Microsoft Word tn292 floor deflection 032109. doc

                                                                            

                   

Technical Note

 

 

 

 

9

 

FIGURE 2   ILLUSTRATION OF EFFECTIVE MOMENT OF INTERTIAL 

IN A PARTIALLY CRACKED SLAB 

 

The value of cracking moment of inertial Icr and the geometry of the section depends on the 

location and amount of reinforcement. For rectangular sections with single reinforcement (Fig. 2 

)  the value is given by: 

 

 

 

 

Icr = (bk3d3)/3 + nAs(d-kd)2 (5) 

Where, 

  kd 

= 

[(2dB+1)1/2 – 1]/B 

(6) 

 

d = distance from compression fiber to center of tension reinforcement 

 

B = b/(nAs)  

 

n = Es/Ec 

   

Es = modulus of elasticity of steel 

Ec = modulus of elasticity of concrete 

   

 

For more details and treatment of other cross-sections refer to ADAPT Technical Note TN293.   

 

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Technical Note

 

 

 

 

10

 

 

 

FIGURE  3  

  

 

In the simplified method an average value of Ie  is used for the entire span. For spans, the 

average value is calculated  

 

 

I

e, av  

= 0.5 [ (I

e,left support 

+ I

e,right support

 )/2  + I

e, midspan 

]  

 

(7) 

 

 

 

EXAMPLE 2 

 

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

identified under the same loading and conditions expressed in Example 1, using the simplified 

option of ACI-318 for equivalent moment of inertia Ie 

 

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 (29558 MPa) 

Other details of the slab are given in Example 1 and the Appendix A 

 

Required 

Determine the deflection at the center of the panel identified in Example 1 due to the sum of 

dead and live loads. 

 

Calculate Cracking Moment Mcr 

 

 

Ig  

= 15,360 in

4

 (6.40e+10 mm

4

)           

 

yt    

=  4 “ ( 101.60 mm) 

 

fr  

=  7.5√f’c = 7.5*√5000 = 530.33 psi   (3.66 MPa) 

 

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Technical Note

 

 

 

 

11

Mcr      = frIg /yt    =  530.33*15,360/(4*12000) = 169.70 k-ft  (230 kNm)    

 

To determine the deflection at center, the applied moment (Ma) for the “design strip” associated 

with the panel in question must be determined. Refer to Fig. EX2 – 1a 

 

Ie = (Mcr / Ma)

3

 * Ig + [1-(Mcr / Ma)

3

] * Icr ≤ Ig 

(1) 

 

The design strip associated with the panel under consideration is shown in Fig. EX-2a. It 

connects the line of columns and extends on each side to the midspan line of the adjacent 

panels.  

 

The design strip extracted from the floor system is shown in its idealized form in Fig. EX-2b. 

Using a computer program, the applied moment Ma in the idealized design strip is calculated. 

 

 

 

(a) Plan of slab showing the design strip associated with the panel under consideration 

 

 

(b) View of the design strip extracted from the floor system 

 

FIGURE EX2 -1   PLAN OF TYPICAL FLOOR HIGHLIGHTING THE DESIGN STRIP OF THE 

SPAN UNDER CONSIDERATION 

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Technical Note

 

 

 

 

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A solution obtained from  the computer program ADAPT-RC

3

 [ADAPT RC, 2008] gives the 

following values: 

 

 

 

            FIGURE EX2-2   DISTRIBUTION OF MOMENTS DUE TO DEAD PLUS  

                                                           LIVE LOAD 

 

 

The computed deflection for the first span without accounting for the crack formation associated 

with moments of Fig. EX2-2 is 0.231 in. (5.9 mm). 

 

It is noteworthy that the strip method, as outlined herein, provides the deflection value in the 

direction of analysis, not accounting for the deflection in the transverse direction. For a complete 

analysis of mid-panel displacement, the deflection in the transferse direction must also be 

calculated and added to the deflection calculated for this direction (Fig. EX2-3). For panels that 

are fairly square, it is acceptable to multiply the deflection calculated for one direction by a 

factor of 2. For this example, the total deflection is estimated as: 

 

 

Total deflection = 2 * 0.231 = 0.462 in. (11.7 mm) 

 

 

 

 

    ADD 

FIGURE 

??? 

 

 

FIGURE EX2-3  COMBINATION OF DEFLECTIONS FROM 

 ORTHOGONAL DIRECTIONS 

on the assumption that there is no transv that is representative of both the midpoint of the panel 

and midpoint of the line of support. 

                                                 

3

  ADAPT-RC is a computer program for design and analysis of conventionally reinforced beam frames and slabs. 

It is based on Equivalent Frame Method (www.adaptsoft.com).. 

 

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