同济大学 材料力学 习题解答1(练习册P61-P64).pptVIP

* * 习题解答(一) P61 34-1(1) F F 2F F F F 1 1 F F 2F FN1 1 1 F 2F FN2 2F FN3 F FN1 1 1 2 2 2 2 3 3 3 3 F F FN2 F F F FN3 ＋ ＋ ＋ 4F 3F 2F FN图 F － ＋ F FN图 ?Fy = 0 FN1 = 4F ?Fy = 0 FN2 = 3F ?Fy = 0 FN3 = 2F ?Fy = 0 FN1 = F ?Fy = 0 FN2 = 0 ?Fy = 0 FN3 = － F P61 34-1(4) 六要素(阴影线方向) R FN3 截面法 分段相等 正向假设 P61 34-2 A B C D F4 F5 F2 F1 F3 E ＋ ＋ ＋ ＋ 500 N 920 N 640 N 240 N F1 = 500 N F2 = 420 N F3 = 280 N F4 = 400 N F5 = 240 N FN 图 s = A FN sAB = 1.25 MPa sBC = 2.3 MPa sCD = 1.6 MPa sDE = 0.6 MPa 应力单位 ：MPa A = 400 mm2 与点有关 P62 34-3 F3 F1 F2 A B C D 1 1 2 2 3 3 FN1 F3 ?Fx = 0 FN1 = F3 F3 F2 FN2 F3 F1 F2 FN3 = 30 kN ?Fx = 0 FN2 = F3 + F2 = 40 kN ?Fx = 0 FN3 = F3 + F2 – F1 = 30 kN ＋ ＋ ＋ 30 kN 40 kN 30 kN FN图 s1 = A1 FN1 A1 = 4×10 –2 m2 = 0.75 MPa s2 = A2 FN2 A2 = 2×10 –2 m2 = 2 MPa s3 = A3 FN3 A3 = 4×10 –2 m2 = 0.75 MPa P62 34-4 F F sa = s cos2a 2 s ta = sin2a a = 0o s = A FN FN = –F = –5 kN = – 50 MPa a = 30o a = 45o a = 60o a = 90o – 50 MPa 0 – 37.5 MPa – 21.65 MPa – 25 MPa – 25 MPa – 12.5 MPa – 21.65 MPa 0 0 a n sa ta s s s s 力的单位 ：kN 正负 P63 35-1 FA FB FD FC A B C D a a a FNi EA FNi a Dli = Dl = S Dli – 4 kN 1 kN 3 kN – 0.05 mm 0.0125 mm 0.0375 mm 0 e 呢? e 与 Dl 区别? Se ? 讨论： CD BC AB P63 35-2 l 2l 2h h A B C D E H 1 2 3 FN1 FN2 F = FN3 FN1 = – 80 kN FN2 = – 40 kN FN3 = 120 kN E1A1 FN1 2h Dl1 = = – 0.4 mm E2A2 FN2 2h Dl2 = = – 0.4 mm E3A3 FN3 h Dl3 = = 0.2 mm dC dH Dl1 Dl2 dC =│Dl1│ dH = dC + Dl3 = 0.4 mm = 0.6 mm (1) (2) dC dH Dl1 Dl2 Dl1 = – 0.2 mm Dl2 = – 0.4 mm Dl3 = 0.2 mm dH = dC + Dl3 = 0.267 mm = 0.467 mm dC = │Dl1│+ │Dl2│ 3 1 3 2 位移分析! P64 35-3 A B C l = 4m h = 3m 5m q = 10 kN/m SMA = 0 FBC FAy FAx FBC = kN 3 100 FN = (1) 斜杆用钢丝索 A = n 4 pd2 s = A FN ≤ [s ] ∴ 取 n = 67 或 (2) 斜杆用两根等边角钢 A = 2A1 s = A FN ≤ [s ] ∴ 取 ∟ 20×3 ( A1 = 1.13 cm2 ) 结论! n ≥ 66.3 66 A1 ≥ 1.044 cm2 平衡计算 用型号表示 设计截面尺寸 P64 35-4 F A C B ① ② 45o 60