# No shear connection

We assume first that there is no shear connection or friction on the interface AB. The upper beam
cannot deflect more than the lower one, so each carries load w/2 per unit length as if it were an
isolated beam of second moment of area bh3/12, and the vertical compressive stress across the
interface is w/2b. The midspan bending moment in each beam is wL2/16. By elementary beam
theory, the stress distribution at midspan is as in Fig. 2.2.(c), and the maximum bending stress in
each component, σ , is given by

There is an equal and opposite strain in the top fibre of the lower beam, so that the difference between the strains in these adjacent fibres, known as the slip strain, is x 2ε ,

It is easy to show by experiment with two or more flexible wooden laths or rulers that under load, the end faces of the two-component beam have the shape shown in Fig,2.3(a).The slip at the interface, s, is zero at x = 0 is the only one where plane sections remain plane. The slip strain, defined above, is not the same as slip. In the same way that strain is rate of change of displacement,
slip strain is the rate of change of slip along the beam. Thus from(2.4),

The constant of integration is zero, since s = 0 when x = 0, so that (2.6) gives the distribution of slip along the beam.
Results(2.5)and(2.6)for the beam studied in Section 2.7 are plotted in fig.2.3.This shows that at midspan, slip strain is a maximum and slip is zero, and at the ends of the beam, slip is a maximum and slip strain is zero. From (2.6), the maximum slip (when x = L / 2 )is wL3 / 4Ebh2 .Some idea of the magnitude of this slip is given by relating it to the maximum deflection of the two beams. From (2.3), the ratio of slip to deflection is3.2h / L , The ratio L / 2h for a beam is typically about 20,so that the end slip is less than a tenth of the deflection. We conclude that shear
connection must be very stiff if it is to be effective.

# Shear Stresses in Beams

In addition to normal stresses (Art. 3.16), beams are subjected to shearing. Shear stresses vary over the cross section of a beam. At every point in the section, there are both a vertical and a horizontal shear stress, equal in magnitude [Eq. (3.37)].

# Principal Stresses and Maximum Shear Stress

When stress components relative to a defined set of axes are given at any point in a condition of plane stress or plane strain (see Art. 3.10), this state of stress may be expressed with respect to a different set of axes that lie in the same plane. For example, the state of stress at point O in Fig. 3.15a may be expressed in terms of either the x and y axes with stress components, ƒx, ƒy, and vxy or the x and y axes with stress components ƒ , ƒ , and v x y xy (Fig. 3.15b). If stress components ƒx, ƒy, and vxy are given and the two orthogonal coordinate systems differ by an angle  with respect to the original x axis, the stress components ƒ , x ƒ , and v can be determined by statics. The transformation equations for stress are

This equation indicates that two perpendicular directions, p and p  (pi / 2), may be found for which the shear stress is zero. These are called principal directions. On the plane for which the shear stress is zero, one of the normal stresses is the maximum stress ƒ1 and the other is the minimum stress ƒ2 for all possible states of stress at that point. Hence the normal stresses on the planes in these directions are called the principal stresses. The magnitude of the principal stresses may be determined from

where the algebraically larger principal stress is given by ƒ1 and the minimum principal stress is given by ƒ2.
Suppose that the x and y directions are taken as the principal directions, that is, vxy  0. Then Eqs. (3.43) may be simplified to

By Eq. (3.46c), the maximum shear stress occurs when sin 2 alpha = pi /2, i.e., when   45 . Hence the maximum shear stress occurs on each of two planes that bisect the angles between the planes on which the principal stresses act. The magnitude of the maximum shear stress equals one-half the algebraic difference of the principal stresses:

If on any two perpendicular planes through a point only shear stresses act, the state of stress at this point is called pure shear. In this case, the principal directions bisect the angles between the planes on which these shear stresses occur. The principal stresses are equal in magnitude to the unit shear stress in each plane on which only shears act.