I have two 2x10’s 18 feet long supported in the middle by one
2x4. Windows are now in these openings. I’d like to remove the
center 2x4 and replace the windows with a 16’x 7’ garage door.
What size (thickness) flitch plate would be needed to strengthen
the 2x10’s to support the open span? (Keeping within the 9¼”
height.) Single story, 4/12 roof, rafters are perpendicular to
header. No live load, just ceiling joist. Snow load for western
NY. Next wall parallel to header is 22’feet away.

Thanks
TP

TP wrote:

I have two 2x10’s 18 feet long supported in the middle by one 2x4.

If the 2x4 post supports its maximum load, say 1.5x3.5x1000 = 5250 pounds,
the beam might have an equivalent uniform load of 583 lb/ft, according to

http://www.toolbase.org/Docs/MainNav...umentID=3D2947

What size (thickness) flitch plate would be needed to strengthen
the 2x10’s to support the open span?

You may be out of flitch territory and into something like a 10"x5" I-beam.

Single story, 4/12 roof, rafters are perpendicular to header. No live load,
just ceiling joist. Snow load for western NY. Next wall parallel to header
is 22’feet away.

Hard to picture that. What's the load on the beam?

You might seek professional help...

Nick

"TP" wrote in message
news
I have two 2x10’s 18 feet long supported in the middle by one
2x4. Windows are now in these openings. I’d like to remove the
center 2x4 and replace the windows with a 16’x 7’ garage door.
What size (thickness) flitch plate would be needed to strengthen
the 2x10’s to support the open span? (Keeping within the 9¼”
height.) Single story, 4/12 roof, rafters are perpendicular to
header. No live load, just ceiling joist. Snow load for western
NY. Next wall parallel to header is 22’feet away.

Thanks
TP

The calculations to determine the individual tributary loads and the
resultant header load are pretty straightforward, the flitch plate
calculations a little more involved. I don't think you are going to
find anyone on a newsgroup (even a PE) who will be willing to give
you an absolute answer.

I'd suggest you contact a local engineer...

"Rick" wrote in message
nk.net...

"TP" wrote in message
news
I have two 2x10's 18 feet long supported in the middle by one
2x4. Windows are now in these openings. I'd like to remove the
center 2x4 and replace the windows with a 16'x 7' garage door.
What size (thickness) flitch plate would be needed to strengthen
the 2x10's to support the open span? (Keeping within the 9¼"
height.) Single story, 4/12 roof, rafters are perpendicular to
header. No live load, just ceiling joist. Snow load for western
NY. Next wall parallel to header is 22'feet away.

Thanks
TP

The calculations to determine the individual tributary loads and the
resultant header load are pretty straightforward, the flitch plate
calculations a little more involved. I don't think you are going to
find anyone on a newsgroup (even a PE) who will be willing to give
you an absolute answer.

I'd suggest you contact a local engineer...

And a more complete deescription would be needed-any roof overhangs,
roof style, stick or truss framed roof, etc. To many thngs not
mentioned.

If I'm not mistaken, a ceiling live load has to be applied to any
space with a headroom greater than 42 inches. I would imagine snow
load is at least 50 PSF in western NY. The local authority would tell
you that value.

Again, call a local...

This is the same info given to the Glulam (paralam) company.
They spec’d 3, 1 1/2”x 9 1/2” GluLam beam. I would just like to
use steel to bring the 2x10’s to the same spec.


Rick wrote:
"Rick" wrote in message
nk.net...

"TP" wrote in message
news
I have two 2x10's 18 feet long supported in the middle by one
2x4. Windows are now in these openings. I'd like to remove the
center 2x4 and replace the windows with a 16'x 7' garage door.
What size (thickness) flitch plate would be needed to strengthen
the 2x10's to support the open span? (Keeping within the 9¼"
height.) Single story, 4/12 roof, rafters are perpendicular to
header. No live load, just ceiling joist. Snow load for western
NY. Next wall parallel to header is 22'feet away.

Thanks
TP

The calculations to determine the individual tributary loads and the
resultant header load are pretty straightforward, the flitch plate
calculations a little more involved. I don't think you are going to
find anyone on a newsgroup (even a PE) who will be willing to give
you an absolute answer.

I'd suggest you contact a local engineer...


And a more complete deescription would be needed-any roof overhangs,
roof style, stick or truss framed roof, etc. To many thngs not
mentioned.

If I'm not mistaken, a ceiling live load has to be applied to any
space with a headroom greater than 42 inches. I would imagine snow
load is at least 50 PSF in western NY. The local authority would tell
you that value.

Again, call a local...

"TP" wrote in message
...
This is the same info given to the Glulam (paralam) company.
They spec’d 3, 1 1/2”x 9 1/2” GluLam beam. I would just like to
use steel to bring the 2x10’s to the same spec.



OK, at first glance I'd say they calculated the load capacity of the
existing configuration and used that to spec a clear span. Maybe I'll
run some numbers to check that....

Rick wrote:

...at first glance I'd say they calculated the load capacity of the
existing configuration and used that to spec a clear span.

Well, if two 9' 2x10s have S = 3x9.25^2/6 = 42.8 in^3 and M = 1000S
= 42800 in-lb = W9x12/8, W = 3169 pounds, it looks like the 2x4 post
(it might support 5250 pounds) isn't the limiting factor. The uniform
load on an 18' beam would be 2x3169/18 = 352 pounds per foot.

The 1981 NAHB ap note says an 18' 2x10 Hem Fir beam with a 1/2"x9" plate
can support 215 lb/ft of uniform load, so a 0.5x352/215 = 0.82" plate
(or 2 246 pound 7/16" plates) might support 352 lb/ft. They suggest
1/2" bolts every 20" near the top and every 40" near the bottom and
something like 2 2x6 jack posts with 3/4" plywood plates at each end.

Please apply the usual disclaimers.

Nick

wrote in message
...
Rick wrote:

...at first glance I'd say they calculated the load capacity of the
existing configuration and used that to spec a clear span.

Well, if two 9' 2x10s have S = 3x9.25^2/6 = 42.8 in^3 and M = 1000S
= 42800 in-lb = W9x12/8, W = 3169 pounds, it looks like the 2x4 post
(it might support 5250 pounds) isn't the limiting factor. The
uniform
load on an 18' beam would be 2x3169/18 = 352 pounds per foot.

This was my initial approach, too.

When you calculate the bending stress (S=67.7) and deflection on the
glulams specified, (E=1.8E06) with a 352 pounds/foot load, Fb is
about 2000 psi (2400 psi allowed) but the deflection is about 0.9
inches, assuming a 16 foot span (garage door width).

If you use the 1/360th of span as the limit, and 16 foot unsupported
span you get about 210 pounds/foot.

The 1981 NAHB ap note says an 18' 2x10 Hem Fir beam with a 1/2"x9"
plate
can support 215 lb/ft of uniform load, so a 0.5x352/215 = 0.82"
plate
(or 2 246 pound 7/16" plates) might support 352 lb/ft. They suggest
1/2" bolts every 20" near the top and every 40" near the bottom and
something like 2 2x6 jack posts with 3/4" plywood plates at each
end.

Please apply the usual disclaimers.

Ditto


Nick

"Rick" wrote in message
ink.net...

wrote in message
...
Rick wrote:

...at first glance I'd say they calculated the load capacity of
the
existing configuration and used that to spec a clear span.

Well, if two 9' 2x10s have S = 3x9.25^2/6 = 42.8 in^3 and M =
1000S
= 42800 in-lb = W9x12/8, W = 3169 pounds, it looks like the 2x4
post
(it might support 5250 pounds) isn't the limiting factor. The
uniform
load on an 18' beam would be 2x3169/18 = 352 pounds per foot.

This was my initial approach, too.

When you calculate the bending stress (S=67.7) and deflection on the
glulams specified, (E=1.8E06) with a 352 pounds/foot load, Fb is
about 2000 psi (2400 psi allowed) but the deflection is about 0.9
inches, assuming a 16 foot span (garage door width).


It does meet 1/180 of the span. Still, even 352 pounds per foot is low
if the header is supporting a roof. If the opposite wall is 22 feet
away, that's only a 32 psf total load.

Oh well....


If you use the 1/360th of span as the limit, and 16 foot unsupported
span you get about 210 pounds/foot.

The 1981 NAHB ap note says an 18' 2x10 Hem Fir beam with a 1/2"x9"
plate
can support 215 lb/ft of uniform load, so a 0.5x352/215 = 0.82"
plate
(or 2 246 pound 7/16" plates) might support 352 lb/ft. They
suggest
1/2" bolts every 20" near the top and every 40" near the bottom
and
something like 2 2x6 jack posts with 3/4" plywood plates at each
end.

Please apply the usual disclaimers.

Ditto


Nick

Thanks Rick,
I was looking at the other side of the building. It has an 18’
opening with the same room dimensions. Of course same roof load.
There are three 2x12’s supporting the same ceiling attic
space. What would be the psf on that span? It has been there
for 17 years. I’m leaning toward the 3 LVL’s (glulam) because
I can’t seem to get a straight answer on a flitch plate.
Seems costly 4.45 per beam, per foot.

TP


Rick wrote:
"Rick" wrote in message
ink.net...

wrote in message
...

Rick wrote:


...at first glance I'd say they calculated the load capacity of

the

existing configuration and used that to spec a clear span.

Well, if two 9' 2x10s have S = 3x9.25^2/6 = 42.8 in^3 and M =

1000S

= 42800 in-lb = W9x12/8, W = 3169 pounds, it looks like the 2x4

post

(it might support 5250 pounds) isn't the limiting factor. The

uniform

load on an 18' beam would be 2x3169/18 = 352 pounds per foot.

This was my initial approach, too.

When you calculate the bending stress (S=67.7) and deflection on the
glulams specified, (E=1.8E06) with a 352 pounds/foot load, Fb is
about 2000 psi (2400 psi allowed) but the deflection is about 0.9
inches, assuming a 16 foot span (garage door width).



It does meet 1/180 of the span. Still, even 352 pounds per foot is low
if the header is supporting a roof. If the opposite wall is 22 feet
away, that's only a 32 psf total load.

Oh well....


If you use the 1/360th of span as the limit, and 16 foot unsupported
span you get about 210 pounds/foot.


The 1981 NAHB ap note says an 18' 2x10 Hem Fir beam with a 1/2"x9"

plate

can support 215 lb/ft of uniform load, so a 0.5x352/215 = 0.82"

plate

(or 2 246 pound 7/16" plates) might support 352 lb/ft. They

suggest

1/2" bolts every 20" near the top and every 40" near the bottom

and

something like 2 2x6 jack posts with 3/4" plywood plates at each

end.

Please apply the usual disclaimers.

Ditto


Nick

Hey Rick, all this time I was comparing between a LVL and
adding a flitch plate. Wouldn’t three 2x12’s support everything
and cost one third the price of LVL?
Where is the LVL advantage?

TP

TP wrote:
Thanks Rick,
I was looking at the other side of the building. It has an 18’
opening with the same room dimensions. Of course same roof load. There
are three 2x12’s supporting the same ceiling attic space. What would be
the psf on that span? It has been there for 17 years. I’m leaning
toward the 3 LVL’s (glulam) because I can’t seem to get a straight
answer on a flitch plate.
Seems costly 4.45 per beam, per foot.

TP


Rick wrote:

"Rick" wrote in message
ink.net...

wrote in message
...

Rick wrote:


...at first glance I'd say they calculated the load capacity of


the

existing configuration and used that to spec a clear span.


Well, if two 9' 2x10s have S = 3x9.25^2/6 = 42.8 in^3 and M =


1000S

= 42800 in-lb = W9x12/8, W = 3169 pounds, it looks like the 2x4


post

(it might support 5250 pounds) isn't the limiting factor. The


uniform

load on an 18' beam would be 2x3169/18 = 352 pounds per foot.


This was my initial approach, too.

When you calculate the bending stress (S=67.7) and deflection on the
glulams specified, (E=1.8E06) with a 352 pounds/foot load, Fb is
about 2000 psi (2400 psi allowed) but the deflection is about 0.9
inches, assuming a 16 foot span (garage door width).




It does meet 1/180 of the span. Still, even 352 pounds per foot is low
if the header is supporting a roof. If the opposite wall is 22 feet
away, that's only a 32 psf total load.

Oh well....


If you use the 1/360th of span as the limit, and 16 foot unsupported
span you get about 210 pounds/foot.


The 1981 NAHB ap note says an 18' 2x10 Hem Fir beam with a 1/2"x9"


plate

can support 215 lb/ft of uniform load, so a 0.5x352/215 = 0.82"


plate

(or 2 246 pound 7/16" plates) might support 352 lb/ft. They


suggest

1/2" bolts every 20" near the top and every 40" near the bottom


and

something like 2 2x6 jack posts with 3/4" plywood plates at each


end.

Please apply the usual disclaimers.


Ditto


Nick