Chapter 2, Problem 2.1P

To determine

Loading diagram for the members BE and FED.

Answer to Problem 2.1P

   $\begin{array}{l}\underset{\_}{\text{One -way slab}}\\ \underset{\_}{\text{Total distributed load =14}\text{.24 KN/m}}\\ \underset{\_}{B_y=E_y=35.6KN}\end{array}$

Explanation of Solution

Given information:

Concept used:

One-way slab is a slab which is supported by beams on two opposite sides to carry the load along one direction. In one way slab, the ratio of longer span (b) to shorter span (a) is equal or greater than 2.

Calculation:

Beam BE:

Since $\frac{b}{a}=\frac{5m}{2m}=2.5$, therefore the slab will behave as a one-way slab.

Thus, the tributary area for this beam is rectangular as shown in Fig. (a) and the intensity of the uniformly distributed load is as follows:

Dead load for 200mm concrete slab $=(23.6KN/m^2)(0.2m)(2m)=9.44KN/m$

Live load for office $=(2.40KN/m^2)(2m)=4.8KN/m$

Total uniform distributed load $\text{= 9}\text{.44 + 4}\text{.8 = }14.24KN/m$

Due to symmetry the vertical reactions at B and E are

   $B_y=E_y=\frac{\text{(14}\text{.24 KN/m})(5)}{2}=35.6KN$

The loading diagram for beam BE is shown in Fig. (b)

Beam FED:

The load supported by this beam is the vertical reaction of beam BE at E which is $E_y=35.6KN$

The loading diagram for this beam is shown below:

Fig. (a)

Fig. (b)

Fig. (c)

Conclusion:

The loading diagrams are shown in the figures above and it is a one-way slab.

   $\text{Total distributed load }=14.24KN/m\text{and}\text{the}\text{reactions}\text{are}\text{as}{B}_y=E_y=35.6KN$