Video 1
Hello so far we have discussed about the design bending strength calculation of a flexural member and we have seen when calculating the design bending strength we need to know the plastic section modulus of the section. And in case of built-up sections we do not have the plastic section modulus in the code so we need to calculate the plastic section modulus and to know the plastic section modulus of a section we need to know what is the theory behind that and that will be discussed today. So first I will go through one workout example and I will show how to calculate the plastic section modulus of a section. Suppose if load is very high on the beam and the member is very long then bending moment is quite high. So the required section modulus will be very high. So when the required section modulus is comparatively high then the available sectionthen we have to go for built-up section. We have to use certain plates on the top and bottom section to increase the section modulus and to take care that much load. So how to increasethe section modulus and how to decide the thickness and width of the plate to withstand the required amount of load that will be demonstrated through an example, so today we will go through these two examples.
Example: Determine the plastic section modulus of ISLB 300 @ 0.369 kN/m about thestrong and weak axis (neglecting the fillets)Solution:For symmetrical I-section the equal area axis zz and yy will pass through the centroid of thesection.Zpz=2(bf tf )× (D−2 tf )+2[{tw×(D2 −tf)}×(D2 − 2 tf)]¿bf tf (D−tf )+ tw(D−4 2tf)2Zpy=4×[(b2f ×tf)× b4f]+2×[{(D−2tf)× t2w}× t4w]¿ tf bf22+(D−2tf)tw 24
The relevant properties of ISLB 300 @ 0.369 kN/m.D = 300 mm, bf = 150 mm, tw = 6.7 mm, tf = 9.4 mm,Zpz=bf tf (D−tf )+ tw(D−4 2tf)2¿150×9.4× (300−9.4 )+6.7×(300−2× 9.4)24= 542.2 × 103 mm3Zpy=tf bf22+( D−2tf)tw
¿ 9.4×15022+(300−2×9.4)× 6.724= 108.9 × 103 mm3
Video 2
Example: Steel beams having a clear span of 8 m are resting on 200 mm wide endbearings. The beams spacing is 3 m and the beams carry a dead load of 4.5 kN/m2including the weight of the section. The imposed load on the beam is 13.25 kN/m2. Thebeam depth is restricted to 500 mm and the yield strength of the steel is 250 N/mm2 and islaterally supported.Solution:Factored loads:Total (Dead Load + Imposed load) = (4.5+ 13.25)=17.75 kN/m2The beams are spaced at 3 m intervals, therefore the load per meter= 17.75 × 3 = 53.25 kN/m2Total factored load = 1.5× 53.25 = 80 kN/mEff. Span = 8 + 2×0.1 = 8.2 mMid span moment = 80 × 8.22/8 = 672.8 kN-mReactions at support = 8.2 × 80/2 = 328 kNSelection of section:Plastic section modulus required
Zp=M ×γm0 f y =672.8×106 ×1.1 250=2960.32×103 mm3
The section with largest plastic modulus under 500 mm depth restriction is ISHB 450 @0.907 kN/m with plastic section modulus 2030.95 × 103 mm3 which is less than requiredvalue. The section must be strengthened with additional plates to provide the required plasticsection modulus.The stiffness required to be provided can be calculated as follows:
Max. deflection = Eff. span/360 = 8200/360 = 22.78 mmSo, required moment of inertia of the beam due to un-factored imposed load,IZ=5384×53.25×8200 42×105×22.7 8 =68807×104 mm4Additional plastic section modulus to be provided by the plate =(2960.32×103 – 2030.95×103) = 929.37×103 mm3
Assume thickness of the plate is 14 mmThus, the total depth of the beam = 478 mm.Distance between the c/c of the plates = 464 mm.So, required area of plate = 929.37×103 /464 = 2003 mm2So provide area of plate = 2200 mm2.Thus the width of plate = 2200/14 = 158 mmThus let provide plate of size 200×14
Thus plastic section modulus of the built up section = 2030.95×103 +200×14×464/2×2 =3330×103 mm3This is greater than required plastic section modulus 2960.32× 103 . OK
Video 3
Check for deflection:Maximum Iz required is 68807×104 mm4Iz provided by ISHB 450, 40349.9×104 mm4Iz provided by plate = 2×200×14×(225+7)2 = 30141× 104 mm4Total Iz provided = (40349.9×104 + 30141× 104 ) = 70490.9×104 mm4 greater than Iz required(= 68807×104 ) OK
Moment capacity of the beam ISHB 450,M = 2030.95 × 103 × 250/1.1 = 461.58 kN-mAt any point distance x from the support,461.58×106 = 328×103x – 80x2/2or, x = 6396.5 , 1803.05Hence the theoretical cut off point is 1800 mm from either side.
Check for Shear:Shear capacity of section,Vd=f yγm0×√3 × D ×tw=1.1250 ×√3 ×450× 11.3=667.23 kN0.6Vd = 0.6×667.23 = 400.33 kN > 328 kN. Low shear OKOkay that means the cut-off point I can make upto 1800 mm say upto this I can providesimply I section and after that we have to provide the built up section, so in this way I canmake use of the steel economically, right. So in plate girder in fact this concept will be usedso that at a particular section we will provide the required amount of steel to withstand thebending moment and we will be go on increasing or decreasing the section size according tothe requirement of the bending moment. So let us conclude with this thank you.
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