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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
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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 - Excessive deflection - Undesirable vibrations - Fatigue (ULS)
Crack Width Control Cracks are caused by tensile stresses due to loads moments, shears, etc..
Crack Width Control Cracks are caused by tensile stresses due to loads moments, shears, etc..
Crack Width Control • Heat of hydration cracking
Crack Width Control Bar crack development.
Crack Width Control Reasons for crack width control? • Appearance (smooth surface > 0.25 to 0.33mm = public concern) • Leakage (Liquid-retaining structures) • Corrosion (cracks can speed up occurrence of corrosion)
Crack Width Control Corrosion more apt to occur if (steel oxidizes rust ) • Chlorides ( other corrosive substances) present • Relative Humidity > 60 % • High Ambient Temperatures (accelerates chemical reactions) • Wetting and drying cycles • Stray electrical currents occur in the bars.
Limits on Crack Width 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)
Example 1 (9-4) 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.
Other important issues for crack control 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 simplified analysis: 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 184.108.40.206 which is based on empirical results.
Deflection Control Reasons to Limit Deflection (Table 9-3) (1.) Visual Appearance ( 7.5m. span 30mm ) Damage to Non-structural Elements - cracking of partitions - malfunction of doors /windows (2.)
Deflection Control (3.) 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) (4.)
Allowable Deflections ACI Table 9.5(a) = min. thickness unless d’s are computed
Allowable Deflections • ACI Table 9.5(b) = max. permissible computed deflection
Deflection Response of RC Beams (Flexure) The maximum moments for distributed load acting on an indeterminate beam are given.
Deflection Response of RC Beams (Flexure) 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
“Moment Vs Slope” Plot The cracked beam starts to lose strength as the amount of cracking increases
To avoid complexity in calculations, an overall average effective moment of inertia
Cracking Moment = Gross moment of inertia of rc cross-section Modulus of rupture = Mcr = Ig = fr = Moment of Inertia for Deflection Calculation For (intermediate values of EI) Branson derived If Ma / Mcr > 3, the cracking will be extensive, Ie = Icr If Ma / Mcr < 1, no cracking is likely and Ie =Ig
yt = Distance from centroid to extreme tension fiber maximum moment in member at loading stage for which Ie ( d ) is being computed or at any previous loading stage Ma = Moment of Inertia for Deflection Calculation
Deflection Response of RC Beams (Flexure) ACI code ACI Com. 435 Weight Average
Definition of Ig 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 ).
Definition of I 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.
Uncracked Transformed Section (n-1) is to remove area of concrete Note:
Solve for the quadratic for Cracked Transformed Section Finding the centroid of singly Reinforced Rectangular Section
Note: Cracked Transformed Section Singly Reinforced Rectangular Section
Note: Cracked Transformed Section Doubly Reinforced Rectangular Section
Note: Uncracked Transformed Section Moment of inertia (uncracked doubly reinforced beam)
Example 2 (9-1) For the shown beam of 28MPa concrete, Find: • Moment of inertia of uncracked section. • Moment of inertia of cracked section.
Example 3 (9-2) 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
Cracked Transformed Section Finding the centroid of doubly reinforced T-Section
Cracked Transformed Section Finding the moment of inertia for a doubly reinforced T-Section
Calculate the Deflections (1) Instantaneous (immediate) deflections (2) Sustained load deflection Instantaneous Deflections due to dead loads( unfactored) , live, etc.
Calculate the Deflections Instantaneous Deflections Equations for calculating Dinst for common cases
Sustained Load Deflections Creep causes an increase in concrete strain Curvature increases 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 Deflections Sustained load deflection = l Di Instantaneous deflection ACI 220.127.116.11 at midspan for simple and continuous beams at support for cantilever beams
5 years or more 12 months 6 months 3 months Sustained Load Deflections x = time dependent factor for sustained load Also see Figure 18.104.22.168 from ACI code
The total long time deflection where 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:
Calculation of long time deflection 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.
Example 4 (9-5) 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.
Example 5 (9-5) 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?
Problem 1 (9-8 + 9-9) 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 • the immediate dead load deflection. • the immediate dead-plus-live load deflection • the deflection occurring after partitions are installed. Assume that the partitions are installed two months after shoring for the beam is removed and assume that 20 percent of the live load is sustained. 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.
Problem 9-10 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 • the immediate dead-load deflection. • the immediate dead-plus-live load deflection. (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.