# 28-Solved problem 4-14 part 2 for effective length factor.

## 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.

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