0 N,-Pcos 45=0 ∑Y=0Psin45-N2=0 解上式得N, P, N P=10V2kN由于 B 结构对称,故有N2=N4=10V2kN No N=N2P=10v2 kN B铰链受力如图,由平衡条件 ∑ X=0Ncos45°-M1=0 解得N5=P=20kN 4 DL=N.a/EA=10×103.a/200×10× 500=√2×107.a ∠L5=20×10×√2.a/200×10×500 =2√2.×10-.a ∠LAc=2√(a+∠L)2-(√2.a/2-4L5/2)2-√2a=6.83×10.a 2√(a+√2×10a)2(√2a/2-√2×10a)2-√2.a =(2√1+2√2×10-+2×1081/2+2×10-4-2×103-√2)a (2√0.5+2×(√2+1)×104-√2)a =(2√0.5+0.0004828-√2)a =(2×0.70744809-1.414213562).a (1.41489618-1.414213562).a 0.0006826. 6.83×10a 杆系的总变形能为U=4xN+N√a 2EA 2EA P2a(2+ 2EA 应用卡氏定理,A、C两点的距离改变为 20×103a 200×109×500×10 0683×10-3a5 X = 0 1 N Pcos 45 0  − = Y = 0 2 P N sin 45 0  − = 解上式得 1 2 2 N P = , 2 2 2 N P = =102kN 由于 结构对称，故有 N N 3 4 = =102kN = N1 2 2 = P =102 kN B 铰链受力如图，由平衡条件 X = 0 5 1 N N cos45 0  − = 解得 N P 5 = =20kN DL1=N1.a/EA=102 × 103 .a/200 × 103 × 500=√2×10-4 .a ⊿L5=20×103×√2.a /200×103×500 =2√2.×10-4 .a ⊿LAC=2√(a +⊿L1) 2 -(√2.a/2-⊿L5/2)2 - √2 a = 6.83×10-4 .a =2(a+√2×10-4 a)2 -(√2a/2 - √2×10-4 a)2 -√2.a = (2√1+2√2×10-4 +2×10-8 -1/2 +2×10-4 -2×10-8 - √2 )a = (2√0.5+2×(√2+1)×10-4 -√2)a =(2√0.5+0.0004828 - √2)a =(2×0.70744809 – 1.414213562).a =(1.41489618 -1.414213562 ）.a =0.0006826.a = 6.83×10-4 a 杆系的总变形能为 U 2 2 1 5 2 4 2 2 N a N a EA EA =  + 2 (2 2) 2 P a EA + = 应用卡氏定理，A、C 两点的距离改变为 A U P   =  (2 2) Pa EA = + 3 9 6 20 10 (2 2) 200 10 500 10 a −  = +    3 0.683 10 a − = 