How Do I figure the angle of a ridge cap along a hip ridge 3

[rsteel13]

Does anyone know what combinations of formulas I need to use to figure the angle for a bent plate to cap a hip ridge? Currently I have been using Acad to draw in the roof planes and then do some measuring to get the angle. But I am looking for something a little easier, and of course something cool that works for any combination of roof pitches. I also need this infomartion so I can figure the structural property of the angle once it is bent.

 

[JAE]

Well, manually, here's how I would do it (its hard to describe so I'll go step by step):

1. Looking down in plan view, draw your ridge line at its proper plan angle and indicate on your drawing the slope directions.
2. Place a dot on the ridge line - Point A.
3. Draw a line, perpendicular to the ridge line, through point A. Call this line LL.
4. From point A, draw a short 1' line parallel with the sloped roof on one side of the ridge - call this line L1. Place a dot at the end of Line L1. Call this Point B. Point B will have the same elevation as Point A.
5. Do the same for the other side - Line L2 and Point C.
6. From Point B, draw a line parallel with the slope, down the roof slope, to a point that intersects Line LL. Draw a dot at the intersection. this is Point D.
7. Do the same on the other side of the ridge line for Point E.
8. Calculate the distances of the lines and then calculate the elevations of each point D and E.
9. From this you have the distances from Point A to Point D and from Point A to Point E. You also have the elevations of A, D and E. From this you can calculate the angle from vertical on each side.

All this could be put together in a spreadsheet to use again in the future.

 

[rsteel13]

Just to clarify:
1. Are you addressing a HIP ridge? I think you are, I just wanted to make sure.
2. I got lost on item 4. "draw a short 1' line parallel with the sloped roof on one side of the ridge". After this line, later you say that eventually I will need to draw a line that intersects LL. I am confused. I know...

 

[rsteel13]

Never mind. I found an awsome site

http://www.Josephfusco.com

. Under the "Articles" link, the term "DIHEDRAL" applies to the item I am looking for. Excellent explanation.

 

[JAE]

When I say "parallel with the sloped roof" I mean to draw a line from Point A along the roof such that this line stays at the same elevation - just like a contour line along your sloped roof.

 

[Kapitan]

If A and B are the angles (in degrees) of the two roof planes to the horizontal and Theta is the Hip Ridge Cap included Angle then

Theta = arccos((-cos A) * (cos(B))

 

[Kapitan]

I eventually found the website that rsteel13 mentioned, it appears to be,

http://josephfusco.org/Articles/Dihedral/Dihedral.html

and not the .com version.
The method described to find the angle in question is a long winded one, whereas the one I've used is from the coordinate geometry way of finding the angle between two planes P1 and P2;

cos(Theta) = ((i1*i2)+(j1*j2)+(k1*k2))

Where i,j,k are the direction cosines of the Normals to the planes. Assuming the building to be orthogonal in plan view then most of the variables cancel out and the equation reduces to the one in the previous post.

This was described in another post, thread684-57983 in the Iron, steel & metal manufacturing engineering Forum on May30, 2003 in reply to a question about hopper angles.

 

[ERV]

Here is a very simple and accurate (to the minute) method to determine angles:
You will need: 1. horizontal distance of the slope and 2. height of the slope. 3. A calculator (I use a TI-30Xa scientific).

Example: Lets say we have horizontal distance of 24' and a total height of 6' (assume this is a hip roof, run and height, but can be used for any slope).

Divide the height 6' by the run 24' = 0.25 (tangent)
Hit (inv) (tan) = 14.03 degrees (this is to a vertical element, add 90 degrees to get the angle to a horizontal element).

14.03 + 90 = 104.03 degrees

To convert to degrees and minutes multiply 0.03 x 60 =1.8

Thus the angle is 104 degrees, 2 minutes.

Print and save this reference, it has served me well.

 

[ERV]

Oops, correction:

in my last post, I said " this is to a vertical element, add 90 degrees to get the angle to a horizontal element" this is not correct. Subtract from 180 degrees, this gives the proper angle.
Thus: 14 degrees 2 minutes - 180 degrees = 165 degrees 58 minutes.

 

[rsteel13]

Kapitan,
Can you please explain further the variables i,j, & k?
For example a bay window where the walls are on a 45^ angle from each other.
cos(Theta) = ((i1*i2)+(j1*j2)+(k1*k2))

I really liked your simplified formula. I have tried it several times and works very well. But now you have me interested in the more involved equation.

 

[Kapitan]

To solve the problem for the case you have suggested means that the directions of the Normals to the two Planes have to be found, so I will try and give some background.

For any any line or vector in space there are three angles between itself and the axis system that it is in. The term Direction Cosines refers to the cosines of each of these angles. Here I call them i,j,k but they may also be called l,m,n. It doesn't matter what they are called, the important thing about them is that the sum of their squares is 1. The cosines of the angles between the line and the axes X,Y,Z are i,j,k respectively.

To find the Direction Cosines of a Line that starts at (x1,y1,z1) and finishes at (x2,y2,z2) then the distance (D) between the two Points is divided by itself to become a unit length (1). This means that the orthogonal virtual 'box' that encloses the line has dimensions, and if these are divided by D then these are equal to the Direction Cosines.

So, if a line of length 1 (inches, metres, etc.) goes from one point to another in space and the values of the start point are considered to be (x1,y1,z1) and the finish point (x2,y2,z2) then the length of the line is;
D = [(x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2]^0.5

Then the Direction Cosines (i,j,k) are;
i = (x1-x2)/D
j = (y1-y2)/D
k = (z1-z2)/d
And
i^2 + j^2 + k" = 1

In the original Thread the two roof slopes were at rightangles to each other - the gutter of one being along X of a local axis system and the other along Y. Therefore the Normals to the roof planes both lay on two of principal planes (Xz and YZ) which means that in each case one of the angles is 90 degree, the cosine of which = 0.

The Origin of the axis system is at the lower end of the Hip, which is also the intersection of the gutters.
Giving some values to the roof slope angles will make it easier to see what is happening;

Roof (1) with gutter in X direction = 35 degrees = Angle (A)
Roof (2) with gutter in Y direction = 60 degrees = Angle (B)
Therefore Normal (1) has angles of 90,55 & 35 to X,Y & Z
and Normal (2) has angles of 30,90 & 60 to X,Y & Z

The Cosines of these angles will values of i,j & k in each case;

Normal (1) : i1=0.000000 , j1=0.573576 , k1=0.819152
Normal (2) : i2=0.866025 , j2=0.000000 , k2=0.500000

Putting these values into equation to find Theta (Hip Ridge Angle)

cos(Theta) = [(i1*i2)+(j1*j2)+(k1*k2)]

= [(0)+(0)+(0.409576)]

Theta = 65.82 degrees or 114.18 degrees included angle.

( Theta is the angle between the two Normals
so 180 - Theta is the angle bewteen the planes )

In the case of the window bays being at 135 degrees then the Roof (2) is rotated about Z by 45 degrees = Angle (C)
The Direction Cosines can be rotated directly by;

i3 = i2 * cos(45) = 0.612372
j3 = j2 * sin(45) = 0.612372
k3 = k2 = 0.500000

Note : the sum of each of these values squared is 1

Putting these values into equation gives;
# using (i1,j1,k1)&(i3,j3,k3)

cos(Theta) = [(0)+(0.351242)+(0.409576)]

Theta = 40.46 or 139.54 degrees included angle.

This can all be compressed and re-written as;

cos(Theta) = (sin(A)*sin(B)*sin(C))+(cos(A)*cos(B))]

Where A & B are Roof Angles and C is the Bay Window Angle,
(C = the rotation out of the rectangular bay shape).

 

[theclipper]

Great thread (especially since I'm designing my first hipped roof system, and I don't fully understand this).

Here's my scenario though:

I have a rectangular maintanence bldg with a steel frame w/ metal deck and joist roof system with a 4:12 pitched roof in the long direction and an 11:12 pitch in the short direction of the bldg. (matching a historic building on site) The change in elevation from eave t/steel to ridge t/steel is 14'-0". I have a W27x84 acting as the hip beam.

I'm having a hard time deciding how to provide the required information to the contractor so that he can construct good flat bearing for the joist seats.

Plugging my figures into the following equation found in this thread:
theta =arc cos((-cosA)x(cos(B)))

theta = 134.38 degrees.

What is this theta? I guess I'm not sure how to translate this information to something useful to the contractor. I visioned a bent plate connected to the top flange of the W27 that would provide a flat seat for the joists.

Any suggestions on how this connection is made to the flange?

Any personal pitfalls that I should keep an eye out for?

Thanks for any input and advise on this topic. (any resources to find pictures of one of these dihedral plates would be great too!)

 

[Kapitan]

theclipper - Theta is the true angle between the two roof planes. A strip of metal bent along its centreline to give the included angle of 134.38 degrees (on your roof), lays exactly on each slope when fitted to the hip ridge. This was in reply to rsteel13's Thread about Hip Ridge Cap Plates.

It's not clear, to me, what useful information your contractor needs, it should be all on the your drawings.

A piece of metal bent into a saddle, that fitted over the hip, with bottom flanges bent outwards to give the included angle Theta would provide a plane for the highest end of the jack rafters to sit on. But if the two roof pitches are different then another calulation needs to done as the included angled Theta tilts to one side when looked at along the hip centreline. It sounds like you need this angular offset for making some type of connection fitting(s), is this so?
I didn't mention joists as they are horizontal beam elements.

This website may be of interest ;

http://www.tfguild.org/tools/hipart.html

Lots of stuff about hip/valley roof angles.

 

[ERV]

theclipper,

In roof design, particularly hip roofs, the pitch of lets say 4:12 or 11:12 along the hip is actually 4:17 or 11:17 respectively. This is so due to the fact that the hip member is at 45 degrees (1.414X longer) to the common members and the rise stays the same. Thus:

12 X 1.414 = 16.97" ( 1.414 is the secant of 45 degrees).

4 / 17 = 0.235(tan) = 13.24d (top angle of member)- 180d (ridge)= 166.76d (ridge to down slope of hip)

11 / 17 = 0.647(tan) = 32.90d (top angle of member)- 180d (ridge) = 147.10d (ridge to down slope of hip)

where d = degrees

 

[Kapitan]

ERV - This is an extract from theclipper's post;

I have a rectangular maintanence bldg with a steel frame w/ metal deck and joist roof system with a 4:12 pitched roof in the long direction and an 11:12 pitch in the short direction of the bldg.

This is why the hip is not at 45 degrees to the common (wallplate) members, when looked at in plan view. Only if the two roof pitches are identical can this happen, in theclipper's roof this is not so, therefore any calculations based on this misapprehension will be, unfortunately, meaningless.

 

[ERV]

Kapitan,

Thank you. I misread the post thinking that there was a difference in plate height creating the 11:12 pitch on the short direction. The hip sits at 70 degrees to the long wall.

 

[rsteel13]

Check out the handy work this guy did. He wrote a book about hip and valleys, but better yet, he has a free download program. Works great. His web site is:

http://www.Hipandvalleybook.com

 

[Lutfi]

I used to detail steel many years ago. In doing these types of details I used Smoleys hand book and Martindale’s. Martindale’s has the most comprehensive sets of formulas for hips and valleys. I have a copy of both. Please contact me and I will scan the needed pages.

Leave me your e-mail address here with the needs besides what have been stated and I will be happy to fax you a copy.

 

[rsteel13]

Kapitan,
You stated in your thread:
cos(Theta) = (sin(A)*sin(B)*sin(C))+(cos(A)*cos(B))
However, if A=B and C=90, then no matter what angle is entered, this produces the same answer. The simplified equation you gave works great for walls intersecting at 90. But does the other equation work for any wall angle?
Also, in your Dec 2nd response, in your explaination you stated:
j3 = j2 * sin(45) = 0.612372
However, j2 = 0. How did you arrive at o.612372?
I know it has been a while, and this problem still pleagues me. Please help.

 

[rsteel13]

Lufti,
Thank you, for responding. My e-mail address is:
rsteel13@yahoo.com