Specification as following In fig3-2, considering the unit width beam, named the moment of double force on per unit width M. The order of here is[force][length]/[length], the result is [force]M On the left or right, the two-dimensional force should be combined to a double force, as the moment of double force is M, this request that h 6ah ydy=0,6ahy dy=M Put inequation (a) into the formula above, then h h o,dy=0, ho,ydy=M 2 The first one al ways can meet, and the second one request that: 2M C h3 12M Put into formula a), then Ox x=3y,O,=0,x=r1=0 h 11
11 Specification as following: In fig3-2,considering the unit width beam ,named the moment of double force on per unit width .The order of here is[force][length]/[length] M ，the result is [force] . M On the left or right，the two-dimensional force should be combined to a double force，as the moment of double force is M ，this request that： − − = = 2 2 2 2 2 6 0,6 h h h a ydy a h y dy M Put into formula（a），then： , 0, 0 12 3 x = y y = xy = yx = h M − − = = 2 2 2 2 0, h h h x dy h x ydy M Put in equation x （a）into the formula above，then： The first one always can meet ,and the second one request that ： 3 2 h M a =
平冒的立么解答 具体解法如下: 如图3-2,取单位宽度的梁来考察,并命每单位宽度上力偶的矩 为M。这里M的因次是[力[长度]/[长度],即[力]。 在左端或右端,水平面力应当合成为力偶,而力偶的矩为M, 这就要求 J2,, dy=0, 3,o,ydy=M 将式(a)中a的代入,上列二式成为: h h 6a, ydy=0, 6a dy= 2 2 2M 前一式总能满足,而后一式要求:a h3 12M 代入式(a),得:、yy=0,x=r=0 12
12 具体解法如下: 如图3-2,取单位宽度的梁来考察,并命每单位宽度上力偶的矩 为 M 。这里 M 的因次是[力][长度]/[长度]，即[力]。 在左端或右端，水平面力应当合成为力偶，而力偶的矩为 ， 这就要求： M − − = = 2 2 2 2 0, h h h x dy h x ydy M 前一式总能满足，而后一式要求： 3 2 h M a = 代入式（a），得： , 0, 0 12 3 x = y y = xy = yx = h M − − = = 2 2 2 2 2 6 0,6 h h h a ydy a h y dy M 将式（a）中 x 的代入，上列二式成为：
g SO 12 ormula al ove can be writ tten y,O,=0,==0 yx The result is same with corresponding part in material mechanics Note The result above is useful to the beam which the length I is greatly more than the depth h to the beam which the length I is equal to the depth h, this result is useless
13 Because the torque of the beam’s cross section is ，so the formula above can be written into ： 12 1 3 h I = x = y, y = 0, xy = yx = 0 I M The result is same with corresponding part in material mechanics Note： The result above is useful to the beam which the length is greatly more than the depth ; to the beam which the length is equal to the depth , this result is useless. l h l h
平动血解答 因为梁截面的惯矩是/1×h,所以上式可改写为: 12 M o,=0,7m,=1m=0 结果与材料力学中完全相同。 注意: 对于长度l远大于深度h的梁,上面答案是有实用价值 的;对于长度l与深度h同等大小的所谓深梁,这个解答是 没有什么实用意义的
14 因为梁截面的惯矩是 ，所以上式可改写为： 12 1 3 h I = x = y, y = 0, xy = yx = 0 I M 结果与材料力学中完全相同。 注意： 对于长度 远大于深度 的梁，上面答案是有实用价值 的；对于长度 与深度 同等大小的所谓深梁，这个解答是 没有什么实用意义的。 l h l h
83-2 Determination of Displacements Take the pure bending of rectangular for example to explain how to determine the displacement by the stress components 1.In the State of two-dimensional stress M Put the stress component yo=0,=x=0 into the physical equation (o-Ho E E,=-(0,-uo E 2(1+4) E 15
15 §3-2 Determination of Displacements Take the pure bending of rectangular for example to explain how to determine the displacement by the stress components. 1.In the State of Two-dimensional Stress Put the stress component into the physical equation x = y, y = 0, xy = yx = 0 I M + = = − = − xy xy y y x x x y E E E 2(1 ) ( ) 1 ( ) 1