解:由位移互等定理知,①杆的伸长量等于 ②杆直径的减小量 △0=△2=|e1d=Ae,d P up 4E丌dE解：由位移互等定理知，①杆的伸长量等于 ②杆直径的减小量 l d d ① ② = =    =   d =  P AE d = 4  P d E