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Technical Note

 

 

 

 

5

Closed Form Formulas 

Closed form formulas are readily available for beams and one-way slabs. The variables that 

describe the geometry of a two-way panel within a floor system, however, are so extensive that 

it becomes impractical to compile a meaningful set of tables or relationships without extensive 

approximation.  For non-cracked sections, compilations such as the one listed in Table 5 are 

readily available in the literature [Bares, 1971]. 

 

In the application of data, such as those given in Table 5, the design engineer must use 

judgment regarding the degree of fixity of the support. 

 

 

TABLE  5  DEFLECTION COEFFICIENTS k 

 

γ

 

           a    

 

 

       b 

 

   

1 

2 

 

 

 

3 

              

4 

1 0.0457 

0.0143 

0.0653 

0.0491 

1.1 0.0373 

0.0116 

0.0548 

0.0446 

1.2 0.0306 

0.0094 

0.0481 

0.0422 

1.3 0.0251 

0.0075 

0.0436 

0.0403 

1.4 0.0206 

0.0061 

0.0403 

0.0387 

1.5 0.0171 

0.0049 

0.0379 

0.0369 

2.0 0.0071 

0.0018 

0.0328 

0.0326 

Notes: 

Poisson’s ratio conservatively assumed 0.25 

γ  = a/b  (aspect ratio) 

Boundary conditions 

 

 

1 = rigid supports; rotationally free; 

 

2 = rigid supports; rotationally fixed; 

 

3 = central panel from an array of identical panels supported on columns;  

                deflection at center; and 

 

4 = similar to case 3, but deflection at center of long span at support line 

 

w = k (a

4

*q / E*h

3

 ) 

 Where, 

 

w = deflection normal to slab; 

 

a  = span along X-direction; 

 

E = Modulus of elasticity; and 

 

h = slab thickness.  

 

 

 

 

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Technical Note

 

 

 

 

6

EXAMPLE 1 

 

Consider the floor system shown in Fig. EX-1. Estimate the deflection of the slab panel 

identified in part b of the figure under the follwoing conditions. Other particulars of the floor 

system are noted in Appendix A. 

 

Given: 

Span length along X-X direction 

 

= 30’  (9.14 m) 

Span length along Y-Y direction 

 

= 26.25’  (8.0 m) 

Slab thickness  

 

 

 

= 8 in  (203 mm) 

Ec (modulus of elasticity) 

 

 

= 4.287 * 10

6 

psi  ( 29,558 MPa) 

 

Superimposed dead load 

 

 

= 25 psf  ( 1.2 kN/m

2

) 

Live load 

 

 

 

 

= 40 psf  (1.9  kN/m

2

) 

 

Required  

Deflection of the panel at midspan for the following load combination 

 

 

1*DL  + 1*LL 

 

Aspect ratio γ = 30/26.25 = 1.14 

 

 

Total service load q = [(20+5+40) + 150*8/12]/144 = 1.146 lb/in

2 

(7.9*10

-3 

N/mm

2

) 

 

 

Using closed form formulas  (Table  5) 

 

 (a

4

*q / E*h

3

 )  = [(30*12)

4

 * 1.146 / (4.287*10

6

 * 8

3

) = 8.77 in  (222.76 mm) 

 

For mid-panel deflection, consider case 3 from Table 5 

 

 

k  = 0.0548 

 

 

Deflection, Δ = k (a

4

*q / E*h

3

 )= 0.0548*8.77 = 0.48 in  (12.20 mm) 

 

For deflection at midpoint of column lines in X-direction, from Table 5 

 

 

 

k  = 0.0446 

 

 

Deflection, Δ = k (a

4

*q / E*h

3

 )= 0.0446*8.77 = 0.39 in  (9.91 mm) 

 

 

 

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Technical Note

 

 

 

 

7

 

 

(a)  3D View of typical floor system 

 

 

(b) Illustration of design panel 

 

FIGURE EX-1     TYPICAL FLOOR HIGHLIGHTING THE SPAN UNDER CONSIDERATION 

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Technical Note

 

 

 

 

8

Using equivalent moment of inertia (Ie) and (ACI-318’s simplified procedure) 

 

In this method allowance is made for crack formation in the slab. The reduction in flexural 

stiffness due to cracking is accounted for by substituting the otherwise gross moment of inertia 

“Ig” used in the calculation with a reduced effective moment of inertia “Ie.”  The equivalent 

moment of inertia Ie can be applied to the entire span through the “simplified procedure of ACI-

318), or applied locally along the entire length of a member for a detailed procedure.  

 

The calculation of the effective moment of inertia Ie will be described next.  

 

Using ACI-318 

 

Ie = (Mcr / Ma)

3

 * Ig + [1-(Mcr / Ma)

3

] * Icr ≤ Ig 

(1) 

Where,  

Ig 

=      Gross moment of inertia; 

Icr 

= 

Moment of inertia of cracked section; 

Ie 

= 

Effective moment of inertia; 

Ma 

= 

Maximum moment in member at stage deflection is computed; and, 

Mcr = 

Cracking 

moment. 

The applied moment, Ma, is calculated using elastic theory and the gross moment of inertia 

(Ig) for the uncracked section. The change in distribution of moment in indeterminate 

structures resulting from cracking in concrete is generally small, and is already accounted 

for in the empirical formula (1) for equivalent moment of inertia.  The cracking moment is 

given by: 

Mcr = frIg /yt (2) 

Where, 

fr   =  Modulus of rupture, flexural stress causing cracking. It is given by: 

fr      = 7.5 f’c

1/2

    

(3) 

y

t

      = distance of section centroid to farthest tension fiber 

 

For all-lightweight concrete, fr is modified as follows: 

 

fr = 0.75 * 7.5 f’c

1/2

 

  (4) 

 

 

Figure 2  illustrates the equivalent moment of inertia Ie for a simply supported concrete slab that 

is partially cracked. 

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