a28c1ff5-e7b3-4941-938a-acac115e571a 40 2 5 decimal

&[DATE] &[TIME] - &[FILENAME]

4.2.5 Steel temperature development according to Eurocode 1993-1-2:2005

b 170 :

h 360 :

A.m 4 b * 2 h * + : 1400

Section is IPE360

A.m.b 2 b h + (* : 1060

4.26a k.sh 0.9 A.m.b V / A.m V / / * : 0.68

V 72701000 / :

A.m V / 192.57

θ.est 500 :

3.2a/b/c/d c.a θ.est 600 < 425 7.73 101^/ θ.est * + 1.69 103^/ θ.est 2^* - 2.22 106^/ θ.est 3^* + θ.est 735 < 66613002738 θ.est - / + θ.est 900 < 54517820 θ.est 731 - / + 650 if if if :

c.a 666.5

ρ.a 7850 :

3.1 t h.net t h.net.c t h.net.r + :

3.2 t h.net.c α.c t Θ.g t Θ.m - (* :

α.c 25 :

3.4 t Θ.g 203458 t * 1 + log10 * + :

t Θ.m 20 :

I just realized Θ.m is actually θ.a(t) which is what I'm trying to find. Does it complicate the matter when θ.a(t) is fed into the DE again? I guess that explains why the slope flattens in the Robot Structural Analysis snap, while the graph ploted in Smath did not.

What would be the correct way to define Θ.m(t) given it is actually θ.a(t)?

3.3 t h.net.r Φ ε.m * ε.f * σ * t Θ.r 4^t Θ.m 4^- (* :

Φ 1.0 :

ε.m 0.8 :

ε.f 1.0 :

σ 5.67 108^/ :

t Θ.r t Θ.g :

Problem 1: How can I solve the differential equation 4.25 with respect to θa(t)? I want to be able to calculate the development of temperature in a steel member as a function of time.

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Problem 2: How can I solve the following equation 4.22 with respect to μ0 using the built in solvers in Smath? Below I have solved the equation manually. Is there a way to do that with a solver in Smath?

μ.0 0.8 :

4.22 θ.a.cr 39.19 1 0.9674 μ.0 3.833^* / 1 - ln * 482 + (: 496.05

4.22 solved by μ0 manually μ.0 1 e θ.a.cr 482 - 39.19 /^1 + / 0.9674 / 3.833 nthroot : 0.8

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4.25 t Δθ.a k.sh c.a ρ.a * / A.m V / * t h.net * Δt * :

k.sh c.a ρ.a * / A.m V / * 100 h.net * UnitsOf 1

Units are removed

t θ.a t diff k.sh c.a ρ.a * / A.m V / * t h.net * ≡

Differential equation?

0 θ.a 500 ≡

k.sh c.a ρ.a * / A.m V / * 1 h.net * UnitsOf 1

What about units?

Units are removed

t θ.a D k.sh c.a ρ.a * / A.m V / * t h.net * :

00 D 0

θ.a.init 20 :

from 0 to 60 min

t0 0 :

tmax 60 :

res θ.a.init 060720 t θ.a D Rkadapt :

Solving ODE

res 020 0.0833 20.0022 0.1667 20.0076 0.25 20.0153 0.3333 20.0249 0.4167 20.036 0.5 20.0484 0.5833 20.062 0.6667 20.0767 0.75 20.0924 0.8333 20.1089 0.9167 20.1264 1 20.1446 1.0833 20.1636 1.1667 20.1833 1.25 20.2037 1.3333 20.2247 1.4167 20.2464 1.5 20.2688 1.5833 20.2917 1.6667 20.3151 1.75 20.3392 1.8333 20.3638 232 mat

The code EN 1993-1-2 states that Δt≤5seconds 60*60/5=720 steps. Would that be correct?

First column time in 'minutes, second temperature in 'K

res