Question

x H Neutral Line Point A L2 [ f(t) L/2 + A double-clamped polyethylene cantilever has length L=5X cm and circular cross section H=4mm. An Impulse force f(t) = 1048(t) applied to the beam at point A. The maximum beam deflection Xmax (at point A) can be obtained from the equation Xmax 192E1 where fis the force, E=2GPa and is the cross-section moment of inertia. Assume the material has internal damping property and system damping ratio is 5 = 0.5. 1- If we model the beam a 1DOF spring (k), mass (m), damper (b) system at point A, Find equivalent model parameters. Polyethylene density is 1200kg/m3. (17 points) (assume equivalent mass of the system in the model is equal by total beam mass and x is measuring at the center of the beam).

Answer 1

Given H-4mm, 20 1/2 @ 니2 k ft L=5 cm = 0.05m H=4 mm= o.ooym fløj = 10Y SCH = 713 192 EL E = 2G pa g= 1200 kg/m3 2 @ 4 Xmax Ze=0.5 solution 1. Free Body Diagram of the beam assuming it to 1 DOF, spring maits damper system be Kuct cacts xia om where fit) ma equivalent mass of the beau Ko squiualent stiffness of the beam c→ Equivalent damping coefficient of the bram → from Newtons a z (forces) = m.ar → from - Kach - crct + f(t) = mult and Law of of mohon the F.B.D.

» much + củlt + kalt act+ fl8 7 (wy i C#+K & CA ( ) x Ct м Governing £.Om for the system, → Now the damping factor (5)= G = amon c= 2zon And ont =k m 3 → Then £.0.7. is tik sich + ( en sect China y ect)= f CH u dcb+lez@) (+) CH = CH - با Modified governing com. (equation of Motion) ► Now, matural paequency (in scadlo) of the system a on g Xmax- 192 EI 1926L 192.617 ing ² P.A.L. 11 192x 2X10 XA (0.00442 64 1200x 2 (oooy Bé xo.os4 9 Or 755.41 sead/8

be rewaelten B.AL → Now Modified Com. (eqh 4 can e (H+ (2x0,5%#$Soul) éch+#15.415 z«= 10YSCH (H+ +55oul it + 512X10 xlt=loy Set 1200xñ (0.0043x y Do rectit 1554) & (Alt st2x10 x (H)=13262.91Y SA Ly Required Model of the given 1 Dof, spring- mass- damper system,