User:Eml4500.f08.gravy.jad/Notes3

Lecture 15
HW problem goes here. (closing the loop between Finite Element Method and statics: virtual displacement.)

Two-Bar Truss System

There are two different ways on solving these types of problems: using FEM, or using statics.



(insert pictures)

These pictures are FBD's of 2 force bodies of element 1 & 2. By Statics reactions are known and therefore the member forces $$\ [P_{1}^{(1)}, P_{2}^{(2)}] \,$$ are equal and opposite in direction. We need to compute the axial displacement dofs ( amount of extension of bars):

$$\ q_{2}^{(1)}= P^{(1)}/k^{(1)}= \underline{AC} \,$$ . ( $$\ q_{1}^{(1)}=0 \,$$ because it's fixed at global node 1.

$$\ q_{1}^{(2)}= -P^{(2)}/k^{(2)}= -\underline{AC} \,$$ . ( $$\ q_{2}^{(2)}=0 \,$$ because it's fixed at global node 3.

Q : How to back out from above results of displacement of dofs of global node 2?

insert picture here triangle thing

insert matlab portion



$$\ u_{y}=R*sin(\alpha) \simeq R*\alpha \,$$

$$\ u_{x}=R*(1-cos(\alpha)) \simeq 0 \,$$ $$\ (cos(\alpha) \simeq 1) \,$$

hence

Lecture 16
Closing the loop continued Infinitesimal displacement (relationship to virtual displacement)



$$\ AC= \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{3/4} = 6.8324 \,$$

insert AB HW

insert xb,yb HW



- P is fixed, Q is "sliding" on the line

- the two lines are parallel $$\ \bar{PQ} =(PQ)*\vec {\tilde i} = (PQ)[cos\theta \vec {i} + sin\theta  \vec {j}] = (x-x_p)\vec i + (y-y_p)\vec j \,$$

$$\ \Rightarrow x-x_p=PQ cos\theta \,$$

$$\ \Rightarrow y-y_p=PQ sin\theta \,$$

$$\ \frac{y-y_p}{x-x_p}= tan\theta \,$$

$$\ (y-y_p)= (tan\theta)(x-x_p) \,$$

eq for a line perpendicular to the line, passing P: $$\ (y-y_p)= \tan(\theta+\frac{\pi}{2})(x-x_p) \,$$

insert HW xD,yD

since $$\ x_A=0 and y_A=0 \,$$ then $$\ \vec {AD} =(x_D-x_A)\vec i + (y_D - y_A)\vec j = x_D-\vec i + y_D\vec j \,$$

By definition: $$\ \vec{AD} = d_3 \vec i + d_4 \vec j \,$$ (these are displacement vectors of A from FEM

///end closing loop

3-Bar Truss Syst:



$$\ E^{(1)} =2, E^{(2)} =4, E^{(3)} =3 \,$$

$$\ A^{(1)} =3, A^{(2)} =1, A^{(3)} =2 \,$$

$$\ L^{(1)} =5, L^{(2)} =5, L^{(3)} =10 \,$$

$$\ P=30, \theta^{(1)}=30^\circ = -\theta^{(2)}, \theta^{(3)}=45^\circ \,$$

Local node numbering is equivalent to the convenience in assembly k.

Convient local node numbering