Identificaiton of Beam requirements 2

[qems]

I am trying to calculate the size of a Universal Beam that I will need to replace an old wooden beam.

It is a simply supported beam that needs to hold up a uniformly distributed load of around 1.5 tonnes, over a length of 14.5ft. The beam is supported at either end by an upright column.

I have calculated this weight using the following:
Surface Area of Slate & Wood Roof = 103 sq ft @ 16lbs / sq ft
= 1624 lbs

Surface area of Wood Panal and Partially Glazed wall = 100sq ft @ 6lbs / sq ft =609lbs

TOTAL Weight = 2233 lbs

Safety Factor = 2233*1.5
= 3349lbs

I have tried using BM = wl^2/8 and D= 5/384 x wl^3/EI , but I am not having much luck as I am getting figures that say a 203 x133 will be sufficient but this seems light.

Any help would be greatly appreciated. If you could show me the calculations it would be fantastic so I can see where I am going wrong.

Rob

 

[BAretired]

There is still a problem with your calculation. You stated that the building size is 4.5m x 2.13m. So the tributary width of floor and roof going to the beam should be 2.13/2 = 1.07m should it not?

When you calculated your roof load and floor load, you assumed the entire area was carried by the beam.

When you considered snow and live load, you did not take into account 1.9 kPa for the floor.

BA

 

[qems]

Hi Kslee1000

Not a problem, it's OK, I wasn't taking it personally, I agree with all of the comments, just a little confused about the "hire and Engineer" one. I am finding it frustrating stumbling at such an early hurdle when I feel I should be able to remember this (even if it was 15 years ago;-) ). SUggesting that I hire an engineer implies that I should not be asking this question on the forum.

That said, I have taken on board your suggestions and tried to do this with a more methodology and structure, please see below:

Roof :
Total surface area: 4.5*1.6=7.2m^2
SA attributed to beam: 7.2m^2/2 = 3.6m^2

Floor:
Total surface area: 4.5*2.13 = 9.6m^2
SA attributed to beam: 9.6m^2/2 = 4.8m^2
Wall:
Total Surface Area: = 9.6m^2

Joists = 5.3m^2 / 2 = 2.65m^2

DEAD LOADS
Roof : 766.08N/m^2 8 3.6m^2 = 2.758kN
Floor: 167.58N/m^2 * 4.8m^2 = 0.85kN
Wall : 287.287 N/m^2 * 9.6m^2 = 2.75kN
Joists: 301.64 * 2.65m^2 = 0.8kN

LIVE LOADS
Wind: 430.92N/m2 * 3.6m^2 = 1.55kN
Snow: 700N/m^2 * 3.6m^2 = 2.52kN
Floor Live Load: 1900N/m^2 * 4.8 = 9.12kN (I am not sure what this is but I have added it under your instruction – I am assuming it is for furniture / people etc)

Total Deal Load = 7.133kN
Total Live Load = 13.19kN

Dead Load Safety Factor = 1.4
Live Load Safety Factor = 1.6

Total UDL = 1.4*7.113kN + 1.6 * 13.19kN
=31.104kN

=6.912kNm for the span of 4.5 m.

Maximum Moment
M=wl/8 (where w is the total UDL)
M= (31.104 * 4.5) /8 = 17.45kNm

Maximum Shear
Shear = (6.912kN * 4.5)/2 = 15.55kN

Moment Capacity for a fully restrained beam is equal to pySx so the Sx I require is 17.45*10^3/265 (py=265N/mm^2) ( I am not sure if I am using the correct “ strength value)

Sx required = 65.84 mm^3.

I now need to select a beam that has an Sx greater than 65.84mm^3.

Now to check the Deflection
Maximum Deflection must not be greater than span/250 = 4500/250= 18mm

D = 5/384 *W * L3/EI
I= (5*(31.104)*4.5^3)/(384*205*18) *10^5
I = (14171.76 / 1416960) *10^5 = 1000.15mm^4

So technically if I choose a section which has an I-value greater than 1000 mm^4 and check that its Sx > 65.84mm^3 then this should be the beam for the job. (assuming I have not made any more school-boy errors)

The parts I am unsure of are:

If I am using the correct Youngs Modulus value, and the correct strength value.

If I use the UDL with the safety factor when calculating the Inertion.

I am then having problems reading the steel charts to identify what I-beam would be sufficient.

Thanks for your persistence guys.

 

[BAretired]

Roof :
Total surface area: 4.5*1.6=7.2m^2 why 1.6? It must be more than 2.13
SA attributed to beam: 7.2m^2/2 = 3.6m^2

Floor:
Total surface area: 4.5*2.13 = 9.6m^2
SA attributed to beam: 9.6m^2/2 = 4.8m^2
Wall:
Total Surface Area: = 9.6m^2

Joists = 5.3m^2 / 2 = 2.65m^2

DEAD LOADS
Roof : 766.08N/m^2 8 3.6m^2 = 2.758kN
Floor: 167.58N/m^2 * 4.8m^2 = 0.85kN
Wall : 287.287 N/m^2 * 9.6m^2 = 2.75kN
Joists: 301.64 * 2.65m^2 = 0.8kN

Unit Dead Loads (kPa)
Roof: 0.77 * 1.3 = 1.0 (assuming 40 degree slope)
Floor: 0.7 (to account for joists, drywall, electrical, etc.
Wall: 0.5 (glass can be pretty heavy)
Joists: included in floor DL

LIVE LOADS
Wind: 430.92N/m2 * 3.6m^2 = 1.55kN
Snow: 700N/m^2 * 3.6m^2 = 2.52kN
Floor Live Load: 1900N/m^2 * 4.8 = 9.12kN (I am not sure what this is but I have added it under your instruction – I am assuming it is for furniture / people etc)

Unit Live Loads (kPa)
Wind: not considered
Snow: 1.0 (minimum snow load in code)
Floor: 1.9 (this is a building code value for residential occupancy in my code...it could be different in yours)


Dead Load Safety Factor = 1.4
Live Load Safety Factor = 1.6
1.25 and 1.5 respectively in my code

The parts I am unsure of are:

If I am using the correct Youngs Modulus value, and the correct strength value. 200,000 MPa

If I use the UDL with the safety factor when calculating the Inertion. (what is inertion?)

I am then having problems reading the steel charts to identify what I-beam would be sufficient. My steel handbook lists all of the beams with their moment capacities for various values of unbraced length.

The calculation in blue is more or less the way I would record calculations for this beam.

Beam Design
Span 4.5m; tributary width = 2.13/2 = 1.07m
wd = 1.07(1.0 + 0.7) + 2*0.5 (wall) = 2.82 kN/m
wl = 1.07(1.0 + 1.9)................= 3.10

wt (total load per meter) ..........= 5.92
wf = 1.4*2.82 + 1.6*3.10............= 8.91 (factored load)

Mf = 8.91(4.5)^2/8 = 22.5 kN-m
Try W200x19 and check for deflection (I = 16.6e6 mm^4)
Mr = 58.1 when top flange is continuously supported
Mr = 21.9 when top flange braced at 4m centers, so make sure you brace at midspan (I would go for 1.5m spacing of braces)
Delta (total) = 5/384 * 5.92(4500)^4/(200000*16.6e6) = 9.5mm
Total deflection = L/472 which is acceptable.





BA

 

[kslee1000]

I think civilperson is as frustrated as you

I will print the above Monday to have a closer look. At the meantime, can you upload a simple floor plan and elevation (scanned hand sketch will do). Also, please list and indicate the material for each element.

 

[qems]

Hi Guys, thanks for this it really is appreciated, (much to my wife's horror as I have spent way to much time at the desk over the weekend!).

BAretired. Fantastic, thank you. In answer to your questions, the roof was a mistake on my part (once again) I took the dimensions of one side, hence I did not have to divide it by two. "Inertion" should be "inertia" = I.

This is where I was getting confused, working out "I", with a maximum Deflection of 4%. Thanks again. The safety factors are from BS5400 (if my memory serves me correctly)

Kslee100, I have drawn a basic layout including materials.

Apologies if it is a bit rough.

 

[BAretired]

qems,

How do you put the picture in with the text? The only way I know how to do it is to upload the file which can then be downloaded by the reader.

This seems like a much better way. Since joining EngTips, I have wondered about this. Is it a trade secret?

BA

 

[kslee1000]

BA:

Pay up to get the trade secret out
Everybody knows something better.

 

[qems]

Hi not a problem, you have to store the image online somewhere. I uploaded it to our business site, then just type the following.

img

http://www.qems.biz/daphnia/house/SUMMERHOUSE

002.jpg]

I have missed out the first [ which should go before the "img" to make the text show up in the post. ie. the statement

img

http://www.qems.biz/daphnia/house/SUMMERHOUSE

002.jpg

should be enclosed in square brackets []

Here is a good page which is linked to the site that give some handy hints on formatting.

http://www.tipmaster.com/includes/tgmlinfo.cfm?w=450&h=450

Not sure if I explained that clear enough, but try it out and have a look at the link, there is some really use tips.

Rob

 

[BAretired]

Rob,

Sorry to be so thick, but how do I store an image online somewhere?

BA

 

[kslee1000]

gems:

Can you follow through on BA's cal? He's pinned everything down, except the roof slope (40 assumed, 27 frm sketch). However, it would have been well covered by the safety factors.

Note, you will need to check wind induced uplift on the floor deck/joist, and the roof, to ensure proper tie down.

 

[qems]

Hi BAretired

There are two ways that you can do this

The easiest way is if you own a website. Simply upload the photo and then take note of the url that the phot appears online and place it in the tags.

If you do not have this ability then you can use sites like

[URL unfurl="true"]http://www.photobucket.com[/URL]

Register for an account and upload your photo.

Then click on the "blogger icon" (the orange one) below the photo next to the word "share".

Next click on the

"Get Link Code"

[img]http://i615.photobucket.com/albums/tt235/hypancistrus046/share.jpg

Next - copy the code in the second box down "Direct Link for layout pages"

Then simply paste this code into the

tags in your post.

Seems complicated but once youh ave done it a few times it will only take a second. If only Beam calcs were this easy ;-)

Hope this helps.

Rob

 

[qems]

Hi Kslee1000

Thanks for that, yes I should be able to follow through with BA's calculations. I have spoken to my contractor today and he has suggested using the UK equivalent to a flanged W6 x 20.

As I am using Uk based standards, (for example a UB203 x 127) is there link to a table that includes the "I" values and Mr values. That way I can do teh same calcs for a W6 x 20 equivalent.

Thanks again

Rob

 

[kslee1000]

Try the link below. I believe BA has suggested a W8x..

http://www.engineeringtoolbox.com/american-wide-flange-steel-beams-d_1318.html

 

[kslee1000]

Also this one. Both tables list properties only good for allowable stress design rather than strength design, please select a section with Ix & Wx at least equal to, or comparable with, the section calculated by BA to assure strength and deflection performances.

http://www.engineeringtoolbox.com/british-universal-steel-columns-beams-d_1316.html

 

[BAretired]

Rob,

The shape which I suggested was a W200x19 (metric designation) which is equivalent to a W8x13 (Imperial designation). This was in keeping with your proposal to use a Universal beam, which I believe is similar to a Wide Flange.

Its properties are:
d = 203 mm
b = 102 mm
tf = 6.5 mm (flange thickness)
w = 5.8 mm (web thickness)
Ix = 16.6 x 10^6 mm^4
Sx = 163 x 10^3 mm^3
Zx = 187 x 10^3 mm^3

We have a W6x20 (Imperial) in Canada which is stronger and slightly stiffer than the W8x13.

You might want to mention to your contractor that a pair of channels is another option. They would be a bit heavier but you would install one each side of the existing wood beam and bolt through thereby eliminating the need to shore the existing structure.



BA

 

[racookpe1978]

You might want to mention to your contractor that a pair of channels is another option. They would be a bit heavier but you would install one each side of the existing wood beam and bolt through thereby eliminating the need to shore the existing structure.


--

Definitely like the bolt-on-a-pair-of-double-channels solution:

You add strength and rigidity while NOT needing to shore up the whole structure while pulling and removing the old wood beam.

Shoring can be done, but ANY problem or slipping threatens the whole building since replacement of the original beam requires (by definition) removing ALL the original structural strength BEFORE being able to put back the new beam. Too many chances for something to go wrong and destroy your building. Paint your new channels: use a good primer plus two cover coats.

Bolt one end of the new channels to the current beam, jack up the other end of the two channels and bolt them in position, making (as said) a sandwhich with the old wood structure in the middle of the two channels: You will almost certainly see the "sag" in the middle in the current wood beam.

Then press up the middle of the old sagging beam until it matches the "straight" new channels. Drill your new holes through the channels and through the old truss (at the bottom) - no smaller than 1/2 dia (12 or 13 mm) high grade bolts every 150-200 mm - and clamp firmly.

Result will be a safer, easier installation; prettier than a WF that will be stronger as well.

 

[qems]

HI Guys

Kslee1000: Thanks for the charts, excellent, and greatly appreciated.

BAretired:

The W6 x 20 is the equivalent of what he is wanting to use, looking at the charts.

As for the pair of channels, all of the joists are rotten and having to be removed along with the floor, the only original part of the structure that will remain will be the roof. Hence we are not worried about having to shore, as it all has to come out anyway.

How did you get on with the image inserting?

Rob

 

[BAretired]

Rob,

If the floor is rotten, then I agree with you. Use the Universal beam.

As for the imaging, I don't have my own web site. I have successfully loaded an image into photobucket, but so far I haven't had any occasion to send it to anyone. One concern is that I seem to be getting a lot spam popping up from that site. I also have some concerns as to the safety of the site as everyone and his dog seems to have access to everyone else's photos. I will mull it over for a while, but thanks for the info. At least I know how to do it now.

BA

 

[qems]

Hi Racookpe1978

Thanks for that.

The only load on the old beam now is the existing floor, which is coming out. see below:

The scaffolding you see going through the floor is load bearing and taking the weight of the roof on teh Right hand side. The other side is anchored onto the existinghosue wall.

The hopefully simple plan is that the contractor removes the floor and Joists. Inserts the Upright column from ground to Roof, bolts up the new beams and re-instates the floor. Then it is a case of putting in the intermediate supports and filling the blank spaces. ;-) All steelwork will be covered with hardwood as per my wifes specification! ;-)

If it goes up as easily as it came down I will be happy. That said, we have been doing this house up for 7 years and it always throws little surprises at us to make life interesting!

Thanks again.

rob

 

[qems]

Hi BA

I am not sure, but I think you can set your preferences on the site so that no-one else can see your images. That said if you are putting them in a forum then they are in the public domain anyway.

When you register, so long as you didn't tick the boxes saying send me info on..... etc, you should be ok for spam.

Failing that, if you ever need images posted, drop me a line and I can upload them to our server for you. It is the least I could do for all your help.

Rob