钢板弹簧课程设计第三组

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贵州大学机械工程学院

3.6 钢板弹簧各片实际长度的计算(即计算弧长的各片曲率半径)

图3.1

1)主片1

①满载时弧高fa1=fa+(i-1)h =28(mm) ②弦长L1=1630(mm) ③计算弦长Ll1?(L1?ks)2?1300?0.5*110?682.5(mm)

2fa1?622.52/2/28?28/2?23295.31mm 2 ④曲率半半径Rm1?Ll1/2/fa1?2 ⑤夹角?1?arcsin(Ll1/Rm1)?arcsin(622.5/6933.75)? 0.0293(rad) ⑥弧长L11=2?Rm1=2?0.0899?6933.75=1365.19(mm) ⑦实际弧长L12 2)主片2 ①满载时弧高 ②弦长L2?L11?ks?1246.68?0.5?110?1420.195(mm)

fa2?fa?h?28?10?38mm

?1420mm

(l2?ks)2?1300?0.5?110?622.5mm

2 ③计算弦长Ll2? 13

贵州大学机械工程学院

④曲率半径Rm2?Ll22/2/fa2?fa2622.52/2??38/2?5117.766mm 382 ⑤夹角?2?arcsin(Ll2/Rm2)?0.121937 (rad)

⑥弧长L21?2?2Rm2?2?0.121937?5117.766?1248.091(mm) ⑦实际弧长L22?L21?ks?1248.091?0.5?110?1303.091(mm) 3)片3

①满载时弧高fa3?48mm ②弦长L3?1200mm ③计算弧长Ll3?537.5mm ④曲率半径Rm3?Ll3/2/fa3?2fa3?3033.44mm 2 ⑤夹角?3?arcsin(Ll3/Rm3)?0.17813(rad) ⑥弧长L31?2?3Rm3?1080.71mm ⑦实际弧长L32?L31?ks?1135.71mm 4)片4

①满载时弧高fa4?58mm ②弦长L4?980mm ③计算弧长Ll4?(L4?ks)2?452.5mm

fa4?1794.14mm 2 ④曲率半径Rm4?Ll42/2/fa4? ⑤夹角?4?arcsin(Ll4/Rm4)?0.25496(rad)

⑥弧长L41?2?4Rm4?2?0.25496?1794.14?914.88mm ⑦实际弧长L42?L41?ks?914.88?0.5*110?969.88mm 5)片5

①满载是弧高fa5?68mm

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贵州大学机械工程学院

②弦长L5?760mm

③计算弦长Ll5?(L5?ks)/2?367.5mm ④曲率半径Rm5?Ll52/2/fa5?fa5?1027.061mm 2 ⑤夹角?5?arcsin(Ll5/Rm5)?0.365929(rad)

⑥弧长L51?2?5Rm5?2?0.365929?1027.061?751.6632mm ⑦实际弧长L52?L51?ks?751.6632?0.5*110?806.6632mm 6)片6

①满载时弧高fa6?78mm ②弦长L6?550mm

③计算弦长Ll6??L6?ks?/2?282.5mm ④曲率半径Rm6?Ll62/2/fa? ⑤夹角?6fa6?550.579mm 2?arcsin(Ll6/Rm6)?0.53879 (rad)

⑥弧长 L61?2?6Rm6?2?0.53879?550.579?593.291mm ⑦实际弧长 L62?L61?ks?593.291?0.5*110?648.291mm 7)片7

①满载是弧高fa7?88mm ②弦长L7?320mm

③计算弦长Ll7?(L7?ks)/2?197.5mm ④曲率半径Rm7?Ll7/2/fa7?2fa7?265.626mm 2 ⑤夹角?7?arcsin(Ll7/Rm7)?0.83833 (rad)

⑥弧长L71?2?7Rm7?2?0.83833?265.626?445.364mm ⑦实际弧长L72?L71?ks?445.364?0.5?110?500.364mm

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贵州大学机械工程学院

9)主片1

1 卷耳内径 D=35(mm) ○

2 卷耳长度L卷 = ?(D+h) ○?

1313 =3.14 ? (35+10) ? =115.4685 (mm) 883 滑动处延伸度 L延 = 50(mm) ○

同理: L2伸 =1528(mm) L3伸 =1130(mm) L4伸 =960(mm) L5伸 = 790(mm) L6伸 =620(mm) L7伸 =450(mm) 3.7 钢板弹簧总成在自由状态下各片的曲率半径 1、未夹紧”U”型螺栓的m空载弧高F=fc+fa+? 1 动拢度fd=50 ○

2 预压缩产生的塑性变形 ? ○

? = (0.055~0.075)(fc+fd)=0.055*(90+56)=7.425(mm) 3 F=90+38+8.03=109.425(mm) ? ○

④ 型螺栓夹紧时对弧高的影响?H

s ?H= ???3L?S??fc?fa? ???=15.54494(mm) 2 2L 2、求自由状态下的总弧高F0

F0=F+?H=136.03+15.54494=123.0781(mm) 3、求主片的曲率半径R01

(R)=1-F0 由公式F/R01?cos[L12/(2R01)]?0

F(R)=0 (计算到小数点后5位)。 1)设Rx1=1200 Rx2=4300;

则F(Rx1)= -0.00839?0 ;F(Rx2)=-0.02556?0 2)令Rx3=(Rx1+Rx2)/2=2750; 则F(Rx3)=-0.01825?0 3)Rx4= (Rx1+ Rx3)/2=1975; 则F(Rx4)=-0.01414?0

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