# 12-Solved problem 4-6-block shear for a C-Channel-1/2.

## Solved problem 4-6-block shear for a C-Channel-1/2.

Topics included in our discussion are shown in the next slides.

Now while dealing with the solved problem- 4-6, the solved problem 4-6  is the same as the solved problem 4-4, but includes the block shear.
The solved problem is from Prof.  Abi O. Aghayere’ss handbook, it is required to determine if the channel and gusset plate are adequate for the applied tension force but we have to consider the block shear.

### Estimation of gross tension area-Agt and net tension area Ant.

For the Solved problem 4-6-P-ult with block shear.

Examine the connection in solved problem  4-6 to ensure the channel and gusset plate are appropriate for the tension load. Consider block shear demands that the plate’s breadth is such that block shear, combined with the failure plane depicted in Figure 4-12, governs the plate’s design.

If we look closely at the gusset plate after it has been split, we can see that the hatched shape is torn from the original shape, there are four bolts, and that a route has been created for the separated component.

1- We have one plane acted upon by tension force for which, Agt is the gross area for tension, and the area Ant is the net area for tension. the section which is perpendicular to the direction of the applied force. the height of this section=4″ and the web thickness =3/8″.

For the estimation of gross area under tension written as Agt=4*3/8=1.50 inch2.
For Ant, which is the net area under tension, deduct two halves, which is one diameter of (5/8″+1/8″)=6/8″, and then multiply by the web thickness. The net area for C channel (6/8)*(3/8)=Agt- sum of holes *t-web=1.50-((6/8)*(3/8))=1.219 inch2.

For the details of these calculations, please refer to the next two slide images, the number of hole diameters for both tension and shear is shown.

2- We have two planes acted upon by the shear force component for which, Agv is the gross area for shear, Anv is the net area for shear, for the estimation of  Agv=(2*5.50*3/8)=4.12 inch2, for the net area in tension Ant deducts 3 diameters of 3*(6/8)*(3/8)=4.12-(54/64)=3.276 inch2.

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

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