Яблонский С1 вариант 29
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), ) ) ( .1) , –
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P = 25 , q = 1 
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.1
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(Fi ) = 0; q ×0.5× 22 - 2q ×0.5× 22 + P sin 30 × 4 - M = 0
M = q ×0.5× 22 - 2q ×0.5× 22 + P sin 30 × 4 = 1×0.5× 22 - 2 ×0.5× 22 + 25×0.5× 4 = 48 × .
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X i = 0; |
P sin 30 - X B = 0 |
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X B = P sin 30 = 25×0.5 = 12.5 . |
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(Fi ) = 0; |
q ×0.5× 22 - 2q ×0.5× 22 + P sin 30 × 4 - × 2 - M = 0 |
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M = q ×0.5× 22 - 2q ×0.5× 22 + P sin 30 × 4 - × 2 = 1×0.5× 22 - 2 ×0.5× 22 + 25×0.5× 4 -12.5× 2 = 23 .
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= 0; |
P sin 30 - = 0 |
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= P sin 30 = 25×0.5 = 12.5 |
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= 0; |
- q × 2 - 2q × 2 - P cos30 = 0 |
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= q × 2 + 2q × 2 + P cos30 = 1× 2 + 2 ×1×1+ 25×0.5 |
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» 27.651 . |
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(Fi ) = 0; |
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q ×0.5× 22 - 2q ×0.5× 22 + P sin 30 × 4 - × 2 + M = 0 |
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M = -q ×0.5× 22 + 2q ×0.5× 22 - P sin 30 × 4 + × 2 = |
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= -1×0.5× 22 + 2 ×0.5× 22 - 25×0.5× 4 + 27.651× 2 » 7.301 . |
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: ): |
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M A |
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min = 7.301 ; |
R = = 12.5 ; RB = = 27.651 . |
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