- Solved problem 4-14 part 2 for the effective length factor.
- The video I used in the illustration.
- Check whether the column is elastic or non-elastic.
- Get the Area, moment of inertia and radius of gyration for the column.
- Use the alignment chart to get the K value for column AB.
- Use the french equation to match the K value for column AB.
- LRFD design procedure for τb.
- Use Table 4-13 to get τb -LRFD design.
- ASD design procedure for τb.
Solved problem 4-14 part 2 for the effective length factor.
The video I used in the illustration.
For the content of this post, please continue from time 5:00 till the end. The video has a subtitle and a closed caption in English.
You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.
Check whether the column is elastic or non-elastic.
The third step for a solved problem 4-14 part 2 will be to check, if K*L/r<4.71*sqrt(E/fy) or not, if the column is the inelastic column, then we continue to estimate τb from the general equation or using the table.
Get the Area, moment of inertia and radius of gyration for the column.
For column AB, W10x33, from the excel sheet, we get Ag=9.71 inch2, Ix=171.0 inch4, and rx=4.19 inch2. at joint A, we have two beams each has W12x14, Ag=4.16 inch2, Ix=88.60 inch4, and rx=4.62 inch2.
The length for the beam is L=18′ and L=18′. For column length =12′, considering E is elastic for the column and beams, beams are always considered elastic.
The terms E, and E will cancel each other, the numerator =170 /12 /∑EI/L for beams, which will be (88.60)* (88.60)* (1/18), one column length =18′, while the other column L=20*1/(Ix*sum(L/18 +L/20), the value will be (14.25/9.3522) GA=1.52.
At joint B, we have two beams with section W14x 22 for each we estimated the area of the beam section as, Ag =6.49 inch2, Ix=199 inch4,rx=5.54 inch2, the inertia of the column is > inertia of the beam.
The same procedure is to be done for the Gb, but consider that at B, we have two columns. The length of each column =12′, so the sum ∑EI/L column AB=2*171.0/12 again to be /(199/18+199/20)=GB= 1.36. This column AB is an inner column.
Use the alignment chart to get the K value for column AB.
We are going to use the alignment chart for the un-braced frame in the solved problem 4-14 part 2 to estimate the K value, at the left we select GA=1.52, and Gb =1.36 at the right side of the graph, joining the two values by a line we get Kx is approximately=1.45.
To use the french equation, then k= SQRT(1.60*GA*Gb+4(GA+GB)+7.50)/SQRT(GA+GB+7.50), we make a substitution for the different values, and we get k=1.466.
Use the french equation to match the K value for column AB.
We proceed, having k=1.45, then KL/r at x=1.45*12*12/rx of the column which=4.19 inch2, K*L/r at x=49.83 which is < Kl/r at fcr=Fy/2.25,=113.43, estimated from 4.71*sqrt(29000/50), E is still=29000 ksi for comparison.
The column is a short or intermediate column.
LRFD design procedure for τb.
We need to make the reduction of stress and use the τb. first, we start with the LRFD, estimate Pultimate as as=1.20 D+1.60L, D=35.50 kips and L=142.0 kips, 1.2D+1.60L=1.2*35.5+1.6*142= 269.80 kips.
First, check that α*Pr/Pns <=0.50, we have τb=1, else then τb <1.0, α =1.0 for the case of LRFD, Pr=Pult=269.80 kips. Py=Ag*Fy=9.71*50=485.50 kips, αPr/Pns will be=(1*269.80/485.50) =0.557 >0.50, then τb<1, but if αPr/Pns<0.50, then τb=1.
But in our case τb<1, then use this equation τb= 4(αPr/Pns)*(1-(αPr/Pns)).
We substitute the values as τb= 4*1*269.80/485.50)*(1-(1*269.80/485.50)), τb=0.9865.this is the the reduction stress factor. Prof. Segui has estimated the value as=0.9877.
Use Table 4-13 to get τb -LRFD design.
The next slide image shows how to derive the value of τb, we have pult/A=27.80, while for Fy=50.0ksi, τb is between 0.986 and 0.994.
We need to adjust the k value, this is done by re-adjustment of GA, Gb, Ga=1.52*0.9877=1.5013, while Gb=1.36*0.9877=1.3432.
ASD design procedure for τb.
Due to inelastic behavior, GA=1.5013 in place of 1.52 and Gb=1.3432 instead of 1.36, then from the chart, we get Kx final=1.43, while from kx french=1.46,
Thus we have completed the LRFD estimation part, k value at the x-direction =1.43 and Ky=1, since it is braced in the perpendicular direction.
In the ASD calculation, we have D=35.50 kips and L=142.0 kips. Pt= 35.50+142.0=177.50 kips, we check for but if αPr/Pns<0.50, then τb=1, else τb<1.
But α=1.60, pr=177.50 kips , Pns=485.50 kips. Then αPr/Pns=0.585>0.50, τb<1, then use the equation τb= 4(αPr/Pns)*(1-(αPr/Pns)).
We make a substitute by the known values, we get τb=0.9703,
From Prof. Segui’s calculation τb=0.9703, which is extracted from table 4-13 as shown in the next slide image. how to use table, 4-13 is shown in full detail for the ASD design.
We need to adjust the k value, this is done by re-adjustment of GA, Gb, Ga=1.52*0.9703=1.47, while Gb=1.36*0.9703=1.32.
The value GA inelastic will be equal to 0.9703*1.52=1.47, While Gb is inelastic=0.9703*1.36=1.32, which means that G values are reduced due to inelastic behavior. k x =1.43 at inelastic buckling. while ky=1, thus we have come to an end for the solved problem 4-14 part 2.
This is the pdf file used in the illustration of this post and the previous post.
This is the previous post, solved problem-4-14 part-1.
A very useful external link.A Beginner’s Guide to the Steel Construction Manual, 15th ed.