工程力学A参考习题之平面任意力系习题及解答(最新整理)

1、第三章平面任意力系习题及解答构架如图，不计各杆自重,已知力 f求铅直杆 ab 上铰链 a、d 和 b 所受的力。解：1.取整体，画受力图mc (f ) = o- fby .2a = 0解得： fby = 02. 取 def 杆，画受力图me (f ) = oa.fdy- a.f = 0解得： fdy = fmd (f ) = oef sin 450.a - f.2a = 0eef sin 450 = 2f fx = of cos 450- fdx = 0解得：fdx= fecos 450 = 2f3. 取 adb 杆，画受力图m a (f ) = ofdx .a + fbx .2a = 0fb

2、x = -f fy = ofay+ fdy+ fby = 0解得：fay= -f图示构架中，物体重 1200n，尺寸如图，不计杆和滑轮的重量。求：a、b处的约束反力及杆 bc 的内力解：1.整体受力如图（a），有 fx = ofax- ft = 0 fy = ofay- p + fnb = 0mb (f ) = o p(2 - r) - 4fay- ft (1.5 - r) = 0式中 r 为轮的半径,ft=p,解得：fax = 1200nfay = 150nfnb= 1050n2.取 adb 为研究对象:如图(b) md (f ) = o2fbc sinq+ 2fnb- 2fay = 0解得

3、： fbc已知 r=a ,p=2f , co=od, q 。= -1500n (压力)求：支座 e 及固定端 a 处的约束反力。解:1.取 cod 及滑轮为研究对象，如图(b)m c (f ) = o3 2afrd- 3af +3ft r - ( 2 a+ r) = 0解得：fre= frd =2f2. 取 abcod 为研究对象，受力如图(a),由 fx = 0fax + 6aq - frd cos 45o = 0 fy = ofay- p - f + frd sin 45o = 0m a (f ) = oma - 6aq.3a - 5.5ap - 6af + 62afrd = 0解得：fa

4、x= f - 6qafay = 2fm a = 5af+ 18qa2图示构架中，已知: f=200n,m=100n.m,尺寸如图，不计各杆自重, 求：a、b、c 处的约束反力。解：1.整体受力如图（a），有me (f ) = o-1.6fay - m- f (0.6 - 0.4) = 0解得： fay= -87.5n2. 取 bd 为研究对象，如图(b)md (f ) = ofnb0.8sin 300 - m- 0.6f = 0解得： fnb = 550n3. 取 abc 为研究对象，如图(c) mc (f ) = oax1.6sin 600.f- 0.8fay- 0.8fnb = 0 fx

5、= ofax- fnbcos 300 + f= 0 fy = ocxfay+ fnb sin 30o + fcy = 0解得：fax= 267n , fcx= 209n , fcy= -187.5n图示桁架中，.abc 为等边三角形,ad=bd=bf=cf=ce=ae 不计各杆自重, 求：cd 杆的内力。解：1.整体受力如图（a），有m a (f ) = ofab -10 = 0nbfabsin60 .32解得：fnb =4 f2 将桁架截开,取右边为研究对象，如图(b)nbmd (f ) = offc.db.sin 600 + f.db - f.df.sin 600 = 0解得：ffc= 1

6、 f2 fx = o3. 取节点 c 为研究对象，如图(c)(fcf - fce ) sin 30o = 0 fy = o- (fcf解得：+ fce ).cos 30o - fcd = 03fcd = - 2f = -0.866f (压)已知：o1a = 0.1m ,o1o2 = ad = 0.05m,o a,图示瞬时w= 2rad / s1j= 300 ；求 ： 三角板 abd 的角速度w和点 d 的速度vd 。解: 如图所示, 三角板 abd 的速度瞬心为点 c，故w= vaca=o1 a.wo a.1o1 a + o1o2 .cotj= 1.072rad/s已知： oa = bd =

7、de = 0.1m ,ef = 0.13m,woa = 4 rad / s ;求ef 杆的角速度w和滑块 f 的速度vf 。解: 各点速度分析如图所示, ab 杆为瞬时平动,故vb = va = oa.woa = 0.4m/sbc 杆的速度瞬心为点 d，三角形 dec 绕 d 点作定轴转动，得v= de. vcedc= de. v bbd= v bfefe由vw= vw+ vw解出v f=vecos300= 0.462m/s，wef= vfeef= 1.333rad/s“”“”at the end, xiao bian gives you a passage. minand once said

8、, people who learn to learn are very happy people. in every wonderful life, learning is an eternal theme. as a professional clerical and teaching position, i understand the importance of continuous learning, life is diligent, nothing can be gained, only continuous learning can achieve better self. only by constantly learning and mastering the latest relevant knowledge, can employees from all walks of life keep up with the pace of enterprise development and innovate to meet the needs of the market. this