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24 Cards in this Set
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When a component of displacement or rotation is restrained at a node of a finite elementmodel, both a row and a column are omitted from the stiffness matrix of a structure usedto solve for the remaining displacements or rotations. Why? 
We have one less displacement unknown, we need solve one less equation to find the displacements/rotations at nodes, so it is not necessary to find the reaction. This equation is discarded, preserving symmetry of the stiffness matrix. 

Why must fixedend reactions on the nodes at each end of a beam be added to the forces computed at the nodes from its element stiffness matrix, if intermediate or distributed loads are present? 
This is done, as when the fixedend case is superimposed, these loads are cancelled out leaving the real loading. The fixedend case can be superimposed after the nodal displacements are found, as it only changes the displacements inbetween the nodes, and only affects the stressing of the one beam with the intermediate/distributed loading. 

Why are local axes used for a beam element, as well as the global axes that describe deflections at nodes of a structure? Give two reasons. 
Local axes are needed to describe the stiffness terms simply, in terms of bending about principal axes of a beam, or twisting, or stretching the beam. They are also needed to report results, as we need to know the axial force, shear forces, the torque present and bending moments defined in local axes 

Rotation degrees of freedom are used for beam and plate elements, but not used for planar elasticity elements or for solid elements. Why? 
A rotation implies a variation of displacement through the depth of a beam or through the thickness of a plate. Similarly, a plate is just a plane (the midplane) and has thickness as a property. Where the geometry is explicitly defined by the shape of the elementas differences between displacements at nodes then imply that rotations must have occurred. 

The interpolation functions Ni that relate displacements at an arbitrary location within a finite element to those at its nodes, must sum to one at all points in an element. Why? 
A rigid body mode that gives equal displacements at all nodes has to be represented correctly. That is,displacement everywhere in the element has to match that at the nodes. This is the case if the interpolation functions sum to one at every location in the element. 

The matrix B is created to relate strains at a point inside a finite element, to the displacements of its nodes. For quadratic tetrahedral solid element, how many rows and how many columns will there be in this matrix? Explain why. 
A quadratic tetrahedral solid element has midedge nodes. As there are 4 corners and 6 edges, the total is 10 nodes. There are 3 components of displacement at these nodes, giving [B] 30 columns. For a solid,there are 3 direct strains and 3 shear strains, so the matrix has 6 rows. 

How many degrees of freedom n are used by the following elements? (e.g. what is the size of the n x n element stiffness matrix?)(a) A spring in 3D?(b) A beam in 3D?(c) A linear interpolation quadrilateral in 2D (eg a plane stress element)?(d) A pentahedral (i.e. 5 sided) solid element with quadratic interpolation? 
(a) A spring in 3D links x, y and z displacements at nodes at each end to the corresponding forces, giving 6 degrees of freedom, and hence a 6 by 6 element stiffness matrix (b) A beam does transmit moments as well as forces and has 3 displacement components and 3rotations at each end, hence 12 degrees of freedom (c) A linear quadrilateral deforming only inplane has x and y displacements only at each of 4corner nodes, hence 8 degrees of freedom. (d) A pentahedron has 5 faces. One with 2 triangular faces has 6 corners and 9 edges, with 3 displacement components associated with corner and midside nodes giving (6 + 9)*3 = 45degrees of freedom. 

The D matrix relating stress to strain varies in size in different elements, depending on how many components of strain need to be related to displacements at nodes. How big is this matrix for (a) a plane strain element, (b) an axisymmetric element, (c) a solid element? 
A plane strain element has inplane strains εx , εy and γxy, hence a 3 by 3 D matrix. An axisymmetric element has also circumferential or hoop strain out of the crosssectional plane modelled causing a 4 by 4 D matrix. A solid element has all possible direct and shear strains and hence a 6 by 6 D matrix. 

Give two reasons why local axes are used to examine results from a beam model? 
Local axes distinguish bending from twisting, and axial deformation from transverse deformation They also enable results to be presented in principal axis directions, and enable stiffness terms to be written more simply 

When solving problems by hand using Castigliano’s theorem, we differentiate energy per length of a beam to find the contribution per length to the displacement or rotation to be calculated, and then integrate along the beam to get the total. Why is this done rather than finding the total energy stored and then differentiating that? 
Differentiating energy per length first avoids the need to integrate algebraic terms found by squaring an expression for bending moment, that would then disappear again upon differentiation of the algebraic expression for the total energy with respect to a particular force or moment, as these terms are not dependent on that force or moment.This reduces the algebra required to obtain an expression for a deflection or rotation. 

The interpolation functions Ni that relate displacements at an arbitrary location in an element to that at the nodes of an element sum to one at any location within the element.Why must this be the case? 
The interpolation functions must sum to one to represent a rigid body translation correctly. One example of a rigid body displacement is where all nodes move together through a unit displacement.In this case all points in the element should have the same unit displacement, which will be so if the interpolation functions sum to one. 

The matrix B is created to relate strains at a point inside a finite element, to the displacements of its nodes. This matrix typically has more columns than rows. Why? 
B has more columns as there are more unknown displacements or rotations at nodes than there are types of strain. The number of columns = the number of nodal degrees of freedom (up to 60 for a hexahedral solid element with quadratic interpolation). The number of rows = the number of types of strain modelled (6 maximum). 

When a component of displacement or rotation is restrained at a node of a finite element model,both a row and a column are removed from the stiffness matrix of a structure. Why? 
The column is removed from [K] as it multiplies a zero displacement or rotation.The row is the coefficients of a redundant equation for an unknown reaction, that is not needed when solving for the displacements, but which can be solved separately later, if reactions are of interest. 

Why does applying minus the fixedend reactions to the nodes at each end of a beam that are caused by a distributed load, result in the correct nodal deflections? 
A distributed load is modelled by superposition of a fixed end solution on the deflections predicted at the nodes. Minus the fixedend reactions are applied to the finite element model so they cancel when the fixedend solution is superimposed. As the fixed end case does not contribute to the nodal deflections, those found from minus the fixed end reactions must be correct. 

Why are local axes used for a beam element, as well as the global axes that describe deflections at nodes of a structure? Give two reasons. 
Local axes are used to both orient the principal axes of a beam, to determine the stiffness terms of a beam, and also to identify output quantities eg to distinguish a bending moment from a twisting moment or to distinguish a shear force from an axial force. 

Displacement interpolation over a 2D planar elasticity finite element is done such that the displacements on element boundaries agree with those of neighbouring elements. Strains and stresses, however, do not agree. Why? 
While displacements are continuous at an element boundary, the spatial derivatives of displacement are not continuous (eg with linear interpolation, adjacent elements can have different linear variations of displacement). Hence strains, which are found from derivatives of displacement, are not continuous, and stresses, which are found from strains, are not continuous. Stresses could also be discontinuous, due to a change of material from one element to the next. 

Describe two properties of the interpolation functions Ni that are used to relate displacements at an arbitrary location within a finite element to those at its nodes. 
Individual interpolation functions Ni of a finite element are equal to one at a particular node, and die away to zero at the other nodes of an element. They form a partition of unity, meaning that the sum ΣNi is one at every point in an element. This property is necessary, if interpolation functions are to capture rigid body motion correctly.They also have the property that a displacement on the edge of an element only depends on the values at nodes on that edge. 

Conceptually, element stiffness matrices are assembled by expanding them to the size of the full stiffness matrix of the structure with rows and columns of zeros, so they can be added together.Comment on how this is coded in practice. 
In practice a vector is stored listing the mapping from the local order in which the equations for an element are written, to the corresponding equation numbers in the full set of equations. This is called an element destination vector. It enables a stiffness term to be put in the correct location in the assembled stiffness matrix, as the transformation in numbering applies to both rows and columns of [K]. The global matrix is also stored in a compacted form, but that is another story. 

In a direct stiffness approach to analysing a structure, element stiffness matrices are expanded with rows and columns of zeros and the expanded matrices are added. Why is this done? 
To achieve equilibrium, the forces and moments on each element at a node mustsum to equal the applied loading (or zero if there are no loads). If we treat connections as rigid, so that all elements joined displace and (if relevant) rotate the same, then summing forces implies summing stiffnesses. To do this in a matrix notation, we need to have matrices defined consistently referring to the complete set of displacements/rotations in global axes at the nodes, and to the corresponding complete set of forces/moments at nodes. This is achieved by expanding the element matrices with rows of zeros,corresponding to equations for forces or moments at other nodes in the structure, and by adding columns of zeros, corresponding to displacements or rotations at other nodes not in the element. Adding the expanded matrices corresponds to summing the forces (and moments if relevant) acting on each element. 

Why do we use minus the fixedend reactions on a beam to represent the effect on the nodes of intermediate or distributed loading? 
The fixedend problem only affects one beam, causing motion between its nodes,so it can be superimposed, to correct the solution for a beam loaded inbetween its nodes,after solving for motion at the nodes of a structure. For the superposition to work, the nodes must be loaded with minus the fixedend reactions, so that they cancel when the fixedend solution is superimposed. 

Why are local axes used of a beam element needed for looking at results of an analysis? 
One local axis must be aligned with the axis of a beam, just to distinguish a bending moment from a twisting moment, or an axial force from a shear force. In addition, if they are aligned with the principal axes of a crosssection, they are convenient for finding stresses. 

Rotation degrees of freedom are not used for planar elasticity elements or for solid elements.Why? 
In a 2D element or a solid element, there is no missing geometry implied by geometric properties, except the thickness of a plane stress element. This differs from abeam, which is geometrically a line with the motion of the crosssection is implied by rotations, or a plate which is geometrically a surface, where the motion of a line normal to the surface is implied by rotations. Knowing the nodal displacements is sufficient to define the deformed shape or a planar or 3D element, as the interpolation functions describe this explicitly. 

The interpolation functions Ni that relate displacements at an arbitrary location within a finite element to those at its nodes, are designed to equal 1 at one node and 0 at the other nodes.Why? 
Interpolation functions give the influence of each node on the deflection of an element. We aim to have the displacement at a node equal to that of the node. This is achieved if the function for node i is 1 at node i and zero at other nodes. 

The matrix B is created to relate strains at a point inside a finite element, to the displacements of its nodes. For quadratic pentahedral solid element, like that shown below,how many rows and how many columns will there be in this matrix? Explain why. 
There are 6 corners and 9 edges, and hence 15 nodes, with 3 displacement components at each node. Hence the B matrix will have 45 columns. The number of rows is the number of types of strain that we need, which is 6 for a solid element (3 direct and 3 shear strains). 