Mechanical Engineering Design Notes



Design Contents

FUNDAMENTALS
Preliminary Matters
Design Methodology
..brain storming
..evaluation matrix
..QFD
Statistical Considerations
..variability in materials
..variability in dimensions
..variability in loading
..preferred sizes
Tolerances
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
..fatigue
..fracture mechanics
Instability - Buckling
Concentrically Loaded Strut
..slender columns
..Euler formula
..effective length
..short and intermediate columns
Eccentrically Loaded Strut
.. theory
Shock Loading
..deflection
..stress








FAILURE OF AN ECCENTRICALLY LOADED STRUT - THE SECANT FORMULA

1. Introduction
In practice struts will either be designed to withstand a compressive load applied at a specified eccentricity, or else for a nominally concentric load which will inevitably be applied slightly non-concentrically due to manufacturing and/or assembly tolerances.

A concentrically loaded strut remains straight under an increasing axial compression until the critical load is reached when it suffers sudden collapse. A strut subject to an increasing eccentric load starts to bend immediately and the deflection will increase as the load increases. In this situation failure is reached when the maximum stress or the maximum deflection exceeds that allowable, rather than the sudden buckling collapse of a concentrically loaded strut.
This section deals with the eccentrically loaded strut (the case of a concentrically load strut has been covered in a previous section, See this link).

Assuming the strut is subjected to a compressive load 'P' at a distance 'e' (the eccentricity) from the axis of the strut, this system can be represented by the equivalent system consisting of a concentric load 'P' and moments at each end of M = Pe. This will cause some bending of the beam, see diagrams below.



From the free body diagram of the portion of the strut AQ, the bending moment at Q is given by:

M = - Py - MA = - Py - Pe From the the equation relating bending moment to curvature: d2y/dx2 = M/EI = - Py/EI - Pe/EI, where E is Young's modulus and I is the second moment of area of the section. Letting p2 = P/EI . . . (a), then re-arranging: d2y/dx2 + p2 = - p2e The general solution of the above equation is: y = A.sin(px) + B.cos(px) - e . . . (b) where e is a particular solution. The constants A and B are obtained from the boundary conditions.

At end A, x = 0 and y = 0 and substituting in eqn. (b) gives B = e

Putting x = L and y = 0 and substituting in (b) gives:

A.sin(pL) = e{1 - cos(pL)} . . . (c) Noting that: sin(pL) = 2sin(pL/2).cos(pL/2) and 1 - cos(pL) = 2sin2(pL/2) and substituting into eqn (c) and rearranging givesL A = e.tan(pL/2) Substituting for A and B in eqn. (b) gives the equation of the elastic curve: y = e {tan(pL/2).sin(px) + cos(px) - 1} . . . (d) The value of the maximum deflection is determined by putting x = L/2 in eqn. (d). ymax = e {tan(pL/2).sin(pL/2) + cos(pL/2) - 1}

ymax = e {[sin2(pL/2) + cos2(pL/2)]/cos(pL/2) - 1}

ymax = e {sec(pL/2) - 1} . . . (e)

Replacing p with (P/EI)0.5 and substituting in (e):

ymax = e {sec([P/EI]0.5L/2) - 1} . . . (f)

From (f): ymax theoretically becomes infinite when [P/EI]0.5L/2 = 3.14159/2 . . . (g)

While the deflection does not become infinite, it will be unacceptably large and P should not be allowed to reach the critical value which satisfies eqn. (g). Solving eqn. (g) for P gives:

Pcr = 3.141592EI/L2 or = 3.141592EI/Le2 where Le is the effective length. See this link

The maximum stress sigmamax occurs in the section of the strut where the bending moment is maximum, ie in the transverse section through the midpoint for the above configuration and can be obtained by adding the normal stresses due to the axial force and the bending momentexerted on that section.

sigmamax = P/A + MmaxC/I . . . (h)

From the free body diagram on the right (above):

Mmax = Pymax + MA = P(ymax + e)

Substituting this value into (h) and noting that I = Ak2 gives:

sigmamax = (P/A)[1 + {(ymax + e)C}/k2] Substituting ymax from (f) gives: sigmamax = (P/A)[1 + (eC/k2)sec({P/EI}0.5{l/2})]

Link to applet to carry out calculations

David J Grieve Revised: 7th October 2013.

Contact the Author:
Please contact me for comments and / or corrections or to purchase the book, at: davejgrieve@aol.com