结构力学 静定梁和静定刚架习题.ppt

静定梁与静定刚架 Valadon onry. ch Asposeslides for NET 4o dient P Evaluation only. Created with Aspose Slides for NET 4.0 dient Profilo 71 Copyright 2004-2017Aspose Pty L 静定梁和静定刚架习题 习题1]作弯矩图 取整体为研究对象, ∑Mn=0,VAL+M=0 得:VA=-ML Ha ∑Y=0,VA+VB=0,得: VDE ML L/2 L/2 取左半为研究对象, ∑Mc=0,VL2=HAL, 得:H=-M2L(向左) ∑Ⅹ=0,HA-H3=0 得:H3=HA=M2L(向右) Ha Valadon onry. ch Asposeslides for NET 4o dient P Evaluation only. Created with Aspose Slides for NET 4.0 dient Profilo 71 Copyright 2004-2017Aspose Pty L 取AD为研究对象,∑M=0, 得:MDA=HAL=-M2(右侧受拉)H 同理:MB=H1L=M/2(左侧受拉) Valadon onry. ch Asposeslides for NET 4o dient P Evaluation only. Created with Aspose Slides for NET 4.0 dient Profilo 71 Copyright 2004-2017Aspose Pty L 集中力矩的右侧作截面,取 BEC为研究对象, ∑M(=0, McE+H1LVBL/2=0,得:bA Mc=M(下侧受拉) MDc与MEc可由结点D和E 的平衡条件得到。 Valadon onry. ch Asposeslides for NET 4o dient P Evaluation only. Created with Aspose Slides for NET 4.0 dient Profilo 71 Copyright 2004-2017Aspose Pty L 20 kN/m 工习题2]作弯矩图 几何构造分析 ABCDE是附属部分, 41 BE是基本部分 F 2.先计算附属部分,是简支刚架 取整体为研究对象, ∑M=0, HR=45KN 得:V ∑ 得:VB=135kN ∑X=0,得:HB=0 Mc=V×3=135kNm,下侧受拉 *CD、AB是悬臂杆,直接写出弯矩 M3A=90kNm,MCD=90kNm,上侧受拉 Valadon onry. ch Asposeslides for NET 4o dient P alveton o Created with Aspose Slides for NET 4.0 dient Profilo 71 Copyright 2004-2017Aspose Pty L 3基本部分的计算,为悬臂杆。 Vp=135 M=135×3=405kNm,左侧受拉 4.作出弯矩图。 405 135 45 Valadon onry. ch Asposeslides for NET 4o dient P Evaluation only. Created with Aspose Slides for NET 4.0 dient Profilo 71 Copyright 2004-2017Aspose Pty L 习题3仵弯矩图,剪力图,轴力图。 4 kN 1取整体为研究对象, ∑MA=0,V(×9 2 kN/m 4×5-2×5×2.5=0, 解得Ⅴ=5kN, ∑Y=0,V=5kN m栏m4m ∑X=0,HA=8kN 8 kN 6 KN 5=10kN Valadon onry. ch Asposeslides for NET 4o dient P Evaluation only. Created with Aspose Slides for NET 4.0 dient Profilo 71 Copyright 2004-2017Aspose Pty L 2.取BC为研究对象, ∑M3=0,得:MBc=22kNm (下侧受拉) 4 kN ∑Y=0,得:VBC=-1kN ∑X=0,N=0 BC杆弯矩图的做法 22kNm 16/3kNm 0 Valadon onry. ch Asposeslides for NET 4o dient P Evaluation only. Created with Aspose Slides for NET 4.0 dient Profilo 71 Copyright 2004-2017Aspose Pty L AB