Part-2-4 for the Solved Problem 9-9-6, How To Find LL?
Part 2 fo the solved problem video.
This is the second video, Part-2-4 for the Solved Problem 9-9-6. We continue discussing the solved problem of Prof. Salmon, which includes How to determine the value 0.70*Fy*Sx, the lower point in the inclined line? 0.30 Fy*Sx accounts for the residual stresses.
This is the right-side graph for the local buckling for the web for the build-up of the I-beam section.
This x-axis is for the lambda value λ for the web, while for the left-side graph, the x-axis is for λ flange, Sx, which is the First Moment of area. Sx=Ix/y or the statical moment of area.
We want to estimate Mn for both the flange and web based on the limit- state of local buckling. By using the lambda values we can get the Mn for both flange and web, This is a part of the video that has a subtitle and a closed caption in English.
Sx for the built-up section.
This is part-2-4 of the solved problem 9.9.6 from Prof. Charles Salmon’s book. We are going to estimate the elastic section modulus Sx for the built-up section.
Ix for this section can be estimated for the whole rectangular section. For the Ix, where the x-axis is the major- axis, the easiest method to evaluate the Ix for the section is by considering the whole rectangle section (16″x27.25″), then deducting the two void rectangles of dimensions (7.843″x26″).
The moment of inertia can then be estimated as Ix=(16*27.25^3/12)-(2*7.843*(26)^3/12))=4002.8116 inch4. the distance from the N>A to the top of the upper flange Y value =27.25/2=13.625″. the section modulus Ix/y=Sx=(4002.8116/13.625)=293.80 inch3.
Zx should be > Sx, we have estimated Zx=319.06 inch3 which is > 293.80 inch3. we can proceed to estimate the Nominal moments.
Mn-based on the local buckling of the flange.
In part-2-4 of the solved problem 9.9.6. Two graphs are drawn the left-side graph is for the local buckling for the flange, and the values are as follows, in which the x-axis represents lambda while the vertical axis is the nominal moment value. Both flange and web are non-compact sections, The Mn value will be >0.70 Fy*Sx but less than Fy*Zx.
λp=8.026 For which the Mn=Fy*Zx=65*319.06/12=1728 Ft- kips. λr=15.90 For which the nominal moment Mn=0.70FySx=0.765*293.806/12=1114 Ft- kips. The Nominal moment of the flange will be with a value between 1728 and 1114 Ft.kips.
When the factor λflange is equal to 12.80 for which the nominal moment value Mn=1728-(1728-1114)*(12.80-8.026)/(15.90-8.026)=1355.70 Ft-kips, this represents the Mn-based on the limit state for the local buckling of the flange.
The nominal moment Mn-based on the local buckling of the Web.
In part-2-4 of the solved problem 9.9.6. Two graphs are drawn from the right-side graph that represents the local buckling for the web, the values are as follows, The plastic lambda value λp=79.46 for which the nominal moment The value of the nominal moment Mn=Fy*Zx=65*319.06/12=1728 Ft- kips.
λr=120.40 For which the Mn=0.70Fy*Sx=0.7*65*293.806/12=1114 Ft- kips.
When the factor λweb=83.20 for which the nominal moment can be estimated as Mn=1728-(1728-1114)*(183.20-79.46)/(120.40-79.46)=1627.0 Ft-kips, this is the Mn-based on the limit state for the local buckling of the web. The detailed calculations for the nominal moment are shown in the next slide image.
This is the pdf file used in the illustration of this post.
The next post will be Part 3/4 for the Solved problem 9-9-6, how to find LL?