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

## Solved problem 12-1 for bearing connections.

In this video. We have a solved problem 12-1 from prof Mccormac’s book, in which a given connection it is required to find out the nominal strength of this connection under a tension load, all the failure modes to be checked and at the end, we will select the failure mode that has the lowest value of nominal strength. The video has a subtitle and a closed caption in English.

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

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

Our new topic will be the discussion of the solved problem 12-1 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 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.

### 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

### 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.

### 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″

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″.

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

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.

The next step is to estimate Lci and Lco. For lci=spacing -diameter of the hole=2/1/16″, while for Lco=2.53″. The calculations are shown in the next slide image.

As shown, the bearing stress distribution from 0 to max at the middle span, acts on an area of db*t, with the stress of C*Fult.

For the upper limit C value =2.4, this force is resisted by shear on the two side planes of 1.2 of length lc at the inner distance and width =t plate, this is applied for the inner spacing and also for the external edge space.

Let us estimate the edge distance which is=3″ -half of the hole,3-0.50*(15/16)=2.53″ can be approximated to 2.5″.
For the external Rn estimation, for one bolt with two shear planes=2*0.50*lc*Fult, Fult=58 ksi, Lc=2.50″.

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 LH S value is 87.0 kips, while the upper limit is 60.90 kips, it will be selected. We have two edge bolts, final Rn=2*60.90=121.80 kips.

While  for inner spacing,=3″lc=3-(7/8+1/16) not 1/8″=2.06″ approximated to 2″. The same rule will be applied as 1.20*Lc*t*Fult=1.2*2.00*1/2*58=69.6 kips to be compared with (2.4*db*t*fult which is 60.90 kips, selected as a min value then multiplied by 2. The ultimate road value due to bearing will be equal to 243.60 kips.

Rn total is for the four bolts.Rn edge+Rn inner=60.90*2+60.90*2=243.60 kips.
For the LRFD Φ=0.75 and  ASD Ω=2.00. So for the LRFD Φ*Rn.=0.75*243.20=182.40 kips.
For ASD (1/Ω)*Rn=1/2*243.20=121.60 kips.

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

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

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