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9/22/2017 12:11:25 AM
Posted: 12/21/2002 8:56:18 AM EDT
How much weight can a 48" span of 2x12 support? I am asking before I pile weight on it. I would think hundreds of pounds but I'd like to get a ballpark, thanks
Link Posted: 12/21/2002 9:58:21 AM EDT
It depends on how much deflection you are willing to live with.
Link Posted: 12/21/2002 10:09:32 AM EDT
16" or 24" on center? What are the 2x12 sitting on? 1st floor on a foundation or 2nd floor on a wall? If 16" on center and proper support under, then you could park a small car on it, only spanning 48". I can't remember the exact formula for figurin' load. But I would say...Alot[:D] Maybe an engineer can tell ya the exact amount. Not much help, am I?[:I]
Link Posted: 12/21/2002 11:17:15 AM EDT
This question is sort of like asking how far a rifle can shoot. Wood is generally a crappy building material, form an engineering standpoint, because of the variability in mechanical properties. However, there are some guidelines and rules-of-thumb. No. 1 Douglas Fir (Southern) has an extreme fiber stress (Fb) of about 1300 psi. Fb is the quotient of Moment divided by Section Modulus (Fb=M/S). The section modulus of a 2x12 dimensional lumber section is 31.64 in^3, so the maximum allowable moment is 1300lb/in^2 x 31.64in^3 = 41,132in-lb. For a uniformly loaded, simply supported span, M=(1/8)w*l^2, where w=load per unit length and l=length of span. We can solve this equation for w, since we know l=48". In this case w=142.8 lb/in. If the beams are spaced at 12" on center, then the maximum allowable floor load would be 142.8lb/in*(144in^2/ft^2)/(12"spacing) = 1714 pounds per square foot! However, the maximum compression perpendicular to the grain, Fc, is 625 psi. Assuming 3.5" of end bearing, the bearing area would be 1.5"X3.5"=5.25in^2 * 625psi = 3281 lb * 2 ends = 6563 lb total load. Divide that by 4 feet, again assuming 12" spacing of the beams, and the maximum allowable load is 1630 psf. But wait, there's more! A previous poster mentioned deflection, and rightly so. For a uniformly loaded, simple span, the maximum deflection, at the center of the span, is the quantity: d=(5 * w * l^4)/(384 * E * I), where w and l are as above, E is the Modulus of Elasticity of the material (we will use 1,500,000 psi), and I is the Moment of Inertia of the section, which for a 2x12 is 178.0 in^4. Deflection is generally limited to l/360, but for temporary loading, l/240 is OK. So 48"/240 = 0.20". We can now back-solve for w in the previous formula. In this example, w=773 lb/in. This value is much greater than previous results, so deflection does not govern the design. Now the practical consideration of transferring such massive loads is really the limiting factor. A 2x12 is really quite a beefy member, and as such, will have much greater stiffness than any deck or secondary members that frame into it. So while the 2x12 itself can support a great amount of load, it's going to be tough to really get that amount of load to the 2x12. Also, such high loading would probably cause local crushing of fibers at the point of loading, well in excess of the compression limits previously stated. This is a good example of where the formulas sort of break down, primarily because of the very short span relative to the depth of the member. I would probably limit the load to 300 pounds per square foot. The actual allowable load is dependant upon many factors, including type of lumber, lateral bracing, connections, bearing details, manner and duration of loading, condition of lumber and other things.
Link Posted: 12/21/2002 11:27:02 AM EDT
Originally Posted By DzlBenz: This question is sort of like asking how far a rifle can shoot. Wood is generally a crappy building material, form an engineering standpoint, because of the variability in mechanical properties. However, there are some guidelines and rules-of-thumb. No. 1 Douglas Fir (Southern) has an extreme fiber stress (Fb) of about 1300 psi. Fb is the quotient of Moment divided by Section Modulus (Fb=M/S). The section modulus of a 2x12 dimensional lumber section is 31.64 in^3, so the maximum allowable moment is 1300lb/in^2 x 31.64in^3 = 41,132in-lb. For a uniformly loaded, simply supported span, M=(1/8)w*l^2, where w=load per unit length and l=length of span. We can solve this equation for w, since we know l=48". In this case w=142.8 lb/in. If the beams are spaced at 12" on center, then the maximum allowable floor load would be 142.8lb/in*(144in^2/ft^2)/(12"spacing) = 1714 pounds per square foot! However, the maximum compression perpendicular to the grain, Fc, is 625 psi. Assuming 3.5" of end bearing, the bearing area would be 1.5"X3.5"=5.25in^2 * 625psi = 3281 lb * 2 ends = 6563 lb total load. Divide that by 4 feet, again assuming 12" spacing of the beams, and the maximum allowable load is 1630 psf. But wait, there's more! A previous poster mentioned deflection, and rightly so. For a uniformly loaded, simple span, the maximum deflection, at the center of the span, is the quantity: d=(5 * w * l^4)/(384 * E * I), where w and l are as above, E is the Modulus of Elasticity of the material (we will use 1,500,000 psi), and I is the Moment of Inertia of the section, which for a 2x12 is 178.0 in^4. Deflection is generally limited to l/360, but for temporary loading, l/240 is OK. So 48"/240 = 0.20". We can now back-solve for w in the previous formula. In this example, w=773 lb/in. This value is much greater than previous results, so deflection does not govern the design. Now the practical consideration of transferring such massive loads is really the limiting factor. A 2x12 is really quite a beefy member, and as such, will have much greater stiffness than any deck or secondary members that frame into it. So while the 2x12 itself can support a great amount of load, it's going to be tough to really get that amount of load to the 2x12. Also, such high loading would probably cause local crushing of fibers at the point of loading, well in excess of the compression limits previously stated. This is a good example of where the formulas sort of break down, primarily because of the very short span relative to the depth of the member. I would probably limit the load to 300 pounds per square foot. The actual allowable load is dependant upon many factors, including type of lumber, lateral bracing, connections, bearing details, manner and duration of loading, condition of lumber and other things.
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You sir, are an asshole! LOL I actually thought I had a decent response, but it would look quite pathetic next to that eloquent destruction of my ego. I feel utterly useless, and question the meaning of my obviously pathetic existence.
Link Posted: 12/21/2002 11:41:50 AM EDT
Originally Posted By BustinCaps: You sir, are an asshole! LOL I actually thought I had a decent response, but it would look quite pathetic next to that eloquent destruction of my ego. I feel utterly useless, and question the meaning of my obviously pathetic existence.
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I made all that stuff up. I just wanted to post something that said "beefy member" on the Internet!
Link Posted: 12/21/2002 11:43:45 AM EDT
Originally Posted By BustinCaps:
Originally Posted By DzlBenz: This question is sort of like asking how far a rifle can shoot. Wood is generally a crappy building material, form an engineering standpoint, because of the variability in mechanical properties. However, there are some guidelines and rules-of-thumb. No. 1 Douglas Fir (Southern) ............... lumber and other things.
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You sir, are an asshole! LOL I actually thought I had a decent response, but it would look quite pathetic next to that eloquent destruction of my ego. I feel utterly useless, and question the meaning of my obviously pathetic existence.
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Ditto!
Link Posted: 12/21/2002 12:01:47 PM EDT
Originally Posted By Sniper_Wolfe: Ditto!
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"Ditto?!? Ditto?!? You provincial putz!" Hedy Lamarr "That's Hedley!" Another thread hijacked by Blazing Saddles!
Link Posted: 12/21/2002 6:12:53 PM EDT
Depends on if the boards are flat or do you have the edge sticking up? Pile shit on until you see the wood start to bend or creak and then take a couple things off.
Link Posted: 12/21/2002 6:21:58 PM EDT
Originally Posted By DzlBenz: This question is sort of like asking how far a rifle can shoot. Wood is generally a crappy building material, form an engineering standpoint, because of the variability in mechanical properties. However, there are some guidelines and rules-of-thumb. No. 1 Douglas Fir (Southern) has an extreme fiber stress (Fb) of about 1300 psi. Fb is the quotient of Moment divided by Section Modulus (Fb=M/S). The section modulus of a 2x12 dimensional lumber section is 31.64 in^3, so the maximum allowable moment is 1300lb/in^2 x 31.64in^3 = 41,132in-lb. For a uniformly loaded, simply supported span, M=(1/8)w*l^2, where w=load per unit length and l=length of span. We can solve this equation for w, since we know l=48". In this case w=142.8 lb/in. If the beams are spaced at 12" on center, then the maximum allowable floor load would be 142.8lb/in*(144in^2/ft^2)/(12"spacing) = 1714 pounds per square foot! However, the maximum compression perpendicular to the grain, Fc, is 625 psi. Assuming 3.5" of end bearing, the bearing area would be 1.5"X3.5"=5.25in^2 * 625psi = 3281 lb * 2 ends = 6563 lb total load. Divide that by 4 feet, again assuming 12" spacing of the beams, and the maximum allowable load is 1630 psf. But wait, there's more! A previous poster mentioned deflection, and rightly so. For a uniformly loaded, simple span, the maximum deflection, at the center of the span, is the quantity: d=(5 * w * l^4)/(384 * E * I), where w and l are as above, E is the Modulus of Elasticity of the material (we will use 1,500,000 psi), and I is the Moment of Inertia of the section, which for a 2x12 is 178.0 in^4. Deflection is generally limited to l/360, but for temporary loading, l/240 is OK. So 48"/240 = 0.20". We can now back-solve for w in the previous formula. In this example, w=773 lb/in. This value is much greater than previous results, so deflection does not govern the design. Now the practical consideration of transferring such massive loads is really the limiting factor. A 2x12 is really quite a beefy member, and as such, will have much greater stiffness than any deck or secondary members that frame into it. So while the 2x12 itself can support a great amount of load, it's going to be tough to really get that amount of load to the 2x12. Also, such high loading would probably cause local crushing of fibers at the point of loading, well in excess of the compression limits previously stated. This is a good example of where the formulas sort of break down, primarily because of the very short span relative to the depth of the member. I would probably limit the load to 300 pounds per square foot. The actual allowable load is dependant upon many factors, including type of lumber, lateral bracing, connections, bearing details, manner and duration of loading, condition of lumber and other things.
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Ahh but why is the deflection limit places at 1/360, or 1/240? hmmm? Hint- has to do with things not made of wood.
Link Posted: 12/21/2002 6:22:12 PM EDT
Another thread hijacked by Blazing Saddles! More beans Mr. Taggart?
Link Posted: 12/21/2002 6:24:57 PM EDT
Originally Posted By LE6920: How much weight can a 48" span of 2x12 support? I am asking before I pile weight on it. I would think hundreds of pounds but I'd like to get a ballpark, thanks
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456 LBS
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