Problem 1
Question
Let uniform traction act normal to a surface of a quadratic solid Lagrange element that is rectangular and has uniformly spaced nodes. In the consistent element load vector \(\left[\mathbf{r}_{e}\right]\), what fraction of the total force appears at each of the nine nodes on the surface?
Step-by-Step Solution
Verified Answer
Each of the nine nodes on the surface of the quadratic solid Lagrange element bears 1/9th of the total force.
1Step 1: Understand Load Distribution
The given quadratic Lagrange element is a 2D element and has uniformly distributed load on its surface. Therefore, the load at each node will be equal, since the traction is said to act normal and uniformly.
2Step 2: Calculate Total Force
To calculate the total force, integrate the normal traction over the surface area of the element. However, the exact value of this force is not needed for this problem since we are asked to find the fraction of total force at each node.
3Step 3: Find the Fraction of Force at Each Node
As the traction is distributed equally and acts uniformly on all nodes, each node will bear 1/9th of the total force.
Key Concepts
Quadratic Lagrange ElementLoad DistributionElement Load VectorTraction on Finite Elements
Quadratic Lagrange Element
A quadratic Lagrange element is a type of finite element characterized by a quadratic variation of displacement within the element. It's often used in finite element analysis (FEA) because it can more accurately model curved boundaries and varying material properties as compared to linear elements. This element utilizes a set of polynomial basis functions, typically Lagrange polynomials, to define the displacement field within the element.
In the case of a solid rectangular element, it will have nine nodes if we employ quadratic interpolation: four at the corners, four along the sides, and one in the center. A key feature of using quadratic elements is their increased accuracy over linear elements, as they can capture the changes in deformation more precisely.
In the case of a solid rectangular element, it will have nine nodes if we employ quadratic interpolation: four at the corners, four along the sides, and one in the center. A key feature of using quadratic elements is their increased accuracy over linear elements, as they can capture the changes in deformation more precisely.
Load Distribution
Load distribution in the context of finite element analysis refers to how external forces, such as pressure or traction, are shared among the nodes of an element. Uniform load distribution means that each node receives an equal fraction of the total force applied to the element. This is a common assumption for simplifying calculations, especially when the traction or load is specified to act normally (perpendicular) and uniformly across the surface.
When evaluating the forces on a quadratic Lagrange element, one must keep in mind that a uniform external load will not necessarily result in linearly distributed internal stresses due to the element's quadratic nature. However, for computation of external forces, such as in the exercise provided, uniform load distribution simplifies to equal force allocation per node.
When evaluating the forces on a quadratic Lagrange element, one must keep in mind that a uniform external load will not necessarily result in linearly distributed internal stresses due to the element's quadratic nature. However, for computation of external forces, such as in the exercise provided, uniform load distribution simplifies to equal force allocation per node.
Element Load Vector
The element load vector, commonly denoted as \(\left[\mathbf{r}_{e}\right]\), is a vector representing the force contributions at each node of a finite element due to external loads. This load vector is crucial for assembling the global force vector that drives the deformation in finite element analysis. To construct this vector, the total external force needs to be distributed among the nodes according to the shape functions and the manner in which the load is applied.
In the example exercise, uniform traction acting normally on the surface implies that the load will be evenly distributed among the nine nodes of the rectangular quadratic element. Therefore, the element load vector will have equal entries representing each node’s share of the total external force.
In the example exercise, uniform traction acting normally on the surface implies that the load will be evenly distributed among the nine nodes of the rectangular quadratic element. Therefore, the element load vector will have equal entries representing each node’s share of the total external force.
Traction on Finite Elements
Traction is a term used in finite element analysis to describe a distributed force applied over the surface of an element. It is essentially the surface equivalent of stress and is measured in force per unit area. When traction is applied to finite elements, it is the job of the analyst to ensure that this distributed load is correctly translated into nodal forces which can be used in the calculation of displacements and stresses within the material.
As per the exercise under consideration, traction that acts normally to a surface will induce forces that are oriented perpendicular to that surface. For a quadratic solid Lagrange element with a rectangular shape and uniform traction, each node on the surface will experience the same magnitude of force, assuming that the load distribution is indeed uniform. This results in each node bearing an equal portion of the total applied force, which is especially relevant when creating the element load vector for further analysis.
As per the exercise under consideration, traction that acts normally to a surface will induce forces that are oriented perpendicular to that surface. For a quadratic solid Lagrange element with a rectangular shape and uniform traction, each node on the surface will experience the same magnitude of force, assuming that the load distribution is indeed uniform. This results in each node bearing an equal portion of the total applied force, which is especially relevant when creating the element load vector for further analysis.
Other exercises in this chapter
Problem 1
Derive locations and weights of an order 2 Gauss rule by requiring that it integrate exactly the polynomial \(\phi=a_{1}+a_{2} \xi+a_{3} \xi^{2}+a_{4} \xi^{3}\)
View solution Problem 2
If element thickness can vary and is computed as \(t=\sum N_{i} t_{i}\) from nodal values \(t_{i}\), what order of Gauss quadrature is needed to compute the exa
View solution Problem 2
Let the three-node bar element of Fig. 6.1-la be uniform, with niode 2 at the midpoint. The first row of its stiffness matrix \([\mathrm{k}]\) is \((A E / 3 L)\
View solution