Serviceability. Introduction. Recall:. Ultimate Limit States Lead to collapse Serviceability Limit States Disrupt use of Structures but do not cause collapse. Introduction. Types of Serviceability Limit States - Excessive crack width
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Ultimate Limit States Lead to collapse
Serviceability Limit States Disrupt use of Structures but do not cause collapse
Types of Serviceability Limit States
- Excessive crack width
- Excessive deflection
- Undesirable vibrations
- Fatigue (ULS)
Bar crack development.
Reasons for crack width control?
Corrosion more apt to occur if (steel oxidizes rust )
ACI Code’s Basis Prior to 1999
max.. crack width =
0.40 mm for interior exposure
0.33 mm for exterior exposure
Now ACI handles crack width
indirectly by limiting the bar spacings and bar cover for beams and one way slabs ACI 10.6.4.
Bar spacings must also satisfy ACI 7.6.5 (3t or 450mm)
A 20cm thick slab has 12mm diameter bars. The bars have 420MPa yield stress and a minimum clear cover of 20mm. Compute the maximum value of s.
1. Negative moment regions of T-beams.
2. Shrinkage and temperature reinforcement: is intended to replace the tensile stresses in the concrete
at the time of cracking, using the following
For grade 60 steel and 28MPa concrete,
Steel ratio is between 0.004 and 0.005.
This limit is about three times that specified by ACI code 126.96.36.199 which is based on empirical results.
Reasons to Limit Deflection (Table 9-3)
( 7.5m. span 30mm )
Damage to Non-structural Elements - cracking of partitions - malfunction of doors /windows
Disruption of function - sensitive machinery, equipment - ponding of rain water on roofs
Damage to Structural Elements - large d’s than serviceability problem - (contact w/ other members modify load paths)
ACI Table 9.5(a) = min. thickness unless d’s are computed
The maximum moments for distributed load acting on an indeterminate beam are given.
A- Ends of Beam Crack
B - Cracking at midspan
C - Instantaneous deflection under service load
C’ - long time deflection under service load
D and E - yielding of reinforcement @ ends & midspan
Note: Stiffness (slope) decreases as cracking progresses
The cracked beam starts to lose strength as the amount of cracking increases
ACI Com. 435
ACI code: Ig is the moment of inertia of the gross concrete section neglecting area of tension steel.
Ig might be more accurate if it includes the transformed area of the reinforcement.
Ig is the moment of inertia of the uncracked transformed section. The transformed section consists of the concrete area plus the transformed steel area(=the actual steel area times the modular ratio n = Es / Ec : Es = 200GPa , Ec =4700fc ).
Once a beam has been cracked by a large moment, it can never return to its original uncracked state; therefore, the effective moment of inertia Ie that should be used in deflection computations must always be equal to the effective moment of inertia associated with the maximum past moment to which the beam has been subjected. Often this moment is impossible to determine for most beams.
(n-1) is to remove area of concrete
For the shown beam of 28MPa concrete, Find:
For the shown beam of 31.5MPa concrete, Find steel stress at service loads if the service live-load moment is 70kN.m and the service dead load moment is 96kN.m
Finding the centroid of doubly reinforced T-Section
Finding the moment of inertia for a doubly reinforced T-Section
(1) Instantaneous (immediate) deflections
(2) Sustained load deflection
due to dead loads( unfactored) , live, etc.
Equations for calculating Dinst for common cases
Creep causes an increase in concrete strain
Increase in compressive strains cause increase in stress in compression reinforcement (reduces creep strain in concrete)
Compression steel present
Helps limit this effect.
Sustained load deflection = l Di
at midspan for simple and continuous beams
at support for cantilever beams
δL = immediate live load deflection
δD = immediate dead load deflection
δSL = sustained live load deflection (a percentage of the immediate δL determined by expected duration of sustained load)
λ = time dependant multiplier for infinite duration of sustained load
λt = time dependant multiplier for limited load duration
To calculate δL (or δSL) due to the live loads, the following procedure has been found to be generally satisfactory:
1. Calculate the deflection δD+L due to dead and live loads acting simultaneously. For this calculation Ie is found using Eq. 9.8 and the moment Ma is the one produced when both dead and live loads are acting simultaneously.
2. Calculate the deflection δD due to the dead load acting alone. For this calculation Ie is found using Eq. 9.8 and the moment Ma is the one produced when the dead load acts alone.
3. Subtract the deflection δD from the deflection δD+L to obtain the desired deflection δL.
If the long time deflections exceeds the value permitted, the designer may either increase the depth of members, or add additional compression steel. If the sag produced by the long time deflections is objectionable from an architectural or functional point of view, forms may be raised (cambered) a distance equal to that of the anticipated deflection.
The T-beam shown in Fig. is made of 28MPa concrete and supports unfactored dead and live loads of 13kN/m and 18kN/m. Compute the immediate midspan deflection. Assume that the construction loads did not exceed the dead load.
If the beam in the previous example is assumed to support partitions that would be damaged by excessive deflections. If 25% of the live load is sustained. The partitions are installed at least 3 months after the shoring is removed. Will the computed deflections exceed the allowable in the end span?
9-8 A simply supported beam with the cross section shown in Figure next page has a span of 7.5m and supports an unfactored dead load of 22.5kN/m, including its own self-weight plus an unfactored live load of 22.5kN/m. The concrete strength is 31.5MPa. Compute
9-9 Repeat Problem 9-8 for a beam having the same dimensions and tension reinforcement, but with two No. 25mm bars as compression reinforcement.
The beam shown in Figure next page is made of 28MPa concrete and supports unfactored dead and live loads of 15kN/m and 17kN/m respectively. Compute
(c)the deflection occurring after partitions are installed. Assume that the partitions are installed four months after the shoring is removed and assumed that 10 percent of the live load is sustained.