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Module 1: Purlins and Gantry Girders

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Video 1
This lecture will cover a workout example on design of purlin, in last lecture we have discussed about the design procedure of purlin members for both, channel or I sections and angle sections. So here we will go through one example and we will try to understand how to design a member using I sections.
Example: Design an I-section purlin, for an industrial building situated in the outskirt ofKolkata, to support a galvanized iron sheet roof for the following data:Slope of truss = 30oSpacing of truss c/c = 5.0 mSpan of truss = 12.0 mspacing of purlins c/c = 2 mwind speed = 50 m/sWeight of galvanized sheets = 120 N/m2Grade of steel = Fe 410
Solution:For steel of grade Fe 410: fy = 250 MPaWeight of galvanized corrugated iron sheets = 120 × 2 = 240 N/mAssume dead load of purlin = 100 N/mTotal dead load = 240 + 100 = 340 N/mThe dead load acts vertically downwards.The component of dead load parallel to roof = 340sin30° = 170 N/mThe component of dead load normal to roof = 340cos30° = 294.5 N/mWind pressure = pz=0.6V 2❑z =0.6 ×502=1500 N /m2Wind load is assumed to act normal to the roof.Wind load = 1500×2×1 = 3000 N/mTotal load on purlin normal to roof = 3000+294.5 = 3294.5 N/mFactored load normal to roof,P = 1.5×3294.5 = 4941.75 N/mFactored load parallel to roof,H = 1.5×170 = 255 N/m
Maximum moment,Muu = Mz = PL/10 = (4941.75×5)×5×10-3 /10 =12.35 kNmMvv = My = HL/10 = (255×5)×5×10-3 /10 =0.6375 kNmLet us try a section with flange width bf = 75 mm and depth, d = 125 mm.Plastic section modulus required,Zpz , reqd=Mzγm 0f y +2.5(d b)(M y γfmy0)Zpz, reqd=12.35× 106× 250 1.1 +2.5(125 75 )(0.6375×106× 250 1.1)= 66 × 103 mm3
Select a section ISLB 150 with Zpz = 104.5 × 103 mm3A= 1808 mm2, D = 150 mm, bf = 80 mm, tf = 6.8 mm, tw = 4.8 mmR1 = 9.5 mm,d = 150 – 2(6.8+9.5) = 117.4 mm
Iz = 688.2 × 104 mm4 , Iy = 55.2× 104 mm4Zez = 91.8 × 103 mm3, Zey = 13.8 × 103 mm3Section classificationϵ=√250 f y =√250 250=1b/tf = 40/6.8 = 5.88 < 9.4d/tw = 117.4/4.8 = 24.5 < 84Hence the section is plastic.
Video 2
Check for bending strengthMdz=Zpzf yγm0=104.5 × 103× 2501.1× 10−6=23.75 kN-m< 1.2Zezf yγm0=1.2×91.8× 103× 2501.1×10−6=25.04 kN-mWhich is alright.Mdz=23.75 kN-m > Md = 12.35 kNm; OK
Mdy=Zp y ×f yγm0≤ γf Zeyf yγm0Zpy=4×[(b2f ×tf)× b4f]+2×[{(D−2tf)× t2w}× t4w]Zpy=tf bf 22+( D−2tf)tw 24=6.8×80❑ 22+(150−2×6.8)4.8❑ 24=22546 mm3Mdy = 22546 × 250/1.1 × 10-6 = 5.12 kN-m< 1.5 × 13.8 × 103 × 250/1.1 × 10-6 = 4.7 kN-m(1.2 is replaced by γf =1.5 since Zpy/Zey (=1.6)> 1.2)Hence, Mdy = 4.7 kN-m > Md = 0.6375 kN-m; OK
Check for overall member strength (local capacity)MzMdz+MyMdy≤ 112.3 523.75+0.63754.7=0.66<1 ; OKCheck for deflectionδall = 5000/180 = 27.78 mmΔ = 1/384 × wl4/EI= (3294.5 × 10-3 × 50004)/ (384 × 2 × 105 × 688.2 × 104)= 4 mm < δall ; OK.
So deflection in both the directions also we have to check and in other direction deflection isquite less because the load is very less so that will be also okay. So this is how one can designthe purlin by checking the deflection and prior to that checking the moment.So in today’s lecture we could see that how to find out the section of a purlin member whileusing I section and how to calculate the biaxial bending moment and check the biaxialbending moment interaction formula. So using this we can design a purlin. Next we willdiscuss about the Gantry Girders because Gantry Girder is nothing but one type of flexuralmember. So how to design that Gantry Girder that will be discussed in next class, thank you.