2105 ENG, Mechanics of Materials 2, Semester 2, 2016 SPACE GASS Project Instructions Before you start get to know these useful keyboard shortcut keys in SAPCEGASS
2 | P a g e
Q.1 Beam (50 Marks) All dimensions are in meters. For example d=8 m Total length L=d+e+f+g=8+3+4+7=22m All distributed loads are in kN/m. For instance d=8 kN/m All concentrated loads are in kN. For instance f=4 kN
Part 1-Using SPACE GASS: (1) Illustration of the model. You need to show the end fixities, loads and dimensions (similar to the Figure above). (10 Marks) Hint- Insert a snapshot of the beam model from SPACEGASS here. Use the “PrtScn” button on your keyboard to take a snapshot of the
computer screen, then edit and crop the picture in paint or similar software. The snapshot should clearly show the end restraints (fixed and roller), the distributed force and the concentrated load. Also show the dimensions between the loads which correspond to your student
number. To add dimensions to your model, right click on the beam and from the menu that pops up select “measure/dimension”, then “add
dimension”. See the sample below which corresponds to the student number shown above:
d e f g
d f
3 | P a g e
(2) Show the maximum deflection in the beam using 200 UB 25.4 Aust300 Universal Beam section. In SPACE GASS use ‘show envelope’ and ‘absolute maximum’ setting. (5 Marks) Hint- To assign 200 UB 25.4 profile to all elements, right click on them, then from the right tab select Properties on the top right of the screen,
under section click on the book symbol and select the section. The snapshot from SPACEGASS below shoes the window that will pop up.
4 | P a g e
To show the section name on each element, from the left bar click on the ‘T’ button (shown below) to enable it.
Then click on the arrow next to it and in the window that pops up, tick “show section names”. The ‘T’ button is used to show the values of the
load, deflection and shear force and bending moment diagrams, etc.
Run the model from the top menu “Analysis”, “Linear elastic analysis” with program defaults. Click on the ‘show displacements’ button on the
left tab right under the ‘T’ button to see the deflection in the beam. If you have defined more than one load case (for example one for the
distributed force and one for the concentrated force) then you need to press on the “Envelope” button on the left tab to show the minimum
(negative deflection) or maximum (positive deflection). If you only have one load case, then simply show the displacements. Below is a sample from the minimum displacements of the envelope.
5 | P a g e
(3) Show maximum reactions and the bending moment diagram of the beam in separate images. (5 Marks) Hint- To show the reaction click on the reaction button from the left tab.
To show the bending moment diagram click on the reaction BMD button from the left tab.
6 | P a g e
BMD under distributed load
BMD under concentrated load
BMD under both distributed and concentrated loads
7 | P a g e
(4) From the SPACE GASS library, select Aust300 Universal Beam section such that the absolute maximum deflection in the beam is smaller than (L/300). You need to show the iterations in Tabular format like the one shown below. (5 Marks) Hint- In the first iteration show the maximum displacement (deflection) of the beam using 200 UB 25.4 from previous section. If that
deflection is larger than L/300 then you need to use a bigger profile to reduce the maximum deflection to a value smaller than L/300. You
need to show some iteration how the deflection is reduced by changing the profile section in a table. See the sample snapshots and table
below:
8 | P a g e
Iteration Section Deflection
(mm) L/300 (mm)
Criteria
1 200 UB 25.4 516.94 =22000/300=73 NG 2 310 UB 46.2 122 73 NG 3 360 UB 50.7 85.91 73 NG 4 410 UB 53.7 64.89 73 OK
9 | P a g e
Part 2-Using Analytical Equations:
(You have to show the entire procedure. No marks are given if only the final answer is shown without the procedure) Instructions- Do your hand calculations on a plain paper, then scan and attach to the space provided here.
(5) Remove the roller support at the right end of the beam. Find the deflection at the right end of the beam under combined action of distributed and concentrated loads. You can use either the double integration method or the virtual work method. Use properties of 200 UB 25.4 Aust300 Universal Beam section. (15 Marks) Hint- Remove the roller support and analyse the cantilevered beam. Use the methods taught in class to find the deflection of the beam.
Compare your results with SAPCEGASS and identify the error. (6) Draw the Shear Force and the Bending Moment Diagram of the beam for the previous case (5). (10 Marks) Hint- Draw the SFD and BMD from your knowledge of the analytical procedure and compare with the SFD and BMD from SPACEGASS.
10 | P a g e
Q.2 Truss (50 Marks) For instance 2968347 then a=2, b=9, c=6, d=8, e=3, f=4, g=7 All dimensions are in meters. Use the Structure wizard and Cross Brace Truss in SPACE GASS: L=d+f+e+g=22m and a=2m P1=a kN=2 kN P2=f kN=4 kN P3=g kN=7 kN
11 | P a g e
Part 1-Using SPACE GASS:
(1) Show the SPACE GASS model with dimensions and member cross section annotations. Use Aust300 Square Hollow Sections (SHS) for all the members. (5 Marks)
Hint- Show all the dimensions and loads. See the sample below. You can pick any SHS section at this stage.
12 | P a g e
(2) Show horizontal and vertical deflections in all nodes. (2 Marks) Hint- See the sample below. You can use any square hollow section (SHS) that you like at this stage to begin with.
(3) Show axial forces in all the members. (2 Marks) Hint- See the sample below. Click on the ‘T’ button below to untick the ‘section names’. Then click on ‘axial force diagram’ button shown
below.
13 | P a g e
(4) Using Aust300 Square Hollow Sections (SHS) design the lightest truss such that the maximum vertical deflection is smaller than (L/300). You need to show at least 3 iterations. In each iteration, show an image of the Truss with member cross sections, vertical deflections in nodes and total truss weight next to it. (6 Marks) Hint- See the sample below. Use any configuration of SHS to start with. Plot the deflection at nodes. If the maximum vertical deflection is
larger than L/300, then use other SHS profiles to reduce the deflection. For each iteration write the total weight of the truss next to the image.
You can either find the total weight by hand or use SAPCEGASS to get it. In SPACEGASS click on “output” from the top bar, then click on “view text report” and press “OK” twice. In the “report viewer” window, select “Analysis results” then “bill of materials”. Total mass is given in
tons. See the image below, the total mass is 0.251 tons equal to 251 kg. This is the optimisation procedure that engineers do a lot to justify
their design to the client.
14 | P a g e
15 | P a g e
251 kg Hint-Be smart! Investigate where the maximum deflection occurs and change the profiles of the elements next to it. In this sample problem, total length is L=22m, so the maximum allowable vertical deflection is L/300=73 mm.
I have only shown a single iteration here and the maximum deflection is 13.93 mm which is much lower than 73mm. So this is not an
optimised design. In your project you should come up with an optimised design.
16 | P a g e
(5) Check all members (SHS) in compression and tension. All members in compression should satisfy a factor of safety in buckling of SFc=2.0 and all members in tension and compression should comply with a factor of safety in yield of SFT=1.5. Use yield stress of 300 MPa. (10 Marks) Hint-Below is a sketch of the truss showing the member numbers. To show the member numbers, go to “T” symbol on the left bar and tick
“show member numbers”.
Go to “Analysis” from the top menu, and click on “buckling analysis”. Pick the load case with all the loads (if you have defined more than one load case) and press Ok. On the “output” option from the top menu, pick “view text report” and click OK twice. In the window that pops up, go to “buckle factors”. You should see a window like the one below:
17 | P a g e
As you see in the window above, the buckling load factor is 1.809 for load case 4 (all loads applied together). This is not acceptable since 1.809 < 2.0. You need to increase the SHS profile dimensions and run “Buckling analysis” again until the load factor becomes larger than two.
18 | P a g e
To check the yield stress in the members, first run a “Linear static analysis”. Then from the left bar click on “σ” and tick the “axial stress” button as shown in
the image below. That will show the axial stresses in the members. Make sure the maximum stress in members is less than the yield stress divided by the of SFT=1.5. that means the maximum axial stress in members should be less than 300/1.5=200 MPa.
19 | P a g e
Part 2- Using Analytical Equations: (You have to show the entire procedure. No marks are given if only the final answer is shown without the procedure) Instructions- Do your hand calculations on a plain paper, then scan and attach to the space provided here.
(6) Find the horizontal deflection at the node where P1 is applied. (10 Marks) Hint- Use the virtual work method to find the horizontal deflection at that node. Use any SHS for the members that you like. Compare your results with SAPCEGASS.
(7) Find the vertical deflection at the node where P3 is applied. (10 Marks) Hint- Use the virtual work method to find the horizontal deflection at that node. Use any SHS for the members that you like. Compare your
results with SAPCEGASS.
(8) Find the member with largest axial compressive force and design a SHS for that member. Use a factor of safety in buckling of SFc=2.0 and a factor of safety in yield of SFT=1.5. Use Le=L for the buckling design. (5 Marks) Hint- In SAPCEGASS fond the member with largest compressive (negative) axial force. For instance in the sample problem, it is the member
with axial force +26.23 in the image below (In SpaceGass compression is shows as positive).
Use the Euler buckling equation with (Le=L) to design a section that can carry that load (26.23 kN) without buckling with a factor of safety of
2.0. Now check the same member and ensure it does not yield using a factor of safety of 1.5.
20 | P a g e