greggery
Bioengineer
- Dec 9, 2018
- 2
Eccentric loading on a timber wall
I am presently working on my own home conversion/extension and framing a timber gable wall which will have an exterior covering of vertically hung tiles.
Last week I started wondering (panic) if I had monumentally underestimated the impact of the eccentricity of the load on this external stud wall (and the torque it creates). Having attempted to work it out I was surprised by how little the force is.
However, It’s a very long while since I did (what at the time was quite basic) mechanics and statics - and so, I was hoping, a more experienced eye might have an opinion on the way I have tried to resolve this.
The stud is 0.14m wide. The gable is 3.6m at its peak. The wall only has an internal finish on battens to 2.4m height.
I have tried to resolve this in a similar way to the ‘leaning ladder’ (with no friction at the wall (taking the fulcrum as being the center of the stud at the bottom of the wall) . I have also tried to resolve it as a simple moment midway up the wall (where the center of gravity for the applied load exists)
I have attached a sketch of the wall.
Thoughts, corrections, jibes or input of any kind would be hugely appreciated.
My workings
Loads on gable wall (I have excluded the studwork itself as its all plumb – I built it!)
External rotation
Vertically hung clay tiles 0.55 kN/m2 (3.6 m height) = 2 kN/ linear m of wall . Center of gravity applied at 1.8m. The load is 0.07m from the face of the stud (at 0.140m from the center line at the base of the wall)
Angle ᴓ 4.5 degree (0.16kN)
12mm ply sheath 0.2 kN/m2 and Batten 0.1 kN/m2 (3.6 m height) = 1 kN / linear m of wall at applied 1.8m at 0.08m from center line
Angle ᴓ 2.5 (0.044kN)
Sum at 1.8m = 0.2kN = 0.36kN m
Internal rotation
Plasterboard and batten 0.18 kN/m2 (2.4m height) = 0.45 kN/m2 at 1.2m 0.2m from centerline
Angle ᴓ -9.5
Sum at 1.2m = 0.08 kN = 0.1 kN m
Residual torque = 0.26 kN m
Resisted by F at 3.6m (top of the wall) = force = 0.07kn per m wall.
Thanks
I am presently working on my own home conversion/extension and framing a timber gable wall which will have an exterior covering of vertically hung tiles.
Last week I started wondering (panic) if I had monumentally underestimated the impact of the eccentricity of the load on this external stud wall (and the torque it creates). Having attempted to work it out I was surprised by how little the force is.
However, It’s a very long while since I did (what at the time was quite basic) mechanics and statics - and so, I was hoping, a more experienced eye might have an opinion on the way I have tried to resolve this.
The stud is 0.14m wide. The gable is 3.6m at its peak. The wall only has an internal finish on battens to 2.4m height.
I have tried to resolve this in a similar way to the ‘leaning ladder’ (with no friction at the wall (taking the fulcrum as being the center of the stud at the bottom of the wall) . I have also tried to resolve it as a simple moment midway up the wall (where the center of gravity for the applied load exists)
I have attached a sketch of the wall.
Thoughts, corrections, jibes or input of any kind would be hugely appreciated.
My workings
Loads on gable wall (I have excluded the studwork itself as its all plumb – I built it!)
External rotation
Vertically hung clay tiles 0.55 kN/m2 (3.6 m height) = 2 kN/ linear m of wall . Center of gravity applied at 1.8m. The load is 0.07m from the face of the stud (at 0.140m from the center line at the base of the wall)
Angle ᴓ 4.5 degree (0.16kN)
12mm ply sheath 0.2 kN/m2 and Batten 0.1 kN/m2 (3.6 m height) = 1 kN / linear m of wall at applied 1.8m at 0.08m from center line
Angle ᴓ 2.5 (0.044kN)
Sum at 1.8m = 0.2kN = 0.36kN m
Internal rotation
Plasterboard and batten 0.18 kN/m2 (2.4m height) = 0.45 kN/m2 at 1.2m 0.2m from centerline
Angle ᴓ -9.5
Sum at 1.2m = 0.08 kN = 0.1 kN m
Residual torque = 0.26 kN m
Resisted by F at 3.6m (top of the wall) = force = 0.07kn per m wall.
Thanks