Structural Analysis and Design ofResidential Buildings Using Staad.Pro, Orion, and Manual Calculations This should be a long post, but I am going to try and keep it as brief as possible. This post is more like an excerpt from the publication 'Structural Analysis and Design of Residential Buildings using Staad Pro V8i, CSC Orion, and Manual calculations'.... (See link below). Here, we are going to briefly present some practical analysis and design of some reinforced concrete elements using Staad Pro software, Orion software and manual calculations. Ultimately, we are going to make some comparisons of the results obtained based on the different methods adopted in the analysis and design. To learn how to model, design, and detail buildings from the scratch using Staad Pro, Orion, and manual methods, see the link at the end of this post. To show how this is done, a simplified architectural floor plans, elevations, and section, for a residential two storey building have been presented for the purpose of structural analysis and design (see the pictures below). Fig 1: Ground Floor Plan Fig.2: First Floor Plan 4: Back View .Fig.3: Front View Fig. from architectural drawings. and their interaction at the floor level under consideration. To have a good idea on how this can be done. For the architectural drawings above.A. .A.A. the columns. is shown in Figure 7. see the link at the end of the post.A.6: Left View The first step in the design of buildings is the preparation of the 'general arrangement'.5: Right View Fig. The G. the floor beams.A. below. Fig. the adopted G. There are no spelt out rules about how to prepare G. but there are basic guidelines that can guide someone on how to prepare a buildable and structurally efficient G. is a drawing that shows the disposition of the structural elements such as the slabs and their types. popularly called the G. Dimension of columns = (230 x 230mm) DESIGN OF THE FLOOR SLABS PANEL 1: MANUAL ANALYSIS The floor slab (PANEL 1) is spanning in two directions. since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.7: General Arrangement Design data: Fck = 25 N/mm2. Cnom (foundations) = 50mm Thickness of slab = 150mm. Cnom (slabs) = 25mm. Dimension of floor beams = 450mm x 230mm. Fyk = 460 N/mm2. Cnom (beams and columns) = 35mm. .Fig. 882k)] = z = d[0.042 Continuous edge = 0.0273)] = 0.034 Continuous edge = 0.87fyk z) As1 = (6. b = 1000mm (designing per unit width) k = MEd/(fckbd2 ) = (6.0475 × 106)/(25 × 1000 × 1192 ) = 0.6252 = 6.056 Long Span Mid-span = 0. Result from Orion showing the Short Span (mid span) design moments (Wood and Armer effects inclusive) (PANEL 1) .055 (say 1.0.(0.5+ √(0.825/3. Short Span Mid-span = 0.m MEd = 6. see the the link at the bottom of this post.25 .25 .625 = 1.0475 × 106)/(0. To see how to carry out deflections and crack control verifications.Hence.668 mm2/m Provide Y12mm @ 250mm c/c BOT (ASprov = 452 mm2/m) A little consideration will show that this provided area of steel will satisfy serviceability limit state requirements.95 × 119) = 133.882 × 0.1) Moment coefficients (α) for two adjacent edges discontinuous (pick from table).042 × 10. k = Ly/Lx = 3.9575 × 3.87 × 460 × 0.0171 Since k < 0.95d As1 = MEd/(0.0475 KNm Effective Depth (d) = h – Cc – ϕ/2 Assuming ϕ12mm bars will be employed for the construction d = 150 – 25 – 6 = 119mm.167 No compression reinforcement required z = d[0.045 Design of short span Mid span M = αnLx2 = 0.0475 KN.5+ √(0. Result from Staad showing the Short Span (mid span) design moments (Wood and Armer effects inclusive) (PANEL 1) . The full detailing of the floor slabs is as shown below.A little observation will show that the design moment values from the different methods are very similar. Figure 9: Bottom Reinforcement Detailing . . To see how to manually calculate the loading on beams. The internal forces from the loading is as shown below.Figure 10: Top Reinforcement Detailing Figure 11: Section of the floor slab DESIGN OF THE BEAMS Let us take Beam No 1 from our GA as a design case study: The loading of the beam has been carried out as shown below. follow the link at the end of the post. The beam is primarily subjected to load from slab. weight of wall. and its own self weight. . Load decomposition using finite element analysis was used for the load transfer.The internal forces from Orion software for Beam No 1 is as shown below. The internal forces from Staad software for Beam No 1 is as shown below. 87fykz) = (36.66 × 106)/(0.667 mm2 Provide 2Y16 mm BOT (ASprov = 402 mm2) The detailing of Beam No 1 is as shown below. .95 × 399) = 241.87 × 460 × 0.As1 = MEd/(0. 87 KN Load from floor beams = 105.14 KN Load from roof beams = 13.49 KN).49 = 88. and Beam No 2 (Support Reaction VA = 42.684 KN COLUMN A3 Total Columns Self weight = 12.41 + 11. Normally.99 = 26.6 KN Axial Load from Staad (A1) = 130.19 KN Axial Load from Orion (A3) = 202.27 KN) and Beam No 3 (Support Reaction VA = 12.8 of EC2. Let us use column A1 as example. uniaxial.46 = 46.85 = 166. For intermediate supports. Another method of calculating Column Axial Load is by Tributary Area Method.14 KN .33 + 60. COLUMN A1 Total Columns Self weight = 12.26 KN Load from floor beams = 46. In typical cases. the column is supporting beam No 2 (Support Reaction V1 = 13. At the first floor level (see Analysis and Design of Beam No 1 and 2). columns are usually rectangular or circular in shape. This method has not been adopted in this work.14 KN Load from roof beams = 35. or biaxial loads depending on the location and/or loading condition.13 KN Axial Load from Orion (A1) = 126. including the self weight of the column. Eurocode 2 demands that we include the effects of imperfections in structural design of columns.18 KN Total = 225.21 + 42. Therefore the summation of all these loads gives the axial load transferred from the beams. Columns are either subjected to axial. The column axial loads have been obtained by summing up the reactions from all the beams supported by the columns. At the roof level. and this in turn influences their mode of failure. the column is supporting Beam No 1 (Support Reaction V1 = 41.27 + 12.632 KN COLUMN A5 Total Columns Self weight = 12.99 KN). Column design is covered in section 5. they are usually classified as short or slender depending on their slenderness ratio. note that the summation of the shear forces at the support gives the total support reaction (neglect the signs and use absolute value.38 KN).DESIGN OF THE COLUMNS Loads from slabs and beams are transferred to the foundations through the columns.70 KN Total = 127.3 KN Axial Load from Staad (A3) = 201. 55 KN Total = 156. follow the link at the end of the post.64 + 37.15 + 9.10 NEd/fyd = (0.2. Design of Column E5 Reading from chart. 20ϕ.Load from roof beams = 17.25ϕ = 0.48 = 52.26 + 62.45 = 100.48 KN Axial Load from Orion (A7) = 133.9 KN Axial Load from Staad (A7) = 140.392 KN As you can see.9 KN Axial Load from Staad (A5) = 163.75 mm2 As.25 × 16 = 4mm < 6mm We are adopting Y8mm as links Spacing adopted = 200mm less than min{b.05 Area of longitudinal steel required (As) = (0.19 + 5.099 mm2 < 0.1 × 399.89 KN Load from floor beams = 83.14 KN Load from roof beams = 43.002 × 230 × 230 = 105.91 = 121.70 = 22. the axial loads from the three methods are very comparable.63 KN Load from floor beams = 38.207 KN COLUMN A7 Total Columns Self weight = 12. for design purposes.03288 NEd/(fckbh) = (399. h.05 × 25 × 230 × 230)/460 = 143. 400mm} Result from Orion for column E5 .58 KN Axial Load from Orion (A5) = 155.887)/400 = 0.71 KN Total = 165. To see how to obtain the column design moments from the use of sub-frames.002 × 106)/(25 × 230 × 2302 ) = 0.min = 0.302 From the chart: (AsFyk)/(bhfck ) = 0.8 mm2 Provide 4Y16mm (Asprov = 804 mm2) Links Minimum size = 0.88 × 103)/(25 ×230 × 230) = 0. d2/h = 0. MEd/(fck bh2 ) = (10. Result from Staad for column E5 . Staad Provided Y8@225mm links The column detailing is as shown below. . It is therefore imperative that adequate care be taken in the design of foundations. Foundation design starts from detailed field and soil investigation. Analysis and Design of footing E8 Bearing Capacity of the foundation = 150 KN/m2. so that the performance of the foundation can be guaranteed. . It is very important to know the index and geotechnical properties of the soil. including the soil chemistry.DESIGN OF FOUNDATIONS All loads from the superstructure of a building are transferred to the ground. If the foundation of a building is poorly designed. then all the efforts input in designing the superstructure is in vain. d = 400 – 50 – 6 = 344mm The ultimate limit state design moment can be obtained by considering the figure below.Effective depth Concrete cover = 50mm AssumingY12mm bars. . 0013 × b × d (447.167 = 3.95d As1 = MEd/(0.25 .887 × 103)/(4(230) × 344) = 1.518 × 106)/(0. ASmin = 498.7 mm2 Check if ASmin < 0.5+ √(0.8√[(16.1 EC2) ASmin = 0. Provide Y12 @ 200mm c/c (ASprov = 565 mm2/m) each way Shear at the column face Ultimate Load on footing from column = 399.6[1 – (25/250) ] = 0.max = 0.25 .167 No compression reinforcement required z = d[0.5649 N/mm2 (Table 3.87 × 460 × 0.5 = 14.15 × 399.3 × (fck)(2⁄3) = 0.2 mm2) Since.518 × 106)/(25 × 1000 × 3442 ) = 0.146 Where uo is the column perimeter and d is the effective depth vEd = βVEd/(u0d) = (1.4. fctm = 0.max = 0.54 N/mm2 fcd = (αcc fck)/γc = (0.5 × 0.(0.k = MEd/(fck bd2) = (37.6[1 – (fck/250) ] = 0.26 × 2.99/230)2] = 1.5649/460 ×1000 × 344 = 498.26 × fctm/Fyk × b × d = 0.85 × 25)/1. the provided reinforcement is adequate.max.887 kN Design shear stress at the column perimeter vEd = βVEd/(u0d) β is the eccentricity factor (see section 6.5vfcd v = 0.95 × 344) = 286.87fykz) = (37.54 × 14.5+ √(0.882 × 0.825 N/mm2 vEd < VRd.3 of EC2) β = 1+ 1.167 N/mm2 VRd.48/230)2+(8.869 mm2/m To calculate the minimum area of steel required.7 mm2.0.01268 (designing per metre strip) Since k < 0.3 × 25 (2⁄3) = 2. This is very ok Transverse shear at ‘d’ from the face of column .0273)] = 0.452N/mm2 VRd.882k)] = z = d[0. ΔVEd = (189. No further check required.σcp] × (2d/a) ≥ (Vmin + k1. d = a Hence. No shear reinforcement is required. k = 1. Punching Shear at 2d from the face of column Punching shear lies outside the footing dimensions.7624 Vmin = 0.7624 (100 × 0.1503 KN).15325 N/mm2 VRd.σ.12 × 1.6772) > VEd (0.4094 N/mm2 ρ1 = As/bd = 565/(1000 × 344) = 0. VRd.c = [0.7624)(3/2) × (25)(1/2) = 0.035k(3/2) fck(1/2) = Vmin = 0.4365 m2 = 79.c = 0.077 KN vEd = VEd/bd = (79.c = 2 × 0. VRd.c = [CRd.4365 m2 Therefore.386 + 175.0.001642 < 0. therefore.18/γc = 0.077 × 103)/(1500 × 344) = 0.635 – 0.d CRd.035 × (1.001642 ×25 )(1/3) ] = 0.c k (100ρ1 fck )(1/3) + k1.3386 = 0.3386 N/mm2 × (2d/a) < Vmin But in this case.291m Area of shaded area = (1.12 k = 1+√(200/d) = 1+√(200/344) = 1.7624 > 2.Width of shaded area = a – d = 0.18/1. Design Result from Orion .02.5 = 0.344 = 0.sub>cp) bw.c (0.291m) = 0.6772 N/mm2 Since VRd.939)/2 × 0.5m × 0. The detailing of the footing is as shown below. . The structural analysis and design of all members have been fully done. . including a step by step tutorial on how to model and design on Orion and Staad Pro. and how to manually design. See the completed models below. .Fully completed model on Orion Fully completed model on Staad Pro.
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