... upright and perhaps later a 6/7 ft grand. The room will have a wooden floor and I am given to understand that a normal wooden floor is designed to take a deadload of 1.5 kN/m² (kN is kilo Newtons) I have no clear idea of what this means. Does anybody happen to know how many kN/m² might be exerted by a grand of this size?
I am not a construction engineer so the following is strictly a simplistic novice application of Physics to the complex computations of Load Distribution. It may help you explain your concerns to the engineer and ask him/her to provide you with a convincing design that mitigates your concern.
1 Newton (N) is 0.101972 kgf.
Kilo (k) = 1000.
1.5 kN/m-2 = 1.5 x 1000 x 0.101972 kgf / m-2 = 152.96 kgf / m-2
Let us assume that the 152.96 kgf
is distributed evenly over 1 square meter, then we have approx. 0.0153 kgf / cm-2
Let us now apply this to a hypothetical piano:
1. Steinway Model B, 211 cm, 345 kg (kgf).
2. Length is 211cm, width is 148cm.
3. Approx. area is 31228 cm-2 or 3.12 m-2 (square meters).
4. Given shape of a grand, it is possible this area is evenly distributed under each foot.
Let us assume, simplistically, that the piano's weight is distributed to three legs evenly; then, each leg of the Model B will exert 115 kgf
down through the wheel to the floor. Exactly how this load is distributed by Building Engineering is beyond my expertise. If this force is evenly distributed to a square meter
, then the the floor deadload specification of 152.96 kgf / m-2
is larger than the piano load of 115-kgf / m-2
For the sake of this discussion, let us further assume, simplistically, that the contact surface of the wheel is 5mm wide by 10 cm long yielding a contact surface area of 5 square centimeter (cm-2).
This means we have a theoretical 115-kgf down through 5 cm-2 of contact surface, or 23 kgf / cm-2
Each contact point of the Model D is exerting a deadload force of 23 kgf / cm-2
on the floor. If this force is not evenly distributed to a square meter
, then the piano load of 23 kgf / cm-2
is much much larger than the floor deadload specification of 0.0153 kgf / cm-2
Again, exactly how this load is distributed by Building Engineering is beyond my expertise.
Hopefully, this can help you start a conversation with the Building Engineer.
Good luck and I hope this is useful.