东南大学机械原理课件作业3

hπ π ds d = sin( ) 2Φ Φ
π hπ 2 2 2 d s d = cos( ) 2 2Φ Φ hπ π Φ ( ds d ) max = sin( ) = h = 50( mm rad ) 2Φ Φ 2 π hπ 2 2 2 ( d s d ) max = cos( 0) = 2h = 100( mm rad 2 ) 2Φ 2 Φ
正弦加速度
1 2π s = h sin( ) Φ Φ 2π
2π h ds d = [1 cos( )] Φ Φ
2π 2 hπ d s d = sin( ) 2 Φ Φ
2 2
( ds d ) max
h 2π Φ 200 = [1 cos( )] = ( mm rad ) Φ Φ 2 π
2π Φ 400 2 hπ = sin( ) = ( mm rad 2 ) Φ2 Φ 4 π
( d s d ) max
2 2
题3-4
s0 = rb2 e2 = 502 122 = 48.54
从动件运动规律: 从动件运动规律: 升程 ∈ [0, π ]
2h 2 80 2 s = 2 = 2 , ∈ [0,π 2] Φ π 2h 80 2 2 s = h 2 (Φ ) = 40 2 (π ) , ∈ [π 2 ,π ] Φ π 回程 ∈ [π ,11π 6]
计算时角度用负值代入! 计算时角度用负值代入! =π/2时,x=-70,y=-25.48
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2h 2 4h s = 2 , ds s = 2 Φ Φ 2h 4h ∈ [π 2 , π ], S = h (Φ ), ds d = 2 (Φ ) Φ Φ 4h Φ 80 lmax = (ds d )max = 2 = = 25.4mm Φ 2 π
ds d cos + (rb + S ) sin xB = y B ds sin + (r + S ) cos b d =π/6时,x=-33.46,y=40.96
2 2 OA 2
已知Φ π π2, 题3-2 已知Φ=π ,h=50mm 等速运动
h S = Φ
ds d = h Φ
d 2 s d 2 = 0
(ds d ) max = h Φ = 100 π ( mm rad )
( d s d ) max = 0
2 2
等加速等减速运动 等加速∈[ Φ2] ∈[0, 等加速∈[ ,Φ ]
4h 2 S = 2 Φ
ds d = ( 4h Φ 2 )
d 2 s d 2 = 4h Φ 2
( ds d ) max
2 2
Φ 200 = ( 4h Φ ) = ( mm rad ) π 2
2
2
( d s d ) max = 4h Φ =
800
π2
( mm rad 2 )
余弦加速度 s = h 1 cos( π ) 2 Φ
题3-1
ω R
h
B
O
rb
rb=R-lOA h=2R-2 rb=2 lOA
A
R 2 = OA2 + OB 2 2 OA OB cos(π ) OB = lOA cos + R 2 l 2OA sin 2
s = OB rb = lOA (1 cos ) + R l sin R
h π s = 1 + cos( ( Φ )) = 20[1 + cos(1.2( π )) ] 2 Φ′
S 题3-4 -ω ω
0 1 7 h 2 1 s2 0 s1 3 4 5 6 Φ' 7
1
2π π
Φ
ΦS'
(0)
1
6
rb ω
s2 5 3 4 h
O
题3-7
B R
1
= -
0
1
C ω O
ω O1
对心平底移动盘形凸轮机构.已知: , 对心平底移动盘形凸轮机构.已知:rb=50mm, 题3-13 凸轮转速ω 顺时针),从动件h=40mm,推 ),从动件 凸轮转速ω(顺时针),从动件 , 程为等加速等减速运动, 程为等加速等减速运动,Φ=180.求lmax, =π/6及=π/2时凸轮 廓线上点的直角坐标. π 及 π 时凸轮 廓线上点的直角坐标. 解: ∈ [0, π 2]
0
a B
0
ψ
l AAΒιβλιοθήκη 0题3-7B R ω C0 C O ψ a l A
题3-8
α 150°
s B0
A r O b
ω
B0
Φ Φ'
A O
h
题3-8
B0 B
A r O b
O
0
Φ' A
B0 1500
ω
h
O0 Φ s
ω
α
题3-9
45°
α
ω
O1
s
e
α ψ
A
45°
ω O1
A
题3-10
B1'
10°
A
B1