 # 4-Nominal shear strength and bearing- tear out made simple.

## Nominal shear strength and bearing- tear out for bearing connections.

### The video used in the illustration.

We are going to start a new subject, which is the relation between the ultimate strength of bolt Fu and the Nominal shear strength. This is the first item of discussion. The second subject is the bearing and the tear-out.

The Nominal shear strength, which is denoted as Fnv is based on Ultimate strength. the relation in between can be that Fnv=Fult* 0.625.

We had before discussed the shear subject while discussing the shear for beams and we consider the shear stress factor is approximate 0.60. That was a part of the video that has a subtitle and a closed caption in English.

If you prefer to check the pdf data and view pictures, please continue to read.

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

The content of the lecture has two items the first one is the relation between ultimate stress and the Nominal shear strength, while the second item is the Bearing and tear out description. These items are shown in the next slide image.

### The nominal shear strength for a bolt Fnv.

The Nominal shear strength, which is denoted as Fnv is based on Ultimate strength. The relation stated that Fnv=Fult* 0.625. We had before discussed the shear subject while discussing the shear for beams and we consider the shear stress factor is approximate 0.60.

But regarding the bolts, the Code considers. That the Nominal strength =Fu*0.625 but again multiplied by two factors, for the first factor it is Ca, Ca depends on the length of the connection.

While Cb, for the types of threaded whether included or excluded. For high-strength tensile bolts. Whether X or N.

For the case of Ca, it is equal to 0.75, if the connection is bigger than 38″. Ca is = 0.90 if the connection =38″. For the case of Cb, it is equal to 1, if the threaded is excluded or the shear line does not pass by the thread of the bolt.

For the case of Cb, it is equal to 0.80, if the threaded is included, which means that the shear plane cuts the threaded portion of bolts.

Let us refer back to the equation, the shear If Fult=120 kSI. while the shear factor 0.625*ca=0.90 if the connection is >38″, cb=1 for threaded portion is excluded, the multiplication will give =120*0.625*0.9*1=67.50 KSI.

But if we consider the threaded is included, then multiply again by 0.80=67.5*0.80=54 ksi.

Let us refer back to the equation, the shear If Fult=120 kSI. while the shear factor 0.625*ca=0.90 if the connection is >38″, cb=1 for threaded portion is excluded, the multiplication will give =120*0.625*0.9*1=67.50 KSI.

But if consider the threaded is included, then multiply again by 0.80=67.5*0.80=54 ksi. Let us have a look at table J3.2.

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The Fult=120.0 ksi, for the case of group A, ASTM A 325. While for group B where ASTM A 490, then Fu=150 ksi, the table J3.2 is divided into two parts, in which the first part Nominal tensile strength=0.75*Fult, whether threaded included or excluded

Let us have a look at table J3.2.The Fult=120.0 ksi, for the case of group A, ASTM A 325. While for group b, where  ASTM A 490, then Fu=150 ksi, the table J3.2 is divided into two parts, in which the first part Nominal tensile strength=0.75*Fult., whether threaded included or excluded.

For group A where ASTM A-325, the F tensile Fnt=0.75*120=70 ksi. While for group B where ASTM A-490, the F tensile Fnt=0.75*150=112.50=113 ksi.

Let us discuss Fnv  .The nominal strength for shear as agreed= 120*0.625*0.90=67.50=68 ksi , for connection >38″, for the x or the threaded is excluded.

Let us check the table back, we will find that 68 ksi is in the table at Group A, at the x, excluded. Group A e.g, X, when threads are excluded from the shear planes. if again multiply by 0.80, for the case of the threaded portion is in the shear plane, then the Fnv=54.4 ksi=54 ksi.

Again if we have for group b, we have Fult=120 ksi multiply by 0.625*0.90=68 ksi. Fnv=Fult*0.625*0.90=0.563*Fult. For Fult=150 ksi, then Fnv=0.563*150=84 ksi.While for N, then Fnv=150*0.563*0.80=67.50=68 ksi.

### The values of nominal shear strength and nominal tensile strength for bolts in the RCSC.

Refer to the bolt council table, Research Council on Structural Connections, for ASTM A 325 or group A, the static tension =90 ksi. While the static tension in group B=113 ksi, while for the case of Fatigue for repeated loads like in bridges, refer to sec 5.5.

Here is the shear condition for connection<=38″, then we will have fnv=68 ksi and 54 Ksi, refer back to table J3.2. The same figures are there. If the connection is >38″, then Fnv will be 45 ksi and 56 ksi for the threaded included in the shear plane.

Here in the case of threaded is excluded, the first case for connection <=38″, we have Fnv=68 ksi, and 84 ksi. Back to J3.2, we will find the two figures Fnv =68 ksi and 84 ksi. The last case for the included thread with connection>38″.

### Bearing and tear -out of a simple connection.

As for bearing it depends on the hole elongation, this is different than the slip connection, which we will discuss later on the critical slip, elongation is prevented.

I quote, when deformation at the hole elongation will not exceed 1/4″. When high tensile stress occurs on the net section.If it is not a design criterion, then other relations can be estimated.

Then a discussion of the tear-out is shown, the tear-out for Lh external, one part will split the same as wedge shape. The inclination angle is based on the external clear distance.If the deformation is of design criteria, or deformation 0.25″ or not, then equation, to check out the upper limit which is 2.4*d *t*Fult.

For the bearing and tear out LRFD and ASD values φ=0.75 while Ω=2.00, for the nominal strength of the connected material

The different equation presented in the specs is presented herewith. For case ( i )of bearing, when elongation of 0.25″ is a design criterion then Rn=2.40*d*t*Fult.

For case ii of bearing, when elongation of 0.25″ is not a design criterion a higher value for Rn is given, Rn=3*d*t*Fult.

For the tear-out case( i), for deformation at the bolt hole is a design criterion, Rn=1.2Lc*t*Fult, While Rn=1.5*Le*t*Fult, If the deformation is not design criteria.

Again long slotted connection with the slot perpendicular to the direction of the force. Again bearing value is Rn=2.30*d*t*fult, while for tear-out Rn=1.0*Lc*Fult*t, there is a condition c for connections made using bolts that pass completely through an un-stiffened box member.

This is table J3.3, if the bolt, that gives the hole dimension for various values of bolts, whether is oversized, short slot, long slot

The hole diameter is bigger than the diameter of the bolt by 1/16″ for the standard as shown in the table.

This is table 3.1 nominal bolt hole dimensions for standard diameter or oversized or shot slotted or long slotted. The bearing value is based on the slot condition.

This is table J3.4, which we have introduced earlier for the minimum of the edge distance from the centerline of the bolt.

The change, where there is a change between the specification 2010 and specification 2016 in the values of Rn, Rn upper value=2.4d*t*Fult and  Rn=1.2*lc*t*Fult.

The next post is a solved problem 10-1 for shear force value.

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