Equilibrium (Plasticity) Truss Model - Unified Theory of Concrete Structures - page 33

Figure 2.5 Equilibrium in beam shear

2.1.3 Equilibrium in Beam Shear
A beam subjected to a concentrated load 2 V at midspan is shown in Figure 2.5(a). Since the
reaction is V , the shear force is a constant V throughout one-half of the beam, and the moment
diagram is a straight line. When a beam element of length d v is isolated and the moment on
the left face is defined as M , then the moment on the right face is M

− Vd v . The shear forces

on both the left and right faces are, of course, equal to V .
A model of the isolated beam element is shown in Figure 2.5(b). The top and bottom
stringers are separated from the main body of the beam element, so that the mechanism to
resist shear can be separated from the mechanism to resist moment. The stringers are resisting
the bending moment and the main body is carrying the shear force. Two assumptions are made
in the establishment of the model:

1. The shear flow q in the main body is distributed uniformly over the depth d v (i.e. in the

= q d( d v )

=

qd v .
2. The shear flow q in the main body is also distributed uniformly along the length of the
main body (i.e. in the longitudinal direction). Hence, the transverse steel stresses ( f t ) and
the stresses in the diagonal concrete struts (

transverse direction). Since q is a constant over the depth, V

σ d ) vary uniformly along their lengths.

Based on these two assumptions the main body can now be treated as a large shear element
discussed in Section 2.1.2. The only difference between the shear element in Figure 2.2 and the
main body of the beam in Figure 2.5(b) is that the longitudinal steel is uniformly distributed
in the former, but is concentrated at the top and bottom stringers in the latter. This difference,