§2-5拉(压杆内的应变能 Strain Energy of Axial Forced Bar 例:图a所示架,杆1和杆2均为钢杆,弹性模量E=20GPa,横截面面积分别 为A1=200mm?2,A2=250mm2,荷载P=10kN,l1=2m。试求节点A的垂直位移。 解∷W=U,而:W=P B 2 30 U=(N1△l1+N2△l2) Al 30 N =2P=20kN sin 30 P N2=M1COs30=1.732P=1732kN 4N20×10×2×10=10mm(伸长)N2 EA4200×103×200 M2N2l21732×10°×1.732×10 =06mm(缩短) EA 200×103×250 61=(N1△4+N2A2/P=(20×1+1732×06)/10 =3.04mm§2－5拉(压)杆内的应变能 Strain Energy of Axial Forced Bar 例: 图(a)所示架，杆1和杆2均为钢杆，弹性模量E=200GPa，横截面面积分别 为A1=200mm2 ，A2=250mm2 ，荷载P=10kN，l1=2m。试求节点Ａ的垂直位移。 解: m m N l N l P 3.04 ( 1 1 2 2 ) (20 1 17.32 0.6) 10 =  ⊥ = D + D =  +  P k N P N 2 20 sin 30 1 = = =  N2 = N1 cos30 =1.732P =17.32k N  1.0 ( ) 200 10 200 20 10 2 10 3 3 3 1 1 1 1 mm 伸长 EA N l l =      D = = 0.6 ( ) 200 10 250 17.32 10 1.732 10 3 3 3 2 2 2 2 mm 缩短 EA N l l =      D = = ( ) 2 1 2 1 , : 1 1 2 2 U N l N l W U W P = D + D = =   而  ⊥