Super-elevation, raising the outside rail relative to the inside like a banked racetrack (eg Brooklands). This reduces the tendency of the loco to topple over on corners. The centripetal force required to make a mass go in a curve is mass * velocity squared / radius (F1 = M*V^2/R). The ^ sign means 'raised to the power of'. The force acting downwards on the mass is mass* gravity (F2 = M*g). If we add these forces together we get a resultant force Fr = sqrt(F1^2 + F2^2) and the angle of this force from vertical is arcTan(F1/F2). If you tip the track at this angle the passengers will not perceive any side force, like leaning a push bike on a corner. However if you come to a standstill there will be be a tendency to topple inwards, so it seems reasonable to use half this angle, then the perceived out-force when going at speed round the bend is the same as the perceived in-force when stationary.
Speed in metres per second, radius in metres mass in kg, although this actually cancels out. g is 9.81 m/s/s. If you really want to do it in mph and feet we need to introduce all sorts of silly factors.
If a 14st man sat in the middle of your proposed span the track would deflect 20mm, which I suggest is quite a lot, and the stress would be 266 N/mm^2 which is definitely a lot, and we've not allowed for the loco , truck and passengers yet. Here's screen dump of the sums
| h |
|
|
25.0 |
mm |
|
| b |
|
|
6.0 |
mm |
|
| second moment of area |
|
|
7812.5 |
Mm^4 |
|
| section modulus |
|
|
625.0 |
|
|
| young’s mod |
|
|
207000.0 |
|
|
| span |
60 |
in |
1524.0 |
mm |
|
| load |
98 |
lb |
44.5 |
kg |
|
| |
|
|
436.5 |
N |
|
| |
|
|
|
|
|
| deflection |
|
|
19.9 |
mm |
|
| bending moment |
|
|
166306.5 |
Nmm |
|
| stress |
|
|
266.1 |
N/mm^2 |
|
Edited By duncan webster on 04/11/2019 13:20:53
Edited By duncan webster on 04/11/2019 13:22:30
Edited By duncan webster on 04/11/2019 13:23:59