 # 12a-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 a c channel.

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 which 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.

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-block shear strength.

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.

This is the general equation to apply for the two cases which we have.

### Estimation of factored Ultimate tension force for Plate-Block shear strength.

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, Ubs=1. This is the equation, the ultimate stress accompanied by the net area in shear Anv.

First, we have φ, which is equal to 0.75 fracture combined with tension and shear, the Fult equals 58 KSI.*Anv, which is 3.276+Ubs, which is =1,*Fy, which is 36,*Ant, which is 1.219 inch2.

Evaluate, the upper limit is equal to 0.75*(0.6*58*3.276+1*58*1.219)= 138.53 Kips, shear yielding, and tension rupture.
while for shear rupture and tension rupture, we have φ=0.75*( 0.6*Fy*Agv+Ubs*Fu*Ant)=0.75*( 0.6*36*4.12+1*58*1.219)=119 kips

We will select the minimum value that will be equal to 119 kips. The gusset plate under the condition of block shear can carry 119 kips. according to block shear strength. This is the calculation for the Guest plate in solved problem 4-6. since the ultimate load is less than 119 kips so the plate is adequate for block shear.

### Estimation of factored Ultimate tension force for the channel -Block shear strength.

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, 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 φ*Pn, which we revise as φ*Rn, is the min of ( 81.34,70.31) 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.

### Estimation of factored nominal load -ASD design-block shear strength.

This is a detailed estimation for the ASD design load based on block shear for both plate and channel. We can find that the channel is not adequate for block shear based on total load Pt=54 kips.
The gusset plate is adequate based on its block shear strength.

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

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