COLLEGE OF
ENGINEERING
Department of Civil & Geological Engineering
CE 463.3 – Advanced Structural Analysis Lab – SAP2000 – Tips for Assignment T.A: Ouafi Saha
Professor: M. Boulfiza
Question 1: http://www.youtube.com/watch?v=tYOByVFtGRU
12m
8m
6
4
Material E=200 000 MPa, Section I=2000 10 mm , find vertical displacement at the concentrated load. The model definition is done following the same steps seen in previous examples. Two points need more details. Material definition: Go to menu Define > Materials … then click on Add New material Material name
Based on Other Put 0 to avoid self-weight
E=200 000 MPa Same as 200e9 n m
1
Section definition: Menu Define > Section Properties > Frame Sections … then click on Add New Property …
Section type Other General
All values can be 0 except 6 4 I33=2000 10 mm
Change default name FSEC1 to your preferred Select Material MAT defined in previous step
2
Make sure your beam has the section SEC defined above and that this section is made of the our particular material MAT. Pay attention to not include self-weight, either by putting material weight = 0 or by putting sel-weight multiplier = 0 in the DEAD load pattern. Run your Analysis. Then display the moment diagram. Show Forces/Stresses > Frames/Cables …
Choose Moment 3-3
Location 12m
Deflection 0.0384m
Usually we use Absolute. But for this example there is no difference
A better way to do it is by defining our structure with nodes at the desired locations if necessary. In this case add a node at the location 12m. 3
Question 2: 4m
4m
C 2 m
m 4
A Material E=200 000 MPa 6 4 Geometrie I = 500 10 mm 2 A = 2000 mm
At least two ways to define our structure are possible, with and without a node at point C. The steps to define the structure are almost the same as previous examples. Unit and Grid; Material and Section (don’t forget to add the cross section A); Draw the structure either with a step at point C or without; Support and Loads; Analysis of Plan Frame. For the horizontal displacement at point A click on the Show Deformed Shape button then put the mouse pointer close to the point A, a bubble appears showing information about that point, the horizontal displacement parallel to X axis is U1 = -0.008 (in this case in the opposite X direction) because local and global axis have the same orientation for the nodes. You can also right click on the node to see these values more precisely in a box. These values are given relatively to the undeformed shape.
Point A (U1=Ux=-0.008, U2=Uy=0, U3=Uz=-0.09117, R1=Rx=0, R2=Ry=0.012, R3= Rz=0) Displacements are given in the distance Unit, Rotations are given in radian according to right hand rule. With a node at C We can use the grid system X (0, 4, 8) and Z (0, 2, 4) we are not using Y axis so fare.
The steps to find vertical displacement at point C are the same as those for point A, the only difference is to look for U 3 = -0.04317 in the negative Z direction.
4
Without a node at C We can use the grid system X (0, 8) and Z (0, 2, 4) In this case we do not have a node at point C and the determination of the vertical displacement is done by displaying the information of the horizontal element (8m beam). For this purpose show the moment 3-3 diagram.
Then right click on the horizontal element. To show the detailed results
Select mid-span
Select Absolute to get vertical displacement with respect to the global axis (XYZ)
Note: positive in -2 direction
If you get a little difference with the hand calculated values try to explain why.
5
Truss Structures: http://www.youtube.com/watch?v=lHS17C5Ntf8
E = 200 GPa 2 A = 10 cm Neglect self-weight What is the vertical displacement at P?
m 2
P = 15
kN
2m
3m
The same steps: Units (kN, m, C) Grids X (0,2,5), Y(0), Z (0, 2) Choose XZ view Define Materials we need only E=200 GPa, 2 Elements properties we need only A = 10 cm Draw the model Assign Supports Assign Loads, one concentrated node at point P •
•
• • •
Here comes the difference. Many ways to do this analysis, the easiest way is: To put all other geometric properties = 0 except Cross Section (axial) Area.
Only Cross section different from 0
To avoid having warnings in the solver module, keep only UX and UZ DOF in the Analysis Options Keep only UX and UZ. No rotation
6
• • •
SOLVE Display Output Displacement and forces Analyze results
Deformed shape
We can have the displacement of any node (joint) by right clicking on it.
The vertical displacement of point P is then U 3 = -5.784 10 Internal forces
In this case only the Axial Forces are present
7
-4
m
