Brief content of steel tension members -post -11.

11-A Solved problem 4-4, Pult without block shear.

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, while Fult=58 ksi.

Solved problem 4-4

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.

Solved problem 4-4 estimate the U value.

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-ult due to yielding will be equal to ty*Agross*Fy=0.90*3.37*36=109.188 kips. can be taken approximately Pult=109 kips.                     

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

Estimation of C channel capacity and compare with the given ultimate load.

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.

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