Analysis and Design of Structures Using Struds Software

April 2, 2018 | Author: Ramachandra Sahu | Category: Truss, Trigonometric Functions, Structural Engineering, Beam (Structure), 2 D Computer Graphics


Comments



Description

Global J. of Engg. & Appl.Sciences, 2012: 2 (3) Research Paper: Suresh, 2012: Pp.275-277 ANALYSIS AND DESIGN OF STRUCTURES USING STRUDS SOFTWARE Suresh, B Dept. of Civil Engineering, Adama Science and Technology University, Ethiopia. ABSTRACT Currently there are several commercially available softwares in market for analysis and design of Civil Engineering structures like ETABS, STAAD Pro and STRUDS etc. Among these STRUDS is an ideal software solution for the usage of structural engineers for the analysis of 2D and 3D structures and design of different R.C.C and Steel components such as slabs, beams, columns, footings and trusses with design sketches. Struds has inbuilt graphical data generator to model the geometry of building structure. It performs analysis using stiffness matrix method and finite element method for maximum solution, accuracy and reliability. Struds performs the integrated design by limit state method of all R.C.C. components of the structure by directly reading the analysis results. If any component fails the program gives you warning messages and suggests you the possible alternative for design. Struds prepares graphical outputs in the form of drawings and diagrams. Design results in the text form of schedules, quantities and details are produced. The design process is highly interactive and extremely user friendly. It facilitates to change the design parameters anywhere in between the design process and redesign the structure. These changes are automatically reflected in graphical and numerical output form. Struds also enables to produce the working drawings in AUTOCAD. Keywords: Software, struds and structural engineer. INTRODUCTION User can idealize a building structure in the form of Plane Truss -> Create -> Predefined Truss -> plane grid, plane frame or space frame. Plane grid is Standard Truss is a two dimensional structure having no horizontal -> Select Truss type -> Here we take Fan - Truss -> (Global X, Y) movement of the structure. The OK degrees of freedom available are Fz, Mx, My. Geometrical Parameters Dialog Box appears. Set the Columns are modeled as supports with only Fz properties as desired Let us set restraint (Fig. 1). A Plane frame structure is bound  Span of Truss = 10 m by a global X-Z or Y-Z coordinate system with loads  Rise of Truss = 4 m -> OK in the same plane. It has three degrees of freedom Property -> Fx, Fz and My in X-Z plane and Fy, fz and Mx in Y-Z -> Create -> Section -> create a section say sec1. plane (Fig. 2). Space frame is a three dimensional -> Load -> Nodal Load -> we create two Nodal structure with loads applied in any plane and has Loads six degrees of freedom Fx,Fy,Fz,Mx,My and Mz(Fig. N1 = 10 kN and N2 = 20 kN 3) (Johansson and Veljkovic, 2001 and Karlstrom, -> Attach -> material (steel) to all elements. 2004). And plane truss consists of truss members -> Section to all elements. which have only axial member forces no bending. It -> Load -> Nodal Load (N2 to Node 16 and N1 to has two degrees of freedom that is transitional in Node 14 and 18). that plane (Fig. 4). In Struds user can start floor grid Support -> In predefined trusses supports are structure and he can transform the same as plane already there. If one wants to change he can. frame or space frame after attaching columns. View The default supports are hinged at Node 1 and and space frame or any of the plane frame Roller at Node 11. generated (Ng and Gardner, 2007 and Wu, 2011). Save -> the truss (say fan.bld) Truss -> Analysis Files -> Current Truss -> save Problem: files in the same folder. Type: Plane Truss Analysis with Nodal Loads (Fan - Close the Truss and save the building (Fig 6). Truss) Purpose: Compare Theoretical results with STRUDS Results: STRUDS. Axial force in the element:Element No:25 Problem: Determine the Axial Forces in all the members of Plane Fan - Truss as shown in figure-5. Distance 1 Find also the reactions at support 0.000 -20.000 4.000 -20.000 Modeling: M factor 1.000 For modeling open a new building file in the Elemental results(Axial,B.M)-Plane struss Preprocessor structure. Preprocessor -> Building -> New Load combination: 1.0 D.L+1.0 L.L 275 ISSN 2249-2631(online): 2249-2623(Print) - Rising Research Journal Publication 47 0.000 -20.98 0.6598 Length of element 25.46 0.39 0.918 -24. Sciences.451 7. Axial Force in Theoretical STRUDS % Deviation Element Results Results P2-13 = 0 1 and 10 32.39 0. For €H = 0. 2012: 2 (3) El. 2 and 9 32.P3-4 * cos (theta) = 21 STRUTS 2-12 0.01 0.47 0. 16 Main-Tie 16-17 1.093 m 12 Main-Tie 12-13 0.02 Negligible For €H = 0.0 2 RAFTER 2-3 1.08 13 and 18 25 25 Nil For €V = 0.0 On solving these two equations (1) and (2).05 Negligible For €V = 0.00 0. L14-15 = 1.No Group Node Length Axial B.400 27. L4-14 = 2. P6-15 = P4-15 * cos (Ø) / cos (Ø1) = 7. P15-16 = P14-15 .01 P2-3 = P1-2 = 32.P4-5 * sin (theta) = 17 Main-Tie 17-18 1.0 0 --------(2) 23 STRUTS 4-14 2.44 0. 14 and 17 25 25 Nil P3-13 = 0 15 and 16 19.0 14 Main-Tie 14-15 1.204 -10.03 Negligible At Node 14: 34 and 32 7.4687 kN 26 STRUTS 7-17 3.46 0.176 32. P4-15 = 7. The Fan . P2-12 = 0 (As there is no vertical force / load at Node P6-16 = 20 (Applied Nodal Force at Node 16) 12) For €H = 0.0 At Node 5: Theoretical results: Resolving along the inclined element 5.98 0.152 m cos (theta) = 5 / sqrt (41) Angle Ø1 is given as sin (theta) = 4 / sqrt (41) tan (Ø1) = 1.0 10 RAFTER 10-11 1.00 0. P16-17 = P15-16 = 19.04 Resolving along the inclined element 2. P4-13 = 0 8 RAFTER 8-9 1.98 0.01 At Node 13: 21 and 29 0 0 Nil 22and28 0 0.015 32. Global J.176 32. 25 20 20 Nil 30 and 36 7.04 0.152 / 4 => Ø1 = 16.000 kN Comparison between theoretical and struds At Node 2: results: Resolving perpendicular to the inclined element 2.02 0.000 kN = 19. For €V = 0.079 -0.00 0.00 For €H = 0. this gives theta = 38. 20 Main-Tie 20-11 0.4031 m Force in member 24.0 18 Main-Tie 18-19 0.Rising Research Journal Publication .00 0.176 32. For €V = 0.0 P4-15 * sin (Ø) + P4-5 * cos (theta) . L6-16 = 4 m (degrees) Length of element 15. of Engg.0 25 STRUTS 6-16 4.00 0.01 0.0 10 --------(1) 19 Main-Tie 19-20 0. we get 24 STRUTS 5-15 3.015 32.093 -25.152 -19.10 For €V = 0.469 27.015 kN 3 and 8 32.0 Force in member 31. 11 and 20 25 24.176 32.00 0.3776 degrees 15 Main-Tie 15-16 1.000 0.Truss taken is symmetrical in loading P5-6 = P4-5 = 27.918 -25.00 At Node 3: 5 and 6 27.0 For €H = 0.000 kN 31 and 37 0 0. P5-15 = 0 The angle theta is given as: Force in member 34.015 kN 12 and 19 25 24.98 0.475 27.15 P13-14 = P12-13 = 25.9916 kN 27 STRUTS 8-18 2.47 0. 3 RAFTER 3-4 1. Resolving perpendicular to the inclined element 5.M P4-14 = 10 (Applied Nodal Force at Node 14) 1 RAFTER 1-2 1.093 / 2. P4-14 = 10 kN 9 RAFTER 9-10 1.01 0.00 0.735 0.98 0. & Appl. so we will analyze the half truss only.0 6 RAFTER 6-7 1.204 -10.0 At node 4 there are 5 elements 7 RAFTER 7-8 1.918 -25.400 27. L15-16 = 1.015 32.07 4 and 7 27.46 0.9916 kN tan (theta) = 4 / 5. Total vertical load = 10 + 20 + 10 = 40 kN P5-15 = 0 Reaction at both Node = 40 / 2 = 20 kN At Node 15: The inclined length = sqrt (4 * 4 + 5 * 5)= sqrt (41) At node 15 there are 5 elements = 6.04 0.00 0.919 -24.388 19.204 m 11 Main-Tie 1-12 0.39 0.475 27.0 tan (Ø) = 1.0 Length of element 14.015 kN For €H = 0.469 27.050 0.P4-15 * sin (Ø) .04 0.0 At Node 4: 5 RAFTER 5-6 1.0 Angle Ø is given as 13 Main-Tie 13-14 0.093 -25.204 => Ø = 26.919 -24.98 0. 35 and 33 0 0.992 8.020 0.3875 kN P1-12 = P1-2 * cos (theta) = 25.152 -19.04 Negligible 276 ISSN 2249-2631(online): 2249-2623(Print) .0 For €H = 0.079 -0.0 P4-15 = 7.0 Length of element 23.0 22 STRUTS 3-13 1.P6-15 * sin (Ø1) P1-12 = P1-2 * cos (theta) = 25. For €H = 0.000 kN 4 RAFTER 4-5 1.176 32.0664 degrees At Node 1: For €V = 0.176 32.08 P3-4 = P2-3 = 32.0 P14-15 = P13-14 = 25.0 P4-5 = 27.00 0.00 0.45057 kN P1-2 = 20 / sin (theta) = 32.918 -24. 23 and 27 10 10 Nil P4-13 = 0 24 and 26 0 0.4687 kN also.0 P4-15 * cos (Ø) + P3-4 * sin (theta) .0 For €V = 0.3875 kN At Node 12: At Node 16: For €V = 0.469 0.00 0.0 Force in member 23. Earth quake The analysis and design by using STRUDS software loads and wind loads. components — A consistent approach. 2. KT and L. Technology. Three dimensional structure model Fig. User can Karlström. Fz restraint model Fig. Progress in Structural Engineering given. Luleå University of grade steel in design of slabs. trapezoidal and Materials. In relation to other consider secondary moments due to eccentric soft wares ETABS file and it’s analysis file can be seating beams . 2004. 5.Truss Fig. 2007. And also Floating columns can be advantages to the users as specified. Perform. P. 4. 3.Use of Fe 550 Licentiat Thesis. 2004 and 2006). 3(1):13–27. 2006. Veljkovic. Journal of account by STRUDS by using Master-slave concept. ”Special Issue on Protection of Reinforced concrete columns. Facility to view shear wall detail for selected floors Johansson. Gardner. The user can account for the REFERENCES soft story effect in Struds for seismic analysis. I and tubular sections. Thus the software is good for defined detailing one option to decide maximum using analysis and design of structures. 130(10):1586– located on sloping hill sides is possible. Numerical However the storey which is to be decided as a soft modelling of stainless steel structural storey needs to be defined explicitily by the user. But It is STRUDS file in ETABS. Stress Ng. 2004. Journal Modeling and analyzing buildings which may be of Structural engineering. B and M. Which 601.A. Beam design is done for worst of all load CONCLUSION combinations including pattern loads. provides an unique feature to modify the footing Gardner. 5. 62:532–43. Engineering including principal and von mises stresses is Structures. 2001. Global J. shear walls and footings. Fig. shear walls and footings is implemented. 2011. Transitional plane Fig.Rising Research Journal Publication . basic sections such as equal and unequal angles. of Engg.columns. Pp. Sweden. & Appl. Analysis File . slabs and flat slabs also can be designed. Sciences. The effect of shear walls can be taken into design of stainless steel structures. columns. And also providing other laps in detailing.” J. 1. L and N. Buckling of stainless contours to visualize the behavior of structure steel columns and beams in fire. Analysis and design recommendations.Truss ************** 277 ISSN 2249-2631(online): 2249-2623(Print) . 2012: 2 (3) Other advantages: Though it is not possible to channels. Facil. L and D. considered in Struds. possible. simple and bar length is given which will be helpful in deciding also user friendly. Struds analyze a truss supported by Wu.And no stress concentration effect imported the same in STRUDS and also export consideration due to cut outs in slabs. Gardner. beams. Nethercot. In user manual calculations. Thin-walled steel studs in fire: specify drawing parameters for slabs. ASCE. is possible to create compound sections from the Constr. Baddoo. Constructional Steel Research.R. It is possible to design beams of inclined in plan. Along these triangular slabs . Plane Fan . In beam design parameters is given results with negligible difference with user defined detailing option is given. Steel plated by selecting default levels for that particular floor is structures. C. And in steel trusses it Structures against Blast Loading. beams . 29(7):717–30. Also possible to import possible to analyze beams of irregular sections that geometry from AUTOCAD to STRUDS (Gardner and is concrete I beams and beams having hexagonal Nethercot. 73. section. Three degree freedom model Fig. 25: 358–359. Fire testing and level.
Copyright © 2021 DOKUMEN.SITE Inc.