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  • 25-Two Practice problems for inertia for trapezium.

    25-Two Practice problems for inertia for trapezium.

    Two Practice problems for inertia for trapezium.

    We will introduce practice problems for trapezium; there are two practice problems for inertia for trapezium, which are 8.41 and 8.47. These problems are quoted from Engineering mechanics statics by BEDFORD.

    The first practice problem for inertia is 8.41. Find Ix and kx for a given trapezium.

    A quick estimate of Ix for a given trapezium.

    For the first practice problem of the two Practice problems for inertia, we have a trapezium of 7 ft base length, and the upper part length is 3 Ft. It is required to determine the moment of inertia Ix and the radius of gyration Kx.

    We have a quick answer for the moment of inertia, as we can see from the upper right corner of the next slide image, that the trapezium consists of a triangle of base (b-a) equals 4 feet with height h equals 3 feet and a rectangle of the base a equals 4 feet and a height h=3 feet.

    From our previous posts, we have the expression for triangle inertia about the base as Ix = base *height^3/12, which is equal to (4*3^3/12)=5 Ft4.

    At the same time, the inertia for a triangle equals the base *height ^3/3 which equals (3^4)/3=81/3=27 ft4. Then, the moment of inertia for the trapezium is 9+27=36 feet4.

    Ix for a given trapezium by integration using a vertical strip.

    As we can see from the next slide image, we can use a vertical strip of y height and of width equals dx. The inertia of this strip about the X-axis is the sum of two inertias. The first inertia is dIx1 plus d Ix2

    To find the value of y, we can determine the equation of the triangular line as y = 3/4 x, while for the rectangular section, y equals h, which is 3 feet. The expression for dIx1 is dIx1 = (dx * y^3 / 3). This expression can be rewritten as dIx1 = 1/3 * (3/4 x^3) * dx.

    The integration is from x=0 to x=4. The Ix1 value is equal to 9 ft4, which matches the previously estimated inertia for the triangular part.

    Two Practice problems for inertia for trapezium.

    The value of Ix2 for the rectangular portion can be found by integrating the vertical strip from x=4 feet to x=7 feet.

    The inertia of the strip about the X-axis is dIx2=(dx)*(h^3/3), the h value is 3 Ft, and the final value is Ix2, which will be found to be equal to 27Ft4.

    The Total Ix value is 36 Ft4. The square of the radius of gyration equals Ix/A=36/15. Then the radius of gyration kx equals 1.55 Ft.

    Detailed solution for ix and kx for a trapezium.

    The second practice problem for inertia is 8.47. Find Ix and kx for a given trapezium about the Cg.

    For the second practice problem, 8.47, of the two practice problems for inertia, it is required to find the moment of inertia for the trapezium at the Cg and determine the kx value.

    The same Trapezium of 7 ft base length and the upper part length is 3 Ft is dealt with.

    Determine the x-bar for the Trapezium.

    We need to find the value of the area and x bar or the horizontal distance from the Cg to the vertical axis y1 passing by the left point of the trapezium, later we will use it for the forthcoming practice problems.

    Use the expression of A(X bar)=(A1*X1+A2*X2), A1=6 ft2, while A2 equals 9 ft2. The X1 is the horizontal distance from the Cg of the triangular part to the axis y1. The value is (2/3)*4=8/3 Ft.

    While X2 is the horizontal distance from the Cg of the rectangular part to the axis y1.

    The x2 value equals (7-1.50)=5.50 ft. Using the expression A(X bar)=(A1*X1+A2*X2), and substitute for the known values we can get x bar as equal to 4.3667 Ft. Please refer to the details of calculations in the next slide image.

    Derive a general expression for Ix value.

    If you wish to derive a general expression for Ix of the trapezium. The Ix is the sum of Ix1 and Ix2. For  Ix1 we can write its expression as equal to the integration from 0 to (b-a) of x^3*dx*(h^3)/3*(b-a)^3.

    Proceeding with the integration will give us h^3/12*(b-a). While for the integration of Ix2, or the rectangular portion, is the integration from (b-a) to b of (dx*(h^3/3), and its final value is a*h^3/3.

    Adding Ix1 to Ix2 will give us (h^3/12)*(3a+b). Substitute to get the value of Ix equal to 36 Ft4.

    Determine the y-bar for the Trapezium.

    We need to find the value of the area and Y bar or the vertical distance from the Cg to the trapezium base. Use the expression of A(y bar)=(A1*y1+A2*y2), A1=6 ft2, while A2 equals 9 ft2.

    The y1 is the vertical distance from the triangular part’s Cg to the trapezium’s base. The value is (1/3)*3=1 Ft.

    The y2 is the vertical distance from the rectangular part’s Cg to the trapezium’s base. The value is (1/2)* 3=1.5 Ft. We can get the value of y bar as equal to 1.3 ft.

    The vertical distance from the Cg of the Trapezium to the base.

    Determine the final value for Ix at the Cg for the Trapezium.

    The final value of Ix at the cg is equal to the value of ix about the base minus the product of the trapezium area by the square of y bar distance.

    From the estimated data, we have Ix equaling 36 ft4, while the area of the trapezium equals 15 ft2. The y bar of the trapezium is equal to 1.3 ft. Apply the expression for Ixcg, and we can get the value of 10.65 ft4.

    The radius of gyration at the Cg is equal to the square root of Ixcg/area and can be found as equal to 0.843 ft.

    The value of Ix at the Cg for the trapezium.

    Derive a general expression for Ix at the Cg value.

    If you wish to derive a general expression for Ix at the Cg of the trapezium, we can use the general expression of Ix for the trapezium. As a reminder, the Ix value equals h^3/12*(3a+b) and deducts the product of area *y bar cg ^2.

    Please refer to the next slide image for more details. We can get the  final expression as Ixcg =(h^3/(36)*(a^2+4*a*bn+b^2)/(a+b).

    Derive a general expression for ix at the Cg.

    We can use the known values as a=3 feet, b=7 ft, and h equals 7 feet and substitute in the final expression for Ix cg. Finally, we get the value of Ixcg equals 10.65 ft4.

    The final value of ix at the Cg from the general expression for trapezium.

    Thanks a lot and see you in the next post-Two Practice problems for polar of inertia for trapezium.

    Please refer to post 22 for more information about Iy for a trapezium.

    Please refer to post 19 for more information about Ix for a trapezium.

    The next post is post 26-Two, Practice problems for the polar of inertia for trapezium.

    A useful external resource for the area, Cg, and inertia for the parallelogram, please refer to this link.

  • 42s-Nominal shear strength for M10x7.50.

    42s-Nominal shear strength for M10x7.50.

    Nominal shear strength for M10x7.50-use CM-15.

    Compute the Nominal shear strength for M10x7.50.

    We have a given M10x7.50 steel section of A572 grade 65; it is required to compute the nominal shear strength. We will apply CM#15 and the related specification 2016 in this example.

    First, we will use Table 2-4 to find the yield stress Fy value for A572 steel grade 65. This type of steel has a yield stress Fy=65 ksi and an ultimate stress Fu=80 ksi.

    Nominal shear strength for M10x7.50

    The next question concerns the table used to obtain the complete data for M10x7.5; in part 1, we will find that Tables 1-2 are used to get the information for sections.

    What is the table to be used for M section?

    The data for M10x7.50 Can be obtained from Table 1-2 from Part 1. The flange width bf is 2.69 inches, and the thickness is 0.173 inches.

    The depth is d=9.99 inches, and the web thickness is 0.13 inches. The most critical parameter is h/tw, which we will find in part 2 of the same table.

    Part 1 of Table 1-2 for M10x7.50

    Find h/tw from Table 1-3, part 2.

    The ratio for h over the thickness of the web is equal to 71. The distance h is between the web’s two filleted portions. The following slide image shows part 2 of Table 1-2.

    Part 2-table 1-2, h/tw=71.

    Find the limiting h/w based on Fy.

    The limiting h/tw and the yield stress are related, as shown in the next slide image. Based on the Fy of the given M10x7.5 for grade 65, the limiting h/w equals 47.30.

    The relating h/tw based on yield stress

    The following slide image shows the relation of h/tw and the nominal shear value. There are three zones; the first zone is from zero to 47.30, marked as point 1 on the x-axis, which is obtained from the relation of 2.24*sqrt(E/Fy); in this zone, Cv1, unlive cv in Cm#14, equals 1, and the φv equals 1. The equation gives the nominal shear value for nominal shear Vn=Cv*0.60Fy*d*tw and can be written as 1*0.60*Fy*Aw.

    The second Zone is from h/tw equals 47.30 to 53.692, marked as point 2 on the x-axis, and is obtained from the relation of 1.1*sqrt(kv *E/Fy); in this zone, Cv equals 1, and the φv equals 0.90. the Kv factor equals 5.34 in specification 2016. The nominal shear value is given by the equation Vn=Cv*0.60Fy*d*tw and can be written as 1*0.60Fy*Aw.

    The third Zone is from h/tw, which equals 53.69 to the end; Cv1 is less than 1 in this zone, and the φv equals 0.90. The nominal shear value is given by the equation Vn=Cv1*0.60Fy*d*tw; Cv1 equals 53.69/h/tw. For our section, h/tw=71.0, and the value of cv1=53.69/71=0.756. Please refer to the next slide image.

    The relation between h/tw and nominal shear for Fy=65 ksi.

    The nominal shear strength for M10x7.50.

    The area of the web for M10x7.50 equals the product of d*tw or (9.99*0.13) or 1.2987 inch2; the cv1 value equals 0.7562, and the nominal shear equals 0.7562*(0.60*Fy)*(Aw)=0.7562*0.6*65*1.2987=38.30 kips.

    The final value of nominal shear.

    Excel plot for h/tw versus Vn.

    I have used an Excel plot for the relation between h/tw, and the nominal shear Vn shows the zones for shear. Zone 1 starts h/tw value from 0 to 47.30, and the nominal value equals 50.65 Kips

    . Zone 2 begins for h/tw from 47.30 to 53.69.and the nominal value equals 50.65 Kips. Zone 3 starts from h/tw bigger than 53.69; The Nominal shear Vn equals 50.41 kips for h/tw equals 71; the section has a nominal shear Vn value from 38.30 kips for h/tw=71.0.Thanks a lot.

    Excel plot for h/tw virsus Vn.

    As an external resource for the shear stress limit state.

    The previous post is on nominal shear strength for S41x121.

  • 41s-Nominal shear strength for S41x121.

    41s-Nominal shear strength for S41x121.

    Nominal shear strength for S41x121-use CM#14.

    Compute the Nominal shear strength for S41x121.

    We have a given S41x121 steel section of A572 grade 65; we must compute the nominal shear strength. In this example, we will apply CM#14 and the related specification 2010.

    First, we will use Table 2-4 to find the yield stress Fy value for A572 steel grade 65. This type of steel has a yield stress Fy=65 ksi and an ultimate stress Fu=80 ksi.

    Nominal shear strength for S41x121.

    The next question concerns the table to get the full data for S41x121; in part 1, we will find that Tables 1-3 are used to get the information for sections.

    What is the table to be used for S sections?

    The data for S21x121 can be obtained from Table 1-3 from part 1. The flange width bf is 8.05 inches, and the thickness is 1.09 inches.

    The depth is d=24.50 inches, and the web thickness is 0.80 inches. The most critical parameter is h/tw, which we will find in part 2 of the same table.

    part 1 of table 1-3 for S24x121.

    Find h/tw from Table 1-3, part 2.

    The ratio for h over the thickness of the web is equal to 25.90. The distance h is between the web’s two filleted portions. The next slide image shows part 2 of Table 1-3.

    Part 2-table 1-3, h/tw=25.90.

    Find the limiting h/w based on Fy.

    The limiting h/tw and the yield stress are related, as shown in the next slide image. Based on the Fy of the given S41x121 for grade 65, the limiting h/w equals 47.30.

    The relating h/tw based on yield stress

    The following slide image shows the relation of h/tw and the nominal shear value. There are four zones; the first zone is from zero to 47.30, marked as point 1 on the x-axis, which is obtained from the relation of 2.24*sqrt(E/Fy); in this zone, Cv equals 1, and the φv equals 1. The nominal shear value is given by the equation for nominal shear Vn=Cv*0.60Fy*d*tw and can be written as 1*0.60Fy*Aw.

    The second Zone is from h/tw equals 47.30 to 53.69, marked as point 2 on the x-axis, and is obtained from the relation of 1.1*sqrt(kv *E/Fy); in this zone, Cv equals 1, and the φv equals 0.90. the Kv factor equals 5.34 in specification 2016. The nominal shear value is given by the equation Vn=Cv*0.60Fy*d*tw and can be written as 1*0.60Fy*Aw.

    The third Zone is from h/tw equals 53.69 and up , Cv1 is less than 1, and the φv equals 0.90. The nominal shear value is given by the equation Vn=Cv1*0.60Fy*d*tw; the Cv1 equals 1.1*sqrt(kv *E/Fy)/h/tw or 53.69/(h/tw).

    For section M10x7.50, h/tw is 71; in zone 3, the Cv1 value will equal 53.69/71=0.756. Please refer to the next slide image.

    The relation between h/tw and nominal shear for Fy=65 ksi.

    The nominal shear strength for M10x7.5.

    The area of the web equals d*tw or 9.99*0.13 or 1.2987; the cv1 value equals 0.7562, and the nominal shear equals 0.7562*(0.60*Fy)*(Aw)=0.7562*0.6*65*11.2987=38.30 Kips.

    The final value of nominal shear.

    Excel plot for h/tw versus Vn.

    An Excel plot for the relation between h/tw and the nominal shear Vn shows the four zones for shear. Zone 3 starts from Vn equals 50.41 kips to Vn equals 13.60 for h/tw=200; our section is shown in the graph with H/tw=71.0 and vn=38.30 kips. Thanks a lot.

    Excel plot for h/tw virsus Vn.

    As an external resource for the shear stress limit state.

    The previously solved problem was checked based on CM#14.

    The next post is a solved problem-Nominal shear strength for M10x7.50.

  • List of Steel Beam Posts-part-4a.

    List of Steel Beam Posts-part-4a.

    List of Steel Beam Posts-part-4a.

    Solved Problem 10-1-Design Of Steel Section For Continuous Beam Part-1/4.

    This is the 38th post of the Steel beam Posts-part-4a, which includes a solved problem 10-1 for a given continuous beam, it is required to select the lightest W-section available using the plastic analysis, and assuming full lateral support is provided for both the flanges.

    Solved Problem 10-1

    Two methods are used, the first method is the statical method or the lower bound method and the second method is the Mechanism or the upper bound method. both methods will lead to the same results.
    This is the link for post-38: Solved problem 10-1 for Design Of Steel Section For Continuous Beam – Part-1/3.

    Solved problem 10-1-design of steel continuous beam-2/4.

    This is post number 38a one of the Steel beam Posts-part-4a which includes how to derive a general expression for the moment values for a continuous beam by using the three-moment equations. This is the second part of the solved problem 10-1.

    Solved problem 10-1-design of steel continuous beam-2-4.

    This is the link for post-38a: Solved problem 10-1-design of steel continuous beam –Part 2/4

    39-Solved problem 10-1-use three-moment equations 3/4.

    This is the 39th post of the Steel beam Posts-part-4a, which includes how to derive a general expression for the moment values for a continuous beam using the three-moment equations. This is the third part of the solved problem 10-1—a step-by-step guide to using the three-moment equation.

    Solved problem 10-1-use three-moment equations 3/4.

    This is the link for post-39: Solved problem 10-1-use three-moment equations 3/4.

    40-Solved problem 10-1-design of steel sections 4/4.

    This is the 40th post of the Steel beam Posts-part-4, which includes the design of the beam, using elastic analysis with service loads and the 0.90 rule, and assuming full lateral support is provided for both the flanges.

    This is the third part of the solved problem 10-1. The next slide image shows the final bending moment diagram after using the 0.90 rule, in which the negative moments at the supports are reduced. A W24x76 is selected, and the section has a Zx value of 200 inch3.

    Solved problem 10-1-design of steel sections 4/4.


    This is the link for post-40: Easy Design of steel section for continuous beam-4-4.

    A very useful external resource A Beginner’s Guide to Structural Engineering

  • List of Steel Beam Posts-part-4.

    List of Steel Beam Posts-part-4.

    List of Steel Beam Posts-part-4.

    Moment Redistribution For Continuous Steel Beams.

    This is the 29th post of the steel beam posts-part-4, which includes a brief discussion of the statically indeterminate beams, where we will have both positive and negative moments and the methods we will use. We will review table-3-23 for continuous beam shear and moment values, redistribution of the moment, elastic redistribution, and 0.90 of the largest—ve moment.

    Table 3-23 for shear and moment values.

    This is a link to post 29-Moment Redistribution.

    Introduction to design of continuous beam, problem 4-15.

    This is the 30th post of the steel beam posts-part-4, which includes a solved problem from Prof. Alan Williams’s book. Solved problem 4-15 Moment Redistribution. The uniform distributed loading, including the beam self-weight, acting on a two-span continuous beam, is shown in Fig. 4.16. Continuous lateral support is provided to the beam.

    Solved problem 4-15 for continuous beams.

    Determine the lightest adequate W10 section using steel with a yield stress of 50 ksi.

    This is the link for post 30: Introduction to design of continuous beam, problem 4-15.

    Collapse load for a simply supported beam.

    This is the 31st post of Steel beam Posts-part-4, which includes a solved problem, how to determine the plastic moment, and the collapse load value for a beam that is fixed at one end and has roller support at the other end. The beam is loaded by a concentrated load.

    Introduction to plastic theory-part of list of steel beam posts-part-4.

    Easy illustration for collapse load for indeterminate beam.

    This is the 32nd post of the Steel beam Posts-part-4, which includes a solved problem, how to determine the plastic moment, and the collapse load value for a beam fixed at one end and with roller support at the other end, loaded by a concentrated load.

    Collapse load for a Built-in and roller support beam.

    This is the link for post 32: the Collapse load for the indeterminate beam. 

    Upper bound and lower bound definitions.

    Here are two posts,33 and 33a-The two posts are among the Steel beam Posts-part-4, which includes the definitions of the lower bound & upper bound and the unique theorem. How to determine the relationship between the load factor and the distance from support for the same solved problem included in the previous post.

    Upper bound and lower bound definitions.

    Load factor and distance from the support relation.

    This is the link for post 33: the Upper and lower bound definitions.

    This is the link for post-33a: Step-by-19 step introduction to lower bound and uniqueness.


    Plastic Moment For Continous Beams.

    This is the 34th post of Steel beam Posts-part-4, which includes determining the plastic moment and the collapse UL load for a continuous beam with equal spans. Two methods are used: the statical method, or the lower bound method, and the Mechanism method, or the upper bound method.

    Plastic Moment For Continous Beams.

    This is the link for post 34: Plastic Moment For Continous Beams.

    For posts from post 35 to post 37, A mastan2 solution is included to verify our estimations.


    Solved Problem 8-22 For Plastic Moment.

    This is the 35th post of Steel beam Posts-part-4, which includes a solved problem: How to determine the collapse load value for a beam that is fixed at both ends and loaded by a concentrated load? Two methods are used: the first method is the statical method or the lower bound method, and the second method is the Mechanism or the upper bound method.

     Solved Problem 8-22 For Plastic Moment.

    This is the link for post 35: Solved Problem 8-22 For Plastic Moment.


    Solved problem 8-32 for the plastic Nominal Uniform load.

    This is the 36th post of Steel beam Posts-part-4, which includes a solved problem: How can the collapse uniform load value for a given section of a simple beam partially loaded and fixed at both ends be determined?

    Using Mastan#2 to find the location of plastic hinges and lambda value.

    The problem was further modified; the uniform load was relocated, and the two methods again estimated the plastic nominal uniform load. This is the link for post 36: Solved problems 8-32.


    Solved problems 8-33&34 for the nominal uniform load.

    There are two posts,37 and 37a. These two posts are part of the steel beam Posts-part-4, which includes two solved problems, 8.33 and 8-34.

    Solved problem 8-33 for the nominal uniform load.

    . Using the given section, all of the A992 steel and the plastic theory,  for the first problem 8.33- determine the value of Wn as indicated for a continuous beam with varying loading.

    For the second problem, #.8.34, determine the value of Wn as indicated for a continuous beam fixed from both ends under Wn loads.

    Solved problem 8-34 for the nominal uniform load.

    Using Mastan-2 for locations of plastic hinges.

    This is the link for post 37: A Solved problem, 8-33, for a nominal uniform load.

    This is a link to post 37a-Solved problem-8-34 for a nominal uniform load.

    This is a link to the next list of steel beam posts.

    A very useful external resource, A Beginner’s Guide to Structural Engineering

  • List of steel Beam Posts-part-3b.

    List of steel Beam Posts-part-3b.

    List of Steel Beam Posts-part-3b.

    Deflection of steel beams.

      

    The post is part of a list of steel Beam posts-part-3b, including deflection disadvantages.
    2-The permitted limits of deflection.
    3- What are the Deflection limitations in the unified building code?
    4- What is the C1 coefficient as a deflection parameter by the AISC?

    The required deflection values as a ratio of the span length.

    C1 value used by the CM#15 for deflection equation.

    Here is the link for post 24: Deflection of steel beams.

    A solved problem 5.1 from Prof. Fredrick Roland’s book. Example 5.1-part 1.

    The 30-foot beam is laterally supported for its entire length. It supports a uniform dead load, including a beam weight of 0.30 kip/ft and a uniform live load of 0.70 kip/ft. In the first part, we will find the required moment of inertia value Ix based on the Total load and Ix based on the LL.

    Solved problem 5.1-estimate the ultimate moment.

    Find the requied Ix.

    Here is a link to post 24a– Easy approach to the solved problem-5-1-part 1-2.

          

    A solved problem 5.1 from Prof. Fredrick Roland’s book- part 2. 

    In part 2, we will estimate the ultimate moment in the LRFD design. We will use Table 3-2 to select a W section and check the Ix for the chosen section against Ix values based on deflection. We will use Table 3-3 to select the proper section based on Ix. Check the section chosen for The ASD design.

    Use table 3-3 for Ux to select a W section.

      Here is a link to post 24b– Easy approach to solved problem-5-1-part 2-2.

    Part 1/4 Solved problem 9-9-6: how to find LL?

    Part 1 includes a solved problem,9-9-6, from Prof. Charles G Salmon’s book. Determine the service live load the beam may carry if the Dead load is  0.15 kip/ft, including the beam weight. Use load and factor design. The steel has Fy=65ksi. The given beam is a built-up beam. We will use Table b4.1B to determine the limiting value for flange and web parameters for local buckling.

    The first part of the solution includes determining the lambda coefficients λp-F λr-F, λp-w λr-w for Local buckling for the Flange and the web of the Built-up section.

    Both the flange and web sections are Non-Compact sections since the limiting parameters are less than those of lambda Plastic.

    estimate of lambda plastic for flange,lambda r for flange.

    Here is the link to post 25 –How to find LL- Part 1/4?

    Part 2/4 for the Solved problem 9-9-6, how to find LL?

    The second part of the solution includes determining the section modulus of elasticity for the Built-up section. Two nominal moments: one is based on the limit state of local buckling for Flange, and the other is based on the local buckling of the web section. After checking the lateral torsional buckling, the final decision on the final value of the nominal moment will be made, which will be included in the next two parts.

    The nominal moment for the Built-up section.

    Here is the link to post 26- Part-2-4 of the Solved problem 9-9-6, How To Find LL?

    Part 3/4 for the Solved problem 9-9-6, how to find LL?

    The third part of the solution includes the lateral-torsional buckling parameters Lp and Lr for the built-up section, estimating the value of limiting length lr.

    The value of Lr.

    Here is the kink to post 27: Part 3/4 of the Solved problem 9-9-6: How do I find LL?

    Part 4/4 for the Solved problem 9-9-6, how do we find LL?

    The fourth part of the solution includes evaluating the nominal moment for the section based on the Lateral torsional buckling and then finalizing the LRFD value of the Final nominal moment to estimate the permitted Live load value.

    The final live load  for the Built-up section.

    Here is the link to post 28:  Part 4/4 Solved problem 9-9-6. How do you find LL?

    The next post will be the list of steel beam posts, part 4.

    A very useful external resource, A Beginner’s Guide to Structural Engineering