Exercise 4.9.5|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 4.9.4 Exercise 4.9.6 »

Exercise 4.9.5

Given : Covering load g_επ=1.5 kN/m², live load q=5.0 kN/m², concrete quality C40/50.

Question : Perform static analysis of the slabs illustrated in the figure (shears forces, bending moments, deflections) using the finite element method with (a) global loading and (b) unfavourable loadings.

Figure 4.9.5-1: project <B_464>

Solution:

Slabs are identical to those of the example of §4.6.4.

g=0.15x25.0+1.5=5.25 kN/m²

q=5.0 kN/m²

In ULS, the maximum design load is equal to:

p_d=1.35g+1.50q=1.35x5.25+1.50x5.0=

=14.59 kN/m²

Software requires only characteristic loads of the structure to be inserted while the rest are obtained through internal calculations.

Due to both structure and loading being symmetric, results will also be symmetrical for both slabs.

In pi-FES, analysis is performed selecting 'SLABS' module and Adverse [*] Note NoteThe software performs all calculations for all loading cases and then combines these to produce the results. Results are then retrieved either for any individual loading case or for any selected combination, e.g. 1.35g+1.50q or g+q etc., or unfavourable loadings. The user is allowed to see only the results summary of unfavourable loadings. loadings.

RESULTS FOR GLOBAL LOADING 1.35g+1.50q

Figure 4.9.5-2: Bending moment diagrams Figure 4.9.5-2: Bending moment diagrams	Figure 4.9.5-3: Shear force diagrams Figure 4.9.5-3: Shear force diagrams
M_x=13.9 kNm, M_y=12.4 kNm, M_yerm=-34.6 kNm	V_xr=23.6 kN, V_yr=24.1 kN, V_yerm=45.5 kN
Figure 4.9.5-4: Deflection diagrams Figure 4.9.5-4: Deflection diagrams	Figure 4.9.5-5: Deflection diagrams and contours Figure 4.9.5-5: Deflection diagrams and contours
y_max=4.03 mm

RESULTS FOR UNFAVOURABLE LOADINGS

Figure 4.9.5-6: Bending moment diagrams Figure 4.9.5-6: Bending moment diagrams	Figure 4.9.5-7: Shear force diagrams Figure 4.9.5-7: Shear force diagrams
M_x=15.5 kNm, M_y=12.4 kNm, M_yerm=-34.6 kNm	V_xr=24.8 kN, V_yr=24.5 kN, V_yerm=45.5 kN
Figure 4.9.5-8: Deflection diagrams Figure 4.9.5-8: Deflection diagrams	Figure 4.9.5-9: Deflection diagrams and contours Figure 4.9.5-9: Deflection diagrams and contours
y_max=4.39 mm	[]Note*Since in “Adverse” state analytical diagrams are not provided, in ‘Element Input’, live load and covering load of the 2nd slab are eliminated, thus approaching the necessary difference of (1-1/1.35)g=0.26g. In this case, the slabs loading is represented by the respective unfavourable loading on their spans.

Notes:

The resulting span moments for unfavourable loadings are higher than those for global loading, only in the span along direction x.
Support moment remains the same, because unfavourable and global loadings are identical.

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