... + 32 (1 + 2U2 + U3 )U2 − 32 (U2 + U3 )U3 = − , − 32 (U2 + U3 ) + 32 (U2 + 2U3 )U3 = − Solving these yields U 22 = 49 , U 32 = 19 21 2 The Finite Element Method Just as in Example 5.5, the positive ... after the conditions U11 = 0, U21 = 29 , U31 = 89 , U41 = 2, U 12 = 1, U 42 = are enforced, becomes ⎤ ⎥ ⎢ ⎢ U 22 ⎥ + ⎣ U 32 ⎦ ⎡ which yields U 22 = 11 , U 32 = 17 29 46 29 29 46 29 ⎡ ⎤ ⎡ ⎤ 0 2 ... Performing the integrations, ⎤ ⎡ 156 22 h 54 −13h kh ⎢ 4h2 13h −3h2 ⎥ ⎥ + EI ⎢ ke = ⎣ 156 22 h ⎦ h3 420 sym 4h2 fe = fh [−6 12 −h h dξ, 4EI N N h2 i j ⎡ ⎢ ⎢ ⎣ 12 − 12 −6h 12 6h 4h2 sym ⎤ 6h 2h2 ⎥ ⎥,...