Hi, I'm in the process of rebuilding my woodshed, and the beam in the middle of the house where the floor joists for the second floor rest needs to be replaced (bearer for the floor structure). Since the first floor is going to be a garage, I was planning to install a HEA beam and have a small crane on it that can lift up to 1 ton.
Now it’s time to calculate the whole thing. The free length of the beam is 5m and on each side of the beam there will be 3.5m of floor. The upper floor will also serve as storage/workshop.
I can get a 200x100x? rectangular profile, but I would prefer to use a HEA or HEB beam so I can install a rail for the lifting block.
Using Moelven's calculation program, I've determined that I would need an L40 90x405 Glulam beam. What would that translate to in HEA? Unfortunately, I couldn't account for the lifting block as an additional load.
The second option is to install three smaller beams across the length, allowing me to use the lifting block in more places. This would divide the 7m x 5m floor area into three sections. However, I plan to install a car lift, and I need to keep free space above the car, which makes me prefer a single beam in the middle.
Anyone willing to help me?
Apologies for any spelling or language errors, I'm not Swedish ;-)
Now it’s time to calculate the whole thing. The free length of the beam is 5m and on each side of the beam there will be 3.5m of floor. The upper floor will also serve as storage/workshop.
I can get a 200x100x? rectangular profile, but I would prefer to use a HEA or HEB beam so I can install a rail for the lifting block.
Using Moelven's calculation program, I've determined that I would need an L40 90x405 Glulam beam. What would that translate to in HEA? Unfortunately, I couldn't account for the lifting block as an additional load.
The second option is to install three smaller beams across the length, allowing me to use the lifting block in more places. This would divide the 7m x 5m floor area into three sections. However, I plan to install a car lift, and I need to keep free space above the car, which makes me prefer a single beam in the middle.
Anyone willing to help me?
Apologies for any spelling or language errors, I'm not Swedish ;-)
If I understand you correctly, the beam is supposed to support the second floor and be loaded with a trolley block/weight of approximately 1 ton? The beam is 5 m long and the building is 7 m wide?
What is the roof pitch and what type of truss do you have?
What is the load on the upper floor?
How will the beam be supported at the ends, on what, and how high is it to the foundation?
Get back to me with the answers and preferably a cross-section sketch through the house and a floor plan, so I can calculate it for you.
__________________
Byggaren
What is the roof pitch and what type of truss do you have?
What is the load on the upper floor?
How will the beam be supported at the ends, on what, and how high is it to the foundation?
Get back to me with the answers and preferably a cross-section sketch through the house and a floor plan, so I can calculate it for you.
__________________
Byggaren
Hi, thank you for the offer of help. I'll do some sketching and get back to you.
For the upper floor, I was thinking of supporting 2-3 tons. I have no idea what a common load is. I can't put any cars there, so it will be some car parts, a few woodworking machines (no massive cast iron blocks, just a construction saw, etc.) and a rack for timber. Maybe 1-2 snowmobiles.
The ends of the beam can be placed on whatever I/you decide. Right now, I have 2.5m of air to the granite foundation. I had planned to frame the wall with 145x45 c-c 600, or alternatively, build a steel structure that I can bolt to the granite.
The roof construction is an A truss and I would suggest a slope of 35 to 40 degrees. I'll need to measure that out tonight.
Nisse
For the upper floor, I was thinking of supporting 2-3 tons. I have no idea what a common load is. I can't put any cars there, so it will be some car parts, a few woodworking machines (no massive cast iron blocks, just a construction saw, etc.) and a rack for timber. Maybe 1-2 snowmobiles.
The ends of the beam can be placed on whatever I/you decide. Right now, I have 2.5m of air to the granite foundation. I had planned to frame the wall with 145x45 c-c 600, or alternatively, build a steel structure that I can bolt to the granite.
The roof construction is an A truss and I would suggest a slope of 35 to 40 degrees. I'll need to measure that out tonight.
Nisse
Now, I'm no expert in Sketchup, but I think you can see what needs to be built. It's those two framed walls that are missing right now, the HEA beam in the middle, and the floor beams. The room is 5x7m in size and the HEA beam runs across the short side (hence 5m). I've only sketched the roof truss in place. Between the side walls, there's an iron rod that holds the house together. The beams that the upper floor will be nailed onto are 200x50 (came across a batch). The floorboards are also somewhere around 35 or 40mm thick. I'm also wondering if 1200cc is sufficient instead of the usual 600cc, or is that being penny-wise and pound-foolish? Does the roof construction have any significance specifically for the HEA beam?
Best regards, Nisse
Best regards, Nisse
As the truss is drawn, it does not touch the beam. The diagonal brace goes down to the floor where it meets the raised wall life. The weight of snow and wind then lands on the outer wall. So, only self-weight and useful load to consider. But...
It seems like you might have a lot of junk
up there. 2-3 tons on 5x7=35 sqm. Additionally, perhaps very unevenly distributed? and with that, a rather hefty self-weight of floor joists (not beams) and thick floorboards.
Statically speaking, from a constructive viewpoint, considering the beam's length, it will need to be significantly larger than usual. (Normally, one calculates with 250 kg/sqm floor area in useful load.) Additionally, a moving point load of 1 ton should be considered, hanging in the lower edge of the beam (in fact, one should also consider the braking force for the movement, which adds about 10-20% parallel to the beam, causing bending with simultaneous pressure, but we can probably skip that?).
Using the strip method (which means calculating the load area that touches the beam to half the area on each side of it up to the outer wall along the entire length of the beam) and seeing the beam as a simply supported 2-support beam, the result is as follows:
The maximum moment lands around 450 kNm, which means you need one of the following profiles to find a bending resistance (Wx) large enough to handle it. INP 240 (not so suitable for the truss considering the flanges incline, but is stronger than a) IPE 240 (which has parallel flanges), HE200A, or a HE160B (both with parallel flanges).
From a deformation consideration in the long term and the requirement that the maximum deflection (Ymax) in the middle should not be greater than 1/400 of the span (500/400=1.25 cm), you should increase with one profile number. That is, INP 260, IPE 260, HE220A, or HE180B.
At both outer ends of the beam, about 850 kg will come down. You should therefore use a square profile with a base plate to distribute the pressure against the granite foundation and cast underneath it with expanding mortar. You can weld the base plate to the square profile and equip it with four holes (one in each corner) for bolts that you drill into the granite. It is then possible to adjust the bearings level before casting under. A nut below and a nut above the base plate. At the top, you can weld or bolt the beam to the square profiles. With a buckling length of 2.5 m, you should consider a VKR or KKR in dimensions around 80x80x5 mm.
Since I have not accounted for braking forces on the truss, you should add a short diagonal brace in each corner between the beam and square profile to make the corner moment-rigid.
If you then go to the scrapyard and buy your beam, it should have the quality designation SIS 1312 according to the old standard, which means an allowable stress of 1470 kp/cm2. The new one is called S235JRG2. (Why do they always have to complicate things so, you might wonder. Well, that's because this number combination includes a description of the steel in the beam.)
The iron rod, you write about, is a tension rod to prevent the pressure from the truss from pushing the outer walls apart. Take care of it.
___________________
The Builder
It seems like you might have a lot of junk
up there. 2-3 tons on 5x7=35 sqm. Additionally, perhaps very unevenly distributed? and with that, a rather hefty self-weight of floor joists (not beams) and thick floorboards.Statically speaking, from a constructive viewpoint, considering the beam's length, it will need to be significantly larger than usual. (Normally, one calculates with 250 kg/sqm floor area in useful load.) Additionally, a moving point load of 1 ton should be considered, hanging in the lower edge of the beam (in fact, one should also consider the braking force for the movement, which adds about 10-20% parallel to the beam, causing bending with simultaneous pressure, but we can probably skip that?).
Using the strip method (which means calculating the load area that touches the beam to half the area on each side of it up to the outer wall along the entire length of the beam) and seeing the beam as a simply supported 2-support beam, the result is as follows:
The maximum moment lands around 450 kNm, which means you need one of the following profiles to find a bending resistance (Wx) large enough to handle it. INP 240 (not so suitable for the truss considering the flanges incline, but is stronger than a) IPE 240 (which has parallel flanges), HE200A, or a HE160B (both with parallel flanges).
From a deformation consideration in the long term and the requirement that the maximum deflection (Ymax) in the middle should not be greater than 1/400 of the span (500/400=1.25 cm), you should increase with one profile number. That is, INP 260, IPE 260, HE220A, or HE180B.
At both outer ends of the beam, about 850 kg will come down. You should therefore use a square profile with a base plate to distribute the pressure against the granite foundation and cast underneath it with expanding mortar. You can weld the base plate to the square profile and equip it with four holes (one in each corner) for bolts that you drill into the granite. It is then possible to adjust the bearings level before casting under. A nut below and a nut above the base plate. At the top, you can weld or bolt the beam to the square profiles. With a buckling length of 2.5 m, you should consider a VKR or KKR in dimensions around 80x80x5 mm.
Since I have not accounted for braking forces on the truss, you should add a short diagonal brace in each corner between the beam and square profile to make the corner moment-rigid.
If you then go to the scrapyard and buy your beam, it should have the quality designation SIS 1312 according to the old standard, which means an allowable stress of 1470 kp/cm2. The new one is called S235JRG2. (Why do they always have to complicate things so, you might wonder. Well, that's because this number combination includes a description of the steel in the beam.)
The iron rod, you write about, is a tension rod to prevent the pressure from the truss from pushing the outer walls apart. Take care of it.
___________________
The Builder
THANK YOU!
I feel a bit embarrassed when I say that I studied mechanical engineering, but the house calculations are not my thing ;-) It's been a while since I dealt with calculations as well. However, I think S235 is a reasonable designation In Germany, it used to be called ST 37, but when I started studying, it was already 235 that was applicable.
No junk will end up up there, only good stuff to have! ;-) But I think it's okay to start from the usual 250kg/sqm and a point load of maybe 500kg. Heavier stuff I shouldn't lift around in the garage. And if, against all odds, it happens, I can always place a temporary column under the beam However, it would be about 8 tons with 250kg/sqm and 35sqm or 4 tons with half the load area, so how do you calculate further then?
You don't need to take into account the braking force for the movement, as you said. It's not daily that I will lift something around. But one never knows, fixing the tractor a bit....
I've already cast a smooth concrete surface on the granite, where I was thinking of taking a longer HEA profile (like 1.5m, or maybe the entire 7m?) and placing it along. Then weld a KKR 80x80 in the middle of it and the rest as you described. Is this an alternative? To the casting you describe?
I was the one who put the tie rods, so I care for them ;-) It used to be a timber house that was unfortunately used for pigs and horses and rotted away completely. So I had to saw off most of it.
Would it also be sufficient with a KKR Rectangular 200x100x5? I saw that it had a slightly smaller Wx. It's one I can get for just over 200SEK/m. I was thinking of maybe welding a HEA underneath for safety.
And once again, a big thank you for offering your time!
Best regards, Nisse
I feel a bit embarrassed when I say that I studied mechanical engineering, but the house calculations are not my thing ;-) It's been a while since I dealt with calculations as well. However, I think S235 is a reasonable designation
In Germany, it used to be called ST 37, but when I started studying, it was already 235 that was applicable.No junk will end up up there, only good stuff to have! ;-) But I think it's okay to start from the usual 250kg/sqm and a point load of maybe 500kg. Heavier stuff I shouldn't lift around in the garage. And if, against all odds, it happens, I can always place a temporary column under the beam
However, it would be about 8 tons with 250kg/sqm and 35sqm or 4 tons with half the load area, so how do you calculate further then?You don't need to take into account the braking force for the movement, as you said. It's not daily that I will lift something around. But one never knows, fixing the tractor a bit....
I've already cast a smooth concrete surface on the granite, where I was thinking of taking a longer HEA profile (like 1.5m, or maybe the entire 7m?) and placing it along. Then weld a KKR 80x80 in the middle of it and the rest as you described. Is this an alternative? To the casting you describe?
I was the one who put the tie rods, so I care for them ;-) It used to be a timber house that was unfortunately used for pigs and horses and rotted away completely. So I had to saw off most of it.
Would it also be sufficient with a KKR Rectangular 200x100x5? I saw that it had a slightly smaller Wx. It's one I can get for just over 200SEK/m. I was thinking of maybe welding a HEA underneath for safety.
And once again, a big thank you for offering your time!
Best regards, Nisse
St 37 is/was the tensile strength in kp/cm². It applies to SIS 1311, which is a standard. (Swedish Industry Standard). S235 indicates it is a structural steel with an upper yield limit of 235N/mm².Koebes said:THANK YOU!
I am a bit embarrassed when I say that I studied mechanical engineering, but calculations for houses are not my thing ;-) It has been a while since I did any calculations as well. S235, on the other hand, seems like a reasonable designation
No junk will end up there, just good things to have! ;-) But I think it would be fine to start with the regular 250kg/m² and a point load of maybe 500kg. I probably won't be carrying heavier stuff around in the garage. And if it should happen against all odds, I can always put a temporary pillar under the beam
You don't need to consider the braking force for the movement, as you said. It's not like I'll be moving things around daily. But you never know, fixing the tractor a bit....
I've already poured a smooth concrete surface on the granite, where I intended to take a longer HEA profile (like 1.5m, or maybe a full 7m?) and lay it along it. Then weld a KKR 80x80 in the middle of it and the rest as you describe. Is this an alternative? To the casting that you describe?
I was the one who put in the tie rods, so I'm careful with them ;-) It used to be a timber house that unfortunately was used for pigs and horses and decayed completely. So, I had to cut off most of it.
Would it be enough with KKR Rectangular 200x100x5? I saw that it had a slightly smaller Wx. It's something I can get for about 200SEK/m. I thought maybe welding an HEA underneath for the truss.
And once again, many thanks for sacrificing your time!
Regards, Nisse
I calculated with 250kg/m². The heaviness of timber (pine and spruce) is set to 500kg/m³. I estimated the beam to 60 kg/m, which turned out to be double in the end. But better that way than the other way, i.e., under-dimensioning.
The formula for the beam looks like a standard 2-support beam. That is, uniformly distributed load for qn and qe and point load for the overhead trolley. The load area is 5 m times half the house's width. Then you get (replace spv on the beam with L and read 2 as 'raised to the power of'): Mmax = (q times L²) divided by 8 plus (P times L) divided by 4.
Then use the old permissible stresses instead of the new building regulations' characteristic strength values (which are 95% of empirically determined fracture limits and require reduction with partial coefficients to get permissible strength values). That is 1470 kp/cm² for steel SIS 1311/S235JRG2 and 80 kp/cm² for wood in the most common construction class (it corresponded to T80 at the time).
Then divide the total Mmax by these strength values to get the required Wx on the beam.
Subsequently, you need to check the deflection with respective cases for distributed load and point load=P. Sum up the deflections (which should not exceed L/400). The formula for the first is: L/400= (5 times q times L⁴) divided by (384 times E times I) and for the second: (P times L³) divided by (48 times E times I). (Read 3 and 4 as 'raised to the power of'.) E is the modulus of elasticity which for steel is: 210000kp/cm². And I you can get from the table for the steel profile you end up with.
For the pillars, Euler's case 2 applies as usual (or possibly case 3 since the upper end can be made moment-stiff and thus counted as fixed).
All according to Hooke's law and elasticity theory (where the partial coefficient method also lands after reducing the characteristic strength values).
Of course, you don't need to use a sledgehammer for tacks. A base plate of 150x150 is sufficient to transfer the force from the posts under the pillars to the granite. Especially if you have leveled the surface with concrete. If you use an HE-A beam, 1.5 meters is more than enough. The main thing, in both cases, is that you ensure they are anchored with drilled bolts into the granite so that the block doesn’t overturn and the post can slide off the support.
If I recalculate Mmax considering a lighter beam, I land on a required Wx=300 cm³. All else being equal. But without taking deflection into account. If you weld an I-beam for the truss under your hollow section and do the welding work professionally (i.e., intermittently along both flanges throughout the length and the last meter fully welded without root defects, etc.) you can also account for Wx in the total section according to Steiner's theorem. The load due to deflection is on the underside. Then the value of I is higher, which means the beam bends less.
_________________
The Builder
Hello again,
I wrote down the first formula here: http://www.techniker-forum.de/testboard-17/test-22876.html#post119716, and there's also a sketch on how I intended to build with the existing material. I'm going to calculate if a square profile 120x120x6.3 welded to an HEA 120 is sufficient. I don't think it's enough, but for practice's sake.
However, I can't quite follow your formulas, and if you're willing, could you point out where I've calculated incorrectly?
The normal load is 250kg/m^2, and I'm using half of the area to determine the load on the Beam
=(5m x 7m / 2 ) *250kg/m^2 = 4375kg or 43kN
Distributed over a 5m beam, it becomes:
qn=43kN/5m=8.6kN/m
Self-weight is 60kg/m or 0.6kN/m. Is this your qe?
Self-weight of the Joist with 200x50 c-c60 = around 8 beams
qb=0.2m*0.05m*3.5m*8*500kg/m^3/5m=28kg/m or 0.28kN/m
Self-weight of the Floor with qg=17.5m^2*0.035m (floor thickness)*500kg/m^3 / 5m=61.25kg/m=0.61kN/m
Without point load, Mbmax1=(qe+qn+qb+qg)*L^2/8=[(8.6kN/m+0.6kN/m+0.28kN/m+0.61kN/m)*25m^2]/8 =31.53kNm and is naturally in the middle of the beam
Point load is:
P=1000kg =10kN
The formula is: Mmax2=P*L/4 =10kN*5m/4=12.5kNm
Mbmax total= Mbmax1+Mbmax2=31.53kNm+12.5kNm=44kNm or 44000kNmm
(you write 450kNm in your first post. My comma error or yours?!? or what is the mistake?)
1470kp/cm^2 is around 144N/mm^2
Wx=44000kNmm/(144N/mm^2) =305555mm^3=306cm^3
And here we seem to agree again with your Wx at 300cm^3
Questions/Remarks:
1. I've included the weight of the joist and floor, or is that included in the 250kg/m^2?
2. Does my calculation agree?
3. How did you calculate 850kg/pillar when just the load from the floor area with 250kg/m^2 is already 4375kg spread over two pillars?
Thanks, Nisse
I wrote down the first formula here: http://www.techniker-forum.de/testboard-17/test-22876.html#post119716, and there's also a sketch on how I intended to build with the existing material. I'm going to calculate if a square profile 120x120x6.3 welded to an HEA 120 is sufficient. I don't think it's enough, but for practice's sake.
However, I can't quite follow your formulas, and if you're willing, could you point out where I've calculated incorrectly?
The normal load is 250kg/m^2, and I'm using half of the area to determine the load on the Beam
=(5m x 7m / 2 ) *250kg/m^2 = 4375kg or 43kN
Distributed over a 5m beam, it becomes:
qn=43kN/5m=8.6kN/m
Self-weight is 60kg/m or 0.6kN/m. Is this your qe?
Self-weight of the Joist with 200x50 c-c60 = around 8 beams
qb=0.2m*0.05m*3.5m*8*500kg/m^3/5m=28kg/m or 0.28kN/m
Self-weight of the Floor with qg=17.5m^2*0.035m (floor thickness)*500kg/m^3 / 5m=61.25kg/m=0.61kN/m
Without point load, Mbmax1=(qe+qn+qb+qg)*L^2/8=[(8.6kN/m+0.6kN/m+0.28kN/m+0.61kN/m)*25m^2]/8 =31.53kNm and is naturally in the middle of the beam
Point load is:
P=1000kg =10kN
The formula is: Mmax2=P*L/4 =10kN*5m/4=12.5kNm
Mbmax total= Mbmax1+Mbmax2=31.53kNm+12.5kNm=44kNm or 44000kNmm
(you write 450kNm in your first post. My comma error or yours?!? or what is the mistake?)
1470kp/cm^2 is around 144N/mm^2
Wx=44000kNmm/(144N/mm^2) =305555mm^3=306cm^3
And here we seem to agree again with your Wx at 300cm^3
Questions/Remarks:
1. I've included the weight of the joist and floor, or is that included in the 250kg/m^2?
2. Does my calculation agree?
3. How did you calculate 850kg/pillar when just the load from the floor area with 250kg/m^2 is already 4375kg spread over two pillars?
Thanks, Nisse
Hello Nisse!
I'm going crazy with this changed forum. It's impossible to write a mathematical formula correctly without * and ^ etc., for example, subscript index. Clearly worse than the previous one in other words. I tried all evening to give you the calculations as inserted comments in your text, but that didn't work either. Used Italy and everything looked fine until the preview when everything became Italy.
I will take more time to respond in the coming days now that my wife's 70th birthday party is over.
____________________
Byggaren
I'm going crazy with this changed forum. It's impossible to write a mathematical formula correctly without * and ^ etc., for example, subscript index. Clearly worse than the previous one in other words. I tried all evening to give you the calculations as inserted comments in your text, but that didn't work either. Used Italy and everything looked fine until the preview when everything became Italy.
I will take more time to respond in the coming days now that my wife's 70th birthday party is over.
____________________
Byggaren
Koebes said:
Since you mentioned in PM that you are moving away from the beam and placing five joists instead, I won't spend more time calculating it. You have an exponent error in your calculation.
______________
Byggaren
Question for imported_Byggaren!
I have read the conversation here and would need help with a follow-up question.
We are going to tear down a 5-meter-long wall between the kitchen and the "old" dining room. The wall currently rests on the heart wall in the basement. A structural engineer has calculated, based on the house plans from 1968, that we need an HEA200. I will be installing an HEA220 to have some margins. The deflection is now calculated to be 8.3 mm. The normal load is estimated to be around 800 kg/meter and the maximum (in cases of snow and similar) to about 1800 kg/meter.
So, the beam will now be supported at both ends. Normally, we can expect about 2000 kg of pressure on each column. The columns are intended to be of the type glulam beam 225x115. I'm just wondering if one needs to consider the basement walls.
At one end, the column will rest on the basement's gable wall, which consists of 30 cm wide concrete blocks. On these lies a brick facade and a sill of the type 15 cm wide and 22 cm high that supports the interior wall.
Question 1: Can I place the glulam column with the normal load of 2000 kg directly on the sill on the basement wall? The basement wall should hold, right? Or should I place a 1-meter-long, heavy steel profile on top of the sill to further distribute the pressure?
At the other end, the column will rest on the basement's heart wall, which consists of 14 cm wide concrete blocks. On the heart wall lies a 14 cm wide and 22 cm high sill, to which the floor joist is attached.
Question 2: Can the heart wall made of 14 cm wide concrete blocks withstand a 2000 kg load? I can also lay down a type of 1-1.5 meter long, heavy steel profile here. The weight distribution would then be about 133 kg per dm of length. This should be totally okay, right?
Grateful for a quick response =)
I have read the conversation here and would need help with a follow-up question.
We are going to tear down a 5-meter-long wall between the kitchen and the "old" dining room. The wall currently rests on the heart wall in the basement. A structural engineer has calculated, based on the house plans from 1968, that we need an HEA200. I will be installing an HEA220 to have some margins. The deflection is now calculated to be 8.3 mm. The normal load is estimated to be around 800 kg/meter and the maximum (in cases of snow and similar) to about 1800 kg/meter.
So, the beam will now be supported at both ends. Normally, we can expect about 2000 kg of pressure on each column. The columns are intended to be of the type glulam beam 225x115. I'm just wondering if one needs to consider the basement walls.
At one end, the column will rest on the basement's gable wall, which consists of 30 cm wide concrete blocks. On these lies a brick facade and a sill of the type 15 cm wide and 22 cm high that supports the interior wall.
Question 1: Can I place the glulam column with the normal load of 2000 kg directly on the sill on the basement wall? The basement wall should hold, right? Or should I place a 1-meter-long, heavy steel profile on top of the sill to further distribute the pressure?
At the other end, the column will rest on the basement's heart wall, which consists of 14 cm wide concrete blocks. On the heart wall lies a 14 cm wide and 22 cm high sill, to which the floor joist is attached.
Question 2: Can the heart wall made of 14 cm wide concrete blocks withstand a 2000 kg load? I can also lay down a type of 1-1.5 meter long, heavy steel profile here. The weight distribution would then be about 133 kg per dm of length. This should be totally okay, right?
Grateful for a quick response =)
a response look at www.stålbyggnadsintitutet.se …..forward it to others who are wondering about buildings for liftingK Koebes said:
imported_Byggaren said: