Solved problem 4-6-block shear for a C-Channel-1/2.
Brief content Content of the two videos.
This is the Solved problem 4-6-P-ult with block shear. It is required to estimate the block shear of the bolted channel and continue to check the adequacy of both the Gusset plate and the channel section for the applied Tension ultimate load of 75 kips.
The video includes the necessary information on how to estimate both the gross and net area for the channel regarding the block shear. This post regarding solved problems 4-6 is covered in this video from time 9.01 to end. and the whole of the second video. This video has a subtitle and closed caption in English.
This is the link for the second video in which we continue the review of the same solved problem 4-6 and determine the different parameters for Agv, Ant for both channel web and Guesset plate.
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.
Let us review The connection shown in solved problem 4-6, and determine if the channel and gusset plate are adequate for the applied tension load. Considering block shear assumes that the plate’s width is such that the block shear along with the failure plane shown in Figure 4-12 controls the plate’s design.
Now the hatched shape is torn from the original shape if we have a close look at the gusset plate after torn, we have the 4 bolts, and we create a path for the separated part
1- We have one plane acted upon by tension force for which, Agt is the gross area for tension, 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 deduct 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.