Category: Elastic and plastic sections.

All problems that deals with the estimation of elastic and plastic section Modulus.

  • 6d-Practice problem 6-17-11-find Sx and ZX for WT5x22.50.

    6d-Practice problem 6-17-11-find Sx and ZX for WT5x22.50.

    Practice problem 6-17-11-find Sx and ZX for WT5x22.50.

    Practice problem 6-17-11-find Sx and ZX for WT5x22.50

    Determine the elastic neutral axis, elastic section modulus, plastic neutral axis, and plastic section modulus for a WT5×22.5 modeled as two rectangles forming the flange and the stem. Compare the calculated values to those given in the Manual. This practice problem is quoted from the Unified Design of Steel Structure Book.

    We will use Table 1-8 to get the information for WT5x22.5, which includes the flange width and thickness and the overall height and width of the stem.

    The sketch shows that the flange width is 8.02 inches, while the thickness is 0.62 inches. The stem width is 0.35 inches, The total depth is 5.05 inches, and the area is 6.63 square inches.

    Please refer to the next slide image for more details about WT5x22.50 in part 1 of Table 1-8.

    Practice problem 6-17-11-find Sx and ZX for WT5x22.50.

     

    How do we find the values of the elastic section modulus and plastic section modulus, using Table 1-8?

    In the next slide, refer to part 2 of Table 1-8; the value of the elastic section modulus is 2.47 inch3, while the plastic section modulus zx is 4.65 inch3.

    The value of elastic section modulus and plastic section for Wt section

    In the third slide, the section of WT5x22.5 is drawn as a T shape with the different items explained for flange and stem. Next to the WT section is an assembly of two rectangles representing the section.

    The upper rectangle, which has a width of 8.02 inches and a width of 0.62 inches, represents the upper flange. The second rectangle, the stem part, has a height of (5.05+0.62)=4.43 inches and a width of 0.35 inches. We need to find the moment of inertia value for the T section.

    The location of the elastic neutral axis from the top.

    To determine the distance from the top of the WT section to the Neutral axis, we need to estimate the area of the upper flange and the stem and find the product of areas by the cg distance about the top of the upper flange. The area of the upper flange equals 8.02*0.62=4.97 square inches, while the area of stem=0.35*(4.97)=1.74 square inches. The total area is 6.55 square inches. The Y1 distance can be estimated by dividing the sum of (A*Y) by the total area. The sum of the moment of area equals 5.935 inch3.

    The y1 distance from the top of the flange to the E.N. axis is 5.935/6.522 = 0.91 inches. The distance from the E.N. axis to the bottom is 4.14 inches.

    Modeling of WT5x22.5 as two rectangles find neutral axis &y1 distance.

    The moment of inertia of the WT 5×22.50 about the E.N .axis is the sum of the inertia of the top flange and the inertia of the stem part. The moment of inertia Ix1 for the flange part equals 1.949 inch4. The moment of inertia for the stem part Ix2 equals 8.279 inch4. Thus, Ix for the WT section equals 10.228 inch4.

    The elastic section modulus equals Ix divided by the bigger values of y1 and y2; we find that y2 is >y1 and will be selected. The Sx value equals 10.228/4.14=2.47 inch3. Please refer to the next slide image for more details.

    Detailed estimate of the moment of inertia of WT section.

      

    Verify Zx of WT5x22.5 by considering areas of flange and stem.

    Again, we will use the two rectangles representing the WT section to find the value of the plastic section modulus. The Plastic Neutral axis, or P.N. axis, cuts the section so that the area above and below it must be equal. The value of At/2 equals 3.261 inch2.

    Let y1p be the distance from the top of the flange to the required P.N . axis, the area enclosed by that axis, and the upper flange equals 3.261 inch2. The value of y1p equals 3.261/(8.02*y1p), which will lead to y1p equals 0.4066 inches. We can get the vertical distance from The P.N axis to the lower part of the stem to equal 4.643 inches.

    The Zx value is the sum of the product of A*y; we have a flange cut into two parts: the Sum of A*y equals 4.615 inch3. Please refer to the next slide for more details.

    How do we find Zx of WT section?

     The next slide shows the estimated value of Zx for the Wt section versus the tabulated value for the Wt section based on Table 1-8. the estimated value of Zx based on calculation equals to 4.65 inch3, while the tabulated value equals 4.615 inch3. Thanks a lot.

    The difference between the value of Zx.

    The previous post is 6c-Practice problem 6-17-5-find Sx and ZX for W18x35.

    For bending members, please refer to Chapter 8-A Beginner’s Guide to the Steel Construction Manual, 15th ed.

    The following post is “A Guide to Local Buckling Parameters for Steel Beams.”

    There is a newly added post, post-9b-Practice problem 5-4-1-Check compactness for Fy=60 ksi.

  • 6c-Practice problem 6-17-5-find Sx and ZX for W18x35.

    6c-Practice problem 6-17-5-find Sx and ZX for W18x35.

    Practice problem 6-17-5-find Sx and ZX for W18x35.

    Practice problem 6-17-5-find Sx and ZX for W18x35.

    Determine the elastic and plastic section modulus for a W18×35 modeled as three rectangles forming the flanges and the web. Compare the calculated values to those given in the Manual. This practice problem is quoted from the Unified Design of Steel Structure Book.

    We will use Table 1-1 to get the information for W18x35, which includes the flange width and thickness and the overall height and width of the web.

    The sketch shows that the flange width is 5 inches, while the thickness is 0.425 inches. The web width is 0.30 inches, The total depth is 17.70 inches, and the area is 10.30 square inches.

    Please refer to the next slide image for more details.

    Practice problem 6-17-5-find Sx and ZX for W18x35.

     

    The value of elastic section modulus and plastic section modulus is shown in Table 1-1.

    In the next slide, refer to part 2 of Table 1-1; the value of the elastic section modulus is 57.60 inch3, while the plastic section modulus zx is 66.50 inch3.

    The value of elastic section modulus and plastic section modulus from table 1-1.

    In the third slide, the section of W18x35 is drawn as an I shape with the different items explained for flange and web. Next to the W section is an assembly of three rectangles representing the section.

    The upper rectangle, which has a width of 6 inches and a width of 0.425 inches, represents the upper flange. The second rectangle, the web part, has a height of 16.85 inches and a width of 0.425 inches. The last rectangle, which represents the lower flange, has similar dimensions to the first rectangle. We need to find the values of the moment of inertia for the Flanges and the web.

    Modeling of W18x35 by three rectangles.

    The value of the moment of inertia Ix of W18x35 by estimation.

    The overall depth of the W section equals 17.70 inches, and due to symmetry, the elastic section modulus will pass by the middle portion of the W section.

    The moment of inertia of the upper flange and the lower flange about the elastic section modulus is the same. It equals the Inertia about the cg of the flange plus the product of the flange area by the square of the distance from flange Cg to the elastic section modulus.

    The moment of inertia of the upper flange to the neutral axis Ix1 equals 190.284 inch4. The moment of inertia of the web to the elastic neutral axis Ix2 equals the width by height^3/12, which is 119.60 inch4. The final moment of inertia of the three rectangles about the neutral axis equals 500.171 inch4.

    Detailed estimate of the moment of inertia of W section.

     

    The value of the elastic section modulus Sx for W18x50.

    After we have estimated the moment of inertia, we need to find the value of Sx, the elastic section modulus, and the distance from the E.N axis to the top, which is 8.85 inches. Sx equals Ix/y, which is 500.171/8.85=56.52 inch3. The exact value for Sx from Table 1-1 equals 57.60 inch3.

    The value of the elastic section modulus Sx for W18x50.

     

    Verify Zx of W18x35 by considering areas of flange and web.

    Again, we will use the three rectangles representing the W section to find the value of the plastic section modulus. The Plastic Neutral axis or P.N axis cuts the section in the middle due to symmetry. The Plastic section modulus Zx is the sum of Area products by the distance to the P.N axis.

    We will consider the upper rectangle and half of the vertical rectangle. The upper rectangle area, A1, equals width by thickness, which is 6*0.425=2.5 inches. The perpendicular distance from Cg to the plastic axis is 8.85-0.5*0.425=8.6375 inches. The area of half of the web equals 0.5*16.85*0.30=2.528 inch2. The Cg distance to the P.N axis equals 19.85*0.25=4.1225 inch2.

    Please refer to the next slide for more details.

    How do we find Zx of W section?

     

    The next slide shows the steps to get the Zx or the plastic section modulus for W18x35. By estimation, the Zx value equals 65.35 inch3, while the exact value for the plastic section modulus from Table 1-1 equals 66.50 inch3. We could use a quicker way to estimate Zx by considering the section as one big rectangle of b1=6 and d1=17.70 inches, and an inner rectangle of width equals 5.70 and depth d2=16.85 inches. Zx can be estimated as (b1*d1^2/4-b2*d2^2/4), which equals 65.345 inch3. Thank You.

    The final value of Zx.

    The previous post is 6b-Practice problem 5-2-3-verify Zx for W18x50.

    The following post is 6d-Practice problem 6-17-11-find Sx and ZX for WT5x22.50.

    For bending members, please refer to Chapter 8-A Beginner’s Guide to the Steel Construction Manual, 15th ed.

    A newly added post, 6d-Practice problem 6-17-11-find Sx and ZX for WT5x22.50.

  • 6b-Practice problem 5-2-3-verify Zx for W18x50.

    6b-Practice problem 5-2-3-verify Zx for W18x50.

    Practice problem 5-2-3-verify Zx for W18x50

    Practice Problem-5-2-3-verify Zx for W18x50.

    Verify the value of Zx of the W section W18x50, which is tabulated in the dimensions and properties Tables in Part 1 of the manual. We will consider two ways to estimate the Zx of W18x50.

    The first method considers the W section as composed of two Wt sections. each Wt section is WT 9×25.
    The second method considers the W section as an assembly of three plates and finds The Yct and the At/2 values.
    We can represent the W18x50 as two Wt sections assembled.

    As seen in the sketch, each Wt is 9×25. Table 1-8 of the manual has two parts. Part 1 gives us the dimensions of the Wt section. The breadth and thickness of the flange, as well as the overall height and thickness of the stem, are important.

    From part 1, I need to know the area for Wt 9×25, equal to 7.34 inch2, and the depth of the wt section, which is 9 inches.

    Practice problem 5-2-3-verify Zx for W18x50

     

    The distance Y bar from the top of the WT9x25.

    In the next slide, there is part 2 of Table 1-8, from which I need only the value of the y bar, which is the distance from the top of the flange to the Cg of the section. The value is 2.12 inches.

    Y bar value for Wt section.

    I drew the W section on the third slide, assembled from the Two WT sections. In this section, we can find the value of the YCt or the distance from the compression and tension forces, and each force acts on the CG.

    The Yct value equals ( d-2y bar), where d is the overall depth of the W18x50 section, the y bar we got from the previous slide. Then, the distance YCT value is equal to 18-2*2.12=13.76 inches. The area of At/2 equals 7.34 inches2, which we obtained from Table 1-8. We apply the equation of Zx equal At/2*(yct)=7.34*13.76=101.0 inch3.

    How  do we find Zx value for W18x50?

    The next slide shows two parts of Table 1- 1; the first part gives the dimensions of the W section. The flange’s breadth is 7.50″, and the overall depth is 18 inches. The area is 14.70 inch2.

    The dimension of W18x50 from part 1

     

    Find the Zx for W18x50 from Table 1-1.

    In the next slide, we view the second part, which gives the value of Zx, 101.0 inch3. The same is the value estimated via Table 1-8. It matches our previous calculations as a two-weight section. The validation is okay.

    Find the Zx for W18x50 from Table 1-1.
    Zx for W18x50 from Table 1-1.

     

    Verify Zx of W18x50 by considering areas of flange and web.

    If we consider the W section composed of areas, the first is the flange, which is 7.50″ by 0.57″.The second is the web, with a height of (18-2*0.57)= 16.86 inches. The last is the lower flange area, which is 7.50″x0.57″ inches.

    The total area is equal to 14.535 inch2. Consider the plastic section modulus equal to Mp/Fy, which can be rewritten as C*yct/Fy=At/2*yct.

    The area(At/2)=0.5*14.70=7.35 inch2. use the sum of areas*y distance and divide by At/2 to get the H-y bar distance.

    Verify Zx of W18x50

     

    The next slide shows that we get the (h-y )bar value, the distance from the point of application of At/2 top of the Plastic neutral axis, equal to 6.8622 inches, and the Zx value of 99.742 inches, less than 101.0 inch3.

    The final value of Zx for W section.
    The final value of Zx for the W section.

    The previous post is “6A-Practice problem 5-2-2 Find y bar, Zx, and Zy for the un-symmetric section.

    For bending members, please refer to Chapter 8-A Beginner’s Guide to the Steel Construction Manual, 15th ed.

    The following post is “6c-Practice problem 6-17-5-find Sx and ZX for W18x35.

  • 6A-Practice problem 5-2-2 Find y bar, Zx, and Zy.

    6A-Practice problem 5-2-2 Find y bar, Zx, and Zy.

    Practice problem 5-2-2: find y bar, Zx, and Zy for the unsymmetric section.

    Practice Problem-5-2-2 from the steel design handbook.

    There is an unsymmetric flexural member consisting of three parts. The upper flange is 12×0.50 inches, the web part is 3/8”x 16”, and the lower flange is 7×1/2”. There are three requirements: a), b), and c).
    For Part a), the y-bar or the distance from the top of the upper flange to the plastic neutral axis must be found.

    First, we estimate the total area of the three parts: the area of the upper flange equals 12×1/2, the area of the web equals 16×3/8, and finally, the area of the lower flange equals 7×1/2. The total area is equal to 15.50 inch2.

    First, we estimate the total area of the three parts: the area of the upper flange equals 12×1/2, the area of the web equals 16×3/8, and finally, the area of the lower flange equals 7×1/2. The total area is equal to 15.50 inch2.

    Practice problem 5-2-2 find y bar, Zx, and Zy for the unsymmetric section.

     

    The distance Y bar from the top of the shape to the P.N.axis.

    In the next slide, since the area of the top flange is less than At/2, the plastic neutral axis cannot cross the upper flange so that it will be located below the upper flange by a distance; we call it y1.

    After checking it, we can find the y1 value by equating the difference between At/2 minus the upper flange area by the product of (3/8 y1). y1 will be equal to 4.667 inches.

    Y1 distance below the upper flange

    In the third slide, we will estimate the y-bar distance, which is the distance from the top of the upper flange to the P.N axis.

    If we add ½ inches to y1, we will have a y-bar equal to 5.166 inches, approximated to 5.17 inches. We could check the lower portion’s area to ensure it equals At/2.

     

    Y bar distance from the top of the upper flange.

    From the next slide, as required by item b), we need to find the Plastic moment in case A572 steel grade 50 is used. The yield stress Fy equals 50 ksi. We will estimate the compression force C, which equals At/2*Fy=7.75*50=387.50 kips.

    The value of the compression force acting of the section.

     

    The distance C1 from the compression force to the P.N axis.

    We will estimate the C1 distance, which is the distance from the point of application of force c to the P.N.A. The value of c1 can be estimated by taking the first moment of area for the upper half of the section about the P.N.A.The moment of area equals 33.609 inch3.

    When we divide the first moment of the area by At/2, we find that the value of C1 equals 4.34 inches.

    The distance of application of C force to neutral axis.

     

    The distance T1 from the Tension force to the P.N axis.

    We move to the next slide to estimate the distance T1, which is the distance from the point of application of Force T to the P.N.A. Since we do not have symmetry, we expect a different value than c1.
    Again, we will estimate the moment of area of the lower half of the section, which is the sum of the moment of area of the lower flange and the web area part below the P.N axis. The value of T1 equals 8.34 inches.

     In the next slide, we have the values of the C and T forces and c1 and T1 the two distances. Mp equals C*(C1+T1) or C*yct, which is the vertical distance between force C and force T.

    The Plastic Moment about P.N axis-part b.

    The Mp value equals (387.50×12.68); we divide by 12 to find the value of Mp in Ft-kips. The Mp value is equal to 409.46 Ft-kips, which is the requirement for item b).

    The value of the plastic moment Mp about The P.N axis.

     

    The value of a section modulus about the minor axis y.

    Due to symmetry about axis y, we will deal with the C-shape section, find the first moment of area about the y-axis, and divide by the sum of the three shapes. We will find that X—X-bar equals 1.592 inches.

    Finally, we will estimate the yielding MomentMy (C*2x bar) and divide by Fy. The final value of the plastic section modulus is about Y, and Zy equals 24.676 inch3.

    For bending members, please refer to Chapter 8-A Beginner’s Guide to the Steel Construction Manual, 15th ed.

    The next post-Post 6b- is Practice problem 5-2-3-verify Zx for W18x50.

    The next post is “A Guide to Local Buckling Parameters for Steel Beams.

  • 6-Solved problem 5-2 for the plastic moment value for W10x60.

    6-Solved problem 5-2 for the plastic moment value for W10x60.

    Solved Problem 5-2 For the plastic moment value for W10x60.

    Estimating the plastic moment Mp for the given section, W10x60 of A992, is required.

    Using the first option by considering the W section as a group of plates.

    As a first option, I have considered the W section to consist of three plates, for the sake of comparison, after evaluating the Zx and Mp. for the first option as an assembly of plates for the solved problem 5-2. The upper and lower plate dimension is (10.10″)0.68″, while the intermediate plate dimension is 8.84″0.42″.

    The following pictures show the sequence for that calculation in detail.

    A solved problem 5-2 how to get the plastic moment for W10 x 60?

    Due to the section’s symmetry, the P.N.A will be in the middle. For A1+A2 areas, where A1 is the upper plate area, its value is 6.868 inch2, while A2 is the area of half of the web. Adding them together will give AT/2, where AT is the total area. The value of AT/2 is 8.7244 inch2.

    The next step is to estimate the Cg distance For A1+A2 from the first moment of areas about the P.N.A. The next step is to estimate the Cg distance For A1+A2 from the first moment of areas about the P.N.A.

    Divide the area into two halves and get Y bar.

    When we consider the formula AT/2y bar=A1y1+A2y2, we have = Ybar= (6.8684.76+1.8564*2.21)/8.7244, which will be Y bar =4.22″, as we can see from the next slide.

    Divide the area into two halves and get Y bar.

    Zx can be estimated as Zx=AT/2(Y1bar+Y2 bar), the AT/2=8.7244 inch2, while Y1 bar=y2 bar=4.22″, then Zx=8.7244(2*4.22)=73.63 inch3.

    The plastic moment based on the first option.

    For the formula Mp=Zx*Fy, the Plastic moment can be calculated as Mp=73.63*50=3681.50 inch-kips. To convert that value to Ft-kips, just divide by 12. Ultimately, the plastic moment value MP=307.0 Ft- kips for the solved problem 5-2 as an option—a).

    Can we use Table 1.1 to get Zx and Mp?

    Using Table 1-1 to get the plastic moment will give us only the Zx value and the dimensions of the sections, such as the W10x60 section from AISC Table 1-1.

    The flange width is 10.10″, the tf is 0.68″, the overall depth is 10.20″, and the web thickness is 0.42″. The Zx value is 74.60 inch3. We need y plastic to estimate the Mp value, and we cannot find the Y plastic in that table.

    Use Wt 5x30 instead of W10x60 to get the y bar value from table 1.8.

    The only solution is to consider W10x60 composed of two Wt sections, each 5×30. We need to proceed to another table.

    Use Wt 5x30 instead of W10x60 to get the y bar value from table 1.8.

    The Z value for the W10 x 60 section.

    The plastic section modulus or Z value of w10x60 can be found in Table 1-1. The Z value =74.60 inch3.

    Using the second option by considering the W section as two Wt sections.

    The data for the needed Wt section of WT5x30 can be found in Table 1-8. We can get the Y bar from the section property of WT5x30, the value Y bar=0.884″, which is the distance from the plastic neutral axis to the upper top flange, as seen from the next slide.

    The properties of wt 5 x 30 from Table 1.8.

    The total area of the Wt 5×30 =8.84 inch2.

    Y bar value for the Wt section.

    To get the plastic moment, we join the two Wt 5×30 to create W10x60. The compression force C is acting in the plastic Neutral axis of Wt, which is = 0.884″, as we get from the previous data of WT5x30. The acting Tensile force T, due to symmetry, acts at the plastic neutral axis but at a distance =0.844″ from the bottom.

    The distance between C and T, which we call Yct, is the whole depth(-) 2*yp. the overall depth we can get from table1-1 fr W10x60, which is=10.20″. The compression force C=T=At/2*Fy=(17.70/2)*50==442.50 kips, yct=10.20″-2*0.884=8.432″.

    The Plastic moment=442.50*8.432/12=310.93 Ft.kips and can be approximated to 311.0 Ft.kips.

    Another way is by considering Mp=Zx*Fy=At/2*yct, which gives us the same result as Mp=311.0 ft. kips.
    The following calculation, accompanied by a sketch from the next slide, shows the value of Zx and Mp. For the sake of comparison between options a and b, the values of Mp are written side by side.

    Reassemble the two Wt sections to estimate the plastic moment for W10 x 60.

    For the buckling concept, please refer to this link from Prof  T. Bart Quimby, P.E., Ph.D., F.ASCE site.

    For more solved problems, please refer to post 6A-Practice problem 5-2-2, Find y bar, Zx, and Zy.

    This is another post: 6b-Practice problem 5-2-3-verify Zx for W18x50.
    This is the link for the next post, 7-Local buckling parameters.

  • 5- A Solved problem 5-1 for Sx & Zx-elastic-plastic moduli.

    5- A Solved problem 5-1 for Sx & Zx-elastic-plastic moduli.

    A Solved problem 5-1 for the elastic and plastic Moduli.

    In the solved problem 5-1, a built-up shape is given. The top and bottom flange widths are 8 inches, and their thicknesses equal 1″. The web height is 1 foot, and its thickness is 1/2 inch.

    The first requirement is to get the elastic section modulus Sx=Ix/y, where Sx is the elastic section modulus, Ix is the moment of inertia about the x-axis, and y is the distance from the Elastic neutral axis to the upper fiber, which is =Ix/y max.

    We have converted the inner height of 1 ft =12″. For the Ymax=1/2(overall depth)=1/2*(12+1+1)=7″.

    Solved problem 5-1-elastic section modulus for built-up section.

    Ix of the built-up section =sum of (Ix upper plate+Ix web+Ix of the lower plate); due to symmetry, it can be rewritten as Ix=2*(Ix-upper plate at the Cg+Ix -plate at CG). The Ix value =749.33 inch4; please refer to the next slide for more details of the calculations. Sx=Ix/y max, since Ix=749.333 inch4 and y max=7″, then Sx=749.333/7=107.05 inch3.

    The elastic section modulus for the built-up section.

    The second requirement is yield moment My, which is My=Sx*Fy. Based on the steel section’s material, ASTMA572 grade 50, the yield stress Fy=50 ksi.

    The yield moment My=107.0*50=5352.0 inch. Kips. To convert to ft-kips, we will divide by/12—finally, the yield moment My=446 ft-kips. The details of the calculations are shown in the next slide image.

    The value of the moment My in the y-direction.

     

    How do we evaluate yield Moment My in a section by the first principles?

    We can get the value of the yield moment My from the first principles by drawing a stress diagram, for which Fy=50 ksi, as shown in the next slide image. We equate the tension and compression forces as C=T=(1/2)Fy*(h/2), y ct is the arm distance between the compression force C and the tension Force T we call it yct=h-(2/6)h=2/3*h.

    The value of the stress at the upper fiber F1 for the I section.

    The compression force can be estimated as the sum of two compressive forces, C1 and C2, where C1 is the force acting due to the trapezium stress distribution. At the same time, C2 is the force acting due to the triangular stress distribution portion.

    C1 = 371.428 kips, while C2 = 64.285 kips. The following slide image shows the application distance from the neutral axis for each force.

    The detailed calculations for C1 and C2 forces.

    A tabulation of the compression forces and Tension forces is shown. C1 is the force acting on the Upper plate, and C2 is the compression force acting on the middle web for a height of 8″. The corresponding tensile forces T1 and T2 are included.

    We can estimate the Yield moment value for the solved problem 5-1 by summing the two moments due to C1, T1, and C2 and T2 forces.

    The value of the moment My in the y-direction from the first principles.

     

    How do we evaluate Zx for part b of solved problems 5-1?

    To evaluate the plastic section modulus Zx, we must evaluate AT/2, half of the total area, and find the axis that divides the whole section equally. Due to the symmetry of the built-up section, we will get the P.N.A at 7″ from the bottom, thus coinciding with the elastic neutral axis previously estimated.

    Estimation of the total area, find half of this total area.

    We get the y bar for At/2, half of the total area about the P.N.A, and estimate its value from the first-moment area.

    y bar distance to the plastic neutral axis for half of the area.

    The plastic section modulus can be estimated as Zx=AT/2*(ybar1+y2); the full calculation is shown in the next slide.

    The plastic section modulus for Solved problem 4-3.

     How to evaluate Mp for part b of solved problem 5-1?

    Mp=Zx*Fy, the plastic moment can be estimated as the product of Zx by the yield stress Fy.Zx=121.957 inch3, while Fy=50 ksi. The final estimated value is shown in the next slide. The Mp value for a solved problem 5-1 is 508.0 ft. kips.

    The plastic moment value Mp.

    For the buckling concept, please refer to this link from Prof  T. Bart Quimby, P.E., Ph.D., F.ASCE site.
    This is a link for the next post, Solved problem 5-2 for Sx&Zx and shape factor.