Drag Force Diagrams from Hiner Seismic Review Course

[migpics]

This question is for those of you who are taking the Hiner course.

I am working on the latest book and trying to figure out Drag Force Diagrams.

My question is regarding Problem #17, Page 2-41, the solution for question 5 shows a drag force diagram.

For Wall line A, it shows the Drag force to be Fd = (262 plf)(20')=5240 lbs.

On the solution to the next problem, (page 2-43) it shows a drag force diagram on Wall Line 1.

Fd= (69 plf)(7')-(343 plf)(7')=-1920 lbs max.

My question is, why is the wall shear not subtracted from the roof shear on the solution for 2-41?

I have the same question regarding problem 9.8 in the sample problems on the back.

It gives a solution of 4800 lbs but does not subtract the roof shear.

What am I missing here?

Thanks in advance for your response.

Miguel

 

[okutlu]

This question is for those of you who are taking the Hiner course.

I am working on the latest book and trying to figure out Drag Force Diagrams.

My question is regarding Problem #17, Page 2-41, the solution for question 5 shows a drag force diagram.

For Wall line A, it shows the Drag force to be Fd = (262 plf)(20')=5240 lbs.

On the solution to the next problem, (page 2-43) it shows a drag force diagram on Wall Line 1.

Fd= (69 plf)(7')-(343 plf)(7')=-1920 lbs max.

My question is, why is the wall shear not subtracted from the roof shear on the solution for 2-41?

The maximum drag force will be at the point where the wall starts. There is only unit roof shear from the intersection point of Line 1 and Line A down to the wall. That is why, it isn't subtracted.

If you were to calculate starting from southern point of line 1 to the end of the wall. The equation would be Fd=262*20-525*20=-5260 lbs.

In the end, it depends on your start point.

I have the same question regarding problem 9.8 in the sample problems on the back.

It gives a solution of 4800 lbs but does not subtract the roof shear.

What am I missing here?

Same concept applies here as well. If you start from top of the wall,

Once you make the calculation, unit wall shear will be approximately 400 plf.

Then, Fd= 240*30-400*30= -4800 lbs = -4.8 kips.

One last note, sign convention is not important in drag force calculation questions. (i.e. whether it is -4.8 kips or 4.8 kips).

I hope this helps.

 

[migpics]

Thanks! That clears it up.

 

[ptatohed]

This question is for those of you who are taking the Hiner course.

I am working on the latest book and trying to figure out Drag Force Diagrams.

My question is regarding Problem #17, Page 2-41, the solution for question 5 shows a drag force diagram.

For Wall line A, it shows the Drag force to be Fd = (262 plf)(20')=5240 lbs.

On the solution to the next problem, (page 2-43) it shows a drag force diagram on Wall Line 1.

Fd= (69 plf)(7')-(343 plf)(7')=-1920 lbs max.

My question is, why is the wall shear not subtracted from the roof shear on the solution for 2-41?

I have the same question regarding problem 9.8 in the sample problems on the back.

It gives a solution of 4800 lbs but does not subtract the roof shear.

What am I missing here?

Thanks in advance for your response.

Miguel

mig,

Why is this question open only to those taking Hiner's course??

Actually, I have never taken his course but I do have his (older) workbook. The problem you are referring to appears to be #15 pg 2-31, question A.5., in my book.

Anyway, I think oku said it well. Sign convention does not matter and it just depends on which end you start analyzing from.

You just need to make sure you have an Fd of 0 on the ends:

Drag force for Line A can be checked by (262)(40) - (525)(20) = -20 (~0) ,or, (525)(20) - (262)(40) = 20 (~0)

Drag force for Line 1 can be checked by (69)(100) - (343)(7) - (343)(13) = 40 (~0) ,or, (343)(7) + (343)(13) - (69)(100) = -40 (~0)

I hope that helps.