Chamfered joists

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I'm looking at doing a loft conversion. The floor joists will need to cover a 4m odd span from the spine wall to the wallplate. Unfortunately a pretty severe chamfer will be needed to fit the joists onto the wallplate. I gather the concern is shear stress, but can't find what the limit would be and if there are any ways to mitigate this issue. Thanks in advance for any information.

IMG20231211235138.jpg
 
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So doing a simple analysis of the shear at the support - effectively at your support you have a beam to resist shear of 70mm depth, allowable shear stress is 0.74 or 0.78N/mm2 depending on wood grade. Assuming your joist spacing is 400mm are 50mm wide and standard floor loading then the actual stress will be 0.61N/mm2 so all would be ok
 
You could double up use C24 and glue and screw the floor to strengthen it overall

Blup
 
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Or you could bolt some 4x2 to the wall face in between your ceiling joists and increase the bearing.
 
So doing a simple analysis of the shear at the support - effectively at your support you have a beam to resist shear of 70mm depth, allowable shear stress is 0.74 or 0.78N/mm2 depending on wood grade. Assuming your joist spacing is 400mm are 50mm wide and standard floor loading then the actual stress will be 0.61N/mm2 so all would be ok
Useful thank you. Would you mind sharing how you get to 0.61N/mm2? The clear span will be 4.65m if that impacts it
 
Useful thank you. Would you mind sharing how you get to 0.61N/mm2? The clear span will be 4.65m if that impacts it
Imposed loading on floor 1.5kn/m2 , dead loading 0.25Kn/m2. Load on a joist per metre run when spaced at 400mm (1.5+0.25) x 400/1000 = 0.7kn/m run. Total load acting at one end of 4m joist = 4/2 x 0.7 = 1.4kn. Shear stress 1.4x1000/(50X75) = 0.4N/mm2, This figure then multiplied by 1.5 FOS to give 0.6N/mm2 . My fig of 0.61 took SW of joist into account. 4.65m -actual stress would then increase by a factor of 4.65/4.00
 
Or you could bolt some 4x2 to the wall face in between your ceiling joists and increase the bearing.
That would only make a slight difference as the joist would then span 3.9m instead of 4m, so would reduce its load and hence shear value and associated stresses by a factor of 3.9/4.
 
Imposed loading on floor 1.5kn/m2 , dead loading 0.25Kn/m2. Load on a joist per metre run when spaced at 400mm (1.5+0.25) x 400/1000 = 0.7kn/m run. Total load acting at one end of 4m joist = 4/2 x 0.7 = 1.4kn. Shear stress 1.4x1000/(50X75) = 0.4N/mm2, This figure then multiplied by 1.5 FOS to give 0.6N/mm2 . My fig of 0.61 took SW of joist into account. 4.65m -actual stress would then increase by a factor of 4.65/4.00
Thanks for the explanation. Interesting to see how the final figure can be altered by using smaller centres and /or wider joists.
 
That would only make a slight difference as the joist would then span 3.9m instead of 4m, so would reduce its load and hence shear value and associated stresses by a factor of 3.9/4.
I think the idea here was to increase the portion of the joist sitting on the support, which would mean the joist depth over the support would increase
 

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