Simply-supported composite slabs and beams - Composite Structures of Steel and Concrete - page 82

the number in each shear span is n , then the degree of shear connection is
defined by:

degree of shear connection = η = n / n f = N c / N c,f (3.63)

where n f is the number of connectors required for full shear connection.
The plastic moment of resistance of a composite slab with partial shear
connection had to be derived in Section 3.3.1(3) by an empirical method
because the flexural properties of profiled sheeting are so complex. For
composite beams, simple plastic theory can be used [31].
The depth of the compressive stress block in the slab, x c , is given by

x c = N c / (0.85 f cd b eff ) (3.64)

and is always less than h c . The distribution of longitudinal strain in the
cross-section is intermediate between the two distributions shown (for
stress) in Fig. 2.2(c), and is shown in Fig. 3.15(d), in which C means
compressive strain. The neutral axis in the slab is at a depth x n , slightly
greater than x c , as shown.
In design of reinforced concrete beams and slabs it is generally assumed
that x c / x n is between 0.8 and 0.9. The less accurate assumption x c

x n is
made for composite beams and slabs to avoid the complexity that otherwise
occurs in design when x c ≈

=

h c or, for beams with non-composite slabs,

x c h t . This introduces an error in M pl that is on the unsafe side, but is

negligible for composite beams. It is not negligible for composite columns,
where it is allowed for (Section 5.6.5.1).
There is a second neutral axis within the steel I-section. If it lies within
the steel top flange, the stress blocks are as shown in Fig. 3.15(c), except
that the concrete block for the force N c,f is replaced by a shallower one of
depth x c , for force N c .
Resolving longitudinally,

≈

N ac = N a,pl − N c

The depth of the neutral axis in the steel is found from

x a = h t + N ac / (2 b f f yd ) (3.65)

Taking moments about the line of action of N c ,

M Rd = N a,pl ( h g + h t − x c /2) − N ac ( x a + h t − x c )/2 (3.66)

If the second neutral axis lies within the steel web, the stress blocks are
as shown in Fig. 3.15(e), and M Rd can be found by a method similar to
that for Equation 3.66.