Interpolation
To check for maximum bending stress in flat plates — this for a hanger clamp, Roark’s formulas are handy. Here’s a table that offers a recipe.
Manually, this is straightforward: First, I calculate the aspect ratio (a / b) of the flat plate, and then pick a value of β_{1} that corresponds to the ratio. If I do not find a ready match, then I perform a linear interpolation between the two values that form the intermediate lower and upper boundaries of (a / b).
For example, if my plate size is 200×100, then (a / b) is 2.0 (in the sixth column), whose corresponding β_{1} would then be equal to 1.226.
But if my plate size is 220×100, then (a / b) becomes 2.2. Now, the nearest lower and upper boundary values in the table are 2.0 and 3.0 respectively. Since Roark’s table does not readily offer a corresponding β_{1} value, this further requires interpolation — let’s stick to linear, for simplicity.
$$ \frac{p}{q} = \frac{P}{Q} $$
$$ p = \frac{P \cdot q}{Q} $$
$$ = \frac{(2.106  1.226) \cdot (2.2  2.0)}{(3.0  2.0)} $$
$$ \therefore \beta_1 = 1.226 + p $$
How to automate this in python? Here’s one way:
from bisect import bisect
# Roark's Table 11.4, Case 10, with three edges fixed
# a / b, and b1 are defined as lists below:
raob = [0.25, 0.50, 0.75, 1.00, 1.50, 2.00, 3.00]
b1 = [0.020, 0.081, 0.173, 0.321, 0.727, 1.226, 2.105]
a, b = input("Enter a, b: ")
a, b = float(a), float(b)
aob = a / b
try:
i = raob.index(aob)
beta1 = b1[i]
except ValueError:
# aob did not match any value in raob
i = bisect(raob, aob)
j = i  1
P = b1[i]  b1[j]
q = (aob  raob[j])
Q = (raob[i]  raob[j])
beta1 = b1[j] + (P * q / Q)
finally:
print "a/b : ", round(aob, 3)
print "Index of a/b ratio: ", i
print "beta1 : ", round(beta1, 3)
pass
And here’s how the above code works:
 First, I import
bisect
standard library for referencing across two lists (raob
, andb1
).  Get user input for plate size in terms of
a
(length) andb
(width).  Then, in the
try
statement, I attempt to find the index,i
(column number) that matches the aspect ratio (a / b).^{1} Finding the index is key, which lets me refer to the corresponding value in the second list.  If an exact numerical match of (a / b) ratio is not found in the
raob
list, then the control jumps to process theexcept ValueError
part.  Using
bisect
, I try and find the nearest next numerical value and its corresponding index number. So to take the aforementioned example, if my (a / b) ratio turns out to be 2.2, then the next numerical value that bisect finds for me is 3.0, and its corresponding i value, which is 6 in this case.^{2}  Rest of the code thereafter deals with linear interpolation between two limits (2.0 and 3.0 in this case, and their corresponding β_{1} values) to get the β_{1} value (corresponding to 2.2 in this example); and then print those values.
The only shortcut I’ve taken in this crude — doesn’tcatchallerrors — code yet I think, is in the exception part. I’ve used that to move the control from simply throwing up ValueError
to the next part of the code where the interpolation occurs. (Note: I haven’t addressed the IndexError
that would occur — when the ratio is either lower than 0.25 or greater than 3.0 — in this code.) Until I work out a more elegant way to do the above, this I think will do for now.

For instance, the index of 0.75 in the
raob
list is 2. (Index numbers in python start with 0, not 1.) ↩ 
For tracing index, and the subsequent interpolation, I preferred using
bisect
from the standard library (instead of using a custom package likenumpy
that comes with better interpolation tools), so I could keep the number of dependencies (and package installation requirements) to a minimum in order to run this code. Without having to resort to using a sledge hammer to kill a mosquito, I’m happy with it so far. ↩