During the virtual displacement, the portions of the slab bounded by the yield lines and
the supports rotate as planes. Refer to Fig. 9b and c: B 1 = A/6; O 2 = 2B 1 = A/3 = 0.333A; B 3
= A/Z>; O 4 = A/c; O 5 = A(l/6 + lie) = [M(AE)}(6la + a/6).- Select a trial value of a, and evaluate the distances and angles
Using a = 4.5 ft (1.37 m) as the trial value, we find AE = (a^2 + 62 )^0 -^5 = 7.5 ft (2.28 m);
b = 5.63 ft (1.716 m); c = 10 ft (3.0 m); O 5 = (A/7.5)(6/4.5 + 4.5/6) = 0.278A. - Develop an equation for the external work WE performed by the
uniform load on a surface that rotates about a horizontal axis
In Fig. 10, consider that the surface ABC rotates about axis AB through an angle B while
carrying a uniform load of w lb/ft^2 (kPa). For the elemental area dAS9 the deflection, total
load, and external work are d = jc0; dW= w dA\ dWE = 8 dW= xOw dA. The total work for
the surface is WE = w0 / x dA, or
WE = WBQ (20)
where Q = static moment of total area, with respect to the axis of rotation.
- Evaluate the external and internal
work for the slab
Using the assumed value, we see a = 4.5 ft (1.37
-Area * dA m)? EF = 16 - 9 = 7 ft (2.1 m). The external work
for the two triangles is 2wI<(A/4.5)(^1 /6)(12)(4.5)^2 =
)
18wwA. The external work for the two trapezoids
C is 2wM(A/6)0/6)(16 + 2 x 7)(6)^2 = 60wwA. Then WE
= wMA(18 + 60) = 78wttA; W 1 = mu(lB 2 + 4 x 7.50 5 )
= 10.67mwA.- Find the value of mu
corresponding to the assumed value
Equate the external and internal work to find this
value of mu. Thus, 10.67wMA = 78wMA; mu =
7.3 IMV- Determine the highest value of mu
Assign other trial values to a, and find the corre-
sponding values of mu. Continue this procedure
until the highest value of mu is obtained. This is
FIGURE 10 me *ruQ value of the ultimate unit moment.
Design of Flexural Members by the Working-Stress MethodAs demonstrated earlier, the analysis or design of a composite beam by the working-stress
method is most readily performed by transforming the given beam to an equivalent homo-
geneous beam. In the case of a reinforced-concrete member, the transformation is made
by replacing the reinforcing steel with a strip of concrete having an area nAs and located
at the same distance from the neutral axis as the steel. This substitute concrete is assumed
capable of sustaining tensile stresses.
The following symbols, shown in Fig. 11, are to be added to the notational system giv-
en earlier: kd = distance from extreme compression fiber to neutral axis, in (mm); jd =
distance between action lines of C and T 9 in (mm); z = distance from extreme compres-
sion fiber to action line of C, in (mm).