工程力学习题全解第章力系的等效与简化习题解.pdf
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2 1
M ∑M (F ) −150×0.25 ×2i +150×0.15j (−75,22.5,0)N ⋅m
2 1 2 2
2 2 A B C yz
MA = 1, 0, 0MA =3.6 1, 0, 0kN ·m
MB = 0, sin40 ,cos40MB =6 0, sin40 ,cos40kN ·m
MC = 0, sin40 ,cos40 MC =6 0, sin40 ,cos40 kN ·m
M = M = M + M + M = 3.6, 12sin40 , 0 kN ·m
i A B C
2 3
2 3 F,2F d
A F A
(1) (a) d
2F d ′′
∑M C (F ) 0 B F B F
−F (d +x) +2F ⋅x 0 x
F
∴x d R C FR
C
∴FR 2F −F F a b
2 4 A 3,0 B 0,4 C 4.5,2
M = 20kN ·m M = 0 M = 10kN ·m
A B C
M = 0 F B M = 20kN ·m M = -10kN ·m F
B R A C R
A C
AG 2CD a
a OF = d
d 4 cot θ
(d +3sinθ) AG 2CD 1
d
CD CE sinθ (4.5 − ) sinθ 2
2
6
2 4
d
(d +3) sinθ 2(4.5 − ) sinθ
2
d +3 9 −d d 3
F 3, 0
a B F
y
4
tan θ G 4
3
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