Exercises|Exercise 4.9.1|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

« Slabs supported on three or two adjacent edges Exercise 4.9.2 »

Exercises

Exercise 4.9.1

Figure 4.9.1-1: Project <B_49-1>

Given: Covering load g_e=1.00 kN/m², live load q=5.00 kN/m², concrete quality C30/37 (E=32.8 GPa).

Question : Perform static analysis to determine bending moments, shear forces, support reaction forces and elastic deflection.

Solution:

Self-weight: g_o=0.17m·25.0kN/m³= 4.25 kN/m²

Covering load: g_e= 1.00 kN/m²

Total dead load: g= 5.25 kN/m²

Total live load: q= 5.00 kN/m²

The design combination in ultimate limit state (ULS) is obtained by load p=γ_g·g+γ _q·q equal to:

p=1.35g+1.50q=1.35x5.25+1.50x5.00=14.59 kN/m²

Three different methods for the calculation of the two-way slab, by means of Marcus, Czerny and finite element method are presented below.

Marcus method:

l_x= 4.00 m,l_y=5.00 m → ε=l_y/l_x=5.00/4.00=1.25

Table b5.1 for ε=1.25 gives:

Bending moments :

k_x=0.709 , k_y=0.291, v_x=v_y=0.622 →

p_x=k_x·p·1.0m=0.709x14.59kN/m²x1.0m=10.34 kN/m

p_y=k_y·p·1.0m=0.291x14.59kN/m²x1.0m=4.25 kN/m

M_x=v_x·p_x · l_x²/8=0.622x10.34x(4.0²/8)=12.86 kNm

M_y=v_y·p_y · l_y²/8=0.622x4.25x5.0²/8=8.26 kNm

Shear forces :

ρ_xr=ρ_yr=0.500 → V_xr=V_yr=0.500·1.0m·p·l_x=0.500x1.0mx14.59 kN/m²x4.00m=29.18 kN [*] Note NoteThe values of shear forces Vxr, Vyr refer to 1.00 m wide strip and their calculation depends only on dimension lx.

Deflection:

a_m=0.0689 → f_m=a_m·p·l_x⁴/(E·h³)= 0.0689x14.59x10³N/m²x4.0⁴m⁴/(32.80x10⁹N/m²x1.0mx0.17³m³)

→ f_m =1597x10^-6m=1.597 mm

Equivalent uniform reaction forces :

υ_xr=0.300, υ_yr=0.250

p_xr=υ_xr·p·l_x=0.300x14.59 kN/m²x4.0m=17.51 kN/m [*] Note NoteThe values of reaction forces pxr, pyr depend only on dimension lx.

p_yr=υ_yr·p·l_x=0.250x14.59 kN/m²x4.0m=14.59 kN/m [*] Note NoteReaction forces pxr, pyr correspond to loadings transferred from slabs onto beams. The load applied on the two beams along direction y is equal to 2·pxr·ly=2x17.5kM/mx5.0m=175.0 kN and along x is 2·pyr·lx=2x14.6kM/mx4.0m=116.8 kN. Therefore the total load applied on all four beams is equal to 175.0+116.8=291.8 kN, exactly the same as the total load applied on the slab p·lx·ly=14.59x4.0x5.0=291.8 kN.

Czerny method:

Table 46, for ε=1.25 gives:

Bending moments:

m_x=17.80 , m_y=29.90

M_x=p·l_x²/m_x=14.59x4.00²/17.80=13.11 kNm

M_y=p·l_x²/m_y=14.59x4.0²/29.90=7.81 kNm

Shear forces :

ρ_xr=0.39, ρ_yr=0.36

V_xr=ρ_xr·p·l_x=0.39x14.59kN/m²x4.00m=22.76 kN/m

V_yr=ρ_yr·p·l_y=0.36x14.59kN/m²x4.00m=21.01 kN/m

Deflection:

100a=7.28 → a=0.0728

f_m=a·p·l_x⁴/(E·h³)=0.0728x14.59x10³N/m²x4.0⁴m⁴/(32.80x10⁹N/m²x0.17³m ³)=1.687 mm

Equivalent uniform reaction forces:

Equivalent uniform reaction forces are determined in exactly the same way as in Marcus method.

Finite element method:

In project < B_49-1>, the following results are obtained by activating module "SLABS" in pi-FES:

Figure 4.9.1-2: Results summary from "Slab Results" module in pi-FES Figure 4.9.1-2: Results summary from "Slab Results" module in pi-FES
M_x=12.9 kNm, M_y=7.6 kNm	V_x=22.1 kN, V_y=20.1 kN	y=-1.73 mm

Figure 4.9.1-3: Analytical results from "FEM Results" of pi-FES Figure 4.9.1-3: Analytical results from "FEM Results" of pi-FES
Bending moment distribution M₁₁	Shear force distribution V₁₁	Deflections

Equivalent uniform reaction forces are determined in exactly the same way as in Marcus method.

Summarizing,

Marcus method gives:

M_x / M_y=12.86 / 8.26 kNm

V_xr / V_yr=29.18 / 29.18 kN

f=1.60 mm

Czerny [*]NoteMethod of Czerny is used by the Standard Solver of the related software. method gives:

M_x / M_y=13.11 / 7.81 kNm

V_xr / V_yr=22.76 / 21.01 kN

f=1.69 mm

Finite element method gives:

M_x/M_y=12.9 / 7.6 kNm

V_xr/V_yr=22.1 / 20.1 kN

f=1.73 mm

Notes :

All three methods give practically identical values for bending moments and deflections. Also practically identical shear forces are obtained by Czerny and finite element methods, whereas according to Marcus method result to higher values [*]NoteAccording to the design of slabs presented in volume C’, in general, sufficiency in shearing is ensured by the required slab thickness against slenderness, which depends on the deflection. As a consequence, shear reinforcement is not provided and shearing is ignored. Sufficiency check in shearing with values higher than the required ones, in the worst case, may lead to excessive slab thickness (thus avoiding hard to be placed shearing reinforcement). .
Note that the maximum bending moment of the two-way slab, i.e. M=13.00 kNm, is lower by 50% than the one of the one-way slab of identical area, i.e. M=29.18 kNm (§3.3.1, §3.3.2).
From §4.6.2.1 it is obvious that the deflection of the one-way slab (ε=l_y/l_x=?), k_x=1.0, ν_x=1.0 → a _m=12·c_x·k_x·ν_x=12x5/384=0.156, is 0.156/0.0689 = 2.26 times greater than the deflection of the two-way slab.

« Slabs supported on three or two adjacent edges Exercise 4.9.2 »