Tag: solved problem 12-1 for bearing connections.

  • 7A- Solved problem 12-1-part 2-connection nominal load.

    7A- Solved problem 12-1-part 2-connection nominal load.

    Solved problem 12-1-part 2-connection nominal load.

    We can find the bearing Nominal load from Table 7-4 for the available bearing strength for the external bolts. No provision in Table 7-5 for external bolts of the outer distance of the bolt with the value of le=3″. Later we will estimate the nominal shear of fasteners by using Table J3.2

    Finally, we find out the design strength of the connection after comparing the least load of the different nominal loads for tensile failure, bearing strength, and shear strength.

    We continue our discussion to the Solved problem 12-1-part 2 for bearing connection, quoted from Prof. McCormack’s book.

    Determine the design strength φ*Pn and the allowable strength Pn/Ω for the bearing type connection shown in Fig.12-5. The steel is A36-Fy=36 KSi and Fu=58 ksi.

    The bolts are 7/8″ in A325, the bolts are standard sizes, and the threads are excluded or X from the shear plane. Assume that deformations at bolt holes are a design consideration.

    Solved problem 12-1-part 2

    Solved problem 12-1-part 2-Bearing Failure nominal load by table 7-4.

    There is a table 7-4 for the available bearing strength at the bolt, based on the diameter of the bolt and type of bolt,Fu, and inner spacing.

    We have 7/8″ and Fult=58 ksi and a standard hole STD. The spacing between bolts is 3 “.

    Move horizontally at the spacing of 3″, intersecting with the vertical line from the nominal diameter of 7/8″ at Φ*rn. We get the value of Φ*rn= 91.40 kips/inch of the plate thickness. Since the plate thickness is 1/2″. Then multiply by (1/2).

    We get the design value ASD value of rn from the same table when selecting Move horizontally at the spacing of 3″, which intersects with the vertical line from the nominal diameter of 7/8″ at (1/Ω) *rn.

    We get the value of (1/Ω) *rn =60.90 kips/inch of the plate thickness. Since the plate thickness is 1/2″. Then multiply by (1/2). For a single bolt but/inch of plate thickness, our plate thickness is 1/2″.

    Available bearing strength at bolt holes based on bolt spacing.

    LRFD value for bearing.

    From Table 7-4 we have Φ*rn=91.4 kips/inch, then multiply by 1/2″, the final Φ*Rn=45.70 kips for one bolt.

    ASD value for bearing.

    For the (1/Ω)*rn=60.90 kips/inch, then multiply by 1/2″, the final (1/Ω)*Rn=30.45 kips for one bolt. The calculations are very close to the previous calculations done by using equations.

    Final value of nominal load for bearing.

    Solved problem 12-1-part 2. Shear Failure nominal load by table J3.2.

    The next step will be the shear estimation. For shear, we need table J3.2. For ASTM A325, we have two figures the first one is for shear which is 54 ksi, and 68 kips, which value to choose from?

    The solved problem mentioned that bolts are excluded, and the thread will not help in the shear strength Fnv=68 ksi, to estimate the shear. The shear calculation is checking how many planes and calculating the area, which is given. For shear estimation.

    We have two methods In the first method, we have table 7-1 available shear strength for bolts.

    We have N and X, we check the diameter, and also the table gives the area of the bolt check diameter. The table will give the LRFD and ASD values.

    We have d=7/8″ for x, we have S and D, where s is the single shear and D is the double shear Φ*Fnv=51 ksi/inch2 of diameter area or Fnv/Ω =34.0 ksi/inch2 of diameter area for LRFD, which can give for 7/8″bolt Φ*Fnv=51*0.601=30.65 kips for single shear. While for 7/8″ bolt. Fnv/Ω*Fnv=34*0.601=20.43 kips for a single shear.

    Solved problem 12-1-part 2. Using table J3.2 for shear strength of bolts.

    A Solved problem 12-1-part 2.Shear Failure nominal load by table 7-1.

    We can go to Table- 7-1 for the available shear strength for bolts. Select the column of 7/8″ with the thread condition x as a horizontal line. The intersected value will be Φ*Rn =30.70 kips for one bolt. For the ASD we have (1/Ω)*Rn =20.40 kips.

    How many bolts do we have? we have 4 bolts, then multiply by 4 For Φ*Rn*n=4*30.70 =122.80 kips.

    For the ASD we have (1/Ω)*Rn*n=20.40*4=81.60 kips.ASD value=81.60 kips. The block shear was not included in the given solved problem 12-1-part 2.4*0.601=20.43 kips for a single shear.

    LRFD and Asd design values for Shear strength from table 7-1.

    Solved problem 12-1-part 2—the final summary for the nominal load.

    We take all the estimated values and put these values in a form of a table, which is shown herewith. This is the summary for yielding for LRFD Φ*Rn=194.40 kips, for fracture Φ*Rn=261 kips. In the case of shear Φ*Rn=122.60 kips. , you can double-check between the values by diving the LRFD /1.5 to get the corresponding ASD values. Ω =1.50. Which are the lowest values? the lowest value is 122.60 kips for the LRFD from shearing.

    The list of all nominal loads for bolts LRFD design.

    The final value of Φ*Rn=122.60 kips and the values for THe ASD design nominal loads are added for each limit state and the final ASD value selected is 81.74 kips.

    The final values in ASD design .

    Thanks a lot, and see you in the next post.

    This link for the previous post-A solved problem 12-1 for bearing connections.

    This is a link to a Solved problem 5-part 1 for design shear strength for bolts.

    This is a very useful source for designing various Steel elements: A Beginner’s Guide to the Steel Construction Manual, 15th ed, Chapter 4—Bolted Connections.

  • 7- A solved problem 12-1 for bearing connections.

    7- A solved problem 12-1 for bearing connections.

    Solved problem 12-1 for bearing connections.

    A Solved problem for bearing connection 12-1-Part 1.

    Our new topic will be the discussion of the solved problem 12-1 from Prof. McCormack’s book.

    Determine the design strength φ*Pn and the allowable strength Pn/Ω for the bearing type connection shown in Fig.12-5. The steel is A36 where (Fy=36 KSi) and Fu=58 ksi, the bolts are 7/8″ in A325, the bolts are standard sizes, and the threads are excluded or X from the shear plane. Assume that deformations at bolt holes are a design consideration.

    The overlap distance is length =9″, where the CL distance=3″ between bolts. The length is in the direction of the load, while the transverse cL distance =6″. The edge distance is 3″ and 3″ from each side.

    solved problem 12-1 for bearing connections.

    Tensile Failure nominal load by Yielding for solved problem 12-1.

    For this connection we have two modes of failure, the first mode of failure is due to yielding as shown in sec 1-1 and the mode of failure is to rupture as shown in sec II-II please refer to the next image for section 1-1.

    We have a single-shear plane. Two plates are placed above each other and connected by 4 bolts. The bolts are 7/8″ where the area for each bolt=0.60 inch2 as given. And the upper plate is 1/2″, the lower plate thickness is 1/2″ and the plate width is given as 12″.

    For the tensile yielding Fy to be multiplied by Ag, or the gross area. In the second case by tensile rupture, the area net is to be multiplied by Fult.

    We are aware that, due to placing bolts during installation, we add 1/16″+1/16″ to be added to the diameter of one bolt which is 7/8″, then the final value considered for diameter=7/8+2*(1/16)=1″.

    For the first case, the area gross=12*1/2=6 inch2. Then the tensile yielding =Ag* Fy, we have Fy=36 KSI and Fult=58  KSI, Ag=6 inch2, then Rn=Ag*Fy=36*6=216 kips, For the LRFD φ=0.90 φ*Rn=0.90*216=194.4 kips.

    While for allowable design (1/Ω)*Rn, where Ω=1.67

    Solved problem 12-1, Tensile failures calculations - yielding.

    Tensile Failure nominal load by rupture for solved problem 12-1.

    For case b) where Area gross- 2*diameter*thickness of the plate. Anet =12*1/2- 2*(1))*1/2.Anet=5 inch2, since we are dealing with the net area then we use Fult, which is=58 ksi.Rn=5*58=290 kips.

    Remember from the tension area lecture that Aeffective=U*Anet, U=1 for plates, then Aeff=1*5=5 inch2. But now φ=0.75, and Ω=2 .φ*Rn=0.75*290=217.50 kips. While Rn/Ω=290/2=145 kips.

    Solved problem 12-1, Tensile failure by rupture.

    The calculation of bearing for solved problem 12-1.

    Prior to the procedure of checking the bearing capacity for the bolts, we need to have a look at the list of the necessary tables that we need to use in our estimation.

    The diameter of bolts=7/8″, the hole diameter is the bolt diameter plus 1/16″, and the diameter hole equals 15/16″. The inner spacing=2 2/3 DB.

    To check the requirement of the edge distance, there is Table J3.4 for a 7/8″ bolt, the required edge distance is 1/18″

    Minimum spacing requirements.

    To check the requirement of the edge distance, there is table J3.4 for dia 7/8″ bolt, the required edge distance is 1/18″.

    Minimum edge distance clause from the specification.

    This is a sketch showing the bolt spacing and edge spacing for a bearing connection.

    Spacing and edge distance.

    We have a distance from cl to the outer edge=3″, the case is satisfied since this distance is >1 1/8″ as given by the table.

    for the spacing requirements for inner bolts, the spacing value is 3 inches which is bigger than 2.33 inches which is the minimum requirement.

    Check specification requirements for bolt distances.

    Nominal bearing value for the external bolts.

    The next step is to estimate the nominal bearing value for the external bolts, we estimate the clear distance and it was found to be equal to 2.531 inches. We compare this value with the value of (2db). we find that the outer clear distance for the exterior bolt is bigger than 2db, which indicates that the nominal value is governed by the upper limit equation which is(2.4*d*t*fu). The details of the estimation can be found on the next slide.

    Nominal bearing load for exterior bolts.

    Nominal bearing value for the inner bolts.

    We will estimate the nominal bearing value for the inner bolts, we estimate the clear distance and it was found to be equal to 2.06 inches. We compare this value with the value of (2db). We find that the outer clear distance for the exterior bolt is bigger than 2db, which indicates that the nominal value is governed by the upper limit equation which is(2.4*d*t*fu). The details of the estimation can be found on the next slide.

    The upper limit is 2.4d*b*t*Fult, db=7/8″ *t=1/2 for a plate, which is common between the left and right sides of the equation.

    The upper limit value will be equal to 2.4*7/8*1/2*58=60.90 kips. The nominal load based on the equation (1.2*lci*t*fu) is shown and as we can see it is higher than the upper limit and is not considered in the design.

    Nomial load Rn for the connection for inner bolts.

    Nominal strength value for bolts.

    The next step is to estimate the shear strength of the bolts. There are two ways to estimate the first way is to use table J3.2 and get the Fnv based on the type of bolt.in our problem, the bolt is 7/8 inch X type A. Fnv value is 68 ksi as single shear value for one bolt.

    What is the value of shear strength for bolts?

    I have included two sketches as a comparison between the bolt’s nominal loads in bearing and the corresponding bolts but with the shear strength and it can be seen that the shear strength of bolts is less than the nominal load due to bearing and shear strength control the design.

    Comparison between bearing and shear of bolts.

    We will continue the next post to finalize the calculations for the given connection. Thank you.

    The next post will be A solved problem 12-1-part 2-connection nominal load.

    A useful link-A Beginner’s Guide to the Steel Construction Manual, 15th ed, Chapter 4 – Bolted Connections