工程力学吉林大学经典课件

第一部分第一部分 静力学静力学8/2/20241KylinsoftKylinsoft次数次数 章节章节题目题目一一1(S)P23-1.1(l)(g)、P24-1.2(k)、P27-1.3(c)(e)二二2(S)P58-2.3、P60-2.14、P61-2.20三三2(S)P63-2.28、P63-2.30、P64-2.32、P64-2.34四四3(S)P85-3.2、P86-3.11、P87-3.18、P90-3.28(f)、P91-3.40、P92-3.43五五4(S)P111-4.2、P113-4.12、P113-4.13 六六4(S)P114-4.14(b)、P115-4.24、P116-4.27七七5(S)P131-5.2、P132-5.12、P133-5.18、P133-5.19、P134-5.22作业题目作业题目作业题目作业题目8/2/20242KylinsoftKylinsoftP23-1.1(l):8/2/20243KylinsoftKylinsoftP23-1.1(g):8/2/20244KylinsoftKylinsoftP24-1.2(k):8/2/20245KylinsoftKylinsoftP27-1.3(c):8/2/20246KylinsoftKylinsoftP27-1.3(e):8/2/20247KylinsoftKylinsoftP58-2.3:Solution:由此得到如下方程组：解得：8/2/20248KylinsoftKylinsoftP60-2.14:Solution:按定义有:So,expressions(a),(d),(e)are valid.8/2/20249KylinsoftKylinsoftP61-2.20:Solution:(a)Vector form:8/2/202410KylinsoftKylinsoft(b)Vector form:8/2/202411KylinsoftKylinsoftP63-2.28:+Solution:8/2/202412KylinsoftKylinsoftP63-2.30:+（a)Solution:8/2/202413KylinsoftKylinsoft(b)+8/2/202414KylinsoftKylinsoftOOriginal forces and couple can be reduced to a force and a couple acting at point O:P64-2.32:Solution:于是得到合力的大小、方向和作用线位置（如图所示）：(或,或)8/2/202415KylinsoftKylinsoftP64-2.34:+Solution:8/2/202416KylinsoftKylinsoftP85-3.2:Solution:RAyRByFrom the FBD of the beam in Fig:+RBx3kN2m+8/2/202417KylinsoftKylinsoftP86-3.11:Solution:GPGL把人和梯子作为研究对象，作出FBD(a)x=1.5m+RBRAyRAx+8/2/202418KylinsoftKylinsoft+GPGLRBRAyRAx(b)8/2/202419KylinsoftKylinsoftP87-3.18:Solution:300N0.5mRDRE(1)Studying entire structureFrom the FBD of entire structure in Fig:+(2)Studying member ABFrom the FBD of member AB in Fig:RBxRByRAxRAy300N0.5m1.6m+8/2/202420KylinsoftKylinsoft(3)Studying member ACEFrom the FBD of member ACE in Fig:0.8mRERAxRAyAECRCxRCy0.8m0.6m0.6m+8/2/202421KylinsoftKylinsoftP90-3.28(f):Solution:(1)Studying member CDFrom the FBD of member CD in Fig:显然，本题中在约束A、B、C、D处没有水平方向的作用力RCRMRCR2qC3m1m+8/2/202422KylinsoftKylinsoft(2)Studying member ABCRARCRB2qAC1m2mRARCRB1mFrom the FBD of member CD in Fig:+8/2/202423KylinsoftKylinsoftP91-3.40:Solution:(1)From the FBD of joint A in Fig:APFABFAD12+8/2/202424KylinsoftKylinsoft(2)From the FBD of joint B in Fig:BPFBCFBDFAB+8/2/202425KylinsoftKylinsoft(3)From the FBD of joint D in Fig:DFBD+FDEFADFCD1212128/2/202426KylinsoftKylinsoft(4)From the FBD of joint C in Fig:CRCFCEFBCFCD12+8/2/202427KylinsoftKylinsoftP92-3.43:Solution:(1)From the FBD of joint D in Fig:DPFCDFDH+(1)From the FBD of joint in Fig:HFDHFGFC2323+8/2/202428KylinsoftKylinsoftP111-4.2:Solution:+8/2/202429KylinsoftKylinsoftP113-4.12:Solution:RAzRAyRByRBxRCxRCzFrom the FBD of bent pipe and cable in Fig:8/2/202430KylinsoftKylinsoft无解！？8/2/202431KylinsoftKylinsoftP113-4.13:Solution:8/2/202432KylinsoftKylinsoftP114-4.14(b):Solution:S令形心坐标为(xc,yc),按定义有:8/2/202433KylinsoftKylinsoftP115-4.24:Solution:已知crucible的密度rc=2000kg/m3,外径R=0.28m,内径r=0.2m;iron的密度ri=7200kg/m3OStudying crucible，Drawing FBD in Fig.WiWcGiGcFyFx+8/2/202434KylinsoftKylinsoftOWiWcGiGcFyFx8/2/202435KylinsoftKylinsoftP116-4.27:Solution:x=0 8/2/202436KylinsoftKylinsoftxyO2xyO1P131-5.2:解：3036lb20lb18in假设系统是静力平衡的先研究圆柱，有P1G1G2N1N2f2f1P1得再研究方块，有得而因，所以系统不能保持静力平衡8/2/202437KylinsoftKylinsoftP132-5.10:Solution:总体分析:分析木条:圆盘分析:解之得：解之得：NfffCBA.=22.06=达到最大静摩擦力时:解之得：解之得：8/2/202438KylinsoftKylinsoft9000N/mBCDBCDP132-5.12:解：P假设系统是静力平衡的先研究桩子CBD，有NBNBNAfAfBRxRy再研究杆AB，有P9000N/m1.5m2m3m0.8mABCDABfBAB8/2/202439KylinsoftKylinsoft只有当fAfAmax或fBfBmax时，杆AB才能移动，由以上计算结果可知，当P6300(N)，杆AB在A点处开始移动8/2/202440KylinsoftKylinsoftP133-5.18:Solution:因此因此:8/2/202441KylinsoftKylinsoftP133-5.19:Solution:木块翻转临界时：滑动时：8/2/202442KylinsoftKylinsoftP134-5.22:当P=600 lb时：当P=200 lb时：Solution:8/2/202443KylinsoftKylinsoft