4扭矩图:表示沿杆件轴线各横截面上扭矩变化规律的图线。 目①扭矩变化规律 的②Th值及其截面位置强度计算(危险截面)。 T 16
16 4 扭矩图：表示沿杆件轴线各横截面上扭矩变化规律的图线。 目 的 ①扭矩变化规律； ②|T|max值及其截面位置 强度计算（危险截面）。 x T
TORSION Example 1 A transmission shaft is rotating at n=300r/min Knowing its input power is P1=500kW, and its output are P2=150kW, P3=150kW P4=200kW, Try to plot the internal torque diagram Solution:① Calculate 3 the external torque m1=9.55 A B C D 500 =9.55 =159(kNm) 300 m2=m3=9.5532=9.55:500=4.78(kN·m) 300 200 m2=955=9553m0≈637(kN·m) 12
17 Example 1 A transmission shaft is rotating at n =300r/min. Knowing its input power is P1=500kW，and its output are P2=150kW，P3=150kW， P4=200kW，Try to plot the internal torque diagram. n A B C D Solution：①Calculate m2 m3 m1 m4 the external torque 15.9(kN m) 300 500 9.55 9.55 1 1 = = = n P m 4 78 (kN m) 300 150 9 55 9.55 2 2 3 = = = = . n P m m . 6 37 (kN m) 300 200 9 55 9.55 4 4 = = = . n P m
「例1已知:一传动轴,n=300r/min,主动轮输入P1=500kW, 从动轮输出P2=150kW,P3=150kW,P2-200kW,试绘制扭矩 图 no mi m 解:①计算外力偶矩 m1=95521=95900 300 A B C D =159(kNm) m2=m3=9.5532=9.55:500=4.78(kN·m) 300 m9554=955.200 300 =637(kNm 12 18
18 [例1]已知：一传动轴， n =300r/min，主动轮输入 P1=500kW， 从动轮输出 P2=150kW，P3=150kW，P4=200kW，试绘制扭矩 图。 n A B C D m2 m3 m1 m4 解：①计算外力偶矩 15.9(kN m) 300 500 9 55 9.55 1 1 = = = n P m . 4 78 (kN m) 300 150 9 55 9.55 2 2 3 = = = = . n P m m . 6 37 (kN m) 300 200 9 55 9.55 4 4 = = = . n P m
TORSION 2 Determine the internal torque (suppose it is positive) n 4 ∑mn=0,F+m2=0 T1=-m2=-4.78kN.m A1 D 72+m2+m2=0, 72=-m2-m3=-(4.78+478)=-956kNm T3+m4=0, T2=m4=6.37kN.m 19
19 n A B C D m2 m3 m1 m4 1 1 2 2 3 3 ②Determine the internal torque（suppose it is positive） 4.78kN m 0 , 0 1 2 1 2 = − = − = + = T m m T m x (4 78 4 78 9 56kN m 0 , 2 2 3 2 2 3 = − − = − + = − + + = T m m . . ) . T m m 6.37kN m 0 , 2 4 3 4 = = − + = T m T m x
②求扭矩(扭矩按正方向设) n 4 ∑mn=0,F+m2=0 T1=-m2=-4.78kN.m A1 D 72+m2+m2=0 72=-m2-m3=-(4.78+478)=-956kNm 73 +m4= 0 2=m4=637kN·m 20
20 n A B C D m2 m3 m1 m4 1 1 2 2 3 3 ②求扭矩（扭矩按正方向设） 4.78kN m 0 , 0 1 2 1 2 = − = − = + = T m m T m x (4 78 4 78 9 56kN m 0 , 2 2 3 2 2 3 = − − = − + = − + + = T m m . . ) . T m m 6.37kN m - 0 , 2 4 3 4 = = + = T m T m x