Category: Steel Beams

The steel beam posts have included different categories, for instance:

1-local buckling.

2-Lateral torsional buckling.

3-The coefficient of bending.

4-How do Deflection control the design?

5- continuous beam.

6-Plastic analysis theory.

All these categories have practice problems.

  • 41-Practice problem 5-6-1-find the total service load for W12x65.

    41-Practice problem 5-6-1-find the total service load for W12x65.

    Practice problem 5-6-1-find the total service load for W12x65.

    Practice problem 5-6-1. A W12x65 is used as a supported, uniformly loaded beam with a span length of 50 feet and continuous lateral support. The yield stress, Fy, is 50 ksi. If the ratio of live load to dead load is 3, compute the available strength and determine the maximum total service load, in kips/ft, that can be supported. a. Use LRFD. b. Use ASD. I will include only the LRFD design in this post. Practice problem 5-6-1 is from the Steel Design Handbook.

    Check the local buckling parameter for W12x65.

    For the given W12x65, which has fy=50 ksi and continuous lateral support, we need to find the local buckling parameters bf/2tf and h/tw to confirm whether the given section is compact or non-compact.

    Practice problem 5-6-1-find the total service load for W12x65.

    We checked Table 1-1 for W12x65 and found that the section has a symbol f, which indicates that W12x65 doesn’t conform to the local buckling parameters.

    Table 1-1 for W12x65.

    We get the data required to solve the practice problem 5-6-1: the values of bf/2tf=9.52 and h/tw=24.90. The plastic section modulus Zx=96.80 inch23, and the elastic section moduls=87.90 inch3.

    Table 1-1 part 2 find bf/2tf nd h/tw for W12x65.

    As a reminder, please find the local buckling parameters for W sections for Fy=50 ksi.

    Based on item 10 in Table B4.1b, the flange λFp=0.38*sqrt(E/Fy), since we have E29000 ksi and the given Fy=50 ksi, then λFp=0.38*sqrt(29000/50)=9.15. λFr=1.0*sqrt(E/Fy)=1*sqrt(29000/50)=24.08.

    Local buckling parameters based of Fy=50 ksi.

    For the web compactness ratio, the flange λwp=3.76*sqrt(E/Fy), λWp=3.76*sqrt(29000/50)=90.55. λwr=5.70*sqrt(E/Fy)=5.70*sqrt(29000/50)=137.27

    Since bf/2tf is bigger than λFp, the flange is noncompact; hence, W12x65 is a non-compact section.

    Find the values for Mpx and 0.7*FySx for W12x65.

    The Plastic moment value equals Fy*Zx, which equals 50×96.8=4840 inch. Kips. The term (0.7*Fy*Sx) value equals 0.7*50*87.90=3076.50 Inch. Kips. The first term corresponds to λFp=9.15; the second corresponds to lλFr=24.08.

    Find Mpx and 0.70*Fy*Sx for W12x65.

    Find the values for φ*Mn for W12x65 based on local buckling.

    To find the values for Mn for W12x65 based on local buckling, we plot the relation between λF and Mn; the Mn can equal 4749 inch. Kips can be approximated to 396 Ft. For the available strength φ*Mn, multiply by 0.9; the final answer is 356 ft. kips. This is the answer for part a.

    Find the values for φ*Mn for W12x65.

    Press on any image, and you will see a slide show of pictures from 1 to 6, which are included above.

    What are the non-compact W sections based on Fy=50 ksi?

    The following slide, from the companion of The AISC steel construction Manual volume 1-design examples, lists Nine non-compact sections starting from W21x48 and ending with W6x8.50. The %centage reduction of the nominal moment is included for each W section.

    List of Non compact W sections via AISC design examples data.

    Create an Excel sheet for noncompact sections based on Fy=50 ksi.

    I have sorted the W sections with bf/2tf bigger than 9.15 using an Excel sheet quoted from The AISC w sections V15; the result is 10 W sections. Please refer to the next slide for more details.

    An excel sheet for non compact W sections.

    Adjustment in the Lp distance based on Mn.

    The bracing Length Lp, which is equal to 10.70 feet, corresponds to Mpx equal to Fy*Zx, which is equal to 403 Ft.kips, and the Bracing length lr 35.10 ft corresponds to 0.7*Fy*Sx equals 256 Ft.kips. But our W12x65 has an Mn equal to only 396 ft. kips, an extra length to add to lp.

    This distance equals the difference between (403-396)/B.F value. The BF value is the slope of the line joining Lp and Lr and equals to 6.02.

    The Final L’p=11.86 ft approximated to 11.90 Ft.

    Adjustement of Lp value based on Mn .

    Check the values for Lp’,lr, and φ*Mn from Table 3-2.

    To check Lp’, lr values, and φ*Mn for W12x65, we find the New Lp value equals 11.90 ft, lr=35.10Ft, and φ*Mn=356 Ft.kips.

    Check Lp and lr and φ*Mn.

    Check the values for the maximum total service load.

    We will equate Mult to φ*Mn. We have l/D=3, the uniform ultimate Load Wul=1.2*D+1.6*3D=6D, the Mu=6D*(50)^2/8=356, then D=0.19 K/Ft, the value of L=3*D=3*0.17=0.51 k/F.

    The value of D and L .

    We will add D+l to get the Maximum Total service load equal to 0.76 K/Ft. Thanks a lot.

    The total service load value.

    Press on any image, and you will get a slide show for pictures from 7-13, which are included above.

    Here is the link for Chapter 8 – Bending Members.
    For more information about LP and Lr, please see the post: Step-by-step guide to Lateral-torsional buckling.

  • Practice problem 5-5-8-Compute Mn.

    Practice problem 5-5-8-Compute Mn.

    Practice problem 5-5-8-Compute Mn for W18x71 with lb=9 feet.

    Practice problem 5-5-8 A W18 x 71 is used as a beam with an unbraced length of 9 feet. Use Fy=65 ksi and Cb = 1 and compute the nominal flexural strength. Compute everything with the equations in Chapter F of the AISC Specification. Practice problem 5-5-8 is from the Steel Design Handbook.

    How do we estimate Lp for W18x71 with Fy=65 ksi?

    In the absence of design aids, Table 3-3, where W sections are sorted by their plastic section Zx, and with different Fy, which equals Fy=65 ksi, we need to estimate Lp and lr values by using the equations given by the specifications.

    The following slide shows The detailed estimate of the Lp value; Lp is the value of the un-braced length that the plastic hinge should be developed, which is given by equation F2-5 as equal to 1.76*rysqrt(E/Fy).

    From Table 1-1, we will get the radius of gyration ry for W18x71. The ry value is 1.70 inches from the data that Fy=65 ksi. The E value is 29000 ksi. The value of Lp will be 63.20 inches and can be rounded to 5.30 feet.

    What is the value of lp for  W18x71-Fy=65 ksi?

    How do we estimate Lr for W18x71 with Fy=65 ksi?

    We use table 1-1 for W section W18x71 to get the necessary data for estimating the Lr. Lr represents the unbraced length at which lateral-torsional buckling transitions from the inelastic range to the elastic range. These items are ry, rts, Zx,SX, Cw,C, J, H0. The Sx value equals 127 inch3 while Zx=146 inch3.

    We can get the value of bf/2tf and h/tw. Please refer to the next slide image for details.

    Use Table 1-1 part2-to get the data for lr value.

    What is the value of lr?

    We use the equation F2-6 provided by the AISC specification. The value of lr for W18x71 equals 196 inches and is rounded to 16.33 feet. Please refer to the next slide image for more details.

    What is the value of lr?

    Check buckling parameters for W18x71 where Fy=65 ksi.

    We will find the local buckling parameters for the flange and web-based Fy=65 ksi. Based on item 10 in Table B4.1b, the flange λFp=0.38*sqrt(E/Fy), since we have E29000 ksi and the given Fy=65 ksi, then λFp=0.38*sqrt(29000/65)=8.03. λFr=1.0*sqrt(E/Fy)=1*sqrt(29000/65)=21.12.

    For the web compactness ratio, the flange λwp=3.76*sqrt(E/Fy), λWp=3.76*sqrt(29000/65)=79.42. λwr=5.70*sqrt(E/Fy)=5.70*sqrt(29000/65)=120.4. Please refer to the next slide image for the detailed estimate of compactness ratios.

    Local buckling parameters based of Fy=65 ksi.

    Check whether W18x71 is compact or non-compact.

    In the next slide image, we have the bf/2tf and h/tw values of W18x71 against the required local buckling parameters based on Fy=65 ksi. We will find out that the flange and web are compact, so the whole section of W18x71 is compact.

    Check  whether W18x71 is compact section or not.

    Find the equation of Mn-based on Fy=65 ksi.

    The Plastic moment value equals Zx*Fy, which is equal to 65×146=9490 inch. Kips. The term (0.7*Fy*Sx) value equals 0.7*65*127=5778.50 Inch. Kips. The first term corresponds to bracing length Lp=5.30 ft; the second corresponds to lr=16.33 feet. We have a given bracing lengthlb=9 ft.

    We get the values of Mp and 0.70*Fy Sx in Ft.kips by using the conversion factor of 1Ft=12 inches. Mp value equals 791 Ft. kips, while 0.70*Fy Sx=482, we could round these values to the nearest ones.

    There is a linear relation for the required Mn with Mp and(0.7*Fy*Sx) together with Lp and Lr. The BF factor is the slope value, which equals (Mp-0.7*Fy*Sx)/(Lr-Lp)=28.01.

    The relation between reuired Mn and lb, BF values.

    What is the value of Mn for W18x71 at lb=9 feet?

    The relation of 791-28.01*(Lb-Lp) represents the nominal moment value. The Lb value equals 9 feet, and the Lp value equals 5.30 feet. The Final value for Mn equals 687 Ft. kips.

    Mn value for W12x30 for a bracing length of 10 ft.

    We could use an Excel sheet to graph the relation between Lb and the Flexural design strength-Mn in ft. kips for W18x71 with Fy=65 ksi. We can see the linear portion and the Value of Mn at 9 feet equals 687 ft. kips. Thanks a lot.

    Graph betwen lb and Nominal moment for W18x71.

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

    A new post is added 41-Practice problem 5-6-1-find the total service load for W12x65. The section is non-compact.

  • Practice problem 5-5-6-Compute Lp and Lr and φb*Mn.

    Practice problem 5-5-6-Compute Lp and Lr and φb*Mn.

    Practice problem 5-5-6-Compute Lp and Lr, φb*Mn and Mn/Ωb for lb=10 feet.

    Practice problem 5-5-6 A W12 x 30 of A992 steel has an unbraced length of 10 feet. Using Cb = 1.0, a. Compute Lp and Lr. Use the equations in Chapter F of the AISC Specification. Do not use any of the design aids in the Manual.
    b. Compute the flexural design strength, φb*Mn.
    c. Compute the allowable flexural strength Mn/Ωb. Practice problem 5-5-6 is from the Steel Design Handbook.

    How do we estimate Lp for W12x30?

    In the absence of design aids, Table 3-3, where W sections are sorted by their plastic section Zx, in which we can easily find the values of lp and lr and other values of flexure design moments, We need to estimate Lp and lr values by using the equations given by the specifications.

    The following slide shows The detailed estimate of the Lp value; Lp is the value of the un-braced length that the plastic hinge should be developed, which is given by equation F2-5 as equal to 1.76*rysqrt(E/Fy).

    From Table 1-1, we will get the radius of gyration ry for W12x30. The ry value is 1.52 inches. From the data on A992 steel, we have the yield stress Fy=50 ksi. The E value is 29000 ksi. The value of Lp will be 64.43 inches and can be rounded to 5.40 feet.

    Practice problem 5-5-6-Compute Lp and Lr, φb*Mn.
And Mn/Ωb for lb=10 feet.

    How do we estimate Lr for W12x30?

    We use table 1-1 for W section W12x30 to get the necessary data for estimating the Lr. Lr represents the unbraced length at which lateral-torsional buckling transitions from the inelastic range to the elastic range. These items are ry, rts, Zx,SX, Cw,C, J, H0.

    We can get the value of bf/2tf and h/tw. Please refer to the next slide image for details.

    Use Table 1-1 part2-to get the data for lr value.

    What is the value of lr?

    We use the equation F2-6 provided by the AISC specification. The value of lr equals 187.24 inches and is rounded to 15.60 feet. Please refer to the next slide image for more details. Thus, we have completed the required part a) in the practice problem 5-5-6.

    What is teh value of lr?

    Check buckling parameters for W12x30 where Fy=50 ksi.

    We will find the local buckling parameters for the flange and web-based Fy=50 ksi. Based on item 10 in Table B4.1b, the flange λFp=0.38*sqrt(E/Fy), since we have E29000 ksi and the given Fy=50 ksi, then λFp=0.38*sqrt(29000/50)=9.15. λFr=1.0*sqrt(E/Fy)=1*sqrt(29000/50)=24.08.

    for the web compactness ratio, the flange λwp=3.76*sqrt(E/Fy), λWp=3.76*sqrt(29000/50)=90.55. λwr=5.70*sqrt(E/Fy)=5.70*sqrt(29000/50)=137.27. Please refer to the next slide image for the detailed estimate of compactness ratios.

    Local buckling parameters based of Fy=50 ksi.

    Check whether W12x30 is compact or non-compact.

    In the next slide image, we have the values of bf/2tf and h/tw of W12x30 against the required local buckling parameters based on Fy=50 ksi. We will find out that the flange and web are compact, so the whole section of W12x30 is compact.

    Check  whether W12x30 is compact section or not.

    Find the equation of Mn-based on Fy=50ksi.

    The Plastic moment value equals Zx*Fy, which is equal to 50×43.10=2155 inch. kips. The term (0.7*Fy*Sx) value equals 0.7*50*38.60=1352 inch. kips. The first term corresponds to bracing length Lp=5.40 ft; the second corresponds to lr=15.60 feet. We have a given lb=10 ft.

    We get the values of Mp and 0.70*Fy Sx in Ft.kips by using the conversion factor of 1Ft=12 inches. Mp value equals 179.583 Ft. kips, while 0.70*Fy Sx=112.583, we could round these values to the nearest ones.

    There is a linear relation for the required Mn with Mp and(0.7*Fy*Sx) together with Lp and Lr. The BF factor is the slope value which equals (Mp-0.7*Fy*Sx)/(Lr-Lp)=6.5686.

    The relation between reuired Mn and lb, BF values.

    What is the value of Mn at lb=10 feet?

    The relation of 179.583-6.53*(Lb-Lp) represents the nominal moment value. The Lb value equals 10 feet, and the Lp value equals 5.40 feet. The Final value for Mn equals 149.367 Ft. kips.

    Part b) Compute the Flexural design strength-φb*Mn.

    For part b of the practice problem 5-5-6, for Flexural design strength-φb*Mn, we multiply the Nominal value by the factor φb, which equals 0.90. We get the value of 134 ft.kips.

    Part c. Compute the allowable flexural strength Mn/Ωb.

    For part b of practice problem 5-5-6, for allowable flexural strength Mn/Ωb, we multiply the Nominal value by the factor (1/ Ωb), which equals (1/1.67). We get the value of 89 Ft.kips.

    Mn value for W12x30 for a bracing length of 10 ft.

    We could use an Excel sheet to graph the relation between Lb and the Flexural design strength-φb*Mn for W12x30 Fy=50 ksi. We can see the linear portion and the Value of φb*Mn at 10 feet equals 134 ft. kips.

    Graph betwen lb and design strength flexture strength for W12x30.

    We could use an Excel sheet to graph the relation between Lb and allowable flexural strength Mn/Ωb for W12x30 and Fy=50 ksi. We can see the linear portion and the Value of Mn/Ωb at bracing length 10 feet equals 89 ft. kips. Thanks a lot.

    Graph betwen lb and  allowable flexture strength for W12x30.

    Here is the link for Chapter 8 – Bending Members.
    This links to the next post, Practice problem 5-5-8-Compute Mn.

  • 9b-Practice problem 5-4-1-Check compactness for Fy=60 ksi.

    9b-Practice problem 5-4-1-Check compactness for Fy=60 ksi.

    Practice problem 5-4-1-Check compactness for Fy=60 ksi.

    Practice problem 5-4-1 For W-, M-, and S-shapes with Fy = 60 ksi: the first part a. List the noncompact shapes in Part 1 of the Manual (when used as flexural members). State whether they are noncompact because of the flange, the web, or both.
    b. List the shapes in Part 1 of the Manual that are slender. State whether they are slender because of the flange, the web, or both. Practice problem 5-4-1 is from the Steel Design Handbook.

    List the W, M, and S shapes based on compactness when Fy=60 ksi.

    The three steel sections, W, M, and S shapes, follow item 10 in Table B4.1b for the lambda value for the unstiffened flange. For the web compactness h/tw, these sections follow item 15 in the same table,

    The following slide shows the main difference between W, M, and S shapes in the profile as I section. W shape has a slope of 2:12 and a broader flange width. The S shape has a slope of 6:1 and is available in a smaller range. The M stands for Miscellaneous beams.

    Practice problem 5-4-1-The differences between W, M and S shapes.

    Determine the values of compactness ratios for Fy=60 ksi.

    Based on item 10 in Table B4.1b, the flange λFp=0.38*sqrt(E/Fy), since we have E29000 ksi and the given Fy=60 ksi, then λFp=0.38*sqrt(29000/60)=8.35. λFr=1.0*sqrt(E/Fy)=1*sqrt(29000/60)=21.964.

    for the web compactness ratio, the flange λwp=3.76*sqrt(E/Fy), λWp=3.76*sqrt(29000/60)=82.66. λwr=5.70*sqrt(E/Fy)=5.70*sqrt(29000/60)=125.31. Please refer to the next slide image for the detailed estimate of compactness ratios.

    The value of compactness ratios for fy=60 ksi.

    Table B4.1b details stiffened and unstiffened elements.

    The next two slides show the details of items 10 and 15 for compactness ratios for members subjected to flexure.

    Part 1 of The detailed B4.1B Table .

    Part 2 of The detailed B4.1B Table .

    Sort W sections based on bf/2tf>λFp but less than λfr.

    We will use an Excel sheet for Table 1-1 for W sections and sort W sections with bf/2tf bigger than 8.35, which is the value of λFp, but smaller than 21.848, the value for λFr. These are the non-compact sections for steel, with Fy=60 ksi.

    The total number of W sections is 20, starting from W30x90 and ending with W6x0.50.

    The next slide image shows the detailed dimension of these non-compact W shapes.

    The details of the 20 non-compact W sections for Fy=60 ksi.

    Find the non-compact web for W sections based on Fy=60ksi.

    There is no non-compact web for W sections based on Fy=60 ksi; the maximum h/tw for W30x90 is 57.50, smaller than the value of λwp, 82.66.

    Non compact  W section data for Fy=60 ksi.

    What are the tables for properties of M and s sections?

    The next slide shows the tables used to find the properties of W, M, and S shapes. We use Table 1-1 for W sections, and for M sections, we use Table 1-2. For S shapes, we use Table 1-3.

    The Tables used for data for W,M and S shapes.

    Sort M sections based on bf/2tf>λFp but less than λFr.

    We will use an Excel sheet for Table 1-2 for M sections and sort M sections with bf/2tf bigger than 8.35, the value of λFp, but smaller than 21.848, the value for λFr. These are the non-compact sections for steel, with Fy=60 ksi. We have only five non-compact sections, starting from M12x10 and ending with M 3×2.9.

    The list of Non compact M shapes for Fy=60 ksi.

    Find the non-compact web for M sections based on Fy=60ksi.

    There is no non-compact web for M sections based on Fy=60 ksi; the maximum h/tw for M12.5×12.40 is 74.80, smaller than the value of λwp, 82.66.

    Non compact web section data for M shape with Fy=60 ksi.

    Sort S sections based on the flange and web compactness ratio.

    e will use an Excel sheet for Table 1-3 for S sections and sort S sections with bf/2tf bigger than 8.35, which is the value of λFp but smaller than 21.848, the value for λFr. There are no non-compact S sections for steel, with Fy=60 ksi. As for the ratio h/tw, there are no non-compact S shapes for the web.

    Are there Non compact sections for web for S section with fy=60 ksi?

    Part b-Sort W, M, and S sections are based on slender sections.

    Based on Fy=60 ksi, the W, M, and S sections do not have slender sections. We have reached the end of our post. Thanks a lot.

    What are the slender sections for W,M and S section for Fy=60 ksi?

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

    There is a newly added post, which is a practice problem 5-5-6-Compute Lp and Lr and φb*Mn.

  • 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.