Notes
Study Reminders
Support
Text Version

### Design of Laterally Unsupported Beams

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

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

Video 1
Hello today I am going to discuss about the design steps for Laterally Unsupported Beams. In fact in last lecture we have discussed how to calculate the Lateral Torsional Buckling moment and what will be the lateral torsional buckling stress and then we have found how to calculate the design bending strength. Today with the based on the last days lecture we will follow certain design steps and after that we will go through one example.Now in design steps basically we will try to find out first a appropriate section based on the approximate section modulus and then we will check whether the assumed section is safe against the bending forces coming on to the member then we will check for shear whether it is safe or not, then we will go for checking deflection, web buckling and web crippling.
So coming to design steps first we will calculate what will be the service load acting on the beam. So once we calculate the service load then we can find out the factored load. So after calculation of factored load we can find out the factored maximum bending moment and shear force.
After that we can start with a trial plastic section modulus means we can find out a plastic section modulus based on this formula that, Zp=Mdf y/γm0 . Remember this is considering the section to be laterally supported. But in case of laterally unsupported beam a major amount of stress is reduced due to lateral torsional buckling. So the section modulus whatever coming here will not be sufficient, wehave to increase substantially. So that the chosen section is safe against this bending moment due to lateral torsional buckling. So what we can do we will choose a higher plastic section modulus which is necessary to account for lateral torsional buckling. Actually we do not know exactly what percentage of increase is required. So we can try with 40 percent or 50 percent increased value of plastic section modulus, however it is a trial and error process, so finally we have to do the trial method and we have to find out the actual requirement. So after finding an appropriate section modulus we can choose a suitable section, based on that plastic section modulus.
And then with that section we will check whether the section is capable of taking that much moment due to lateral torsional buckling or not, if that is fine then we can go to step 4, wherewe have to check the beam for shear. So we will check for shear, if the design shear stress is more than the shear force coming on to the beam then it is fine or we have to increase the again we have to increase the section size to take care the shear.
In step 5, we will check for deflection as per Table 6 where the limiting deflection is given and we know, what is the maximum deflection of that particular beam, based on the loading condition and support condition. So based on the loading condition and support condition we can find out the maximum deflection on that particular beam and we will check whether the maximum deflection is exceeding the limit permissible limit or not. If maximum deflection is exceeding the permissible limit then again we have to increase the section size to accommodate this otherwise if it is not then the section is safe from serviceability point of view then what you can do we can go for next step.Next step is the web buckling, so we know the beam web may buckle due to the consistent load acting on the member or at the support. So we have to check the web buckling and if the buckling strength is more than the force coming on that particular place then it is fineotherwise we have to increase the section size or we can increase the bearing length, if we increase the bearing length then also we can increase the buckling strength and we can keep it safe. Once it is done then we will go for web crippling. So this is the process which we have to follow to design a laterally unsupported beam. So the process is basically a trial and error process.
Video 2
Example: Design a simply supported steel joist of 5 m effective span, carrying a uniformlydistributed load 12 kN/m if compression flange of the joist is laterally unrestrained.SolutionStep-1: BM & SF on beam
Load on the beam = 12 kN/mFactored load = 12 × 1.5 = 18 kN/mMax. B. M. = 18 × 52/8 kN-m= 56.25 kN-mMax S. F. = 18 × 5/2= 45 kNStep-2: Selection of initial section,Zp=Mf y/γm0=56.25×106250/1.1 =247.5× 103kNmIncreasing 50%, the required Zp will be 1.5 ×247.5×103= 371.25 ×103kNm
Step-3 : Calculate bending strength of section,Select ISHB 200 @ 40 kg/mD = 200 mm ry = 44.2 mm Zpz = 414.23 × 103 mm3bf = 200 mm R1 = 9 mm Zez = 372.2 × 103 mm3tf = 9.0 mm tw = 7.8 mm d = 200 – 2×(9+9) = 164 mmAfter calculation it is seen that the section is not adequate and need to increase the sectionsize to ISLB 325…
Select ISLB 325D = 325 mm ry = 30.5 mm Zpz = 687.76 × 103 mm3bf = 165 mm R1 = 16 mm Zez = 607.7 × 103 mm3tf = 9.8 mm Ixx = 9870 × 104 mm4
tw = 7.0 mm Iyy = 510.8 × 104 mm4d = 325 – 2×(9.8+16) = 273.4 mm
Section classification:b/tf = 82.5/9.8 = 8.41 < 9.4d/tw = 273.4/7.0 = 39 < 84Hence, section is plastic.Calculation of bending strength:KL/ry = 5000/30.5 hf/tf = (325 – 9.8 )/9.8= 164 = 32.16From Table 14, IS 800: 2007fcr,b = 122.82 N/mm2From Table 13(a), IS 800: 2007,fbd = 93.17 N/mm2So, Md = 1×687.76×103×93.17= 58.57 kN-m > 56.25 kN-mOK.
Step-4: Check for shear:Design shear strength of the section,Vd=f y√3γm0 D tw=√3250 ×1.1 ×3 25×7× 10−3=299 kN > V=45 kNStep-5: Check for deflection:δ= 5w l4384 EI=5×12×50004384×2×105× 9870× 104=4.9 mmAllowable maximum deflection, L/300 = 5000/300 = 16.67 mm.Hence, safe against deflection.
Video 3
Step-6: Check for web buckling:Assuming stiff bearing length 100 mmn1 = D/2 = 325/2 = 162.5 mmC/S area for web buckling Ab = (b + n1) × tw
= (100 + 162.5) × 7.0 = 1837.5 mmEffective length of web, leff = 0.7 × 273.4 = 191.38 mmI = 100 × 73/12 = 2858.33 mm
A = b × tw = 100 × 7 = 700 mm2rmin = √2858.33 700 =2.02λ = leff/rmin = 191.38/2.02 = 94.742From Table 9(c), IS 800: 2007, fcd = 114.364 N/mm2Capacity of the section, Ab × fcd = 1837.5 × 114.364 = 210 kN> 45 kNHence, the section is safe against web buckling.Step-7: Check for web crippling:Fcrip=(b1+n2 ) ×tw× f yγm0n2=2.5 (16+9.8)=64.5 mm
Fcrip=(100+64.5) ×7 × 2501.1=261.70 kN > 45 kNHence, the section is safe against web crippling
Now I will go through a few slides that is the design of beam with GUI based MATLABalgorithm. So here what we have seen that especially when the design has to be done for unsupported beam we need to do lot of iteration. So manually to do all these things is veryhectic therefore what we suggest that if we can develop a GUI based algorithm then once it is developed, it can be very useful. So my students has developed a software. I am demonstrating here where the beam design is done for two cases, one is for laterally supported beam and another one is laterally unsupported beam, right. So by choosing aparticular type we can go to the design of that case. So once we choose say for this case we have chosen unsupported beam.
Then if we go to next then we can provide these requirement like the maximum factored shear, maximum factored bending moment, the effective span etc. These are the input which we need to take from the user and maximum deflection coming on the on the beam. Somaximum deflection we can calculate from the formula and this also can be done through algorithm but that will be very complicated because in that case we have to find out what is the support condition, what is the loading condition and because of loading and support condition deflection we have to find out the deflection. Therefore we can calculate manually for particular case and we can enter this value what is the maximum deflection. Then we have to assign the properties of steel, like yield stress, we can give the command to take the default value of grade Fe410 or we can insert the yield stress value manually. In a similar way we can give the command to the input of partial safety factor and young’s modulus.
So once we go to next it will show whatever data we have entered. Next there is two option here one is design for economic section and check for a particular section. So if user want to find out an economical section the program will find lowest size of the section which will be safe under the given condition. In other option we can chose a particular section and we can check whether that section is safe or not.
So after that if we click on proceed then it will show the results. So in the results we will see that the software is giving ISJC 150 as economical section. So now if we click on output.doc then the intermediate calculations are given in the file which can be seen and if one user want to check the intermediate calculations he can cross check from the output file.
And if the user want to redesign, that option is also there, he has to click on redesign then it will go back to the first slide and it will ask for data. So the option of redesign is there which can be used. We can check for a particular section also. So for that we have to click onparticular section and then we will see whether it is okay or not, say for this case the selected beam was not okay so that is shown and there also the user can redesign and can find out some other section which may be safe.
So this is how one can develop the logic and make a flow chart and then can find out algorithm. So I would suggest the viewers to make their own algorithm according to their own logic, and write a program whatever language they are comfortable may be Python, may be C, may be MATLB, may be Fortran and this way he can develop the software and that will be customized software he can use forever.So this is all about the design of beam. In next class we will discuss certain other things like the how to calculate the plastic section modulus that we will see because that I have not covered in this class also purlin and gantry girder will be taught in next few classes