10.6. Slender beams

Bernoulli kinematic assumption.

\[\begin{split}\begin{aligned} 0 & = s'_{Px} - \theta_y \quad && \rightarrow \qquad \theta_y = s'_{Px} \\ 0 & = s'_{Py} + \theta_x \quad && \rightarrow \qquad \theta_x =-s'_{Py} \\ \end{aligned}\end{split}\]

As an example, the constitutive law for bending in a structurally decoupled elastic beam becomes

\[\begin{split}\begin{aligned} M_x & = EJ_x \theta'^{\ mech}_x = - EJ_x s''^{\ mech}_{Py} \\ M_y & = EJ_y \theta'^{\ mech}_y = EJ_y s''^{\ mech}_{Px} \\ \end{aligned}\end{split}\]

Example 10.2 (Clamped beam)

Internal equilibrium equations read

\[\begin{split}\begin{aligned} 0 & = F_z' + f_z && \text{(axial)} \\ 0 & = F'_y + f_y && \text{(shear)} \\ 0 & = M'_x - T_y + m_x && \text{(bending)} \\ \end{aligned}\end{split}\]

with axial force \(F_z = EA s'_{Pz}\), shear force \(F_y = \chi^{-1} GA ( s'_{Py} + \theta_x )\), and bending moment \(M_x = EJ \theta'_x\). The beam has length \(b\), it’s clamped in \(A: z=0\), s.t. essential boundary conditions in \(A\) read

\[s_{Pz}(z=0) = 0 \quad , \quad s_{Py}(z=0) = 0 \quad , \quad \theta_{x}(z=0) = 0 \ ,\]

and loaded lumped force and moment in \(B: z=b\), s.t. natural boundary conditions in \(B\) read

\[F_z(z=b) = Z \quad , \quad F_y(z=b) = Y \quad , \quad M_x(z=b) = M\]

and no distributed actions \(f_y = f_z = 0\), \(m_z = 0\), s.t. equilibrium equation becomes

\[\begin{split}\begin{aligned} 0 & = F_z' && \text{(axial)} \\ 0 & = F'_y && \text{(shear)} \\ 0 & = M'_x - F_y && \text{(bending)} \\ \end{aligned}\end{split}\]

From the equilibrium equations and the boundary conditions in \(z=b\), it immediately follows the distribution of the internal actions along the beam

\[F_z(z) = Z \quad , \quad F_y(z) = Y \quad , \quad M_x(z) = M + Y ( z - b ) \ .\]

Using the constitutive laws and the essential boundary conditions in \(z=0\), it immediately follows the displacement field along the beam,

\[\begin{split}\begin{aligned} s_{Pz}(z) & = \frac{Z}{EA} z \\ s_{Py}(z) & = - \frac{1}{EJ} \left( Y \frac{z^3}{6} + (M-Yb) \frac{z^2}{2} \right) + \frac{Y}{\chi^{-1} GA} z \\ \theta_x(z) & = \frac{1}{EJ} \left( Y \frac{z^2}{2} + (M-Yb) z \right) \end{aligned}\end{split}\]

The displacement of the extreme point \(B\) thus reads

\[\begin{split}\begin{aligned} s_{Bz} & = s_{Pz}(z=b) && = \frac{Z b}{EA} \\ s_{By} & = s_{Py}(z=b) && = -\frac{M b^2}{2 EJ} + \frac{Y b^3}{3EJ} + \frac{Y b}{\chi^{-1} GA} \\ \theta_{Bx} & = \theta_x(z=b) && = \frac{M b}{EJ} - \frac{Y b^2}{2 EJ} \end{aligned}\end{split}\]

Let’s discuss the order of magnitude of the two terms in \(s_{By}\) due to \(Y\).

Transverse displacement: bending and shear contributions. For an elastic medium, the shear modulus \(G\) can be written as a function of the elastic modulus \(E\) and the Poisson ration \(\nu\),

\[G = \frac{E}{2 (1 + \nu)} \ .\]

While the value of the Poisson ratio i limited to \(-1 \le \nu \le 0.5\), it’s usually in the range \([0,0.5]\). If Poisson ratio belongs to the latter range, the ratio \(\frac{G}{E}\) belongs to the range \(\left[ \frac{1}{3} , \frac{1}{2} \right]\), and thus \(G\) has the same order of magnitude as \(E\).

The properties of a square section are

\[A = a^2 \quad , \quad J = \frac{1}{12} a^4 \quad , \quad \chi = \frac{6}{5} \ .\]

Thus the ratio of the two contributions to transverse displacement has order of magnitude

\[\dfrac{\frac{Y b^3}{3 EJ}}{\frac{Yb}{ \chi^{-1} G A}} = \dfrac{G}{E} \chi \dfrac{ b^2 A }{ J } = \dfrac{G}{E} \frac{6}{5} 12 \dfrac{b^2}{a^2} = \dfrac{24}{5} \div \dfrac{36}{5} \dfrac{b^2}{a^2} = 4.8 \div 7.2 \left( \dfrac{b}{a} \right)^2 \ .\]

It immediately follows that the displacement of \(B\) due to shear deformation for a beam with square section with side \(a\) and length \(b = 10 a\) is \(480 \div 720\) times larger than the contribution due to bending (and the following transverse displacement of the axis of the beam associated with the rotation of tits sections).

Transverse and axial displacement. The comparison of the transverse displacement and the axial displacement gives the ratio

\[\dfrac{\frac{Y b^3}{3 EJ}}{\frac{Z b}{E A}} = \dfrac{Y}{Z} \dfrac{ b^2 A }{ J } = \dfrac{Y}{Z} 12 \left(\dfrac{b}{a}\right)^2 \ ,\]

and if the components of the force have similar magnitude \(Y \sim Z\), the order of magnitude of the ratio becomes \(12 \left( \frac{b}{a} \right)^2\), so that axial displacement of slender beams \(\frac{b}{a} \gg 1\) becomes negligible if compared with transverse displacement.