Me350 note5

1) To determine the bending moment and shear force on the beam.

2012

Shear Stress in Beams: When a beam is subjected to nonuniform bending, both bending moments, M, and shear forces, V, act on the cross section. The normal stresses, σ x , associated with the bending moments are obtained from the flexure formula. We will now consider the distribution of shear stresses, τ, associated with the shear force, V. Let us begin by examining a beam of rectangular cross section. We can reasonably assume that the shear stresses τ act parallel to the shear force V. Let us also assume that the distribution of shear stresses is uniform across the width of the beam. Shear stresses on one side of an element are accompanied by shear stresses of equal magnitude acting on perpendicular faces of an element. Thus, there will be horizontal shear stresses between horizontal layers (fibers) of the beam, as well as, transverse shear stresses on the vertical cross section. At any point within the beam these complementary shear stresses are equal in magnitude. The existence of horizontal shear stresses in a beam can be demonstrated as follows. A single bar of depth 2h is much stiffer that two separate bars each of depth h.

For torsion of rectangular sections the maximum shear stress tmax and angle of twist 0 are given by T tmax = ~ kldb2 T-e L k2db3G kl and k2 being two constants, their values depending on the ratio d l b and being given in Table 5.1. For narrow rectangular sections, kl = k2 = i. Thin-walled open sections may be considered as combinations of narrow rectangular sections so that 3T-_ _ _-T Ckldb2 Cdb2 rmax = 3T-T-0-L Xk2db'G GCdb' The relevant formulae for other non-rectangular, non-tubular solid shafts are given in For thin-walled closed sections the stress at any point is given by Table 5.2. T 2At r =-where A is the area enclosed by the median line or mean perimeter and t is the thickness. The maximum stress occurs at the point where t is a minimum. The angle of twist is then given by-e =-/ d s T L 4A2G t which, for tubes of constant thickness, reduces to T s rs-e L 4A2Gt 2AG

Journal of Constructional Steel Research, 1992

Large diameter fabricated steel tubes subjected to transverse shear forces and bending moments can fail either in a local compression buckling mode because of flexure or in a diagonal buckling failure mode as a result of shear forces. Drawing on recent work by the writers at the University of Alberta, the ultimate shear strength is investigated through a parametric study to determine the factors affecting failure. A nonlinear regression analysis is then carried out using the available experimental data to derive an empirical relation for the ultimate shear strength. The formula is compared to the current design approach and found to perform better throughout the range of applicability.

Proceedings of the ICE - Structures and Buildings, 2008

Following the introduction of hot-finished elliptical hollow sections to the construction industry, recent research has been performed to develop supporting structural design guidance. This paper focuses on shear resistance. Twenty-four shear tests were performed on hot-finished steel elliptical hollow section members. The shear tests were arranged in a three-point bending configuration with span-to-depth ratios ranging from 1 to 8. This enabled the study of cross-section resistance in shear and the interaction between shear and bending. Measurements were taken of cross-section geometry, local initial geometric imperfections and material properties in tension. Test results, including moment– rotation histories, are presented. Additional results were generated numerically. These results have been used to verify proposed design expressions for shear resistance and resistance to combined shear and bending. Further investigations into design criteria for shear buckling are currently und...

We have talked about internal forces, distributed them uniformly over an area and they became a normal stress acting perpendicular to some internal surface, or a shear stress acting tangentially, in plane. Up to now, we have said little about how these normal and shear stresses might vary with position throughout a solid. 1 Up to now, the choice of planes, their orientation within a solid, was dictated by the geometry of the solid and the nature of the loading. We have said nothing about how these stress components might change if we looked at a set of planes of another orientation. Now we consider a more general situation, an arbitrarily shaped solid. We are going to lift our gaze up from the world of crude structural elements such as truss bars in tension, shafts in torsion, or beams in bending to view these " solids " from a more abstract perspective. They all become special cases of more general stuff we call a solid continuum. We will address two questions: • How might stresses vary from one point to another throughout a continuum; • How do the normal and shear components of stress acting on a plane at a given point change as we change the orientation of the plane at the point. The first bullet introduces the notion of stress field; the second concerns the transformation of components of stress at a point. To begin with the first bullet we reexamine the case of a bar suspended vertically and loaded by its own weight, a case considered in section 3.2, page 62. (Note, I have changed the orientation of the reference 1. The beam is the one exception. There we explored how different normal stress distributions over a rectangular cross-section could be equivalent to a bending moment and zero resultant force.