Hip beam hand calcs

• Jun 2, 2022

• #2

jayrod12

Structural

Mar 8, 2011

6,170

At 55 psf, their math checks out to me? 55*14=770 plf. Tapering down to zero at the wall corner.

 

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• Jun 2, 2022

• Thread starter
• #3

Marc Rogue

Structural

Jul 30, 2021

19

@jayrod, my problem number one is that the shaded area isn’t a right angle triangle so to assume it is makes the dist load greater than it really is, No. 2 and my biggest concern is that based on the way they approached the dist load they end up with more load than 55x14x14=10780lbs

 

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• Jun 2, 2022

• #4

XR250

Structural

Jan 30, 2013

5,415

It should be 55*14 x 0.707 = 544 plf tapering down to zero on a 19.8 ft long member = 5385 lb total or 3590 at the top and 1795 at the bottom - so basically what the OP said.

 

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• Jun 2, 2022

• Thread starter
• #5

Marc Rogue

Structural

Jul 30, 2021

19

@xr250 that’s a way of doing it 55x14x14/19.8 to lower the load onto a longer span

 

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• Jun 2, 2022

• #6

jayrod12

Structural

Mar 8, 2011

6,170

Ugh, I see my mistake. The spacing of the rafter reactions on the beam isn't the same as the rafter spacing. I couldn't understand where the 0.707 came in until the lightbulb went on.

That being said, I'd have no arguments with being overly conservative on these things and use the higher amount. I've seen many hip beams sagging.

 

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• Jun 2, 2022

• Thread starter
• #7

Marc Rogue

Structural

Jul 30, 2021

19

The .707 it’s just reducing the 770 over 14’ to 19.8’
14/19.8=.707. So are we in agreement that the book may be wrong

 

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• Jun 2, 2022

• #8

kipfoot

Structural

Oct 25, 2007

492

As I see it, the book is wrong. The load it calculates is appropriate for the 14' span, but the same load is distributed over 19.8ft on the hip. On the hip, it's not 55psf * 14 = 770 lb/ft, it's 770 lb/ft * cos45

 

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• Jun 2, 2022

• #9

ProgrammingPE

Structural

Apr 12, 2017

249

An easy way to see what's going on is to look at the beam that is horizontal in plan which spans 14 feet and has rafters spanning perpendicular to it. They calculated the load correctly for that beam using a tributary width that varies from 7' to 0' (7'*55psf = 385plf). The loading for the hip beam should be double that (since it is loaded on two sides), but it has a span that is 19.8' instead of the 14', so just adjust accordingly (2*385plf * 14'/19.8' = 544plf).

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• Jun 2, 2022

• #10

jayrod12

Structural

Mar 8, 2011

6,170

I agree the book is wrong. But wrong on the right side of wrong.

 

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• Jun 2, 2022

• #11

BAretired

Structural

Nov 16, 2008

10,866

Fig. 2.14 shows a dimension of 14'-0" o/o wall, which should be 28'=0".
Fig. 2.16 shows a point load of 10,164# plus a uniform load of 110#/' on the main rafter.
I believe that the point load should be 14*14*55(2/3) = 7187# and the uniform load should be 55#/', representing one foot of tributary area, because one foot has already been included in the point load.

BA

 

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• Jun 2, 2022

• #12

dik

Structural

Apr 13, 2001

25,792

Sorry BART... I didn't run through the numbers. I was only looking at the triangular load distribution, which I've always used.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Do you feel any better?

-Dik

 

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• Jun 2, 2022

• #13

RontheRedneck

Specifier/Regulator

Jan 1, 2014

233

Something you may or may not know - The pitch of a hip ridge board is the main pitch divided by the square root of 2.
Knowing that can sometimes be handy when figuring things.

 

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