Analysis of a Reinforced Concrete Foundation Slab for Normal Crack Opening
Figure 1. Design crosssection of the element
Objective: Check the calculation of the crack opening width.
Task: Verify the correctness of the analysis of normal crack opening.
References:
1. Guide on designing of concrete and reinforced concrete structures made of heavyweight concrete (no prestressing) (to SP 521012003), 2005, p. 155157.
2. M.A. Perelmuter, K.V. Popok, L.N. Skoruk, Calculation of the Normal Crack Opening Width for SP 63.13330.2012, Concrete and Reinforced Concrete, 2014, №1, p.21,22
Initial data file:
Example 43.SAV
report – Arbat 43.doc
Compliance with the codes: SP 521012003, SP 63.13330.2012.
Initial data:
b×h = 1150×300 mm  Slab section sizes 
а = 42 mm  Distance from the center of gravity of the reinforcement to he compressed edge of the section 
A_{s }= 923 mm^{2} (6Ø14)  Crosssectional area of reinforcement 
M_{l} = 50 kH∙m  Moment in the design section from permanent and longterm loads 
M_{sh} = 10 kH∙m  Moment from shortterm loads 
Concrete class 
В15 А400 
ARBAT initial data:
Importance factor γ_{n} = 1
Importance factor (serviceability limit state) = 1
Member length 1 m
Effective length factor in the XoY plane 1
Effective length factor in the XoZ plane 1
Random eccentricity along Z according to SNiP 52012003 (Russia)
Random eccentricity along Y according to SNiP 52012003 (Russia)
Structure is statically indeterminate
Limit slenderness  200
Section:
b = 1150 mm

S1  6Ø14

Reinforcement 
Class 
Service factor 
Longitudinal 
A400 
1 
Transverse 
A240 
1 
Concrete:
Concrete type: Heavyweight
Concrete class: B15
Service factor for concrete 


γ_{b1} 
allowance for the sustained loads 
1 
γ_{b2} 
allowance for the failure behavior 
1 
γ_{b3} 
allowance for the vertical position during concreting 
1 
γ_{b4} 
allowance for the freezing/thawing and negative temperatures 
1 
Humidity of environmental air  4075%
Crack resistance:
Limited crack opening width
Requirements to crack opening width are based on the preservation of reinforcement
Allowable crack opening width:
Shortterm opening 0,4 mm
Longterm opening 0,3 mm
Forces:
N = 0 kN
M_{y} = 60 kN*m
Q_{z} = 0 kN
M_{z} = 0 kN*m
Q_{y} = 0 kN
T = 0 kN*m
Factor for sustained load 0,83333
Theoretical solution:
Diagrams of the strain ε and stress σ distribution in concrete for the determination of the stress σ_{s} obtained in the theoretical calculation [2] based on the nonlinear deformation model are shown in Fig. 2. The following values of the internal longitudinal force N and bending moment M correspond to these diagrams
N = 0,00439 kN ≈ 0;
M = 50,096 ≈ 50 kNm.
There is a balance between internal and external forces. In this solution the stress in the tensile reinforcement is σ_{s}=236,692 MPa.
Figure 2. Strain e and stress s diagrams (for the determination of σ_{s})
Similarly, solving the problem of determining the cracking moment, we obtain the following diagrams (Fig. 3), which satisfy the requirements of Sec. 8.2.14 of SP 63.13330.2012.
Figure 3. Strain e and stress s diagrams (for the determination of σ_{s,crc})
In accordance with these diagrams M_{crc}= 36,244 kN∙m, σ_{s,crc} = 22,651 MPa.
On the basis of formula (1) (formula (8.128) SP 63.13330.2012) we obtain a_{crc}=0,306 mm.
\[ a_{crc} =\varphi_{1} \cdot \varphi_{2} \cdot \varphi_{3} \cdot \psi_{s} \cdot \frac{\sigma_{s} }{E_{s} }\cdot l_{s} . \]  (1) 
Comparison of solutions:
Check 
crack opening width (longterm) 
Theory 
0,306/0,3 = 1,02 
ARBAT 
0,974 
Deviation, % 
4,51% 
Comments:
 The member length and the class of transverse reinforcement have to be specified in ARBAT. Since they are not determined in the problem, the following data are used 1 m and А240, respectively.
 The value of the concrete cover is equal to a – d/2 = 42 – 14/2 = 35 mm.
 The value of the total moment acting in the section, М = M_{l} + M_{sh} = 50 + 10 = 60 kN∙m, factor for sustained load is equal to M_{l} /М = 50/60 = 0,833.
 The crack opening width obtained in the Guide [1] is equal to 0.227 mm. Such a significant discrepancy with the above theoretical solution is due to the use of the approach based on the ultimate forces, instead of the nonlinear deformation model (see [2]).
 The deviation of the results of ARBAT from the theoretical solution is due to the fact that in order to provide computational stability, diagrams in which the horizontal part of the graph σ(ε) has a small slope are used in ARBAT instead of the perfect diagrams of the material behavior.