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Analisa struktur dibawah ini dengan menggunakan metoda "kolom analogi" serta gambarkan bidang M, D, dan N !
| Diket : | L1 = | 4 | m |
| L2 = | 5 | m | |
| P = | 4 | t | |
| q = | 2 | t/m |
| Mc = Px1/2L2 = | 10 | tm | |||
| Mb={P(L1x1/2L2)}+{1/2xqxL1xL1}= | 42 | tm | |||
| Ma = Mb = | 42 | tm | |||
| A I = 1/2xMcx1/2L2 = | 12.5 | ||
| A II=0ᶴ4 (2x2-16x+42)dx = | 82.66 | ||
| A III = Ma x L1 = | 168 | ||
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| Diagram Ms tekan |
 |
| Titik berat |
| A I = L1 x 1 = | 4 | m2 | |
| A II = L1 x 1 = | 4 | m2 | |
| A III = L1 x 1 = | 4 | m2 | |
| A IV = L2 x 1 = | 5 | m2 | |
| A V = L2 x 1 = | 5 | m2 | |
| ∑A = | 22 | m2 | |
| y=∑Ai.yi/∑Ai | x=∑Ai.xi/∑Ai | |||||||
| 22 y =(4x3)+(4x5)+(4x3)+(5x5)+(5x2.5) | 22 x =(4x0)+(4x2)+(4x4)+(5x6.5)+(5x9) | |||||||
| y = 3.6 | x = 4.64 | |||||||
BAT A y Iox Ay2 Ioy x Ax2
I 4 0.7 5.33 1.96 0 4.64 86.1184
II 4 1.3 0 6.76 5.33 2.64 27.8784
III 4 0.7 5.33 1.96 0 0.64 1.6384
IV 5 1.3 0 8.45 10.42 1.86 17.298
V 5 1.2 10.42 7.2 0 4.36 95.048
| ∑Iox = | 21.08 | m4 | ∑Ioy = | 15.75 | m4 | |
| ∑Ay2 = | 26.33 | m4 | ∑Ax2 = | 227.98 | m4 | |
| Ix = | 47.41 | m4 | Iy = | 243.73 | m4 |
| Ixy = | 15.126 |
I'x = Ix(1-I^2xy/IxIy)= 46.47 m4
I'y = Iy(1-I^2xy/IxIy)= 238.91 m4
P x y My=Px Mx=Py
-12.5 0.19 1.3 -2.38 -16.25
-82.66 -3.16 1.3 261.21 -107.46
-168 -4.64 -0.7 779.52 117.60
Jumlah :
-263.16 1038.35 -6.11
M'x=Mx-My x Ixy/Iy = -70.55
M'y=My-Mx x Ixy/Ix = 1040.30
Titik x y Ms P/A M'yX/I'y M'xY/I'x Mi M
A -4.64 -2.7 42 -11.96 -20.20 4.10 -28.07 70.07
B -4.64 1.3 42 -11.96 -20.20 -1.97 -34.14 76.14
C -0.64 1.3 10 -11.96 -2.79 -1.97 -16.72 26.72
D 4.36 1.3 0 -11.96 18.99 -1.97 5.05 -5.05
E 4.36 -3.7 0 -11.96 18.99 5.62 12.64 -12.64
F -0.64 -2.7 0 -11.96 -2.79 4.10 -10.65 10.65
Bidang Momen
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