yennguyenportfolio.files.wordpress.com€¦  · Web viewOpen MATLAB. Navigate to proper directory. Run the MATLAB file “Pipe_all.m” by clicking “Run” or type “Pipe_all.m”

Open MATLAB. Navigate to proper directory. Run the MATLAB file “Pipe_all.m” by clicking “Run” or type “Pipe_all.m” in the command window.

Clicking

d,e,f,g. Entering

h.Pressing

b.Sliding

c.Sliding

a.Sliding

Entering

Figure 1. 2D view of the first frame: Geometry and Load inputs. Note: the plot in MATLAB is 3D and can be freely rotated.

0a. Pipe geometry. (found in the input file ``Pipe_geometry.txt’’)

· L1 : length of the first pipe.

· L2 : length of the second pipe.

· R : pipe’s radius.

1a. Figure 1. Pipe problem.

Input: giving input information for loads and cut section location. In this pipe system, the position of the concentrated load P is fixed (at x=L1-R, y=L2, z=0) as shown in Figure 1. Other variables listed as a, b, c, d, e, f, g below can get various values.

How to enter input: Solid boxes, labelled with “a”, “b”,” c”,… are mandatory. Dashed boxes are optional, usually for displaying results.

Hint: try Pz=0 and observe the output in the 3rd frame and 4th frame.

· a : xc (sliding) : horizontal or x-location of the cut section.

· b,c : x_q1, x_q2 (sliding) : x-locations of the distributed load q0 (x_q2 > x_q1), thus their difference x_{q2}-x_{q1} is the span of the distributed load.

Restriction: 0

· d : q0 (entering a value): distributed load magnitude. Sign convention:

q0>0, upward

q0<0, downward.

· e,f,g : Px, Py, Pz (entering values): x, y, z magnitudes of the concentrated load P, respectively. (P is applied on the top of the vertical pipe aligned with y-axis). Sign convention:

Px>0, left toward right

Px<0, right toward left

Py>0, upward

Py<0, downward

Pz>0: out of plane

Pz<0: into plane.

Output: Plots of cut section and loads in the same frame with the pipe system.

2. Figure 2.

Purpose: Plotting either by choosing one of the three buttons sxx, sxy or sxz and press “Plot”.

Figure 2b. Stress distribution sigma_xx plot at x=xc. Drag the curso to select y&z for next figure.

c.Pressing

b.Sliding

a.Sliding

Results

Pressing

Pressing

Figure 3a. Choosing y & z locations for one stress element. Displaying stress element in its principal directions.

3. Figure 3

Input: Entering the y and z locations of one material point that is represented by the stress element.

· a : Press “Enter”.

Output: Pictures of stress element (light blue color) in x, y, z (original coordinate system) and black labelled arrows indicating stress values

Displaying principal axis if it is among one of the original coordinate axis x, y, z (P axis) and the angle of rotation (theta_p)

Pictures of stress element (light grey color) in its principal directions and green arrows indicating principal directions and principal stress values.

Note: Arrow length is rescaled by the stress magnitude.

Figure 3b. Stress element (light blue) in the original coordinate x,y,z; black arrows : stresses on this element.

Stress element (gray) in the rotated coordinate. Green arrows : principal stress directions.

4. Figure 4.

Input: Entering axis of rotation and angle of rotation theta.

· a : Theta (slider): angle of rotation, 0

· b : x, y, z (button): axis of rotation.

· Write2File (pressing button): write result to file (in progress).

Output: Pictures of stress element (light blue color) in x, y, z (original coordinate system).

Pictures of the rotated stress element (light grey color) and green arrows indicating stress values in the new coordinates.

Hints:

try theta = 0. Determine if this stress state is in: original coordinate system/principal coordinate system/rotated coordinate system? Observe the stresses indicated by the arrows and see if: are positive or negative, which one has highest/lowest value.

try theta_p and P axis from the third frame as input of the fourth frame.

Pressing

c.Pressingg

b.Choosing among

a.Sliding

Pressing

Figure 4a. Stress element in rotated coordinate system. The value theta = 0 corresponds to the original coordinate.

Figure 4b. Stress element in rotated coordinate system. Green arrows stresses in new coordinate system. Example: rotate the original stress element 36 degree about z axis.

a.Choosing

b.Pressing

k. Pressing

e,f.Sliding

g,h.Sliding

i,j.

Entering values

c,d.Sliding

Figure 5. C beam with supports and loads. Note: the MATLAB plot is 3D.

B. Beam problem.

In this beam problem, load P can be applied at any arbitrary point along the upper surface of the beam but P is restricted to y direction only, Px=Pz=0. First support is pin type, 2nd support is roller. Recommend: not use Ibeam option because it is incomplete work.

0b. Beam geometry.

· L: beam length.

· R: beam radius, circular beam.

· H: beam heigh/depth & W: beam width, for I beam or rectangular beam.

· Ia: web thickness, Ib: thickness; for I beam only.

1b. Beam input.

Input: giving input information for loads and cut section location.

· Select one of three buttons: C for circular, S for rectangle, I for I beam.

· c : x_s1 (sliding) x-locations of the pin support.

· d : x_s2 (sliding) x-locations of the roller support.

· e,f : x_q1, x_q2 (sliding) x-locations of the distributed load q0 (x_q2 > x_q1).

· h : xc (sliding) horizontal or x-location of the cut section.

· g : lp (sliding) x-location of P.

· i : q0 (entering a value) distributed load magnitude.

· j : Py (entering values) y component of the concentrated load P. (Px=Pz=0).

Note: it does not requires that x_s2>x_s1. The condition x_q2>x_q1 is the only restriction.

Output: plot of beam with supports and loads.