Impact. m k. Natural Period of Vibration τ. Static load Gray area Impact load t > 3 τ. Absorbing energy. Carrying loads

Transcription

1 Impact also called shock, sudden or impulsive loading driving a nail with a hammer, automobile collisions. dashpot a) Rapidly moving vehicles crossing a bridge To distinguish: b) Suddenly applied c) Direct Impact Combustion in pile drive, engine cylinder drop forge, car crash Compare Time of Load Application t with Natural Period of Vibration τ = 2π Static load Gray area Impact load t > 3 τ 3 τ > t > 1/2 τ t < 1/2 τ Carrying loads Design for m k Absorbing energy

2 Impact material properties vary with loading speed ksi S y /S % orks favorably u Promotes brittle fracture S u elongation S y In praxis: Static properties known Impact Factor x4 for automobile suspension parts Loading rate only approximated

3 Linear & Bending Impact k h Guide rod k F e Elastic-strain energy stored in structure = 0.5F e ork of falling weight = (h+) st h Assumptions: No mass of structure No deflection within impacting mass No damping dynamic = static deflection curve only multiplied by impact factor all energy goes into structure most severe impact (h + ) = 0.5 F e F e = k k = / st Impact Deflection Impact Load Impact Stress F = σ e e st = = σ st F e = / st 2h st 2h Impact Factor st 2h st

4 Linear & Bending Impact k h Guide rod (h + ) = 0.5 F e F e = k k = / st F e = / st k F e st For Elastic-strain energy stored in structure = 0.5F e ork of falling weight = (h+) h >> st h v 2 = 2gh h= v 2 /2g h= 0 or v=0 suddenly applied load Impact Factor = 2 (Double Safety Factor) Impact Factor

5 Linear & Bending Impact k h Guide rod (h + ) = 0.5 F e F e = k k = / st k F e st For Elastic-strain energy stored in structure = 0.5F e ork of falling weight = (h+) h >> st h F e = / st Vertical movement Impact Factor Horizontal movement = k st = g m Impact kin. Energy U = 0.5 m v 2 As greater Stiffness and kin. Energy as greater Equivalent static force

6 Linear Impact of Straight Bar σ = F e / A k = A E / L σ Impact Energy Capacity of a straight rod Basic relationship Gives optimistic results Calculates stress lower than actual stress because of - Non-uniform loading & stress distribution - Mass of the impacted rod

7 Linear Impact of Straight Bar How compare Energy-Absorbing Capacaties? Set = S y But V b = 2½ V a mass specific energy capacity Shows Importance of Section Uniformity

8 Impact Absorption Capacity = F e L / AE U = 0.5 F e F e = S e A A bumper mass specific energy capacity S e elastic limit

9 Bending Impact Scan p285 u equations Scan p286 o equations

10 Effect of Stress Raisers on Impact Ratio 2.25 : 1 : 1/64 225% : 100% : 1.56%

11 Effect of Stress Raisers on Impact Due to combination of Impact and Stress Raiser Brittle fracture is promoted with stress concentration factor almost equal K t

12 Effect of Stress Raisers on Impact pilot Improved Design Reduce Stress Concentration and design for Uniform Stress Distribution < Head/Pilot Shank

13 E f f e c t o f St r e s s R a i s e r s o n I m p a c t pilot Improved Design Reduce Stress Concentration and design for Uniform Stress Distribution Raiser pilot Axial hole

14 Torsional Impact Linear deflection [m] F e eq. static force [N] m mass [kg] k spring rate [N/m] v velocity [m/s] U kin. energy [Nm] Torsional θ deflection [rad] T e eq. static Torque [Nm] I moment of inertia [kgm 2 ] K spring rate [Nm/rad] ω velocity [rad/s] U kin. energy [Nm]

15 Torsional Impact ρ=2000kg/m 3 G=79GPa 2400rpm sheave suddenly stopped U=0.5 I ω 2 kin. energy I=0.5 m r 2 moment of inertia of wheel m= π r 2 b ρ mass of wheel U=0.25 π r 4 b ρω 2 U=0.25 π (0.060) 4 (0.002) (2000) (2400x2π/60) 2 Nm U=25.72 Nm T=τJ/r

16 Exam Monday April 18 Chapter 5: 5.1-2, 5.5-7, Chapter 6: 6.1-2, 6.5-8, Chapter 7: 7.1-2, 7.4 Chapter 8: Elastic Strain Deflection Stability Failure Theories Safety Factors Impact Fatigue