If you were Registered and logged in, you could reply and use other advanced thread options
|
Posted by JJ on April 22, 2008, 2:47 pm
I would like to construct a very simple footbridge over our creek. The
distance is around 12-14 feet from bank to bank. I figured on using two
16' pieces of lumber as the support (about 18-20 inches apart), with 1x6
decking boards along the top. About 2.5' to 3' wide.
My question is what should I use for the support beams. I don't want it
to get too high off the ground, because we would like to easily ski over
it in the winter.
I was conteplating either using 4x6 lumber or 2x8 lumber. I'm not sure
which will provide a stiffer bridge - which is what I want. How much
weight can I expect such a bridge to hold? Should I be using a third
support beam?
Thanks.
-Jonathan
|
|
Posted by ransley on April 22, 2008, 4:08 pm
> I would like to construct a very simple footbridge over our creek. =A0The
> distance is around 12-14 feet from bank to bank. =A0I figured on using two=
> 16' pieces of lumber as the support (about 18-20 inches apart), with 1x6
> decking boards along the top. =A0About 2.5' to 3' wide.
>
> My question is what should I use for the support beams. =A0I don't want it=
> to get too high off the ground, because we would like to easily ski over
> it in the winter.
>
> I was conteplating either using 4x6 lumber or 2x8 lumber. =A0I'm not sure
> which will provide a stiffer bridge - which is what I want. =A0How much
> weight can I expect such a bridge to hold? =A0Should I be using a third
> support beam?
>
> Thanks.
>
> -Jonathan
a 4x4 will hold a truck, a 4x4 will hold a house if it aint got no
wind, Blow a bit, and call back.
|
|
Posted by Wayne Whitney on April 22, 2008, 4:26 pm
>> I would like to construct a very simple footbridge over our creek. The
>> distance is around 12-14 feet from bank to bank.
>
> a 4x4 will hold a truck
A 4x4 spanning 12-14 feet won't hold a truck.
To the OP, the stiffness of a rectangular beam of depth d and width
varies as bd^3 (that's b*d*d*d). For a 4x6, that's 3.5 * 5.5^3 = 582,
and for a 2x8 that's 1.5 * 7.25^3 = 571. So they are almost
equivalent in stiffness.
Cheers, Wayne
|
|
Posted by JJ on April 22, 2008, 4:47 pm
Thanks for the info Wayne,
A friend has a bridge of similar length (but just over a marshy
depression) that is built using 4x4x16s, and that thing would b a bit
too bouncy for comfort when the drop is closer to 2.5'.
I'm not sure what the units are (distance^4 ?) or how the length of the
span figures in, but from the equation, it looks like using 4x6s should
provide almost four times the stiffness of 4x4s (3.5 * 3.5^2 = 150).
Anyone know how to calculate the weight that can be supported if the
weight is centered over a 14' span?
Thanks.
-Jonathan
Wayne Whitney wrote:
> A 4x4 spanning 12-14 feet won't hold a truck.
>
> To the OP, the stiffness of a rectangular beam of depth d and width
> varies as bd^3 (that's b*d*d*d). For a 4x6, that's 3.5 * 5.5^3 = 582,
> and for a 2x8 that's 1.5 * 7.25^3 = 571. So they are almost
> equivalent in stiffness.
>
> Cheers, Wayne
|
|
Posted by Wayne Whitney on April 22, 2008, 7:59 pm
> I'm not sure what the units are (distance^4 ?)
Yes, the units are distance^4. The relevant formula for the moment of
inertia is actually I = 1/12 * b * d^3, but for comparison the factor
of 1/12 is unimportant.
> or how the length of the span figures in
For a point load of fixed weight, deflection at midspan will vary as
L^3. For a distributed load, where the total load increases as the
length increases, deflection will vary as L^4.
> but from the equation, it looks like using 4x6s should
> provide almost four times the stiffness of 4x4s (3.5 * 3.5^2 = 150).
Yes, the exact ratio is (5.5/3.5)^3 = 3.88.
> Anyone know how to calculate the weight that can be supported if the
> weight is centered over a 14' span?
I'm afraid answering that question would take a bit longer than space
permits. If you know what you are doing, the span calculators at
www.awc.org are very useful, although they are for distributed loads,
not point loads. You should understand that allowable load may be
limited by deflection (where the deflection criteria is specified as
e.g. L/360, that midspan deflection should be less that 1/360 times
the span) or may be limited by strength.
Cheers, Wayne
|
Page 1 of 3 1 2 3 > last >>
| Similar Threads | Posted | | Concrete question (regarding previous plumbing question) | October 17, 2006, 2:29 pm |
| AC and fan question | July 17, 2005, 6:59 pm |
| Yet another A/C question............... | July 19, 2005, 11:33 pm |
| AC question | August 21, 2005, 10:29 pm |
| A/C Question | August 24, 2005, 10:21 pm |
| a/c question | September 13, 2005, 8:37 am |
| UBC Question | September 26, 2005, 6:02 pm |
| Gas Log Question | December 10, 2005, 3:45 pm |
| Another A/C question | June 15, 2005, 10:06 pm |
| Well Question | March 13, 2006, 11:50 am |
|
|
Yahoo! 
Google 
Windows Live 
del.icio.us 
digg 
Netscape 
XML Sitemap