Mathematics and Origami

The resolution of the already mentioned differential equation bears on the formula that
gives the maximum sagita (at its free end) of a cantilever beam that has embedded the other end
and is subjected to a uniform load along its entire length:

EIz

ql

f

8

4

= ; from which we get E:

I f

l q

E

(^8) z
4

with:
l, free paper length in mm.
q, kilograms / mm as paper ́s unitary own weight along the l dimension.
Iz, the momentum of inertia of the paper beam section with respect to its axis z, ex-
pressed in mm^4 (see in Fig. 3 the section of the paper sheet).
f, the paper ́s free flexing sagita expressed in mm.
Figs. 1 and 2 show the approach of both experiments, in accordance with each of the
axes of the Din A4 rectangular paper.
We have to take into consideration that, in any piece of paper it is predominant the di-
rection in which that paper has been rolled: i.e. the paper is an anisotropic material because it
has different molecular orientation and therefore different mechanical behaviour depending on
the rolling direction. Hence, we shall have two different E values.
Let ́s see how q and Iz have been obtained, since they are not shown in Figs. 1 and 2.
To obtain the paper thickness (Fig. 3) we took 900 grams of A4 sheets that came out to
be in a quantity of 181 with a pile height of 20 mm; hence e 0 , 11 mm
181
20
= =.
At the same time we ́ll have:
q 2 , 3677979 10 Kg/mm
181 210
0 , (^95)
1
= × −
×

q 1 , 6742006 10 Kg/mm
181 297
0 , (^95)
2
= × −
×

According to Fig. 3, the momentum of inertia of that paper section is:
12
be^3
Iz=
Therefore:
4
3
1 0 ,^0329422
12
297 0 , 11
Iz = mm
×
= ;^4
3
2 0 ,^0232925
12
210 0 , 11
Iz = mm
×

Obviously, the value of sagitas is:
f 1 = 64 − 28 = 36 mm ; f 2 = 64 − 24 = 40 mm
Substituting values we have:
b
3
y axis
z axis e