4-1b)求形心的位置 yc=0 500 A1=500×600mm,C=500mm 36 A2=(500-36×2)×(e 260 ZC2=480mm 600 (b) A3=200×600mm.zC3=100mm ∑A aizi+a2z2+A =260mm ∑A A1+A2+A
4-1 b）求形心的位置 yC = 0 Ⅰ Ⅱ A1 = 500 × 600mm , z 2C = 500mm 1 A2 = (500 − 36 × 2 ) × (600 − 40 )mm 260 y 2 2 Ⅲ zC 2 = 480mm A3 = 200 × 600mm , zC3 = 100mm zC ∑Az = ∑A i Ci i A1 z1 + A2 z2 + A3 z3 = = 260mm A1 + A2 + A3 1
4-4b)试计算平面图形对形心轴yc的惯性矩 500 「500×600 +500×600×(500-260)mm4 12 263×1010mm4 (500-72)×5603 +(500250×(480-260)h 12 260 79×1010mm4 600 600×2003+200×600×(260-100)1m mm 4 0.347×1010mm4 y=1yc+ly-ly=(263+0.347-179)×10mm1.19×10mm10,4
4-4 b）试计算平面图形对形心轴yC的惯性矩。 I yC1 ⎡ 500 × 6003 2⎤ =⎢ + 500 × 600 × (500 − 260) ⎥ mm4 ⎣ 12 ⎦ =2.63×1010mm4 I yC 260 2 ⎡(500− 72)×5603 2⎤ =⎢ + (500− 72)×560× (480− 260)⎥mm4 ⎣ 12 ⎦ =1.79×1010mm4 ⎡ 600 × 200 3 2⎤ I yC3 = ⎢ + 200 × 600 × (260 − 100 ) ⎥ mm 4 ⎣ 12 ⎦ =0.347×1010mm4 I yC = I yC + I yC − I yC = (2.63 + 0.347 −1.79)×10 mm = 1.19×10 mm 2 10 4 10 1 3 2 4
4-1c)求形心的位置 ArC 查型钢表,槽钢No.14b,z01=1.67cn No.b A1=21.3c c=(20+167)cm 工字钢A2=396cm2,=0cm 14.1 No 20b Zc= A1Zci+A2 ZC2 141cm Al+a 4-4c)试计算平面图形对形心轴yc的惯性矩。 槽钢Iy=61.1cm,玊字钢Iy=2500cm I yc=Ily+ Ai(h+ A 4.45×10
4-1 c）求形心的位置 C1 yC=0 查型钢表，槽钢No.14b，z01 = 1.67cm y1 A1=21.3cm2，zC1 = (20 + 1.67)cm 工字钢 A2 =39.6cm2， h zC2 = = 10cm C2 2 14.1 y2 zC = A1 zC1 + A2 zC2 A1 + A2 = 14.1cm 4-4 c）试计算平面图形对形心轴yC的惯性矩。 槽钢 I y1 = 61.1cm , 工字钢 4 I y = 2500cm 2 4 I yC = I[y1 + A1(h + z01 − zC ) 2 ] ⎡ ⎛ h⎞ + ⎢I y2 + A2 ⎜ zC − ⎟ ⎢ ⎝ 2⎠ ⎣ 2 ⎤ −5 4 ⎥ = 4.45 × 10 m ⎥⎦ 3
51试求梁中截面1-1、2-2、3-3上的剪力和弯矩。设P、q、a 均为已知 I 9 P=qa Fsi=P+ ga=2ga m=44- B M1=-P Fs2=P+ga= 2qa M2=-Pa-qa+m=-ga 2
5.1 试求梁中截面1-1、2-2、3-3上的剪力和弯矩。设P、q、a 均为已知。 (a) MA FS1 = P + qa = 2qa M 12 32 1 = − Pa − qa = − qa 2 2 RA FS 2 = P + qa = 2qa M 12 12 2 = − Pa − qa + m = − qa 2 2 4
D 51试求梁中截面1-1、2-2、3-3上的剪力和弯矩。设P、q、a 均为已知。 2=10kN/m B D RA 200+200200-B ∑Mn=0.,RAx06-10×04×02=0,RA=133N ∑F、=0,RA+RB-10×04=0,RB=267kN Fs1=RA=1.33kN,M1=RA×0.2=267N·m Fs2=RA-10×0.2=-0.67kN M2=RA×0.4-10×0.2×0.1=333N
5.1 试求梁中截面1-1、2-2、3-3上的剪力和弯矩。设P、q、a 均为已知。 (c) RA RB ∑M B ∑F = 0, R A × 0.6 − 10 × 0.4 × 0.2 = 0, R A = 1.33kN = 0, R A + RB − 10 × 0.4 = 0, RB = 2.67 kN y FS 1 = R A = 1.33kN, M 1 = RA × 0.2 = 267 N ⋅ m FS 2 = R A − 10 × 0.2 = −0.67 kN, M 2 = RA × 0.4 − 10 × 0.2 × 0.1 = 333N ⋅ m 5