Question 13.17: A cantilever beam of length L has a rectangular cross sectio...

A cantilever beam of length L has a rectangular cross section of breadth B and depth D and carries a vertical concentrated load, W, at its free end. Determine the deflection of the free end, including the effects of both bending and shear. The flexural rigidity of the cantilever is EI and its shear modulus G.

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Using Eq. (13.21) we obtain the form factor β for the cross section of the beam directly.
Thus

\beta =\frac{A}{I_z^2} \int_{y_1}^{y_2}{\frac{(A^\prime\overline{y} )^2}{b_0} } \ dy                                                     (13.21)

\beta =\frac{BD}{(BD^3/12)^2} \int_{-D/2}^{D/2}\frac{1}{B} {\left[B\left(\frac{D}{2} – y \right) \frac{1}{2} \left(\frac{D}{2} + y \right) \right]^2 } \ dy       (see Ex. 10.1)

which simplifies to

\beta =\frac{36}{D^5} \int_{-D/2}^{D/2}{\left(\frac{D^4}{16} – \frac{D^2y^2}{2} + y^4 \right) } \ dy

Integrating we obtain

\beta =\frac{36}{D^5} \left[\frac{D^4y}{16} – \frac{D^2y^3}{6} + \frac{y^5}{5} \right]_{-D/2}^{D/2}

which gives

\beta =\frac{6}{5}

Note that the dimensions of the cross section do not feature in the expression for β. The form factor for any rectangular cross section is therefore 6/5 or 1.2.

Let us suppose that v_s is the vertical deflection of the free end of the cantilever due to shear. Hence, from Eq. (13.22) we have

v_s = \frac{\beta }{G} \int_{L}{\frac{S_y}{A} } \ dx                                                         (13.22)

v_s = \frac{6 }{5G} \int_{0}^{L}{\left(\frac{-W}{BD}\right) } \ dx

so that

v_s = – \frac{6WL}{5GBD}                                              (i)

The vertical deflection due to bending of the free end of a cantilever carrying a concentrated load has previously been determined in Ex. 13.1 and is −WL³/3EI. The total deflection, v_T , produced by bending and shear is then

v_T = – \frac{WL^3}{3EI} – \frac{6WL}{5GBD}                                                (ii)

Rewriting Eq. (ii) we obtain

v_T = – \frac{WL^3}{3EI} \left[1+\frac{3}{10} \frac{E}{G} \left(\frac{D}{L} \right)^2 \right]                                                            (iii)

For many materials (3E/10G) is approximately unity so that the contribution of shear to the total deflection is (D/L)² of the bending deflection. Clearly this term only becomes significant for short, deep beams.

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