Project Submission 1

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1046

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Jan 9, 2024

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ECOR 1046 Mechanics Submission 1 1 ECOR 1046 Mechanics Roof Truss Design (Submission 1) Submission 1 shall be submitted by 11:30 am November 24 th . It shall consist of: 1. Calculations of the point loads acting on the Truss 2. An Engineering drawing showing the geometry of the truss, the point loads acting on the truss, and clearly labeled joints. 3. A full analysis showing the forces acting on each truss member. All calculations and Free body diagrams of each analysis must be shown. 4. A table and a second engineering drawing summarizing all the forces acting on each truss member. Drawing Guidelines: Two engineering drawings are required and must be completed using drawing tools such as AutoCAD or a similar software (SketchUp, LibreCAD, Google Documents drawing tool). All the drawings must show the dimensions of the truss. The two required engineering drawings are: I. The truss with the external applied loads and reactions; and II. The truss with the internal tension and compression forces acting on the members See the example drawings in Appendix B of this document. A title block is not required. Format the engineering drawings on 8.5x11” paper in landscape orientation . Drawings should be laid out to make effective use of the entire page . The drawings should include the drawing name, the course name and number, the names of group members, and the group name. Group Specified Variables and Input Parameters Each group will be assigned a different combination of 5 parameters: span length, spacing between trusses, spacing between open web steel joists, concrete slab thickness and snow accumulation on the roof. It is important that you design for the assigned variables and parameters. The assigned group parameters will be available on Brightspace. Figure 1: Standard Hockey Arena

ECOR 1046 Mechanics Submission 1 2 Figure 1 above shows that the standard width of ice in a hockey arena is 25.9 m. The span of the roof truss you will be designing must be greater than the width of the hockey rink and bleachers, to prevent obstructions on the rink and prevent obstructions to the view of spectators. For this project, different spans will be assigned to each group. The standard length of the ice surface in a hockey arena is 61 m. The length of the roof depends on the spacing between the 11 trusses supporting the roof. The spacing between the roof trusses will be different for each group. Open web steel joists span the spacing between the trusses. A plan view of the structure supporting the roof is shown in Figure 2 below. Your objective is to design the interior truss along column line D. Figure 2: Plan View of Arena Technical Considerations Truss Design During the design process, designers often estimate the results of calculations before performing the calculations. Critical thinking of the outcomes helps designers achieve the most effective design faster and allows them to notice errors. Geometry The variables that determine the geometry of the truss include the truss type, the support orientation, the span, and spacing between trusses. The support orientation, span length, spacing between trusses and spacing between open web steel joists will be provided. Each group must independently determine which truss type is optimal for their design. Each group will design one of the several trusses supporting the roof of the hockey arena. No group will have the same design, although multiple groups will be designing a truss with Truss Open web steel joist

ECOR 1046 Mechanics Submission 1 3 some of the same variables. Refer to Lectures on methods of truss analysis (Method of Joints and Method of Sections) to analyze and determine the forces in the members of your truss. Supports For this project, assume the roof truss is simply supported , i.e. one end of the roof truss is supported by a pin-support, and the other end is supported by a roller-support. Loads There are several different loads that act on structures. The different loads can be categorized as follows: • Dead Loads: Permanent loads that act on the structure over its entire lifetime. Dead loads are the self-weight of structural members. • Live Loads: Live loads are due to the use and occupancy of structure. • Snow Loads: Are due to the accumulation of snow or ice on structure. • Wind Loads: Loads applied to structures due to high wind pressures. • Earthquake Loads: These loads are due to seismic and earthquake events. Snow Load The snow load is calculated based on the anticipated maximum snow load that may occur every 1 in 50 years. The NBCC specifies the weight of snow to be approximately 3.2 ?𝑁 ? 3 ⁄ . For this project, the anticipated snow accumulation on the roof will be provided as one of the design variables for the project. The NBCC also states that rainwater accumulation in snow must be accounted for as well, as wet snow is heavier. Wet snow has contributed to the collapse of structures in the past, such as the Listowel Arena in Orangeville (Figure 3). Figure 3: Listowel Arena Collapse.

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ECOR 1046 Mechanics Submission 1 4 The NBCC accounts for the accumulation of rainwater in snow. For this this project the additional load due to rain will be 0.4 ?𝑁 ? 2 ⁄ . The load per unit area of snow can be calculated using the following equation: 𝑆 ? = 𝛾 ? ℎ ? Where: • 𝑆 ? : Snow load due to accumulation of snow (KPa). • 𝛾 ? = 3.2 ?𝑁 ? 3 ⁄ : Unit weight of snow. • ℎ ? : Anticipated height of snow accumulation (m). The total snow load per unit area acting on the roof of the arena is a combination of the accumulation of snow and the additional load due to rain: 𝑆 = 𝑆 ? + 𝑆 ? Where: • 𝑆 : Total snow load (kPa). • 𝑆 ? : Snow load due to accumulation of snow (kPa). • 𝑆 ? = 0.4 ?𝑁 ? 2 ⁄ : Load due to accumulation of water (kPa). Concrete Slab, Steel Deck, and Roof Insulation A corrugated steel deck is used to support roof finishes and the snow load on the roof. Sometimes, the steel deck on also has a layer of concrete as shown in Figure 4. This layer of concrete is known as a concrete slab. The concrete slab increases the strength of the roof but adds cost and weight. For this project, a steel deck with a concrete slab will be used. The concrete slab thickness will be assigned to each group as a design variable. Figure 4: Steel Deck with Concrete Slab.

ECOR 1046 Mechanics Submission 1 5 To calculate the weight of the concrete slab and the steel deck, the slab-deck system can be divided into three separate components as follows: 1. Steel Deck: Steel decks are often light due to the thickness of the steel deck being between 0.76 mm – 1.21 mm. The weight of the steel deck to be used in this project can be taken as 𝑊 ?? = 0.10 ?𝑁 ? 2 ⁄ . 2. Concrete Cover: Is the portion of concrete that is above the top of the steel deck profile (i.e. steel deck flute) highlighted in Figure 5. Figure 5: Concrete Slab and Steel Deck Dimensions. The weight of the concrete slab can be calculated as follows: 𝑊 ?? = (? ? − ℎ ? ) × 𝛾 ? Where: • 𝑊 ?? : Weight of concrete cover (?𝑁 ? 2 ⁄ ) • ? ? : Thickness of the concrete slab in (mm). • ℎ ? = 38?? : Height of the steel deck flute • 𝛾 ? = 24 ?𝑁 ? 3 ⁄ : The unit weight of concrete 3. Concrete Between Flutes: This is the portion of the slab that is contained in the flutes highlighted in Figure 6. Figure 6: Concrete Contained in Flute. A good approximation to obtain the weight per unit area of concrete contained in the flutes is to use half the height of the flute. 𝑊 ?𝑓 = ( ℎ ? 2 ) × 𝛾 ? Where: • 𝑊 ?𝑓 = Weight of concrete between steel deck flutes (?𝑁 ? 2 ⁄ ) . • ℎ ? = 38?? : Height of the steel deck flute • 𝛾 ? = 24 ?𝑁 ? 3 ⁄ : The unit weight of concrete

ECOR 1046 Mechanics Submission 1 6 In addition to the steel deck and concrete slab weights, we must also account for the weight of roof finishes. Roof finishes are used to protect the roof from the elements (snow, rain, wind, UV, etc.). For this project, assume the use of built-up roof (Figure 7) with a weight of 𝑊 ? = 0.31(?𝑁 ? 2 ⁄ ) . Figure 7: Built-up roof The total weight of the concrete slab and steel deck is the summation of all the above components: 𝑊 ? = 𝑊 ?? + 𝑊 ?? + 𝑊 ?𝑓 + 𝑊 ? Where: • 𝑊 ? = The total weight of all roof components (?𝑁 ? 2 ⁄ )𝑜? ?𝑃𝑎 . • 𝑊 ?? = Weight of steel deck (?𝑁 ? 2 ⁄ )𝑜? ?𝑃𝑎 . • 𝑊 ?? = Weight of concrete cover (?𝑁 ? 2 ⁄ ) 𝑜? ?𝑃𝑎 . • 𝑊 ?𝑓 = Weight of concrete between steel deck flutes (?𝑁 ? 2 ⁄ ) 𝑜? ?𝑃𝑎 . Load Factors To account for the variability in loads acting on structures, the National Building Code of Canada (NBCC) uses load factors to magnify applied loads. This design philosophy is known as limit states design. Limit states used to prevent the collapse or failures of structures are known as ultimate limit states (ULS). Structural failures may include rupture, instability, or buckling. To determine the internal loads developed in the truss members, the snow and dead loads, calculated as shown above, need to be multiplied by load factors to determine the total factored loads. The NBCC outlines several load factors and load combinations which can be used in structures depending on different loading conditions. For this project we will only

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ECOR 1046 Mechanics Submission 1 7 be focused on the load combination that magnifies dead and snow loads. In this load combination, the dead and snow loads applied to the truss are multiplied by the following factors to determine the factored load per unit area 𝑤 𝑓 : 𝑤 𝑓 = 1.25𝐷 + 1.5𝑆 • 𝑤 𝑓 : Factored load per unit area applied on the roof • Dead load (D): 1.25 (self-weight of concrete slab and other roofing materials) • Snow Load (S): 1.5 (accumulated snow and rain) Since the design of the roof truss is an iterative process, do not account for the weight of the truss members or the open web steel joists in this iteration of your design. The self-weight of the truss and the weight of the open web steel joists is often accounted for in a later iteration of the design, and will be outside the scope of this project. Load Path The load path is the sequence in which loads are transferred through various members from the location where they are applied until they reach the foundation of the structure. The load path for the arena roof is as follows: The snow rests on the built-up roof. The weight of the snow and built-up roof is supported by the presence of a steel deck with a concrete cover. The steel deck distributes the loads to the open-web steel joists. Open web steel joists are supported at each end by roof trusses. There elements are the main components of the roof system shown in Figure 8 below. Roof trusses are supported at their ends by columns which transfer that load to the foundation of the structure. Figure 8: Components of Arena Roof. Loads Transferred to Open Web Steel Joists and Tributary Area Each open web steel joist is designed to resist loads applied to the area that it directly supports. This area is referred to as the “Tributary Area” of the joist. Consider the floor plan shown in Figure 9 where the open web steel joists span east to west and the main trusses

ECOR 1046 Mechanics Submission 1 8 span north to south. The spacing between joists is 4 m and the spacing between trusses is 9 m. Figure 9: Arena Floor Plan. The typical tributary area supported by each joist is as shown in Figure 10 below. Figure 10: Tributary Area Acting on Open Web Steel Joist. To design the open web steel joists as 2D members, the loads applied on this tributary area need to be converted to a uniformly distributed load (UDL). The uniformly distributed load can be calculated by multiplying the load per unit area by the width of the area that the joist supports, referred to as the tributary width. In this project, the tributary width of the joists is simply the spacing between the joists ? 𝑗 . This gives us the following equation for calculating the factored uniformly distributed load acting on the joists: 𝑈𝐷𝐿 𝑓𝑗 = 𝑤 𝑓 × (𝑆 𝑗 ) Where: • 𝑈𝐷𝐿 𝑓𝑗 : Factored uniform distributed load acting on the joist (?𝑁 ? ⁄ ) • 𝑤 𝑓 : Factored load per unit area applied on the roof (?𝑁 ? 2 ⁄ ) • ? 𝑗 : Spacing between joists (?)

ECOR 1046 Mechanics Submission 1 9 The free body diagram of the open web steel joist is shown in fugure 11 below. Figure 11: FBD of open web steel joist Load Transfer from Open Web Steel Joists to Truss Open web steel joists are supported by a truss at each of its ends. Since joists are symmetric about their midspan, the factored support reaction at each joist end 𝑃 𝑓𝑗 is equal to the factored uniformly distributed load on the joist 𝑈𝐷𝐿 𝑓𝑗 , multiplied by the length of each joist and divided by 2. Since the length of the joist is equal to the spacing between the trusses ? ? , we can arrive at the following expression for calculating the factored reaction at joist ends 𝑃 𝑓𝑗 : 𝑃 𝑓𝑗 = 𝑈𝐷𝐿 𝑓𝑗 × ? ? /2 𝑃 𝑓𝑗 = 𝑤 𝑓 ? 𝑗 ? ? /2 The load applied at the connection between the OWSJ and the truss is equal to 2 times the factored support reaction of the joists 𝑃 𝑓𝑗 , and that is simply because interior trusses support pairs of open web steel joists, a joist to the left and a joist to the right of the truss. This is shown in Figure 12 below. Figure 12: Interior trusses often support OWSJ in pairs

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ECOR 1046 Mechanics Submission 1 10 Accordingly, the factored load applied to the truss from the joists 𝑃 𝑓 can be calculated as follows: 𝑃 𝑓 = 2 × 𝑃 𝑓𝑗 = 2 × 𝑤 𝑓 ? 𝑗 ? ? /2 𝑃 𝑓 = 𝑤 𝑓 ? 𝑗 ? ? Where: • 𝑃 𝑓 : Factored load applied to the truss from the joists • 𝑤 𝑓 : Factored load per unit area applied on the roof (?𝑁 ? 2 ⁄ ) • ? 𝑗 : Spacing between joists (?) • ? ? : Spacing between trusses (?) The open web steel joists should be placed at the joints of the main truss. This is done to ensure the truss members do not have to resists any internal shear or moment. Accordingly, the FBD of a truss would be similar to the one shown in Figure 13 below. Figure 13: FBD of truss supported by columns at each end Truss properties (Truss span, truss spacing, truss depth and joist spacing) For this project different groups will be assigned different span lengths varying between 35 m and 42 m, and different truss spacings ranging from 7 m to 11 m. The height of the truss, also commonly referred to as the depth of the truss, is dependent on the total span of the truss and how much load the truss carries. A good estimate for the design depth of the truss ranges between 𝒍 ?? ≤ 𝒉 ≤ 𝒍 𝟗 (shown in Figure 14) . For heavy loaded trusses use a depth closer to 𝑙 9 , for lighter loads use a depth closer to 𝑙 12 .

ECOR 1046 Mechanics Submission 1 11 Figure 14: Truss depth Another important truss property is the spacing of joists, which in turn sets the points of applications of the load on the truss. In structural design, it is efficient to space the joists relatively close to each other. Another reason to keep joist spacing low is to ensure steel decking is long enough to span between the joists. For this project, use an equal spacing between joists equal to 𝒍 ?𝟒 where ? is the span of the truss. This should result in 13 equally spaced point loads along the span of the truss. Truss Types Each group must determine an appropriate truss type for their design. Some truss types to choose from include Pratt, Howe, warren without verticals and Warren-with-verticals as shown in Table 1 below. Table 1: Some suggested truss types Pratt Truss Howe Truss Warren without verticals Warren with verticals Truss Analysis Internal Forces Determine the internal forces acting on the truss members based on the loads you determined previously. Choose either the Method of Joints or the Method of Sections to analyze your truss.

ECOR 1046 Mechanics Submission 1 12 Draw a free body diagram of the internal forces acting on the members of the truss, showing which members are in tension and which members are in compression. See the example engineering drawing in Appendix B. Recall: Internal tension forces pull on the member ends while internal compression forces push on the member ends as shown in Figure 15. Figure 15: Sign convention for internal forces in truss members.

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ECOR 1046 Mechanics Submission 1 13 Appendix A: Example Results Table Member Factored Internal Force (kN) Selected Member Designation Member Resistance (kN) AB ex: 570 kN (T) ex: HSS 152 X 102 X 11 ex: 1100 kN (T) AD ex: 470 kN (C) ex: 500 kN (C) BD BC CD CE DE

ECOR 1046 Mechanics Submission 1 14 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) Figure 1A: Sample Internal Forces Diagram without a title block