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

footing loads and horizontal support

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 $∼0.13B$ So the bridge should be secure without additional horizontal support, when placed on concrete.

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

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

What about if placed on soil? boussinesq solutions?

simple beam-component strength estimate

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

based on 3MPa modulus of rupture, the splitting tension test calculation gives a maximum supported force of

uconvert(u"N",3u"MPa"*2*75u"mm"*(50u"mm")^2/(3*725u"mm"))

with rebar, assuming cracking

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

@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])

31.1520370152924

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

342599.125813200

for a 30MPa crush strength concrete, by elastic flexure formulas

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

1293.24279495813 N

the stress in the steel is

uconvert(u"MPa", (crushload*725u"mm"/4)*8*(y0v-18)*1u"mm"/(Iv*1u"mm^4"))

71.9871270820728 MPa

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

This tells me that the bridge would benefit from using fibre reinforcement to improve the crush strength.