Solved Problem 4-6-How To find The Available Flexure Strength?
Content of the video.
We will continue the previously solved problem 4-5, but we will increase the bracing distance to 6 feet.
As a reminder of the example, we have quoted this problem from Prof. Alan Williams’s book.
It is required to design W-section based on the given M ultimate =270.0 Ft-kips, Fy=50 KSI.
The bracing distance=4′.
We have solved problem 4-6, in which we have obtained Lp= 5.55′, which is > bracing distance Lb which is 4′, then the nominal available flexure strength Fy* Zx, then the LRFD value, Φb*Mn=Mult= Φb*Fy*Zx or the LRFD, while for the ASD, Mn/Ωb=Mt=Fy*Zx/Ωb, now we have Lb=6′.
As a reminder, the Lp distance between bracings Lp= ry* 300/sqrt(Fy), ry from the section. =1.57 inch2, where- ry is the radius of gyration about the y-axis, then Lp=1.57*300/sqrt(50)=66.60 inch.
To convert to ft, to be divided by 12 so Lp=5.55′.
As per the given solved problem, 4-6. Our given bracing distance is Lb=6′ >Lp >5.55′, what about the Lr distance, and how to estimate it?, this will be our next step, by God’s will, to estimate Lr from the equation.
We need to find some data from table 1-1 for w-shapes Sx=64.70 inch2, this is highlighted by the yellow color Zx=73.0 inch2, plastic section moduli. This is a part of the video, which has a subtitle and closed caption in English.
You can click on any picture to enlarge then press the small arrow to review all the other images as a slide show.
A solved problem 4-6 for lateral-torsional buckling, when lb>Lp but<Lr.
From Prof. Alan Williams’s book Structure -Reference manual, solved problem 4.6 A simply supported W16 x40 beam of grade 50 steel is laterally braced at 6 ft intervals and is subjected to a uniform bending moment with Cb =1.0. Determine the available flexure strength of the beam.
A solved problem 4-6 from Prof. Alan Williams‘s Structural Engineering Reference Manual, The solved problem is similar to solved problem 4-5, except that, the Lb is increased to 6 ‘.
Topics included in our discussion are shown in the next slides.
This graph represents the relation between Lb and the nominal moment Mn, it has three zones based on the value of Lb and its relation with Lp &Lr.
Analysis for the given section by the LRFD design.
Solved problem 4-6 is an analysis problem, the section is given for which, the distance between bracing for a beam is Lb > Lp, but Lb <Lr, for the LRFD design.
The steps for the solution are as follows:
1-get the Zx value for the given W section from table 1-1.
3-From table 1-1 get the Sx value, ry for the selected section, and either estimate Lp, from the equation LP= ry*300/ sqrt(Fy) or from Table 3-2, as we can see from the table Lb=5.55 FT.
These are the values of Sx and Zx from table 1-1 for the w sections.
4-Check Lr value either from the following equation or get the value from table 3-2 for the selected section, which is 15.90′.
From the previous equation F2-6, we need the following values from Torsional properties: J, CW, and other properties, tf, Sx, rts, ho, selected from table 1-1.
This is the equation for Lr value for the bracing distance.
This is the detailed reference equation number as presented in the AISC code.
This is the detailed estimation of the value of lr by using the equation.lr=15.9′.
This is the value of lr by using table- 3-2.
5- For the given Lb check if Lb>Lp and Lb <Lr, then the section is not compact, the value of φbMn is < φb*(Mpx), but φbMn > φb*(Mrx), where Mpx=Fy*Zx, while Mrx= (0.70*Fy*Sx).
6-Estimate φb*Zx*Fy for Lb and φb*Fy*Sx for Lr.
7- Estimate the value of φb *BF.
8– The final φb*Mn= φb (Zx*Fy)- φb *BF*(Lr-Lb), this is explained as per the next picture.
The available flexure strength is based on the LRFD, φb *Mn=269.50 Ft.kips.
This is the value of φb*Mp and φb*Mr by using table 3-2
These are the detailed calculations for the LRFD value of moment by using BF as shown in the next slide image.
The analysis for the given section by ASD.
For the ASD design, to get the flexure strength, follow the next steps:
1-Get a preliminary Zx value by considering that φb*Mn=Mult, since Mn=Zx*Fy. We can get Zx from the relation Zx*Fy = Mt / Ω.
2-From table 3-2, select the lightest w section, that gives Zx>Zx preliminary.
3-From table 1-1 we get the Sx value, ry for the selected section, and either estimate Lp. LP= ry*300/ sqrt(Fy), from the Table 3-2.
4-Check Lr value either from the following equation:
Lr =1.95*rts*(E/0.7 Fy)*SQRT(((J*C/(Sx*ho)+SQRT(J*C/(Sx*ho)^(2)+6.76*(0.7Fy//E)^(2))), or from table 3-2 for the selected section.
If Lr by the equation. We need the following values from Torsional properties: J, CW.
5- For the given Lb check is Lb>Lp and Lb <Lr, then the section is not compact, the value of Mn/Ω is < (Mpx)/ Ωb, > (1/Ωb)*(Mrx)
Mpx=Fy*Zx, Mrx= (0.70*Fy*Sx).
6-Estimate (1/Ωb)*Zx*Fy for Lb and (1/Ωb)* 0.70*Fy*Sx for Lr.
7- The final (1/Ωb)*Mn= (1/Ωb) (Zx*Fy)- (1/Ωb) *Bf *(Lr-Lb).
The available flexure strength based on the ASD, (1/Ωb)*Mn=179.00 Ft.kips.
This is the value of (1/ Ωb )*Mp and ( 1/Ωb )*Mr by using table 3-2
These are the detailed calculations for the ASD value of moment by using BF as shown in the next slide image.
This is a link to download the pdf file used for the illustration of this post.