102年高考三級工程力學第五題 - 土木

※ 引述《mountain161 (山豬)》之銘言：
: 題目：
: http://farm6.staticflickr.com/5478/11319942594_ff39502bcb_c.jpg

圖五

────────────┐ ─
│ ↑
│ │
│ │
│ │ -
│ │ x = 2 + 1 * (1 - cosθ) = 3 - cosθ
│ │
│ │
σ __-─┘ │
╲ ─ | │
↘╱ \ | ↓
╱ \ | ￣
/ ＼ dθ\ |
/ ＼ \ θ|
/ ＼ \ |
｜ ＼ \ |
────┘----------------
(1)
-
σ = γw * x = 9.81 * (3 - cosθ)

σv = σ * cosθ = 9.81 * (3cosθ - 1/2 - 1/2*cos2θ)

σh = σ * sinθ = 9.81 * (3 - cosθ) * sinθ

(2)

dF = σ * dA = σ * (1 * dθ) = σdθ

π/2 π/2
Fv = ∫ σvdθ = 9.81 * ∫ (3cosθ - 1/2 - 1/2*cos2θ)dθ
0 0

π/2
= 9.81 * [3sinθ - θ/2 - 1/4*sin2θ]
0

= 9.81 * [3 - π/4]

= 21.73 kN/m


π/2 π/2
Fh = ∫ σhdθ = 9.81 * ∫ [(3 - cosθ) * sinθ]dθ
0 0

π/2
= 9.81 * ∫ (3 - cosθ)(-dcosθ)
0

π/2
= 9.81 * [1/2*cosθ^2 - 3cosθ]
0

= 9.81 * (-1/2 + 3)

= 24.53 kN/m

合力F = √(Fv^2 + Fh^2) = 32.77 kN/m (── )........................(Ans.)
↘ β

β = atan(Fv/Fh) = 41.54°..........................................(Ans.)


(3)

圖六類似圖五 只是應力作用方向為↗

↗ β
合力F = 32.77 kN/m (── )..........................................(Ans.)

β = 41.54°........................................................(Ans.)


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我果然是吃飽太閒....

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