For this question, I was wondering where the 0.15 came from in the dead load equation because when I looked at table 1.2, I saw it was 144. Also, for the Dead load equation, I thought it was DL=q*A*thickness so I was wondering why aren’t they doing 30*10 for the area 2-3. The floor system used in a school classroom consists of a4-in. reinforced stone concrete slab. Sketch the loading thatacts along the joist BF and side girder ABCDE. Seta = 10 ft, b = 30 ft. Hint: See Tables 1.2 and 1.4.SOLUTIONJoist BF. SinceaThus, the tributary area for this joist is the rectangular area shown in Fig. a, and theintensity of the uniform distributed load isb30 ft= 3, the concrete slab will behave as a one-way slab.10 ft44-in-thick reinforced stone concrete slab: (0.15 k/ft³) (ft)(1ft (10 ft) = 0.5 k/ft12

Answer We know that Dead load(DL) from slab can be calculated using DL = q*A*t ...say equation 1…

Answered: For this question, I was wondering…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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For this question, I was wondering where the 0.15 came from in the dead load equation because when I looked at table 1.2, I saw it was 144. Also, for the Dead load equation, I thought it was DL=q*A*thickness so I was wondering why aren’t they doing 30*10 for the area

2-3. The floor system used in a school classroom consists of a
4-in. reinforced stone concrete slab. Sketch the loading that
acts along the joist BF and side girder ABCDE. Set
a = 10 ft, b = 30 ft. Hint: See Tables 1.2 and 1.4.
SOLUTION
Joist BF. Since
a
Thus, the tributary area for this joist is the rectangular area shown in Fig. a, and the
intensity of the uniform distributed load is
b
30 ft
= 3, the concrete slab will behave as a one-way slab.
10 ft
4
4-in-thick reinforced stone concrete slab: (0.15 k/ft³) (ft)(1
ft (10 ft) = 0.5 k/ft
12

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