User:Eml4500.f08.bottle.butler/HW 2

9/3 Given: Truss with elastic bars When faces with a problem as presented above in Statics, the boundary conditions are removes and replaced with forces. Certain problems are left statically indeterminate. The flowing steps are used to solve a statically indeterminate problem using matrices.

First, draw a global free body diagram. The axis for the FBD are chosen based on the direction of the reaction forces. The FBD is then labeled using the following notation: 1.	Nodes are numbered within circles 2.	Elements are then numbered within triangles 3.	Unknown reaction forces are labeled 4.	The three equations of quality are written.

Then, a FBD for each element is drawn. The following notation is used: 1.	Each node is labeled within a box 2.	Element is numbered within a triangle 3.	The reaction at each node is labeled in the following way a.	f degree of freedomelement #, d degree of freedomelement # numbering begins with 1 on X axis and then moves to Y axis, then proceeds to next node.

Once the elements are labeled the force displacement relationship is written. An example used to clarify the method is a spring example where f=kd Forces are arranged in a matrix: where k is the stiffness matrix

9/5 The big picture: Steps to solve a simple truss problem: 1.	Global Picture (Description) a.	At structure level i.	Global degrees of freedom (displacement, dof’s) ii. Global forces 2.	Displacement degrees of freedom are partitioned into a.	A known part (e.g. fixed dof, constraint/applied force) b.	An unknown part (solve using FEM/resultant forces) 3.	Element picture a.	Element DOF’s b.	Element faces

4.	Global Force Displacement relationship a.	Element stiffness matrices in global coordinates b.	Element force matrices in global c.	Assembly of element stiffness matrix and element force matrix in global FD rel i.	Kd=F 5.	Elimination of known dof’s to reduce global relationship a.	Stiffness matrix is non singular 6.	Compute forces from known d= element stress 7.	compute reactions

Specific example:

Element Length L(1)= 4 L(2)=2 Youngs modulus E(1)= 3 E(2)=5

(1)=30 (2)=-45

Matrices: {F}=[k]{d}

~Remember to use convenient coord system and apply load to only one node.

k(e) = K

where l(e)= cosθ(e) and m(e)= sinθ(e)

9/10

The focus of the lesson is to continue with the computation of the simple truss problem presented in lecture 4. We know that θ(1) = 30°. Therefore, l(1)= lnθ(1)= cos 30° = And m(1) = sinθ(1) = sin30° = Also, K(1) = =   =

Therfore, we can compute the stiffness matrix k(1), which is calculated in the following form:

K matrix with super and sub scripts

Using the numbers computed above we calculated k(1) to be the following:

Image of K stiffness matrix completed

When looking at the stiffness matrix we are able to observe that only 3 values need to be calculated: l(e)m(e), m(e), and l(e). The absolute values of these numbers form the stiffness matrix with only the sign changing. We are also able to observe that the matrix is symmetric, or the transpose of k(e) is equal to k(e).

We can use the same method described above to find the stiffness matrix for element 2. θ(2)=   l(2)= cos – 45° =

m(2)= sin -45° =

A brief description of the force-displacement relationship:

Elemental FD: k(e)d(e)= f(e) d(e) =  and f (e)=

Global FD: Kd = F

9/12

The global notation can be simplified into the form [Kij]{dj} = {Fi} where K= [Kij] is the global stiffness matrix, d= {dj} is the global displacement matrix and F= {Fi} is the global force matrix. The elemental notation can be simplified into the form k(e)d(e)=f (e), where k= [k(e)ij] is the element stiffness matrix, d= {d (e)j} is the element displacement matrix and f= {f (e)i} is the element force matrix.

One can move from the element to the global matrices through an assembly process. The correspondence between the element degrees of freedom and the global displacement degrees of freedom can be seen below for element 1 and 2:

For node 1: d1 = d(1)1 d2 = d(1)2

For node 2: d3 = d(1)3= d(2)1 d4= d(1)4= d(2)2

For node 3: d5 = d(3)4 d6 = d(2)4

When fully completed we can see that the topography of K resembles the figure below.

Picture of stiffness matrix with K1 and K2 blocks

9/15

Therefore, for example 1, K 6x6 =

Picture of K

Where K11= k(1)11 = ;  K33= k(1)33 = k(2)11 = 3.0625; etc. At this step we can eliminate the known DOF’s. Because the truss example we are calculating is static, we know d1=d2=d5=d6= 0. When we apply these boundary conditions we can delete the corresponding columns in the stiffness matrix and by using the Principle of Virtual Work we can also eliminate the corresponding rows. We are left with the following equation:

Kd=