Faculty of Built Environment and Engineering School of Civil and Environmental Engineering CIVN 4015: Civil Engineering Design Final Year Design Project: Structures Group 4 Internal Supervisor: Dr. Ryan Bradley External Supervisor: Mr. Daniel Surat, PREng Compiled by: Student Number 908593 1391559 1612449 1631963 1633852 1669086 Student’s Name and Surname Zamanguni Ntozakhe Musa Mdhluli Makhosonke Mhlanga Nonhlanhla Mhlongo Maxwell Mohau Majoro Jackson Mahlaule Executive Summary The purpose of this report is to detail the design and analysis of a warehouse. This warehouse needs to have a bay which will allows for circulation of vehicles and have an overhead crawl beam that can be used to remove equipment whenever necessary. The warehouse is designed as a duo-pitched portal frame. The software which was used to determine the design loads as well as deflections was Prokon. The loads combinations which were used for the software analysis were determined using the Ultimate Limit State and the Serviceability Limit State. The Ultimate Limit State was used to ensure that the sections used in the design can carry the applied loads. The Serviceability Limit State was used to ensure that the sections used do not experience excessive deformations when the structure is occupied. Table of Contents Executive Summary .......................................................................................................................... 2 List of figures .................................................................................................................................... 5 List of tables ...................................................................................................................................... 6 1.Introduction .................................................................................................................................... 7 1.1 Background ................................................................................................................................. 7 1.2 Scope of the project ..................................................................................................................... 7 1.3 Methods used to achieve the design outcomes .............................................................................. 8 1.4 Role of each group member ......................................................................................................... 9 2. Reference documents ................................................................................................................... 10 3.Preliminary Design Considerations ............................................................................................... 11 3.1 Preliminary Design 1 – Latticed Portal Frame ............................................................................ 11 3.2 Preliminary Design 2 – Trussed Portal Frame ............................................................................ 12 3.3 Preliminary Design 3 – Rolled Section Portal Frame .................................................................. 14 3.4 Preliminary Design 4 – Pre-engineered built-up Portal Frame .................................................... 15 4.Structure geometry ....................................................................................................................... 17 4.1 General arrangement .................................................................................................................. 17 4.1.1 General arrangement considerations ........................................................................................ 18 4.1.2 Design solution for the 15m bay .............................................................................................. 19 5.Analysis assumptions ................................................................................................................... 20 5.1 Wind loads ................................................................................................................................ 20 5.2 Crane loads ................................................................................................................................ 21 5.3 Imposed Loads .......................................................................................................................... 21 5.4 Dead Loads ............................................................................................................................... 21 5.5 3-D Structural steel framework .................................................................................................. 22 5.6 Members and connection conditions .......................................................................................... 22 5.7 Support conditions ..................................................................................................................... 23 5.8 End stop .................................................................................................................................... 23 6.Loads ........................................................................................................................................... 24 6.1 Dead loads ................................................................................................................................. 24 6.2 Wind loads ................................................................................................................................ 24 6.3 Imposed loads ............................................................................................................................ 26 6.4 Crane load ................................................................................................................................. 26 6.5 Geotechnical considerations ....................................................................................................... 26 7.Load combinations ....................................................................................................................... 27 8.Analysis results ............................................................................................................................ 31 8.1 Prokon Analysis ........................................................................................................................ 31 8.1.1 Member Selection ................................................................................................................... 31 8.1.2 Displacements ........................................................................................................................ 31 8.1.3 Reactions ................................................................................................................................ 32 8.1.4 Code checking ........................................................................................................................ 33 8.1.5 Connection design .................................................................................................................. 33 Rafter to Column: ........................................................................................................................ 33 Bracing beam to rafter connection: .............................................................................................. 34 Summary of connection details .................................................................................................... 35 Column to Rafter Connection:.................................................................................................. 35 8.2 Foundation ................................................................................................................................ 36 8.3 Ground slab design .................................................................................................................... 36 Conclusion ...................................................................................................................................... 37 References....................................................................................................................................... 38 Acknowledgements ......................................................................................................................... 39 Appendix 1 – Meeting Minutes ........................................................................................................ 40 Appendix 2 – Cross-disciplinary and stakeholder engagement GA 8 form........................................ 68 Appendix 3 – Preliminary Design 1 (Latticed Portal Frame) ............................................................ 69 Appendix 4 – Preliminary Design 2 (Trussed Portal Frame) ............................................................. 70 Appendix 5 – Preliminary Design 3 (Rolled Section Portal Frame) .................................................. 71 Appendix 6 – Preliminary Design 4 (Pre-engineered built-up Portal Frame) ..................................... 72 Appendix 7 – Dead Loads ............................................................................................................... 73 Appendix 8 – Wind Loads ............................................................................................................... 81 Appendix 9 - Imposed Loads ........................................................................................................... 82 Appendix 10 – Member designs....................................................................................................... 83 Appendix 11 – Foundation Design ................................................................................................... 84 Appendix 12 – Ground slab design .................................................................................................. 85 List of figures Figure 1 Latticed portal frame ............................................................................................. 11 Figure 2: Trussed frame ...................................................................................................... 12 Figure 3: Rolled section ...................................................................................................... 14 Figure 4: Pre-engineered built-up Portal Frame ................................................................... 15 Figure 5: Front view of the structure ................................................................................... 18 Figure 6 Plan view of the structure ...................................................................................... 18 Figure 7 Elevation of the structure....................................................................................... 19 Figure 8 Plan view of the structure ...................................................................................... 20 Figure 9 Elevation of the structure....................................................................................... 20 Figure 10: summary of the connection details ...................................................................... 35 List of tables Table 1 details the role which each group member plays in the completion of the project ...... 9 Table 2 contains a list of documents used to design the structure. ........................................ 10 Table 3 contains the dead loads which are used to analyse the structure. .............................. 24 Table 4 contains the imposed loads which act on the structure. ............................................ 26 Table 5 contains the forces that act on the crawl beam. ........................................................ 26 Table 6 contains the legend for the symbols used to determine the load combinations. ........ 27 Table 7 contains the load combinations considered for the analysis of the structure. ............ 27 Table 8 contains the load combination cases for ULS and SLS ............................................ 29 Table 9 contains the allowable deflections for the structural members ................................. 32 Table 10 contains the dominant ULS load combination ....................................................... 32 Table 11 contains the dominant SLS load combination ........................................................ 32 Table 12 contains the reaction forces used to design the gable column foundation. .............. 32 Table 13: code checking of structural members ................................................................... 33 Table 14 contains the dimensions of the foundation as well as the loads acting on it. ........... 36 Table 15 contains the loads that are used for the ground slab design .................................... 36 1.Introduction Final year civil engineering students at the University of the Witwatersrand were tasked to design a warehouse. The end-span of the structure needs to be 15m wide on both sides to allow for circulation of vehicles and all other bays will be 7.5m wide. The warehouse is to be framed using portal frames of constant width and height. All fire, acoustic, mechanical, fabrication costs, site erection costs on optimization, and electrical requirements for the warehouse must be ignored. 1.1 Background Portal frames are generally low-rise structures, comprising columns and horizontal or pitched rafters, connected by moment-resisting connections. Resistance to lateral and vertical actions is provided by the rigidity of the connections and the bending stiffness of the members, which is increased by a suitable haunch or deepening of the rafter sections. This form of continuous frame structure is stable in its plane and provides a clear span that is unobstructed by bracing. Pitched portal frames are commonly used in single storey buildings. 1.2 Scope of the project The project aims to produce a design for a 1200m2 portal framed warehouse, with specified parameters detailed in the general arrangement section. The design will be done in compliance with the South African national standards and relevant building codes. Listed below are the deliverables for the project: Detailed design of the warehouse, member sizes specified, connection details specified Detailed drawings of the design, including sketches of the preliminary structures and the final structure. Design of a crane runway. Listed below are the items which are outside of the scope of the project: Gutters are not accounted for. Fabrication costs and site erection costs are not taken into consideration in the report. Load calculations do not account for thermal and seismic loading. Acquiring materials and erection of structure. 1.3 Methods used to achieve the design outcomes The structure is designed to meet the Ultimate Limit State requirements, and these requirements are for checking the strength of the members selected in designing the structure. The structure is also designed to meet the Serviceability Limit State requirements, and these requirements are for checking that when the structure is in use, the deflections are not too excessive. Listed below are the steps which are taken to design the structure: 1. Selecting the type of portal frame which is to be used: This step involves conducting research on the types of portal frames (only four types are considered) which can be used to design the warehouse. One of the four options presented is to be selected to design the warehouse. 2. Determine the general arrangement of the structure: A decision is taken on the type of bracing, which is used for the structure, and where it is best (the bay) to brace the structure. The orientation of the columns is also determined in this step. Positioning of the gable columns is also determined in this step. 3. Determine the loads acting on the structure: Using the project brief, the loads acting on the structure are determined. The SANS code is used to compute the wind loads. Research is conducted to determine the best sheeting to use for the structure. 4. Determine the load combinations: The SANS code is used to determine the load factors as well as the load cases for the structure. 5. Analyze the structure on a design software: The structure is analyzed on Prokon to determine if the selected members (sections for rafters, columns etc.) can withstand the loads acting on the structure, and if the deflection that occur are within the limits stated in the SANS code. It is also used to determine if the connections and the foundation can carry the loads. 6. Hand calculations to determine if the members selected, connections and foundation meet the SANS code requirements: Using the checks stipulated in the code, the members selected, the connections and foundation are verified. 7. Produce engineering drawings: AutoCAD is used to produce all the drawings necessary for the project. 1.4 Role of each group member Each group member is expected to contribute towards the completion of this project. Table 1 details the role which each group member plays in the completion of the project Group member Zamanguni Ntozakhe Role played by the group member Research advantages and disadvantages for Preliminary design 2. Determine the load cases that would be analyzed on the software Design the purlins, girts and connections Researched advantages and disadvantages for Preliminary design 3 Musa Mdhluli Makhosonke Mhlanga Nonhlanhla Mhlongo Design gable columns Model the structure on Prokon Determine the dead loads acting on the structure. Write the introductory chapter (excluding scope) Design the ground slab, concrete foundation, and tie beams Research advantages and disadvantages for Preliminary design 1. Design the crawl beam, and door frames. Compile the report. Determine the loads acting on crawl beam. Determine the wind load Mohau Majoro Jackson Mahlaule Responsible for Autocad drawings Discuss the scope of work for the project Responsible for the Autocad drawings Determine the live loads acting on the structure. 2. Reference documents The table below contains a list of documents which are used in the design process for the project. These codes are prescribed in the project brief. Table 2 contains a list of documents used to design the structure. Scope Code/ Handbook Description Loading SANS 10160-1:2018 Ed1.2 Basis of structural design and actions for buildings and industrial structures Part 1: Basis of Structural Design SANS 10160-2:2011 Ed1.1 Part 2: Self-weight and imposed loads SANS 10160-3:2018 Ed2.0 Part 3: Wind actions SANS 10160-5:2011 Ed1.1 Part 5: Basis for geotechnical design and action SANS 10162-1:2011 Ed2.1 The structural use of steel Part 1: Limitstates design of hot-rolled steelwork The Red Book Southern African Steel Construction Handbook The Blue Book Design of Structural Steelwork to SANS 10162 The Green Book Structural Steel Connections Concrete SABS 0100-1:2000 Ed2.2 The structural use of concrete Part 1: Design Foundation Design Concrete Industrial Floors on Ground Provides a guide for designing ground slabs Structural Steel 3.Preliminary Design Considerations There are many types of portal frames that can be used. There are four types of portal frames that have been selected and discussed below are the advantages and disadvantages of each type of portal frame. 3.1 Preliminary Design 1 – Latticed Portal Frame (Ebid,M.(2021.Portal Frame Truss [Image]. University of Cairo) Figure 1 Latticed portal frame A Lattice Portal frame is constructed by using angles or tube steel members (Andrews, n.d.). The structural integrity of the latticed portal frames comes from the moment resistance of the connections and its ability to carry bending moment (Andrews, n.d.). Written below are the advantages and disadvantages of using latticed portal frames (Andrews, n.d.): Advantages: They are considered a good option for longer than spans. They are not expensive to construct, as they can be prefabricated. They are lightweight structures which make them easy to erect. Disadvantages: There are more factors that need to be considered in the design of the latticed portal frame before it is deemed structurally sound. Each member in the structure must be checked extensively for: 1. The effects of shear forces as well as axial forces 2. Buckling: the local buckling of the member, lateral bucking as well as column bucking 3. Deflection that occurs under service loads. All the connections for the structure need to meet the following conditions: economy, strength, rotational capacity, and stiffness. Lattice portal frames have a low fire resistance, and this is problematic because the stability of the structure is compromised quickly when a fire occurs. The graphical representation (sketch) of a latticed portal frame can be found in Appendix 3. 3.2 Preliminary Design 2 – Trussed Portal Frame (Ebid,M.(2021.Portal Frame Truss [Image]. University of Cairo) Figure 2: Trussed frame A truss is a structure formed of rod members organized in one or more triangle-shaped configurations. The members must be triangulated because the joints are pinned (do not transmit moments). On the other hand, a frame is a structure made up of beam members that can be placed in any direction and are joined rigidly or with pins at joints. The members support bending as well as axial loads (Doyle, 1991). Written below are the advantages and disadvantages of using the Trussed Portal Frame: Advantages: High strength and resistance to tension and pressure During production there is quality control of steel unlike concrete structures Possibility of structural developments Prefabricated parts can be used to construct Structural development High installation speed Less space needed to occupy material Able to connect numerous pieces using nuts, bolts, and welds Can be used at high altitudes Disadvantages: There could be poor connections between welds, bolts, and nuts Requires a lot of material not so economical Angles should be precise to avoid failure 3.3 Preliminary Design 3 – Rolled Section Portal Frame (Ebid,M.(2021.Portal Frame Truss [Image]. University of Cairo) Figure 3: Rolled section Generally fabricated from UB sections with a substantial eaves' haunch section, which may be cut from a rolled section or fabricated from plate (Mohammed, 2021). Written below are the advantages and disadvantages of using a Rolled section Portal Frame. Advantages: - - Use of uniform sections, standard universal beams Less members used, reduces cost Reduced complexity of fabrication, time of fabrication reduced The use of a haunch at the eaves reduces the required depth of rafter by increasing the moment resistance of the member where the applied moments are highest (institute, 2022). The haunch also adds stiffness to the frame, reducing deflections, and facilitates an efficient bolted moment connection (Institute, 2022) (HRS) frame is the optimum system for low and short frames because its high constructability and low unit price compensate any waste of material due to using standard sections (Mohammed, 2021). Disadvantages: - Members are too heavy More time is needed to develop the structural designs Of site fabrication not possible (Geometry), due to difficulty in transportation - More accuracy is required to build these types of structures. The portal frames are placed at regular intervals We cannot construct any structure above the portal frames. It is very difficult to construct structure above the portal frame, due to its sloppy head 3.4 Preliminary Design 4 – Pre-engineered built-up Portal Frame (Ebid,M.(2021.Portal Frame Truss [Image]. University of Cairo) Figure 4: Pre-engineered built-up Portal Frame Pre-engineered built- up Portal Frames are used because they are roughly 30% lighter than other systems because steel sections are used efficiently (Ebid, et al., 2021). Written below are the advantages and disadvantages of using the Pre-engineered built- up Portal Frame (Ebid, et al., 2021): Advantages: Quicker to build Easy to construct Has the ability to Build Long Spans It Is Possible to Use It for Temporary Construction It is Reliable Disadvantages: High Maintenance Costs Less Flexibility on-site Potential for Leaks due to many joints Design Difficulties Transportation cost to on-site Poor Fire Resistance 4.Structure geometry The preliminary design which is to be designed is the Preliminary Design 3 which is the Rolled Section Portal Frame. The reasons for selecting this preliminary design are stated below. Uniform (same size) structural members ensure ease of designing connections and joining the members. The design is not complex, this reduces the overall cost of the structure. The structure is easy to construct due to minimal use of members. The overall disadvantages of the structure are ‘better’ as compared to those of the other structures put into consideration for adoption. 4.1 General arrangement The structure is 60m long and 20m wide. A duo-pitch portal frame is used, and the roof slope is 10 degrees. The eaves height is 8m and the apex height is 9.76m. The end span of the structure is 15m on both sides of the structure. There are six 7.5m wide bays on both sides of the structure. The 15m bay is fitted with two roller shutter doors that are 4m high and 4m wide. The two doors have a 4m distance between them. The structure does not have gutters. The steel super structure is supported by concrete plinths which extend 300mm above a ground slab. An overhead crawl beam is needed, and it runs along the length of the structure. The roof system which consists of external sheeting and purlins needs to be able to carry imposed loads and wind loads to the structural frame of the building (Mahachi, 2004). The purlins support the external sheeting on the roof. The sagbars are used to provide lateral support to the bottom flange of the purlins, when it goes into compression due to reverse loading caused by wind suction. The sagbars are also used to decrease the in-plane deflections of the roof (Mahachi, 2004). The rafters support the purlins and sagbars and are also used to transfer the loads from the purlins to the columns. For the wall system, girts are used to carry external sheeting. Even on the walls, the sagbars are used to reduce the vertical in-plane deflections due to the self-weight of sheeting and girts (Mahachi, 2004). The columns support the girts and sagbars and are used to transfer the loads from the rafters and girts to the concrete plinths. The bracing system of the structure consists of tie beams, gable columns and diagonal bracing elements. The sides of the structure are braced using the tie beams and the diagonal bracing elements (Mahachi, 2004). This is used to carry the wind loads and horizontal cranes loads that act on the structure. The gable columns on the gable wall which is on the front and rear end of the structure are used to carry the wind load to the foundation (through the concrete plinths). The gable columns are used to brace the front and last portal frame, and the wind load on the gable wall are carried along the purlins towards the bracing systems. The steel superstructure is supported by concrete plinths which transfers the loads from the columns to the foundation. The foundation transfers the load from the concrete plinths to the ground below the structure. 4.1.1 General arrangement considerations The gable columns are placed 5m apart to provide support/ brace the front and rear end of the structure. The image shown below displays the front view of the structure. Figure 5: Front view of the structure The structure is braced at the end bay. The roof is also braced on the 15m bay. The images shown below display the elevation and plan view of the structure. Figure 6 Plan view of the structure Figure 7 Elevation of the structure 4.1.2 Design solution for the 15m bay It is not possible to use an eaves beam that is longer that is longer than 12m. A truss is used to span the 15m bay. Two rafter beams are placed within the 15m bay and they are placed there to provide support for the purlins that support the roof cladding. The roof within this 15m bay, is braced to provide a medium for the transfer of loads. The images shown below detail the side view and the top view of the structure. Figure 8 Plan view of the structure Figure 9 Elevation of the structure 5.Analysis assumptions 5.1 Wind loads These are the assumptions that are made regarding the computation of wind loads. Determining the internal pressure: The structure that needs to be designed does not have a dominant wall, because both sides of the structure have a 15m bay that accommodates two doors that have the same dimensions. Note 2 of clause 8.3.9.6 of SANS 10162-3 is used to determine the internal pressure coefficient. It states that in a situation where it is not justified or possible to estimate the opening ratio, the internal pressure coefficient can be taken as +0.2 (Anon., 2018). Written below are the assumptions that are made as the wind loads are inputted onto Prokon. Wind loading effects the entire shell of the structure and the loads are applied to the purlins, girts, columns and gable columns to model its effect. On the roof wind loading is assumed to act as a pressure onto the sheeting which in turn is distributed to the purlins as a line load and they are transferred to the rafters as distributed load. Therefore, the wind load is modelled as a line load on the purlins and transferred to the rafters as distributed load along the member. On the walls where sheeting occurs, it is assumed that wind loading acts as a pressure onto the sheeting which is then distributed to girt as a line load and then taken up by the columns to the foundation. Therefore, the wind load is modelled as a line load to the girts. The same principle applies to the girts on the gable columns. 5.2 Crane loads These are the assumptions that are made regarding the computation of crane loads. Crawl beam loads: For the overhead crawl beam, the loads are classified into two categories. The self-weight of the crawl beam is considered as a dead load since the beam itself is a permanent fixture. The effect of the loads that act on the crawl beam due to the crab (the self-weight of the crab and the hoist capacity) is considered using the crane load factor. Written below are the assumptions which are made to account for the loads due to the crane for the Prokon analysis. The loading from the crane was applied to the structure in the frame analysis as follows. Reactions from the crane beams at the supports which were applied to the crane supports as loads in the main structure were determined in a separate analysis. This analysis (which is described in detail under the crane beam section) used influence lines to determine what the worst case of the reactions at the crane beam supports would be when the crane moves along the crane runway girder, and since the crane runway girders were simply supported between the crane beams supports, this occurred when the moving crane was directly over the crane supports. The reactions, applied as point loads in the structure at the crane beam supports were excluded the self-weight of the crane runway girders because the runway girders were modelled in the main frame analysis too, where their self-weight was accounted for. 5.3 Imposed Loads Written below are the assumptions that are made to input the imposed loads onto Prokon for the analysis of the structure. Live load (imposed loads) for the portal frame occurs on the roof only. Therefore, the same assumptions that apply to wind loading on the roof, apply to the live load modelling. Live loading is assumed to act as a pressure onto the sheeting which in turn is distributed to the purlins as a line load. Therefore, the live load is modelled as a line load on the purlins. 5.4 Dead Loads Written below are the assumptions that are made to input the dead loads onto Prokon for the analysis of the structure. Dead loads (self-weight) acting on the structure are taken into account by the analysis by adding the self-weight of the elements to the permanent load case. Roof sheeting, insulation, and services not part of the self-weight of the structural elements are assumed to be a pressure on the purlins and girts (no services) which are then distributed to the purlins and girts as live loads. These loads included the sheeting and insulation and services which were prescribed in the design criteria. Written below are the assumptions that are made for the 3-D model of the structure for the Prokon analysis. 5.5 3-D Structural steel framework The 3-dimensional structural frame analysis of the warehouse, including the columns, gable columns, rafters, crane beam supports, bracing beams, tie beams, purlins, girts and sag bars was carried out using structural analysis software, Prokon. For the structural analysis the following aspects of the structural arrangement and the loading onto the structure were considered to best model reality and abide by the relevant codes of practise and are discussed here in this section. Members and connection conditions Support conditions Loading Load combinations Connection design 5.6 Members and connection conditions The connection types chosen between members were chosen based on conditional requirements such as stability, load paths, economy, and deflection limits. The following connection conditions were chosen between connecting members. Column to rafter connection is assumed to be completely rigid, a condition which is necessary to have stability in the plane of the portal frame. Rafter to rafter (apex) connection is assumed to be completely rigid, a connection which is necessary to have stability in the portal frame and have continuity in the rafters. Tie beams to columns will be modelled as pin connections as it is only necessary to transfer lateral (axial) loads through the tie beams for stability in the direction perpendicular to the portal frame. Truss to column connection is modelled as pin connections as it is only necessary to transfer lateral (axial) loads through the truss Bracing members will be modelled as pin connections as it is only necessary to transfer lateral (axial) loads through the bracing members which transfers the loads perpendicular to the portal frames into the foundation supports. Where bracing in a bay cross over each other the one beam is modelled as continuous while the other is modelled as two beams pin jointed to either side of the continuous member of bracing. This ensure the effective length of the bracing is to where the beams cross over in the centre of the bay. Gable column to rafter connection can be simply connected (pinned) or rigidly connected to the rafter. In our model it is assumed to be pinned to the rafter because this is a simpler connection to design and there won’t be moment forces transferred to the rafter form the gable column and it is not necessary to have a rigid connection at this point. Crawl beam support to apex connection is simply supported at this point when designing as to create a moment of zero at this point Purlins are modelled as continuous over the rafters to limit deflection and will be rigidly connected to the rafter. Where spans exceed transport limits for length (12 m). The girts are modelled as continuous over columns and gable columns where possible. In some instances, on the gable columns, girts are simply supported between gables and columns because they are only one span long. 5.7 Support conditions The support conditions for the columns and gable columns were altered during the design of the structure to achieve certain outcomes. Column to foundation connection will be designed and built to resist rotation in the direction of the portal frames only, i.e., the connection will be fixed in the plane of the portal frame. The following instances describe what support conditions were chosen under certain circumstances and the reason they were selected. The supports were modelled as pinned connections when designing the members and checking for lateral stability in the structure. These conditions ensure that there is no resistance to rotation at the base of the building and if the building can resist the lateral loads under these conditions, it is effectively braced. Furthermore, under these conditions members are designed to effectively redistribute the loads if settlement or failure of one or more of the foundations occurs. It must be noted that since the support conditions modelled here do not match reality, the deflections, particularly in the plane of the portal frame, are exaggerated and can be ignored. Deflections were checked when support conditions more accurately matched what the as built conditions would be. The support conditions were modelled as fixed in the direction of the portal frames and pinned in the transverse directions to best model the actual built conditions. Under these conditions, the deflections were checked to be within the limits under serviceability limit state and the reactions (with the correct loading combinations) were used to design the foundations. An additional check was done to ensure that all the member sizes selected were sufficiently strong under the fixed support conditions too. 5.8 End stop The End stop is put in place to prevent the crib from sliding off the crawl beam. The End stop is assumed to be a horizontal roller because the only force it transmits is the buffer force. It is designed as a simple connection. The section that is used for the end stop is the same as the section which is used for the crawl beam. Since this is a large section, the only check that needs to be considered is for “Bolts in shear for supporting member”. There are several assumptions that have been made to determine the loads that act on the structure. All the assumptions that have been made are detailed below. 6.Loads Loading has been applied according to the design criteria: SANS10160-2 (Self-weight and imposed loads), SANS 10160-1 (Basis of structural design), SANS10160-3 (Wind loading) and SANS10160-6 (Cranes and machinery) to all structural members. The loads shown below are used to design the warehouse. Thermal and seismic loads do not need to be considered for the design of the structure. Actions during execution of the project do not need to be accounted for in the design of the structure. 6.1 Dead loads The dead load of the structure consists of the weight of the structure itself and all the materials or finishes on the structure which cannot be removed from the structure (if they are removed the structural integrity of the structure would be undermined) (Parrot, 2005). Table 3 contains the dead loads which are used to analyse the structure. Specific Use Sheeting Crawl Beam Door Dead Load 6.28 𝑘𝑔/𝑚2 0.452 𝑘𝑔/𝑚2 9 𝑘𝑔/𝑚2 The hand-calculations are presented in Appendix 7. 6.2 Wind loads The wind can cause an uplift of the roof of a structure, as well as push on the sides of the structure. Quantifying the wind load is important because in most case, wind loads cause structural failure for roofs (Parrot, 2005). Listed below are the parameters which are considered to compute the wind loads: Fundamental value of basic wind speed (Vb,()): 36 m/s (3s gust) Mean return period = 50 years Upstream terrain category = B (Low vegetation or separation of at least 20 obstacle heights.) Topography factor c0(z) = 1.00 Site height Above Mean Sea Level = 1600m The hand-calculations are presented in Appendix 8. 6.3 Imposed loads Imposed loads are induced when the structure is occupied, these loads are produced by people, equipment, or vehicles (Parrot, 2005). Table 4 contains the imposed loads which act on the structure. Category Specific use Example E2 F Industrial use Parking areas for light vehicles of <=25kN gross vehicle weight Inaccessible roofs Ground slab/ floor Garages, parking areas and parking halls H1 Inaccessible roof during construction Imposed Load (kN/m2) 5.0 2.0 0.25 The hand-calculations are presented in Appendix 9. 6.4 Crane load The loads shown in the table are loads that act on the structure (crawl beam) due to crab. Table 5 contains the forces that act on the crawl beam. Description of the force (load) Self-weight of the crab Hoist capacity Horizontal skewing force Buffer force Force (kN) 3 20 5 5 6.5 Geotechnical considerations The allowable bearing pressure is assumed to be 150 kPa and the founding horizon is 1m below the ground slab. Listed below are ground soil properties: The angle of shearing resistance is 30 degrees (unfactored) The dry density is 18 kN/m3 The saturated density is 20 kN/m3 7.Load combinations The table below shows all the load cases that are used to analyse the structure on Prokon. The load cases are determined for the Ultimate Limit State as well as the Serviceability Limit State. The table below provides an explanation of the symbols used in the table that contains the load combinations used for the analysis of structure. Table 6 contains the legend for the symbols used to determine the load combinations. DEFINITIONS Structural internal failure STR STR-P Dominate Structural internal failure GK Permanent Load QK,1 Imposed Load QK,2 Imposed Load as crane W1 Wind load at 0 degrees W1 Wind load at 90 degrees ULS Ultimate limit state SLS Serviceability limit state The table below contains all the load combinations which can be used to analyse the structure. These load combinations are determined in accordance with the Ultimate Limit State as well as the Serviceability Limit State. Table 7 contains the load combinations considered for the analysis of the structure. Load Combination Leading Factor Permanent Load Gk Imposed Qk,1 SLS LC1 GK 1,1 LC2 GK 1,1 1 LC3 GK 1,1 1 LC4 GK 1,1 1 LC5 GK 1,1 1 LC6 GK 1,1 1 1,0X0,6 LC7 GK 1,1 1 1,0X0,6 LC8 GK 1,1 1 LC9 GK 1,1 1 LC10 GK 1,1 LC11 GK 1,1 LC12 GK 1,1 1,0X0,6 1 LC13 GK 1,1 1,0X0,6 1 LC14 GK 1,1 LC15 GK 1,1 0,6 Impose d Qk,2 Wind W1 1,0X0,6 0,6X0,3 0,6X0,3 1 0,6X0,3 0,6X0,3 0,6X0,3 1 0,6X0,3 0,6X0,3 0,6X0,3 0,6 0,6 Wind W2 0,6 LC16 GK 1,1 1,0X0,6 0,6 LC17 LC18 GK GK 1,1 1,1 1,0X0,6 1,0X0,6 0,6 LC19 GK 1,1 0,6 LC20 GK 1,1 LC21 GK 1,1 1,0X0,6 LC22 LC23 GK GK 1 1 1 LC24 GK 1 1 LC25 GK 1 1 LC26 GK 1 1 LC27 GK 1 1 1X0,6 LC28 GK 1 1 1X0,6 LC29 GK 1 LC30 GK 1 LC31 LC32 GK GK 1 1 LC33 GK 1 LC34 GK 1 LC35 GK 1 LC36 LC37 GK GK 1 1 0,6 LC38 GK 1 0,6 LC39 GK 1 LC40 GK 1 LC41 LC42 GK GK 1 1 STR-P LC25 GK 1,35 LC26 LC27 GK GK 1,35 1,35 LC28 GK 1,35 LC29 GK 1,35 LC30 GK 1,2 LC31 QK 1,2 1,6 LC32 QK 1,2 1,6 LC33 QK 1,2 1,6 LC34 LC35 QK,2 QK,2 1,2 1,2 0,6 0,6 1,0X0,6 1,0X0,6 0,6 0,6 0,6 1,0X0,6 0,6X0,3 0,6X0,3 0,6X0,3 0,6X0,3 1 0,6 1 1 1 0,6X0,3 0,6 1 0,6X0,3 0,6 1 0,6X0,3 0,6 0,6 0,6 0,6 0,6 0,6 0,6 0,6 0,6 0,6 0,6 0,6 0,6 0,6 ULS 1 1 1 1 STR 1,6X0,6 1,6X0,6 1,6X0 1,6 1,6 1,6X0 LC36 QK,2 1,2 1,6 LC37 LC38 W1 W1 1,2 1,2 LC39 W1 1,2 LC40 W1 1,2 LC41 W2 1,2 LC42 LC43 W2 W2 1,2 1,2 1,6X0,6 LC44 W2 1,2 1,6X0,6 1,6X0 1,6 1,6 1,6X0,6 1,6X0,6 106X0 1,6X0,6 1,6 1,6X0,6 1,6 1,6 1,6X0,6 1,6 1,6 1,6X0,6 1,6 The table below contains the load combinations which are used to analyse the structure. These load combinations are determined in accordance with the Ultimate Limit State as well as the Serviceability Limit State. Table 8 contains the load combination cases for ULS and SLS Load combinations DL LL ULS 1.2 1.6 SLS 1.1 1.0 LC2 DL LL CL 1.2 1.6 0.96 1.1 1.0 0.6 LC3 DL CL 1.2 1.6 1 1.1 LC4 DL LL CL 1.2 0.96 1.6 1.1 0.6 1 LC5 DL WL_0 1.2 1.6 1.1 0.6 LC6 DL WL_90 1.2 1.6 1.1 0.6 LC7 DL CL WL_0 1.2 0.96 1.6 1.1 0.6 0.6 Load cases LC1 LC8 DL LL WL_90 1.2 0.96 1.6 1.1 0.6 0.6 LC9 DL WL_0 0.9 1.6 1 1 LC10 DL WL_90 0.9 1.6 1 1 LC11 DL WL_270 1.2 1.6 1.1 0.6 LC12 DL CL WL_270 1.2 0.96 1.6 1.1 0.6 0.6 LC13 DL LL WL_270 1.2 0.96 1.6 1.1 0.6 0.6 8.Analysis results 8.1 Prokon Analysis The Prokon outputs from the models analysed are presented in appendix C. the results were used for the design of the various members, connections and supports. The members were designed to meet ULS and SLS criteria of SANS 10162-1:2011 Ed 2.1. 8.1.1 Member Selection Member selection was carried out to the ability to resist the forces experienced by the members under the ultimate limit state conditions and the ability to limit deflection under serviceability limit state conditions. SANS 10162-1:2011Ed2.1 was the code used to determine that members satisfied the requirements for strength and deflections. All of the members seen in Table 5 below can resist the factored loads applied to them without failing. In some cases lighter sections could have been used, but in order to remain within the deflection limits, the member size was increased. The table below details the final members to be used in the structure. Table 5: Members selected Columns Rafters Purlins Girts Tie Beams Sag Tie Bracing Gable Columns Crane Beam Sag Rods Truss Member 1 Truss Member 2 356x171x67 356x171x67 125x75x20x3 125x75x20x3 139.7x4.0 80x80x6 139.7x5.0 356x171x67 305x165x40 70x70x6 70x70x6 125x75x20x2 I - Section I - Section Cold formed Section Cold formed Section Hollow Section Cold formed Section Hollow Section I - Section I - Section Angle Angel Channel The hand-calculations are presented in Appendix 10. 8.1.2 Displacements Under serviceability conditions it is important that the structural members of the warehouse do not deflect more than what is allowed for their purpose. Excessive deflections can lead to improper functioning of the building and can lead the occupants of the building to feel unsafe in its vicinity. According to SANS10162-1 Table (D.1) structural members should not exceed deflections of more than the span of the member divided by a limit value. The maximum vertical and horizontal displacement were assessed against the code limits. The limits are in accordance to SANS 10162-1:2011 Ed 2.1, Limit states design of hot rolled steelwork. The full analysis output of the displacement can be found in appendix. Table 9 contains the allowable deflections for the structural members Structural Member Deflection type Deflection limit SANS10162-1 Table (D.1) Span - L - (mm) Limit value (mm) Maximum deflection (mm) L/limit (mm) Door Column Lateral 200 4000 20 -1.84 Rafter both side Vertical 240 10154 42 25.66 Tie beam Vertical 300 7500 25 -18.85 Bracing Crane runway beam Column Vertical 300 8500 28 -17.91 Vertical 240 200 7500 8000 31 40 -17.91 8.1.3 Reactions When designing for the foundations, the support conditions of the model on Prokon was changed from pinned to a fixed support. The highest reaction forces and the relevant load combination obtained from Prokon are tabled below. The full reaction force output is found in appendix. The largest ULS load combination is used to design flexural steel while the largest SLS load combination is used to verify bearing pressure and overturning. Table 10 contains the dominant ULS load combination Load Dominant combinations Reaction Ultimate Limit State Horizontal Vertical Moment LC11 LC2 Horizontal -52.26 27.55 Vertical -33.02 148.42 157.3 -1.06 LC11 Moment 157.3 -33.02 -52.26 Table 11 contains the dominant SLS load combination Load Dominant combinations Reaction Serviceability Limit State Horizontal Vertical Moment LC11 LC2 Horizontal -14.11 117.04 Vertical LC11 Moment 77.5 13.01 21.17 31.32 -69.66 34.21 25.38 The following reaction forces were used to design the gable column foundations. Table 12 contains the reaction forces used to design the gable column foundation. Load Dominant combinations Reaction Ultimate Limit State Horizontal Vertical LC11 Horizontal -5.44 3.63 LC1 Vertical 0.47 24.72 Load Dominant combinations Reaction SLS Horizontal Vertical LC11 Horizontal -3.32 9.02 LC2 Vertical 0.31 20.84 8.1.4 Code checking In accordance to the code, the minimum slenderness ratio for tension is 300 and 200 for compression. The ratio indicates the tendency of members to fail in buckling in that direction. The following table lists the worst code ratios for each member section. Table 13: code checking of structural members Element Column Gable column Rafter Tie beam Vertical Bracing Section Size 356x171x67 I 356x171x67 I 356x171x67 I 139.7x4.0 O 139.7*5 O L/r 170 200 57 133 97 Critical axis y-y axis y-y axis Fails in y-y axis x-x axis x-x axis 8.1.5 Connection design The column to rafter connection at the eaves beam level. At this point, the following members connect: the rafter connects rigidly to the column flange, tie beams connection simply to the web of the column, the column (wall) bracing connects to the web of the column and the roof bracing connects to the web of the rafter. Connection design for each of the elements connecting to this point was carried out as follows. Rafter to Column: The design link function in the Prokon software was used to design a rigid connection for the column to rafter connection. By inspection, the connection with the highest moment forces, shear forces and axial forces was used to design the connection. Furthermore, the connection was designed where support conditions of the columns were modelled as pinned and as fixed, because this slightly changed the forces being transferred through the connection. It was decided that a flush end plate with a haunch welded to the rafter in the shop and then bolted onto the column on site would be used as the connection. A haunch was used to limit deflection as much as possible in rafter, even if this means overdesigning the connection. A haunch of the same size as the rafter beam is used and extends 400 mm from the centre of the rafter and 2000 mm away from the column. Under these conditions, the design link programme was then used to optimize the connection for plate thicknesses, welds and the number of M20 bolts to be used. The limit state checks for this design under fixed support conditions for the highest loaded connection can be found in appendix. Bracing beam to rafter connection: The rafter bracing beams do not clash with tie beams thus this the connection is simpler to design. The same principles of the tie beam and bracing beam to the column were followed for this connection. A circular hollow section bracing is capped with a plate and a fin plate which is bolted to a fin plate on the web at the base of the rafter beams. The connection consists of 10 mm plates with M20 bolts and 6 mm fillet welds. The limit state checks of this connection are also found in the appendix. Summary of connection details A summary of the connection details above are as follows: Column to Rafter Connection: Figure 10: summary of the connection details 8.2 Foundation The foundation was designed in accordance with SANS 10160-5: 2021 Edition 1.2, Basis of geotechnical design and actions. It is designed as a rectangular base with a concrete plinth. The reaction forces obtained from the structural analysis were used to design foundation for the portal frame columns and the gable columns. Although the columns are carrying different quantities of load, all portal frame column foundations are the same. This also applies to gables foundations. The table below summaries the foundation designed. Foundation and reinforcement details are found in Appendix 11. Table 14 contains the dimensions of the foundation as well as the loads acting on it. Variable Quantity a 3m b 2.75m P 41.36kN Hx 22.81kN Mx 104.33kNm LC 7 was found to be the worst-case scenario after our analysis 8.3 Ground slab design The suspended floor slab of the office building is designed according to the SABS 01001:2000 Ed2.2 and SANS 10160 codes. The suspended slab was designed for thickness and cracking. The loads on the slab were analysed using Prokon. The self-weight of the concrete is added by the software and the load combinations are used from table. A 50mm Screed layer is also assumed on top of the concrete slab to be able to achieve a smooth and level surface. A finite element shell analysis model was analysed to get the maximum positive moment, maximum negative moment, maximum shear forces in the x and y directions as well as the maximum deflection for the slab. The dead and live load used to calculate the ULS and SLS are tabulated below. The hand-calculations are presented in Appendix 12 Table 15 contains the loads that are used for the ground slab design Type of load Quantity of load DL 4.65kN/m LL 7kN/m The ULS was calculated and found to be 16.78kNm The SLS was calculated and found to be 12.12kNm Conclusion The aim of the project was to design a warehouse in accordance with South African standards. The selection of all structural members was optimized by weight only. The project was successfully completed with use of computer software (AutoCAD and Prokon), industry consultations (Mr Daniel Surat and Mr Thamsanqa Shangase), and relevant documents (indicated under references and acknowledgements). The design of the warehouse facilitated a learning of fundamentals of structural design and significant knowledge on technical insights in designing a portal frame. References Andrews, N., n.d. Home Sweet home. [Online] Available at: https://www.ehow.com/info_10075491_advantages-disadvantages-latticedportal-frames.html [Accessed 6 October 2022]. Anon., 2018. South African National Standard Basis of structural design and actions for buildings and industrial buildings Part 3: Wind loads. 2 ed. Groenkloof: South African Bureau of Standards. Doyle, J. F., 1991. Springer link. [Online] Available at: https://link.springer.com/chapter/10.1007/978-94-011-34200_4#:~:text=A%20truss%20is%20a%20structure,or%20by%20pins%20at%20joints. [Accessed 10 October 2022]. EL-Alghoury, M.A., Ebid, A.M. and Mahddi, 2020. Decision support system to select optimum steel portal frame coverage system.. Ains Shams Engineering journal, Volume 12, pp. 73-82. institute, S., Available at: [Accessed 10 october 2022]. 2022. SteelConstruction.info. [Online] https://www.steelconstruction.info/Portal_frames Mahachi, J., 2004. Design of structural steelwork to SANS 10162. Pretoria: CSIR . Miller, B., 2019. GreenGarage. [Online] Available at: https://greengarageblog.org/15-arch-bridges-advantages-and-disadvantages-tiedthrough-and-truss Mohammed A, E.-. A. A. M. E. I. M. M., 2021. ScienceDirect. [Online] Available at: https://www.sciencedirect.com/science/article/pii/S2090447920301829 [Accessed 10 October 2022]. Parrot, G., 2005. Structural Steel Design to SANS 10162 - 1: 2005. 2nd ed. Durban: SHADES Technical Publications. PR Salter, A. M. C. K., 2004. Design of a single span steel portal frames to BS 5950- 1:2000, Silwood Park: The Steel Construction Institute. Acknowledgements CIVN4015A: Civil Engineering Design Project: Structures, 2022 Elvin A. (2021). CIVN3010: Structural Steel Design lecture notes. Johannesburg: University of the Witwatersrand Rathod G.W. (2022). CIVN4004: Geotechnical Engineering 2 – Foundations. Johannesburg: University of the Witwatersrand SANS Codes Appendix 1 – Meeting Minutes School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group …4... of ……Structural Design…. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… Date: ………03/10/2022….. Time: 12:00 – 13:00…… Computer Lab…………….. Name & Student No. of Chair & signature: Nonhlanhla Mhlongo 1631963 Name & Student No. of Secretary & signature: Makhosonke Mhlanga 1612449 Names & Student Nos. of those Present & signatures: Makhosonke Mhlanga 1612449 Nonhlanhla Mhlongo 1631963 Zamanguni Ntozakhe 908593 Jackson Mahlaule 1669086 Musa Mdhluli 1391559 Names & Student Nos. of those Absent: Mohau Majoro 1633852 Venue: …Hillman S/N Item Matters from Previous meeting 1. Election of group member Items for consideration of current meeting 1. Word document 2. WhatsApp Group Action Responsible person Nonhlanhla Mhlongo Present was automatically members chosen as the group leader. A word document Nonhlanhla that tracks changes Mhlongo and contribution of each group member. A whatsapp group Nonhlanhla was created by Mhlongo Nonhlanhla Mhlongo for easier and faster communication amongst members. 3. Google Drive for collection of materials 4. Discussion of strengths and weaknesses 5. Report writing Nonhlanhla Mhlongo Nonhlanhla created a google Mhlongo drive for group members to upload any material necessary for the project. Present group Present members outlined members their strengths and weaknesses which gave us an idea as to how we will tackle the project. Tasks to draft the Present introduction part of members the project were delegated to group members 6. Preliminary designs A task to come up Present with preliminary members designs was delegated to group members School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group …4.. of ……Structures Design…. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… Date: …07/10/2022……….. Time: …11:00 – 12:00… Teams…………………….. Name & Student No. of Chair & signature: Mohau Majoro 1633852 Name & Student No. of Secretary & signature: Jackson Mahlaule 1669086 Names & Student Nos. of those Present & signatures: Mohau Majoro 1633852 Jackson Mahlaule 1669086 Makhosonke Mhlanga 1612449 Nonhlanhla Mhlongo 1631963 Musa Mdhluli 1391559 Names & Student Nos. of those Absent: Zamanguni Ntozakhe 908593 Venue: ……MS S/N Item Action Responsible person Matters from Previous meeting 1. Report Writing Work on the Present introduction is members underway. -Feedback on the introduction was given by Jackson and Makhosonke - Possibilities of changes as we go along with the project are to be expected, as we get more information 2. Preliminary Design considerations Nonhlanhladid Present research on a latticed members portal frame and does not recommend it since it contains a lot of truss members, as such many connections will need to be considered, increasing the possibility of calculation errors. It is not too safe under fire. Advantages: light weight, and easy erection, not costly to erect. Musaconducted research on the rolled section portal frame. Advantagescontains all sections of the required beams, no tapered beams, uses less members. Analysis on this beam is easier. Disadvantagesdesign with the lowest section possible, but you might find that you need stronger member size and offsite fabrication is not possible. Musa conducted research on the Bus portal frame. Advantagesprefabricated at the factory, take all primary and secondary members to site, it is very cost effective and time effective, can expand a span about 100 meters or more, it beats truss on the erection time and members are labelled. Items for consideration of current meeting 1. Choosing the design Musa’s design was Present chosen by the group. members 2. Uploading on the master document It is important that Present every one of us members uploads on the master document to prevent the loss of sections of work. 3. Report writing Tasks to draft the Present introduction part of members the project were delegated to group members 4. Preliminary designs A task to come up Present group with preliminary members designs was delegated to group members School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group …4.. of …Structures Design ……. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… Date: ……10/10/2022…….. Time: …12:00 – 13:30… ……………Hillman H205…………….. Name & Student No. of Chair & signature: Nonhlanhla Mhlongo 1631963 Name & Student No. of Secretary & signature: Musa Mdhluli 1391559 Names & Student Nos. of those Present & signatures: Makhosonke Mhlanga 1612449 Nonhlanhla Mhlongo 1631963 Jackson Mahlaule 1669086 Musa Mdhuli 1391559 Mohau Majoro 1633852 Names & Student Nos. of those Absent: Zamanguni Ntozakhe 908593 Venue: S/N Item Action Responsible person Matters from Previous meeting 1. Report writing Ongoing process Present Makhosonkemembers finalizing reasons for selecting the rolled section portal frame as the frame that the group would design. 2. Discussing issues about group members Participation in Present group meeting- it members was requested that people should notify the group if they are not able to attend meetings and give reasons 3. Clarity in preliminary designs Items for consideration of current meeting 1. Work distribution The need for Makhosonke preliminary designs for drawings Dead load calculations Wind load calculations Imposed load calculations Finalizing reasons for selecting for rolled section portal frame as the main frame to be designed. Crane load calculations Load combinations Musa Majoro Jackson Makhosonke Nonhlanhla Zamanguni 2. Submissions Submissions of Present calculations by all members group members next week Monday 3. Consultation Clarity needed on the Present design of the 15m members end bay span and the roller shutter doors School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group 4….. of ……Structures Design …. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… Date: ……14/10/2022…….. Time: …11:19 – 11:58… ……………………Hillman Computer Lab…….. Name & Student No. of Chair & signature: Makhosonke Mhlanga 1612449 Name & Student No. of Secretary & signature: Nonhlanhla Mhlongo 1631963 Names & Student Nos. of those Present & signatures: Makhosonke Mhlanga 1612449 Nonhlanhla Mhlongo 1631963 Zamanguni Ntozakhe 908593 Jackson Mahlaule 1669086 Musa Mdhuli 1391559 Mohau Majoro 1633852 Names & Student Nos. of those Absent: None Venue: S/N Item Action Responsible person Matters from Previous meeting 1. Allocation of tasks Makhosnke Makhosonke reminded every one of the tasks which were allocated to them in the previous meeting. Items for consideration of current meeting 1. Load combinations Zamanguni informed Zamanguni the group that she experienced difficulty with using the new edition of the code to determine the load combinations. She informed the group that she would consult with the lecturer. 2. Imposed loads Jackson informed the Jackson group that he had done his calculations but he had not completed the equipment loading calculations. Makhosonke offered to help him with the calculation. Musa also stated that he would help Jackson with determining the nodes which would be used for calculations 3. Paragraph explaining preliminary design was chosen which Makhosonke Makhosonke informed the group that he was done with the paragraph. He said he would add more comments as we continue to make progress project. with the 4. Dead loads Musa informed the group that he had not started with the calculations because he needed the group’s input on which sheeting to select for the project. Musa 5. Wind loads School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group …4... of …Structures Design……. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… H208……………. Date: …17/10/2022………. Time: …12:00 – 15:00… Name & Student No. of Chair & signature: Jackson Mahlaule 1669086 Name & Student No. of Secretary & signature: Mohau Majoro 1633852 Names & Student Nos. of those Present & signatures: Makhosonke Mhlanga 1612449 Mohau Majoro 1633852 Zamanguni Ntozakhe 908593 Jackson Mahlaule 1669086 Musa Mdhluli 1391559 Venue: Hillman Names & Student Nos. of those Absent: Nonhlanhla Mhlongo 1631963 S/N Item Action Responsible person Matters from Previous meeting 1. Loads on the structure Every member Present presented their members allocated work on loads acting on the structure. 2. Software to use Musa informed the Musa group that Prokon will not be used for the analysis of the structure. Items for consideration of current meeting 1. Wind loads Majoro still needs to Majoro do some work on the wind loads and consult with Dr. Bradley and others regarding the internal pressures when there is no dominant wall. 2. Load combinations Zamanguni needs to Zamanguni consult with Dr Bradley regarding loading combinations using the new SANS code and watch YouTube video on load combinations and upload the loads by Wednesday midnight. 3. Design for 15m bay Every member needs Present to do research on members how we are going to design the 15m bay. 4. 2 Portal Frames A group needs to Present produce 2 frames of members which we are to compare and choose one for the design. School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group …4.. of …Structure Design……. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… Date: 28/10/2022………….. Time: …12:00 – 13:00… …………Hillman H205……………….. Name & Student No. of Chair & signature: Makhosonke Mhlanga 1612449 Name & Student No. of Secretary & signature: Jackson Mahlaule 1669086 Names & Student Nos. of those Present & signatures: Makhosonke Mhlanga 1612449 Jackson Mahlaule 1669086 Nonhlanhla Mhlongo 1631963 Musa Mdhluli 1391559 Mohau Majoro 1633852 Zamanguni Ntozakhe 908593 Names & Student Nos. of those Absent: None Venue: S/N Item Action Responsible person Matters from Previous meeting 1. Progress on finalizing calculations Outstanding calculations: Present members Wind load Mohau calculations- clause 8.3.9.6 gives clarity on calculation when there is a dominant wall is absent. We are only waiting for a response on the 90 degrees part. Tables on load Zamanguni combinations are yet to be uploaded but the load combinations are completed. Calculations need to Present be complete so that members the analysis can begin. Items for consideration of current meeting 1. Door frames 2. Crawl beam 2. Roller shutter doors Door frames need to be braced. The group will consider asking Dr. Bradley on the way in which the door bracing could be incorporated into the structure. Present members Present members Crawl beam will not Present be extended, a members member will support it just beneath the apex. Need to find the Present weight members measurements for the type of door we are designing for, so far, the websites visited do not have the measurements for this specific door type. School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group …4.. of …Structures Design……. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… Date: …2/11/2022……….. Time: … 12:00 – 13:00… ………Hillman H205………………….. Name & Student No. of Chair & signature: Nonhlanhla Mhlongo 1631963 Name & Student No. of Secretary & signature: Musa Mdhluli 1391559 Names & Student Nos. of those Present & signatures: Nonhlanhla Mhlongo 1631963 Musa Mdhluli 1391559 Makhosonke Mhlanga 1612449 Zamanguni Ntozakhe 908593 Jackson Mahlaule 1669086 Mohau Majoro 1633852 Names & Student Nos. of those Absent: None Venue: S/N Item Action Responsible person Matters from Previous meeting 1. Wind load calculations Majoro explained the Majoro challenges he still has with wind calculations. 2. Meeting attendance The group discussed Present the issue of time members keeping and attendance of meetings Items for consideration of current meeting 1. Communication issues We had Present communication members issues amongst group members by not letting us know that you will not be available for the meeting and telling us late that you had an emergency. 2. Fixing wind calculations Everyone must give Present input on what they members thinks should be done to get better results. 3. Time Discussed that we are Present running out of time to members finish the design project. 4. Warnings Issued out warnings Present that not being part of members the meeting will decrease your overall contribution towards the project. Simply reading the meeting minutes does not excuse your absence. 5. Writing minutes Everyone was told to Present write the minutes and members upload them no later than a day after the meeting. School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group ….. of ………. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… Date: 07/11/2022………….. Time: 12:00 – 12:30…… Computer Lab………………………….. Name & Student No. of Chair & signature: Nonhlanhla Mhlongo 1631963 Name & Student No. of Secretary & signature: Zamanguni Ntozakhe 908593 Names & Student Nos. of those Present & signatures: Makhosonke Mhlanga 1612449 Nonhlanhla Mhlongo 1631963 Zamanguni Ntozakhe 908593 Jackson Mahlaule 1669086 Musa Mdhluli 1391559 Names & Student Nos. of those Absent: Mohau Majoro 1633852 Venue: Hillman S/N Item Matters from Previous meeting 1. Update on tasks allocated in the previous meeting. Action Responsible person Musa had a problem Musa with the wind calculations not being clear and was going to seek clarity from Majoro. 2. Prokon results The deflections from Musa & the prokon results Zamanguni were too high and not within the code limits. Musa told the group that he would consult with Dr. Bradley. Items for consideration of current meeting 1. Meeting the professional in industry 2. Tabulate Loads A profession Civil Nonhlanhla Engineer is willing to meet the group on Ms Teams to discuss the progress of the group. Time TBA on WhatsApp. Makhosonke will Makhosonke, tabulate the dead Nonhlanhla & loads, Nonhlanhla Jackson. will tabulate the crane loads and Jackson will tabulate the imposed loads. School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group …4.. of …Structures Design……. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… Date: …10/11/2022……….. Time: 12:00 – 13:00…… group chat………………………….. Name & Student No. of Chair & signature: Mohau Majoro 1633852 Name & Student No. of Secretary & signature: Jackson Mahlaule 1669086 Names & Student Nos. of those Present & signatures: Makhosonke Mhlanga 1612449 Nonhlanhla Mhlongo 1631963 Jackson Mahlaule 1669086 Musa Mdhluli 1391559 Mohau Majoro 1633852 Zamanguni Ntozakhe 908593 Names & Student Nos. of those Absent: None Venue: Whatsapp S/N Item Action Responsible person Matters from Previous meeting 1. Prokon analysis The analysis will be Musa done soon then everyone will be able to start with the design calculations. Items for consideration of current meeting 1. Member design calculations Tie beams – Nonhlanhla & Makhosonke Ground slab – Makhosonke Concrete foundation – Makhosonke Door frameNonhlanhla Crawl beamNonhlanhla Gable columns – Musa ConnectionsZamanguni Purlins and girts Zamanguni Makhosonke, Nonhlanhla, Musa & Zamanguni School of Civil and Environmental Engineering University of the Witwatersrand Minutes of meeting of Group …4.. of …Structures Design……. Project (These minutes are taken as part of fulfilment of GA 8 – Individual, team and multidisciplinary working: Demonstrate competence to work effectively as an individual, in teams and in multidisciplinary environments.) Important Instructions with compiling the minutes 1. There should be minutes for each of week of the Design Project 2. The Chair and Secretary should be rotated every week among members of the group 3. The minutes must be signed by all present at the meeting 4. The minutes should be incorporated in the Appendix of the Design Report Week: ……… Date: …14/11/2022……….. Time: …12:18 – 13:11… Computer Lab………………………….. Venue: Hillman Name & Student No. of Chair & signature: Musa Mdhluli 1391559 Name & Student No. of Secretary & signature: Nonhlanhla Mhlongo 1631963 Names & Student Nos. of those Present & signatures: Makhosonke Mhlanga 1612449 Nonhlanhla Mhlongo 1631963 Zamanguni Ntozakhe 908593 Jackson Mahlaule 1669086 Musa Mdhluli 1391559 Mohau Majoro 1633852 Names & Student Nos. of those Absent: None S/N Item Action Responsible person Matters from Previous meeting 1. Prokon analysis Musa was reminded Present that he said he would members present the prokon results to the group today. Items for consideration of current meeting 1. Prokon analysis 2. Purlins Upon running the Musa Mdhluli analysis of the structure on Prokon. Musa discovered that the deflections were not within the deflection limits stated in the code. It was suggested that we have a meeting with the industry advisor so solve the issue. Zama discussed her Zamanguni progress on the purlin design calculations and informed the group that she was unable to finish the calculations because there were some uncertainties. However, she received assistance from the group. Appendix 2 – Cross-disciplinary and stakeholder engagement GA 8 form Appendix 3 – Preliminary Design 1 (Latticed Portal Frame) Appendix 4 – Preliminary Design 2 (Trussed Portal Frame) Appendix 5 – Preliminary Design 3 (Rolled Section Portal Frame) Appendix 6 – Preliminary Design 4 (Pre-engineered built-up Portal Frame) Appendix 7 – Dead Loads Appendix 8 – Wind Loads Appendix 9 - Imposed Loads Appendix 10 – Member designs Appendix 11 – Foundation Design Appendix 12 – Ground slab design University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design As partial fulfilment of GA 8 (Individual, team and multidisciplinary working), cross-disciplinary interactions and stakeholder engagement of students are assessed in the form of a log-report which must be completed to the satisfaction of the Supervisor. Date Name 07/11/202 Thami Langa 2 Profession or Affiliation Structural Issues Addressed Engineer uncertain about where the peak Wind calculations: The group was pressure must be calculated ( the eaves height or the apex height). Mr Langa advised that the apex height be used since more wind pressure is experienced as you go higher (wind curve). After investigating the wind calculation, it was discovered that the way in which the coefficients were calculated was erroneous (the values which we got were too low). Mr Langa provided pointers for rectifying the errors. Written below are the pointers which we were given: ● Mr Langa advised that the group use clause 8.3.9.6 Note 2. ● Mr Langa informed the group that when all doors Signature University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design are open, that is considered as an accidental load condition (clause 8.3.9.3). Therefore the case that should be considered is one where all doors are closed. ● Total Cp is determined by subtracting the Cpi from the Cpe. The total Cp is multiplied by the peak pressure to obtain the net wind pressure on the face of the wall or the roof. ● The wind load udl is the resultant wind pressure. ● According to the code the, edges are where the higher pressures will occur so it must be used when designing the purlins. ● When the wind hits the walls at 0 degrees and 180 degrees, the sway movement will mirror each other (for the walls). University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design ● When the wind hits the walls at 90 and 270 degrees, the pressure decreases as you move away from the face where the wind acts. Prokon analysis: The group showed Mr Langa the 3-D model which was used on Prokon. Mr Langa then suggested that the group first analyze the structure in 2-D to obtain the correct section sizes and strength. The loads should first be applied on a single frame to determine the lateral deflections as well as the sway on the structure. Once the sizing is correct, a 3-D model is used to determine the stability of the structure. After investigating the Prokon analysis, it was discovered that the approach used by the group was erroneous. Mr Langa provided pointers for rectifying the errors. Listed below are the pointers which the group was given: University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design ● The columns were supposed to be pinned at the bottom in order to determine the highest moment that would occur at the connection between the beam (rafter beam) and the column. This is the moment that is used for the design calculations. ● The columns were supposed to be fixed in order to determine the moment at the base of the column, this is what will be used for the foundation calculations. ● The members used for the columns and rafters are too light, therefore stiffer members. ● The deflections were too high because there were no haunches at the apex. ● It was recommended that the group should use SANS University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design 10162- Part 2 to determine the deflection limits. ● It was recommended that a I- beam or Circular hollow section be used for the Eaves Beam. Bracing: The group decided on bracing three bays. Mr Langa then told the group to conduct research on the limits of the number of bays that should be braced on the structure. Load combinations: After investigating the load combinations, it was discovered that they were not well-written.The main issue was with the load factors (they need to be re-calculated). Mr Langa advised that they be revised because they were the reason why the Prokon analysis yielded incorrect results. For all the load cases with wind load, the wind loads need to be considered in all directions. When the wind hits the structure at 90 & University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design 270 degrees, it will not behave in the same manner because the bay spaces are not the same. When the load factors for the ULS and SLS need to go hand in hand when it is inputted into Prokon. 2022/11/22 09/11/202 2 Thami Langa Structural Wind calculations: The group Engineer showed Mr Langa the adjustments made to the wind calculations following the meeting we had with him (initially, the Cpi and Cpe calculations were wrong) and the adjustments were done to his satisfaction. Mr Langa recommended that an accurate sketch that shows how the wind pressure decrease further away from the face at which the wind acts) Load combinations (and factors): The group showed Mr Langa the adjustments made to the load combinations (the ULS and SLS load combinations were not corresponding with each other) but University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design he noted that there were still errors which needed to be corrected, (Load case 1,2,3 and 6 were not done correctly) . Bracing: The group discussed how they intended to brace the structure (the blue book suggests that both end bays should be braced if the structure is more than 50m long) and Mr Langa was satisfied with the group’s decision. There are two bays which will be braced, the second and the seventh bay. Prokon analysis: The group showed Mr Langa the results obtained on Prokon (which were based on the incorrect load combination) so it was noted that the results would need to be revised. 15/11/202 2 Thami Langa Structural Load combinations: Mr Langa Engineer raised the concern that of the load combinations presented by the group, there’s no load case that caters to the wind uplift. The other 2022/11/22 University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design problem is that for all load cases where there are wind loads, they need to be considered for the angles at which the wind acts on the structure (0 , 90, and 270). Prokon analysis: Due to the fact that the wind loads and the load combinations are still incorrect, the results obtained from Prokon were incorrect. Wind load: The height used to calculate the pressure was incorrect and it needed to be changed. 17/11/202 2 Thami Langa Structural Ground slab design: The group was Engineer under the impression that the ground slab would be designed as a two way slab, however Mr Langa informed the group that two-way or one-way slabs only apply to suspended slabs. He further explained that the ground slab can be considered as a flat slab. He recommended that the group should use a book by Brian Perry 2022/11/22 University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design called ‘Concrete industrial floors on the ground’ to get more insight on how to design the ground slab. He also informed the group that the slab dimensions are actually the width and length of the whole structure (20m x 60m), but there are limits regarding how long and wide the slab will be because shrinkage affects the slab. The slab needs to be broken down so as to avoid cracks within the slab by splitting it into panels, with saw cut joints in between the panels which are used to control where cracking occurs due to shrinkage and isolation joints on the edges to allow for movement in the three directions (contraction and expansion). The joints are there to enable the slab to release internal stresses to prevent the occurrence of cracks. He recommended that the panels (ratio of 1:1 - width to length) should be square in shape. University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design The group was also under the impression that the thickness of the slab would be assumed in order to carry out the design calculations, but Mr Langa corrected the group by stating that the thickness is determined by carrying out the design calculations. Foundation design: Mr Langa informed the group that the founding level is determined by taking the top of the slab as the reference height, then subtracting the slab thickness and 800. Concrete plinths: The group lacked knowledge on what concrete plinths are and so Mr Langa informed the group that they are used to transfer the load from the structure to the foundation. The plinth goes from the foundation and it ends 300mm above the top of the slab. Connection design: The group informed Mr Langa that the only software they are familiar with for University of the Witwatersrand School of Civil & Environmental Engineering CIVN4015A – Civil Engineering Design designing connections is the eToolkit which is no longer available to students at the computer lab. Mr Langa then showed the group how to use Prokon to design the connection. 2022/11/22 I confirm that the CIVN4015A Group 4 had consultations and received guidance from me with regards to the above mentioned for their group project. T. Langa - BScEng (Civil) Structural Engineer, WSP Africa Email: shangase.thami@gmail.com Cell: 071 214 5686 2022/11/22 Wind loads - CIUNQOIS Reference : SANS 10160-3 roughness and Obstructions Terrain * - (tow category Bheights) vegetation obstacle Terrain 20 Height to the top of - Structure = = Cr Gm) i. Cr (9,76 1.36 fÉ¥É 1.36 (§÷→)"° -0,96£ m ) = 1.36 (390%-50)%0%0,981 = : Wind * The i. C building will prob = have so - ≥ 36 mfs year mean return period 1.0 where : Nb, peak = 1.0 ✗ 36 Up ( 8m ) Vigo a Peak Wind Speed : Vp (e) i. 1.0 Speed Fundamental wind speed : → = " _- Topography Factor Co G) → " Cm Topography * → 9,761s roughness factor : Craig Terrain - of at least 0,09s 2 0 300 B separations to tan lot 81-0,00156 Category Zgccn) Z.cm) Zccm ) a - or = Up (9.76m) = = 0,964×1 ✗ 36 = Creed ✗ Co G) ✗ Vb peak , 36m Is = 0,982 ✗ 1- ✗ 36 34,700mi = 3F35mls- ✓ Wind Pressure * Altitude ÷ - : ,_ = ) (1600-1500) [ (%::÷÷)4a• isao " ◦◦ density Gloom =P Air → 1600M i. + + 2000 - P's " - 1500 - _Ém: Peak Wind Pressure : i. qp( 9,76m ) Cfp (som) * External = { P✗VbKJ2 £6,985)( 3%3532--0,6151511 = , = ¥6198s ) (34/704) Pressure Roof Slope For wind qpcas = 0 at =Q593KPa_ Coefficients 10° ( normal long sided - d- 20,2m µ ≥ * * e=min{b ; 2h } i. - F - = Min = min Wind direction > { D G J I E. % e ee = {60,2 { 60,2 ; 219,76)} ; 19,52 } 19,52m b i. @ = 19.52m Cd C.Fig 8 b) F • B A * els * keys C * d- I - 20 in Sizes Zone 45 = . 3,9m ; % = 1. 95m pressures : G- - Zone A B C D E f cpe.io = ; 445=15 62m ; % 0,48m Correlation Factor -1,2 - - 0,8 0,5 0,73 0,85 0,36 0,85 - - 1,3/1-0,1 § -1,0/+0,1 It -0,45/+91 I -0,5/-0,3 I -0,8/-0,3 = 4,88m For wind 90 at ( Normal to gable End ) ' E e=min{b ; 2h } * =min{20,2 ; 2×9,76} C e=↑9,52m_ C I I s É It It g F g A F & D K b = 20,2m A → Side walls b- 20,2m Cfig 8 b) B B A G- 19,52ms & * wind direction Roof Pressures : % = 0,162 I 0,25 Zone A Cpe 10 , ' Correlation Factor -1,2 B - c - D 0,5 0,71 0 0,31 0,85 E - t - g 0,8 l 85 - 45 1,3 - It , , -0,6s I - g- 0,55 _ * Internal Pressure Coefficients Check for dominance with - Area of roller shutter roller Shutter doors : doors on Windward wall = 2 (4×4) =32_m} Area of roller shutter doors on Leeward wall = = • • . There's no dominant wall . 2 14×4 ) 32m£ (Area of openings are equal] iii - µ G- = = ( Normal Long Sides ?EÉ = as 0,48 By interpolation : i. - 0° Wind at Wind at 90 Cpi }÷%= as G- = 0,162 = 0 , Is (From the graph ) - = Cpi 0,14 ( Normal to gable Ends µ i. = C 0,2 (From the graph] ( Normal to long Side) Wind at Zone Gp ftp.ascpe.eocp.icpr-cpe-cpiwk-qpxcpr A - 0° 0,62 -1,2 0,2 -1,4 -0,87 B 0,62 -0,8 0,2 -1.0 -0,62 C 0,62 -0,5 0,2 -0,7 -0,43 ☐ 0,62 0,73 012 0,53 0,33 E 0,62 -0,36 0,2 -0,5-6 -0,35 F 0,62 -1,3 0,2 -1,5 -0,93 9 0,62 -1,0 0,2 -1,2 -0,74 It 0,62 -0,450,2 -0,65 -0,40 I 0,62 -0,5 0,2 -0,7 -0,43 J 0,62 -98 0,2 -1.0 -0,62 Wind at - 90° (Normal to gable """ "" " "" A -1,2 C • E , 9 It I - End ) " " """ "" "" " / / [Y f | 0,62 B - ' " " 0,2 - 1,4 -0,87 0,62 -0,8 0,2 -1.0 -0,62 0,62 -0,5 0,2 -0,7 -0,43 "" "" " "" "" 0,62 -93 -0,2 -0,5 . %, % -0,31 ,• + ,, . . -0,93 962 -143 0,2 -1,5 0,62 -0,65 0,2 -0,85 0,62 -0,5-5 0,2 -0,75 - -0,53 - - -0,47 * Area leads to line wind at (Normal to gable 90° loads - End ) 0,31 KPa 94 I ? 0,43 KPa kPa 7 0,62 0,62 KPa kPa [ 0,87 Kpq 0,87 KPa 0,32 Kpq ^ wind Note : Red = Blue = Pressure Away ( Suction ) Pressure towards For @ 7,5m @ End @ distribute design I 15m - bag ! -0,62×7 s bag : -0,62×75/2 = 0162 ✗ 2,75 - - 0,62 ✗ 4 = = 0,62×7,5 Roof -2.33 KN fun = -1,71 AN / in - 2,48 / § 0,47 kPa "" kPa 5- B- 0,5-35 µ, µ , i÷a A Note : Red _- Pressure Away Wind KN fun -4,65 KN fun - , 0,62hPa -4,65 KN / in . 0 = 0,62 ✗ 11,25=-6,98 AN / in - - For = , bag ! load area . . For design distribute area load = -0 s , - s - kPa @ 7 , Sm i -95-5×715=-4113 kN / in @ End bag : -0,5-5×742=-2,06 KNIM @ 15m bag : 0,5s - - Wind 0,55×6,25 - - 0,58 - ✗ 5- - = = ✗ 2,5 -3,44 KNIM -2,75 KNIM = - ( Normal to long side) at 0,87 KPa -0,62 KPa 0143 " " ÷: 0133 -8° KPa ÷ § it ¥ I i. n et Ñ - É I r 0,87 kPa 0,62 Rpg 0143 kPa 0,3s kPa 1,38 KN 1m A B C ohsuEI.IT? -6,5319N / on 4,65kW / in - -3,2619N / -5,44kW/ m -4,35kW / m -2,33km/ -3,88 Kmfm -3,1 KN / m -3,23 ANIM -1,61 KIM " m - in 2,691%-2,15 KTN -0,88 kN_m -2,623141m • →• 9 -5,55 KIM -2,781¥ -4,63 "Tn -3,7 It -3 KIM -1,5k£ -2,5k£ -2kW / in I -3,23 "Fn Ktn -2,69 Kwan -2,15µm - KIM 1,3111µm -2,19 Afn -1,75 $1m E J -1,08 0,83 KIM 2,48 Fn - -1,5-5 KNIM 1,6s KIM 24ᵗʰ 2,06141m D 1. -2,18 AN /in . -161 .÷→.÷ 4,65k£ -2,3319m -3,88kt .÷. -3, / ᵗᵈm KIM ⇐ -1.85 - ᵗᵈm I kN /us -1,08 RN / us - I ,SSKNlm Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Member Design for Combined Stresses MOMENTS: X-X M max = 154.6kNm @ 8.00m 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 -50.0 50.0 100 Lx Eff = 6.800 m W1x = 1.00 Ly Eff = 6.800 m W1y = 1.00 Le Eff = 8.000 m W2 = 2.37 Fy = 350 MPa Fu = 480 MPa Tension area factor (Ane/Ag) = 1.00 Flange class: 1 Web class : 1 Critical Load Case : LC2 150 MOMENTS: Y-Y M max = -4.889kNm @ 6.99m 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 -4.00 -2.00 Ver W5.0.04 - 13 Sep 2022 Combine Ver W5.0.04 Element 42-38 Evaluate current section Section name COLUMN Section 356x171x67 I-sections (Web vert) SANS 10162-1:2011 13.8.2 : a) Cross-sectional strength (Crit. pos.= 8.000 m) Cu 0.85Mux 0.60Muy 77.6 133 .092 -- + ------- + ------- = ---- + ---- + ---- = 0.38 Cr Mrx Mry 2693 381 76.5 OK 2.00 4.00 AXIAL FORCE P max = 125.4kN @ 0.00m 120 c) Lateral torsional buckling strength Cu 0.85U1xMux ßU1yMuy 125 131 3.53 -- + ---------- + ---------- = ---- + ---- + ---- = 0.75 Cr Mrx Mry 483 295 76.5 100 80.0 60.0 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 40.0 20.0 b) Overall member strength Cu 0.85U1xMux ßU1yMuy 125 131 3.53 -- + ---------- + ---------- = ---- + ---- + ---- = 0.74 Cr Mrx Mry 359 381 76.5 d) Additional check for class 1 I sections Mux Muy 155 4.15 --- + --- = ---- + ---- = 0.58 Mrx Mry 295 76.5 OK OK OK 13.4:Shear Vux<Vrx Vuy<Vry 34.5 < 687.5 4.8 < 471.1 Slenderness Ratio: Lx/rx = 45 Ly/ry = 170 ------------------------------------------------------------ OK OK OK OK Job Number Sheet Job Title Client Your details here Calcs by 7.00 6.00 5.00 4.00 M max = .8284kNm @ 3.75m 3.00 2.00 1.00 MOMENTS: X-X Checked by Combine Ver W5.0.04 Element 47-68 Section name EAVES Date Evaluate current section .600 Lx Eff = 6.375 m W1x = 1.00 Ly Eff = 6.375 m W1y = 1.00 Le Eff = 7.500 m W2 = 1.00 Fy = 350 MPa Fu = 480 MPa Tension area factor (Ane/Ag) = 1.00 Section class: 1 .800 Critical Load Case : LC11 .200 .400 .00500 MOMENTS: Y-Y M max = 0.000kNm @ 0.00m Round hollow sections SANS 10162-1:2011 13.8.3 : a) Cross-sectional strength (Crit. pos.= 3.750 m) Cu Mux Muy 12.6 .888 0.00 -- + --- + --- = ---- + ---- + ---- = 0.06 Cr Mrx Mry 536 23.0 23.0 .00400 .00300 .00200 AXIAL FORCE 7.00 6.00 5.00 4.00 3.00 2.00 1.00 .00100 P max = 12.64kN @ 0.00m 12.0 b) Overall member strength Cu U1xMux U1yMuy 12.6 .828 0.00 -- + ------ + ------ = ---- + ---- + ---- = 0.15 Cr Mrx Mry 111 23.0 23.0 c) Lateral torsional buckling strength Cu U1xMux U1yMuy 12.6 .828 0.00 -- + ------ + ------ = ---- + ---- + ---- = 0.12 Cr Mrx Mry 147 23.0 23.0 10.0 8.00 6.00 OK OK OK 13.4:Shear 7.00 6.00 5.00 4.00 3.00 2.00 1.00 4.00 2.00 Section 139.7x4.0+ Vux<Vrx Vuy<Vry 0.4 < 196.4 0.0 < 196.4 OK OK Slenderness Ratio: Lx/rx = 133 Ly/ry = 133 OK OK ------------------------------------------------------------ 7.00 6.00 5.00 4.00 M max = .8284kNm @ 3.75m 3.00 2.00 1.00 MOMENTS: X-X Combine Ver W5.0.04 Element 26-68 Section name EAVES Evaluate current section .600 Lx Eff = 6.375 m W1x = 1.00 Ly Eff = 6.375 m W1y = 1.00 Le Eff = 7.500 m W2 = 1.00 Fy = 350 MPa Fu = 480 MPa Tension area factor (Ane/Ag) = 1.00 Section class: 1 .800 Critical Load Case : LC12 .200 .400 .00500 MOMENTS: Y-Y M max = 0.000kNm @ 0.00m Round hollow sections SANS 10162-1:2011 13.8.3 : a) Cross-sectional strength (Crit. pos.= 3.750 m) Cu Mux Muy 31.2 .993 0.00 -- + --- + --- = ---- + ---- + ---- = 0.10 Cr Mrx Mry 536 23.0 23.0 .00400 .00300 .00200 AXIAL FORCE 7.00 6.00 5.00 4.00 3.00 2.00 1.00 .00100 P max = 31.18kN @ 0.00m 30.0 b) Overall member strength Cu U1xMux U1yMuy 31.2 .828 0.00 -- + ------ + ------ = ---- + ---- + ---- = 0.32 Cr Mrx Mry 111 23.0 23.0 c) Lateral torsional buckling strength Cu U1xMux U1yMuy 31.2 .828 0.00 -- + ------ + ------ = ---- + ---- + ---- = 0.25 Cr Mrx Mry 147 23.0 23.0 25.0 20.0 15.0 OK OK OK 13.4:Shear 7.00 6.00 5.00 4.00 3.00 2.00 1.00 10.0 5.00 Section 139.7x4.0+ Vux<Vrx Vuy<Vry 0.4 < 196.4 0.0 < 196.4 OK OK Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Slenderness Ratio: Lx/rx = 133 Ly/ry = 133 ------------------------------------------------------------ OK OK Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Design Code: SANS 10162-1:2011 Element : 11-321 Section: 139.7x5.0+ O Lx eff = 4587 Ly eff = 4587 Lz eff = 5397 Le eff = 5397 Section Slenderness: DT_ratio = = Table 4 D. fy T 139.00 ×350 5.00 = 9 730.000 DT_ratio < 13000 => Class 1 section Maximum slenderness ratio X_ratio = = Kx. Lx rx 0.85 ×5 397.00 47 = 97.605 X_ratio < 200 => OK Y_ratio = = Ky. Ly ry 0.85 ×5 397.00 47 = 97.605 Y_ratio < 200 => OK 10.4.2.1 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Axial compression and bending 13.8 Member strength and stability — All classes of sections except class 1 and class 2 I-shaped sections 13.8.3 Cross-sectional strength 13.8.3 a Cr = = f . A . fy 1000 0.9 ×2 100.00 ×350.00 1000 = 661.500 kN Mrx = = f . Zpx. f y 1×106 0.9 ×89 800.00 ×350.00 1×106 = 28.287 kNm Mry = = f . Zpy. f y 1×106 0.9 ×89 800.00 ×350.00 1×106 = 28.287 kNm End moment factors: k x = -1 (no X moments) k y = -1 (no Y moments) w x = 0.6 - 0.4 . k x = 0.6 - 0.4 ×0.00 = 0.6000 w y = 0.6 - 0.4 . k y = 0.6 - 0.4 ×-1.00 = 1.0000 13.8.5 Job Number Sheet Job Title Your details here Client Calcs by Cex = = Checked by Date p2 . E . Ix 1000 . (Kx. Lx)2 p2 ×200 000.00 ×4 730 000.00 1000 ×(0.85 ×5 397.00 )2 = 443.658 kN Cey = = p2 . E . Iy 1000 . (Ky. Ly)2 p2 ×200 000.00 ×4 730 000.00 1000 ×(0.85 ×5 397.00 )2 = 443.658 kN U1x = wx 1- = Cu Cex 1.000 31.43 1443.74 = 1.076 U1y = wy 1- = Cu Cey 1.000 31.43 1443.74 = 1.076 F13.8.3(a) = = Cu U1x. Mux U1y. Muy + + Cr Mrx Mry 31.4 1.076 ×0.53 1.076 ×0.00 + + 661.5 28.29 28.29 = 0.0676 OK Overall member strength Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports and does not refer to restraint conditions) 13.8.3 b Job Number Sheet Job Title Client Your details here Calcs by f ex = = p2 . E Kx. Lx 2 rx p2 ×200000 1 ×5397 2 47.4 = 152.259 MPa f ey = = p2 . E Ky. Ly 2 ry p2 ×200000 1 ×5397 2 47.4 = 152.259 MPa p2 . E . Cw + G. J (Kz . Lz )2 f ez = A . ro2 p2 ×200000 ×0 = (1 ×5397 )2 + 77000 ×9470×103 2100 ×67.034 2 = 77.27×103 MPa fe = min(fex, fey, fez) = 152.3 MPa l= = fy fe 350 152.26 = 1.516 Checked by Date Job Number Sheet Job Title Your details here Client Calcs by Checked by Date 1 Cr = f . A . f y. 1 + l 2. n n 1×103 1 = 0.9 ×2 100.0 ×350 × 1 + 1.5162 2 ×1.34 1×103 -1.34 = 232.881 kN U1x = 1.000 U1y = 1.000 Cu U1x. Mux U1y. Muy F13.8.3(b) = + + Cr Mrx Mry = 31.4 1.000 ×0.53 1.000 ×0.00 + + 232.9 28.29 28.29 = 0.1536 Lateral torsional buckling strength Mp = = f y. Zplx 1×106 350.000 ×89 800.000 1×106 = 31.430 kNm My = = f y. Zex 1×106 350.000 ×68 100.000 1×106 = 23.835 kNm Mr = f . Mp = 0.9 ×31.430 = 28.287 kNm Cr based on weak-axis bending ly= = Ky. Ly. fy 2. ry p E 0.85 ×5 397.00 350 × 2 47.40 p ×200000 = 1.289 OK 13.8.3 c Job Number Sheet Job Title Your details here Client Calcs by Checked by Date 1 Cr = f . A . f y. 1 + l 2. n n 1×103 1 = 0.9 ×2 100.0 ×350 × 1 + 1.289 2 ×1.34 1×103 -1.34 = 293.242 kN F13.8.3(c) = = Cu U1x. Mux U1y. Muy + + Cr Mrx Mry 31.4 1.000 ×0.53 1.000 ×0.00 + + 293.4 28.29 28.29 = 0.1258 OK Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Design Code: SANS 10162-1:2011 Element : 42-38 Section: 356x171x67 I Lx eff = 6800 Ly eff = 6800 Lz eff = 8000 Le eff = 8000 Section Slenderness: Table 4 Flanges Bratio = = B. fy Tf 86.60 × 350 15.70 = 103.193 Bratio < 145 => Class 1 Flange Web Wratio_1 = = 1100 . fy 1100 350 1 - ×1 - 0.39 . Cu f . Cy 0.39 ×131.8 0.9 ×2 992.5 = 57.675 Hw/Tw(34.3) < Wratio_1 => Class 1 Web => Section class = 1 Maximum slenderness ratio X_ratio = = Kx. Lx rx 0.85 ×8 000.00 151 = 45.033 X_ratio < 200 => OK Y_ratio = = Ky. Ly ry 0.85 ×8 000.00 40 = 170.000 Y_ratio < 200 => OK 10.4.2.1 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Axial compression and bending 13.8 Member strength and stability - class 1 and 2 I-shaped sections 13.8.2 Cross-sectional strength 13.8.2 a Cr = = f . A . fy 1000 0.9 ×8 550.00 ×350.00 1000 = 2 693.250 kN Mrx = = f . Zpx. f y 1×106 0.9 ×1 211 000.00 ×350.00 1×106 = 381.465 kNm Mry = = f . Zpy. f y 1×106 0.9 ×243 000.00 ×350.00 1×106 = 76.545 kNm End moment factors: Mux_min kx = Mux_max = 91.21 176.91 = 0.5156 ky = Muy_min Muy_max = 0.00 -1.31 = 0.0000×100 w x = 0.6 - 0.4 . k x = 0.6 - 0.4 ×0.52 = 0.3920 w x = max(w x,0.4) = 0.4 w x = 1 (loads between supports) 13.8.5 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date w y = 0.6 - 0.4 . k y = 0.6 - 0.4 ×0.00 = 0.6000 w y = 1 (loads between supports) Cex = = p2 . E . Ix 1000 . (Kx. Lx)2 p2 ×200 000.00 ×195 000 000.00 1000 ×(0.85 ×8 000.00 )2 = 8 324.277 kN Cey = = p2 . E . Iy 1000 . (Ky. Ly)2 p2 ×200 000.00 ×13 600 000.00 1000 ×(0.85 ×8 000.00 )2 = 580.565 kN U1x = wx 1 - = Cu Cex 1.000 131.77 1 8 324.28 = 1.016 U1y = wy 1- = Cu Cey 1.000 131.77 1580.56 = 1.294 F13.8.2(a) = = Cu 0.85 . U1x. Mux b . U1y. Muy + + Cr Mrx Mry 80.7 0.85 ×1.016 ×176.91 0.6 ×1.294 ×0.00 + + 2 693.2 381.46 76.54 = 0.4305 OK Overall member strength 13.8.2 b Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports and does not refer to restraint conditions) f ex = = p2 . E Kx. Lx 2 rx p2 ×200000 1 ×8000 2 151 = 703.240 MPa f ey = = p2 . E Ky. Ly 2 ry p2 ×200000 1 ×8000 2 39.9 = 49.102 MPa p2 . E . Cw + G. J (Kz . Lz )2 f ez = A . ro2 p2 ×200000 ×4120×108 (1 ×8000 )2 = + 77000 ×560000 8550 ×156.18 2 = 267.688 MPa fe = min(fex, fey, fez) = 49.1 MPa l= = fy fe 350 49.102 = 2.670 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date 1 Cr = f . A . f y. 1 + l 2. n n 1×103 1 = 0.9 ×8 550.0 ×350 × 1 + 2.6698 2 ×1.34 1×103 -1.34 = 358.757 kN U1x = 1.000 U1y = 1.000 Ky. Ly. fy ly = 2 ry p .E = 1.00 ×8 000.0 350 × 2 39.90 p ×200000 = 2.670 b = 0.6 + 0.4 . l y = 0.6 + 0.4 ×2.269 = 1.508 => b = 0.85 13.8.2 F13.8.2(b) = = Cu 0.85 . U1x. Mux b . U1y. Muy + + Cr Mrx Mry 131.8 0.85 ×1.000 ×176.91 0.85 ×1.000 ×6.46 + + 358.7 381.46 76.54 = 0.8334 Lateral torsional buckling strength Mp = = f y. Zplx 1×106 350.000 ×1 211 000.000 1×106 = 423.850 kNm My = = f y. Zex 1×106 350.000 ×1 071 000.000 1×106 = 374.850 kNm OK 13.8.2 c Job Number Sheet Job Title Client Your details here Calcs by Checked by Date w 2 = 1.75 + 1.05 . k x + 0.3 . k x2 = 1.75 + 1.05 ×0.516 + 0.3 ×0.516 2 = 2.372 w 2. p. Mcr = Ke. L E . Iy. G . J + p. E 2 . . Iy Cw Ke. L 1000 2 p×200 2.371 ×p × 200 ×13600000 ×77 ×560000 + ×13600000 ×412000000000 1.000 ×8000 1.000 ×8000 = 1000 = 362.826 kNm Sclass=1 and section is doubly symmetric => Mcr > 0.67Mp => 0.28 . Mp Mr = 1.15 . f . Mp. 1 Mcr = 1.15 ×0.9 ×423.850 × 1 - 0.28 ×423.850 362.841 = 295.200 kNm Cr based on weak-axis bending ly= = Ky. Ly. fy 2 ry p .E 0.85 ×8 000.00 350 × 2 39.90 p ×200000 = 2.269 1 Cr = f . A . f y. 1 + l 2. n -n 1×103 1 = 0.9 ×8 550.0 ×350 × 1 + 2.269 1×103 2 ×1.34 -1.34 = 483.521 kN F13.8.2(c) = = Cu 0.85 . U1x. Mux b . U1y. Muy + + Cr Mrx Mry 131.8 0.85 ×1.000 ×176.91 0.85 ×1.000 ×6.46 + + 483.4 295.20 76.54 = 0.8538 Job Number Sheet Job Title Your details here Client Calcs by Additional check for class 1 and 2 sections F13.8.2 = = Checked by Date OK 13.8.2 Mux Muy + Mrx Mry 176.9 6.46 + 295.20 76.54 = 0.6837 OK Shear 13.4 Shear buckling coefficient kv=5.34 13.3.4.1.1 a) hwtw = hw tw = 312 9.1 = 34.286 Shear buckling factor Sfac = = 13.4.1.1 a) kv fy 5.34 350 = 0.1235 Inelastic critical plate-buckling stress in shear f cri = = 13.4.1.1 b) 290 . f y. k v hw tw 290 × 350 ×5.34 312 9.1 = 365.670 MPa Elastic critical plate-buckling stress in shear 13.4.1.1 d) Job Number Sheet Job Title Your details here Client Calcs by f cre = = Checked by Date 180000 . k v hw 2 tw 180000 ×5.34 312 2 9.1 = 817.688 MPa hw/tw <= 440*sqrt(kv/fy) => 13.4.1.1 a) f sx = 0.66 . f y = 0.66 ×350 = 231.000 MPa Factored shear resistance X-axis: 13.4.1.1 Avx = H . tw = 363.4 ×9.1 = 3 306.940 mm2 Vrx = = 0.9 . Avx. f sx 1000 0.9 ×3 306.9 ×231 1000 = 687.505 kN Vux = 39.6kN Factored shear resistance Y-axis: f sy = 0.66 . f y = 0.66 ×350 = 231.000 MPa Avy = = 2 . B . tf . 5 6 2 ×86.6 ×15.7 ×5 6 = 2 266.033 mm2 OK 13.4.1.1 Job Number Sheet Job Title Your details here Client Calcs by Vry = = Checked by Date 0.90 . Avy. f sy 1000 0.90 ×2266 ×231 1000 = 471.101 kN Vuy = 7.8kN OK Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Design Code: SANS 10162-1:2011 Element : 190-191 Section: 356x171x67 I Lx eff = 3400 Ly eff = 3400 Lz eff = 4000 Le eff = 4000 Section Slenderness: Table 4 Flanges Bratio = = B. fy Tf 86.60 × 350 15.70 = 103.193 Bratio < 145 => Class 1 Flange Web Wratio_1 = = 1100 . fy 1100 350 1 - ×1 - 0.39 . Cu f . Cy 0.39 ×13.2 0.9 ×2 992.5 = 58.685 Hw/Tw(34.3) < Wratio_1 => Class 1 Web => Section class = 1 Maximum slenderness ratio X_ratio = = Kx. Lx rx 0.85 ×4 000.00 151 = 22.517 X_ratio < 200 => OK Y_ratio = = Ky. Ly ry 0.85 ×4 000.00 40 = 85.000 Y_ratio < 200 => OK 10.4.2.1 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Axial compression and bending 13.8 Member strength and stability - class 1 and 2 I-shaped sections 13.8.2 Cross-sectional strength 13.8.2 a Cr = = f . A . fy 1000 0.9 ×8 550.00 ×350.00 1000 = 2 693.250 kN Mrx = = f . Zpx. f y 1×106 0.9 ×1 211 000.00 ×350.00 1×106 = 381.465 kNm Mry = = f . Zpy. f y 1×106 0.9 ×243 000.00 ×350.00 1×106 = 76.545 kNm End moment factors: Mux_min kx = Mux_max = -1.60 -4.31 = 0.3712 ky = Muy_min Muy_max = 0.34 2.84 = 0.1197 w x = 0.6 - 0.4 . k x = 0.6 - 0.4 ×0.37 = 0.4520 w x = 1 (loads between supports) 13.8.5 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date w y = 0.6 - 0.4 . k y = 0.6 - 0.4 ×0.12 = 0.5520 Cex = = p2 . E . Ix 1000 . (Kx. Lx)2 p2 ×200 000.00 ×195 000 000.00 1000 ×(0.85 ×4 000.00 )2 = 33.30×103 kN Cey = = p2 . E . Iy 1000 . (Ky. Ly)2 p2 ×200 000.00 ×13 600 000.00 1000 ×(0.85 ×4 000.00 )2 = 2 322.260 kN U1x = wx 1 - = Cu Cex 1.000 13.20 1 33 297.11 = 1.000 U1y = wy 1 - = Cu Cey 0.552 13.20 1 2 322.26 = 0.5552 F13.8.2(a) = = Cu 0.85 . U1x. Mux b . U1y. Muy + + Cr Mrx Mry 13.2 0.85 ×1.000 ×1.60 0.6 ×1.000 ×2.84 + + 2 693.2 381.46 76.54 = 0.0307 OK Overall member strength Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports 13.8.2 b Job Number Sheet Job Title Client Your details here Calcs by Checked by and does not refer to restraint conditions) f ex = = p2 . E Kx. Lx 2 rx p2 ×200000 1 ×4000 2 151 = 2 812.961 MPa f ey = = p2 . E Ky. Ly 2 ry p2 ×200000 1 ×4000 2 39.9 = 196.406 MPa p2 . E . Cw f ez = (Kz . Lz )2 + G. J A . ro2 p2 ×200000 ×4120×108 (1 ×4000 )2 = + 77000 ×560000 8550 ×156.18 2 = 450.477 MPa fe = min(fex, fey, fez) = 196.4 MPa l= = fy fe 350 196.41 = 1.335 Date Job Number Sheet Job Title Your details here Client Calcs by Checked by Date 1 Cr = f . A . f y. 1 + l 2. n n 1×103 1 = 0.9 ×8 550.0 ×350 × 1 + 1.3349 2 ×1.34 1×103 -1.34 = 1 138.894 kN U1x = 1.000 U1y = 1.000 Ky. Ly. fy ly = 2 ry p .E = 1.00 ×4 000.0 350 × 2 39.90 p ×200000 = 1.335 b = 0.6 + 0.4 . l y = 0.6 + 0.4 ×1.135 = 1.054 => b = 0.85 13.8.2 F13.8.2(b) = = Cu 0.85 . U1x. Mux b . U1y. Muy + + Cr Mrx Mry 13.2 0.85 ×1.000 ×4.31 0.85 ×1.000 ×2.84 + + 1 138.9 381.46 76.54 = 0.0527 Lateral torsional buckling strength Mp = = f y. Zplx 1×106 350.000 ×1 211 000.000 1×106 = 423.850 kNm My = = f y. Zex 1×106 350.000 ×1 071 000.000 1×106 = 374.850 kNm OK 13.8.2 c Job Number Sheet Job Title Client Your details here Calcs by Checked by Date w 2 = 1.75 + 1.05 . k x + 0.3 . k x2 = 1.75 + 1.05 ×0.371 + 0.3 ×0.371 2 = 2.181 w 2. p. Mcr = Ke. L E . Iy. G . J + p. E 2 . . Iy Cw Ke. L 1000 2 p×200 2.181 ×p × 200 ×13600000 ×77 ×560000 + ×13600000 ×412000000000 1.000 ×4000 1.000 ×4000 = 1000 = 865.914 kNm Sclass=1 and section is doubly symmetric => Mcr > 0.67Mp => 0.28 . Mp Mr = 1.15 . f . Mp. 1 Mcr = 1.15 ×0.9 ×423.850 × 1 - 0.28 ×423.850 865.968 = 378.564 kNm Cr based on weak-axis bending ly= = Ky. Ly. fy 2 ry p .E 0.85 ×4 000.00 350 × 2 39.90 p ×200000 = 1.135 1 Cr = f . A . f y. 1 + l 2. n -n 1×103 1 = 0.9 ×8 550.0 ×350 × 1 + 1.135 1×103 2 ×1.34 -1.34 = 1 399.548 kN F13.8.2(c) = = Cu 0.85 . U1x. Mux b . U1y. Muy + + Cr Mrx Mry 13.2 0.85 ×1.000 ×4.31 0.85 ×1.000 ×2.84 + + 1 400.0 378.56 76.54 = 0.0506 Job Number Sheet Job Title Your details here Client Calcs by Additional check for class 1 and 2 sections F13.8.2 = = Checked by Date OK 13.8.2 Mux Muy + Mrx Mry 4.3 2.84 + 378.56 76.54 = 0.0485 OK Shear 13.4 Shear buckling coefficient kv=5.34 13.3.4.1.1 a) hwtw = hw tw = 312 9.1 = 34.286 Shear buckling factor Sfac = = 13.4.1.1 a) kv fy 5.34 350 = 0.1235 Inelastic critical plate-buckling stress in shear f cri = = 13.4.1.1 b) 290 . f y. k v hw tw 290 × 350 ×5.34 312 9.1 = 365.670 MPa Elastic critical plate-buckling stress in shear 13.4.1.1 d) Job Number Sheet Job Title Your details here Client Calcs by f cre = = Checked by Date 180000 . k v hw 2 tw 180000 ×5.34 312 2 9.1 = 817.688 MPa hw/tw <= 440*sqrt(kv/fy) => 13.4.1.1 a) f sx = 0.66 . f y = 0.66 ×350 = 231.000 MPa Factored shear resistance X-axis: 13.4.1.1 Avx = H . tw = 363.4 ×9.1 = 3 306.940 mm2 Vrx = = 0.9 . Avx. f sx 1000 0.9 ×3 306.9 ×231 1000 = 687.505 kN Vux = 3.1kN Factored shear resistance Y-axis: f sy = 0.66 . f y = 0.66 ×350 = 231.000 MPa Avy = = 2 . B . tf . 5 6 2 ×86.6 ×15.7 ×5 6 = 2 266.033 mm2 OK 13.4.1.1 Job Number Sheet Job Title Your details here Client Calcs by Vry = = Checked by Date 0.90 . Avy. f sy 1000 0.90 ×2266 ×231 1000 = 471.101 kN Vuy = 0.8kN OK Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Member Design for Combined Stresses 4.00 3.50 3.00 2.50 2.00 M max = 15.49kNm @ 4.00m 1.50 1.00 .500 MOMENTS: X-X 2.00 4.00 6.00 8.00 10.0 12.0 14.0 4.00 3.50 3.00 2.50 2.00 1.50 1.00 .500 M max = -1.240kNm @ 0.00m .200 .400 -2.00 -3.00 -4.00 -5.00 -6.00 -7.00 4.00 3.50 3.00 2.50 2.00 P max = -7.060kN @ 0.00m 1.50 1.00 .500 AXIAL FORCE -1.00 Lx Eff = 3.400 m W1x = 0.90 Ly Eff = 3.400 m W1y = 1.00 Le Eff = 4.000 m W2 = 1.14 Fy = 350 MPa Fu = 480 MPa Tension area factor (Ane/Ag) = 1.00 Flange class: 1 Web class : 1 Critical Load Case : LC2 MOMENTS: Y-Y -1.20 -1.00 -.800 -.600 -.400 -.200 Ver W5.0.04 - 13 Sep 2022 Combine Ver W5.0.04 Element 192-193 Evaluate current section Section name DOORBE Section 356x171x67 I-sections (Web vert) SABS 0162 - 1993 13.9 : a) Cross-sectional strength (Crit. pos.= 4.000 m) Tu Mux Muy 7.06 15.5 .860 -- + --- + --- = ---- + ---- + ---- = 0.05 Tr Mrx Mry 2693 381 76.5 b) Lateral torsional buckling strength Mux Muy T/A 15.5 1.24 .8257 --- + --- - ------- = ---- + ---- - ----- = 0.06 Mrx Mry Mrx/Zpx 323 76.5 267.1 Slenderness Ratio: Lx/rx = Ly/ry = 23 85 ------------------------------------------------------------ OK OK OK OK Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Design Code: SANS 10162-1:2011 Element : 297-11 Section: 356x171x67 I Lx eff = 8296 Ly eff = 8296 Lz eff = 9760 Le eff = 9760 Section Slenderness: Table 4 Flanges Bratio = = B. fy Tf 86.60 × 350 15.70 = 103.193 Bratio < 145 => Class 1 Flange Web Wratio_1 = = 1100 . fy 1100 350 1 - ×1 - 0.39 . Cu f . Cy 0.39 ×-14.3 0.9 ×2 992.5 = 58.919 Hw/Tw(34.3) < Wratio_1 => Class 1 Web => Section class = 1 Maximum slenderness ratio X_ratio = = Kx. Lx rx 0.85 ×9 760.00 151 = 54.940 X_ratio < 300 => OK Y_ratio = = Ky. Ly ry 0.85 ×9 760.00 40 = 207.400 Y_ratio < 300 => OK 10.4.2.2 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Axial tension and bending 13.9 Cross-sectional strength 13.9 a Cr = = f . A . fy 1000 0.9 ×8 550.00 ×350.00 1000 = 2 693.250 kN Mrx = = f . Zpx. f y 1×106 0.9 ×1 211 000.00 ×350.00 1×106 = 381.465 kNm Mry = = f . Zpy. f y 1×106 0.9 ×243 000.00 ×350.00 1×106 = 76.545 kNm F13.9(a) = = Tu Mux Muy + + Tr Mrx Mry 3.8 22.03 42.22 + + 2 693.2 381.46 76.54 = 0.6108 Lateral torsional buckling strength Mp = = f y. Zplx 1×106 350.000 ×1 211 000.000 1×106 = 423.850 kNm My = = f y. Zex 1×106 350.000 ×1 071 000.000 1×106 = 374.850 kNm 13.9 c Job Number Sheet Job Title Client Your details here Calcs by Checked by Date w 2 = 1.75 + 1.05 . k x + 0.3 . k x2 = 1.75 + 1.05 ×0.992 + 0.3 ×0.992 2 = 3.087 w 2. p. Mcr = Ke. L E . Iy. G . J + p. E 2 . . Iy Cw Ke. L 1000 2 p×200 2.500 ×p × 200 ×13600000 ×77 ×560000 + ×13600000 ×412000000000 1.000 ×9760 1.000 ×9760 = 1000 = 301.641 kNm Sclass=1 and section is doubly symmetric => Mcr > 0.67Mp => 0.28 . Mp Mr = 1.15 . f . Mp. 1 Mcr = 1.15 ×0.9 ×423.850 × 1 - 0.28 ×423.850 301.641 = 266.088 kNm F13.8.9(b) = = Mux Muy Tu. Zpx + Mrx Mry Mrx. A 22 030 000.00 42 220 000.00 14 340.0 ×1 211 000.00 + 266 088 311.94 76 545 000.00 266 088 311.94 ×8 550.00 = 0.6267 Shear 13.4 Shear buckling coefficient kv=5.34 13.3.4.1.1 a) hwtw = hw tw = 312 9.1 = 34.286 Shear buckling factor 13.4.1.1 a) Job Number Sheet Job Title Your details here Client Calcs by Sfac = = Checked by Date kv fy 5.34 350 = 0.1235 Inelastic critical plate-buckling stress in shear f cri = = 13.4.1.1 b) 290 . f y. k v hw tw 290 × 350 ×5.34 312 9.1 = 365.670 MPa Elastic critical plate-buckling stress in shear f cre = = 13.4.1.1 d) 180000 . k v hw 2 tw 180000 ×5.34 312 2 9.1 = 817.688 MPa hw/tw <= 440*sqrt(kv/fy) => f sx = 0.66 . f y 13.4.1.1 a) = 0.66 ×350 = 231.000 MPa Factored shear resistance X-axis: Avx = H . tw = 363.4 ×9.1 = 3 306.940 mm2 13.4.1.1 Job Number Sheet Job Title Your details here Client Calcs by Vrx = = Checked by Date 0.9 . Avx. f sx 1000 0.9 ×3 306.9 ×231 1000 = 687.505 kN Vux = 4.5kN Factored shear resistance Y-axis: OK 13.4.1.1 f sy = 0.66 . f y = 0.66 ×350 = 231.000 MPa Avy = = 2 . B . tf . 5 6 2 ×86.6 ×15.7 ×5 6 = 2 266.033 mm2 Vry = = 0.90 . Avy. f sy 1000 0.90 ×2266 ×231 1000 = 471.101 kN Vuy = 25.0kN OK Job Number Sheet Job Title Client Your details here Calcs by MOMENTS: X-X Checked by M max = -20.71kNm @ 0.00m -20.0 Date Combine Ver W5.0.04 Element 297-11 Section name GABLEC Evaluate curre 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 -10.0 10.0 20.0 MOMENTS: Y-Y M max = -42.88kNm @ 0.00m -40.0 -30.0 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 -10.0 1.00 -20.0 Lx Eff = 8.296 m W1x = 0.40 Ly Eff = 8.296 m W1y = 1.00 Le Eff = 9.760 m W2 = 2.50 Fy = 350 MPa Fu = 480 MPa Tension area factor (Ane/Ag) = 1.00 Flange class: 1 Web class : 1 Critical Load Case : LC11 10.0 20.0 Section 356x171x67 -10.0 -15.0 -20.0 -25.0 -30.0 I-sections (Web vert) 9.00 8.00 7.00 6.00 5.00 4.00 P max = -32.90kN @ 9.76m 3.00 2.00 -5.00 1.00 AXIAL FORCE SANS 10162-1:2011 13.9 : a) Cross-sectional strength (Crit. pos.= 0.000 m Tu Mux Muy 25.0 20.7 42.9 -- + --- + --- = ---- + ---- + ---OK= 0.62 Tr Mrx Mry 2693 381 76.5 b) Lateral torsional buckling strength Mux Muy T/A 20.7 42.9 3.848 --- + --- - ------- = ---- + ---- OK----- = 0 Mrx Mry Mrx/Zpx 266 76.5 219.7 13.4:Shear Vux<Vrx Vuy<Vry 4.2 < 687.5 25.3 < 471.1 Slenderness Ratio: Lx/rx = 55 Ly/ry = 208 OK OK OK OK ------------------------------------------------ Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Member Design for Combined Stresses MOMENTS: X-X M max = 114.7kNm @ 0.00m 10.0 8.00 6.00 4.00 20.0 40.0 60.0 80.0 100 2.00 -60.0 -40.0 -20.0 Lx Eff = 8.632 m W1x = 0.51 Ly Eff = 8.632 m W1y = 1.00 Le Eff = 10.155 m W2 = 1.99 Fy = 350 MPa Fu = 480 MPa Tension area factor (Ane/Ag) = 1.00 Flange class: 1 Web class : 1 Critical Load Case : LC2 MOMENTS: Y-Y M max = -5.100kNm @ 0.00m -5.00 -4.00 -3.00 10.0 8.00 6.00 4.00 2.00 -2.00 -1.00 Ver W5.0.04 - 13 Sep 2022 Combine Ver W5.0.04 Element 32-38 Evaluate current section Section name RAFTER 1.00 2.00 AXIAL FORCE P max = 71.81kN @ 2.03m 70.0 Section 356x171x67 I-sections (Web vert) SANS 10162-1:2011 13.8.2 : a) Cross-sectional strength (Crit. pos.= 0.000 m) Cu 0.85Mux 0.60Muy 70.6 97.5 3.82 -- + ------- + ------- = ---- + ---- + ---- = 0.33 Cr Mrx Mry 2693 381 76.5 b) Overall member strength Cu 0.85U1xMux ßU1yMuy 71.6 97.5 4.34 -- + ---------- + ---------- = ---- + ---- + ---- = 0.63 Cr Mrx Mry 228 381 76.5 c) Lateral torsional buckling strength Cu 0.85U1xMux ßU1yMuy 71.6 97.5 4.34 -- + ---------- + ---------- = ---- + ---- + ---- = 0.76 Cr Mrx Mry 311 206 76.5 60.0 50.0 40.0 OK OK OK 30.0 10.0 8.00 6.00 4.00 10.0 2.00 20.0 d) Additional check for class 1 I sections Mux Muy 115 5.10 --- + --- = ---- + ---- = 0.62 Mrx Mry 206 76.5 OK 13.4:Shear Vux<Vrx Vuy<Vry 37.3 < 687.5 3.5 < 471.1 Slenderness Ratio: Lx/rx = 57 Ly/ry = 216 ------------------------------------------------------------ OK OK OK FAIL Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Design Code: SANS 10162-1:2011 Element : 190-191 Section: 356x171x67 I Lx eff = 3400 Ly eff = 3400 Lz eff = 4000 Le eff = 4000 Section Slenderness: Table 4 Flanges Bratio = = B. fy Tf 86.60 × 350 15.70 = 103.193 Bratio < 145 => Class 1 Flange Web Wratio_1 = = 1100 . fy 1100 350 1 - ×1 - 0.39 . Cu f . Cy 0.39 ×19.6 0.9 ×2 992.5 = 58.631 Hw/Tw(34.3) < Wratio_1 => Class 1 Web => Section class = 1 Maximum slenderness ratio X_ratio = = Kx. Lx rx 0.85 ×4 000.00 151 = 22.517 X_ratio < 200 => OK Y_ratio = = Ky. Ly ry 0.85 ×4 000.00 40 = 85.000 Y_ratio < 200 => OK 10.4.2.1 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Axial compression and bending 13.8 Member strength and stability - class 1 and 2 I-shaped sections 13.8.2 Cross-sectional strength 13.8.2 a Cr = = f . A . fy 1000 0.9 ×8 550.00 ×350.00 1000 = 2 693.250 kN Mrx = = f . Zpx. f y 1×106 0.9 ×1 211 000.00 ×350.00 1×106 = 381.465 kNm Mry = = f . Zpy. f y 1×106 0.9 ×243 000.00 ×350.00 1×106 = 76.545 kNm End moment factors: Mux_min kx = Mux_max = 6.03 -6.56 = -0.9192 ky = Muy_min Muy_max = 0.70 -1.46 = -0.4795 w x = 0.6 - 0.4 . k x = 0.6 - 0.4 ×-0.92 = 0.9680 13.8.5 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date w y = 0.6 - 0.4 . k y = 0.6 - 0.4 ×-0.48 = 0.7920 w y = 1 (loads between supports) Cex = = p2 . E . Ix 1000 . (Kx. Lx)2 p2 ×200 000.00 ×195 000 000.00 1000 ×(0.85 ×4 000.00 )2 = 33.30×103 kN Cey = = p2 . E . Iy 1000 . (Ky. Ly)2 p2 ×200 000.00 ×13 600 000.00 1000 ×(0.85 ×4 000.00 )2 = 2 322.260 kN U1x = wx 1 - = Cu Cex 0.968 19.62 1 33 297.11 = 0.9686 U1y = wy 1 - = Cu Cey 1.000 19.62 1 2 322.26 = 1.009 F13.8.2(a) = = Cu 0.85 . U1x. Mux b . U1y. Muy + + Cr Mrx Mry 19.6 0.85 ×1.000 ×6.03 0.6 ×1.009 ×1.46 + + 2 693.2 381.46 76.54 = 0.0323 OK Overall member strength 13.8.2 b Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports and does not refer to restraint conditions) f ex = = p2 . E Kx. Lx 2 rx p2 ×200000 1 ×4000 2 151 = 2 812.961 MPa f ey = = p2 . E Ky. Ly 2 ry p2 ×200000 1 ×4000 2 39.9 = 196.406 MPa p2 . E . Cw + G. J (Kz . Lz )2 f ez = A . ro2 p2 ×200000 ×4120×108 (1 ×4000 )2 = + 77000 ×560000 8550 ×156.18 2 = 450.477 MPa fe = min(fex, fey, fez) = 196.4 MPa l= = fy fe 350 196.41 = 1.335 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date 1 Cr = f . A . f y. 1 + l 2. n n 1×103 1 = 0.9 ×8 550.0 ×350 × 1 + 1.3349 2 ×1.34 1×103 -1.34 = 1 138.894 kN U1x = 1.000 U1y = 1.000 Ky. Ly. fy ly = 2 ry p .E = 1.00 ×4 000.0 350 × 2 39.90 p ×200000 = 1.335 b = 0.6 + 0.4 . l y = 0.6 + 0.4 ×1.135 = 1.054 => b = 0.85 13.8.2 F13.8.2(b) = = Cu 0.85 . U1x. Mux b . U1y. Muy + + Cr Mrx Mry 19.6 0.85 ×1.000 ×6.56 0.85 ×1.000 ×1.46 + + 1 138.9 381.46 76.54 = 0.0480 Lateral torsional buckling strength Mp = = f y. Zplx 1×106 350.000 ×1 211 000.000 1×106 = 423.850 kNm My = = f y. Zex 1×106 350.000 ×1 071 000.000 1×106 = 374.850 kNm OK 13.8.2 c Job Number Sheet Job Title Client Your details here Calcs by Checked by Date w 2 = 1.75 + 1.05 . k x + 0.3 . k x2 = 1.75 + 1.05 ×-0.919 + 0.3 ×-0.919 2 = 1.038 w 2. p. Mcr = Ke. L E . Iy. G . J + p. E 2 . . Iy Cw Ke. L 1000 2 p×200 1.038 ×p × 200 ×13600000 ×77 ×560000 + ×13600000 ×412000000000 1.000 ×4000 1.000 ×4000 = 1000 = 412.113 kNm Sclass=1 and section is doubly symmetric => Mcr > 0.67Mp => 0.28 . Mp Mr = 1.15 . f . Mp. 1 Mcr = 1.15 ×0.9 ×423.850 × 1 - 0.28 ×423.850 412.238 = 312.393 kNm Cr based on weak-axis bending ly= = Ky. Ly. fy 2 ry p .E 0.85 ×4 000.00 350 × 2 39.90 p ×200000 = 1.135 1 Cr = f . A . f y. 1 + l 2. n -n 1×103 1 = 0.9 ×8 550.0 ×350 × 1 + 1.135 1×103 2 ×1.34 -1.34 = 1 399.548 kN F13.8.2(c) = = Cu 0.85 . U1x. Mux b . U1y. Muy + + Cr Mrx Mry 19.6 0.85 ×1.000 ×6.56 0.85 ×1.000 ×1.46 + + 1 400.0 312.39 76.54 = 0.0481 Job Number Sheet Job Title Your details here Client Calcs by Additional check for class 1 and 2 sections F13.8.2 = = Checked by Date OK 13.8.2 Mux Muy + Mrx Mry 6.6 1.46 + 312.39 76.54 = 0.0402 OK Shear 13.4 Shear buckling coefficient kv=5.34 13.3.4.1.1 a) hwtw = hw tw = 312 9.1 = 34.286 Shear buckling factor Sfac = = 13.4.1.1 a) kv fy 5.34 350 = 0.1235 Inelastic critical plate-buckling stress in shear f cri = = 13.4.1.1 b) 290 . f y. k v hw tw 290 × 350 ×5.34 312 9.1 = 365.670 MPa Elastic critical plate-buckling stress in shear 13.4.1.1 d) Job Number Sheet Job Title Your details here Client Calcs by f cre = = Checked by Date 180000 . k v hw 2 tw 180000 ×5.34 312 2 9.1 = 817.688 MPa hw/tw <= 440*sqrt(kv/fy) => 13.4.1.1 a) f sx = 0.66 . f y = 0.66 ×350 = 231.000 MPa Factored shear resistance X-axis: 13.4.1.1 Avx = H . tw = 363.4 ×9.1 = 3 306.940 mm2 Vrx = = 0.9 . Avx. f sx 1000 0.9 ×3 306.9 ×231 1000 = 687.505 kN Vux = 0.1kN Factored shear resistance Y-axis: f sy = 0.66 . f y = 0.66 ×350 = 231.000 MPa Avy = = 2 . B . tf . 5 6 2 ×86.6 ×15.7 ×5 6 = 2 266.033 mm2 OK 13.4.1.1 Job Number Sheet Job Title Your details here Client Calcs by Vry = = Checked by Date 0.90 . Avy. f sy 1000 0.90 ×2266 ×231 1000 = 471.101 kN Vuy = 1.8kN OK Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Member Design for Combined Stresses MOMENTS: X-X M max = 154.6kNm @ 8.00m 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 -50.0 50.0 100 Lx Eff = 6.800 m W1x = 1.00 Ly Eff = 6.800 m W1y = 1.00 Le Eff = 8.000 m W2 = 2.37 Fy = 350 MPa Fu = 480 MPa Tension area factor (Ane/Ag) = 1.00 Flange class: 1 Web class : 1 Critical Load Case : LC2 150 MOMENTS: Y-Y M max = -4.889kNm @ 6.99m 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 -4.00 -2.00 Ver W5.0.04 - 13 Sep 2022 Combine Ver W5.0.04 Element 42-38 Evaluate current section Section name COLUMN Section 356x171x67 I-sections (Web vert) SANS 10162-1:2011 13.8.2 : a) Cross-sectional strength (Crit. pos.= 8.000 m) Cu 0.85Mux 0.60Muy 77.6 133 .092 -- + ------- + ------- = ---- + ---- + ---- = 0.38 Cr Mrx Mry 2693 381 76.5 OK 2.00 4.00 AXIAL FORCE P max = 125.4kN @ 0.00m 120 c) Lateral torsional buckling strength Cu 0.85U1xMux ßU1yMuy 125 131 3.53 -- + ---------- + ---------- = ---- + ---- + ---- = 0.75 Cr Mrx Mry 483 295 76.5 100 80.0 60.0 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 40.0 20.0 b) Overall member strength Cu 0.85U1xMux ßU1yMuy 125 131 3.53 -- + ---------- + ---------- = ---- + ---- + ---- = 0.74 Cr Mrx Mry 359 381 76.5 d) Additional check for class 1 I sections Mux Muy 155 4.15 --- + --- = ---- + ---- = 0.58 Mrx Mry 295 76.5 OK OK OK 13.4:Shear Vux<Vrx Vuy<Vry 34.5 < 687.5 4.8 < 471.1 Slenderness Ratio: Lx/rx = 45 Ly/ry = 170 ------------------------------------------------------------ OK OK OK OK Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Member Design for Combined Stresses MOMENTS: X-X M max = 154.6kNm @ 8.00m 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 -50.0 50.0 100 Lx Eff = 6.800 m W1x = 1.00 Ly Eff = 6.800 m W1y = 1.00 Le Eff = 8.000 m W2 = 2.37 Fy = 350 MPa Fu = 480 MPa Tension area factor (Ane/Ag) = 1.00 Flange class: 1 Web class : 1 Critical Load Case : LC2 150 MOMENTS: Y-Y M max = -4.889kNm @ 6.99m 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 -4.00 -2.00 Ver W5.0.04 - 13 Sep 2022 Combine Ver W5.0.04 Element 42-38 Evaluate current section Section name COLUMN Section 356x171x67 I-sections (Web vert) SANS 10162-1:2011 13.8.2 : a) Cross-sectional strength (Crit. pos.= 8.000 m) Cu 0.85Mux 0.60Muy 77.6 133 .092 -- + ------- + ------- = ---- + ---- + ---- = 0.38 Cr Mrx Mry 2693 381 76.5 OK 2.00 4.00 AXIAL FORCE P max = 125.4kN @ 0.00m 120 c) Lateral torsional buckling strength Cu 0.85U1xMux ßU1yMuy 125 131 3.53 -- + ---------- + ---------- = ---- + ---- + ---- = 0.75 Cr Mrx Mry 483 295 76.5 100 80.0 60.0 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 40.0 20.0 b) Overall member strength Cu 0.85U1xMux ßU1yMuy 125 131 3.53 -- + ---------- + ---------- = ---- + ---- + ---- = 0.74 Cr Mrx Mry 359 381 76.5 d) Additional check for class 1 I sections Mux Muy 155 4.15 --- + --- = ---- + ---- = 0.58 Mrx Mry 295 76.5 OK OK OK 13.4:Shear Vux<Vrx Vuy<Vry 34.5 < 687.5 4.8 < 471.1 Slenderness Ratio: Lx/rx = 45 Ly/ry = 170 ------------------------------------------------------------ OK OK OK OK Job Number Sheet Job Title Client Your details here Calcs by 7.00 6.00 5.00 4.00 M max = .8284kNm @ 3.75m 3.00 2.00 1.00 MOMENTS: X-X Checked by Combine Ver W5.0.04 Element 47-68 Section name EAVES Date Evaluate current section .600 Lx Eff = 6.375 m W1x = 1.00 Ly Eff = 6.375 m W1y = 1.00 Le Eff = 7.500 m W2 = 1.00 Fy = 350 MPa Fu = 480 MPa Tension area factor (Ane/Ag) = 1.00 Section class: 1 .800 Critical Load Case : LC11 .200 .400 .00500 MOMENTS: Y-Y M max = 0.000kNm @ 0.00m Round hollow sections SANS 10162-1:2011 13.8.3 : a) Cross-sectional strength (Crit. pos.= 3.750 m) Cu Mux Muy 12.6 .888 0.00 -- + --- + --- = ---- + ---- + ---- = 0.06 Cr Mrx Mry 536 23.0 23.0 .00400 .00300 .00200 AXIAL FORCE 7.00 6.00 5.00 4.00 3.00 2.00 1.00 .00100 P max = 12.64kN @ 0.00m 12.0 b) Overall member strength Cu U1xMux U1yMuy 12.6 .828 0.00 -- + ------ + ------ = ---- + ---- + ---- = 0.15 Cr Mrx Mry 111 23.0 23.0 c) Lateral torsional buckling strength Cu U1xMux U1yMuy 12.6 .828 0.00 -- + ------ + ------ = ---- + ---- + ---- = 0.12 Cr Mrx Mry 147 23.0 23.0 10.0 8.00 6.00 OK OK OK 13.4:Shear 7.00 6.00 5.00 4.00 3.00 2.00 1.00 4.00 2.00 Section 139.7x4.0+ Vux<Vrx Vuy<Vry 0.4 < 196.4 0.0 < 196.4 Slenderness Ratio: Lx/rx = 133 Ly/ry = 133 ------------------------------------------------------------ OK OK OK OK Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Design Code: SANS 10162-1:2011 Element : 47-68 Section: 139.7x4.0+ O Lx eff = 6375 Ly eff = 6375 Lz eff = 7500 Le eff = 7500 Section Slenderness: DT_ratio = = Table 4 D. fy T 139.00 ×350 4.00 = 12.16×103 DT_ratio < 13000 => Class 1 section Maximum slenderness ratio X_ratio = = Kx. Lx rx 0.85 ×7 500.00 48 = 132.812 X_ratio < 200 => OK Y_ratio = = Ky. Ly ry 0.85 ×7 500.00 48 = 132.812 Y_ratio < 200 => OK 10.4.2.1 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Axial compression and bending 13.8 Member strength and stability — All classes of sections except class 1 and class 2 I-shaped sections 13.8.3 Cross-sectional strength 13.8.3 a Cr = = f . A . fy 1000 0.9 ×1 700.00 ×350.00 1000 = 535.500 kN Mrx = = f . Zpx. f y 1×106 0.9 ×72 900.00 ×350.00 1×106 = 22.964 kNm Mry = = f . Zpy. f y 1×106 0.9 ×72 900.00 ×350.00 1×106 = 22.964 kNm End moment factors: k x = -1 (no X moments) k y = -1 (no Y moments) w x = 0.6 - 0.4 . k x = 0.6 - 0.4 ×-1.00 = 1.0000 w y = 0.6 - 0.4 . k y = 0.6 - 0.4 ×-1.00 = 1.0000 13.8.5 Job Number Sheet Job Title Your details here Client Calcs by Cex = = Checked by Date p2 . E . Ix 1000 . (Kx. Lx)2 p2 ×200 000.00 ×3 870 000.00 1000 ×(0.85 ×7 500.00 )2 = 187.966 kN Cey = = p2 . E . Iy 1000 . (Ky. Ly)2 p2 ×200 000.00 ×3 870 000.00 1000 ×(0.85 ×7 500.00 )2 = 187.966 kN U1x = wx 1- = Cu Cex 1.000 12.29 1187.97 = 1.070 U1y = wy 1- = Cu Cey 1.000 12.29 1187.97 = 1.070 F13.8.3(a) = = Cu U1x. Mux U1y. Muy + + Cr Mrx Mry 12.3 1.070 ×0.83 1.070 ×0.00 + + 535.5 22.96 22.96 = 0.0616 OK Overall member strength Calculate Cr (Note: K=1 will only be used if K>0.7 and K<1, because K<0.7 implies intermediate lateral supports and does not refer to restraint conditions) 13.8.3 b Job Number Sheet Job Title Client Your details here Calcs by f ex = = p2 . E Kx. Lx 2 rx p2 ×200000 1 ×7500 2 47.8 = 80.179 MPa f ey = = p2 . E Ky. Ly 2 ry p2 ×200000 1 ×7500 2 47.8 = 80.179 MPa p2 . E . Cw + G. J (Kz . Lz )2 f ez = A . ro2 p2 ×200000 ×0 = (1 ×7500 )2 + 77000 ×7740×103 1700 ×67.599 2 = 76.72×103 MPa fe = min(fex, fey, fez) = 80.2 MPa l= = fy fe 350 80.179 = 2.089 Checked by Date Job Number Sheet Job Title Your details here Client Calcs by Checked by Date 1 Cr = f . A . f y. 1 + l 2. n n 1×103 1 = 0.9 ×1 700.0 ×350 × 1 + 2.0893 2 ×1.34 1×103 -1.34 = 111.335 kN U1x = 1.000 U1y = 1.000 Cu U1x. Mux U1y. Muy F13.8.3(b) = + + Cr Mrx Mry = 12.3 1.000 ×0.83 1.000 ×0.00 + + 111.3 22.96 22.96 = 0.1467 Lateral torsional buckling strength Mp = = f y. Zplx 1×106 350.000 ×72 900.000 1×106 = 25.515 kNm My = = f y. Zex 1×106 350.000 ×55 700.000 1×106 = 19.495 kNm Mr = f . Mp = 0.9 ×25.515 = 22.964 kNm Cr based on weak-axis bending ly= = Ky. Ly. fy 2. ry p E 0.85 ×7 500.00 350 × 2 47.80 p ×200000 = 1.776 OK 13.8.3 c Job Number Sheet Job Title Your details here Client Calcs by Checked by Date 1 Cr = f . A . f y. 1 + l 2. n n 1×103 1 = 0.9 ×1 700.0 ×350 × 1 + 1.776 2 ×1.34 1×103 -1.34 = 146.853 kN F13.8.3(c) = = Cu U1x. Mux U1y. Muy + + Cr Mrx Mry 12.3 1.000 ×0.83 1.000 ×0.00 + + 146.9 22.96 22.96 = 0.1199 OK Job Number Sheet Job Title Your details here Client Calcs by Checked by Date S13 Apex Connection - Ver W5.0.00 - 01 Apr 2022 Title : Code of Practice : SANS 10162-1:2011 Created : 2022/11/21 21:39:12 Notes and Assumptions 1 2 3 All bolt holes are assumed to be normal clearance holes. All bolts are assumed to have threads in their shear planes. It is assumed that the connection is deep enough for the flanges to resist the compressive and tensile forces in them. Summary Summary of Forces and Capacities for Design to SANS 10162-1:2011 Check Member Type LC Applied Capacity Units % of Cap. ? 1 Weld Flange LC2 119.7 143.1 kN 83.6 O.K. 2 Weld Web LC10 9.6 630.9 kN 1.5 O.K. 3 Bolts Shear LC10 1 42.2 kN 2.3 O.K. 3b Bolts Slip N/A N/A N/A kN N/A N/A 4 Bolts Combined LC2 0.4 1.4 kN 28.7 O.K. 5 Bolts & Plate Tension & Bending LC2 50.1 151.6 kN 33.1 O.K. 6 Plate Bearing LC10 1 144.6 kN 0.7 O.K. Input General Settings Bolt Tension Analysis Plastic Job Number Sheet Job Title Client Your details here Calcs by Checked by Bolt Type Bearing Bolt Grade 4.8 Member Ultimate Strength 480 Member Yield Strength 350 Weld Ultimate Strength 480 Connection Type Extended End Plate : Top Beam 356x171x67 Beam Angle 10 Haunch Depth (mm) 400 Haunch Length (mm) 2030.6 I1 Ultimate Limit State Loads in Beam Shear (kN) Load Case Axial (kN) Moment (kNm) Divide Factor (to obtain SLS Loads) LC1 5.55 26.51 -74.55 1 LC2 6.82 27.88 -79.32 1 LC3 4.97 13.06 -39.36 1 LC4 6.59 22.50 -65.24 1 LC5 -0.87 -19.25 61.42 1 LC6 0.02 -44.30 44.36 1 LC7 1.11 -43.05 40.45 1 LC8 0.40 -17.88 56.65 1 LC9 1.29 -42.93 39.59 1 LC10 2.39 -41.68 35.68 1 LC11 -1.58 -21.95 69.27 1 LC12 -0.69 -46.99 52.21 1 LC13 0.40 -45.75 48.31 1 End Plate Bolts Rows of Bolts Bolt Offsets Width (mm) 173.3 Extent Above Beam Flange (mm) 51.2 Extent Below Haunch (mm) N/A Thickness (mm) 16 Diameter (mm) 20 Above Top Flange 1 Below Top Flange 2 Above Haunch 2 Below Haunch N/A Row Spacing (mm) 75 Date Job Number Sheet Job Title Your details here Client Calcs by Weld Sizes Checked by Web (mm) 41.2 Flange (mm) 23.1 Above Haunch (mm) 23.1 Beam Flanges (mm) 3 Beam Web (mm) 3 Date Check 1 : Capacity of the Beam Flange Welds The worst load is encountered for Load Case : LC2 when Fmax =119.65 kN The Capacity of the weld is the lesser of : 0.67 . f w. Aw. Xu Vr = 1000 = 13.13.2.2 13.13.2.2b 0.67 ×0.67 ×672.246 ×480 1000 = 144.850 kN 0.67 . fw. Am. f u Vr = 1000 = 13.13.2.2a 0.67 ×0.67 ×950.7 ×480 1000 = 204.849 kN Beam Flange Weld is safe Check 2 : Capacity of the Beam Web Welds The worst load is encountered for Load Case : LC10 when Fmax =9.591 kN The Capacity of the weld is the lesser of : 0.67 . fw. Aw. Xu Vr = 1000 = 13.13.2.2 13.13.2.2 0.67 ×0.67 ×2 927.878 ×480 1000 = 630.876 kN 13.13.2.2 Job Number Sheet Job Title Your details here Client Calcs by Vr = = Checked by Date 0.67 . fw. Am. f u 1000 0.67 ×0.67 ×4 140.644 ×350 1000 = 650.557 kN Beam Web Weld is safe Check 3 : Shear Capacity of the Bolts The worst load is encountered for Load Case : LC10 when Vmax =0.959 kN 13.12.1.2 The resistance of any bolt is : Vr = = 0.60 . f b. m . 0.7 . Ab. f u 1000 0.60 ×0.80 ×1 ×0.7 ×314.159 ×400 1000 = 42.223 kN Bolt shear is safe Check 4 : Shear and Tension Capacity of the Bolts The worst load is encountered for Load Case : LC2 The factor must be less than or equal to 1.4 : Vu Tu Factor = + Vr Tr = 13.12.1.4 0.188 25.057 + 35.362 63.146 = 0.4021 Bolt shear and tension is safe Check 5 : Bolt tension and End Plate Bending The worst load is encountered for Load Case : LC2 Fmax = 50.114 kN Eurocode 1993-1 8 : 6.2.4 8 : Table 6.2 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date The resistance is the smaller of the 3 possible failure modes : Mode 1 : Complete yielding of the End Plate R1 = = 4 . Mpl . 1000 m 4 ×2.83576 ×1000 38.8 = 292.346 kN Mode 2 : Bolt Failure with yielding of the End Plate R2 = = 2 . Mpl . 1000 + n . 2 . Bt m+n 2 ×2.83576 ×1000 + 40.9 ×2 ×78.4 38.8 + 40.9 = 151.627 kN Mode 3 : Bolt Failure only R3 = 2 . Bt = 2 ×78.4 = 156.800 kN Therefore R = R2 = 151.626 Bolt tension and end plate bending is safe Check 6 : Bearing of the End Plate The Bearing Capacity of the Plate at any Bolt is the lesser of : The worst load is encountered for Load Case : LC10 when Bmax =0.959 kN Br = = fbr. t . a . f u 1000 0.67 ×16 ×28.1 ×480 1000 = 144.591 kN End plate bearing is safe 13.10.1c Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces (Forces are given per bolt and not per row) Bolt Forces for Load Case : LC1 0 kN 0 kN 0 kN 23.53 kN 23.53 kN Shear force per bolt : 0.09 kN Bolt Forces for Load Case : LC2 0 kN 0 kN 0 kN 25.06 kN 25.06 kN Shear force per bolt : 0.19 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC3 0 kN 0 kN 0 kN 12.49 kN 12.49 kN Shear force per bolt : 0.26 kN Bolt Forces for Load Case : LC4 0 kN 0 kN 0 kN 20.64 kN 20.64 kN Shear force per bolt : 0.26 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC5 15.8 kN 15.8 kN 15.8 kN 2.39 kN 2.39 kN Shear force per bolt : 0.25 kN Bolt Forces for Load Case : LC6 13.9 kN 13.9 kN 13.9 kN 5.45 kN 5.45 kN Shear force per bolt : 0.77 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC7 12.88 kN 12.88 kN 12.88 kN 5.28 kN 5.28 kN Shear force per bolt : 0.86 kN Bolt Forces for Load Case : LC8 14.57 kN 14.57 kN 14.57 kN 2.19 kN 2.19 kN Shear force per bolt : 0.35 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC9 12.66 kN 12.66 kN 12.66 kN 5.26 kN 5.26 kN Shear force per bolt : 0.87 kN Bolt Forces for Load Case : LC10 11.64 kN 11.64 kN 11.64 kN 5.08 kN 5.08 kN Shear force per bolt : 0.96 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC11 17.85 kN 17.85 kN 17.85 kN 2.74 kN 2.74 kN Shear force per bolt : 0.23 kN Bolt Forces for Load Case : LC12 15.95 kN 15.95 kN 15.95 kN 5.8 kN 5.8 kN Shear force per bolt : 0.75 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC13 14.93 kN 14.93 kN 14.93 kN 5.62 kN 5.62 kN Shear force per bolt : 0.83 kN 3 3 400 824 x 173 x 16 End plate 2031 3 92 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Base Plate Design - SANS 10162 - 2005 Material Strength Properties fcu : 30 MPa Bolt Grade : 8.8 Bolt fy : 640 MPa Bolt fu : 800 MPa fy Baseplate : 355 MPa fu Baseplate : 470 MPa fy Column : 355 MPa fu Column : 470 MPa tu Weld : 483 MPa Column Section I1 356x171x67 Base Plate Design Data: Plate Shape : Height : Breadth : Thickness : Rectangular 600 mm 600 mm 30 mm Weld Properties Size :10 mm Fillet Weld Bolt Properties Diameter : 20 mm Anchor Length : 250 mm Compression not allowed in bolts Bolt End Plate Properties End Type : Dimension : Thickness : Square Plate 100 x 100 mm 30 mm Design Loads Summary Load Case P (kN) Torsion (kNm) Vx (kN) Mx (kNm) Vz (kN) Mz (kNm) LC1 LC2 LC3 LC4 LC5 LC6 LC7 LC8 LC9 LC10 LC11 LC12 LC13 98.07 100.02 61.39 85.35 -5.29 -30.45 -30.82 -3.33 -28.49 -28.87 -19.82 -44.98 -45.36 -0.03 -0.03 -0.02 -0.03 -0.02 0.01 -0.01 -0.02 0.01 -0.01 -0.01 0.01 -0.01 1.14 1.19 0.7 1.02 -3.21 -1.24 0.41 -3.16 -1.19 0.46 -3.37 -1.39 0.25 90.78 94.71 51.46 78.98 62.84 31.72 75.89 66.78 35.66 79.83 51.62 20.5 64.67 27.44 28.6 15.49 23.83 24.89 36.43 41.91 26.06 37.6 43.08 21.5 33.05 38.52 -2.18 -2.31 -1.36 -1.99 6.86 2.68 -0.95 6.72 2.54 -1.09 7.14 2.96 -0.66 Job Number Sheet Job Title Client Your details here Calcs by Checked by Date Factor of Safety Summary Load Case Steps LC1 LC2 LC3 LC4 LC5 LC6 LC7 LC8 LC9 LC10 LC11 LC12 LC13 FOS Concrete 112 112 107 111 111 106 112 111 106 112 111 109 111 FOS Bolt T 5.84 5.55 10.90 6.84 7.62 17.10 7.82 7.25 15.50 7.36 8.94 24.90 9.59 1.68 1.60 3.08 1.94 1.77 3.06 1.49 1.69 2.83 1.43 1.98 3.62 1.66 FOS Bolt C FOS Bolt Shear - 8.01 7.69 14.20 9.22 8.77 6.04 5.26 8.39 5.86 5.11 10.10 6.66 5.72 FOS Base Plate 6.13 5.84 11.20 7.07 6.46 11.10 5.44 6.16 10.30 5.21 7.23 13.20 6.06 FOS Weld 2.87 2.74 5.14 3.30 3.56 6.34 2.88 3.37 5.77 2.75 4.15 8.51 3.28 FOS Bolt T+V 1.95 1.85 3.54 2.24 2.06 2.84 1.63 1.97 2.67 1.56 2.32 3.28 1.80 FOS Bolt Pull-Out 2.54 2.42 4.66 2.93 2.68 4.62 2.26 2.56 4.28 2.16 3.00 5.48 2.51 Bolt Resistance Forces Bolt Net Cross Section 25.2.2.1 p An = . (d - 0.938 . P )2 4 p = ×(20 - 0.938 ×2.5 )2 4 = 244.808 mm2 Tension Resistance Tr = = 25.2.2.1 fb. An. f u 1000 .67 ×244.81 ×800 1000 = 131.218 kN 25.2.2.2 SANS 10100-1 4.11.6.2 Tension Resistance Concrete Trc = 0.28 . f cu. p. d . lb + 0.6 . f cu. (AnchorArea - BoltArea) 1000 = 0.28 × 30 ×p×20 ×250 + 0.6 ×30 ×(10000 - 314.16 ) 1000 = 198.435 kN 6.2.4.4.3 (b) Job Number Sheet Job Title Your details here Client Calcs by Shear Resistance Vr = = Checked by Date 25.2.3.3 0.6 . fb. 0.7 . An. f u 1000 0.6 ×.67 ×0.7 ×244.81 ×800 1000 = 55.112 kN Compression Resistance Cr = = 13.3.1 0.9 . An. f u 1000 0.9 ×244.81 ×516.13 1000 = 113.718 kN Find Effective Compression Area Calculate Zpl b . tp2 Zpl = 4 = 1 ×30 2 4 = 225.000 mm3 Moment Resistance Mr = = f . Zpl . f y 1000 .9 ×225 ×355 1000 = 71.888 Nm Moment Ultimate equation Mu = (c*b)*(c/2)*fcu Through substitution cMax can be calculated Effective Distance from Edge of Section 13.5 (a) Job Number Sheet Job Title Your details here Client Calcs by fy Zpl . 2 . 0.9 . 1.15 cMax = f cu b. 1.5 225 ×2 ×0.9 × = 355 1.15 30 1× 1.5 = 79.064 mm Calculation Sheet for Load Case : LC8 Factored loads P: -3.33 kN Mz : 6.72 kNm Vx : -3.16 kN Mx : 66.78 kNm Vz : 26.06 kN Torsion : -0.02 kNm Checked by Date Job Number Sheet Job Title Your details here Client Calcs by Checked by Find Equilibruim The actual number of Grid Point used for calculation is 1078 Date Job Number Sheet Job Title Your details here Client Calcs by Checked by Moment balancing Sum Of Moments around X-axis = 0.0 kNm Sum Of Moments around Y-axis = 0.0 kNm Axial Force balancing Sum Of Forces in Y-direction = 0.0 kN The Shear Resistance in the Bolts Resists the Following Forces: Forces in X-direction Moments around Y-axis Forces in Z-direction Calculating Factors of Safety in Concrete FOS = StrainMax Strain = .0035 .000483 = 7.246 Calculating Factors of Safety in Critical Bolt Tension in Bolts Critical Bolt Tension FOS = = Tr Tension 131.22 77.638 = 1.690 Critical Bolt Pull-Out FOS = = Trc Tension 198.44 77.638 = 2.556 Shear in Bolts Critical Bolt Shear Date Job Number Sheet Job Title Your details here Client Calcs by FOS = = Checked by Date Vr Shear 55.111 6.5722 = 8.385 Shear and Tension combined in Bolts The factor should be less than 1.4 for bolts in shear and tension The bolt number 4 has the critical shear and tension combination The tension in the bolt is: 77.64 kN The shear in the bolt is: 6.56 kN 13.12.1.4 Tension and Shear Resistance combination combinedfactor = = Shear Tension + Vr Tr 6.5572 77.638 + 55.111 131.22 = 0.7106 0.711 <= 1.4 OK Converted to Factor of Safety relevant to 1 FOS = = 1.4 f actor 1.4 .71066 = 1.970 Bolt BasePlate interaction FOS = Resistance Force = 478.4 77.638 = 6.162 25.2.4 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Welds Since unit values are used for the length and size of the weld, the capacity of this layout is given in kN/mm The capacity, Vr is the lesser of Vr1 and Vr2: Resistance of parent material Vr1 = = 13.13.2.2 (a) 0.67 . fw. 0.707 . Size. f u 1000 0.67 ×.67 ×0.707 ×10 ×470 1000 = 1.492 kN/mm The angle of the axis of the weld conservatively taken as 0° Resistance of weld material Vr2 = = 0.67 . fw. 0.707 . Size. xu. 1 + 0.5 . sin(q)1.5 1000 0.67 ×.67 ×0.707 ×10 ×483 × 1 + 0.5 ×sin(0 )1.5 1000 = 1.533 kN/mm Capacity of 10mm weld is 1.492kN/mm FOS = Resistance Force = 1.4916 .44301 = 3.367 13.13.2.2 (b) 30 250 30 I1 356x171x67 100 16 500 500 I1 356x171x67 Hole Size = 18 769 x 185 x 16 End Plate 100 Haunch cut from Beam section Height : 400 mm Width : 2000 mm All welds are Both Sides unless otherwise indicated All bolts are of Grade 8.8 All bolt diameters are 20 mm 30 400 369 100 30 93 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date S12 Beam - Column Connection - Ver W5.0.00 - 01 Apr 2022 Title : Code of Practice : SANS 10162-1:2011 Created : 2022/11/21 17:02:58 Notes and Assumptions 1 All references are formated "EC3 Part : Section" eg: 8 : 3.6.2(3)a. for Eurocode 1993-1-8 Section 3.6.2(3)a. All bolt holes are assumed to be normal clearance holes. All bolts are assumed to have threads in their shear planes. It is assumed that the connection is deep enough for the flanges to resist the compressive and tensile forces in them. It is assumed that compressive forces in flanges and stiffeners are conveyed through welds and not through bearing. Axial force in the column is not considered in the design. 2 3 4 5 6 Summary Summary of Forces and Capacities for Design to SANS 10162-1:2011 Check Member Type LC Applied Capacity Units % of Cap. ? 1 Weld Flange LC13 250.4 384.1 kN 65.2 O.K. 2 Weld Web LC12 51.5 1261.8 kN 4.1 O.K. 3 Column Web Tension Yielding LC13 250.4 701.3 kN 35.7 O.K. 4 Column Web Compression Crippling LC2 236.7 447.6 kN 52.9 O.K. 5 Column Web Compression Buckling LC2 236.7 812.3 kN 29.1 O.K. 6 Column Web Shear LC13 250.4 491.1 kN 51 O.K. 7 Bolts & Flange Tension & Bending LC13 138.5 234.2 kN 59.1 O.K. 8 Column Flange Bearing LC12 6.4 284 kN 2.3 O.K. 9 Bolts & End Plate Tension & Bending LC13 138.5 217.9 kN 63.6 O.K. 10 End Plate Bearing LC12 6.4 289.4 kN 2.2 O.K. 11 Bolts Shear LC12 6.4 84.4 kN 7.6 O.K. 12 Bolts Shear & Tension LC13 0.6 1.4 kN 43.7 O.K. 13 Bolts Slip N/A N/A N/A kN N/A N/A Input Job Number Sheet Job Title Client Your details here Calcs by Checked by General Settings Bolt Tension Analysis Plastic Bolt Type Bearing Bolt Grade 8.8 Member Ultimate Strength 450 Member Yield Strength 300 Weld Ultimate Strength 480 Connection Type Flush End Plate Column 356x171x67 I1 356x171x67 I1 Beam Column Extent Above Beam Angle (mm) Continuous 10 Haunch Depth (mm) 400 Haunch Length (mm) 2000 Date Job Number Sheet Job Title Client Your details here Calcs by Checked by Ultimate Limit State Loads in Beam Load Case Shear (kN) Axial (kN) Moment (kNm) SLS Factor (Divide to get Loads) LC1 33.36 56.68 149.85 1 LC2 34.40 57.93 154.50 1 LC3 16.26 28.18 72.92 1 LC4 27.57 46.53 123.73 1 LC5 -19.68 -36.94 -127.93 1 LC6 -57.48 -43.30 -116.95 1 LC7 -41.30 -40.15 -152.03 1 LC8 -18.63 -35.69 -123.28 1 LC9 -56.44 -42.05 -112.31 1 LC10 -40.25 -38.90 -147.38 1 LC11 -23.31 -43.47 -144.22 1 LC12 -61.11 -49.82 -133.25 1 LC13 -44.93 -46.68 -168.32 1 Width Extent Above Beam Flange End Plate Extent Below Haunch Thickness Stiffeners Width Top Stiffener Thickness Column Stiffeners Bottom Stiffener Thickness Shear Stiffener Thickness Shear Stiffener Orientation Layout Web Plates Thickness Top Backing Plate Thickness Bottom Backing Plate Thickness (mm) 185.2 (mm) N/A (mm) N/A (mm) 16 N/A (mm) None (mm) None (mm) None (mm) None None None (mm) 5 (mm) None (mm) None Bolts (mm) 20 N/A 2 2 N/A (mm) 100 Rows of Bolts Diameter Above Top Flange Below Top Flange Above Haunch Below Haunch Row Spacing Date Job Number Sheet Job Title Client Your details here Calcs by Bolt Offsets Welds Web Flange Above Haunch Beam & Haunch Flanges Beam Web Top Stiffener Bottom Stiffener Shear Stiffener Checked by Date (mm) 42 (mm) 30 (mm) 30 6 (mm) 6 (mm) N/A (mm) N/A (mm) N/A Check 1 : Capacity of the Beam Flange Welds The worst load is encountered for Load Case : LC13 when Fmax =250.4 kN The Capacity of the weld is the lesser of : 0.67 . fw. Am. f u Vr = 1000 = 13.13.2.2 13.13.2.2a 0.67 ×0.67 ×1 901.4 ×450 1000 = 384.092 kN 0.67 . f w. Aw. xu. 1.5 Vr = 1000 = 13.13.2.2b 0.67 ×0.67 ×1 300.376 ×480 ×1.5 1000 = 420.292 kN Beam Flange Weld is safe Check 2 : Capacity of the Beam Web Welds The worst load is encountered for Load Case : LC12 when Fmax =51.53 kN The Capacity of the weld is the lesser of : 0.67 . fw. Am. f u Vr = 1000 = 13.13.2.2 13.13.2.2a 0.67 ×0.67 ×8 281.289 ×450 1000 = 1 672.862 kN 13.13.2.2b Job Number Sheet Job Title Your details here Client Calcs by Vr = = Checked by Date 0.67 . fw. Aw. xu 1000 0.67 ×0.67 ×5 855.755 ×480 1000 = 1 261.751 kN Beam Web Weld is safe Check 3 : Capacity of the Column web in tension Opposite Top flange of the beam : The worst load is encountered for Load Case : LC2 when Tmax =173.65 kN The Capacity of the web is : 0.9 . tw. leff . f y Tr = 1000 = 0.9 ×9.1 ×285.422 ×300 1000 = 701.282 kN Opposite Bottom flange of the beam : The worst load is encountered for Load Case : LC13 when Tmax =250.4 kN The Capacity of the web is : 0.9 . tw. leff . f y Tr = 1000 = 0.9 ×9.1 ×285.422 ×300 1000 = 701.282 kN Column web is safe in tension Check 4 : Crippling Capacity of the Column web in Compression Opposite Top flange of the beam : The worst load is encountered for Load Case : LC13 when Bmax =196.627 kN The Capacity of the web is : 21.3a Job Number Sheet Job Title Your details here Client Calcs by Br = = Checked by Date fbi . tw. leff . f y 1000 0.8 ×9.1 ×204.942 ×300 1000 = 447.593 kN Opposite Bottom flange of the beam : The worst load is encountered for Load Case : LC2 when Bmax =236.674 kN The Capacity of the web is : Br = = fbi . tw. leff . f y 21.3a 1000 0.8 ×9.1 ×204.942 ×300 1000 = 447.593 kN Column web is safe in compression for crippling Check 5 : Buckling Capacity of the Column web in Compression Opposite Top flange of the beam : The worst load is encountered for Load Case : LC13 when Bmax =196.627 kN The Capacity of the web is : Br = = fbi . 640000 . twc. leff 21.3 hwc 2 . 1000 twc 0.8 ×640000 ×9.1 ×204.942 312 2 ×1000 9.1 = 812.299 kN Opposite Bottom flange of the beam : The worst load is encountered for Load Case : LC2 when Bmax =236.674 kN The Capacity of the web is : 21.3 Job Number Sheet Job Title Your details here Client Calcs by Br = = Checked by Date fbi . 640000 . twc. leff hwc 2 . 1000 twc 0.8 ×640000 ×9.1 ×204.942 312 2 ×1000 9.1 = 812.299 kN Column web is safe in compression for buckling Check 6 : Shear Capacity of the Column Web The worst load is encountered for Load Case : LC13 when Vmax =250.4 kN 0.55 . f . f y. tw. h Vr = 1000 = 13.4.1.2 0.55 ×0.9 ×300 ×9.1 ×363.4 1000 = 491.081 kN Column web shear is safe Check 7 : Bolt tension and Column Flange Bending The worst load is encountered for Load Case : LC13 Fmax = 138.483 kN Eurocode 1993-1 8 : 6.2.4 8 : Table 6.2 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date The resistance is the smaller of the 3 possible failure modes : Mode 1 : Complete yielding of the flange R1 = 4 . Mpl + 2 . Mbp. 1000 m = 4 ×2.37444 + 2 ×0 ×1000 33.84 = 280.667 kN Mode 2 : Bolt Failure with yielding of the flange R2 = = 2 . Mpl . 1000 + n . 2 . Bt m+n 2 ×2.37444 ×1000 + 40.05 ×2 ×156.8 33.84 + 40.05 = 234.248 kN Mode 3 : Bolt Failure only R3 = 2 . Bt = 2 ×156.8 = 313.600 kN Therefore R = R2 = 234.248 Bolt tension and Column Flange bending is safe Check 8 : Bearing on the Column Flange The Bearing Capacity of the flange at any Bolt is the lesser of : The worst load is encountered for Load Case : LC12 when Bmax =6.441 kN 3 . fbr. t . d . f u Br = 1000 = 13.10c 3 ×0.67 ×15.7 ×20 ×450 1000 = 284.013 kN Column flange bearing is safe Check 9 : Bolt tension and End Plate Bending The worst load is encountered for Load Case : LC13 Fmax = 138.483 kN Eurocode 1993-1 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date 8 : 6.2.4 8 : Table 6.2 The resistance is the smaller of the 3 possible failure modes : Mode 1 : Complete yielding of the End Plate R1 = = 4 . Mpl . 1000 m 4 ×2.78953 ×1000 43.2 = 258.290 kN Mode 2 : Bolt Failure with yielding of the End Plate R2 = = 2 . Mpl . 1000 + n . 2 . Bt m+n 2 ×2.78953 ×1000 + 40.05 ×2 ×156.8 43.2 + 40.05 = 217.883 kN Mode 3 : Bolt Failure only R3 = 2 . Bt = 2 ×156.8 = 313.600 kN Therefore R = R2 = 217.883 Bolt tension and End Plate bending is safe Check 10 : Bearing on the End Plate The Bearing Capacity of the Plate at any Bolt is : The worst load is encountered for Load Case : LC12 when Bmax =6.441 kN 3 . fbr . t . d . f u Br = 1000 = 3 ×0.67 ×16 ×20 ×450 1000 = 289.440 kN End plate bearing is safe Check 11 : Shear Capacity of the Bolts The worst load is encountered for Load Case : LC12 when Vmax =6.441 kN 13.10c Job Number Sheet Job Title Your details here Client Calcs by Checked by Date 13.12.1.2 The resistance of any bolt is : Vr = = 0.60 . fb. m . 0.7 . Ab. f u 1000 0.60 ×0.8 ×1 ×0.7 ×314.159 ×800 1000 = 84.446 kN Bolt shear is safe Check 12 : Shear and Tension Capacity of the Bolts The worst load is encountered for Load Case : LC13 The factor must be less than or equal to 1.4 : Vu Tu Factor = + Vr Tr = 4.518 69.242 + 70.724 126.292 = 0.6122 Bolt shear and tension is safe 13.12.1.4 Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces Bolt Forces for Load Case : LC1 47.96 kN 47.96 kN 0 kN 0 kN Shear force per bolt : 2.88 kN Bolt Forces for Load Case : LC2 49.51 kN 49.51 kN 0 kN 0 kN Shear force per bolt : 2.98 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC3 23.26 kN 23.26 kN 0 kN 0 kN Shear force per bolt : 1.39 kN Bolt Forces for Load Case : LC4 39.63 kN 39.63 kN 0 kN 0 kN Shear force per bolt : 2.38 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC5 4.97 kN 4.97 kN 52.49 kN 52.49 kN Shear force per bolt : 1.62 kN Bolt Forces for Load Case : LC6 6.58 kN 6.58 kN 50.02 kN 50.02 kN Shear force per bolt : 6.14 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC7 5.84 kN 5.84 kN 62.31 kN 62.31 kN Shear force per bolt : 4.21 kN Bolt Forces for Load Case : LC8 4.8 kN 4.8 kN 50.59 kN 50.59 kN Shear force per bolt : 1.52 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC9 6.4 kN 6.4 kN 48.12 kN 48.12 kN Shear force per bolt : 6.04 kN Bolt Forces for Load Case : LC10 5.66 kN 5.66 kN 60.4 kN 60.4 kN Shear force per bolt : 4.11 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC11 5.86 kN 5.86 kN 59.43 kN 59.43 kN Shear force per bolt : 1.93 kN Bolt Forces for Load Case : LC12 7.46 kN 7.46 kN 56.95 kN 56.95 kN Shear force per bolt : 6.44 kN Job Number Sheet Job Title Your details here Client Calcs by Checked by Date Bolt Forces for Load Case : LC13 6.72 kN 6.72 kN 69.24 kN 69.24 kN Shear force per bolt : 4.52 kN PRELIMINARY DESIGN 1 INNER PORTALS PLAN VIEW END GABLE SIDE VIEW PRELIMINARY DESIGN 2 INNER PORTALS PLAN VIEW END GABLE SIDE VIEW - All steelwork is of grade: S355JR - Bearing bolts: Grade 8.8 (only M20 in connections) - Welds: Grade E70XX - The steel sheeting to be used to cover the warehouse is: Supa-Clad IBR890 by Global Roofing Solutions (Pty) Ltd. (GRS) - All concrete to have a cube crushing strength of fcu=30 Mpa - All Rebar: Grade Y - Concrete cover to all rebar: 50mm for all foundations and 40 mm for ground slab - Warehouse dimensions: 60m long, 20m wide, 8m eaves height, Roof slope =10 degrees duo-pitch - Doors: 4x4 Roller shutter doors 7500 7500 7500 7500 7500 7500 5000 5000 5000 15000 5000 500 600 500 2750 600 3000 300 BASE PLATE CONCRETE PLINTH 150 1000 GROUND SLAB 400 SOIL BED FOUNDATION 500 600 1875 2750 - All steelwork is of grade: S355JR - Bearing bolts: Grade 8.8 (only M20 in connections) - Welds: Grade E70XX - The steel sheeting to be used to cover the warehouse is: Supa-Clad IBR890 by Global Roofing Solutions (Pty) Ltd. (GRS) - All concrete to have a cube crushing strength of fcu=30 Mpa - All Rebar: Grade Y - Concrete cover to all rebar: 50mm for all foundations and 40 mm for ground slab - Warehouse dimensions: 60m long, 20m wide, 8m eaves height, Roof slope =10 degrees duo-pitch 4 @ 2000 - Doors: 4x4 Roller shutter doors 3 @ 4000 1500 1000 1500 15 @ 1000 scale 1:2 6 @ 7500 - All steelwork is of grade: S355JR - Bearing bolts: Grade 8.8 (only M20 in connections) - Welds: Grade E70XX - The steel sheeting to be used to cover the warehouse is: Supa-Clad IBR890 by Global Roofing Solutions (Pty) Ltd. (GRS) - All concrete to have a cube crushing strength of fcu=30 Mpa - All Rebar: Grade Y - Concrete cover to all rebar: 50mm for all foundations and 40 mm for ground slab - Warehouse dimensions: 60m long, 20m wide, 8m eaves height, Roof slope =10 degrees duo-pitch - Doors: 4x4 Roller shutter doors INNER PORTAL RAFTER (WITH NO CRANE) ROLLER SHUTTER DOORS(CLOSED) INNER PORTAL RAFTER (SHOWING CRANE) ROOF BRACING (X BRACING) DOOR COLUMN GABLE END PORTAL DOOR BEAM WALL BRACING INNER PORTAL RAFTER - All steelwork is of grade: S355JR B - Bearing bolts: Grade 8.8 (only M20 RAFTER: 356x171x67 in connections) (I-SECTION) RAFTER-COLUMN BOLTS: M20 - Welds: Grade E70XX A - The steel sheeting to be used to cover the warehouse is: Supa-Clad IBR890 by Global Roofing Solutions (Pty) Ltd. (GRS) B - All concrete to have a cube crushing strength of fcu=30 Mpa EAVES HAUNCH CLADDING (THICKNESS=0.58mm) - All Rebar: Grade Y A - Concrete cover to all rebar: 50mm for all foundations and 40 mm for ground slab - Warehouse dimensions: 60m long, 20m wide, 8m eaves height, Roof SIDE RAILS 125x75x20x2 (CHANNEL) COLUMN: 356x171x67 slope =10 degrees duo-pitch (I-SECTION) - Doors: 4x4 Roller shutter doors RAFTER-COLUMN CONNECTION 186 728 200 PURLINS: 125x75x20x2 (CHANNEL) 400 200 RAFTER:356x171x67 SECTION A-A SCALE 1:2 (I-SECTION) SECTION B-B SCALE 1:2 - All steelwork is of grade: S355JR - Bearing bolts: Grade 8.8 (only M20 in connections) CLADDING (THICKNESS=0.58mm) RAFTER-RAFTER CONNECTION BY M20 BOLTS - Welds: Grade E70XX RAFTER: 356x171x67 (I-SECTION) - The steel sheeting to be used to cover the warehouse is: Supa-Clad IBR890 by Global Roofing Solutions (Pty) Ltd. (GRS) - All concrete to have a cube crushing strength of fcu=30 Mpa - All Rebar: Grade Y C - Concrete cover to all rebar: 50mm for all foundations and 40 mm for ground slab APEX HAUNCH-CRAWL BRAM CONNECTION BY M20 BOLTS - Warehouse dimensions: 60m long, 20m wide, 8m eaves height, Roof slope =10 degrees duo-pitch - Doors: 4x4 Roller shutter doors C APEX-CRAWL BEAM CONNECTION CRANE TROLLEY CRAWL BEAM: 305x165x40 (I-SECTION) CRANE TROLLEY APEX-CRAWL BEAM CONNECTION SCALE 1:2 TROLLEY DIRECTION OF MOTION SECTION C-C - All steelwork is of grade: S355JR - Bearing bolts: Grade 8.8 (only M20 in connections) CLADDING (THICKNESS=0.58mm) RAFTER-RAFTER CONNECTION BY M20 BOLTS - Welds: Grade E70XX RAFTER: 356x171x67 (I-SECTION) - The steel sheeting to be used to cover the warehouse is: Supa-Clad IBR890 by Global Roofing Solutions (Pty) Ltd. (GRS) - All concrete to have a cube crushing strength of fcu=30 Mpa - All Rebar: Grade Y C - Concrete cover to all rebar: 50mm for all foundations and 40 mm for ground slab - Warehouse dimensions: 60m long, 20m wide, 8m eaves height, Roof slope =10 degrees duo-pitch - Doors: 4x4 Roller shutter doors C APEX-CRAWL BEAM CONNECTION GABLE END COLUMN-CRAWL BEAM: CONNECTION BY M20 BOLTS CRANE TROLLEY CRAWL BEAM: 305x165x40 (I-SECTION) GABLE END COLUMN: 356x171x67 (I-SECTION) GABLE END COLUMN: 356x171x67 (I-SECTION) TROLLEY DIRECTION OF MOTION END STOP BEAM 305x165x46 (I-SECTION) SECTION C-C END GABLE COLUMN-CRAWL BEAM CONNECTION