 # 16a- Solved Problem 5-10 with more iterations-Fcr value.

## A solved problem 5-10 with more iterations to find available strength.

This is the same solved problem 5-10 from the unified design of steel structures by  Louis  F. Geshwendener, it is required to determine the available strength of a compression member with a slender web, the steel section is W16x26 as a column with lcy =5.0 ft.

The modulus of elasticity value E is equal to 29000 ksi, while the yield stress Fy equals 50 Ksi.

Step-1- Check whether the column is long or short, by using the limiting ( kL/r) value for the solved problem 5-10, estimated from the formula (KL/r)=4.71sqrt(E/FY)=4.70SQRT(29000/50)=113.43.

### Confirm that the column is short.

Step-2-Get the Area of the section, ry radius of gyration about y, for the solved problem 5-10 with more iterations.

From the relevant table of W 16×26, we get the data for the area, try,d, h, t-web, the (K*l/r) at the y- direction=(1*5*12/1.12)=53.57, which is < 113.43, then the column is short.

We want to check the slenderness of the flange and web to check if local buckling might occur whether in the elastic or inelastic range might occur.

We will check both the ratios bf/tf and hw/tw for the given W section. The data can be obtained from Table 1-1 for the given section. Bf, tf, h, t web, section W 16×26 has a footnote c, which means that it does not meet the requirement of h/tw.

### Is column elastic or inelastic?

Step-3- Check the column, whether elastic or inelastic column, for both flange and web, from the previous image we get the values of Bf/2tf=7.97.

We will check against the value of 0.56*sqrt(E/fy), which gives 13.49>bf/2tf of the flange for the solved problem 5-10.

Step-4: Make an estimate for λ2, which is Fy/FE, and get the relevant Fcr, consider Q=1, where Q is the reduction factor, Fe value=99.705 ksi, while Fy=50 ksi. λ2=(50/99.705)=0.5014, we will evaluate Fcr by using the equation fcr=0.658^(λ2Q)(Q*Fy).

We will evaluate fcr by using the equation Fcr=0.658^(λ2Q)(QFy)=0.658^(0.50141)(150)=40.50 ksi.

Please for more details about steps 5 to 7, refer to the previous post.

Step-8-For more iterations, use the new value Q=Aef/Agr=6.83/7.68=0.889, from which we can get a new value of Fcr as =0.658^(0.889*)(0.5014) (50*0.889)=36.90 ksi. The calculation is shown in the slide.

### Find a new value of he for solved problem 5-10 with more iterations.

Step 9: Estimating the new he value after introducing fcr at the formula=36.90 ksi, we get he=11.20″<14.206″.

### Find the new area value for solved problem 5-10 with more iterations.

Step-10 estimates the new Area value for the new iteration, Anet= Agross-hwtw+hetweb=7.68-(14.2060.25-11.200.25)=6.93 inch2, We use this value to get a new value for Q.

The final Fcr value for the solved problem 5-10 with more iterations.

The new Q will be=(6.93/7.68)=0.902.

Step-11 Get the revised Q value is used to get the new value for Fcr by using the equation FCr=0.658^(λ2Q)(Q*Fy)=37.33 ksi.

Step-12 Use the value of A eff and the Fcr to get the LRFD and ASD values, we have A eff=6.93, Fcr=37.33 ksi.

The value of compression Strength for the given column W16x26 as LRFD value will be equal to Φc*Pn=Φc*Fcr*A=0.90*37.33*6.93=233.0 kips.

While for the ASD value=A*Fcr/Ω=7.68*40.50/1.67=155.0.0 kips.

This is the full PDF data with iterations.
This is the next post Alignment chart part 2.
This is a link to A very useful external link Chapter 7 – Concentrically Loaded Compression Members. Scroll to Top
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