Mechanical Engineering Design Notes

Design Contents

Preliminary Matters
Design Methodology
..brain storming
..evaluation matrix
Statistical Considerations
..variability in materials
..variability in dimensions
..variability in loading
..preferred sizes
Design Factor
Introduction to Failure
Failure Theories
Application of von Mises
..criterion in 2 D

Stress Concentration
..and notch sensitivity
Failure Under Combined Loading
..combined bending and torsion
Failure Under Cyclic Loading
..fracture mechanics
Instability - Buckling
Concentrically Loaded Strut
..slender columns
..Euler formula
..effective length
..short and intermediate columns
Eccentrically Loaded Strut
.. theory
Shock Loading


1. Theory

2. Comparison Between Slowly and Suddenly Applied Load

3. Worked Example

Assume a mass of 2kg travelling at 0.5 m/s hits the free end of a 3m long steel cantilever beam which has a cross section 50mm deep (in the direction that the load is travelling) by 60mm wide. Assume E for steel is 200GPa.

Determine the maximum dynamic deflection and stress and what the equivalent drop height is.

I = b d3/12 = 0.06(0.05)3/12 = 625.0E-9m4 The static (due to gravity) deflection is given by: y = P(L)3/3 E I = 2 x 9.81(3)3/(3 x 200E9 x 625.0E-9) = 1.41E-3m or 1.41mm The maximum stress = Mmax x half depth of beam/I = 2 x 9.81 x 3 x 0.025/625.0E-9 = 2.35MPa For the impact load:
The maximum deflection = v (m L3 / 3 E I)0.5 = 0.5(2 x (3)3/3 x 200E9 x 625.0E-9)0.5 = 6.0E-3m or 6mm Maximum stress is proportional to maximum deflection and in the dynamic case = 6 x 2.35/1.41 = 10MPa

To determine the equivalent drop height - use the equation of constant acceleration:

v2 = u2 + 2 a s

(0.5)2 = 2 x 9.81 x s

s = (0.5)2/2 x 9.81 = 12.7mm

Dropping a load from a height of just 12.7mm generates a stress exceeding four times the stress generated by the same load applied slowly.

David J Grieve, 1st March 2005.

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