Aluminium Design and Construction John Dwight

Figure 10.1 Symmetric plastic bending.
axis. The half of the section above ss is divided into convenient elements,
and S then calculated from the expression:
(10.1)
where A E =area of element, and y E =distance of element’s centroid E above
ss. The summation is made for the elements lying above ss only, i.e. just
for the compression material (C).
Figure 10.1(b) shows another case of symmetrical bending, in which
M acts about an axis perpendicular to the axis of symmetry. The neutral
axis (xx) will now be the equal-area axis, not necessarily going through
the centroid, and is determined by the requirement that the areas above
and below xx must be the same. S is then found by selecting a convenient
axis XX parallel to xx, and making the following summation for all the
elements comprising the section:
(10.2)
where A E =area of element, taken plus for compression material (above
xx) and negative for tensile material (below xx), and Y E =distance of element’s
centroid E from XX, taken positive above XX and negative below.
The correct answer is obtained whatever position is selected for XX,
provided the sign convention is obeyed. It is usually convenient to
place XX at the bottom edge of the section as shown, making Y E plus
for all elements. No element is allowed to straddle the neutral axis xx;
thus, in the figure, the bottom flange is split into two separate elements,
one above and one below xx.
10.2.2 Unsymmetrical bending
We now consider the determination of the plastic modulus when the
moment M acts neither about an axis of symmetry, nor in the plane of such
an axis. In such cases, the neutral axis, dividing the compressive material
Copyright 1999 by Taylor & Francis Group. All Rights Reserved.