 Notes Study Reminders
Support
Text Version

### Set your study reminders

We will email you at these times to remind you to study.
• Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

Video 1
Based on the last lecture, now we will go to one work out example and we will see how to calculate the design bending strength of a member which is laterally unsupported.
So before going to that example, first I will show the glimpse of the lecture what we have discussed roughly. First we have to find out the XLT value, the bending stress reduction factor XLT, if we can find out the value of bending reduction factor due to lateral torsional buckling then I can find out the design bending compressive stress and once I find out the design bending compressive stress then I can find out the design bending strength.
I can find out XLT from this expression,Χ¿=1∅¿+√∅¿ 2−λ¿ 2 ≤1.0which we have shown earlier and to find out the value of XLT we need to find out value of ∅¿ and where the ∅¿ depends on the imperfection factor for laterally torsional buckling that is α LT and it may be 0.21 for rolled section and 0.49 for welded steel sections and the λLT , the non-dimensional slenderness ratio can be calculated from this formula,λ¿=√βbMZp crf y ≤√1.2 ZMe fcry¿√ffcry, bwhere we need to know again the value of Mcr, which is lateral torsional buckling moment.
So Mcr can be calculated from,Mcr=√{(π(2LE¿)I 2 y)[G It+ π(2LE¿)I2w]}¿ βb Zp f cr ,b
Fcr,b value also can be found from this formula,
f cr, b=1.1 π2 E(rLy¿)2 √1+ 20 1 (L r htfy¿ f )2otherwise as I told we can find out from table 14, in which we do not need to calculate in details through these expressions. So once we find the value of Fcr,b we can find out Fbd correspondingsto Fy and Fcr,b. So once we find Fbd value then we can find out the Md, the design bending strength.
Example: Calculate the design bending strength of ISLB 300 @ 0.369 kN/m considering thebeam to be(a) Laterally supported(b) Laterally unsupportedAssume the design force is less the design shear strength and is of low shear. The effectivelength of the beam (LLT) is 4 m. Assume Fe410 grade of steel
Solution:The relevant properties of ISLB 300D = 300 mm, bf = 150 mm, tw = 6.7 mm, tf = 9.4 mm,R1 = 15.0 mm
rx = 124 mm, ry = 28 mmZpz = 554.32 × 103 mm3, Zez = 488.9 × 103 mm3,Iz = 7333 × 104 mm4, Iy = 376 × 104 mm4d = D – 2(tf + R1) = 300 – 2(9.4 + 15) = 251.2 mmFor rolled section: αLT = 0.21For Fe 410 grade of steel: fy = 250 MPaPartial safety factor: m0 = 1.10
btf =150/29.4=7.98 < 9.4dtw=251.26.7=37.49 < 84Hence, the section is plastic.Since,dtw =251.26.7=37.49 < 67ϵShear buckling check of web will not be required.
(a) Laterally supported beamFor low shear,Md=βb Zpf yγm 0=1.0× 554.32× 103× 2501.1=125.98 kN-m≤1.2 Zef yym0=1.2× 488.9× 103× 2501.1=133.3 4 kN-mHence, design bending strength = 125.98 kN
Video 2
(b) Laterally unsupported beamMcr=√{(π(2LE¿)I 2 y)[G It+ π(2LE¿)I2w]}LLT = 4000 mmG= E2(1+μ)=2×1052×(1+0.3)=76.92×103Torsional constant, It=∑ bi ti 33¿2× 150× 9.433+(300−2× 9.4)× 6.733¿11.12×104 mm4Warping constant, Iw=(1−βf )βf I y hf 2Here, h❑f = c/c distance between flanges =D - t❑f =300-9.4 = 290.6βf=IfcIfc+Ift=0.5 [Since Ifc=Ift ]Thus, Iw ¿(1−0.5)× 0.5× 376 ×104 × 290.62= 7.94 × 1010 mm
Mcr=√{[76.92×103 ×(11.22 π2×2××10 10 4000 45+×π376 22×2××10 104000 45)×7.94 2 ×1010]}= 92.45 kNm
Or,Mcr=π2E Iy hf2L¿ 2 [1+ 20 1 (Lh¿f//r tfy)2]0.5¿ π2×2×105 ×376×104×290.62×40002 [1+ 20 1 (290.6 4000//28 9.4)2]0.5= 96.92 kNm
λ¿=√βbMZpz crf y =√1× 554.32 92.45××10 1036× 250 =¿ 1.22 > 0.4So, the effect of lateral torsional buckling has to be considered.∅¿=0.5 [1+α¿ ( λ¿−0.2)+ λ¿ 2]∅¿=0.5 [1+0.21 (1.22−0.2)+1.222]= 1.35Χ¿=1∅¿+√∅¿ 2−λ¿ 2 ≤1.0¿ 11.35+√1.352−1.222= 0.52
f bd=Χ¿f yγm0¿ 0.52×2501.1= 118.2 N/mm2Md=1×554.32×103 ×118.2×10−6=¿ 65.52 kN-m
Using Table1. KL/ry = 4000/28 = 142.86,hftf=290.69.4=30.9For fy = 250 N/mm2, from Table 14, fcr,b = 155 N/mm2Using equations: f cr, b=1.1π2 E(rLy¿)2 √1+ 20 1 (L r htfy¿ f )2f cr, b=1.1 π2 ×2×105(4000 28 )2 √1+20 1 (290.6 4000 9.4 28 )2 = 153 N/mm2For, fcr,b = 155 N/mm2 and fy = 250 N/mm2 and αLT = 0.21 , fromTable 13(a), IS 800: 2007, fbd = 109.53 N/mm2Md = 1 × 554.32 × 103 × 107.59 = 60.71 kN-mMd=¿ 65.52 kN-m (Using expressions)
So the design bending strength of the member when lateral torsional buckling is considered is65.52 kNm and earlier we found Md as 125.98kNm when the beam is laterally supported.So we could see from this demonstration that the design bending strength of the member isdecreasing to a certain extent if the lateral torsional buckling effect is considered that meanswhen the section is laterally unsupported its bending strength is reduced. The reduction that willdepend on the section shape, the lateral torsional length of LLT and the support condition. In nextclass, we will discuss about the design procedure and we will go through one example to see howto find out a section size for a member when it is laterally unsupported.