Exercise 4.9.6|www.BuildingHow.com

VOLUME A
The art of construction and detailing
Introduction
The structural frame
- Introduction
- Structural frame elements
  - Columns
  - Beams
  - Slabs
  - Staircases
  - Foundation
- STRUCTURAL FRAME LOADING
- BEHAVIOR OF THE STRUCTURAL FRAME
  - The behavior and reinforcement of a slab
  - Behavior and reinforcement of beams and columns
Construction methods
- The materials
- The moulds
- Thermal insulation of structural elements
- Concrete
- The steel
- Reinforcement specifications of antiseismic design
- Industrial stirrups - stirrup cages
- Standardized cross-sections of reinforced concrete elements
Reinforcement
- Columns
- Shear walls
- Composite elements
- Beams
- Slabs
- Staircases
- Foundation
Quantity surveying
- Estimation of the concrete’s quantity
- Estimation of the formworks’ quantity
- Estimation of the spacers’ quantity
- Estimation of the reinforcements' quantity
- Total estimation of the materials’ quantities
- Optimization of the reinforcement schedule
- Estimation of the structural frame’s cost
- Electronic exchange of designs - bids - orders
Detailing drawings
- General
- The drawings’ title block
- Carpenter’s drawings
- Steel fixer’s drawings
Tables
- Table 1
- Table 2
- Table 3
Drawings
- Carpenter’s drawings
- Steel fixer’s drawings
Model (exemplary) construction

« Exercise 4.9.5 Exercise 4.9.7 »

Exercise 4.9.6

Figure 4.9.6-1

Given : Covering load g_επ=1.0 kN/m², live load q= 10.0 kN/m², slab thickness h=140 mm.

Question: Determine bending moments of the slabs in the figure having three spans in x direction and practically infinite number of spans in y direction.

Solution:

The total dead load is equal to: g_d=0.14x25.0+1.0=4.50 kN/m² .

Due to high live load, static analysis will be performed using unfavourable loadings.

The design loads to be used for loading the slabs are g_d and p_d=1.35g+1.50q= g+(0.35g+1.50q) which is analyzed to:

Static analysis

Moments are determined by means of the technique described in §4.6.4. According to that, the response of the slabs is determined by the particular loadings applied. Thus, while there are actually two slab types, one with four fixed edges and one with three fixed and one free edge, two extra slab types are created: one with four free edges and one with three free and one fixed edge (as illustrated in the figures analyzing the loadings). For ε=l_y/l_x=4.0/4.5=0.89:

Figure 4.9.6-2 Figure 4.9.6-2	For type '1' slab from table b5.1 for ε=l_y/l_x=4.5/4.0=1.125 → k_x,1=0.615, k_y,1=0.385, v_x,1=v_y,1=0.595 → p_2x=0.615x8.29=5.10 kN/m, p_2y=0.385x8.29=3.19 kN/m
Figure 4.9.6-3 Figure 4.9.6-3	For type '2' slab from table b5.2 for ε=l_y/l_x=4.5/4.0=1.125 → k_x,1=0.391, k_y,1=0.609, v_x,1=0.744, v_y,1=0.639 → p_2x=0.391x8.29=3.24 kN/m, p_2y=0.609x8.29=5.05 kN/m
Figure 4.9.6-4 Figure 4.9.6-4	For type '2' slab from table b5.2 for ε=l_y/l_x=4.0/4.5=0.89 → k_x,1=0.799, k_y,1=0.201, v_x,1=0.703, v_y,1=0.789 → p_2x=0.799x8.29=6.62 kN/m, p_2y=0.201x8.29=1.67 kN/m
Figure 4.9.6-5 Figure 4.9.6-5	For type '5' slab from table b5.5 for ε=l_y/l_x=4.5/4.0=1.125 → k_x,1=0.445, k_y,1=0.555, v_x,1= 0.835, v_y,1=0.805 → p_1x=0.445x12.79=5.69 kN/m, p_1y=0.555x12.79=7.10 kN/m
Figure 4.9.6-6 Figure 4.9.6-6	For type '6' slab from table b5.6 for ε=l_y/l_x=4.5/4.0=1.125 → k_x,1=0.615, k_y,1=0.385, v_x,1=v_y,1=0.865 → p_1x=0.385x12.79=4.92 kN/m, p_1y=0.615x12.79=7.87 kN/m

minM_s1-s2

≡

Figure 4.9.6-7

≡

minM_s1-s1

Figure 4.9.6-8

minM_s₂_-s₂

Figure 4.9.6-9

maxM_s1

Figure 4.9.6-10

maxM_s2

Figure 4.9.6-11

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