Question

Plates and slabs: In the unenclosed situation, this middle part of this tube can be described by this differential equation in the cylindrical axis system $r$, $\phi$, $z$): $\frac{d^2u_r}{dr^2} + \frac{1}{r}\frac{du_r}{dr} - \frac{1}{r^2}u_r = 0$ With the general solution of the displacement field: $u_r(r) = \frac{A}{r} + B \cdot r + C$ $u_\phi(r) = 0$ with these kinematic equations: $\epsilon_{rr} = \frac{du_r}{dr}$ $\epsilon_{\phi\phi} = \frac{u_r}{r}$ E Modules [N/mm$^2$] 210.000 Transverse contraction coefficient 0,3 Duct yield strength [N/mm$^2$] 235 $r_{inner}$ [m] 0,10 $r_{outer}$ [m] 0,11 Overpressure p [bar] 220 1. What are the tangential stresses at [radius]inner and [radius]outer? 2. What are the radial stresses at [radius]inner and [radius]outer Please calculate by hand so I can follow. Solution would help me so much, thanks! $\sigma_{rr} = E \cdot \epsilon_{rr} = E \cdot \frac{du_r}{dr}$

          Plates and slabs:
In the unenclosed situation, this middle part of this tube can be described by this differential
equation in the cylindrical axis system $r$, $\phi$, $z$):
$\frac{d^2u_r}{dr^2} + \frac{1}{r}\frac{du_r}{dr} - \frac{1}{r^2}u_r = 0$
With the general solution of the displacement field:
$u_r(r) = \frac{A}{r} + B \cdot r + C$
$u_\phi(r) = 0$
with these kinematic equations:
$\epsilon_{rr} = \frac{du_r}{dr}$
$\epsilon_{\phi\phi} = \frac{u_r}{r}$
E
Modules
[N/mm$^2$]
210.000
Transverse
contraction
coefficient
0,3
Duct yield
strength
[N/mm$^2$]
235
$r_{inner}$
[m]
0,10
$r_{outer}$
[m]
0,11
Overpressure
p [bar]
220
1. What are the tangential stresses at [radius]inner and [radius]outer?
2. What are the radial stresses at [radius]inner and [radius]outer
Please calculate by hand so I can follow. Solution would help me so much,
thanks!
$\sigma_{rr} = E \cdot \epsilon_{rr} = E \cdot \frac{du_r}{dr}$

Plates and slabs:
In the unenclosed situation, this middle part of this tube can be described by this differential
equation in the cylindrical axis system r, ϕ, z):
(d^2ur)/(dr^2) + (1)/(r)(dur)/(dr) - (1)/(r^2)ur = 0
With the general solution of the displacement field:
ur(r) = (A)/(r) + B · r + C
uϕ(r) = 0
with these kinematic equations:
ϵrr = (dur)/(dr)
ϵϕϕ = (ur)/(r)
E
Modules
[N/mm^2]
210.000
Transverse
contraction
coefficient
0,3
Duct yield
strength
[N/mm^2]
235
rinner
[m]
0,10
router
[m]
0,11
Overpressure
p [bar]
220
1. What are the tangential stresses at [radius]inner and [radius]outer?
2. What are the radial stresses at [radius]inner and [radius]outer
Please calculate by hand so I can follow. Solution would help me so much,
thanks!
σrr = E ·ϵrr = E ·(dur)/(dr)

Added by Patrick G.

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