By Victor E Sauoma
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Extra info for Finite Elements Part 2 (solid mechanics)
9) where ρ is the mass density, c is the heat capacity per unit mass (amount of heat needed to raise a unit mass by one degree), k is the conductivity, and the temperature θ is the state variable. State also the boundary and initial conditions. Solution: 1. We consider a typical diﬀerential element of the slab. The element equilibrium requirement is that the net heat ﬂow input to the element must equal the rate of heat stored in the element. 10) qA − q + ∂x ∂t 2. 11) 3. Substituting from Eq. 11 into Eq.
F. f. 3 Direct Stiﬀness Method 1–9 2. After all the node boundary conditions have been read, assign incrementally equation numbers (a) First to all the active dof (b) Then to the other (restrained) dof, starting with -1. Note that the total number of dof will be equal to the number of nodes times the number of dof/node NEQA. 3. The largest positive global degree of freedom number will be equal to NEQ (Number Of Equations), which is the size of the square matrix which will have to be decomposed.
The global matrix can be rewritten as √ −P √ 0 R3 ? R4 ? ¾ = 12EI/L2 −6EI/L2 −12EI/L3 −6EI/L2 −6EI/L2 4EI/L 6EI/L2 2EI/L −12EI/L3 6EI/L2 12EI/L3 6EI/L2 −6EI/L2 2EI/L 6EI/L2 4EI/L ¿ ∆1 ? θ2 ? √ ∆3 √ θ4 4. Ktt is inverted (or actually decomposed) and stored in the same global matrix ¾ L3 /3EI L2 /2EI −12EI/L3 −6EI/L2 L2 /2EI L/EI 6EI/L2 2EI/L −12EI/L −6EI/L2 3 Victor Saouma 2 6EI/L 2EI/L 3 12EI/L 6EI/L2 ¿ 6EI/L2 4EI/L Finite Elements II; Solid Mechanics Draft 1–22 PREREQUISITE 5. Next we compute the equivalent load, Pt = Pt − Ktu ∆u , and overwrite Pt by Pt Pt − Ktu ∆u ¾ −P 0 0 0 = − L3 /3EI −12EI/L3 L2 /2EI 2 L /2EI 2 L/EI −12EI/L −6EI/L2 3 −6EI/L2 6EI/L 2 −P 0 0 0 2EI/L 3 6EI/L 2EI/L ¿ 6EI/L2 4EI/L 12EI/L 6EI/L2 −P 0 0 0 = 6.