- Solved problem 4-6-block shear for a C-Channel-2/2.

## Solved problem 4-6-block shear for a C-Channel-2/2.

### Estimation of gross shear area Agv .

For block shear to occur at the web of the channel continued:

3- We have two planes acted upon by the shear force component for which, we will estimate both the Agv is the gross area for shear, Anv is the net area for shear.

4- to get the total number of hole diameters as we can see we have a total of 3 diameters each diameter=6/8″ and the thickness of the web is equal to 0.22”.

The length of shear planes is 2*(4″+1.5″=2*5.50″) .For the estimation of the gross area for shear Agv, the value will be equal to (2*5.50*0.22)=2.42 inch2.

#### Estimation of the net shear area Anv for the channel web.

The length of shear planes is 2*(4″+1.5″)=2*5.50″. For the estimation of Agv=(2*5.50*0.22)=2.42 inch2.

To get the value of Ant, we will deduct the area of 3 diameters as follows: 3 diameters of 3*(6/8)*(0.22) the Anv=Agv-sum d*tw=2.42-(54/64)=0.715 inch2.

### The general equation for the ultimate tension force based on block shear.

Let us review the equation for the nominal rupture, the value of the nominal load Pn=Ubs*Fult*Ant and we have nominal rupture strength in shear Vn=0.6 Fu*Anv. The nominal yield strength in shear Vn= 0.6 Fy*Agv.

T*his is the general equation to apply for the two cases which we have.*

### Estimation of factored Ultimate tension force for Plate based on the Block shear value.

Let us review the data from the gusset plate for the gross areas and net areas, we have the values of Agv=4.12 inch2. Anv=3.276 inch2 and Agt=1.5 inch2, Ant=1.219 inch2, we have a gusset plate,Ubs=1. This is the equation, the ultimate stress accompanied by the net area in shear Anv.

First, we have φ, φ=0.75 fracture combined with tension and shear, φPn=0.75*(0.60*Fult, which is 58,*Anv, which is 3.276+Ubs, which is =1,*Fy , which is 36,*Ant, which is 1.219 inch2.

Evaluate,then 0.75*(0.6*58*3.276+1*58*1.219)= 138.53 Kips, what is the upper limit, for which we cannot exceed.We have φ=0.75*( 0.6*Fy*Agv+Ubs*Fu*Ant)=0.75*( 0.6*36*4.12+1*58*1.219).

We make arrows for the identical figures, the value is =119.77 kips, not to exceed, this means that we are going to select, so the min value for the gusset plate.

The gusset plate under the condition of block shear can carry 119.77 kips. This is the calculation for the Guest plate in solved problem 4-6.

### Estimation of factored Ultimate tension force for channel based on the Block shear value.

Now we have a look at the channel, the same procedure we are going to apply, first, the areas as Agv=2.42 inch2, Anv=1.925 inch2, Agt=0.88 inch2, Ant=0.715 inch2, the Ag to be multiplied by0.60*fy, For which, do not forget to multiply the Anv *0.60*Fu, Ubs still =1.

Let us apply the equation, but this time for the channel, not the plate, 0.75(0.6*58*1.925+1*58*0.715)=81.34 kips, we perform the upper limit check, as follows,0.75(0.6*36*2.42+1*58*0.715)=70.31 kips.

This value is the upper limit, so the min is 70.31 kips, so φPn, which we revise as φRn, is the min of ( 81.34,70.21) gives 70.31 kips, then due to the block shear criteria, the channel can carry 70.31 kips, but the applied force =75 kips.

The applied force Pu=75 Kips is higher than the capacity of the channel to carry; this channel is, therefore, not adequate.

In order to improve the carrying capacity, take the following procedure.

The section is to be increased to improve the nominal strength of the channel and increase the value.

This is the PDf file used in the illustration of this post.

There is a very useful external link-**Block Shear Rupture**.

For the next post Block shear and coped beams,