ECV 432 Steel and Timber Design Lecture 2 PDF

ECV 432 STEEL AND TIMBER DESIGN Lecture 2 Steel Purlins Steel Beams 1  STEEL STRUCTURES The major elements in steel structures are: Purlins - beam members carrying roof sheeting and cladding Beams and girders - members carrying lateral loads in b...

ECV 432 STEEL AND TIMBER DESIGN Lecture 2 Steel Purlins Steel Beams 1  STEEL STRUCTURES The major elements in steel structures are: Purlins - beam members carrying roof sheeting and cladding Beams and girders - members carrying lateral loads in bending and shear Steel and Timber Design ROLLED AND FORMED SECTIONS  PURLINS Sections used for purlins and sheeting rails Steel and Timber Design  PURLINS Roof materials Steel and Timber Design PURLINS Sag rods Roof Elements Steel and Timber Design  PURLINS Sag rods Conservative design details Steel and Timber Design  PURLINS Angle at full plasticity Steel and Timber Design PURLINS Steel and Timber Design  Design of Purlins Question Design the purlin in S275 steel for the single portal frame structure as shown in the diagram. Columns/Rafters are spaced at 6m. Use channel section as purlins at 2m centres and roof angle is 6°. Imposed load = 0.75kN/m2 Dead load = 0.45kN/m2 PURLINS Steel and Timber Design  Solution to purlin design For slopes < 10°, frame may be analysed as a rectangular single portal frame structure. Imposed load Qk = 6m x 2m x 0.75kN/m2 = 9kN Dead load Gk = 6m x 2m x 0.45kN/m2 = 5.4kN Ultimate design Load (w) w = (1.4 x Gk ) + (1.6 x Qk) = (1.4 x 5.4) + (1.6 x 9) = 21.96kN 21.96 UDL = = 3.66𝑘𝑁/𝑚 6 Steel and Timber Design Solution to purlin design cont’d… 𝑤𝑙 2 3.66 × 62 M= = = 16.47𝑘𝑁𝑚 8 8 𝑤𝑙 3.66 × 6 𝐹𝑣 = = = 10.98𝑘𝑁 2 2 Check moment capacity (4.2.5.2) 𝑀 16.47 × 105 3 S> = = 59.89𝑐𝑚 𝑝𝑦 275 × 102 S = Plastic modulus in major axis py = Design Strength M = Design Moment Try 150 x 75 channel section Steel and Timber Design  Solution to purlin design cont’d… Try 150 x 75 channel Sx = 132cm3 Zx = 115cm3 ry = 2.4cm x = 13.1 u = 0.945 D = 150mm B = 75mm t = 5.5mm T = 10.0mm r = 12mm d = 106mm b/T = 7.5 d/t = 19.3 I = 861cm4 Section classification (Table 11) 1 1 275 2 275 2 𝜀= = =1 𝑝𝑦 275 𝑏 𝑑 < 9𝜀 = 7.5 < 9 < 40𝜀 = 19.3 < 40 𝑇 𝑡 Hence section is plastic  Solution to purlin design cont’d… Check Shear capacity (4.2.3) 𝑇ℎ𝑒 𝑠ℎ𝑒𝑎𝑟 𝑎𝑟𝑒𝑎, 𝐴𝑣 = 𝑡𝐷 = 5.5 × 150 = 825𝑚𝑚2 𝑆ℎ𝑒𝑎𝑟 𝑐𝑎𝑝𝑐𝑖𝑡𝑦, 𝑃𝑣 𝑃𝑣 = 0.6 × 𝑝𝑦 × 𝐴𝑣 = 0.6 × 275 × 825 = 136.13𝑘𝑁 Determine Low Shear or High Shear (4.2.5.2 and 4.2.5.3) 0.6𝑃𝑣 = 0.6 × 136.13 = 81.68𝑘𝑁 For low shear, 𝐹𝑣 < 0.6𝑃𝑣 10.98𝑘𝑁 < 81.68𝑘𝑁, 𝑡ℎ𝑢𝑠 𝑳𝑶𝑾 𝑺𝑯𝑬𝑨𝑹 Steel and Timber Design  Solution to purlin design cont’d… Moment capacity (4.2.5.2) 𝐹𝑜𝑟 𝑐𝑙𝑎𝑠𝑠 1 𝑝𝑙𝑎𝑠𝑡𝑖𝑐, 𝑀𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑝𝑐𝑖𝑡𝑦, 𝑀𝑐 𝑀𝑐 = 𝑝𝑦 𝑆𝑥 = 275 × 103 × 132 × 10−6 = 36.3𝑘𝑁𝑚 But 𝑀𝑐 < 1.2𝑝𝑦 𝑍𝑥 (4.2.5.1) 𝑀𝑐 < 1.2 × 275 × 115 × 10−3 = 37.95𝑘𝑁𝑚 Steel and Timber Design  Solution to purlin design cont’d… Deflection (2.5.2 & Table 8) 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑙𝑖𝑚𝑖𝑡𝑠 𝑓𝑜𝑟 𝑝𝑢𝑟𝑙𝑖𝑛𝑠 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑠𝑝𝑎𝑛 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑖𝑛 𝑡𝑎𝑏𝑙𝑒 8, ℎ𝑜𝑤𝑒𝑣𝑒𝑟, 𝑖𝑠 𝑢𝑠𝑢𝑎𝑙𝑙𝑦 𝑢𝑠𝑒𝑑. 200 𝑠𝑝𝑎𝑛 6000 𝛿= = = 30𝑚𝑚 200 200 5𝑊𝐿3 5 × 9 × 103 × 60003 𝛿= = = 14.34𝑚𝑚 384𝐸𝐼 384 × 205 × 10 × 861 × 10 3 4 Hence deflection is checked. Therefore, section is adequate. Steel and Timber Design THIS SLIDE IS INTENTIONALLY LEFT BLANK 18  BEAM DESIGN Beams are usually horizontal structural elements used primarily to resist loads (bending, shear, etc.) applied laterally to the beam’s (major) axis. However, deflections and local stresses are also important. Beams may be cantilevered, simply supported, fixed ended or continuous. Beams are used to: – Support floors and columns – Carry roof sheeting as purlins and side cladding as sheeting rails.  BEAM DESIGN Types of beams BEAM DESIGN  BEAM DESIGN Rectangular Hollow section Typical beam sections  BEAM DESIGN Simply supported beam with UDL Bending moment diagram Simply supported beam  BEAM DESIGN Fixed supported beam with UDL Bending moment diagram Beam with fixed support  BEAM DESIGN The projecting flange of an I-beam will buckle if it is too thin. Webs will also buckle under compressive stress from bending and from shear. Types of buckling  BEAM DESIGN Lateral and torsional restraint  BEAM DESIGN Behaviour in bending STRESS-STRAIN CURVE 28  Question The beam shown below is fully restrained along its length and has stiff bearing of 50mm at the supports and 75mm under the point load. Design the beam in S275 steel, given the serviceability imposed loads as wi = 30kN/m and Wi = 50kN for UDL and point load respectively. The design moment (M) = 585kNm, shear force at end (Fve) = 292kN, shear force at centre (Fvc) = 67.8kN Take Young’s modulus (E) = 205 kN/mm2. Factored (point) load from secondary beam =136 kN  Solution 1. Section Trial - Check moment capacity (4.2.5.2) Steel grade = S275 hence, 𝑝𝑦 = 275𝑁/𝑚𝑚2 𝑀 585 × 105 3 S> = = 2127.27𝑐𝑚 𝑝𝑦 275 × 102 Try 533 x 210 x 92 UB in grade S275 Steel and Timber Design Solution cont’d… Steel and Timber Design  Solution cont’d… 2. Section classification (Tables 9 & 11) Steel grade = S275 T < 16 mm hence, 𝑝𝑦 = 275𝑁/𝑚𝑚2 1 1 275 2 275 2 𝜀= = =1 𝑝𝑦 275 𝑏 𝑑 < 9𝜀 = 6.71 < 9 < 80𝜀 = 47.2 < 80 𝑇 𝑡 Hence section is class 1 plastic Steel and Timber Design  Solution cont’d… 3. Check for shear buckling (4.2.3) If the d/t ratio exceeds 70ε for a rolled section, the web should be checked for shear buckling in accordance with clause 4.4.5 of BS 5950-1:2000. In this case, d/t = 47.2< 70ε, so there is no need to check for shear buckling In this instance, d/t = 47.2< 70ε, so there is no need to check for shear buckling Steel and Timber Design  Solution cont’d… 4. Check Shear capacity (4.2.3) 𝑆ℎ𝑒𝑎𝑟 𝑎𝑟𝑒𝑎, 𝐴𝑣 = 𝑡𝐷 = 10.1 × 533.1 = 5384.31𝑚𝑚2 𝑆ℎ𝑒𝑎𝑟 𝑐𝑎𝑝𝑐𝑖𝑡𝑦, 𝑃𝑣 𝑃𝑣 = 0.6 × 𝑝𝑦 × 𝐴𝑣 = 0.6 × 275 × 5384.31 = 888.41𝑘𝑁 Determine Low Shear or High Shear (4.2.5.2 and 4.2.5.3) 0.6𝑃𝑣 = 0.6 × 888.41 = 533.05𝑘𝑁 For low shear, 𝐹𝑣𝑒 < 0.6𝑃𝑣 292𝑘𝑁 < 533.05𝑘𝑁, 𝑡ℎ𝑢𝑠 𝒍𝒐𝒘 𝒔𝒉𝒆𝒂𝒓 Steel and Timber Design  Solution cont’d… 5.a. Moment capacity (4.2.5.2) Check if there is high or low shear (i.e. Fv > 0.6 Pv ) at the point of maximum moment. At the centre of the beam, Fvc = 67.8kN 0.6𝑃𝑣 = 0.6 × 888.41 = 533.05𝑘𝑁 For low shear, 𝐹𝑣𝑐 < 0.6𝑃𝑣 67.8𝑘𝑁 < 533.05𝑘𝑁, 𝑡ℎ𝑢𝑠 𝒍𝒐𝒘 𝒔𝒉𝒆𝒂𝒓 Steel and Timber Design  Solution cont’d… 5.b. Moment capacity check cont’d (4.2.5.2) 𝐹𝑜𝑟 𝑐𝑙𝑎𝑠𝑠 1 𝑝𝑙𝑎𝑠𝑡𝑖𝑐 with low shear, 𝑀𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑝𝑐𝑖𝑡𝑦, 𝑀𝑐 𝑀𝑐𝑥 = 𝑝𝑦 𝑆𝑥 = 275 × 103 × 2360 × 10−6 = 649𝑘𝑁𝑚 To avoid irreversible deformation under serviceability loads, check limit (4.2.5.1) For a simply supported beam 𝑀𝑐𝑥 < 1.2𝑝𝑦 𝑍𝑥 𝑀𝑐𝑥 < 1.2 × 275 × 2070 × 10−3 = 683.1𝑘𝑁𝑚 𝑀𝑥 < 𝑀𝑐𝑥 = 585 < 683.1𝑘𝑁𝑚 Steel and Timber Design Therefore, the moment capacity is adequate  Solution cont’d… 6.a. Web bearing and buckling under the point load (Bearing capacity of unstiffened web) - 4.5.2.1 𝐵𝑎𝑠𝑖𝑐 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 𝐹𝑥 ≤ 𝑃𝑏𝑤 = 601.33kN 𝐹𝑥 ≤ 𝑃𝑏𝑤 = 136 < 601.33𝑘𝑁 Thus, the bearing capacity of the unstiffened web under the point load is adequate.  Solution cont’d… 6.b. Buckling resistance of the unstiffened web - 4.5.3.1 𝐵𝑎𝑠𝑖𝑐 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 𝐹𝑥 ≤ 𝑃𝑥 25𝜀𝑡 𝑃𝑥 = 𝑃𝑏𝑤 𝜀=1 (𝑏1 +𝑛𝑘) 𝑑 25 × 1 × 10.1 𝑃𝑥 = × 601.33 = 472.73𝑘𝑁 216.5 × 476.5 𝐹𝑥 ≤ 𝑃𝑥 = 136 < 472.73𝑘𝑁 Therefore, the buckling resistance of the unstiffened web under the point load is adequate.  Solution cont’d… 7.a. Web bearing and buckling at the support - Bearing capacity of the unstiffened web (4.5.2.1) Basic Requirement 𝐹𝑥 ≤ 𝑃𝑏𝑤 = 296.08kN 𝐹𝑥 ≤ 𝑃𝑏𝑤 = 292 < 296.08𝑘𝑁 Thus, the bearing capacity of the unstiffened web under the point load is adequate.  Solution cont’d… 7.b. Buckling resistance of the unstiffened web (4.5.3.1) Basic Requirement 𝐹𝑥 ≤ 𝑃𝑥 𝐹𝑥 ≤ 𝑃𝑥 = 292 > 178𝑘𝑁 Load-carrying stiffener Load-carrying stiffener Load-carrying stiffener  Solution cont’d… 8.a. Deflection (2.5.2 & Table 8) The serviceability loads are taken as the unfactored imposed loads, i.e.: Distributed imposed load wi = 30kN/m Imposed point load Wi = 50 kN E = 205 kN/mm2 L = 6.5m Ix = 55200cm4 𝑠𝑝𝑎𝑛 6500 𝑇ℎ𝑒 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑙𝑖𝑚𝑖𝑡 𝑖𝑠 = =18.1mm 360 360  Solution cont’d… 8.b. Deflection (2.5.2 & Table 8) 5𝑤𝑖 𝐿4 𝑊𝑖 𝐿3 1 5𝑤𝑖 𝐿4 𝑊𝑖 𝐿3 𝛿= + = + 384𝐸𝐼 48𝐸𝐼 𝐸𝐼 384 48 1 5 × 30 × 65004 50 × 103 × 65003 𝛿= 3 4 + 205 × 10 × 55200 × 10 384 48 = 8.69𝑚𝑚 8.69 mm < 18.1 mm, hence the deflection is satisfactory. Use 533 x 210 x 92 UB in grade S275 (with stiffeners at the supports)  Homework The diagram below shows a portion of the third floor of a banking hall. The structure is made up of steel columns and beams with 150mm reinforced concrete slab. Design the beam on gridline B in S275 to satisfy the requirements of ultimate and serviceability limits states. The beam is laterally restrained and simply supported at both ends. Floor finishes: 600x600x10 porcelain tiles, with 40mm screed backing. Assume materials and loading for ceiling, MEP, partition, etc. Homework cont’d… Beam-arrangement Homework cont’d… Structural frame ECV 432 STEELAND TIMBER DESIGN Lecture 3 COLUMN DESIGN Steel Column Design ✓ Column subjected to axial load Columns or stanchions, struts, are structural elements subjected to bending and compression. Compression members must resist buckling, so, they tend to be stocky Tubular (hollow) sections are the ideal Universal columns (H-sections) are equally good Compression member sections Beams column joints – Factory or single-storey building Beams column joints - multi-storey building Types of crane columns  Crane beam to column connection Classification of cross-sections The same classification that is set out for beams is used for compression members. That is, to prevent local buckling. Limiting proportions for flanges and webs in axial compression are given in Table 11, BS 5950-1 AXIALLY LOADED COMPRESSION MEMBERS Behaviour of members in axial compression EFFECTIVE LENGTHS Theoretical considerations: The actual length of a compression member on any plane is the distance between effective positional or directional restraints in that plane. A positional restraint should be connected to be capable of bracing the system (4.7.1). The actual column is replaced by an equivalent pin- ended column of the same strength that has an effective length: L = KL ,where L is the actual length, and K the effective length E  SLENDERNESS The slenderness λ is defined in Section 4.7.3 of the code as: λ = Effective length / Radius of gyration about relevant axis λ = L /r E The code states that, for members resisting loads other than wind load, λ must not exceed 180. Radius of gyration The radius of gyration is a measure of the elastic stability of a cross-section against buckling. It is a property of a section and is also a function of the second moment of area. In general terms, the radius of gyration can be considered to be an indication of the stiffness of a section based on the shape of the cross-section when used as a compression member NB: In buckling analysis of a column, the least value of radius of gyration of the column’s cross-section must be used. COMPRESSION RESISTANCE The compression resistance of a strut is defined in (4.7.4) Part 1 as: (1) Plastic, compact or semi-compact sections: P = A p c g c (2) Slender sections: P = A p c eff cs  where Ag is the gross sectional area defined in (3.4.1) A is the effective sectional area defined in (3.6.2) and eff p is the compressive strength (4.7.5) & Tables 27(a)–(d) c p is the value pc from 4.7.5 for a reduced slenderness of cs 0.5 λ(A /A ) in which λ is based on the radius of gyration r eff g of the gross cross-sections. COLUMN DESIGN Column design procedure: 1. Determine effective length (4.7.3 & Table 22) 2. Select the steel grade and section (3.1.1.1, 4.74, Tables 9 & 11 3. Calculate the slenderness λ for the relevant axis (4.7.2) 4. Compression resistance (4.7.4) 5. Select the strut curve from Table 23 6. Read the compressive strength from Tables 24a-d 7. Calculate the compression resistance Pc (7.4.6 ). Question The column shown below is pin-ended about both axes and without intermediate restraint. Verify the suitability of 356 x 368 x 129 UC in grade S275 steel for the designed load 2500kN Solution 1. Determine effective length (4.7.3 & Table 22) The member is effectively held in position at both ends, but not restrained in direction at either end.  Therefore, the effective length L = 1.0L = 6000 mm E Local buckling ratios: Flange b/T = 10.5 Web d/t = 27.9 2. Section classification (3.1.1.1 Tables 9 & 11) Steel grade = S275 T > 16 mm hence, Py = 265𝑁/mm2 1⁄ 1⁄ 275 2 275 2 𝜀 = ( 𝑃𝑦 ) = (265) = 1.02 𝑏 𝑑 𝑇 < 15𝜀 = 10.5 < 15.3 𝑡