Structural Analysis (10th Edition)
Structural Analysis (10th Edition)
10th Edition
ISBN: 9780134610672
Author: Russell C. Hibbeler
Publisher: PEARSON
Question
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Chapter 2, Problem 2.1P
To determine

Loading diagram for the members BE and FED.

Expert Solution & Answer
Check Mark

Answer to Problem 2.1P

   One -way slab_Total distributed load =14.24 KN/m_By=Ey=35.6 KN_

Explanation of Solution

Given information:

Structural Analysis (10th Edition), Chapter 2, Problem 2.1P , additional homework tip  1

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 ba=5 m2 m=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 =(23.6 KN/m2)(0.2 m)(2 m)=9.44 KN/m

Live load for office =(2.40 KN/m2)(2 m)=4.8 KN/m

Total uniform distributed load = 9.44 + 4.8 = 14.24 KN/m

Due to symmetry the vertical reactions at B and E are

   By=Ey= (14.24 KN/m)(5)2=35.6 KN

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 Ey=35.6 KN

The loading diagram for this beam is shown below:

Fig. (a)

Structural Analysis (10th Edition), Chapter 2, Problem 2.1P , additional homework tip  2

Fig. (b)

Structural Analysis (10th Edition), Chapter 2, Problem 2.1P , additional homework tip  3

Fig. (c)

Conclusion:

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

   Total distributed load =14.24 KN/mandthereactionsareasBy=Ey=35.6 KN

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