## A Solved problem 4-4, P Ult Without Block Shear.

### Brief content Content of the video.

The solved problems 4-4, and the other solved problem 4-6 are quoted from Prof. Abi Aghayere’s book structural steel design. Application for Architects.

In the first solved problem 4-4, there is a Channel bolted to a gusset plate, it is required to check the adequacy of the channel under the applied ultimate tension load of 75 kips, in that example the block shear of the **section under the load is not considered.**

It is required to estimate the minimum value of the ultimate load d P ultimate of the section due to yielding and the ultimate load that the section can carry under tensile rupture. then compare this value by the given applied load, this was a part of the video, which has a subtitle and closed caption in English. This post covers the solved problems 4-4, which are included in the video from time 0.00 to time 8.55.

The remaining time is for the discussion of solved problem 4-6, which will be covered in the next post.

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

### A Solved problem 4-4.

In the solved example 4-4, from prof. Abi O. Aghayere handbook, it is required to estimate the ultimate tensile force that a channel C 8×11.50, which is connected to a gusset plate, by two lines of bolts.

Each line has two bolts, that can carry, but neglect the Block shear and compare this value by a given ultimate load of 75 kips, to check the adequacy of the section.

The bolts are 5/8 inch each.Given Fy=36 ksi,Fult=58 ksi.

*For the state of rupture*, we need to estimate the net area for a section, which is perpendicular to the force direction, the deduction of bolts taking into consideration adding 1/8″ to the bolt diameter.Anet=Ag-2*dia*tweb=3.37-2*(5/8+1/8)*0.22=3.04 inch2.

### Limit state of yielding.

For the state of yielding, we need to estimate Fy*Ag, Fy=36ksi, Ag=area of a channel without deduction, from table C 8×11.50=3.37 inch2, tweb=0.22 inch. the necessary data are obtained from table- 1-5 for the C channel.

### Limit state of rupture for the solved problem 4-4.

*For the state of rupture*, of the solved problem 4-4, we need to estimate the net area for a section, which is perpendicular to the force direction, We have a section that passes by the two bolts, so we will deduct the area of bolts as follows:

Anet=Ag-2*dia*tweb=3.37-2*(5/8+1/8)*0.22=3.04 inch2.

*It is required to estimate u factor from the relation U=(1- x̅ /L), we have a plastic distance about the y-axis which is written as x- bar from table 1-5, which is assigned for C channel sections, is taken as x̅=0.572, U=(1-0.572/4)=0.857.*

P-ultim due to yielding =φty*Agross*Fy=0.90*3.37*36=109.188 kips.can be taken approximately =109 kips.

P-ult due to rupture = =φtfr*U*Anet*Fult=0.75*0.857*3.04*58=126.57 kips, approximately 127 kips.

*The Minimum value, of the previous two values, is Pult =109.0 kips, since the acting force, which is=75 kips, is<127.0 kips, then the section is adequate, *

*The detailed estimation of the capacity for the given channel is shown in the previous slide image.*

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

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

This is the next post- Solved problem 4-6-block shear for a C-Channel-1/2, in which we solve a problem and get the ultimate tensile force after taking the block shear into consideration.