学 大小:Mo=∑m(F) 主矩M仿方向:方向规定+F (转动效应)简化中心:(与简化中心有关) (因主矩等于各力对简化中心取矩的代数和) 固定端(插入端)约束 在工程中常见的 雨搭 车刀
16 大小： 主矩MO 方向： 方向规定 + — 简化中心：(与简化中心有关) （因主矩等于各力对简化中心取矩的代数和） ( ) MO =mO Fi （转动效应） 固定端（插入端）约束 在工程中常见的 雨 搭 车 刀
Statics Fixed support (rigid embedding) Explanation: (Consider the forces Fi acting in the same plane 2Reduce F to the point A F get the force and the force couple: R ③ the direction ofr is M uncertain. but it can be determined by its rectangular A components Yand Xa; OY XA and MA are the reaction forces of the fixed support ⑤ Land a restrict the translation of the rigid body. M restrict the rotation
17 Fixed support (rigid embedding) Explanation: ①Consider the forces Fi acting in the same plane; ②Reduce Fi to the point A, get the force and the force couple; ③the direction of RA is uncertain, but it can be determined by its rectangular components YA and XA ; ④YA, XA and MA are the reaction forces of the fixed support; ⑤YA and Xa restrict the translation of the rigid body, Ma restrict the rotation
学 固定端(插入端)约束 说明 ①认为F这群力在同一 平面内; F ②将F向4点简化得一 力和一力偶; R M ③R方向不定可用正交 A 分力Y,X表示; ④Y,X,M为固定端 约束反力; ⑤Y,X限制物体平动, M为限制转动
18 固定端（插入端）约束 说明 ①认为Fi这群力在同一 平面内; ② 将Fi向A点简化得一 力和一力偶; ③RA方向不定可用正交 分力YA , XA表示; ④ YA, XA, MA为固定端 约束反力; ⑤ YA , XA限制物体平动, MA为限制转动
Statics 83-3 Result of reduction of a general coplanar force system. the law of the resultant moment Result of the reduction: principal vector R, principal moment Mo OR=0, Mo=0, the force system is in equilibrium. This case will be further discuss in the follow section 2R=O, Mo+0, the system can be reduced to a force couple with the moment Mo=M. Here, the forces acting on the rigid body are equivalent to the action of a force couple. Because the force couple can be transferred anywhere in its plane of action, the principal moment is independent on the given center of simplification B R#O, Mo=0, the system can be reduced to a resultant foro going through the given center of simplification. The result of the reduction is a resultant force( the resultant force of the force system), R=R(It depends on the given center of simplification, if the given center is changed, the principal moment is not equal to zero. 19
19 §3-3 Result of reduction of a general coplanar force system • the law of the resultant moment ② =0, MO≠0, the system can be reduced to a force couple with the moment MO =M. Here, the forces acting on the rigid body are equivalent to the action of a force couple. Because the force couple can be transferred anywhere in its plane of action, the principal moment is independent on the given center of simplification. R ① =0, MO =0, the force system is in equilibrium. This case will be further discuss in the follow section. R Result of the reduction: principal vector , principal moment R MO. ③ ≠0, MO =0, the system can be reduced to a resultant force going through the given center of simplification. The result of the reduction is a resultant force (the resultant force of the force system), . (It depends on the given center of simplification, if the given center is changed, the principal moment is not equal to zero.) R R =R
学 §3-3平面一般力系的简化结果·合力矩定理 简化结果:主矢R,主矩MO,下面分别讨论。 ①R=0,M=0,则力系平衡,下节专门讨论。 ②R=0,MO≠0即简化结果为一合力偶,MO=M此时刚 体等效于只有一个力偶的作用,因为力偶可以在刚体平 面内任意移动,故这时,主矩与简化中心O无关。 ③R′≠0,MO=0,即简化为一个作用于简化中心的合力。这时, 简化结果就是合力(这个力系的合力),R=R′。(此时 与简化中心有关,换个简化中心,主矩不为零) 20
20 §3-3 平面一般力系的简化结果 • 合力矩定理 简化结果： 主矢 ，主矩 MO ，下面分别讨论。 ② =0,MO≠0 即简化结果为一合力偶, MO =M 此时刚 体等效于只有一个力偶的作用，因为力偶可以在刚体平 面内任意移动，故这时，主矩与简化中心O无关。 R ① R =0， MO =0，则力系平衡,下节专门讨论。 R ③ ≠0,MO =0,即简化为一个作用于简化中心的合力。这时， 简化结果就是合力（这个力系的合力）, 。（此时 与简化中心有关，换个简化中心，主矩不为零） R R =R