Exercise 4.9.4|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 4.9.3 Exercise 4.9.5 »

Exercise 4.9.4

Given: Loads and concrete class are the same as in the previous exercise.

Question : Perform static analysis to determine deflections for the previous exercise slabs, considering the first two as two-way instead of one-way. See the following figure.

Figure 4.9.4-1: Project <Β_494>

Solution:

Figure 4.9.4-2

Due to the unknown influence of cantilever slab s3 on the adjacent one, for safety reasons, slabs s2 and s1 are considered fixed at one end.

s1,s2: consider the slabs rotated by 90º, thus from table b5.2 for ε=l_y/l_x=4.0/5.0=0.80 ak_x=0.141, k_y=0.859, v_x=0.817, v_y=0.742, a_m=0.0180, ρ_xr=ρ_yr=0.293, ρ_yerm=0.507, υ_xr=0.146, υ_yr=0.207, υ_yerm=0.359 a

p_x=k_x·p=0.141x9.75=1.37 kN/m, p_y=k_y·p=0.859x9.75=8.38 kN/m

M_x=ν_x·p_x · l_x²/8=0.817x1.37x5.0²/8=3.50 kNm

M_y=ν_y·p_y · l_y²/14.22=0.742x8.38x4.0²/14.22=7.00 kNm

M_yerm=-p_y·l_y²/8=-8.38x4.0²/8=-16.76 kNm

f_m=0.0180x9.75x10³N/m²x5.0⁴m⁴/(35.20x10⁹N/m²x1.0mx0.16³m ³)==761x10^-6m=0.76 mm

V_xr=V_yr=ρ_xr·p·l_x=0.293x9.75x5.0=14.28 kN

V_yerm=ρ_yerm·p·l_x=0.507x9.75x5.0=24.72 kN

p_xr=υ_xr·p·l_x=0.146x9.75x5.0=7.12 kN/m

p_yr=υ_yr·p·l_x=0.207x9.75x5.0=10.09 kN/m

p_yerm=υ_yerm·p·l_x=0.359x9.75x5.0=17.50 kN/m

Figure 4.9.4-3

Support s1-s2: M_1-2=-(16.76+16.76)/2=-16.76 kNm

Support s2-s3:

Μ_2-3=M_yerm,3=-p·l_y²/2-P·l_y=-14.25x1.45²/2-1.35x1.45=-16.94 kNm

V_yerm,3=p·l_y+P=14.25x1.45+1.35=22.01 kN

The deflection of the cantilever end is determined by the expression

, where the first term is due to the uniform load and the second to the concentrated load applied at its end.

In case of a slab (i.e. a strip of width b=1.0m), where I=b·h³/12, the maximum deflection (at the end) is:

Results are transferred to the initial coordinate system producing the following stress resultants:

M_1-2=-16.76, M_2-3=-16.94, M_x,1=7.00, M_y,1=3.50, M_x,2=7.00, M_y,2=3.50 [kNm]

V_xr,1=14.28, V_xerm,1=24,72, V_yr,1=14.28, V_xerm,2=24.72, V_xr,2=14.28, V_yr,2=14.28, V _xerm,3=22.01 [kN]

the following equivalent uniform support reactions:

p_xr,1=10.09, p_xerm,1=17.50, p_yr,1=7.12, p_xerm,2=17.50, p_xr,2=10.09, p_yr,2=7.12 [kN/m]

and loads on beams:

b1: p=10.09 kN/m, b4 and b6: p=7.12, b2: p=17.50+17.50=35.00, b5 and b7: p=7.12 b3:p=10.44+22.01=32.45 [kN/m]

« Exercise 4.9.3 Exercise 4.9.5 »