# Metode Interpolation

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Penyelesaian perletakan sendi rol dengan metode interpolasi

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Diketahui perletakan seperti gambar: 0,4 2,3 3,7540 1 2 30,44q = 4 t/m 'A i!ang #men  \$% \$ &itung nilai ' pa!a ( = 0,4 m , ( = 2,3 m !an ( = 3,75 m !engan interp#lasi)*a+aban: enghitung m#men terha!ap ) 0 221 =×−×  l ql  Av     l = 4 m 0444  221 =××−  Av % =  Av  t#n1  ∑  (sepan,ang  MB 0 221 =−  qx Avx 02%  2 =−  x x ( = 0    ( = 0( = 1    ( = \$( = 2    ( = %( = 3    ( = \$( = 4    ( = 0*ika !iketahui ( = 0,4 m maka '  = - xy (M)∆ 1  yo∆ 2  yo∆ 3  yo∆ 4  yo 00 \$ 1\$.4 20 2%.40.203\$.4 .\$ 40  /= h xo x −   ( = 0,4  ( 0 = 0  h= 1= 104,0  −  = 0,4(0,4= 030200 321 21  yU U U  yU U  yU  y  ∆×−−+∆×−+∆×+ = 02324,014,04,0 4214,04,0 \$4,00  ××−−+−×−+×+ = 2,4  0,4%= 2,%% t2  *ika !iketahui ( = 2,3 m maka '  = - xy (M)∆ 1  yo∆ 2  yo∆ 3  yo∆ 4  yo 00 \$ 1\$.4 20 2%.40.203\$.4 .\$ 40  /= h xo x −   ( = 2,3  ( 0 = 2  h= 1= 123,2  − = 0,3(2,3= 030200 321 21  yU U U  yU U  yU  y  ∆×−−+∆×−+∆×+ = 02324,014,04,0 4213,03,0 23,0%  ××−−+−×−+−×+ = % . 0,\$  0,42= 7,%2 t*ika !iketahui ( = 3,75 maka ' = - xy (M)∆ 1  yo∆ 2  yo∆ 3  yo∆ 4  yo00 166 282-4 36-2-40 40-6-400 3  /= h xn x −   ( = 3,75  ( n = 0  h= 1= 1475,3  −  = .0,25(3,75= nnnn  yU U U  yU U  yU  y  32 321 21 ∆×+++∆×−+∆×+ = 42125,025,0 \$25,00  −×+−−+−×−+  = 1,5  0,375= 1,%75 t4
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