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∑ M iDI = 0 i ∑ M iEIprawa = 0 .
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VC 2l + HC l –2ql l = 0
VC l + HC 2l –ql l/2 = 0.
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HC= - 1/3 ql i VC=7/6ql.
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„trójprzegubowego”. Do obliczenia reakcji pionowych wykorzystamy równania
∑ M iA = 0 i ∑ M iB = 0 .
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VB 2l +7/6ql 3l –3ql 3/2l = 0 i -VA 2l +3ql 1/2l +7/6ql l –= 0,
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A
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. Równanie
i HC.
∑ M iDII = 0
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l + HA l + ql l/2
LVWG
2
HA = 5/6 ql.
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+B
Z równania
∑
III
M iD
.
= 0 Z\QLND*Hòql l- HB l = 0 i HB = ½ ql.
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2/6 ql
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7/6 ql