A new shear formula for tapered beamlike solids undergoing large displacements

Abstract

In many engineering applications it is often necessary to determine the flow of shear stresses in the cross-sections of beamlike bodies. Taking a cue from Jourawski's well-known formula, several scholars have proposed expressions for evaluating the shear stresses in non-prismatic linear elastic beams, where longitudinal variations in the size and shape of the cross-sections produces complex stress fields. In the present paper, a new shear formula, derived using a mechanical model developed in a previous work, is presented for tapered beams subject to even large displacements and small strains. Numerical examples and comparisons with results obtained using other formulas in the literature and non-linear 3D-FEM simulations show how the new formula constitutes an important generalization of the previous ones and is able to provide particularly accurate results.

Appendix

In the case of rectangular cross-sections, Eqs. (34)–(39) generalize the Jourawski’s solution for prismatic beams [5]. It is worth noting that if the current and reference states of the beam are very close and can be thought to almost coincide, such equations reduce to the formulas made available in the literature by other investigators who have studied the effects produced by the taper (e.g., Refs. [10, 15, 16]). To prove this, it is sufficient to specialize (37) for a flap-wise linearly tapered beam, loaded at the tip by an axial force N, transverse force T, and bending moment M, undergoing small displacements of the centre-line and small rotations of the local triads. Starting with (37), with θ≃0, we obtain:

$$\frac{q}{{2h_{2} }} = \frac{{\left( {h_{3}^{2} - x_{3}^{2} } \right)}}{{2J_{2} }}F_{Z} + \frac{{\Lambda_{3}^{ - 1} \Lambda^{\prime}_{3} x_{3} }}{A}F_{X} + \frac{{\Lambda_{3}^{ - 1} \Lambda^{\prime}_{3} \left( {3x_{3}^{2} - h_{3}^{2} } \right)}}{{2J_{2} }}M_{Y}$$

(46)

The next step is to express coefficient Λ₃ as a linear function of s:

$$\Lambda_{3} = 1 - \frac{\tan \alpha }{{h_{3R} }}s,$$

(47)

where h_3R is the height of the cross-section at the root (i.e., at s = 0), and angle α denotes the slope of the beam extrados. Combining (46)–(47) yields

$$\frac{q}{{2h_{2} }} = \frac{{\left( {h_{3}^{2} - x_{3}^{2} } \right)}}{{2J_{2} }}F_{Z} + \frac{{x_{3} \tan \alpha }}{{Ah_{3} }}F_{X} + \frac{{\left( {3x_{3}^{2} - h_{3}^{2} } \right)\tan \alpha }}{{2J_{2} h_{3} }}M_{Y}$$

(48)

The expressions for the cross-sectional area A and moment of inertia J₃ for the present case are to be substituted into (48), as are those for the cross-section resultants F_X, F_Z, M_Y with those in terms of axial force N, transverse force T, and bending moment M applied at the tip section, i.e. F_X = N, F_Z = T, M_Y = M-T(L − s), where L is the beam length. We finally obtain

$$\frac{q}{{2h_{2} }} = - \frac{{3\left( {x_{3}^{2} - h_{3}^{2} } \right)h_{3} - 3\left( {3x_{3}^{2} - h_{3}^{2} } \right)(L - s)\tan \alpha }}{{8h_{2} h_{3}^{4} }}T - \frac{{x_{3} \tan \alpha }}{{4h_{2} h_{3}^{2} }}N - \frac{{3\left( {3x_{3}^{2} - h_{3}^{2} } \right)\tan \alpha }}{{8h_{2} h_{3}^{4} }}M$$

(49)

which is the expression we can find, for example, in [16].

A similar relation can be obtained for elliptical cross-sections. As an example, still assuming small displacements, when one dimension, say h₂, is kept constant and a linear taper is assumed for the other, h₃, starting with (43), we get

$$\frac{q}{{2d_{2} }} = - \frac{{4\left( {x_{3}^{2} - h_{3}^{2} } \right)h_{3} - 4\left( {4x_{3}^{2} - h_{3}^{2} } \right)(L - s)\tan \alpha }}{{3\pi h_{2} h_{3}^{4} }}T - \frac{{x_{3} \tan \alpha }}{{\pi h_{2} h_{3}^{2} }}N - \frac{{4\left( {4x_{3}^{2} - h_{3}^{2} } \right)\tan \alpha }}{{3\pi h_{2} h_{3}^{4} }}M$$

(50)

Similar expressions can be obtained for other tapered beams, characterized by solid or hollow cross-sections of various shapes, and will be presented in subsequent works.