The type of frame to be considered in this chapter is in fact a plane frame, but loaded transversely. Thus it will be assumed that all members of the frame lie initially in a certain plane (for convenience, the horizontal plane), and that all loads act in a direction perpendicular to this plane (vertically). Bending moments about vertical axes will be taken to be zero, so that shear forces in the plane are also zero. Any member of the frame will then be acted upon by vertical shear forces, and by two moments about horizontal axes, that is, by a bending moment M (flexural couple) and a torque T (twisting couple).

It will be assumed further (as for the case of the simple plane frame) that the presence of shear forces has no effect on the formation of plastic hinges, so that the yield condition f(M, T) = const. of equation (1.14) may be used. (Shear force may have to be taken into account in any practical design, but the adjustment is relatively easy to make.)

The right-angle bent

The problem shown in fig. 2.1 was solved in vol. 1 (chapter 5, fig. 5.1), but without any real explanation of how the actual values of bending moment and torque at the hinges were determined. The right-angle bent lies in a horizontal plane, is continuous at B, and is built into fixed walls at A and C.