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Direct Time Integration (4)<br />

• The final algorithm reads:<br />

u = u +∆t⋅v<br />

n+ 1 n n+<br />

1/2<br />

u M ( f B σ dV )<br />

= −∑ ∫<br />

n+ 1 − 1 ext( n+ 1) T e( n+<br />

1)<br />

e e( n+<br />

1)<br />

V<br />

v = v +∆t⋅<br />

u<br />

n+ 3/2 n+ 1/2 n+<br />

1<br />

A new configuration is obtained first. On this known configuration,<br />

equilibrium is enforced. The new mid-step velocity is obtained last.<br />

• This scheme is explicit.<br />

How does one<br />

( 1)<br />

obtain σ en+<br />

?<br />

(see below)<br />

• If ∆t varies in time, the only change is in the third<br />

equation, which becomes:<br />

n n+<br />

1<br />

n n+<br />

1 n<br />

n+ 3/2 n+ 1/2 ∆ t +∆t<br />

n+<br />

1<br />

∆t ≡t −t<br />

v = v + ⋅ u with:<br />

2<br />

∆t ≡t −t<br />

n+ 1 n+ 2 n+<br />

1<br />

17<br />

Scheme start-up and marching<br />

u 0<br />

0<br />

σ<br />

u , u<br />

0 0<br />

step 0 1 2<br />

0<br />

t<br />

0 0 0<br />

∆t<br />

u , σ , u<br />

, ∆t<br />

are given<br />

n =−1<br />

0 −1 ext int 0<br />

u<br />

= M ( f − f )<br />

u = u +∆t⋅v<br />

1/2<br />

v<br />

n+ 1 n n+<br />

1/2<br />

u<br />

= M ( f − f )<br />

n+ 1 − 1 ext int n+<br />

1<br />

v = v +∆t⋅<br />

u<br />

n+ 3/2 n+ 1/2 n+<br />

1<br />

u 1<br />

1<br />

σ<br />

1<br />

u<br />

t<br />

3/2<br />

v<br />

u 2<br />

2<br />

σ<br />

2<br />

u<br />

1/2 3/2 N<br />

fin<br />

v = u<br />

+ ( ∆t/2)<br />

⋅u<br />

n← n+<br />

1<br />

1 0 1/2<br />

u = u +∆t⋅v<br />

u<br />

= M ( f − f )<br />

1/2 0 0<br />

1 −1 ext int 1<br />

v = v +∆t⋅<br />

u<br />

etc.<br />

3/2 1/2 1<br />

t<br />

18<br />

t<br />

9

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