Project Assignment 2 - ECOR1046A Mechanics (LEC) Fall 2023

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ECOR 1046 Mechanics Submission 2 ECOR 1046 Mechanics Roof Truss Design (Submission 2) Submission 2 shall be submitted by 11:30 am December 8t It shall consist of: 1. A table summarizing the steel sections chosen for each truss member based on their relative resistance and factored load they support; 2. An Engineering drawing showing the geometry of the truss with the selected section size clearly indicated on each truss member; 3. A table summarizing the stress, strain, and length deformation for each truss member; 4. Sample calculations to indicate how the truss members were chosen, and how the stress, strain, and length deformation were calculated. Truss Member Sizes Determine the truss member sizes required to resist the tensile and compressive forces acting on the truss members. To achieve an efficient design, it is important that selected truss members are not significantly over-capacity. Tension Members To determine adequate truss member size capable of resisting the tensile forces in the truss, compare the tensile force acting in each truss member, and select an efficient steel section for each truss member that will resist the associated internal force. Available steel sections can be found in Appendix A of this document. Use the following formula to select the appropriate tension members: T, = ¢ %o, XA where e ¢ =0.9 for Steel; e 0, =370MPa: Yield strength of Steel; A': Cross-sectional area (mm?); T, : Tensile resistance of the steel section (kN) Select the appropriate member so that T, > Ty, where T; is the internal factored tensile force acting on the truss member. Check your units! The final answers for internal forces should be specified in kN. Be sure to select appropriate members that are not significantly over-capacity; too-large members lead to inefficient and expensive designs. Compression Members To determine adequate truss member size capable of resisting the compressive forces in the truss, buckling must be taken into consideration. The strength of the steel is reduced under compressive forces due to the slenderness of the member. The longer a member, the weaker

ECOR 1046 Mechanics Submission 2 it becomes under compression because it is more likely to buckle, which in turn reduces the maximum stress it can resist. To determine the compressive resistance of a steel section, a factor f must be applied that considers the slenderness ratio and the grade of the steel. Figure 1 below shows how the strength of compressive members is reduced with an increase in slenderness ratio KL/r. In the diagram, o, is the yield strength of steel members. The radius of gyration r represents the ratio between the square root of the moment of inertia of the cross section of a member and the cross-sectional area of the member. Ty Short column Intermediate column Longcolumn Figure 1: The effect of slenderness on member strength. The following equations which accounts for slenderness effects is used to calculate the compression resistance of a steel section selected from the table in Appendix A: Cr = ¢fay,A where e ¢ =0.9 for steel; e 0, =370MPa: Yield strength of the Steel; A': Cross-sectional area (mm?); C, : Compressive resistance of the steel section (kN) f: Factor to account for slenderness of steel member calculated as follows: 1 1 (+22m)n where

ECOR 1046 Mechanics Submission 2 E = 200,000 MPa : Modulus of elasticity for Steel; e n = 1.34 : A parameter for calculating compressive resistance, for hot-rolled, fabricated structural sections and hollow structural sections manufactured in accordance with CSA G40.20, Class C; e K =1:Effective length factor; L: unsupported length of the truss member (mm); r: radius of gyration (mm) Because all the members are connected by pins at both ends, K = 1.0. L is the length of each truss member, and r can be found in the table included in Appendix A for each steel section. When selecting a steel section from the table, check that KL/r < 200. Tip: In practice, the appropriate steel section must be determined through trial and error, such that it will resist the internal factored compressive forces. Notice how f reduces the compression resistance of a steel section. Compare the internal factored compressive force C; from your calculations acting on each truss member with the compressive resistances C, of steel sections selected for each truss member. Ensure that C, 2 C; for each member. Check your units throughout your calculations! Example calculation Compare the tensile and compressive resistance of a 5 m long HSS 152 X 102 X 11 section. Section properties can be found from the table in Appendix A. The section has an area of A = 4840 mm?,r = 53.1 mm, and length L = 5m. Tensile Resistance: T, = ¢ %o, XA : T, = QTIPSO 1617 kv Compressive Resistance: n?E _ m2(200,000MPa) g, = = ¢ (KL/)? (1 * 5000mm)2 53.1mm = 222.6 MPa

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ECOR 1046 Mechanics Submission 2 1 1 f= 7= — = 0.443 (A +2207 (14 1.2892+134)T357 Cr = 0.443 x 0.9 x Z0MPa8I0mME _ 1)y 1000 N/kN The section HSS 152 X 102 X 11 with 5 m length can resist 1612 kN in tension, but only 714 kN in compression, a 56% reduction. Once all members have been designed, draw a diagram of your roof truss, labelling all the sections selected for the members of the truss. See the example engineering drawing in Appendix B. Also, summarize your design in a table similar to Appendix C. Stress, Strain and Deformation Based on the internal forces acting in each member of the truss determined in the final iteration, determine the stress, strain and change in length in each member. Recall the equation for normal stress: oc=F/A where e 0 :Internal tensile or compressive stress (MPa) e F :Internal tensile or compressive force (N) e A:Cross-sectional area of the member (mm?) And the equation of the change in length in a member: AL = PLJAE where AL : Change in length (mm) P : Internal tensile or compressive force (N) L : Original length of the member (mm) A': Cross-sectional area of the member (mm?) E = 200,000 MPa : Modulus of Elasticity for steel And the equation for normal strain: £=AL/L where e & Strain in the member (x 10~3mm/mm) e AL: Change in length of the member (mm) e L :Original length of the member (mm)

ECOR 1046 Mechanics Submission 2 Summarize the stress, strain, and change in length of each member of the truss in a table, along with the internal forces in each member. The table must be included in your report. An example can be found in Appendix C.

ECOR 1046 Mechanics Appendix A: Steel Sections and Properties. Submission 2 HSS Shapes Designation Dead Load (kN/m) | Area (mm2) r (mm) ) HSS51X51X6.4 0.079 1030 17.6 HSS 76 X 51 X 6.5 0.104 1350 26.1 HSS 76 X 51 X 8.0 0.155 1600 25.1 HSS 89X 64 X 8.0 0.155 2010 30.5 HSS 102 X102 X 6.4 0.178 2320 38.4 HSS 102 X 76 X 8.0 0.186 2410 35.8 HSS 102 X 76 X 9.5 0.215 2790 35.0 HSS 127 X 64 X 9.5 0.234 3030 41.7 HSS 127 X76 X 9.5 0.252 3280 43.2 HSS 152 X102 X 8.0 0.279 3620 54.8 HSS 152 X102 X 9.5 0.327 4240 54.0 HSS 152 X102 X 11 0.373 4840 53.1 HSS 203 X102 X 9.5 0.401 5210 70.3 HSS 203 X102 X 11 0.460 5870 69.3 HSS 203 X102 X 13 0.514 6680 68.4 HSS 203 X152 X 9.5 0.476 6180 75.1 HSS 203 X152 X 11 0.547 7100 74.2 HSS 254 X 254 X 8.0 0.590 7660 99.9 HSS 254 X 254 X 9.5 0.699 9090 99.1 HSS 254 X 254 X 13 0.912 11800 97.6 HSS 203 X152 X 13 0.614 7970 73.4 HSS 254 X152 X 11 0.634 8230 91.0 HSS 254 X152 X 13 0.713 9260 90.1 HSS 305 X 305 X 8.0 0.714 9280 121 HSS 305 X 203 X 9.5 0.699 9090 113 HSS 305 X203 X 11 0.808 10500 112 HSS 305 X203 X 13 0.912 11800 111 HSS 305 X305 X 11 0.982 12800 119 HSS 305 X 305 X 13 1.11 14400 118 HSS ## X ## X ## are nominal rectangular and square cross sections dimensions (see Figure 17 on the next page) Source: Handbook of Steel Construction, 11th Edition, Canadian Institute of Steel Construction

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ECOR 1046 Mechanics Submission 2 HSSdxbxt S— N R d X e Figure 1A: HSS dimensions

ECOR 1046 Mechanics Submission 2 Appendix B: Example Engineering Drawings (please note each drawing should take up one full page and must have a clear title block at the bottom right corner) N s c o - s " T [ R A [ m e ST BT " - 3 3 . P i B B H B [ 8 : ¢ % 3 9 5 p - 8 E g 5 g : 8 ay E 3 : - : : g 5 : : : E 4m o N L L K 4 [l 2.86 Note: Member selection symmetric along Hl axis. Member AB corresponds with member AB' Red = C Figure 1B: Sample Steel section diagram (this diagram does not have a title block)