陶瓷大學(xué)機(jī)械原理作業(yè)及答案

Name Class Student No. Date Ex.2-1 Draw the kinematic diagrams of the mechanisms shown below.-11223344ABC4scale3:11BC2AName Class Student No. Date Ex.2-2 Draw the kinematic diagram of the mechanism shown below.1233445566ABCDEABCDEFF21Name Class Student No. Date Ex.2-3 Draw the kinematic diagram of the mechanism shown below.8I7B65HA1G2C3E4DF278I156HAGFCB3ED4Name Class Student No. Date Ex.2-4 Calculate the degree of freedom of the mechanisms shown below. Indicate all points for attention before the calculation of the DOF.解：(a) 左右為對稱結(jié)構(gòu)，設(shè)左側(cè)為虛約束。(b) E 為桿 4、5、6 的復(fù)合鉸鏈。 (d) 滑塊 7 與機(jī)架 8 間為移動(dòng)副。F=3n-2PL-Ph=37-210=1解： (1)紅線內(nèi)的構(gòu)件為重復(fù)結(jié)構(gòu)，構(gòu)成虛約束。(2)去掉以上構(gòu)件后，C 仍為構(gòu)件2、3、4 的復(fù)合鉸鏈。(3)滑塊 5 與機(jī)架 6 之間為移動(dòng)副。F= 3n-2PL-Ph =3527=1ABCDFGHIJReduntcosrai231KName Class Student No. Date Ex.2-5 Calculate the degree of freedom of the mechanisms shown below. Indicate all points for attention before the calculation of the DOF.解：(a)兩個(gè)滾子有局部自由度。(b)滾子 D 與凸輪 1 之間只能算一個(gè)高副。F=3n-2PL-Ph =37 29 -2=1解：(1) Link BC is welded to gear 2(2) A is a compound hinge of gear 4、link 1、and frame 5.F= 3n-2PL-Ph =34 25 -1 =1常見錯(cuò)誤：認(rèn)為 B 是復(fù)合鉸鏈，而不認(rèn)為 A 是復(fù)合鉸鏈。3ABCEFGLMNO234568Name Class Student No. Date Ex.2-6 Calculate the degree of freedom of the mechanisms shown below. Indicate all points for attention before the calculation of the DOF.解：(a) C 為構(gòu)件 2、3、4 的復(fù)合鉸鏈。(b) C 處有兩個(gè)轉(zhuǎn)動(dòng)副和兩個(gè)移動(dòng)副。 E 處有一個(gè)轉(zhuǎn)動(dòng)副和兩個(gè)移動(dòng)副。F= 3n-2PL-Ph =37 210=1 注意：E不是復(fù)合鉸鏈！解：當(dāng)構(gòu)件尺寸任意時(shí)，構(gòu)件2 作平面復(fù)雜運(yùn)動(dòng)，而桿 4 與機(jī)架間組成移動(dòng)副，所以桿 4 僅作平動(dòng)。因此，構(gòu)件 2 和構(gòu)件4 之間有相對轉(zhuǎn)動(dòng)。因此，應(yīng)該有構(gòu)件 6，并且構(gòu)件 4 和 6 之間有轉(zhuǎn)動(dòng)副，如右圖所示。當(dāng) ABCD 且 BCAD時(shí)，桿 2 僅作平動(dòng)。桿 4 與機(jī)架間組成移動(dòng)副，所以桿 4 也僅作平動(dòng)。這樣，構(gòu)件 2 和構(gòu)件 4 之間就沒有相對轉(zhuǎn)動(dòng)，只有相對移動(dòng)。即：構(gòu)件 4 和構(gòu)件 6 之間就沒有相對轉(zhuǎn)動(dòng)了，因此，可將構(gòu)件 6 與構(gòu)件 4 焊接起來（去掉構(gòu)件 6） ，如左圖所示。然而，在計(jì)算機(jī)構(gòu)自由度時(shí)，應(yīng)該按一般尺寸情況下進(jìn)行分析，即：應(yīng)該按照右圖情況來分析機(jī)構(gòu)的自由度。F= 3n-2PL-Ph =35 27=1ABD=C1358Name Class Student No. Date Ex.2-7 Shown below is the kinematic diagram of an engine mechanism.(1) Calculate the degree of freedom of the mechanism. Indicate all points for attention before the calculation of the DOF.(2) Carry out the structural analysis for the mechanism.(3) Carry out the structural analysis for the mechanism if link EFG is a driver.Note: During structural analysis, list the assembly order of Assur groups, the type of group, the grade of group, the grade of the mechanism, the link serial numbers, the inner pair and the outer pairs of each group in each mechanism.解：（1） F= 3n-2PL-Ph =37 210=1（2） 當(dāng) AB 為原動(dòng)件時(shí)，（3）當(dāng) EFG 為原動(dòng)件時(shí)，ABCDEFGH4658第 一 桿 組 RP,內(nèi) 副 外 副移第 二 桿 組 類 型 桿 號(hào)第 三 桿 組 轉(zhuǎn) -移轉(zhuǎn)第 一 桿 組 內(nèi) 副 外 副移第 二 桿 組 類 型 桿 號(hào)第 三 桿 組 轉(zhuǎn) 移轉(zhuǎn)I級(jí) 桿 組 46內(nèi) 副 外 副AE移 C類 型 桿 號(hào) BD轉(zhuǎn) 38轉(zhuǎn) HG移 7ABCDEFGH12346578I級(jí) 桿 組RP,內(nèi) 副 外 副移類 型 桿 號(hào) 轉(zhuǎn) -8轉(zhuǎn) 移Name Class Student No. Date Ex.2-8 Carry out the structural analysis for the mechanism(a) if link 1 is a driver. (b) if link 5 is a driver.Note: During structural analysis, list the assembly order of Assur groups, the type of group, the grade of group, the grade of the mechanism, the link serial numbers, the inner pair and the outer pairs of each group in each mechanism.解：(a) if link 1 is a driver,Links 2, 3, 4 and 5 constitute a grade III Assur group.(b) if link 5 is a driver.Links 3 and 4 constitute the first RPR Assur group.Links 1 and 2 constitute the second RPR Assur group.BC6ADE26ACDB5E3BD4FgrdeIV6ACDBEBD3A1524CEAD53B4A6CEName Class Student No. Date 2-9 The schematic diagram of a punch machine designed by someone is shown in Fig2-9. This machine should be able to transform a