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Design of strip footing


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Aslamoalaikum. I am trying to design strip footing for a two storey mosque. I have shared load distribution in the photo. There are two beams in the main hall.

I can't figure out how this load distribution can be used to design strip footing for wall 1 and wall 2.

Can anyone guide me?

Assume net allowable bearing capacity is 0.70 T/sft.

Other assumptions

All masonry walls 13.5" thick & 12'0" high

Slabs: 65+65

Finishes: 50+60 psf

Live: 50+50 psf

 

Please guide.

IMG_20230331_194255.jpg

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Lots of questions...

I cannot tell what the framing for this is.  Are the floors structural concrete? or something else. What is the roof construction?  Is it one way, or two way spanning for the slab?  If slabs, is the span for the upper part in a left-right direction?  Why is the middle panel in the upper part different than the panels at each edge? Wouldn't they all be the same? It looks like the spans are approx 15'; does that mean that you are using a 6" slab or something similar?  I don't know what your numbers 61, 61, and 20 represent. I don't know what your loads 50 + 50psf mean.

Edited by Dik
added questions...
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22 hours ago, Dik said:

Lots of questions...

I cannot tell what the framing for this is.  Are the floors structural concrete? or something else. What is the roof construction?  Is it one way, or two way spanning for the slab?  If slabs, is the span for the upper part in a left-right direction?  Why is the middle panel in the upper part different than the panels at each edge? Wouldn't they all be the same? It looks like the spans are approx 15'; does that mean that you are using a 6" slab or something similar?  I don't know what your numbers 61, 61, and 20 represent. I don't know what your loads 50 + 50psf mean.

Floors/slabs are of RCC.

In main hall, there are two beams which are resting on masonry walls (load bearing walls). In all other rooms, slabs are resting directly on masonry walls (load bearing walls).

In the main hall of size 36'×22' (check dimensions), there are two beams. Portion of slab between two beams is two way and remaining two portions are one way.

No using 5" slab.

Numbers 61, 61, 20 represent magnitude of tributary areas formed using one way & two way load distribution concept.

65+65 means 65psf slab at FF and 65psf slab at Roof. 50+65 means 50psf finishes at FF slab and 65psf at Roof slab. 50+50 means 50psf live load at FF slab and roof slab each.

Edited by M. Asad Warraich
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Just now, M. Asad Warraich said:

You can have a look at this drawing. Green lines are two beams.

IMG_20230405_090125.jpg

Portion between two beams is two way slab (shorter dimension: 15'-0", longer dimension: 22'-0")

In main hall (36'×22'), both slabs at sides of beam is one way (9'-3"×22'-0") i.e. 102+102 tributary areas

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I'm listed as an rookie... but, I have over 50 years of experience.  Some comments...

image.thumb.png.a36530aeb301947001cb106567891a01.png

Edited by Dik
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To design Wall - 1 for two storey building, I adopted this method. Please comment.

LOAD CALCULATION

1- Masonry Walls Load 13.5" Thick 12'-0" High

13.5/12×(12+12)×120/1000=3.24 k/ft

2- Parapet Wall Load

4.5/12×2×120/1000=0.10 k/ft

3- Masonry Walls Load 13.5" Thick under plinth level

13.5/12×(3+3)×120/1000=0.81 k/ft

4- Floor Loads

Both beams will bear load coming from slab and transfer it to their ends resting on walls as point loads. That point load will be divided by the width of the tributary area being born by both beams.

Tributary Area of beams for wall - 1=102/2+109/2+109/2+102/2+56=267

Beam size is 12"×18".

> Slabs Loads=65+65=130 psf

Point load due to slabs' distributed loads=267×130/1000=34.71 kips

Divide by width of tributary area=4+7.5/12+15/12+7.5+15/12+4+7.5/12=19.25 feet

Load per unit length due to self weight of slab=34.71/19.25=1.80 k/ft

> Finishes Loads=50+60=110 psf

Using same procedure

Load per unit length due to floor finishes=1.53 k/ft

> Live Load=50+50=100 psf

Load per unit length due to distributed live loads=1.39 k/ft

> PL due to beam self weight

12×18/144×(22/2+13.5/12)×150/1000=2.73 kips for one beam

For four beams (two at FF slab & two at roof slab)=4×2.73=10.91 kips

Divide by width of tributary area=10.91/19.25=0.57 k/ft

TOTAL LOADS ON WALL PER UNIT LENGTH

• Total unfactored dead load=3.24+0.10+0.81+1.80+1.53+0.57=8.05 k/ft

• Total unfactored live load=1.39 k/ft

FOOTING WIDTH REQUIRED

Wreq=(8.05+1.39) / (0.70×2.2045)=6.11 feet

Edited by M. Asad Warraich
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Thanks for the added information. For personal reasons, I don't want to review your calculations, but it looks like you have things in order...

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6 hours ago, M. Asad Warraich said:

WHY 2 WAY ACTION HERE?

Because dimensions of slab in between two beams is 15'-0"×22'-0".

WHY 1 WAY ACTION HERE?

Because it's a verandah and its dimensions are 33'-6"×10'-0".

Just because the aspect ratio of the slab is greater than 2 or whatever, the flexibility of the beams, as opposed to it being a wall support would tend to cause it to behave as a one way slab. Even your load takedown for the wall, as 2 point loads from the slab indicates the one way action. 

Same with the verandah, it doesn't matter what the use of the slab is, it would tend to distribute some of the load at the ends to the adjacent masonry wall. The aspect ratio is still pretty small.

Edited by Dik
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Thanks for your comments. Actually, my purpose of sharing calculations and this post here was to get experts' opinion on if I am doing the design correctly. Would be really grateful if you can have a look at the calculations. Thanks.

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You don't have to look at all the calculations. Just give your opinion on this part of the calculations.

4 hours ago, M. Asad Warraich said:

4- Floor Loads

Both beams will bear load coming from slab and transfer it to their ends resting on walls as point loads. That point load will be divided by the width of the tributary area being born by both beams.

Tributary Area of beams for wall - 1=102/2+109/2+109/2+102/2+56=267

Beam size is 12"×18".

> Slabs Loads=65+65=130 psf

Point load due to slabs' distributed loads=267×130/1000=34.71 kips

Divide by width of tributary area=4+7.5/12+15/12+7.5+15/12+4+7.5/12=19.25 feet

Load per unit length due to self weight of slab=34.71/19.25=1.80 k/ft

 

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