brief illustration for post 9A-Solved problem-7-4-1-part 2.

9A-Solved problem-7-4-1-part 2.

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solved problem-7-4-1-part 2.聽

This is the second part of the solved problem 7-4-1, where it is required to design a W section for a given beam with a span of 20 Feet. The steel is A572, where Fy=65 ksi.

For Case-3, design a steel beam with Fy=65 ksi. Solve the problem: 7-4-1-part 2.

Find the Ultimate Moment.

The next slide includes the value of the ultimate load without considering the beam鈥檚 weight, which will be added after selecting the section. The ultimate load is 1.52 kips/ft, and the ultimate moment is 76 ft. kips. The beam鈥檚 flange is fully supported against lateral movement.

Solved problem-7-4-1-part 2, The ultimate load for steel beam grade 65.

We will equate the ultimate moment with the 蠁b*Mn or 蠁b*Fy*Zx.the Fy=65 ksi, 蠁b=0.90, Mu=76 Ft.kips.

Determine the required Zx-select a section.

The required Zx value is 15.589 inch3. Using Table 3-2 for W sections sorted based on Zx, we will select a W12x14 with a Zx value of 17.40 inch3, Zx selected > Zx required. Please refer to the next slide image for more details.

Determine the plastic section Zx for Fy=65 ksi.

We need to check whether W12x14 meets the requirement of local buckling parameters. The next slide shows the controlling parameters for the flange and web.

What are the local buckling parameters based on Fy=65 ksi?

Based on Fy=65 ksi, the lambda for flange at plastic stage 位Fp=8.03. The lambda for flange at elastc stage 位Fr=21.12. The lambda for web at plastic stage 位wp=79.42. The lambda for flange at elastc stage 位wr=120.40.

The values for parameters 位Fp and 位r for flange, 位Wp and 位wr for web-Fy=65 ksi.

Check flange 位 and web 位 against local buckling parameters.

In the second part, we use Table 1-1 for W12x14 to find bf/2tf, which equals 8.82, while h/tw equals 54.30. bf/2tf is bigger than 位Fp but smaller than 位Fr, which means that the flange is a non-compact section. For 位w is smaller than 位wp, meaning the web is compact.

The Nominal Moment for W12x14 is between the value of (Fy*Zx) and (0.7*Fy*Sx). The elastic section modulus Sx=14.90 inch3, while the plastic section modulus Zx equals 17.40 inch3.

check whether W12x14 is compact or non compact-Fy=65 ksi.

Find the Nominal moment of W12x14.

For the next slide, the flange lambda value 位F =8.82,Fy=65 ksi.

Our 位f=8.80, in between 位f-p and 位f-r. It is required to get the Mn value.From Equation F3-1, Mn=(Mp-(Mp-0.70Fy*Sx)/(位fr-位fp)). This is the last step of designing a steel beam under Fy=65 ksi.

Mn is the same as the equation for a straight line y=m*x. We will estimate the Mn for the upper聽 Point=Mp=Fy*Zx=65*17.40=1131 inch. Kips. The second point Mn=0.70*Fy*Sx, Sx for the section=14.90 inch2.
0.70*Fy*Sx=0.70*65*14.90= 677.95 inch. Kips.

Find the slope of the Linear portion.

We can find the slope of the linear portion, which equals (Fy*Zx-0.70Fy*Sx)/位fr-位fp). The next slide shows the slope鈥檚 value as 34.61-inch. Kips.

Find the slope of the linear part of M n and lambda for flange.

Now, we can find the Nominal moment equal to Mp-S*(位f-位p)=1103.66 inch. Kips. We can find the value in Feet Kips as 92.0 ft. kips. The 桅b value is 0.90. the factored Mn=82.80 Ft.kips.

Check whether Mult is bigger than 桅b*Mn or not.

We will finalize the ultimate moment value by adding the W section鈥檚 own weight moment. The final Mu value is 77.0 feet. KIPS. The 桅b*Mn=83.00 feet. Kips. The section is safe since 桅b*Mn>Mult. This is the end of the post. Thanks a lot.

Check whether section is adequate.

For the first part of this post, please refer to post 8: How to design a steel beam? Solved problem-7-4-1.

Here is the link for Chapter 8 鈥 Bending Members.
This links to the next post, 10-lateral-torsional buckling for steel beams.

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