Category: Block shear for beams

All solved problems used to illustrate block shear for beams.

  • List of steel Beam Posts-part-3a.

    List of steel Beam Posts-part-3a.

    List of Steel Beam Posts-part-3a.

    1-Solutions for Block Shear-Coped Beam Problem

       This is Post number 19 of the Steel beam Posts, which includes 1—Solutions for the Block Shear-Coped Beam Problem.

    A solved problem 5.13 for Block Shear-Coped Beam-case-1-UBS=1.00.

    The solution includes a detailed description of the forces that affect the beam: shear yielding and tensile rupture, Shear rupture, and tension rupture. How do we estimate the LRFD and ASD for block shear?

    Solved problem  5-13-Determine the resistance to block shear of the coped W16 x 40 grade A36

    The next slide image shows part of the shear yielding and tensile rupture estimation, with the values of the shear rupture and tensile rupture.

    Solved problem 5-13-List of steel Beam Posts-part-3a

           This is the link to post 19.

    AISC Tables 9-3a,b, and C for Block shear-coped beam. 

    This is Post number 20 of the Steel beam Posts, which includes: 1- Solutions for Block Shear-Coped Beam Problem for the solved problem 5.13 for Block Shear-Coped Beam-case-1-UBS=1.00. Use tables Aisc Tables 9-3a for tensile rupture for block shear.

      Table 9-3b or shear yielding component and AISC Table 9-3C for shear rupture component.

    How do we estimate The LRFD and ASD for block shear?

    This is a snapshot for Table 9-3a titled block shear-tension rupture component per inch of thickness kip/inch.

    Table 9-3a titled block shear-tension rupture component
    AISC Tables 9-3a,b, and C for Block shear-list of Steel Beam Posts-part-3a

    This is a snapshot for Table 9-3c titled block shear-shear rupture component per inch of thickness kip/inch.

    Table 9-3c titled block shear-shear rupture component
    List of Steel Beam Posts-part-3a-Table 9-3c.

      This is the link to post 20

    Case 2 for block Shear-Coped Beam Problem.

    This is solved problem #10.8 from the Unified Design of Steel Handbook. It requires determining the resistance to block shear of the coped beam W16x40 grade A992. There are two bolt lines, a typical case -2 where UBs=1/2.

    A solved problem #10.8 where Ubs=1/2.

    This is a snapshot of the shear yielding and tensile rupture calculations versus shear rupture and tensile rupture.

    Calculations of the LRFd value for the coped beam case-2.

    This is the link to post 21-Case 2 for block Shear-Coped Beam Problem.

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



  • 21- Case 2 for block Shear-Coped Beam Problem.

    21- Case 2 for block Shear-Coped Beam Problem.

    Case 2 for block Shear-Coped Beam Problem.

    In this post, I will include a solved problem for Case 2 for block Shear-Coped Beam problem where the UBS=1/2. The solved problem number #10.8 from the Unified Design of Steel Handbook.

    It is required to determine the resistance to block shear of the coped beam W16x40 grade A992, which is the same coped beam in the first solved problem in the previous post But with a different steel grade which is ASTM A992.

    The horizontal distance between the beam edge and the center line of the first line of bolts is 1.25 inches, and the vertical distance between the edge of the beam to the first line of bolts is 2 inches. While the vertical spacing between bolts is 3 inches.

    There are two vertical lines of bolts, the horizontal edge distance to the first line of bolts is 1 1/4″. the horizontal spacing between the lines of bolts is 3 inches. Please refer to the next slide image for more details. The bolt diameter is 5/8 inches. because we have more than one line of bolts this case is case 2, where the UBS value is equal to 1/2.

    Case 2 for block Shear-Coped Beam Problem

    What are the gross sections for the given beam?

    For W16x40, the web width is 0.305 inches, the breadth of the flange is 7 inches, and the flange width is 0.505 inches. The yield stress for A992 is 50 ksi, while the ultimate stress is 65 ksi.

    For the hole diameter, we add 1/8 of inches to the bolt diameter and the hole diameter will be equal to 6/8 inches.

    2-There is one section that is subjected to shear, and that is the vertical section thru bolts. There is also one section that is subjected to tension, and that is the lower section.

    The value of shear yielding.

    3-We estimate the gross area for the vertical section subjected to shear which can be found equal to (11×0.305)=3.355 inch2. We can estimate the shear yielding value which is the product of the gross area by (0.6*Fy), where Fy=50 ksi. The shear-yielding value is equal to 100.65 kips.

    Gross and net area for tension-UBs=1/2 and tension rupture.

    4-We estimate the gross area for the horizontal section subjected to tension which can be found equal to (4.25×0.305)=1.296 inch2.

    To get the net area for tension or Ant we deduct the area for one and a half of a bolt hole, the net area Ant can be found to equal 0.9531 inch2. This is case 2 for the block Shear-Coped Beam Problem. For the tension rupture force multiply (ubs*Ant*Fu) which is (1/2*0.9351*65)=30.976 kips.

    Shear yielding and tension rupture.

    Estimate the shear rupture and tension rupture.

    We can get the value of shear yielding and tension rupture. To estimate the shear rupture, we need to find the net area for shear which is the gross area for shear minus (3.5 holes area). The hole diameter of the bolts is 6/8 inches, the web thickness is 0.305 inches and the height is 11 inches. The net area for shear is equal to 2.554 inch2. Multiply by (0.6*Fy) to get the shear rupture force value which is 99.60 kips. for the tension rupture, we estimated earlier as equal to 30.976 kips.

    Shear rupture and tension rupture.

    Estimate the LRFD value of the block shear force Case 2 for the block Shear-Coped Beam Problem.

    We list all the data which we have obtained we group into two groups. The first group is the case of shear yielding and tension rupture. The second group will be the Shear rupture and tension rupture. We can find that the failure is controlled by shear rupture and tension rupture since their sum is less than the shear yielding and tension rupture.

    We select the minimum 130.576 kips, we multiply by phi which is equal to 0.75 we get the LRFD value which is 98 kips, please refer to the next slide image for more details.

    The LRFD final value for block shear.

    Estimate the ASD value of the block shear force Case 2 for block Shear-Coped Beam Problem.

    We list all the data which we have obtained we group into two groups. The first group is the case of shear yielding and tension rupture. The second group will be the Shear rupture and tension rupture. We can find that the failure is controlled by shear rupture and tension rupture since their sum is less than the shear yielding and tension rupture.

    We select the minimum 130.576 kips, we divide by omega which is equal to 2 we get the ASD value which is 65.30 kips, please refer to the next slide image for more details.

    The ASD value for the block shear for coped beam.

    I have come to an end for Case 2 for block Shear-Coped Beam Problem. Thanks a lot.

    For more details regarding block shear, please refer to the post -Quickstart to the introduction to block shear resistance.

    For an introduction to coped beams, please refer to the previous posts, post 13-The relation between Block shear and coped beams.

    Post-19-Solutions for Block Shear-Coped Beam Problem.

    Post-20-Aisc Tables 9-3a,b, and C for Block shear-coped beam.

    For a more detailed illustration of block shear, there is a very useful external link- Tensile yielding and tensile rupture. A Beginner’s Guide to the Steel Construction Manual, 14th ed.

  • 20-Aisc Tables 9-3a,b, and C for Block shear-coped beam.

    20-Aisc Tables 9-3a,b, and C for Block shear-coped beam.

    AISC Tables 9-3a,b, and C for the Block shear-coped beam.

    In this post, we will verify by the use of AISC tables 9-3a,b, and C for the block shear-coped beam to check the solution of solved problem #5.13 from Prof. Alan Williams‘s handbook.
    The solved problem was included in post 19, for which it was required to find out the block shear for the coped beam where UBS equals 1.00, which is the case where we have a uniform stress distribution for tension force.

    It is required to determine the resistance to block shear of the coped beam W16x40 grade A36.
     The horizontal distance between the beam edge and the center line of the first line of bolts is 1.50 inches, and the vertical distance between the edge of the beam to the first line of bolts is 1.5 inches. While the vertical spacing between bolts is 3 inches as shown in the next slide image. The bolt diameter is 3/4 inches.

    Solved problem 5-13 for block shear for coped beam.

    This is an elevation section to show the beam, the number, and arrangement of bolts, and the vertical and horizontal distance between the edge of the coped beam and the bolts.

    How to estimate the LL for the beamPart 4/4 of the Solved problem 9-9-6?

    AISC Table 9-3a for tensile rupture for block shear.

    We start with AISC table 9-3a, the first AISC Table of the three AISC tables to be used for block shear evaluation.

    The coefficient for block shear UBS equals 1. Aisc table 9-3a determines both the LRFD and ASD value for the block shear tension rupture component of the block shear, but the estimated values are for kips per unit inch.

    We have Ultimate stress equals 58 ksi, log in with the first column of Leh values, and select Leh value of 1.50 inches where leh is the horizontal distance from the edge to the first line of bolts. The bolt diameter is 3/4 inches, and the intersection between the horizontal line from Leh equals 1.50″ line and the two vertical lines from the bolt column of 3/4 inches will give us both the LRFD value and the ASD value for tensile rupture.

    As we can see, the LRFD value of tensile rupture is 46.20 kips/inch, while the ASD value is 30.80 kips/inch.

    Table 9-3A-Block shear -shear rupture component.

    AISC Table 9-3b for shear yielding component.

    Here comes the second of the three AISC tables, Table 9-3b, which is used for estimating the shear-yielding component of block shear. The yield stress is 36 ksi.

    The first column is for Lev, which is the vertical distance between the edge and the first line of bolts. We also need the S value, which is the spacing between bolts.

    Column n specifies the number of bolts, in our example, we have three bolts so n equals 3.

    The horizontal line from Lev=1.50 inches and n=3 is drawn and intersects with the lines from Fy=36 ksi will give the two values for LRFD and ASD of the shear-yielding components for block shear.

    AISC Table 9-3b for shear yielding component.Aisc tables

    As we can see, the LRFD value of the shear-yielding component is 121 kips/inch, while the ASD value for tensile rupture is 81.1 kips/inch.

    AISC Table 9-3C for shear rupture component.

    This is the third AISC table, the last of which is table 9-3C, which is used for estimating the shear rupture component of block shear. The ultimate stress is 58/ ksi.

    The first column is for n, the number of bolts, which equals 3 bolts.

    The second column is the lev column. We draw a line with n = 3 and lev equal to 1.50 inches. We draw two vertical lines from the bolt column where db=3/4 inches. We will get two values for LRFD and ASD for the shear rupture component. The LRFD value equals 139 kips /inch, and the ASD value equals 92.40 kips/inch.

    Block shear -shear rupture.

    We can see the results for The LRFD values for tension rupture, shear yielding, and shear rupture from AISC tables 9-3a, 9.-3b, and 9-3c. We multiply each value by the web thickness of the web tw, which is 0.305 inches.

    Analysis of the values obtained from AISC tables.

    We have two values for the LRFD from two cases. The first case is shear yielding and tension rupture, for which the sum of the two values is approximately 51.0 kips.

    The second case is shear rupture and tension rupture, for which the sum of the two values is approximately 56.49 kips. We will select the smaller value, which is 51.00 kips.

    Select the minimum LRFD value for block shear.

    This is a reminder of the previously estimated LRFD value for block shear based on the calculation. Please refer to the previous post for more detailed information. There is a match between this estimation and the one obtained by using tables for the LRFD value.

    The LRFD value as estimated by calculation for block shear

    We can see the results for The ASD values for tension rupture, shear yielding, and shear rupture from AISC tables 9-3a, 9.-b, and 9-3c. We multiply each value by the web thickness of the web tw, which is 0.305 inches.

    Grouping for the ASD values from the three AISC tables.

    We have two values for ASD from two cases. The first case is shear yielding and tension rupture, for which the sum of the two values is approximately 34.14 kips. The second case is shear rupture and tension rupture, for which the sum of the two values is approximately 37.57 kips. We will select the smaller value, which is 34.14 kips.

    Select the minimum ASD value for block shear.

    This is a reminder of the previously estimated ASD value for block shear based on the calculation. Please refer to the previous post for more detailed information. There is a match between this estimation and the one obtained by using tables for the LRFD value.

    The ASD value for block shear based on calculation.

    Post-19-Solutions for Block Shear-Coped Beam Problem.

    The next post: Post 21-Case 2 for the block Shear-Coped Beam Problem.

    For a more detailed illustration of block shear, there is a very useful external link yielding and tensile rupture. A Beginner’s Guide to the Steel Construction Manual, 14th ed.

  • 19-Solutions for Block Shear-Coped Beam Problem.

    19-Solutions for Block Shear-Coped Beam Problem.

    Solutions for Block Shear-Coped Beam Problem.

    This post will introduce solutions for the Block Shear-Coped Beam Problem. A solved problem is quoted from solved problem #5.13 from Prof. Alan Williams‘s handbook.
    The solved problem shows how to estimate the block shear-coped beam value when the factor for block shear  UBS equals 1.00, which is the case where we have a uniform stress distribution for tension force.

    A solved problem for Block Shear-Coped Beam-case-1-UBS=1.00.

    It is required to determine the resistance to block shear of the coped beam W16x40 grade A36.
     The horizontal distance between the beam edge and the center line of the first line of bolts is 1.50 inches, and the vertical distance between the edge of the beam to the first line of bolts is 1.5 inches.

    The vertical spacing between bolts is 3- inches, as shown in the next slide image. The bolt diameter is 3/4 inches.

    Block Shear-Coped Beam-A solved problem.

    Gross and net area for shear-UBs=1.

    1. For W16x40, the web width is 0.305 inches, the flange breadth is 7 inches, and the flange width is 0.505 inches. The yield stress for A36 is 36 ksi, while the ultimate stress is 58 ksi.

    For the hole diameter, we add 1/8 of inches to the bolt diameter, and the hole diameter will be equal to 7/8 inches.

    2-One section is subjected to shear, the vertical section through bolts, and one section is subjected to tension, the lower section.

    3-We estimate the gross area for the vertical section subjected to shear, which is equal to (7.50×0.305)=2.29 inch2. To get the net area for shear or Anv, we deduct the area for two holes and a half. The net area Anv is equal to 1.625 inch2.

    Estimate gross area and net area for shear.

    Case of shear yielding and tension rupture.

    The slide shows the case of shear yielding, where we have Agv equals 2.288 inch2. The acting shear stress is equal to 0.6*Fy, Since Fy=36 ksi. The shear stress accordingly will be equal to (0.6*36)=21.6 ksi.

     Multiply the shear stress by the gross area for shear to get the shear yielding, equal to 49.42 kips.

    For the case of tension rupture, first estimate the gross tension area equal to width by web thickness or (1.5*0.305)=0.458 inch2. To get the value of the net area for tension, which we call Ant, we will deduct one hole diameter area. As we can see from the slide, the Ant will be equal to 0.3241 inch2. The Ubs value equals 1. The ultimate stress Fult=58 ksi.

    Case of shear yielding and tension rupture

    Case of shear rupture and tension rupture.

    For the case of shear rupture and tension rupture-LRFD, the block shear-coped beam Phi value equals 0.75. We will multiply (0.60*Fult*Anv) plus (Ubs*Ant*Ful) by the Phi value. The result is equal to 56.46 kips, as shown in the next slide image.

    We will select the min value of (51.20,56.42), which can be found to be equal to 51.20 kips, which is the LRFD value for the block shear-coped beam for UBS=1.

    For the ASD value, we can find that it is equal to 34.13 kips for shear yielding and tension rupture and 37.64 kips for shear rupture and tension rupture. We will select the minimum value, which is kips, for UBS=1.

     shear rupture and tension rupture.

    Estimate the final LRFD value for block shear.

    Now, we are ready to proceed to our estimation for block shear. We have a yield stress of 36 ksi and an ultimate stress of 58 ksi. The gross area for shear, Agv, equals 2.228 inch2. The net area for shear is equal to 0.3241 inch2.

    The first case is the case of shear yielding and tension rupture. For that case, the shear stress equals (0.60*Fy), while the tension stress is fult acting on the net area of tension Ant. Please refer to the left side of the slide.

    Shear yielding is estimated as equal to 49.42 kips. The tension rupture equals UBs by Fult by Ant; the UBS value equals 1, Ant equals 0.3241 inch2, and the Fult=58 kips. The total value of tension rupture will be equal to 18.798 kips. Adding the value of shear yielding and tension rupture will give a total value of 68.218 kips.

    The second case is the case of shear rupture and tension rupture.

    Please refer to the right side of the slide. In that case, the shear stress is equal to (0.60*Fult), while the tensile stress is Fult acting on the net area of tension Ant. The shear rupture value is 56.411 kips, and the tensile rupture value is 187.798 kips.

    Adding the value of shear rupture and tension rupture will give a total value of 75.028 kips.

    The LRFD value for block shear.

    If we want to find the LRFD value for block shear for the first case, we multiply phi, which is equal to 0.75. Please refer to the left side of the slide. Phi will be multiplied by opening a bracket (0.60Fy*Agv plus ( 0.60*fult)*ubs*Ant). The final LRFD value for case 1 equals 51.16 kips.

    We will select the minimum values of the two (51.20 and 56.46). The final LRFD value for Block Shear-Coped Beam will equal 51.20 kips.

    That means shear yielding and tension rupture controls.

    Estimate the final ASD value for block shear.

    For the ASD value, consider (1/omega)=(1/2).

    We can find that the final ASD value for the block shear*-coped beam will equal 34.13 kips. We have come to the end of the solved problem.

    The ASD value for block shear-Ubs=1.

    For more details regarding block shear, please refer to the post -*Quickstart to the introduction to block shear resistance.

    Our next post will discuss using AISC tables to get the block shear, yielding, block shear rupture, and tension rupture values and how to develop these expressions. Please find the detailed post.

    A new post: Post 21-Case 2 for the block Shear-Coped Beam Problem.

    For an introduction to coped beams, please refer to the previous post, post 13-The relation between Block shear and coped beams.

    For a more detailed illustration of block shear, there is a very useful external link-Tensile yielding and tensile rupture. A Beginner’s Guide to the Steel Construction Manual, 14th ed.