Microsoft Word tn292 floor deflection 032109. doc

                                                                            

                   

Technical Note

 

 

 

 

13

 It is noteworthy that the strip method, as outlined herein, provides ?? 

 

Using the moment values given in Fig. EX2-2 for the right support of span 1 the Ie is given by: 

 

 

M

a

 = 326.4 k-ft (442.53 kNm) 

 

I

g

 = 17,019 in

4

 (6.40e+10 mm

4

) 

 

M

cr

      = f

r

I

g

 /y

t

 = 530.33*17,019/(4*12000) = 188.04 k-ft   (254.94 kNm)

 

  

 

I

e

 = (M

cr

 / M

a

)

3

 * Ig + [1-(M

cr

 / M

a

)

3

] * I

cr

 ≤ I

g

 

 

 

Where, 

 

I

cr

 = (bk

3

d

3

)/3 + nAs (d-kd)

2

 

 

 

 

kd   = [(2dB+1)1/2 – 1]/B 

 

 

d     = 6.81 in (173 mm) 

 

B    = b/(nAs)  

n     =  Es/Ec = 30000/4287 = 7.0 

As   =  10.12 in

2 

 (6529 mm

2

) 

 

 

B     =  360/(7.0* 10.12) = 5.08 /in (0.2/mm) 

 

 

Kd   =  [(2*6.81*5.08+1)

1/2

 – 1]/5.08 

 

 

        = 1.45 in (36.83 mm) 

 

 

 I

cr

  

= (360*1.45

3

)/3 + 7.0*10.12* (6.81-1.45)

2

 

  = 

2401 

in

4 

(9.99e+8) 

 

I

e

  

= (188.04 / 326.4)

3

 * 17019 + [1-(188.04 / 326.4)

3

] *2401  

  = 

5196 

in

4 

(2.16e+9)= 0.31 Ig 

 

Using the same procedure, the value of Ie at other locations required by the code formula are 

calculated and listed below: 

 

 

Left cantilever: 

 

I

e

 at face of support = I

g

   

 

 

= 1.536e+4 in

4   

(6.39e+9 mm

4

) 

 

First  Span: 

I

e

 at left support centerline    

 

= 1.70e+4 in

4   

(7.08e+9 mm

4

) 

I

e

 at midspan   

  

 

 

= 1.44e+4 in

4   

(5.99e+9 mm

4

) 

I

e

 at right support centerline    

 

= 5.20e+3 in

4   

(2.16e+9 mm

4

) 

 

Second Span: 

I

e

 at left support centerline    

 

=  5.63e+3 in

4   

(2.34e+9 mm

4

) 

I

e

 at midspan   

  

 

 

= 1.536e+4 in

4  

(6.39e+9 mm

4

) 

I

e

 at right support centerline    

 

= 1.702e+4 in

4  

(7.08e+9 mm

4

) 

 

 

Right cantilever: 

---

                                                                            

                   

Technical Note

 

 

 

 

14

 

I

e

 at face of support = I

g

   

 

 

= 1.536e+4 in

4  

(6.39e+9 mm

4

) 

 

Using the averaging procedure suggested by ACI-318, the Ie values to be used for deflection 

calculation are: 

 

Left and right cantilevers  I

e

 = I

g

 

 

First span 

Average I

e

 

 

=  [(1.70*10

4

+5.20*10

3

)/2 +1.44*10

4

]/2  

   = 

12.755e+3 

in

4  

(5.31e+9 mm

4

) 

 

Second span 

Average I

e

 

 

=  [(5.63*10

3

+1.702*10

4

)/2 +1.536*10

4

]/2  

   = 

13.343e+3 

in

4  

(5.55e+9 mm

4

) 

 

 

In order to use the same frame program for the calculation of deflected shape, the calculated 

equivalent moments of inertia are used to determine an equivalent thickness (h

e

) for each of the 

spans. The equivalent thickness is given by 

 

 

I

e

 = b*h

e

3

 /12  

 

 

Where,  b  is the width of the tributary of the design strip. 

 

Left 

cantilever: 

  h

e

 = 8 in      (203 mm) 

First 

span 

   h

e

 = 7.52 in (191 mm) 

Second 

span 

   h

e

 = 7.63 in (193.8 mm) 

Right 

cantilever   h

e

 = 8 in      (203 mm) 

 

Using the same computer program, material values, boundary conditions and loads, but with the 

reduced slab thickness of modified  moment of inertia a new solution is obtained. 

 

The maximum value of deflection for span 1 is 0.264 in (6.71 mm), compared to 0.231 in (5.87 

mm), without allowing for reduction of stiffness due to cracking. Note that the above deflections 

do not account for the flexure of the slab in the orthogonal direction, as indicated in Example 2. 

For engineering design, where panels are fairly square, the calculated values are commonly 

multiplied by 2 to represent the deflection at the middle of panel. Hence, mid-panel deflections 

would be 0.528 in. (13.42 mm) and 0.462 in. (11.74 mm). 

 

 

 

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