d de ax (b) T=G dp d T 故得出:a= (122) T TA TR T (123){m===m|(124 R 单位:mcm1m/ 记扭转截面系数 P R (125)dx d T GI p  = (c) dx d dx d       = = (a) dx d G G    =   =  （b） 故得出： GI p T dx d  = =   (12.2) p p I T GI T G    =  = (12.3) p p Wp T R I T I TR  max = = = (12.4) 记扭转截面系数 R I W p p = 单位：m3 ,cm3 ,mm3 (12.5)