mungkin sketsa soalnya seperti ini,

mungkin sketsanya seperti ini,



diasumsikan kekakuan semua segmen sama.

Segmen AB,

Mab = 4EIΦa/L + 2EIΦb/L - 6EIΔ/L²

Mab = 4EIΦa/4 + 2EIΦb/4 - 6EIΔ/4²

Mba = 2EIΦa/L + 4EIΦb/L - 6EIΔ/L²

Mba = 2EIΦa/4 + 4EIΦb/4 - 6EIΔ/4²



Segmen BC,

Mbc = 4EIΦb/L + 2EIΦc/L

Mbc = 4EIΦb/4 + 2EIΦc/4

Mcb = 2EIΦb/L + 4EIΦc/L

Mcb = 2EIΦb/4 + 4EIΦc/4



Segmen CD,

Mcd = 4EIΦc/L + 2EIΦd/L - 6EIΔ/L²

Mcd = 4EIΦc/2 + 2EIΦd/2 - 6EIΔ/2²

Mdc = 2EIΦc/L + 4EIΦd/L - 6EIΔ/L²

Mdc = 2EIΦc/2 + 4EIΦd/2 - 6EIΔ/2²


Mab = 0

4EIΦa/4 + 2EIΦb/4 - 6EIΔ/4² = 0. . . . . . . . . . . . . . (persamaan 1)

Mba + Mbc = 0

2EIΦa/4 + 4EIΦb/4 - 6EIΔ/4² + 4EIΦb/4 + 2EIΦc/4 = 0 . . . . . . . . . . (persamaan 2)

Mcb + Mcd = 0

2EIΦb/4 + 4EIΦc/4 + 4EIΦc/2 + 2EIΦd/2 - 6EIΔ/2² = 0. . . . . . . . . . (persamaan 3)

Mdc = 0

2EIΦc/2 + 4EIΦd/2 - 6EIΔ/2² = 0. . . . . . . . . (persamaan 4)

ΣF horizontal = 0

F - Ha - Hd = 0

F + (Mab + Mba)/L + (Mdc + Mcd)/L = 0

F + (4EIΦa/4 + 2EIΦb/4 - 6EIΔ/4² + 2EIΦa/4 + 4EIΦb/4 - 6EIΔ/4²)/4 + (2EIΦc/2 + 4EIΦd/2 - 6EIΔ/2² + 4EIΦc/2 + 2EIΦd/2 - 6EIΔ/2²)/2 = 0. . . . . . . . (persamaan 5)

5 persamaan dengan 5 peubah Φa, Φb, Φc, Φd dan Δ

Φa = 128 F/(69EI)

Φb = 8F/(69EI)

Φc = 104F/(69EI)

Φd = 212F/(69EI)

Δ = 352F/(69EI)


substitusikan harga di atas ke persamaan awal,

Mab = 4EIΦa/4 + 2EIΦb/4 - 6EIΔ/4²

Mab = 4EI(128 F/(69EI))/4 + 2EI(8F/(69EI))/4 - 6EI(352F/(69EI))/4²

Mab = 0

Mba = 2EIΦa/4 + 4EIΦb/4 - 6EIΔ/4²

Mba = 2EI(128 F/(69EI))/4 + 4EI(8F/(69EI))/4 - 6EI(352F/(69EI))/4²

Mba = -20F/23

Mbc = 4EIΦb/4 + 2EIΦc/4

Mbc = 4EI(8F/(69EI))/4 + 2EI(104F/(69EI))/4

Mbc = 20F/23

Mcb = 2EIΦb/4 + 4EIΦc/4

Mcb = 2EI(8F/(69EI))/4 + 4EI(104F/(69EI))/4

Mcb = 36F/23

Mcd = 4EIΦc/2 + 2EIΦd/2 - 6EIΔ/2²

Mcd = 4EI(104F/(69EI))/2 + 2EI(212F/(69EI))/2 - 6EI(352F/(69EI))/2²

Mcd = -36F/23

Mdc = 2EIΦc/2 + 4EIΦd/2 - 6EIΔ/2²

Mdc = 2EI(104F/(69EI))/2 + 4EI(212F/(69EI))/2 - 6EI(352F/(69EI))/2²

Mdc = 0

Ha = -(Mab + Mba)/L

Ha = -(0 + (-20F/23))/4

Ha = 5F/23

Hd = -(Mdc + Mcd)/L

Hd = -(0 + (-36F/23))/2

Hd = 18F/23

Va = -(Mbc + Mcb)/L

Va = -(20F/23 + 36F/23)/4

Va = -14F/23

Vd = (Mbc + Mcb)/L

Vd = 14F/23




periksa hasil di atas,

ΣM di titik A = 0

(F)(4) - (Hd)(4 - 2) - (Vd)(4) = 0

(F)(4) - (18F/23)(4 - 2) - (14F/23)(4) = 0

0 = 0

OK





ΣM di titik D = 0

(Va)(4) + (Ha)(4 - 2) + (F)(2) = 0

(-14F/23)(4) + (5F/23)(4 - 2) + (F)(2) = 0

0 = 0

OK




ΣF vertikal = 0

Va + Vd = 0

-14F/23 + 14F/23 = 0

0 = 0

OK



ΣF horizontal = 0

F - Ha - Hd = 0

F - 5F/23 - 18F/23 = 0

0 = 0

OK








silakan agan periksa dengan metoda yang lain...