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Wall Removal Question?
I am planning on removing a 13 foot (plaster) wall on the 1st floor of
my 2 story house that I am going to assume is load bearing. When I look in the basement directly under the current wall, there is a double 13 foot 2X12 with one end sitting on the foundation wall and the other end on a metal i-beam that spans the length of the house. Not sure of the wood type, but late 1940's construction. Directly above the wall I am planning on removing is the center of the floor of a 13X20 bedroom. (also: the joists above the removal wall are perpindicular to the wall.) My first plan was to replace the current wall with a double 13 foot 2x12 header. (possibly triple?) My Question is. 1. Does this sound adequate. 2. What would be the equivelent LVL beam(s) (ie: microllam) to use instead of using wood? I would like the reduce the depth of the header so it will not be as notciable. And yeah, I know, I should consult an engineer. Just looking for opinions here. |
#2
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.. My Question is.
1. Does this sound adequate. No, my SWAG (no calcs) would be a 4 x14 but someone's got the run the numbers 2. What would be the equivalent LVL beam(s) (ie: microllam) to use instead of using wood? I would like the reduce the depth of the header so it will not be as notciable. . Reducing the depth is asking for trouble unless you go to steel. If you really wan to make the header "disappear" completely you could: Shore in the basement & the first floor on either side of the exsisting wall. remove the wall cut second floor joists back to allow for a steel beam ~the depth of the cut joists. Install hangers & the beam. close the ceiling, remove the shoring. no visible header, but a lot of work cheers Bob |
#3
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on second reading of the OP, your double or triple 2x12 would work;
that's what's holding up first floor currently. I wasn't exactly clear on that. How does the first floor feel? my suggest to hide the header still stands cheers Bob |
#4
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I've removed 2 walls in my life but nothing that complicated. I'm 51%
sure you're ok with doing what you describe, one problem is if you're wrong and your house caves in your insurance may not cover it. For example you're counting on the builder following logic, are the floor joists for room above actually able to cover the span without the wall you wanna yank ? Lotta questions. |
#5
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schreef in bericht ups.com... I've removed 2 walls in my life but nothing that complicated. I'm 51% sure you're ok with doing what you describe, one problem is if you're wrong and your house caves in your insurance may not cover it. For example you're counting on the builder following logic, are the floor joists for room above actually able to cover the span without the wall you wanna yank ? Lotta questions. Yep, maybe the floor above the wall isn't continous. |
#6
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That's why I figured I might as well overdo it.
I've now decided to go with a Steel I-beam (looks like it will be the same or cheaper than lumber - but will make a much smaller header.) Question now is, should I go with a 4" 6" or 8" IBeam? What iBeam size would be equivalent to Three 2X12 at 13 feet?? Keep in mind, the only thing holding up the wall I plan on removing is Two 2X12's |
#7
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#8
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#10
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On Sun, 20 Mar 2005 13:22:34 GMT, Rob Munach
wrote: Brian Whatcott wrote: On 18 Mar 2005 08:36:10 -0800, wrote: That's why I figured I might as well overdo it. I've now decided to go with a Steel I-beam (looks like it will be the same or cheaper than lumber - but will make a much smaller header.) Question now is, should I go with a 4" 6" or 8" IBeam? What iBeam size would be equivalent to Three 2X12 at 13 feet?? Keep in mind, the only thing holding up the wall I plan on removing is Two 2X12's A specimen load for a 6X12 in wood beam at 13 feet is 200lb per foot uniform loading, for which the max deflection is 0.1 inch This has a safety rating of X2 to failure for reasonable assumptions: Youngs 1.5E6 psi limit stress 1.2 kips A steel I beam: W8X15 deflects 0.09 inches for this load, a W8X13 deflects 0.11 inches. Youngs 29E6 yield stress 36 ksi Respectfully Brian Whatcott Altus, OK Brian, You may want to re-check your calcs. 6x12 (assuming 5 1/2 x 11 1/4 - which is larger than the (3)2x12 specified) S = 116 For 200 plf over 13', M = 4225 ft-lb M/S = 437 psi which is a little under 1/2 of what is typically used for the design stress (which already has a huge factor of safety). Regards, Rob, the calcs aren't mine, they are the result of Archon's Beams program which also checks against code. I made explicit my inputs, as to Youngs, limit stress and dimensions for both the wood beam and the steel beam. You note that you used commercial "finished" dimensions rather than gross rough-sawn, and that's a not unreasonable basis for the difference in your wood beam calcs, I certainly agree. Respectfully Brian Whatcott Altus, OK |
#11
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Brian Whatcott wrote:
On Sun, 20 Mar 2005 13:22:34 GMT, Rob Munach wrote: Brian Whatcott wrote: On 18 Mar 2005 08:36:10 -0800, wrote: That's why I figured I might as well overdo it. I've now decided to go with a Steel I-beam (looks like it will be the same or cheaper than lumber - but will make a much smaller header.) Question now is, should I go with a 4" 6" or 8" IBeam? What iBeam size would be equivalent to Three 2X12 at 13 feet?? Keep in mind, the only thing holding up the wall I plan on removing is Two 2X12's A specimen load for a 6X12 in wood beam at 13 feet is 200lb per foot uniform loading, for which the max deflection is 0.1 inch This has a safety rating of X2 to failure for reasonable assumptions: Youngs 1.5E6 psi limit stress 1.2 kips A steel I beam: W8X15 deflects 0.09 inches for this load, a W8X13 deflects 0.11 inches. Youngs 29E6 yield stress 36 ksi Respectfully Brian Whatcott Altus, OK Brian, You may want to re-check your calcs. 6x12 (assuming 5 1/2 x 11 1/4 - which is larger than the (3)2x12 specified) S = 116 For 200 plf over 13', M = 4225 ft-lb M/S = 437 psi which is a little under 1/2 of what is typically used for the design stress (which already has a huge factor of safety). Regards, Rob, the calcs aren't mine, they are the result of Archon's Beams program which also checks against code. I made explicit my inputs, as to Youngs, limit stress and dimensions for both the wood beam and the steel beam. You note that you used commercial "finished" dimensions rather than gross rough-sawn, and that's a not unreasonable basis for the difference in your wood beam calcs, I certainly agree. Respectfully Brian Whatcott Altus, OK Actually, Brian, 5 1/2x11 1/4 is the finshed dimensions of a 6x12. You may want to do the calc by hand to verify the results from the program. I imagine you have an input wrong. Your deflections, however, seem to be correct for the steel beam. For 0.1" deflection and an E of 1.5, the moment of inertia of your wood section is 856 in^4 which is the moment of inertia of a true 6x12 (6"x12") The section modulus of this beam is 144in^3. For a 4225ft-lb moment, the bending stress is 352 psi! Regards, -- Rob Munach, PE Excel Engineering PO Box 1264 Carrboro, NC 27510 |
#12
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On Mon, 21 Mar 2005 10:44:00 GMT, Rob Munach
wrote: A specimen load for a 6X12 in wood beam at 13 feet is 200lb per foot uniform loading, for which the max deflection is 0.1 inch This has a safety rating of X2 to failure for reasonable assumptions: Youngs 1.5E6 psi limit stress 1.2 kips A steel I beam: W8X15 deflects 0.09 inches for this load, a W8X13 deflects 0.11 inches. Youngs 29E6 yield stress 36 ksi Respectfully Brian Whatcott Altus, OK Brian, You may want to re-check your calcs. 6x12 (assuming 5 1/2 x 11 1/4 - which is larger than the (3)2x12 specified) S = 116 I made explicit my inputs, as to Youngs, limit stress and dimensions for both the wood beam and the steel beam. You note that you used commercial "finished" dimensions rather than gross rough-sawn, and that's a not unreasonable basis for the difference in your wood beam calcs, I certainly agree. Respectfully Brian Whatcott Altus, OK Actually, Brian, 5 1/2x11 1/4 is the finshed dimensions of a 6x12. Let me say this again, slowly: a finished (planed) 6X12 may measure about 5 3/4 X 11 3/4, I believe depending on its moisture content. A finished composition of 3 off 2 X 12 may measure about 5 1/4 X 11 3/4,I believe. These figures may or may not represent your experience, probably not. Still, I could care less, because it is not material to my response. If I say that a rough sawn 6 X 12 inch measures 6 X 12 inch, (duh) and provide the Code approved results for such a beam, given the Youngs and stress limit I specify, then I have provided an engineering calculation which is explicitly correct. If you wish to say that your planed, finished beams are 5 1/2 X 11 1/4 that's just fine by me too. And your figures may be accurate for these dimensions. Or not. I haven't checked. Wouldn't it be better to ask the questioner what are the ACTUAL dimensions of the beam he has in mind, rather than playing "I'm smarter than you"? [People don't usually win these games with me, Rob!] :-) Sincerely Brian Whatcott Altus, OK |
#13
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In a previous post Brian Whatcott says...
If I say that a rough sawn 6 X 12 inch measures 6 X 12 inch, (duh) and provide the Code approved results for such a beam, given the Youngs and stress limit I specify, then I have provided an engineering calculation which is explicitly correct. Brian: The calculation may be correct, but the answer is wrong. What is the point of doing a calculation on a product that one cannot readily purchase? That is not an "engineering" solution, but simply a mathematical exercise. What Rob is trying to tell you is that if you are going to make a recommendation then base your recommendation on real lumbers sizes with real lumber material properties. It does no one any good to specify some material that cannot be purchased. -- Bob Morrison, PE, SE R L Morrison Engineering Co Structural & Civil Engineering Poulsbo WA |
#14
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On Mon, 21 Mar 2005 15:39:10 GMT, Bob Morrison
wrote: In a previous post Brian Whatcott says... If I say that a rough sawn 6 X 12 inch measures 6 X 12 inch, (duh) and provide the Code approved results for such a beam, given the Youngs and stress limit I specify, then I have provided an engineering calculation which is explicitly correct. Brian: The calculation may be correct, but the answer is wrong. What is the point of doing a calculation on a product that one cannot readily purchase? That is not an "engineering" solution, but simply a mathematical exercise. What Rob is trying to tell you is that if you are going to make a recommendation then base your recommendation on real lumbers sizes with real lumber material properties. It does no one any good to specify some material that cannot be purchased. Ah, now we are getting to the nub, I see. Can you purchase a 13 ft beam in the US? Can you purchase a (metric) 13 ft beam elsewhere? Can you ask a mill to resaw used timber? Can you have beams rough-cut to specified dimensions? Does the allowance for timber "finished four sides" allow a wastage off the rough-cut (nominal) dimensions? We are (presumably) not talking about dashing down to Lowes with a bill of materials, but talking engineering stress. The answers to these questions may help one be better prepared NOT necessarily to suppose a weak hypothetical wood beam, and speculate that it is Southern yellow pine finished four sides, rather than rough cut Douglas Fir on nominal dimension, and so to avoid converting it to a steel beam that is lighter than a conservative conversion would provide. If a home-owner asks for advice, it is far, FAR better to err on the conservative side, wouldn't you say? To put it another way: whose advice would YOU take, if you were not in the field: mine or Rob's? Still, I will keep batting this one around with you, by all means, at least until the ad hominem approach supervenes. This can be educational, after all. Have a happy day! Brian Whatcott, Altus, OK |
#15
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In a previous post Brian Whatcott says...
If a home-owner asks for advice, it is far, FAR better to err on the conservative side, wouldn't you say? To put it another way: whose advice would YOU take, if you were not in the field: mine or Rob's? Brian: You can specify any product you want, including odd sized material. If it suits your fancy call out 3-3/4" x 10-7/8" teak beam, but if you give the homeowner advice about a product he cannot readily buy then you begin to look foolish and the homeowner will begin to question whether or not your advice was accurate. Since "I" know this was not your intent, it is not really a problem for me. But, if the person to whom you are giving advice is not an engineer, then giving misleading or inadequate information can be a problem. This is one of those issues where knowing the materials and their specifications is important. It gives the homeowner a sense of confidence that the engineer knows what he/she is talking about. BTW, there are much better programs than "Archon" for computing stresses in wood members. I'm partial to "Beamchek" which I use daily and is an excellent tool for doing simple beam calculations in wood or steel. One simply selects the material type (Douglas-Fir/Larch, Southern Pine, steel, etc) and the software will input the appropriate "E" and "Fb" values. Members will be then selected by the engineer from "standard" manufactured sizes. This provides economical design. As in any good software you can set your deflection limit to a more strict value than the building code requires, thus giving you stiffer beams. -- Bob Morrison, PE, SE R L Morrison Engineering Co Structural & Civil Engineering Poulsbo WA |
#16
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On Tue, 22 Mar 2005 02:18:45 GMT, Bob Morrison
wrote: In a previous post Brian Whatcott says... If a home-owner asks for advice, it is far, FAR better to err on the conservative side, wouldn't you say? To put it another way: whose advice would YOU take, if you were not in the field: mine or Rob's? Brian: You can specify any product you want, including odd sized material. If it suits your fancy call out 3-3/4" x 10-7/8" teak beam, but if you give the homeowner advice about a product he cannot readily buy then you begin to look foolish and the homeowner will begin to question whether or not your advice was accurate./// Perhaps it would be better to read the thread Bob. I specified some standard steel I beam sections, as did Rob. The ones he suggested to a naive home-owner were weaker than the ones I mentioned. Not a lot weaker. Just low-ball. Did you think the questioner was going to buy wood beams then? Oh well! Brian Whatcott Altus, OK |
#17
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"Bob Morrison" wrote in message k.net... In a previous post Brian Whatcott says... If a home-owner asks for advice, it is far, FAR better to err on the conservative side, wouldn't you say? To put it another way: whose advice would YOU take, if you were not in the field: mine or Rob's? Brian: You can specify any product you want, including odd sized material. If it suits your fancy call out 3-3/4" x 10-7/8" teak beam, but if you give the homeowner advice about a product he cannot readily buy then you begin to look foolish and the homeowner will begin to question whether or not your advice was accurate. Since "I" know this was not your intent, it is not really a problem for me. But, if the person to whom you are giving advice is not an engineer, then giving misleading or inadequate information can be a problem. This is one of those issues where knowing the materials and their specifications is important. It gives the homeowner a sense of confidence that the engineer knows what he/she is talking about. BTW, there are much better programs than "Archon" for computing stresses in wood members. I'm partial to "Beamchek" which I use daily and is an excellent tool for doing simple beam calculations in wood or steel. One simply selects the material type (Douglas-Fir/Larch, Southern Pine, steel, etc) and the software will input the appropriate "E" and "Fb" values. Members will be then selected by the engineer from "standard" manufactured sizes. This provides economical design. As in any good software you can set your deflection limit to a more strict value than the building code requires, thus giving you stiffer beams. -- Bob Morrison, PE, SE R L Morrison Engineering Co Structural & Civil Engineering Poulsbo WA Great response there Bob. Chuck... __________________________________________________ ______ Charles I. Dinsmore, PE SE RA ~ |
#18
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Brian Whatcott wrote:
On Mon, 21 Mar 2005 10:44:00 GMT, Rob Munach wrote: A specimen load for a 6X12 in wood beam at 13 feet is 200lb per foot uniform loading, for which the max deflection is 0.1 inch This has a safety rating of X2 to failure for reasonable assumptions: Youngs 1.5E6 psi limit stress 1.2 kips A steel I beam: W8X15 deflects 0.09 inches for this load, a W8X13 deflects 0.11 inches. Youngs 29E6 yield stress 36 ksi Respectfully Brian Whatcott Altus, OK Brian, You may want to re-check your calcs. 6x12 (assuming 5 1/2 x 11 1/4 - which is larger than the (3)2x12 specified) S = 116 I made explicit my inputs, as to Youngs, limit stress and dimensions for both the wood beam and the steel beam. You note that you used commercial "finished" dimensions rather than gross rough-sawn, and that's a not unreasonable basis for the difference in your wood beam calcs, I certainly agree. Respectfully Brian Whatcott Altus, OK Actually, Brian, 5 1/2x11 1/4 is the finshed dimensions of a 6x12. Let me say this again, slowly: a finished (planed) 6X12 may measure about 5 3/4 X 11 3/4, I believe depending on its moisture content. A finished composition of 3 off 2 X 12 may measure about 5 1/4 X 11 3/4,I believe. These figures may or may not represent your experience, probably not. Still, I could care less, because it is not material to my response. If I say that a rough sawn 6 X 12 inch measures 6 X 12 inch, (duh) and provide the Code approved results for such a beam, given the Youngs and stress limit I specify, then I have provided an engineering calculation which is explicitly correct. If you wish to say that your planed, finished beams are 5 1/2 X 11 1/4 that's just fine by me too. And your figures may be accurate for these dimensions. Or not. I haven't checked. Wouldn't it be better to ask the questioner what are the ACTUAL dimensions of the beam he has in mind, rather than playing "I'm smarter than you"? [People don't usually win these games with me, Rob!] :-) Sincerely Brian Whatcott Altus, OK Brian, Please re-read my post and do these calcs by hand! Your stress results for the wood beam are off by at least a factor of two. Here is my previous post: "Actually, Brian, 5 1/2x11 1/4 is the finshed dimensions of a 6x12. You may want to do the calc by hand to verify the results from the program. I imagine you have an input wrong. Your deflections, however, seem to be correct for the steel beam. For 0.1" deflection and an E of 1.5, the moment of inertia of your wood section is 856 in4 which is the moment of inertia of a true 6x12 (6"x12") The section modulus of this beam is 144in3. For a 4225ft-lb moment, the bending stress is 352 psi!" Based on the deflections you gave, the section properties compute to a "true" 6x12, but your bending stress doesn't even copme close to this section. You have an error! -- Rob Munach, PE Excel Engineering PO Box 1264 Carrboro, NC 27510 |
#19
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On Tue, 22 Mar 2005 11:51:26 GMT, Rob Munach
wrote: Brian Whatcott wrote: On Mon, 21 Mar 2005 10:44:00 GMT, Rob Munach wrote: A specimen load for a 6X12 in wood beam at 13 feet is 200lb per foot uniform loading, for which the max deflection is 0.1 inch This has a safety rating of X2 to failure for reasonable assumptions: Youngs 1.5E6 psi limit stress 1.2 kips A steel I beam: W8X15 deflects 0.09 inches for this load, a W8X13 deflects 0.11 inches. Youngs 29E6 yield stress 36 ksi Respectfully Brian Whatcott Altus, OK Brian, You may want to re-check your calcs. 6x12 (assuming 5 1/2 x 11 1/4 - which is larger than the (3)2x12 specified) S = 116 I made explicit my inputs, as to Youngs, limit stress and dimensions for both the wood beam and the steel beam. You note that you used commercial "finished" dimensions rather than gross rough-sawn, and that's a not unreasonable basis for the difference in your wood beam calcs, I certainly agree. Respectfully Brian Whatcott Altus, OK Actually, Brian, 5 1/2x11 1/4 is the finshed dimensions of a 6x12. Let me say this again, slowly: a finished (planed) 6X12 may measure about 5 3/4 X 11 3/4, I believe depending on its moisture content. A finished composition of 3 off 2 X 12 may measure about 5 1/4 X 11 3/4,I believe. These figures may or may not represent your experience, probably not. Still, I could care less, because it is not material to my response. If I say that a rough sawn 6 X 12 inch measures 6 X 12 inch, (duh) and provide the Code approved results for such a beam, given the Youngs and stress limit I specify, then I have provided an engineering calculation which is explicitly correct. If you wish to say that your planed, finished beams are 5 1/2 X 11 1/4 that's just fine by me too. And your figures may be accurate for these dimensions. Or not. I haven't checked. Wouldn't it be better to ask the questioner what are the ACTUAL dimensions of the beam he has in mind, rather than playing "I'm smarter than you"? [People don't usually win these games with me, Rob!] :-) Sincerely Brian Whatcott Altus, OK Brian, /// For 0.1" deflection and an E of 1.5, the moment of inertia of your wood section is 856 in4 which is the moment of inertia of a true 6x12 (6"x12") The section modulus of this beam is 144in3. For a 4225ft-lb moment, the bending stress is 352 psi!" Based on the deflections you gave, the section properties compute to a "true" 6x12, but your bending stress doesn't even copme close to this section. You have an error! Forgive me for seeming to be picky with this illustrious band of professionals who presumably make a living at getting structures right, but here are the section properties I used (notice I ran them two ways, for a cross-check) I lifted then straight out of Archon Beam. I did not use a LIMIT load, but as specified I used a specimen load of 200 lb/ft. What is the conceptual problem with you fellows? Civils can't ALL be wrong, not all the time! :-) TRAPEZOID, Bending about CG d(in) = 12.00 b(in) = 6.000 b1(in) = 6.000 A(in^2) = 72.00 Ix(in^4) = 864.0 Sx(in^3) = 144.0 Rx(in) = 3.464 Cx(in) = 6.000 RECTANGLE, Bending about CG d(in) = 12.00 b(in) =6.000 A(in^2) = 72.00 Ix(in^4) = 864.0 Sx(in^3) = 144.0 Rx(in) = 3.464 Zx(in^3) = 216.0 Iy(in^4) = 216.0 Sy(in^3) = 72.00 Ry(in) = 1.732 Zy(in^3) = 108.0 Cx(in) = 6.000 Cy(in) = 3.000 So, to summarize: you are absolutely correct as to the timber section properties I used. AND you are absolutely correct that I did not use the max permissible load as a specimen load for either the timber, or the steel section. So what? Please try to make the dialog meaningful, The innocent bystanders may be questioning one's probity. I certainly am! Brian Whatcott Altus OK |
#20
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Brian Whatcott wrote:
On Tue, 22 Mar 2005 11:51:26 GMT, Rob Munach wrote: Brian Whatcott wrote: On Mon, 21 Mar 2005 10:44:00 GMT, Rob Munach wrote: A specimen load for a 6X12 in wood beam at 13 feet is 200lb per foot uniform loading, for which the max deflection is 0.1 inch This has a safety rating of X2 to failure for reasonable assumptions: Youngs 1.5E6 psi limit stress 1.2 kips A steel I beam: W8X15 deflects 0.09 inches for this load, a W8X13 deflects 0.11 inches. Youngs 29E6 yield stress 36 ksi Respectfully Brian Whatcott Altus, OK Brian, You may want to re-check your calcs. 6x12 (assuming 5 1/2 x 11 1/4 - which is larger than the (3)2x12 specified) S = 116 I made explicit my inputs, as to Youngs, limit stress and dimensions for both the wood beam and the steel beam. You note that you used commercial "finished" dimensions rather than gross rough-sawn, and that's a not unreasonable basis for the difference in your wood beam calcs, I certainly agree. Respectfully Brian Whatcott Altus, OK Actually, Brian, 5 1/2x11 1/4 is the finshed dimensions of a 6x12. Let me say this again, slowly: a finished (planed) 6X12 may measure about 5 3/4 X 11 3/4, I believe depending on its moisture content. A finished composition of 3 off 2 X 12 may measure about 5 1/4 X 11 3/4,I believe. These figures may or may not represent your experience, probably not. Still, I could care less, because it is not material to my response. If I say that a rough sawn 6 X 12 inch measures 6 X 12 inch, (duh) and provide the Code approved results for such a beam, given the Youngs and stress limit I specify, then I have provided an engineering calculation which is explicitly correct. If you wish to say that your planed, finished beams are 5 1/2 X 11 1/4 that's just fine by me too. And your figures may be accurate for these dimensions. Or not. I haven't checked. Wouldn't it be better to ask the questioner what are the ACTUAL dimensions of the beam he has in mind, rather than playing "I'm smarter than you"? [People don't usually win these games with me, Rob!] :-) Sincerely Brian Whatcott Altus, OK Brian, /// For 0.1" deflection and an E of 1.5, the moment of inertia of your wood section is 856 in4 which is the moment of inertia of a true 6x12 (6"x12") The section modulus of this beam is 144in3. For a 4225ft-lb moment, the bending stress is 352 psi!" Based on the deflections you gave, the section properties compute to a "true" 6x12, but your bending stress doesn't even copme close to this section. You have an error! Forgive me for seeming to be picky with this illustrious band of professionals who presumably make a living at getting structures right, but here are the section properties I used (notice I ran them two ways, for a cross-check) I lifted then straight out of Archon Beam. I did not use a LIMIT load, but as specified I used a specimen load of 200 lb/ft. What is the conceptual problem with you fellows? Civils can't ALL be wrong, not all the time! :-) TRAPEZOID, Bending about CG d(in) = 12.00 b(in) = 6.000 b1(in) = 6.000 A(in^2) = 72.00 Ix(in^4) = 864.0 Sx(in^3) = 144.0 Rx(in) = 3.464 Cx(in) = 6.000 RECTANGLE, Bending about CG d(in) = 12.00 b(in) =6.000 A(in^2) = 72.00 Ix(in^4) = 864.0 Sx(in^3) = 144.0 Rx(in) = 3.464 Zx(in^3) = 216.0 Iy(in^4) = 216.0 Sy(in^3) = 72.00 Ry(in) = 1.732 Zy(in^3) = 108.0 Cx(in) = 6.000 Cy(in) = 3.000 So, to summarize: you are absolutely correct as to the timber section properties I used. AND you are absolutely correct that I did not use the max permissible load as a specimen load for either the timber, or the steel section. So what? Please try to make the dialog meaningful, The innocent bystanders may be questioning one's probity. I certainly am! Brian Whatcott Altus OK Brian, What is confusing is this paragraph: uniform loading, for which the max deflection is 0.1 inch This has a safety rating of X2 to failure for reasonable assumptions: Youngs 1.5E6 psi limit stress 1.2 kips The original poster wanted to know a steel beam that was equivalent to (3)2x12 at 13'. The ONLY way to do this (other than by testing) is to analytically determine the strength and stiffness of the existing beam and find a substitute beam whose properties exceed those. I don't know how a 6x12 or 200 plf even entered into your calculations. You have proven, however, that you can enter numbers into a computer program: Regards, -- Rob Munach, PE Excel Engineering PO Box 1264 Carrboro, NC 27510 |
#21
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Ok, now I see what you did (I think). You put a 6x12 (which is larger
than the beam in question) in your beam program and placed an arbitrary load of 200 plf on it. You then let it come up with steel beams that had simlar deflections and were not overstressed. The problems with this approach are as follows: 1) This will give conservative stiffness results (not necessarily bad) 2) This *may* give you a steel beam that is overstressed as your 200 plf load may be less than the actual service load. You would have been better off using the correct section ( three 2x12's ) and a load that was at the limit stress for this section. Regards, -- Rob Munach, PE Excel Engineering PO Box 1264 Carrboro, NC 27510 |
#22
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On Tue, 22 Mar 2005 13:39:19 GMT, Rob Munach
wrote: [Days ago, I wrote this - Brian] A specimen load for a 6X12 in wood beam at 13 feet is 200lb per foot // uniform loading, for which the max deflection is 0.1 inch // [Rob] Brian, What is confusing is this paragraph: [brian] uniform loading, for which the max deflection is 0.1 inch This has a safety rating of X2 to failure for reasonable assumptions: Youngs 1.5E6 psi limit stress 1.2 kips Regards, Rob Ah, yes: I suspected that this might be the root of the confusion. 1) For any load at all, on any beam at all, there is a place on the beam that experiences maximal deflection (for that load). 2) For a limit load on a beam, there is a point that shows a maximal allowable deflection for that beam. You know that: I know that: and I possibly made it to easy for you to suppose that the maximal beam deflection of type 1) was referring to a deflection type 2) Moreover I suggested that this specimen load might be half of the load at failure. It probably isn't though, wouldn't you say? Now for one interesting feature of using beam programs (as every professional would be well-advised to do steel beams have pretty uniform parameters, so very much derating is unnecessary. Wood in contrast, has rather variable parameters and as an anisotropic material, there are just plain more of them too! So a program offering wood beams had better provide low-ball values of important parameters. Nobody will disagree with me on this, I don't suppose. Now here's the kicker: Say we want to replace a wood beam with a steel beam, for strength, for reduced deflection, or most likely,for reduced depth wasted. In this situation we DON'T want the low ball safety values, we want the HIGH ball values for the wood, in order to guarantee the same or better performance when we match strength or stiffness in steel. The reason is because the previous (wood) beam may have been a lucky, strong example at the upper end of the distribution: it is vital not to degrade values of that rare HIGH end sample. That has been the other issue: everyone has been so comfortable with their timber or steel beam program, it was easy to forget that the brain has to stay engaged, for the safety of the public, whom one is mandated to keep safe. This has turned out to be a remarkably civilized thread, in the end. I salute you all! Brian Whatcott Altus OK |
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Err...forgive me for harping on the topic, Rob, but it was *I* who
provided the stiffer stronger steel sections. It was *you* who included two sections that were closer to being overstressed. Still, your recommendations below are sound. The only issue left seems to be the rough-sawn versus finished issue which I think we have kicked around pretty well. Regards Brian Whatcott Altus OK On Tue, 22 Mar 2005 16:41:53 GMT, Rob Munach wrote: Ok, now I see what you did (I think). You put a 6x12 (which is larger than the beam in question) in your beam program and placed an arbitrary load of 200 plf on it. You then let it come up with steel beams that had simlar deflections and were not overstressed. The problems with this approach are as follows: 1) This will give conservative stiffness results (not necessarily bad) 2) This *may* give you a steel beam that is overstressed as your 200 plf load may be less than the actual service load. You would have been better off using the correct section ( three 2x12's ) and a load that was at the limit stress for this section. Regards, |
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Brian Whatcott wrote:
On Tue, 22 Mar 2005 13:39:19 GMT, Rob Munach wrote: [Days ago, I wrote this - Brian] A specimen load for a 6X12 in wood beam at 13 feet is 200lb per foot // uniform loading, for which the max deflection is 0.1 inch // [Rob] Brian, What is confusing is this paragraph: [brian] uniform loading, for which the max deflection is 0.1 inch This has a safety rating of X2 to failure for reasonable assumptions: Youngs 1.5E6 psi limit stress 1.2 kips Regards, Rob Ah, yes: I suspected that this might be the root of the confusion. 1) For any load at all, on any beam at all, there is a place on the beam that experiences maximal deflection (for that load). 2) For a limit load on a beam, there is a point that shows a maximal allowable deflection for that beam. You know that: I know that: and I possibly made it to easy for you to suppose that the maximal beam deflection of type 1) was referring to a deflection type 2) Moreover I suggested that this specimen load might be half of the load at failure. It probably isn't though, wouldn't you say? Now for one interesting feature of using beam programs (as every professional would be well-advised to do I use them, however, for simple span beams with uniform or central point loads, I can calc 'em much faster by hand then using a beam program. I probably do 20 a day by hand. steel beams have pretty uniform parameters, so very much derating is unnecessary. Wood in contrast, has rather variable parameters and as an anisotropic material, there are just plain more of them too! True, but if you do this long enough you will have memorized S, E and I for most of the sections. When I go into the field, I rarely need a cheat sheet or beam program. I simply use my calculator and my brain. So a program offering wood beams had better provide low-ball values of important parameters. Nobody will disagree with me on this, I don't suppose. It shouldn't lo-ball anything. It should use the design values set forth in the NDS. Believe me, not only is there a huge factor of safety in the material, there is also a huge factor of safety on the loads. It is rare, if ever, the structure even comes close to seeing its design gravity live loads. Additonally, there is no accounting for the large amount of unintended composite action going on in light frame structures. Although, I am starting to believe that the NDS modulus of elasticity numbers are high. I am seeing in the field, members that are deflecting signifcantly more than they should. Now here's the kicker: Say we want to replace a wood beam with a steel beam, for strength, for reduced deflection, or most likely,for reduced depth wasted. In this situation we DON'T want the low ball safety values, we want the HIGH ball values for the wood, in order to guarantee the same or better performance when we match strength or stiffness in steel. Exactly what you did. While safe and conservative, it won't keep you in business too long as you will get a reputation for being "too conservative" However, with all due respect, you probably dont' do a lot of this type of work anyway. FWIW, serviceability controls most beams designs in light frame structures. I have seen plenty of excessively sagged beams, but have never seen one break. The reason is because the previous (wood) beam may have been a lucky, strong example at the upper end of the distribution: it is vital not to degrade values of that rare HIGH end sample. That has been the other issue: everyone has been so comfortable with their timber or steel beam program, it was easy to forget that the brain has to stay engaged, for the safety of the public, whom one is mandated to keep safe. This has turned out to be a remarkably civilized thread, in the end. I salute you all! Agreed. Brian Whatcott Altus OK Regards. -- Rob Munach, PE Excel Engineering PO Box 1264 Carrboro, NC 27510 |
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Your guys talked a lot about beam strenth. But how do you deal with
longgivity of the beam support? I meanm how do you prevent the rust on the beams, and particularly the support colummns? The columns rest of the comcrete pads, and bottom parts are buried under Before my house was completed in 2000, I noticed the columns rests on the concrete pads that were submerged under water. There are rusts on the columns. After I raised the issue, the builder sprayed some black rustproof stuff on them. I don't think that solves the problem. But I don't see rust develops on the columns above ground. Do you suppose to change support columns every a few decades? I now finished the basment, and all the columns/beams are behind drywalls. |
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In a previous post Brian Whatcott says...
Now here's the kicker: Say we want to replace a wood beam with a steel beam, for strength, for reduced deflection, or most likely,for reduced depth wasted. In this situation we DON'T want the low ball safety values, we want the HIGH ball values for the wood, in order to guarantee the same or better performance when we match strength or stiffness in steel. The reason is because the previous (wood) beam may have been a lucky, strong example at the upper end of the distribution: it is vital not to degrade values of that rare HIGH end sample. That has been the other issue: everyone has been so comfortable with their timber or steel beam program, it was easy to forget that the brain has to stay engaged, for the safety of the public, whom one is mandated to keep safe. This has turned out to be a remarkably civilized thread, in the end. I salute you all! Brian: Here's how I approach this situation if I am unable to determine the anticipated service loads, but want to have the "new" section at least as strong and as stiff as the existing section: 1) computer section properties for existing section 2) Using NDS (assumes 3-2x12) and allowable design stresses for the material, compute an allowable bending moment, then use that to compute an allowable uniform load. 3) Set a defection limit, typically L/360 for live load, but can be conservatively used for DL+LL. Using section properties, deflection limit and "E" for the original material compute an allowable uniform load. 4) Design of new beam is now governed by the both loads. 5) Since the allowable design stress for Douglas Fir is about 1/20 that of steel and the "E" for Douglas fir is conservatively about 1/20 that for steel, you can reasonably design the new steel beam for the lower of the two loads you computed above. Now for a really interesting problem, let's say you want to take out a post and increase the span of the 3-2x12 from 13 feet to 18 feet, but you must leave the 3-2x12 in place. "Flitch" beams are allowed. Do you compute combined section properties of wood and steel or do you simply let the steel carry all the load and the wood becomes "filler"? This problem has a trick in it that I will not divulge. -- Bob Morrison, PE, SE R L Morrison Engineering Co Structural & Civil Engineering Poulsbo WA |
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In article t, Bob
Morrison wrote: Now for a really interesting problem, let's say you want to take out a post and increase the span of the 3-2x12 from 13 feet to 18 feet, but you must leave the 3-2x12 in place. "Flitch" beams are allowed. Do you compute combined section properties of wood and steel or do you simply let the steel carry all the load and the wood becomes "filler"? This problem has a trick in it that I will not divulge. I look forward to the trick. In the meantime we have a page on the approach to flitch beam design we use in our UK SuperBeam program at http://www.sda.co.uk/info/sbw/flitch.htm. One of the crunch things in your question is that the steel plating to an existing beam will do nothing unless you jack the timbers first to remove the load they are carrying. -- Tony Bryer SDA UK 'Software to build on' http://www.superbeam.com |
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In a previous post Tony Bryer says...
One of the crunch things in your question is that the steel plating to an existing beam will do nothing unless you jack the timbers first to remove the load they are carrying. Tony: That certainly would help and would be considered good construction practice. However, if you apply the steel plates BEFORE you take out the post then loads will transfer to the steel plates. BTW, I will post the "trick" part of the question later today. -- Bob Morrison, PE, SE R L Morrison Engineering Co Structural & Civil Engineering Poulsbo WA |
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Bob Morrison wrote:
In a previous post Tony Bryer says... One of the crunch things in your question is that the steel plating to an existing beam will do nothing unless you jack the timbers first to remove the load they are carrying. Tony: That certainly would help and would be considered good construction practice. However, if you apply the steel plates BEFORE you take out the post then loads will transfer to the steel plates. BTW, I will post the "trick" part of the question later today. If you want to real tricky, you can overjack the timbers then even a larger percentage of the load will be transferred to the steel. *May* be helpful if the wood fiber stress is controlling your flitch. -- Rob Munach, PE Excel Engineering PO Box 1264 Carrboro, NC 27510 |
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In article
, Rob Munach wrote: If you want to real tricky, you can overjack the timbers then even a larger percentage of the load will be transferred to the steel. *May* be helpful if the wood fiber stress is controlling your flitch. The latest structural innovation - a prestressed flitch beam! -- Tony Bryer SDA UK 'Software to build on' http://www.superbeam.com |
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On Wed, 23 Mar 2005 12:27:36 GMT, Tony Bryer
wrote: In article t, Bob Morrison wrote: Now for a really interesting problem, let's say you want to take out a post and increase the span of the 3-2x12 from 13 feet to 18 feet, but you must leave the 3-2x12 in place. "Flitch" beams are allowed. Do you compute combined section properties of wood and steel or do you simply let the steel carry all the load and the wood becomes "filler"? This problem has a trick in it that I will not divulge. I look forward to the trick. In the meantime we have a page on the approach to flitch beam design we use in our UK SuperBeam program at http://www.sda.co.uk/info/sbw/flitch.htm. One of the crunch things in your question is that the steel plating to an existing beam will do nothing unless you jack the timbers first to remove the load they are carrying. I am not exactly enthused about the timber beam application code approach to flitch plates. Seems like an inefficient use of material to sandwich a verical plate between two wood beams. I think it might be worth considering "flange doublers" top and bottom - which is an aircraft main spar approach. Leave the timber as shear and anti buckle and apply steel strips top n bottom of each side, bolted through I haven't worked the numbers, it's true. Brian Whatcott Altus, OK |
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In article , Brian
Whatcott wrote: I am not exactly enthused about the timber beam application code approach to flitch plates. Seems like an inefficient use of material to sandwich a verical plate between two wood beams. It's not very efficient, but in certain cases is preferred by contractors as being more buildable. In the UK loft (attic) conversions are a favourite way of creating more space which generally means having to introduce new beams to support the new floor and existing roof sans struts and purlins. The options are basically (a) full length steel beams craned in by mobile crane - not always possible if there are trees or overhead wires; (b) steel beam with splices - high fabrication costs + space taken by splice; or (c) a flitch beam, the plate part of which is generally much lighter than a steel would be. In this context the flanking timbers make it fixing joist hangers and providing fire resistance much easier -- Tony Bryer SDA UK 'Software to build on' http://www.superbeam.com |
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I am not exactly enthused about the timber beam application code approach to flitch plates. Seems like an inefficient use of material to sandwich a verical plate between two wood beams. I think it might be worth considering "flange doublers" top and bottom - which is an aircraft main spar approach. Leave the timber as shear and anti buckle and apply steel strips top n bottom of each side, bolted through I haven't worked the numbers, it's true. Brian Whatcott Altus, OK Ideally - Yes. However, practically, there would be alot more holes to drill - which is usually 50% of the steel cost. Additonally, your holes would be fairly close to the edge of the wood member unless large pieces of steel were used at which point, you might as well use standard vertical plate. I imagine, also, if you worked out the numbers, you would need alot of bolts Most of my clients prefer to use I-beams anyway instead of flitches as do I. They buy blank pieces of steel and ramset the top nailer on. It is a lot cheaper than the flitch - especially for large loads. Additionally, 8" I-beams work nicely as flush beams in 2x10 floor systems. Pocket the joists into the web, block between the joists and shoot a nailer on top - quick and easy and at least 1/2 the cost of a flitch and alot lighter as well. Regards, -- Rob Munach, PE Excel Engineering PO Box 1264 Carrboro, NC 27510 |
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In a previous post Bob Morrison says...
BTW, I will post the "trick" part of the question later today. Okay gang here's the question I posed the other day: Now for a really interesting problem, let's say you want to take out a post and increase the span of the 3-2x12 from 13 feet to 18 feet, but you must leave the 3-2x12 in place. "Flitch" beams are allowed. Do you compute combined section properties of wood and steel or do you simply let the steel carry all the load and the wood becomes "filler"? This problem has a trick in it that I will not divulge. The "trick" in the question is that nearly all wood beams are split over posts making them simple spans. If you want to move the post to increase the span then you must take in account that there is no moment transfer at the existing post location. The easiest way to do this is to ignore the contribution from the wood for section modulus and moment of inertia UNLESS you want to get into a complex shear transfer analysis for those parts of the beam located away from the existing support. In most residential projects this is simply not feasible (and I doubt in most commercial jobs either). So, the short answer is to design the "flitch" beam in such a way that the steel carries all the load for bending and deflection and the wood is simply "filler". It is a different matter when one gets to the supports. The wood may be used as long as allowable shear and bearing stresses are not exceeded. Here's another way of looking at it: If the wood has an E of 1.8x10^6 and you do a section transform to steel for your "composite" beam, then a 6x (5-1/2" wide) will have a transformed width of 0.34". In other words your 6x suddenly becomes a very small part of the beam and can safely be ignored. -- Bob Morrison, PE, SE R L Morrison Engineering Co Structural & Civil Engineering Poulsbo WA |
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In a previous post Brian Whatcott says...
I am not exactly enthused about the timber beam application code approach to flitch plates. Seems like an inefficient use of material to sandwich a verical plate between two wood beams. Brian: You are correct when you say "flitched" beams are not very efficient, but it is generally the labor involved in building them that makes this so. The wood side members allow easier attachment of wood framing at right angles to the "flitch", which was important before joist hangers became common. "Flitch" beams were commonly used on the east coast where steel was easy to get and large timbers were not. Out here on the west coast large timber have almost always been easier to get than steel, so "flitched" beams are not very common, except when needing to retrofit an existing structure. In our example case of moving a post (or more commonly removing a post and doubling the span), there is often a lot of other stuff already attached to the existing wood beam. It becomes more practical to add steel to the sides of the wood than to remove it entirely. I cannot think of a case where I would design a "flitched" beam for new construction, but would instead go with either all wood or all steel. -- Bob Morrison, PE, SE R L Morrison Engineering Co Structural & Civil Engineering Poulsbo WA |
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Rob-
you wrote: Although, I am starting to believe that the NDS modulus of elasticity numbers are high. I am seeing in the field, members that are deflecting signifcantly more than they should. I'm glad I'm not the only one questioning those number; they seem a little high. Maybe the "poor" deflect behavior is due to lower grade timbers being wrongly upgraded? Modulus is a strong function (at least I was told) of density (ring spacing), maybe the currecnt timber is just grown too fast? cheers Bob |
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In a previous post BobK207 says...
I'm glad I'm not the only one questioning those number; they seem a little high. Maybe the "poor" deflect behavior is due to lower grade timbers being wrongly upgraded? Modulus is a strong function (at least I was told) of density (ring spacing), maybe the currecnt timber is just grown too fast? Bob: I think you are also seeing the effect of long term creep due to the wider ring spacing. It might be prudent to use higher than required DL deflection limits, or perhaps take a look at deflections using something like 1.5DL for service loads to account for the long term creep effects. This would be similar in methodology to that used in concrete design. -- Bob Morrison, PE, SE R L Morrison Engineering Co Structural & Civil Engineering Poulsbo WA |
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In article .com, "BobK207" wrote:
Maybe the "poor" deflect behavior is due to lower grade timbers being wrongly upgraded? If structural timber is graded similarly to furniture-grade timber... there's little doubt in my mind that this explanation is correct. All the time, I see furniture lumber being sold as "FAS" that should have been graded #1COM. It's also not unusual to see lumber sold as "#1 COM" that doesn't even meet minimum standards for #2 COM, and should have been graded #3. -- Regards, Doug Miller (alphageek at milmac dot com) Nobody ever left footprints in the sands of time by sitting on his butt. And who wants to leave buttprints in the sands of time? |
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BobK207 wrote:
Rob- you wrote: Although, I am starting to believe that the NDS modulus of elasticity numbers are high. I am seeing in the field, members that are deflecting signifcantly more than they should. I'm glad I'm not the only one questioning those number; they seem a little high. Maybe the "poor" deflect behavior is due to lower grade timbers being wrongly upgraded? Modulus is a strong function (at least I was told) of density (ring spacing), maybe the currecnt timber is just grown too fast? cheers Bob I agree. It most likely has to due with the fast growth timber. I even see this in LVL beams. I got a complaint from a customer last year that his 9' span garage door header that I designed was sagging. It was a 3 1/2x14 LVL with a computed dead load deflection of less than 1/8". It was sagging 3/8" before the sheetrock even got installed. The manufacturer inspected it and did nothing about it. We ended up sistering it with a piece of steel as he did not want to pursue it. Typically, in my designs, I won't even come close to code allowed deflections. It doesn't cost much to go up 2" or 4" on an LVL, but you get a significantly stiffer beam and no call backs. -- Rob Munach, PE Excel Engineering PO Box 1264 Carrboro, NC 27510 |
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In a previous post Rob Munach says...
Typically, in my designs, I won't even come close to code allowed deflections. It doesn't cost much to go up 2" or 4" on an LVL, but you get a significantly stiffer beam and no call backs. And particularly important when the beam is a window or door header! -- Bob Morrison, PE, SE R L Morrison Engineering Co Structural & Civil Engineering Poulsbo WA |
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