Equal constraint

Fig. 1.46 (Up) Constant thickness beams providing both equal constraint and parabolic flexure. For a finite width end the shape is a trapeze and otherwise a triangle. (Down) From segment cuts, as shown in dashed lines, one obtains the classical rearrangement for suspension systems ci = -Fcos do, C2 = 0, c2 = 2EI(F + ci)/F2, C4 = 0, (1.113a)

where the negative sign is taken for the square root term of (1.98b) and the positive for the two other equations. The equation for d0 is fEI rd0 dd £ =J _/_====. (1.113b)

2F cos d - cos d For a small flexure, d0 C 1, the integration leads to arcsin d/d0

hence d0 vanishes. This shows that the flexure only exists if d0 exists, i.e. if

This important result by Euler defines the so-called buckling critical load - here the force FCr = n2 EI/4£2 - where the beam suddenly ceases to be straight. This introduced the first mathematical formalism in the general notion of elastic stability.

Equations (1.100b) and (1.113b) are nearly identical. They are different only by the integration range. One is for the cantilever bars, while the second is for the compressed beams and buckling limit. Other buckling limits of straight beams can be determined from Eqs. (98) and taking into account the boundaries, e.g. for beams with ends either clamped or free to rotate (Fig. 1.47).

Another case of any importance, that of a beam ceasing to be straight under its own weight, was also treated by Euler and Lagrange. In 1773, Lagrange [90] determined a variable cross-section distribution of columns providing the strongest resistance to buckling.

• Torsion of beams of various cross sections: The torsion of rods of constant diameter was first investigated by Coulomb [39] in 1784 for the determination of the attraction laws in electrostatics and electromagnetism. Developing torsion balances for accurate measures of those forces and discovering the inverse square law

Fig. 1.47 Buckling of beams for various boundary conditions. The two beams on the left show the first and second buckling modes

q1 q2/d2, Coulomb showed that the torsion moment Mt around the z-axis of the rod is proportional to its torsion angle t per unit length. Also for cylinders of any uniform cross-section, the total rotation is the product 9 = tL of the angle t and the length L of the bar. The constant c linking the quantities Mt and t, so-called torsional rigidity and of dimension [F][L2], is (cf. for instance Timoshenko [158])

Mt GA4

T 4n2I

p ja where G is the shear modulus of the material, A the area of the cross-section, and Ip the polar moment of inertia around the center of the cross-section for which r is taken as the polar distance to the element area dA.

For a rod of diameter 2a, the polar moment of inertia and the torsional rigidity are

For a cylinder of elliptical cross-section whose lengths of principal axes are 2a, 2b, the polar moment of inertia and the torsional rigidity are

Saint-Venant [91] noticed that the assumption of cross-sections remaining plane during the torsion is perfectly valid only if Coulomb's theory is restricted to the case of a rod. For any other case, he devised the large torsion theory of various cross-section cylinders in the celebrated memoir La Torsion des Prismes (1855). Investigating bars of square (Fig. 1.48), rectangular, elliptical, and various cross-sections expressed in a polynomial form, Saint-Venant derived the axial and transversal displacements. For instance for an elliptical cylinder, he showed that the shape of any cross-section is represented by the local equation

where the x, y coordinates coincide with the principal axes of the elliptic cross-section when distorted. The level curves of this surface are hyperbolas whose asymptotes are the principal axes of the ellipse.

Whatever the magnitude of the torsion angle per unit length, if no axial force is applied to the prism ends, then the deformation of any volume element made of two cross-sections is with a constant volume and is said to be a pure shear deformation (cf. Sect. 1.13.3).

Was this article helpful?

0 0

Post a comment