Problem 8
Question
The sketch represents three nodes on a \(z=\) constant face of an axisymmetric quadratic element. Node 7 is at midside. Determine the consistent nodal load vector for these three nodes if \(z\)-direction surface traction \(\Phi_{z}\) is applied as follows. (a) \(\Phi_{z}\) is the constant value \(p\) over the face. (b) \(\Phi_{z}=\left(\xi^{2}-\xi\right) p_{4} / 2+\left(1-\xi^{2}\right) p_{7}+\left(\xi^{2}+\xi\right) p_{3} / 2\), which is a parabolic variation based on nodal values \(p_{4}, p_{7}\), and \(p_{3}\).
Step-by-Step Solution
Verified Answer
The load vectors for Nodes 4, 7 and 3 under a constant \(\Phi_{z}\) are all evaluated to \(p * \frac{l}{2}\). Under parabolic variation \(\Phi_{z}\), the load vectors are \(\frac{l}{2} p_{4}\), \(l p_{7}\), and \(\frac{l}{2} p_{3}\) for Nodes 4, 7 and 3 respectively.
1Step 1: Calculate load vector for constant \(\Phi_{z}\)
For a constant value of \(\Phi_{z} = p\), each nodal load vector naught in the \(\Phi_{z}\) direction is computed as \(t_a = p * \frac{l}{2}\), where \(l\) is the length of the edge. Therefore, \(t_4 = t_7 = t_3 = p \frac{l}{2}\). This is because for constant surface traction, the load is evenly divided among the nodes.
2Step 2: Calculate load vector when \(\Phi_{z}\) has parabolic variations
For parabolic variation, we substitute \(\xi = -1\) for Node 4, \(\xi = 0\) for Node 7 and \(\xi = 1\) for Node 3 into the parabolic equation. We obtain:\(t_4 = \left((-1)^{2} - (-1)\right) \frac{p_{4}}{2} + (1 - (-1)^{2}) p_{7} + ((-1)^{2} + (-1)) \frac{p_{3}}{2} = \frac{l}{2} p_{4}\)\(t_7 = \left(0^{2} - 0\right) \frac{p_{4}}{2} + (1 - 0^{2}) p_{7} + (0^{2} + 0) \frac{p_{3}}{2} = l p_{7}\)\(t_3 = \left(1^{2} - 1\right) \frac{p_{4}}{2} + (1 - 1^{2}) p_{7} + (1^{2} + 1) \frac{p_{3}}{2} = \frac{l}{2} p_{3}\)These are the values of the nodal load vectors for Nodes 4, 7, 3 respectively when \(\Phi_{z}\) varies parabolically.
Key Concepts
Nodal Load VectorAxisymmetric ElementsParabolic VariationSurface Traction
Nodal Load Vector
In finite element analysis, the nodal load vector represents the distribution of forces or loads at specific points—nodes—of a structural element. The nodal load vector is crucial because it helps engineers understand how external forces translate into stresses and strains within the element.
- A fixed load can be evenly distributed across nodes when the force is constant.
- Complex distributions, such as parabolic loads, require integration to determine how the load affects each node differently.
Axisymmetric Elements
Axisymmetric elements are used in finite element analysis to model structures that have rotational symmetry around an axis. These elements simplify the analysis by reducing three-dimensional problems to two dimensions.
- Commonly used for objects like cones, cylinders, and spheres.
- Effective in analyzing pressures, forces, or tractions that are symmetrical.
Parabolic Variation
Parabolic variation refers to how a variable load is distributed in a curve, resembling a parabola, across a surface. This is more complex than a constant load as it requires consideration of how the load intensity changes across the element.
- In our exercise, \(\Phi_z\) varies across nodes in a parabolic pattern.
- Each node will experience different force intensities, calculated using the parabolic equation.
Surface Traction
Surface traction describes the distribution of force acting across the surface of an element. In finite element analysis, it’s essential to understand how this force is applied to model elements accurately.
- When surface traction is constant, nodal loads are evenly distributed.
- With variable surface traction, each node experiences a different force, reflecting the real distribution pattern.
Other exercises in this chapter
Problem 1
If a problem is to be mathematically two-dimensional, \(\theta\) independence is required of all dependent variables. Explain by example why this requires that
View solution Problem 6
The sketch shows the cross section of a flat element shaped like a metal washer. D.o.f. are radial displacements \(u_{1}\) and \(u_{2}\) at nodal circles 1 and
View solution Problem 10
Show that \(\epsilon_{r}=\epsilon_{\theta}\) at \(r=0\) in an axially symmetric problem.
View solution