工程力学(静力学与材料力学)-3-静力学平衡问题PPT课件

2021/6/712021/6/722021/6/73 2021/6/74FP1FP2FP3M1q(x)xFP6FP5FP4dxM2dxq(x)FQ(x)M(x)2021/6/752021/6/762021/6/77返回返回2021/6/782021/6/79R10FFnii 10FnOOiiMM2021/6/710RRR000 xixyiyzizFFFFFF000FFFOxOxiOyOyiOzOziMMMMMM2021/6/7110zF 00FFxyMM0FzM0FOM2021/6/7122021/6/7132021/6/7142021/6/715 0FAM0 xF0yF2021/6/716 0FAM0sin2TWQlFxFlFBQWQPT2sin2FlxFllFxFFB0 xF0cosTBAxFF23cos302QWQWFxlFllFxFFAx0yF0sinTPQBAyFFFF2QWFFlxlFAy2021/6/717QWQWT2sin2FlxFllFxFFBQWQWQWT22sin302sin22FFFFllFlFFB2021/6/7182021/6/7192021/6/7200 xF0AxF0yF0PFqlFAyqlFAy2 0FAM02PMlFlqlMA225qlMA2021/6/7212021/6/7222021/6/7232021/6/724FP2021/6/725FPFAyFAxFBC2021/6/726FPFAyFAxFBCP2 2BCFF2021/6/727FPFAyFAxFBC2021/6/728FPFAyFAxFBC2021/6/729M=FP l2021/6/730FAxFByFBxFAyM=FP lFBy FBx FCx FCy M=FP l2021/6/731FAxFByFBxFAy考察考察ABD杆杆 的平衡的平衡 MB ( F ) = 0 : MA ( F ) = 0 :FBy= 0FAy= 0Fx = 0 : FBx+ FAx=0 FBx= -FAx2021/6/732M=FP lFBy FBx FCx FCy 考察考察BC杆杆 的平衡的平衡Fx = 0 : FBx - FCx =0 FCx = FBx = FBxFy = 0 : FBy - FCy =0 FCy = FBy =FBy=0 MB ( F ) = 0 : FCx l+M = 0FCx = FBx = -FP2021/6/733M=FP lFPM=FP lFPFPFP2021/6/734FP2021/6/735FPFCyFAFCx2021/6/736FPFCyFAFCx ME ( F ) = 0 : MA ( F ) = 0 : MC ( F ) = 0 :FCx l -FP 2l = 0-FA l - FP 2l = 0-FCy 2l -FA l = 0E2021/6/737M=FP l2021/6/738M=FP lFAFC2021/6/739M=FP lFAFC MC(F) = 0 : FA= FC = FPFAl +M=02021/6/7402021/6/7412021/6/7422021/6/7432021/6/7442021/6/7452021/6/7460 xF0AxF2021/6/7472021/6/7480FBMR202ClFlMqlR24CMqlFl2021/6/749 0yFR20AyCFqlF 0AMR2240ACMqllMFlR24CMqlFllMqlFAy247MqlMA23部分资料从网络收集整理而来，供大家参考，感谢您的关注！