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Solutions to problems related to dynamic force analysis in mechanical engineering. The problems involve finding the equilibrium moments and joint forces in various mechanisms given their dimensions, angular speeds, and external forces or moments. The solutions are based on the principles of mechanics and are presented using kinematic data and free-body diagrams.

Typology: Assignments

Pre 2010

1 / 6

Download Dynamic Force Analysis: Solving Equilibrium Moments and Joint Forces in Mechanisms - Prof. and more Assignments Mechanical Engineering in PDF only on Docsity! Dynamic Force Analysis 1 6 Force Analysis Homeworks: Joint reaction forces and equilibrium moment Problem 6.1: four-bar mechanism The four-bar mechanism shown in Fig. P6.1 has the dimensions: AB=CD=0.04 m and AD=BC=0.09 m. The driver link AB rotates with a constant angular speed of 120 rpm. The links are homogeneous rectangular prisms made of steel with the width h = 0.01 m and the depth d = 0.001 m. The density of the material is ρSteel = 8000 kg/m 3 and the gravitational ac- celeration is g = 9.807 m/s2. The external moment applied on the link CD is opposed to the motion of the link and has the value Mext = −|Mext| ω3 |ω3| , where |Mext| = 500 N·m. Find the equilibrium moment on link AB and the joint forces for φ = 30◦ using: a) free-body diagram of individual links and b) dyad method. For φ=30◦ the kinematics of the mechanism are given by: xB=0.034641 m, yB=0.02 m, xC=0.103859 m, yC=-0.0375222 m, φ2=-39.7274 ◦, φ3=-69.7274 ◦, ẋB=-0.251327 m/s, ẏB=0.435312 m/s, ẍB=-5.47029 m/s 2, ÿB=-3.15827 m/s 2, ẋC=-0.884619 m/s, ẏC=-0.32675 m/s, ẍC=3.84741 m/s 2, ÿC=25.1222 m/s 2, ω2=φ̇2=-11.0095 rad/s, α2=φ̈2=307.84 rad/s 2, ω3=φ̇3=-23.5759 rad/s, α3=φ̈3=307.84 rad/s 2. Dynamic Force Analysis 2 Problem 6.2: R-RRR-RRT mechanism A planar mechanism is shown in Fig. P6.2. The following data are given: AB=0.150 m, BC=0.400 m, CD=0.370 m, CE=0.230 m, EF=CE, La=0.300 m, Lb=0.450 m, and Lc=CD. The constant angular speed of the driver link 1 is 60 rpm. The links are homogeneous rectangular prisms made of steel with the width h = 0.01 m and the depth d = 0.001 m. The steel slider 5 has the width wSlider = 0.05 m, the height hSlider = 0.02 m, and the depth d = 0.001 m. The density of the material is ρSteel = 8000 kg/m 3 and the gravitational acceleration is g = 9.807 m/s2. The external force on 5 is opposed to the motion of the link Fext = F5ext = −|Fext| vF |vF | , where |Fext| = 500 N. Find the equilibrium moment on link AB and the joint reaction forces for φ=φ1=30 ◦. For φ=30◦ the kinematics of the mechanism are given by: xB=0.129904 m, yB=0.075 m, xC=-0.0689445 m, yC=0.422073 m, xE=-0.298288 m, yE=0.404712 m, xF =-0.37 m, yF =0.186177 m, φ2=-1.05052 rad, φ3=0.0755515 rad, φ4=1.25372 rad. ẋB=-0.471239 m/s, ẏB=0.81621 m/s, ẍB=-5.1284 m/s 2, ÿB=-2.96088 m/s 2, ẋC=-0.0788027 m/s, ẏC=1.04105 m/s, ẍC=2.87595 m/s 2, ÿC=1.03567 m/s 2, ẋE=-0.127788 m/s, ẏE=1.68819 m/s, ẍE=4.66371 m/s 2, ÿE=1.67947 m/s 2, ẋF =0 m/s, ẏF =1.64625 m/s, ẍF =0 m/s 2, ÿF =3.29262 m/s 2, ω2=φ̇2=-1.1307 rad/s, α2=φ̈2=-22.33 rad/s 2, ω3=φ̇3=-2.82169 rad/s, α3=φ̈3=-2.20443 rad/s 2. ω4=φ̇3=0.58475 rad/s, α4=φ̈3=-21.453 rad/s 2. Dynamic Force Analysis 5 Problem 6.5 The dimensions of the mechanism shown in Fig. P6.5 are: AB=200 mm, AC=300 mm, CD=500 mm,DE=250 mm, CF = AB+AC, and La=400 mm. The constant angular speed of the driver link 1 is n=40 rpm. The links are homogeneous rectangular prisms made of steel with the width h = 0.01 m and the depth d = 0.001 m. The sliders 2 and 5 have the width wSlider = 0.05 m, the height hSlider = 0.02 m, and the depth d = 0.001 m. The density of the material is ρSteel = 8000 kg/m 3 and the gravitational acceleration is g = 9.807 m/s2. The external force on 5 is opposed to the motion of the link Fext = F5ext = −|Fext| vE |vE| , where |Fext| = 1000 N. Find the equilibrium moment on link AB and the joint reaction forces for φ=φ1=60 ◦. Find the velocities and the accelerations of the mechanism when the angle of the driver link 1 with the horizontal axis is φ=60◦. For φ=60◦ the kinematics of the mechanism are given by: xB=0.1 m, yB=0.173205 m, xD=-0.758831 m, yD=-0.19868 m, xE=-0.7 m, yE=0.0442993 m, φ2=0.408638 rad, φ4=1.33324 rad, ẋB1=-0.72552 m/s, ẏB1=0.418879 m/s, ẍB1=-1.7546 m/s 2, ÿB1=-3.03905 m/s 2, ẋD=0.30661 m/s, ẏD=-0.708086 m/s, ẍD=0.841861 m/s 2, ÿD=1.05257 m/s 2, ẋE=0 m/s, ẏE=-0.633848 m/s, ẍE=0 m/s 2, ÿE=0.846817 m/s 2, ω2=φ̇2=1.54324 rad/s, α2=φ̈2=-1.26276 rad/s 2, ω4=φ̇3=1.26188 rad/s, α4=φ̈3=3.0792 rad/s 2. Dynamic Force Analysis 6 Problem 6.6 The dimensions of the mechanism shown in Fig. P6.7 are: AB=180 mm, AC=90 mm, CD=200 mm, CF = CD, and AG = CD+AC. The constant angular speed of the driver link 1 is n=180 rpm. The links are homogeneous rectangular prisms made of steel with the width h = 0.01 m and the depth d = 0.001 m. The sliders 2 and 4 have the width wSlider = 0.05 m, the height hSlider = 0.02 m, and the depth d = 0.001 m. The density of the material is ρSteel = 8000 kg/m 3 and the gravitational acceleration is g = 9.807 m/s2. The external moment on 5 is opposed to the motion of the link Mext = M5ext = −|Mext| ω5 |ω5| , where |Mext| = 1500 N·m. Find the equilibrium moment on link AB and the joint reaction forces for φ=φ1=60 ◦. For φ=60◦ the kinematics of the mechanism are given by: xB=0.09 m, yB=0.155885 m, xD=-0.16138 m, yD=-0.0281381 m, φ2=0.631914 rad, φ4=0.172624 rad, ẋB1=-2.93835 m/s, ẏB1=1.69646 m/s, ẍB1=-31.9775 m/s 2, ÿB1=-55.3867 m/s 2, ẋD3=3.28823 m/s, ẏD3=-4.4918 m/s, ẍD3=178.405 m/s 2, ÿD3=18.6037 m/s 2, ω2=φ̇2=27.8338 rad/s, α2=φ̈2=451.854 rad/s 2, ω4=φ̇3=30.4604 rad/s, α4=φ̈3=992.942 rad/s 2.