civil engineering report and need the explanation and answer to help me learn.

HI this a steel structure Assignment I ve attached a previous student report for reference and the assignment brief

Requirements: as long as it needs to be

Table of Contents Preliminary Structural Plans ………………………………………………………………………………….. 3 Floor Plan and positioning of structural beams ……………………………………………………………………… 3 Roof Beams and Purl in Design …………………………………………………………………………………………….. 5 Cleat Design ……………………………………………………………………………………………………………………… 6 Design assumptions, codes, assumed loads …………………………………………………………….. 8 Gravity and lateral load path ………………………………………………………………………………… 8 Calculations ……………………………………………………………………………………………………….. 9 Secondary & Primary Floor Beam Design ……………………………………………………………………………… 9 Secondary Beam Analysis ……………………………………………………………………………………………….. 9 Primary Beam Analysis …………………………………………………………………………………………………. 13 Wind Force Calculation …………………………………………………………………………………………………….. 19 Local wind pressure …………………………………………………………………………………………………………. 22 Internal wind pressure -Selection of Cladding and Purlins …………………………………………………… 23 Design of Rafter ………………………………………………………………………………………………………………. 28 Design of beam-column ……………………………………………………………………………………………………. 33 Primary to Column Connection …………………………………………………………………………………………. 38 References ……………………………………………………………………………………………………….. 43 2

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Cleat Design 500 WC 383 _J I I I EB I I I EB I I EB 4 M O GR8.8/s 700 WB 130 12 mm Thick Plate 6
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Design assumptions, codes, assumed loads The building designed in this report is assumed to have a life span of 50 years. This assumption governed the wind speeds and return periods. Other assumptions involved ceiling services, partitions, fire protection spray and floor finishes. Beam self-weights were initially assumed and then checked for adequacy. Concrete self-weight were provided by the manufacturer while imposed load varied for different building usages. Since the building being designed is used as a commercial office building, imposed load was found within AS1170.1 to be 3 kN/m. A number of factors used within the codes were taken with a more conservative approach. For simplicity, wind loads were neglected in the design of structural frame members in the building such as floor beams and columns. However, wind was considered when designing roof rafters in case of suction and uplifts. All design procedure and formulas involved in design calculations were according to the Australian Standards including AS4100, AS1170.0, AS1170.l and AS1170.2. The selection of beams and columns were made using the OneSteel catalogue. The selection of metal cladding and purlins were made using the Lysaght cladding and purlins catalogue. Gravity and lateral load path Different types of loads acting on the building travels down to the foundation through the following paths: Imposed and Dead Loads Slabs "7 Secondary Beam "7 Connections "7 Primary Beam "7 Connections "7 Columns "7 Foundation Lateral loads Slabs "7 Bracing "7 Columns "7 Foundation 8
Calculations Secondary & Primary Floor Beam Design Secondary Beam Analysis 3m PB2 SBl Dead Loads (G) Floor Slab: Ceilings & services: Partitions: Floor Finishes: Fire resistivity: 9m Beam self-weight (assumed): Total Live Loads (Q) Imposed Loads: [AS1170.1 Table 3.l(B)] PBl SBl PBl PB2 3.7 kN/m2x3 m 0.3 kN/m2x3 m 1.5 kN/m2x3 m 0.15 kN/m2x3 m 0.1 kN/m 1 kN/m = 3 kN/m2x3 m Serviceability Limit State: Short term factors [AS1170.0 4.3] G +

Check Shear Capacity Applied Shear V* V* = wl = 34.944X10 = 174_72 kN 2 2 ·check Shear Capacity Vv [AS4100 5.11.2]: dp 82 If -= Rt; then Vu = Vw where Vw = 0.6fyAw tw fy 250 82 52.3 ~ Jffi (= 72.5) OK 320 250 :. Vu= Vw = 0.6x300x106x0.528×0.0096 = 912.4 kN Shear Capacity Check [AS4100 5.11.1] needs to satisfy V* <

Primary Beam Analysis 9m PBl SBl SBl PB2 PBl PB2 SBl SBl PBl Dead Loads (G) Fire resistivity: Beam self-weight (assumed): Total Point load from Secondary Beam: Dead point load: Live Load (Q): From SB: PB2 PB2 0.1 kN/m 1.0 kN/m 1.1 kN/m = 178. 7 kN (9×5)x2 = 90kN 10 m 10 m 13

Deflection [AS4100 Table Bl]: . 9000 Max allowable deflection of beam: /1 ::-; -zso :. 11 ::-; 36 mm Point load and UDL gives different deflections: Point: l l a a 0 = ~[3a _ (~)3] 6Elx 41 I Point loads are greater, use that to find required Ix 36x10_3 = 268.7x103x93 [3×3 _ (~)3] 6X200Xl09Xix 4X9 9 Min req. Ix = 9.66×10-4 m4 = 9.66×10-4x(103)4 mm4 = 9.66×108 mm4 UDL: w 5wl4 8=–384Elx From OneSteel catalogue=> Try 700WB130, Ix= 1400 x 108 mm4 UDL adjusted to 1.4 kN/m (assume self-weight 1.3 kN/m) Deflection Check: Deflection due to UDL: O = Swl4 384Elx 5Xl.40Xl03X94 384x2oox109x1~ox10-4 = 0.000427 m = 0.427 mm 14

Deflection due to point load: 8 = :~:J~~ -(ff] 268.7Xl03X93 [3X3 (3)3] = 6X200x109x14.0X10-4 4X9 -9 = 0.0248 m = 24.8111111 Total deflection: 0.427 + 24.8 = 25.3 mm< 36 mm OK Check Moment Capacity Ultimate Limit State [AS1170.0 4.2.2]: Load combination 1: 1.35G = 1.35x1.40 = 1.89 kN/m = 1.35x178.7 = 241.245 kN 241.245 kN Applied Moment M• UOL: i I I I I 9m M* _ wl2 _ 1.89x92 1--8 8 = 19.14 kN. m I Total M' = 19.14 + 723.74 = 742.88 kN.m 1.89 kN/m I I 241.245 kN 9m l l 1.89 kN/m Point: 241.245 kN 241.245 kN l l 9m M; = 241.245x3 = 723.74 kN. m l 15
Load combination 2: 1.2G + 1.SQ = 1.2xl.40 = 1.68 kN/m Applied Moment M' UDL: 1 I I I I 9m M* _ wl2 _ 1.68x92 1--8 8 = 1.2x178.7 + 1.Sx90 = 349.44 kN 349.44 kN 349.44 kN 1.68 kN/m 9m Point: 349.44 kN 1.68 kN/m I I I l 1 l M; = 349.44X3 9m = 17.01 kN. m = 1048.32 kN. m Total M' = 17.01 + 1048.32 = 1065.33 kN.m Moment Capacity of beam Ms [AS4100 5.2.1]: Ms = fyXmin{l.SZx, Sx} = 300x106xmin{l.Sx3990x103x(10-3)3, 4490x103x(10-3)3} = 300x106x0.00449 = 1347000 N. m = 1347.0 kN. m Capacity Factor: cj>M~ (cp = 0.9 [AS4100 Table 3.4]) ¢Ms = 0.9Xl347.0 = 1212.3 kN.m > M*(= 1065.33 kN.m) OK 349.44 kN 1 l 16

Check Shear Capacity Applied Shear v· Load combination 1: Reaction forces= 1·99×9 + 241.245 = 249.75 kN 2 Max v· = 249.75 kN 241.245 kN 241.245 kN 9m 249.75 kN Load combination 2: Reaction forces= 1.6axg + 349.44 = 357 kN 2 Max v'(ends) = 357 kN 349.44 kN 349.44 kN 9m 357 kN 1.89 kN/m 249.75 kN 1.68 kN/m 357 kN 17

Check Shear Capacity Vv [AS4100 5.11.2): dp 82 _ _ _ If -:::::; g then Vv -Vu where Vu -Vw -0.6fyAw tw !x_ 250 ~= 74.86 (iori -.}250 dp = 66.0 tw d 82 :. __E_(= 66.0) < g (= 74.86) SO Vv = Vw tw !x_ 250 Vu= Vw = 0.6x300x106x(700x10-3)(10x10-3) = 1260000 N = 1260 kN Shear Capacity Check [AS4100 5.11.1) needs to satisfy V* < cpvu ¢Vv = 0.9X1260 = 1134 kN > V*(= 354 kN) OK Primary Beam: 700WB130 18

R 5 4 3 2 1 Wind Force Calculation Building Importance Level: [AS1170.0 -Table Fl] Building Design Life: Annual Probability of exceedance: [AS1170.0-Table F2] Wind region: [AS1170.2 -Fig 3.1(A)] Terrain category: [AS1170.0-4.2.1] Vn z 26 22 18 46 14 10 6 Ms 1.0 M, 1.0 Critical Direction 1: M;, Mzcat 0.78 0.76 0.75 0.75 0.75 0.75 r ‘ 3(High) 50 years 1/1000 AS 4 N 1.00 35.88 34.96 34.5 34.5 34.5 34.5 NE E SE s SW w NW 0.85 0.80 0.80 0.85 0.90 1.00 0.95 ~·it,b 30.498 28.704 28.704 30.498 32.292 35.88 34.086 29.716 27.968 27.968 29.716 31.464 34.96 33.212 29.325 27.6 27.6 29.325 31.05 34.5 32.775 29.325 27.6 27.6 29.325 31.05 34.5 32.775 29.325 27.6 27.6 29.325 31.05 34.5 32.775 29.325 27.6 27.6 29.325 3/.05 34.5 32. 775 Critical Direction 2: N 19

Critical Wind Direction 1 1 -+-R -5 ….. 4 ~3 -….2 27 +——–t—–il——+—-+—–+—-+-“——1 N NE E SE s SW w NW A(11z1J w p Surface vd,s,• Cp,c K,. Kee K, Kp Cr.g,c cd)’II (kPa) (storey (kN) frontal area) ” I • I • • I W (Level 35 0.72 0.53 180 95.26 5) W(Level 4) 34.5 0.72 0.51 180 92.55 W(Level 3) 34.5 0.72 0.51 180 92.55 W(Level 2) 34.5 0.72 0.51 180 92.55 W(Level 1) 34.5 0.72 0.51 225 115.69 L (roof) 36 1.0 1.0 -0.45 1.0 -0.35 90 -31.49 L (Level 5) 35 -0.45 -0.33 180 -59.54 L(Level 4) 34.5 -0.45 -0.32 180 -57.85 L(Level 3) 34.5 -0.45 -0.32 180 -57.85 L(Level 2) 34.5 -0.45 -0.32 180 -57.85 L(Level 1) 34.5 -0.45 -0.32 225 -72.31 s 20

Critical Wind Direction 2 27 N NE E SE W (roof) W (Level 5) 35 0.9 W(Level 4) 34.5 0.9 W(Level 3) 34.5 0.9 W(Level 2) 34.5 0.9 W(Level 1) 34.5 0.9 L (roof) 36 0.9 1.0 1.0 L (Level 5) 35 0.9 L(Level 4) 34.5 0.9 L(Level 3) 34.5 0.9 L(Level 2) 34.5 0.9 L(Level 1) 34.5 0.9 s s SW w NW 0.72 0.53 0.72 0.51 0.72 0.51 0.72 0.51 0.72 0.51 -0.4275 1.0 -0.33 -0.4275 -0.31 -0.4275 -0.31 -0.4275 -0.31 -0.4275 -0.31 -0.4275 -0.31 160 160 160 160 200 80 160 160 160 160 200 1 -+-R -5 ….. 4 ~3 ~2 84.67 82.27 82.27 82.27 102.84 -26.59 -50.27 -48.85 -48.85 -48.85 -61.06 21

Local wind pressure Ultimate Limit Serviceability Limit State (m/s) State (m/s) Reference ULS V2s {SLS) V1000 {ULS) Ratio 36 TOTAL 36.00 Ultimate Limit State: Vdes,s= 36 m/s [ASll 70.0 (Table 3.3)] 50 years SLSl Importance level 3 1/25 return period Serviceability Limit State 36 37 46 0.80 28.96 [AS1170.2: Cl 3.2) [AS1170.2: Cl 3.2) Vs ~ Aregion 1 -7 ~ 37 V1000 ~ Aregion ~ 46 RA1 RA2 RC1 :. Serviceability Limit State = 36 m/s x 37 m//s 46m s -1.3 -1.3 -1.3 Vdes,e = 28.96 m/s 1.0 1.0 1.0 0.9 0.9 0.9 a = min{0.2b, 0.2d, h} = min{0.2×45 m, 0.2×40 m, 26 m} = min{9, 8, 26} a=8m 1.5 2.0 1.0 3.0 -1.76 -2.34 1.0 -2.70 -1.36 -1.82 -2.10 h = 26 m / b=45m d = 40m -0.88 -1.18 -1.36 22

Note: • Product Cp,e K1 has negative limit of-3.0 per clause 5.4.4 (AS1170.2) (where Cp,e is the external pressure coefficient and K1 is the local pressure factor, greater than 1.0 for non-structural elements). • Cp,e taken as -1.3 for conservative measure. It is the highest possible value for external pressure coefficients for rectangular enclosed buildings. In practice this saves time and adds safety to the calculations. Per the code [AS1170.2: Table 5.3(a)) the building would have a h/d ratio of 0.65 and interpolation would be required between values -1.3 and -0.9. • Ka taken as 1.0. This area reduction factor applies to areas equal to or less than 10 m2. The tributary area in this situation is 16 m2 (0.25a2) however 1.0 has been taken conservatively [AS1170.2 Table 5.6). Internal wind pressure -Selection of Cladding and Purlins • Assume no dominate opening • Cp,i Internal pressure coefficient (all walls are equally permeable) Cp,i = -0.3 or 0.0 (Take more severe on combined forces [AS1170.2: Table 5.1 (A))) Scenario 1) = -0.3 Scenario 2) = 0.0 Kc,i [AS1170.2:Table 5.5 (f)J Combination factor Kc,i for applied internal pressures -Take 0.9 for Kc,i :. 1) Cfig,i = 0.9X-0.3 = -0.27 2) Cfig,i = 0.9XO = 0 Cfig is the aerodynamic shape factor Cdyn = 1 [AS1170.2 6.l(a)) Pressure [AS1170.2 2.4.1) 1) Max negative internal pressure (away from surface) ULS (strength): p = 0.5×1.2x362x-0.27×1 2 1 kN = -209.95 N/111111 X 1000 N p = -0.210 kPa SLS: p = 0.5×1.2X28.962x-0.27X1 2 1 kN = -135.87 N/mm X 1000 N p = -0.136 kPa 23

2) Max positive internal pressure {into surface) ULS (strength): p = 0 kPa SLS: p = 0 kPa :. Take Oas worst case scenario for internal pressure Selection of Cladding: Max external pressure Positive internal pressure Negative internal pressure Governing Pressure Manufacturer: Product: Lysaght Custom Orb ULS -2.10 kPa 0 -0.210 kPa -2.10 kPa BMT: Fasteners: 0.48 mm (Cladding catalogue) 5 per sheet SLS -1.36 kPa 0 -0.136 kPa -1.36 kPa Weight: 0.05 kPa (4.93x 9’81) (Cladding catalogue) 1000 Span: Capacity: Serviceability: Strength: 1500 mm [per catalogue) 1.97 KPa (Cladding catalogue) 10.6 kPa (Cladding catalogue) 24

Selection of Purlins (assumed) Product: Lysaght Supa-purlin Span of purlin: 10.0 m Arrangement: Single Spacing: 1.5 m (distance between roof beams) (assumed) (span of cladding) Pressure Catchment (m) Dead Load [G] Metal sheet 0.05 kPa 1.5 Self-weight of purlin Ceiling services 0.3 kPa 1.5 TOTAL ( ~) + 0.12 1.5 10X18 Imposed Load [Q] = 0.21 < 0.25 0.25 kPa TOTAL External wind pressure -2.lkPa 1.5 (-ve) Internal wind pressure 0.0 kPa 1.5 (+ve) Internal wind pressure -0.21kPa 1.5 (-ve) Load Combinations Load (kN/m) Note/Reference 0.073 (calculated, catalogue) 0.123 0.45 (assumed) 0.646 0.375 AS1170.1 Table 3.2 (ii) 0.375 -3.15 0 -0.315 Load combination 1: 1.2G + 1.5Q (inward) (1.2X0.646) + (1.5x0.375) Reference [AS1170.0 Cl4.2.2 (b)] = 1.338 kN/m2 Load combination 2: 1.2G + Wu+ 4JcQ (inward) -Permanent, wind and imposed action 4Jc= 0 [AS1700.0 Table 4.1] (1.2X0.646) + 0.315 + (OX0.25) = 1.09 kN/m2 Reference [AS1170.0 Cl4.2.2 (d)] 25
Load combination 3: 0.9G -Wu (outward) Reference [AS1170.0 Cl4.2.2(c)] wind action reversal Dead load (G) is taken as O because the direction of the load due to gravity counteracts the wind suction = (0.9x0) + (-3.15) = -3.15 kN/m2 Serviceability Limit State Load combination 4: G + lVsO (sagging) -lVs is short term imposed, i.e. workers on roof [AS1170.0 Table 4.1] (0.646 + 0.7x0.375) = 0.909 kN/m2 Load summaries: Ultimate Limit State: Inward: Inward (wind): Reversal (wind outward): Serviceability Limit State Sagging: Selected Purlin: SC/5230030:2 2 bridges Span: 10 m Spacing: 1.5 m Outward capacity Inward capacity Serviceability 3.59 kN/m 3.70 kN/m 2.13 kN/m 1.31 kN/m 1.09 kN/m -3.15 kN/m 0.886 kN/m 26
Check outward: 3.59 kN/m > 3.15 kN/m OK Check inward: 3.70 kN/m > 1.09 kN/m OK Check SLS: 2.13 kN/m > 0.909 kN/m :. OK Self-weight Check Self-weight SZ30030 12.56 kg/m = 12.56 kg x9.8111~ m s = 123.21 kgm ms2 = 123.2 !!._ m kN = 0.123-:. OK m 27

Design of Rafter 9000 RB I LJ RB 1.5 m Spacing Design Information: 9000 RB 0 a 8 ~ Metal cladding: BMT 0.48 mm: 4.93 kg/m2 (Cladding catalogue) Purlin: SZ30030:2 bridges: 12.56 kg/m (Purlin catalogue) Calculations: Dead Loads: Cladding: 4.93 kg/m2 x9.81×10-3×10 m Purlin: 12.56 kg/mx9.81×10-3×10 mx6x¼ Insulation (roof): 0.15 kN/m2x10 Beam self-weight (assumed): 0.6 kN/m Fire Protection: 0.1 kN/m Ceiling services: 0.3 kN/m2x10 m Catchment = 0.483 kN/m = 0.82 kN/m = 1.5 kN/m = 0.6 kN/m = 0.1 kN/m = 3.0 kN/m = 6.5 kN/m 28

Imposed Loads: Roof Imposed loads [AS1170.1 Table 3.2]: = 1.8 + 0 12 A . = 1.8 + 0 12 A , = 0.14 < 0.25 · :. 0.25 kN/m2x10 m = 2. 5 kN/m Wind Load (WuL Pressure calculated using K1 value of 1.0 because it is a structural element [AS1170.2 CIS.4.4] Cng = Cµ,eKaKc,eK1Kp = -1.3x1.0x0.9x1.0x1.0 = -1.17 p = 0.5x1.2x362x-1.17x1.0x10-3 = -0.91 kPa -0.91 kPa x10 m = -9.1 kN/m Load Combinations [AS1170.0 4.2.2]: LC1: 1.2G + 1.SQ ( ,J,) 1.2x6.503 + 1.5x2.5 = 11.55 kN/m LC2: 0.9G +Wu+ iJ,.Q ( 1') 0.9X6.503 + (-9) + Ox2.5 = -3.15 kN/m 29
Design Actions: Load Combination 1: M* = w12 8 11.ssx92 8 = 116.9 kN.m v· = wl 2 11.55X9 = 2 = 52.0 kN Load Combination 2: M* = w12 8 -3.1Sx92 8 = -31.9 kN.m v· = wl = 2 -3.15X9 2 = -14.2 kN Moment Capacity of Beam [AS4100 5.2.lj: Try 360U BSO. 7 self-weight= 0.49 kN/m, therefore assumed self-weight OK Compact=> Ms = fyX min{l.SZx, Sx} = 300x106x min{l.Sx798x103x(10-3)3, 897×103 x(l0-3)3} = 300x106x0.000897 = 269100 N.m = 269.1 kN.m Capacity Factor: cj)M~ (¢ = 0.9 [AS4100 Table 3.4)) cpM5 = 0.9X269.1 = 242.2 kN. m > M* (116.9 kN. m) OK 30

Member Capacity of Beam: Wind suction Fly braces required to stabilise bottom flange to stop LTB Check for full restraint assuming fly brace placed at every purlin (1.5 m spacing). It will be fully restrained if: _.!_ :’S (80 + 50~m) §f50 ry .._J fy [AS4100 5.3.2.4] I 1500 = ry 38.5 = 38.96 (80 + 50~m) §f50 = (80 + 50(-1)) {zso ,j r y ._j 300 = 27.39 < 38.96 :. Not fully restrained, need to check member capacity Mb Fly brace at every second purlin. S,. = 1500 mm St= 3000 mm le= 0.85Sr = 0.85X3000 = 2550 mm [Woolcock, 2011] Member Capacity of Segments without Full Lateral Restraint [AS4100 5.6.1.1] (XS= 1.0 Mo = C21;ly) [ GJ + (n2l~lw)] e e (n2x2oox109x9.6Xl06x(10-3)4) [so X 1Q9x 241 X 103 X cio-3)4 + (n2x2oox109x284Xl09x(10-3)6)] 2.552 2.552 = 554.5 kN.m 31
as= 0.6 [ [(~)2 + 3]-(~) a,= 0.6 [ [(269.1X103)2 + 3] _ (269.1X103) 554.5X103 554.SX103 = 0.788 Mb = amasMs :S Ms = 1.0x0.788x269.7 = 212.1 kN. m

© @ ® Design of beam-column G) @ ® © ® 5000 9000 9000 9000 f r r r f PB1 PB1 PB1 PB1 0 -u en -u en en -u en en en 0 -u en -u 0 OJ ~ OJ ~ ~ OJ ~ ~ OJ ~ OJ ~ “‘ “‘ “‘ “‘ ~ “‘ 5< bo PB1 PB1 PB1 PB1 Cf) -0 0::: -u 0 -u ~ -u en en -u en en -u en en OJ 0 OJ OJ ~ ~ OJ ~ ~ ID ~ ~ "' ~ "' "' 0 "' "' 0 Cf) N .... EN RAt JCE PB1 PB1 PB1 PB1 I . . 9+9 10+10 2 Tributary Area: -x--= 90 m 2 2 Dead Loads: Roof Level Metal Cladding: Insulation: Purlins: Ceiling services: Rafter self-weight: 1-Sth Level: Concrete: PB1: PB2: SB1: Partitions: Ceiling services: Floor Finishes: Column self-weight (500WC383): 4.93 kg/m2 x9.81x10-3x90 m2 0.15 kN/m2x90 1112 12.56 kg/m2 x9.81x10-3x90 0.3 kN/m2x90 m2 ( 50.7 1~~ x9.81x10-3 + 0.1) x9m (3.7 kN/m2x90 m2)x5 (130 kg/m2x9.81x10-3 + 0.1 kN/111)(9 m) x5 (82 kg/m2x9.81x10-3 + 0.1 kN/111)(10 m)x5 (82 kg/m2x9.81x10-3 + 0.1 kN/m)(10 m)x2x5 (1.5 kN/m2x90 m2)x5 (0.3 kN/m2x90 m2)x5 (0.15 kN/m2x90 m2)x5 (383 kg/m2x9.81x10-3 + 0.1 kN/m)x26 -04 -CRITICAL COLUMN = 4.35 kN = 13.5 kN = 11.09 kN = 27kN = 5.38 kN = 61.32 kN = 1665 kN = 61.89 kN = 45.21 kN = 90.44 kN = 675 kN = 135 kN = 67.5 kN = 97.69 kN = 2837. 75 kN Total Dead Load = 2899. 07 kN 33
Imposed Load (Q): Roof: 0.25 kN/m2x90 m2 = 22.5 kN [AS1170.1 Table 3.2] Level 1-5: 3 kN/m2x90 m2x5 = 1350 kN [AS1170.1 Table 3.1] Total Imposed Load= 1372. 5 kN Total area being supported by column= 90 x 6 = 540 m2 Code says if area > 200 m2, apply reduction factor 4la [AS1170.1: 3.4.2]: Reduction factor Reduced imposed load: = 0.5×1372.5 = 686.25 kN 3 3 lVa = 0.3 + . r. = 0.3 + ,-;=-;-;; = 0.43 vA v540 0.43 < 0.5 :. Wa = 0.5 Applied axial compression (N') for column: = 1.2G + 1.5Q = 1.2x2899.07 + 1.5x686.25 = 4508.26 kN [AS4100 4.3.4] Min eccentricity for column= 100 mm M; =My= N*e = 4508.26X0.1 = 450.83 kN. m Nominal section capacity N5-Check 500WC383 [AS4100 6.2.1] N5 = kfAnfy An = Ag = 48800 mm2 N5 = 1.0x48800x280x10-3 = 13664 kN 34
Nominal member capacity Ne (y-axis) [AS4100 4.6.3] [AS4100 T6.3.3 {1)] [AS4100 6.3.3] [AS4100 T6.3.3 {3)] [AS4100 T3.4] le = kel = 1.0x6.0 = 6000 mm % = 0 (kr = 1), ab for hot rolled section per Table 6.3.3(1) An = (~) Ji<; 0y -Slenderness reduction factor. l'y '1m = (6000) .JI /280 124 --Jm = 51.21 ac = 0.836 (reading table where ab = 0, An = 55) Ne = acNs ::; Ns = 0.836X13664 = 11423.1 kN

Section Capacity under biaxial bending [AS4100 8.3.4) Needs to satisfy: ( M~ )Y + ( My )Y ::; 1 ct>Mrx ct>Mry Where y = 1.4 + (!:::…) :S 2.0 ct>Ns ( 4508.26 ) y = 1.4 + 0.9X13664 = 1.77 < 2.0 OK ( 450.83 )1·77 + ( 450.83 )1·77 0.9X1910.69 0.9X970.83 = 0.40 :S 1.0 OK X-axis member compression capacity [AS4100 6.3.3 (1)) [AS4100 6.3.3 (3)) [AS4100 8.4.2] le = 1.0x6000 = 60000 ab= 0 An= (~)Jkr~ = (6000) ',/1 µso 197 ,Jm = 32.22 :. ac = 0.925 Ncx = acNs = 0.925X13664 • In-plane capacity M* :S

Check shear capacity through critical section of primary beam: I I 9 cp cb Uniform Check: [AS4100 5.11.2) [AS4100 5.11.4) Critical Line 022 dp = 700-4X22 = 61,2 tw 10 ~=~= 73.64 0v {llii ~250 ~250 . dp 82 _ .. -:::; ~ ~ Vv -0.6fyAw tw _2’_ 250 Vv = 0.6X310X10X (?00-4×22) 1000 = 1138.32 kN

Bolt Capacity 357 kN ‘ M* = 357 X 0.05 = 17.85 kN.m Assume bolts support vertical loads evenly 170 35 V’ 357 Fv = -= -= 89.25 kN 4 4 Take moment about centroid of bolt group: F1x(105) + F2x(35) + F1X(105) + F2x(35) = 17.85X103 kN. mm -(1) Similar Triangles: 1 F2 = 3 F1 -(2) Sub F2 = 2:. F1 into (1) 3 Bolt in Shear [AS4100 9.3.21] 1 1 3 105F1 + 35x3F1 + 105F1 + 35x3F1 = 17.85×10 kN.mm 700 3 -F1 = 17.85×10 kN.mm 3 F1 = 76.5 kN v; = ../76.52 + 89.252 = 117.55 kN 1 shear plane through core/shank [AS4100 T9.3.1] Vf = 0.62x8301x(O + 1xn:x102)x10-3 = 161.7 kN cpVf = 0.8×161.7 kN = 129.3 kN > W(117.55 kN) OK 40

Check ply in bearing [AS4100 9.3.2.4] Weld Capacity Check: vb = 3.2drtpfup = 3.2x20x10x430x10-3 = 275.2 kN q>Vb = 0.8X275.2 kN = 220.16 kN > v; (117.55 kN) OK [AS4100 Table 9.7.3.10 (1)) Weld Type B-E43XX fuw = 430 MPa 12 +–+ ‘ 17.85 kN.m 357 kN [AS4100 9.7.3.2] Max tw = t-1 mm because t > 6 mm=> 9 mm> 7 mm fillet weld OK t1 = 7 x 0.7 = 4.9 mm bh3 lweld = 212 2 4.9X2803 = x—12 = 17.9×106 mm4 Stress on weld due to eccentricity moment My cr11 = i a = 11.ssx106x14o = 139_6 N. mm2 h l 7.9X106 41

Stress on weld due to vertical shear Weld Capacity [AS4100 9.7.3.10] V* CJ=—v A (weld) = 2X280X4.9 = 130.1 N. mm2 R = resultant stress = CJ = .J 130.12 + 139.62 = 190.8 N. mm2 Vw = 0.6fuwttkr tt tt 0.6X430X4.9Xl.0 4.9 = 258 N.mm2 cjJ Viv= 0.8×258 tt = 206.4N.mm2 > CT(= 190.8N.mm2) OK 42

References Standards Australia Limited 1998, Australian Standard -Steel Structures AS4100-1998, Standards Australia, Sydney NSW. Standards Australia Limited 1998, Structural Design Actions AS1170.0:2002, Standards Australia, Sydney NSW. Standards Australia Limited 1998, Structural Design Actions Part 1: Permanent, imposed and other actions AS1171.0:2002, Standards Australia, Sydney NSW. Standards Australia Limited 1998, Structural Design Actions Part 2: Wind actions AS 1171.2:2011, Standards Australia, Sydney NSW. Woolcock, S 2011, Design of portal frame buildings : including crane runway beams and monorails, 4th edn, Australian Steel Institute Sydney NSW. 43

CIVE1179 – STEEL STRUCTURES 1 DESIGN PROJECT No. 1000 La Trobe Street, Docklands 1. Project brief You are assigned to provide a detailed structural design and documentation for a purposed office building located at 1000 La Trobe Street, Docklands. The site is shown in Figure 1 below. Preliminary Google Map sketches of the project are provided. Figure 1. 1000 La Trobe Street, Docklands & Location Map 2. Building specifications The following specifications are requested by the architect. The structural design must comply with Australian Standards and the manufacturer’s specifications for materials. Item Descriptions Usage Commercial office building and retails on ground For imposed load, see Table 3.1, AS1170.1 Floor elevations (measured at top of floor slab) Elevation Australian Height Datum Roof +10.5 m

Level 2 +7.0 m Level 1 +3.5 m Ground +0.0 m Floor slabs Bondek composite slabs spanning 1-direction Self-weight: 3.8 kN/m2 Columns Column spacing ≤ 10 m Ceilings and services 0.34 kPa Partitions 1.8 kPa on office floors Floor Finishes Ceramic floor tiles or carpets – allow 0.2 kPa Roofs Non-habitable Metal roof system Roof Insulation 0.15 kPa Lateral load resisting system Provided by steel braces Fire resistivity allow for dead weight of fire spray 0.1 kN/m along the length of all beams & columns 3. Design assumptions To simplify your design, it is assumed that wind loads govern the design. For roof beams, wind uplifts will need to be considered. 4. Omit from design • Design of concrete floor slabs • Design of foundation • Design of stairs • Design of ground floor slabs

5. Submission details A final report overall worth 40 MARKS – due on 4th June 2023. Submit report with a cover page (student number, full name and course details), a table of content and a reference list. Task Max mark Structural plans • Sketched neat manually or with AutoCAD • Plans should include: 1. Grid lines and dimensions 2. Label all members 3. Selection of steel section 3 Structural design calculations Part-I • Highlighted secondary beam (SB1) • Highlighted primary beam (PB1) or highlighted rafter (RB) [pick one] • Highlighted column segment • Wind pressure on building surfaces 24 Structural design calculations Part-II (pick either Part-II or Part-III) • Wind pressures on roof metal sheets • Selection of roof metal sheets and purlins 8 Structural design calculations Part-III (pick either Part-II or Part-III) • Design of one bolted connection • Design of one welded connection 8 Design selection and overall presentation (see item 6 for explanation) • Design assumptions, design codes and load paths • Design selections • Overall presentation 5 Submit all the above as one FINAL DESIGN REPORT by 4th June 2023; penalties apply for late submission according to university policies. 6. Design selection and overall presentation Professionalism marking will be based on how well you have liaised with the teaching staff, interest and effort shown in this project, how professional your report is and how well you have followed the instructions – Purely professional judgment. Design selection marks are awarded on the appropriateness of the chosen members. This is again a professional judgment. In simple terms, if you have chosen a member that carries 1000 kNm when the requirement is only 100 kNm, it is not that appropriate. The appropriate member may be the one with a capacity above 100 kNm but less than 150 kNm. 7. Feedback

Students can submit the preliminary work to Canvas for feedback. No mark is allocated to this partial submission and it is not compulsory. You may also utilize discussion board on Canvas for feedback. 8. Late submission Assignments received late and without prior extension approval or special consideration will be penalised by a deduction of 10% of the total score possible per calendar day late for that assessment.

9. Structural plans The plans for ground floor, standard floors and roof are given below. The dimension of the floor plan is dependent on your student number. Assume the last digit of your student number is x, then L1 x+5 For x = 0~4 0.5x+4 For x =5~9 L2 9 x is an odd number 8.1 x is an even number L3 6 x is an odd number 7 x is an even number For example, student number s3750424, L1 = 4+5 = 9 m; L2 = 8.1 m; L3 = 7 m. Ground floor plan

Standard floor plan (1st and 2nd floors)

Roof plan

Examples of common mistakes Typ. Mistakes Design standards used, assumed loads, other assumptions, etc. • Missing info Structural plans • Grid lines missing • Title missing • Dimensions missing • Member labels • Drawing conventions A short description how does your structure resist lateral (i.e. wind) loads • Info missing Critical secondary beam • Missing loads • Deflection check used load factors 1.2G+1.5Q Critical primary beam • M* wrong • Deflection check used load factors 1.2G+1.5Q Critical primary beam to column connection • Welding check wrong • Bolt forces wrong Critical roof beam • Loading wrong • Mb check missing Critical column segment • Loading wrong • Effective length wrong • Nc wrong • Combined action check missing Wind pressure on building surfaces • Cpe wrong • Wu wrong Wind pressures on roof metal sheets & Selection of roof metal sheets / purlins • Internal/external pressure wrong • Bridging missing

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