機動學電腦程式上機測驗卷 (A)
修課班級:_____________________
學號:___________________________________
姓名:______________________________
2021/3/30
Read the problem statement in section I and write a computer program in Scilab 6.0.0 to calculate the required outputs. In Section III, fill in the blanks with computed numerical values. Your final
score will depend on the correctness and accuracy of the numerical values.
I. Problem statement
For the following mechanism, 𝑟1 = 100 cm, 𝑟3 = 40 cm, the length of link 4 (L4) = 200 cm, the mass center 𝐺3 is at the midpoint of 𝑟3 , 𝐺4 is at the midpoint of L4 (𝑟4 ≠ 𝐺4 ), 𝑚2 = 0kg, 𝑚3 = 15 kg,
𝑚4 = 50 kg, 𝐼3 = 0.05 kg-m2, and 𝐼4 = 0.2 kg-m2. When 𝑟2 ≤ 60 cm, the load torque 𝑇4 = 20 N-m CCW. When 𝑟2 > 60 cm, 𝑇4 = 100 N-m CCW.
II Given
(1) For inverse dynamic problems (IDP), 𝑟2 = 50 ~ 150 cm, ∆𝑟2 = 0.1 cm, 𝑟2̇ = 20 cm/s, constant.
(2) For forward dynamic problems (FDP), 𝑟2 = 50 cm, 𝑟2̇ = 0 cm/s, when t = 0, time increment: Δt = 0.005 second, period: t = 0 ~ 3 s. 𝐹2 = 40 N.
(3) Required formulas:
𝑟 𝑟 +𝑟2 𝑟4
𝜃4 = cos −1 ( 1 32
𝑟4 = √𝑟1 2 + 𝑟2 2 − 𝑟3 2
𝑟3 sin 𝜃3
[−𝑟3 cos 𝜃3
1
𝑟4 sin 𝜃4
−𝑟4 cos 𝜃4
−1
𝑟3 +𝑟4 2
− cos 𝜃4 ℎ3
−1
− sin 𝜃4 ] [ℎ4 ] = [ 0 ]
𝑓4
0
0
𝑟3 sin 𝜃3
[−𝑟3 cos 𝜃3
1
1
𝑥𝐺3 = 𝑟4 cos 𝜃4 + 𝑟3 cos 𝜃3
2
𝑥𝐺4 =
𝑟4 sin 𝜃4
−𝑟4 cos 𝜃4
−1
)
𝜃3 = 𝜃4 + 90°
′
− cos 𝜃4 ℎ3
−𝑟3 cos 𝜃3 ℎ3 2 − 2𝑓4 sin 𝜃4 ℎ4 − 𝑟4 cos 𝜃4 ℎ42
′
− sin 𝜃4 ] [ℎ4 ] = [ −𝑟3 sin 𝜃3 ℎ32 + 2𝑓4 cos 𝜃4 ℎ4 − 𝑟4 sin 𝜃4 ℎ42 ]
0
𝑓4 ′
0
1
𝑦𝐺3 = 𝑟4 sin 𝜃4 + 𝑟3 sin 𝜃3
2
1
𝐿 cos 𝜃4
2 4
𝑦𝐺4 =
1
𝑓𝐺3𝑥 = 𝑓4 cos 𝜃4 − 𝑟4 sin 𝜃4 ℎ4 − 𝑟3 sin 𝜃3 ℎ3
2
1
𝐿 sin 𝜃4
2 4
1
𝑓𝐺3𝑦 = 𝑓4 sin 𝜃4 + 𝑟4 cos 𝜃4 ℎ4 + 𝑟3 cos 𝜃3 ℎ3
2
1
1
𝑓𝐺3𝑥 ′ = 𝑓4′ cos 𝜃4 − 2𝑓4 sin 𝜃4 ℎ4 − 𝑟4 cos 𝜃4 ℎ42 − 𝑟4 sin 𝜃4 ℎ4′ − 2 𝑟3 cos 𝜃3 ℎ32 − 2 𝑟3 sin 𝜃3 ℎ3′
1
1
𝑓𝐺3𝑦 ′ = 𝑓4′ sin 𝜃4 + 2𝑓4 cos 𝜃4 ℎ4 − 𝑟4 sin 𝜃4 ℎ42 + 𝑟4 cos 𝜃4 ℎ4′ − 𝑟3 sin 𝜃3 ℎ32 + 𝑟3 cos 𝜃3 ℎ3′
2
2
1
1
1
𝑓𝐺4𝑥 ′ = − 2 𝐿4 𝑐𝑜𝑠 𝜃4 ℎ42 − 2 𝐿4 𝑠𝑖𝑛 𝜃4 ℎ4′
2
2
𝐴 = 𝑚(𝑓𝑔𝑥
+ 𝑓𝑔𝑦
) + 𝐼ℎ2 ;
1
𝑓𝐺4𝑦 ′ = − 2 𝐿4 sin 𝜃4 ℎ42 + 2 𝐿4 cos 𝜃4 ℎ4′
′
′
𝐵 = 𝑚(𝑓𝑔𝑥 𝑓𝑔𝑥
+ 𝑓𝑔𝑦 𝑓𝑔𝑦
) + 𝐼ℎℎ′ ;
𝐹2 + 𝑇4 ℎ4 = (∑ 𝐴)𝑟̈2 + (∑ 𝐵) 𝑟̇2 2
1
(4) Euler’s method: ∆𝜃2̇ = 𝜃2̈ ∆𝑡 and ∆𝜃2 = 𝜃2̇ ∆𝑡 + 𝜃2̈ ∆𝑡 2
2
III Outputs (Note: Use 4 significant digits for all answers, except for 𝒓𝟐 ).
When 𝑟2 = 100 cm, we know that 𝐴4 = 6.2397 kg and 𝐵4 = -7.431 kg.
(1) [IDP, 20] When 𝑟2 = ____50____cm, the value of 𝐴4 is the largest at _____20.9978_____ kg.
(2) [IDP, 10] When 𝑟2 = ____81.2____cm, the positive value of 𝐵3 is the largest at _____0.895______ kg.
(3) [IDP, 10] When 𝑟2 = ____50____cm, the negative value of 𝐵4 is the largest at ______-24.467_____ kg.
(4) [IDP, 20] When 𝑟2 < 60 cm, 𝑟2 = ____50____cm, the absolute magnitude of 𝐹2 is the largest at ______11.9764_____ N.
(5) [FDP, 20] When 𝑡 = 0.485 s, the velocities of link 2 is 0.4383 m/sec. In the specified period (0 – 3 sec), determine the maximum and minimum velocities of link 2 when 𝑟2 > 60 cm: maximum 𝑟2̇
= ____0.8909_____ m/sec; minimum 𝑟2̇ = ____0.1888_____ m/sec.
(6) [FDP, 20] In the specified period (0 – 3 sec), determine the maximum and minimum 𝑟2̈ : maximum 𝑟2̈ = ____1.2536_____ m/𝑠 2 ; minimum 𝑟2̈ = _____-0.4706_____ m/𝑠 2 .