Question

The 3505-kg anvil A of the drop forge is mounted on a nest of heavy coil springs having a combined stiffness of 2.2(106) N/m. The 590-kg hammer B falls 430 mm from rest and strikes the anvil, which suffers a maximum downward deflection of 20 mm from its equilibrium position. Determine the height h of rebound of the hammer and the coefficient of restitution e which applies.

B 430 mm

Answer 1

Given M = 3505 kg stifiness ka 2-2x 10 N/m. hammer = 590kg, Falls from 430mm. Deflection = comm velocity of hammer = J2gh - v = (2x 9.81x 0.43 = 2.904 mis. energy for anuil Equation for change in kinetic AT= 0 - 12 mn mur (-V = Velocity of anvil) elastic potential energy Expression for change in for anvil Ave / Ex + Fox = 1 (kxx+ mgx = kx + mg x gravitational obtain expression for change in potential energy for anvil. Avg = ngx.

From Energy balance equation for anuit AE = AT + Over Aug 0 -4 mm, + } *x'+ mgx-mgx - mut I kx (x= Deflection) = -1 * 3505x4,+ * 2.2x16 x 0.02 VT 6-2510 => V -$80 = 0.5 mis after impact using momentum Velocity of hammer equation AG = my mavom o= mv, - mnu'amnu 590x 2.904 o= 3505x0.5 - 5900 - v'x590 - 3505x0.5-5900 2.904 Jul = 0.0663 mis = Jagh 0.0663 = 2x9.81xh (0.0663)² = 2x9.51xh ha (0.0663) = 2.24x104m 2x9.51 = 0.24 mm

coefficient of estitution e = v'-(-u) V e = 0.0663 - (-0.5) 2.904 = 0.195
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