Question 14.3.3: A simply supported beam of rectangular section b x h is load...
A simply supported beam of rectangular section b × h is loaded beyond the elastic range by a load Q at midspan.
(a) Determine the shape of the plastic zones.
(b) Determine the distribution of curvatures along the beam. What is the curvature at midspan?
(c) Explain how the deflections of the beam may be determined.
(d) If Q is increased to such magnitude that a plastic hinge just forms at the midspan section, repeat the solutions for (a), (b) and (c) above.
(a) Referring to Fig. 14.3-5(a), for M > M_y the shape of the plastic zones is completely defined by the coefficient α. From Eqn 14.3-6
M = M_p(1 – α²/3)
Therefore α = [3 (1 – M/M_p)]^{1/2} (14.3-12)
where, for a typical section X-X, the value of M is given by the bending moment diagram in Fig. 14.3-5(b) and Mp is the plastic moment σybh2/4.
(b) The curvature distribution is as shown in Fig. 14.3-5 (c); at a typical section of the beam, the curvature κ is given by Eqn 14.3-9 :
\qquad \kappa =\frac{1}{\alpha } \left(\frac{2\sigma _{y}}{Eh} \right)=\frac{1}{\alpha } \kappa _{y} (14.3-13)
where κ_y is the curvature at yield and α is as given by Eqn 14.3-12 above. Eqn 14.3-13 defines the curvature distribution along the beam.
At midspan M = QL/4, and from Eqn 14.3-12.
\qquad \alpha =\left[3\left(1-QL/4M_{p}\right) \right] ^{1/2}Hence, the midspan curvature is
\qquad \kappa =\frac{2\sigma _{y}}{Eh\left[3\left(1-QL/4M_{p}\right) \right]^{1/2} }(c) The deflections of an elastic beam may readily be determined by the second moment-area theorem (see Chapter 4: Eqn 4.18-2) which states that the intercept on any vertical line between two tangents to the bent beam is equal to the moment about that vertical line of the “M/EI diagram” between the tangent points. For application to a beam loaded beyond the elastic range, the moment-area theorem is rephrased as a curvature-area theorem8 by replacing the words “M/£7 diagram” by “curvature diagram”. Note that the curvature-area theorem is of general application, provided deflections are small: it holds irrespective of whether the beam is elastic, plastic or elasto-plastic. (Equally, it holds irrespective of whether the curvatures are caused by loading or by, say, differential temperature change; for detailed applications of the curvature-area theorem, see Reference 8.)
(d) If a plastic hinge forms at midspan, then the shape of the plastic zones is as in Fig. 14.3-5(d), where a is still given by Eqn 14.3-12, with M obtained from the bending moment diagrams in Fig. 14.3-5(e).
Having calculated α, the curvature at any point is given by Eqn 14.3-13. Note, however, that α is now zero at midspan (Eqn 14.3-12 with M=M_p) so that κ at midspan is infinity, implying unrestrained rotation at the plastic hinge.
As regards deflection, it should be noted that the curvature-area theorem is applicable only for small deflections. As the plastic hinge forms, the central deflection of the beam increases indefinitely at constant load. Of course, the curvature-area theorem may still be used to find the deflected shapes of those portions of the beam away from the plastic hinge.
\qquad jk=ss^{\prime }-tt^{\prime }=\int_{x_{1}}^{x_{2}}{Mx\frac{dx}{EI} } (4.18-2)
\qquad \kappa =\frac{1}{\alpha }\left(\frac{2\sigma _{y}}{Eh} \right) =\frac{1}{\alpha }\kappa _{y} (14.3-9)
General Reference 8
Livesley, R. K. Matrix Methods of Structural Analysis. Pergamon Press,Oxford, 1975.
