bridge

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da bridge

* version 1
- upper member is 725mm long
- 50mm at the thinnest members
- two 1/2 inch rebar,
  placed ~18mm from the bottom of the top member

[[https://bcitaci.club/catenarybridge_full.svg]]

* footing loads and horizontal support
:PROPERTIES:
:ID:       411fb925-2fe6-418e-9bcb-82bbb5d034cc
:END:
the point of a catenary is that its supposed to be in pure compression,
i.e the force vector at the footings should directed along the angle of the catenary at the footings.

need to get the data from our openscad model on the flash drive,
but it looks like the angle is about 15 degrees.

Then given a supported load $B$,
each footing supports $B/2$,
and a coefficient of static friction $0.6$ implies that it will resist horizontal load $0.3B$

the horizontal force from the angle 15 degrees is $\sim0.13B$
So the bridge should be secure without additional horizontal support, when placed on concrete.

basically we need coefficient of friction > $\tan(\theta)=0.26$

Friction of concrete against steel might be ~0.3-0.7?
but it can be worse ?

What about if placed on soil?
[[id:1d521a37-86e1-4b38-9fca-f6bdd5cd96bf][boussinesq solutions]]?

* simple beam-component strength estimate
:PROPERTIES:
:ID:       93f81d43-4314-43e8-a1ea-124378c14f71
:END:
consider center point-loading of a beam with height 50mm, width 75mm, length 725mm.

suppose the steel reinforcement is 2 rods with a diameter of 12.7mm, 20mm from the bottom of the beam

** pure concrete beam
:PROPERTIES:
:ID:       4c193c21-60d6-45d9-a925-bd12653667df
:END:
based on 3MPa modulus of rupture,
the [[id:57a8c77a-c98f-4a1c-ac57-8a4b356118dc][splitting tension test calculation]]
gives a maximum supported force of
#+begin_src julia
   uconvert(u"N",3u"MPa"*2*75u"mm"*(50u"mm")^2/(3*725u"mm"))
#+end_src

#+begin_example
517.2413793103449 N
#+end_example

** with rebar, assuming cracking
:PROPERTIES:
:ID:       309c3353-64a7-41c7-a45e-48e86acd458c
:END:
Let $y=0$ be the side with rebar,
and let $y=y_0$ be the neutral surface,

Let the concrete crush strength be 30MPa,
the streel's tensile strength be 200MPa,

Assume $n=8$ for the steel's modulus over the concrete
#+begin_src julia
  @syms y y0
  cIcrc(u) =integrate(u,(y,y0,50),(x,0,75))
  cIcrs(u) =0.5*8*(3.14*6.35^2)*(subs(u,x=>25,y=>18)+subs(u,x=>50,y=>18))
  y0v=cIcrc(y-y0)+cIcrs(y-y0) |> simplify |> (u->solve(u,y0)[1])
#+end_src

#+begin_example
31.1520370152924
#+end_example

#+begin_src julia
  Iv=cIcrc((y-y0)^2)+  cIcrs((y-y0)^2) |> subs(y0=>y0v)
#+end_src

#+begin_example
342599.125813200
#+end_example

for a 30MPa crush strength concrete,
by [[id:0c49d881-25e0-4e93-b1f9-a8e8cc3ad365][elastic flexure formulas]]

#+begin_src julia
crushload=uconvert( u"N" , 30u"MPa"*(Iv*1u"mm^4")/((75-y0v)*1u"mm")*4/725u"mm")
#+end_src

#+begin_example
1293.24279495813 N
#+end_example

the stress in the steel is
#+begin_src julia
  uconvert(u"MPa", (crushload*725u"mm"/4)*8*(y0v-18)*1u"mm"/(Iv*1u"mm^4"))
#+end_src

#+begin_example
71.9871270820728 MPa
#+end_example

So we put in way more steel than required,
since it has a yield strength >250MPa.

whether fibre reinforcement would help is more complicated lol

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