Dimensionally-inconsistent Calculations (Emperical formulae)

Sometimes calculations and formulae are based on curve-fitting the results of various tests. The resulting formula from the curve-fitting may assume input units with a particular scaling and produce outputs assumed to be in a different scaling.

import forallpeople as si
si.environment('default')

Dimensionally-inconsistent radicals

Observed in many concrete design codes is a value expressed as \(\sqrt{f'_c}\) where \(f'_c\) is in scaled units of stress (e.g. MPa) and \(\sqrt{f'_c}\) is also in units of stress (MPa in this example).

The intention behind this particular expression is to compute the square-root of 35 and for that value to be in units of MPa.

Strategy 1 - Manually add the units back on

import math
MPa = 1e6 * si.Pa
f_c = 35e6 * si.Pa

math.sqrt(f_c) * MPa

5.916 MPa

Strategy 2 - Define a custom sqrt function

import math
MPa = 1e6 * si.Pa

def sqrt(x):
    MPa = 1e6 * si.Pa
    if (
        isinstance(x, si.Physical)
        and x.dimensions == (1, -1, -2, 0, 0, 0, 0)
    ):
        return math.sqrt(x) * MPa # Special case
    else:
        return x**(1/2) # General case

f_c = 35 * MPa # Example of special case
area = 0.0350 * si.m**2 # Example of general case

display(
    f_c,
    sqrt(f_c)
)

display(
    area,
    sqrt(area)
)

35.000 MPa

5.916 MPa

35000.000 mm²

187.083 mm

Strategy 3

Continuing with the example above, a calculation may intend to compute the square root of 35e6 (instead of 35) and apply the same units to that resulting value.

This strategy can be combined with Strategy 2 to define a custom sqrt function for special cases.

import math
MPa = 1e6 * si.Pa
f_c = 35 * MPa

value, unit = f_c.split()
math.sqrt(value) * unit

5.916 kPa

Hidden units

Another kind of seemingly dimensionally-inconsistent formula may just be a case where units are assumed within a numerical value but not communicated.

An example commonly used in Canadian flexural concrete design:

\[A_s = 0.0015 (f'_c) (b) (d) - \sqrt{d^2 - \frac{3.85 M_f}{(f'_c) (b)}}\]

This formula for calculating the required area of flexural tension steel should result in values of area (say, in mm2). In its current form, however, it is dimensionally-inconsistent because the value of 0.0015 contains hidden units of 1 / MPa.

si.environment('default')
MPa = 1e6 * si.Pa

f_c = 35 * MPa
b = 0.30 * si.m
d = 0.60 * si.m
M_f = 200e3 * si.N * si.m

A_s = 0.0015 * f_c * b * (d - sqrt(d**2 - (3.85 * M_f)/(f_c * b)))
A_s # Resulting units are in N, which are not expected

1.017 kN

Here, we will manually add the units into the calculation as (0.0015 / MPa).

A_s = (0.0015 / MPa) * f_c * b * (d - sqrt(d**2 - (3.85 * M_f)/(f_c * b)))
A_s # These are the correct dimensions

1017.251 mm²

Completely inconsistent formulae

Last, there are times where the formula essentially serves as a numeric transformation and the units simply need to be manually applied.

An example from the Canadian wood design code:

\[QS_i = 14t \sqrt{\frac{d_e}{1 - \frac{d_e}{d}}}\]

This formula for calculating tension resistance in the parallel-to-grain direction of a bolted timber connection is based on empirically-measured fracture test results.

All symbols (t, d_e, and d) are numbers, scaled to units of mm, and the resulting value of QS_i is a number in units of force, N.

Because this formula is a strict numeric computation with no dimensional consistency, it is best to treat it as such by manually removing units from the inputs and adding the units to the output.

si.environment('default')
import math

# Use .prefix to force the numerical value to scale in mm
t = (0.354 * si.m).prefix('m')
d = (0.620 * si.m).prefix('m')
d_e = (0.480 * si.m).prefix('m')

t = float(t)
d = float(d)
d_e = float(d_e)

QS_i = 14 * t * math.sqrt(d_e / (1 - d_e/d)) * si.N
QS_i

228.499 kN