MEE 323 Su21 - Homework #4

pdf

School

Arizona State University *

*We aren’t endorsed by this school

Course

323

Subject

Mechanical Engineering

Date

Jan 9, 2024

Type

pdf

Pages

Uploaded by EarlGuineaPigPerson1061

Summer 2021B MEE 323 – Computer Aided Engineering II Homework #4 Assigned: July 26, 2021 Due Date: August 02, 2021 by 11:59 PM Instructions: C reate a folder named “Homework _4 ” in your U: \ drive. Save all of your model data and a copy of your homework report in this folder. Use this word file as a template for your homework report. Please turn in the assignment on the course Gradescope before the deadline. Homework Objectives: Learn to create and validate line models (beams and trusses). Learn to carry out stress stiffening and buckling analyses in ANSYS Workbench. Reading Assignment: Look through Chapter 6 in the textbook for examples of surface models and simulation techniques. Question 1: The members of the truss shown in the figure below have a square cross-sectional area of 1.96 in 2 and are made of structural steel. The left end is fixed but allowed to pivot, and the right end is allowed to freely move horizontally as well as pivot (no vertical motion is allowed); use appropriate supports to model the two boundary conditions. Model the truss structure in ANSYS by creating lines on a sketch and then using the "Lines from Sketches" tool to generate the line model. Use a mesh "Edge Sizing" control to ensure that each member of the truss consists of only one element (as was done in class).

Summer 2021B 2 (a) Show a figure of the sketch appropriately dimensioned and fully constrained. (b) Show a figure of the mesh with one element for each member of the truss. (c) Show a plot of the total deformation with the undeformed wireframe. (d) Insert a beam tool and show the figure of the direct stress (unaveraged) in the members of the truss. Right-click on the direct stress and choose "Export" to save the stress data in a file. Show the stress data in a table below. Will the truss support the given load without failing? What is the safety factor?

Summer 2021B 3 Node Number Element Number Direct Stress (psf) 1 3 42399 2 4 -42405 3 2 -42401 4 1 -42405 5 1 -42405 1 5 21206 1 6 42399 1 7 21206 2 7 21206 3 3 42399 3 4 -42405 4 2 -42401 4 6 42399 5 5 21206 F.S. = 5.22E6/42405 = 123 (e) Insert a probe and extract the reaction forces at the supports and tabulate them below. Verify these values by carrying out a global verification, i.e. sum of forces in the two directions must be zero, and the sum of the moments taken about any point on the structure must also be zero. Pin Support Moment = 0 Force in +Y direction = 500 Roller Support Moment = 0 Force in +Y direction = 500 (f) Insert a beam result to extract the axial forces in each member of the truss and tabulate them below. Use static equilibrium hand calculation (at node 1) to verify that this is indeed correct

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Summer 2021B 4 for the members M1 and M2 as labeled on the figure (remember to consider the reaction at the node as well). Beam Axial Force – Ansys Axial Force – Hand Calc 12 288.65 14 -577.18 23 288.65 24 -577.12 25 577.12 35 -577.18 45 -577.12 (g) Mesh the truss with multiple elements for each member and carry out an Eigenvalue Buckling analysis. Fix all vertices such that they do not deflect out of the plane of the truss and fix their rotations so that they just pivot about the Z-axis. Show figures of the buckling modes. Identify the critical member and the buckling load multiplier. Is the truss structure safe from a buckling standpoint? a. Buckling mode 1 – Load multiplier =35.462 – Critical member = L_45 b. Buckling mode 2 – Load multiplier = 44.56 – Critical Member(s) = L_14 + L_35 No, the structure is not safe from a buckling standpoint because the F.S. for yield strength of the material is much higher than the load multiplier required for buckling to occur.

Summer 2021B 5 Question 2: The members of the 3-D space truss shown in the figure below are made of aluminum alloy and have a circular cross-sectional area of 2.250e-3 m 2 . The 4 left endpoints are fixed but allowed to pivot (use appropriate supports), and the two endpoints on the right end are subjected to a load of 5200 N oriented at 30 deg. from the vertical as shown in the figure. The node numbers and coordinates are given in the following table. Node Number X Coord. (m) Y Coord. (m) Z Coord. (m) 1 0 0 0 2 0 4 0 3 4 4 0 4 4 0 0 5 0 0 4 6 0 4 4 7 4 4 4 8 4 0 4 9 8 0 0 10 8 0 4 5200 N 5200 N Pinned

Summer 2021B 6 (a) Create the geometry of the truss by creating a "Point" feature in Design Modeler and using the "Lines From Points" to create the truss elements. Specify a rectangular cross section for the truss members. Create a coordinates text file and read that in when you create the "Point" feature. Show your coordinates text file below (copy and paste what you have in your coordinates text file below). Also show a figure of the final geometry. 1 1 0 0 0 1 2 0 4 0 1 3 4 4 0 1 4 4 0 0 1 5 0 0 4 1 6 0 4 4 1 7 4 4 4 1 8 4 0 4 1 9 8 0 0 1 10 8 0 4 (b) Use a mesh "Edge Sizing" control to ensure that each member of the truss consists of only one element. Show a figure of the mesh.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Summer 2021B 7 (c) Insert a beam tool and show the figure of the direct stress (unaveraged) in the members of the truss. Use the "Max" and "Min" probes to show the location of the maximum and minimum stresses. Explain why we use the unaveraged stresses and not the averaged stresses. Use unaveraged stresses because averaged stresses consider the stress of nearby elements which would be incorrect for trusses since we know from mechanics of materials that the stress in each truss is constant. i) Minimum combined stress ii) Maximum combined stress

Summer 2021B 8 (d) Determine the factor of safety considering yielding failure. Comment on the magnitude of the factor of safety – i.e., why do you think this structure warrants such a high factor of safety? Which truss member is the critical one from the standpoint of yielding failure? F.S. = 2.8E8Pa/3.3383E6Pa = 83.87 L_68 is the critical member from a standpoint of yielding failure since it experiences the largest magnitude of stress in the analysis. This structure warrants a high factor of safety because of the un equal loading throughout the members and to minimize movement in the structure by ensuring that the structure is ridged enough to withstand the applied load. (e) Insert probes and extract the reaction forces and moments at the supports and tabulate them below. Do they satisfy static equilibrium conditions? Support Force X Force Y Force Z Moment 6 -9698.6 5239.7 737.77 0 2 -8314 3765.2 -0.35 0 5 7098.6 1.14 -737.7 0 1 5714.8 0.576 0.286 0 All the forces do satisfy static equilibrium conditions. (f) Duplicate the analysis of (a) above and change the mesh so that you now have 25 elements per truss member. Repeat the analysis and check the safety factor of the design. Comment on any differences seen or not seen. F.S. = 2.8E8Pa/3.34E6 = 83.83, is the factor of safety found from running the analysis again with 25 elements per truss. Support Force X Force Y Force Z Moment 6 -9698.6 5239.7 737.77 0 2 -8314 3765.2 -0.35 0 5 7098.6 1.14 -737.7 0 1 5714.8 0.576 0.286 0 Did not notice any differences in the results for the 25 elements per truss analysis as compared to the analysis with one element per truss. All analyses yielded almost identical results with very small changes. (g) Pass the solution of (f) to an eigenvalue buckling block and extract the first 4 positive buckling modes and load multipliers. Show figures of the buckling modes. Considering the first buckling mode, what is the factor of safety of the design from the standpoint of buckling and which is the critical member in the system?

Summer 2021B 9 a. Mode 1 – Multiplier = 7.2332 For the first buckling mode, the factor of safety from the design standpoint of buckling is 20.385, the critical member in the system is the member from point 5 to point 8. b. Mode 2 – Multiplier = 8.0242 c. Mode 3 – Multiplier = 9.5474 d. Mode 4 – Multiplier = 10.351

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Summer 2021B 10 Question 3: Consider the structural steel space load frame shown in the figure below. A frame is different from a truss in that the members are modeled as beams, and the mesh needs to have multiple elements per member. (a) Create a set of points for the frame in Design Modeler (either from a coordinates file or one by one). Use the Lines from Points tool to create a line body of the geometry shown. Use a rectangular cross section of 0.4 x 0.1 m 2 for all the beams. Show a figure of the final line body with the cross section assigned such that the bending stiffness is maximized in the appropriate direction (i.e. in the X-Y plane for the horizontal beam, and in the Y-Z plane for the vertical beam as well as the angled middle beam). Set the View options in the menu bar as shown on the right. (b) Mesh the model with the relevance center set to medium. Apply the loads and boundary conditions shown in the figure. Show plots of the mesh, total deformation, minimum combined stress, and maximum combined stress. Set the display options to also show the undeformed model. What is the factor of safety for this design? You can calculate this by dividing the yield strength of structural steel by the maximum absolute value of the stress results. x y z

Summer 2021B 11 i) Mesh ii) Total deformation iii) Minimum combined stress iv) Maximum combined stress

Summer 2021B 12 v) Factor of safety The factor of safety for the space truss is 2.5E8 Pa / 1.48E8 Pa = 1.68 (c) Extract the force reactions at the supports and verify that global equilibrium of forces is satisfied. i) X Y Z Top Support 4527.1 40541 -26300 Bottom Support -4527.1 -25541 -123700 Applied Forces 0 -15000 150000 Sum 0 0 0 (d) Pass the solution of (b) to an eigenvalue buckling analysis block and extract the first 2 positive buckling modes and load multipliers. Show figures of the buckling modes. Considering the first buckling mode, what is the factor of safety of the design from the standpoint of buckling and which is the critical member in the system? Factor of safety of the design is 16.5 from the standpoint of buckling, the critical member is the middle member between the two supported members. i) Buckling mode 1 – Load multiplier = 16.503 ii) Buckling mode 2 – Load multiplier = 40.577

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Summer 2021B 13 Question 4: Apply a moment on the thin aluminum alloy can shown below. Model the can as a surface body with the appropriate thickness. The can is of length 122 mm with a diameter of 64 mm and a thickness of 0.1 mm. The moment will cause a tensile stress on a principal direction in the can and a compressive stress in another principle direction causing the skin to buckle. Predict the moment that will cause the skin to buckle. (a) Use the model geometry provided. Mesh the model using a face meshing control to get a uniform regular mesh. Make sure your mesh is relatively fine. Show a figure of the mesh. (b) Apply a fixed support on one end. On the other end, create a cylindrical coordinate system. Then apply a displacement support with just the tangential (theta) direction free. Finally apply a moment of 1N-mm to this end. Show a figure of the total deformation.

Summer 2021B 14 (c) Pass the solution of (b) to an eigenvalue buckling block and predict the moment for the first buckling mode. What is the value of the moment required for this buckling to happen? Also show the first buckling mode shape. The value of moment required for buckling to happen is 24kN*m (d) The buckling load prediction also shows convergence based on the size of the mesh. Run a convergence analysis by manually setting the mesh size to 12mm, 10mm, 8mm, 6mm, 4mm, 2mm, and 1mm (you can use parameters to have ANSYS simulate these all at once). Plot the buckling load multiplier vs. mesh size (or number of elements). 0.00E+00 5.00E+09 1.00E+10 1.50E+10 2.00E+10 2.50E+10 3.00E+10 0 2 4 6 8 10 12 14 Load Multiplier