Deflection According to Pascal's Triangle
Pascalin kolmion mukainen taipuma
 This concept involves using Pascal's Triangle to determine the bending moments in beams subjected to various loading conditions. The mathematical second pendulum has the length of 99,5 cm. For the same purpose, we choose the HEB 100 beam length as 99.5 cm and the load in the middle as 1 kN.
1.00 - 1.03 - 1.06 - 1.12
99.5 x 1.03 = 103 cm
Deflection of the girder weight Deflection of the concentrated force
f = deflection f = deflection
f = 5 x F x L3 / 384 x E x I f = F x L3 / 48 x I x E
f = 0.00032 cm f = 0.00246 cm
E= 20600 kN/cm2 Ix = 450 cm4
Total deflection
0,00273 cm => 0,00273/ 0,00249 = 1,0962 = 1,1
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99.5 x 1.06 = 105.5 cm
Deflection of the girder weight Deflection of the concentrated force
f = deflection f = deflection
f = 5 x F x L3 / 384 x E x I f = F x L3 / 48 x I x E
f = 0.00035 cm f = 0.00264 cm
Total deflection
0,00298 cm => 0.00298/ 0.00249 = 1.19876 = 1.21
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99.5 x 1.09 = 108.5 cm
Deflection of the girder weight Deflection of the concentrated force
f = deflection f = deflection
f = 5 x F x L3 / 384 x E x I f = F x L3 / 48 x I x E
f = 0.00039 cm f = 0.00287 cm
Total deflection
0.00326 cm => 0.00326/ 0.00249 = 1.30776 = 1.331
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99.5 x 1.12 = 111.5 cm
Deflection of the girder weight Deflection of the concentrated force
f = deflection f = deflection
f = 5 x F x L3 / 384 x E x I f = F x L3 / 48 x I x E
f = 0.00043 cm f = 0.00311 cm
Total deflection
0.00354 cm => 0.00354/ 0,00249 = 1.4234 = 1.4641
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1.03 => 1.1(2) 1.03(28)3 = 1.12
1.06 => 1.21 1.03(28)6 = 1.21
1.09 => 1.331 1.03(28)9 = 1.331
1.12 => 1.4641 1.03(28)12 = 1.4729
Later, we can combine the above in a greater whole.
21.6.2018*08:00 (36628 - 824)
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