} = 96.52 MN/m^{2}(C), \sigma ^{i}_{3} = 100.42 MN/m^{2}(C) and \sigma ^{o}_{3} = 22.20 MN/m^{2} (T).
