TRANSVERSE SHEAR

In this chapter we will develop a method for finding the shear
stress in a beam and discuss a way to find the spacing of fasteners
along the beam’s length. The concept of shear flow will be
presented, and used to find the average stress within thin-walled
members. The chapter ends with a discussion of how to prevent
twisting of a beam when it supports a load.
In general, a beam will support both an internal shear and a moment. The
shear V is the result of a transverse shear-stress distribution that acts over
the beam’s cross section, Fig. 7–1. Due to the complementary property of
shear, this stress will also create corresponding longitudinal shear stress
that acts along the length of the beam.
To illustrate the effect caused by the longitudinal shear stress, consider
the beam made from three boards shown in Fig. 7–2a. If the top and
bottom surfaces of each board are smooth, and the boards are not bonded
together, then application of the load P will cause the boards to slide
relative to one another when the beam deflects. However, if the boards
are bonded together, then the longitudinal shear stress acting between
the boards will prevent their relative sliding, and consequently the beam
will act as a single unit, Fig. 7–2b.
As a result of the shear stress, shear strains will be developed and these
will tend to distort the cross section in a rather complex manner. For
example, consider the short bar in Fig. 7–3a made of a highly deformable
material and marked with horizontal and vertical grid lines. When the
shear force V is applied, it tends to deform these lines into the pattern
shown in Fig. 7–3b. This nonuniform shear-strain distribution will cause
the cross section to warp; and as a result, when a beam is subjected to
both bending and shear, the cross section will not remain plane as
assumed in the development of the flexure formula.
7.2 THE SHEAR FORMULA
Because the strain distribution for shear is not easily defined, as in the
case of axial load, torsion, and bending, we will obtain the shear-stress
distribution in an indirect manner. To do this we will consider the
horizontal force equilibrium of a portion of an element taken from the
beam in Fig. 7–4a. A free-body diagram of the entire element is shown in
Fig. 7–4b. The normal-stress distribution acting on it is caused by the
bending moments M and M + dM. Here we have excluded the effects of
V, V + dV, and w(x), since these loadings are vertical and will therefore
not be involved in a horizontal force summation. Notice that Fx = 0 is
satisfied since the stress distribution on each side of the element forms
only a couple moment, and therefore a zero force resultant.
Boards not bonded together
(a)
P
Boards bonded together
(b)
P
Fig. 7–2
(a) Before deformation
V
(b) After deformation
V
Fig. 7–3