- Solved problem 12-1-part 2-connection nominal load.
- Solved problem 12-1-part 2 -The calculation of bearing Failure nominal load by table 7-4.
- Solved problem 12-1-part 2.The calculation of shear Failure nominal load by table j3.2.
- Solved problem 12-1-part 2.The calculation of shear Failure nominal load by table 7-1.
- Solved problem 12-1-part 2.The final summary for the nominal load.

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

In this post, we will estimate 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 le=3″.

Later we will estimate the nominal shear of fasteners by using table j.3.2.

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.

You can click on any picture to enlarge then press the small arrow to review all the other images as a slide show.

We continue our discussion to the Solved problem 12-1-part 2 for bearing connection, from Prof. Mcccormac’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** -The calculation of 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 bolt and type of bolt,Fu, and inner spacing.

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

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 value the design value ASD value of rn. from the same table when selecting Move horizontally at the spacing of 3″, 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″

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.

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.

### **Solved problem 12-1-part 2**.The calculation of 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 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 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**.The calculation of 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

**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.

For the bearing Φ*Rn=182.40 kips. In the case of shear Φ*Rn=122.60 kips. And these are the corresponding values for the ASD, 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 final value of Φ*Rn=122.60 kips and the ASD value of (1/ Ω )*Rn=122.60 kips)*Rn=81.74 kips.

This is a link for the pdf file used in the illustration of this post and the previous post.

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

This is a very useful link: **A Beginner’s Guide to the Steel Construction Manual, 15 ^{th} ed, Chapter 4 – Bolted Connections**