deflections of cracked concrete T-section.pdf

deflections of cracked beam sections using ACI

1. Cracking –Concrete structures
Direct Tension Cracks
Bending cracks with and without axial force

2. Cracking in Concrete
Concrete carries stresses between the cracks. The stresses in cracked
elements can be illustrated
Change in tension stresses in concrete with crack formation

3. ACI Committee 224, in a report on cracking, presented a set of
approximately permissible maximum crack widths for reinforced
concrete members subject to different exposure situations.
Cracking in Concrete

4. Gergely–Lutz equation
In applying it to beams, reasonable results are usually obtained if βh is set equal to 1.20. For
thin one-way slabs, however, more realistic values are obtained if βh is set equal to 1.35.
The number of reinforcing bars present in a particular member decidedly affects the value
of A to be used in the equation and thus the calculated crack width.
If more and smaller bars are used to provide the necessary area, the value of A will be
smaller, as will the estimated crack widths.
Crack width –Concrete Beams

5. The best model for crack width is Gergely-Lutz model
d
c
h
1
h
2
N.A
x
x
Tension steel
centroid
A= Shaded area/
number of bars6
1
23
cs 10
h
h
Adf11w

 3
1
23
cs 10
h
h
Adf076.0w


Metric
US
f
sin ksi
The ratio (h
2/h
1) can
be taken 1.2 for
beamsand 1.35 for
slabs
A = 2d
c.s
Crack width –Concrete Beams

6. Example Crack Width
Assuming β
h= 1.20 and fy= 60 ksi, calculate the estimated width of flexural cracks that will occur in the beam
shown below. If the beam is to be exposed to moist air, is this width satisfactory as compared to the values given
in ACI 224? Should the cracks be too wide, revise the design of the reinforcing and recomputethe crack width.

7. ACI Code Provisions Concerning Cracks
The ACI equation for maximum spacing does not apply to beams with extreme exposure or
to structures that are supposed to be watertight. Special consideration must be given to such
situations. It is probably well to use the Gergely–Lutz equation and a set of maximum crack
widths

8. To calculate deflection at service load levels there are two
approaches
M
service< M
cr:Use transformed sectionI
g
M
service> M
cr:Use effectiveI
e
The effective inertia approach
The curvature approach
M
service< M
cr:Use transformed sectionI
g
M
service> M
cr:Interpolate between cracked and non-
cracked curvatures
Deflection of RC beams

9. Deflection: Effective Inertia Approach
Cracked
I
e
Uncracked
I
g
Fully cracked
I
cr
where
Branson (1977) Equation
g
3
a
cr
crgcre I)
M
M
()II(II +

22
s
3
cr dk1An
3
kdb
I + 12
bh
I
3
g nnn2k
2
 +

10. Deflection: Effective Inertia Approach

11. As concrete creeps under sustained loads, time-dependent “creep”
effects will increase the deflection of sustained loads
Two methods to incorporate the time-dependent creep effect
If Branson’s equation –ACI approach is used
The ACI approachuses a multiplier, therefore it has a limited
accuracy.
The curvature approachcan use any classical or recent creep
prediction models. The predicted deflections using the curvature
approach with time-dependent effects are usually accurate.
Deflection: time-dependent effects
If the curvature approach is used

12. •The Value for z is determined from Figure R9.5.2.5 from the ACI318-
08 code and is a function of the duration of load.
•Research experiments showed that this approximate method has a
very poor prediction accuracy within ±30%!!
•To account for the effect of creep and shrinkage, ACI 318-08 for RC section
require multiplying the instantaneous deformations by a factor “l
D”
where
'

x
l
D
501+

 
loadssustainedatnoustaninsshrinkagecreep
dl
D
d 
+









monthst.
monthst.
yearst.
301
1241
502
x
Creep effects: ACI approach

13. Deflection: ACI 318-08 Limits (table 9.5(b))
Member type
Deflection to be
considered
Limit
Elements not supporting
or attached to non-
structural elements likely
to be damaged by large
deflection
Roof or floors supporting or
attached to non-structural elements
likelyto be damaged by large
deflection
Flat roofs
Floors180
span 360
span
Roof or floors supporting or
attached to non-structural elements
not likelyto be damaged by large
deflection480
span 240
span

14. The beam shown below has a simple span of 20 ft and supports a dead load including its own weight of 1 klf and a
live load of 0.7 klf. fc’ = 3000 psi.
(a) Calculate the instantaneous deflection for D + L.
(b) Calculate the deflection assuming that 30% of the live load is continuously applied for 3 years.
By transformed-area calculations, the values of x and Icr can
be determined in previous Example
Deflection: Example

15. This is the live load deflection that would be compared with the allowable limits
for deflection. If the beam is part of a floor system that is ‘‘not supporting or
attached to nonstructural elements likely to be damaged by large deflections’’
then the deflection limit is l/480 = (20 ft) (12
in/ft)/360 = 0.67 in. This limit would easily be
satisfied in this case, as the calculated
deflection is only 0.22 in.
Deflection: Example 360

16. The answer is compared with either l/480 or l/240, depending on whether the structural member
supports elements likely to be damaged by large deflections.
Deflection: Example

17. Example: Serviceability and Deflection
Determine which of the ACI Code deflection criteria will be satisfied by the non-prestressed
reinforced concrete beam having a cross section shown in Figure below and subjected to
maximum moments of MDL = 20 ft.-kips and MLL= 15 ft.-kips. Assume a 50% sustained live
load and a sustained load time period of more than 5 years. Assume normal-weight
concrete. Use f c = 3000 psi and fy= 60,000 psi. The beam is on a simple span of 30 ft.
The maximum service moments at midspan are MDL = 20 ft.kipsand MLL= 15 ft.kips
Check minimum depth = minimum h = l/16 = 30(12)/16 = 22.5 in.
Because 22.5 in. > 16.5 in., Deflection must be calculated.
Practice Problem

18. Continuous-Beam Deflections
ACI Code (9.5.2.4) permits the use of a
constant moment of inertia throughout the
member equal to the average of the Ie values
computed at the critical positive-and negative-
moment sections. The Ie values at the critical
negative-moment sections are averaged with
each other, and then that average is averaged
with Ie at the critical positive-moment section.
It should also be noted that the multipliers for
long-term deflection at these sections should
be averaged, as were the Ie values.

19. Example Continuous-Beam Deflections
Determine the instantaneous deflection at the midspan of the continuous T beam shown in
Figure below. The member supports a dead load, including its own weight, of 1.5 k/ft, and a live
load of 2.5 k/ft. fc’ = 3000 psi and n = 9.

20. Example Continuous-Beam Deflections
Locating the centroidalaxis of cracked section and calculating transformed moment of inertia
Icrfor the positive-moment region.
quadratic equations results in x = 5.65 in.
Since this value of x exceeds hfof 5 in., the
assumption that the neutral axis is in the web is valid.
If x had been smaller than 5 in., then the value we
obtained would not have been valid, and the
preceding equations would have to be rewritten and
solved assuming x<hf.

21. Example Continuous-Beam Deflections
Cracked section and calculating the transformed moment of
inertia Itr for the negative-moment region

22. Continuous-Beam Deflections
using only live loads to calculate deflections

23. Example Continuous-Beam Deflections

24. References
•Textbook, Reinforced Concrete Mechanics and Design: James K.
Wight and James. G. MacGregor (2009)
•Chapter 9
•Ghali, A., Favre R., and Elbadry, M. Concrete Structures: Stresses and
Deformation –Third Edition, Spon Press, 2002.
•CEB-FIP Model code 1990
•ACI 318 –08-Chapter 9