4 Initial Sizing of members

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Beatrice Patrick
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1 Initial sizing if members 7 /8/16 4 Initial Sizing of members See for the method Figure 5 Dimensions of portal (The institutionof Structural Engineers, /8/16) a) L/h = 3/6 = 5 r/l = 1.577/3=26 b) Loading 1) Gravity loading Snow loading =.54 x 7.2 = 3.8KN/m Self weight =.65 x 7.2 = 4.68 KN/m 2) Factored load w= (4.68 x 1.35 ) + (3.8 x 1.5 ) = 12. KN/m c) Finding Mp for the sections 1) Total load on the frame (wl)= 12. x 3 = 36.5KN 2) Parameter wl 2 = 12. x 3 2 = 1816 KNm 3) From Graphs (Figure B2) obtain horizontal force ratio (.36) H=.36 x 36.5 = KN 4) From Graphs (Figure B3) obtain rafter Mp ratio (.34) M rafter,,rd =.34 x 1816 = KNm

2 Initial sizing if members 8 5) From Graphs (Figure B4) obtained column Mp ratio (.63) M column, Rd =.63 x 1816 = KNm. 6) Selecting members a) W pl (rafter),required = (367.7 x 1 6 ) / 275 = 1337 x 1 3 cm 3 Try UB 457x152x74 b) W pl(column),required = (681.4 x 1 6 )/275= 2478 x 1 3 cm 3 Try UB 533 x 21 x 19 Section Tables of Universal Beams Properties Rafter Section UB 457x152x74 G=74.2 Kg/m h= 462mm b=154.4mm t w =9.6mm t f =17mm A=94.48 x 1 2 mm 2 d=428mm I y = 3267 x 1 4 mm 4 W pl,y =1627 x 1 3 mm 3 i y =186 mm i z = 33,3 mm I z = 147 x 1 4 mm 4 W pl,z = x 1 3 mm 3 I t = x 1 4 mm 4 I w = x 1 6 mm 6 EN : 25 (E) Table 3.1 Properties Column Section UB 533x21x19 G=19 Kg/m h= 539.5mm b=21.8mm t w =11.6mm t f =18.8mm A=138.9 x1 2 mm 2 d=51.9mm I y = 6682 x 1 4 mm 4 W pl,y =2828 x 1 3 mm 3 i y =218.7 mm i z = 45.7 mm I z = 2692 x 1 4 mm 4 W pl,z = x 1 3 mm 3 I t = 11.6 x 1 4 mm 4 I w = 1811 x 1 6 mm 6 Steel grade is S275 Assume Sections Class1, then check

3 45 /8/16 B1 Initial sizing using Weller s charts The method described relies for its simplicity on a series of three charts developed by Alan Weller. The chart has been constructed with the following assumptions and which leads to reasonably economic solution (Note. This is not a rigorous design method; it is a set of rules to arrive at initial size). 1) The rafter depth is approximately span / 55 2) The hunch length is approximately span /1 3) The rafter slope lies between o and 2 o. 4) The ratio of span to eaves height is between 2 and 5. 5) The hinges in the mechanism are formed at the level of the underside of the haunch in the column and close to the apex. Each chart requires a knowledge of the geometry of the frame and the design loading as input data in order to determine approximate sizes for the column and rafter members Using of charts Figure B1 Dimensions of portal (The institutionof Structural Engineers, /8/16) a) Calculate the span/height to eaves ratio = L/h b) Calculate the rise/span ratio = r/l c) Calculate the total design load FL on the frame and then calculate FL 2, where F is the load per unit length on plan of span L (e.g. F =qs, where q is the total factored load per m 2 and s is the bay spacing). d) From figure B2 obtain the horizontal force ratio H FR at the base from r/l and L/h e) Calculate the horizontal force at the base of span H=H FR W L.

4 46 /8/16 f) From figure B3 obtain the rafter M p ratio M PR from r/l and L/h. g) Calculate the M p required in the rafter from M p (rafter) = M PR x W L 2. h) From figure B4 obtain the column M p ratio M PL from r/l and r/h. i) Calculate the M p required in the rafter from M p (rafter) = M PL x W L 2. j) Determine the plastic moduli for the rafter W pl,y,r and the column W pl,y,c from W pl,y,r =M p,(rafter) /f y W pl,y,c = M p,(column) /f y Where f y is the yield strength. Using the plastic moduli, the rafter and column sections may be chosen from the range of plastic sections as so defined in the section books H/wL Figure B2- Horizontal force at the base (The institutionof Structural Engineers, /8/16)

5 47 /8/ M pr / wl² Figure B3- M p ratio required for the rafter (The institutionof Structural Engineers, /8/16) M pl / wl² Figure B4- M p ratio required for the column (The institutionof Structural Engineers, /8/16)

Plastic Design of Portal frame to Eurocode 3

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