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Introduction Ultimate Limit States Lead to collapse Serviceability Limit States Disrupt use of Structures but do not cause collapse Recall:

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Presentation on theme: "Introduction Ultimate Limit States Lead to collapse Serviceability Limit States Disrupt use of Structures but do not cause collapse Recall:"— Presentation transcript:

1 Introduction Ultimate Limit States Lead to collapse Serviceability Limit States Disrupt use of Structures but do not cause collapse Recall:

2 Introduction Types of Serviceability Limit States - Excessive crack width - Excessive deflection - Undesirable vibrations - Fatigue (ULS)

3 Crack Width Control Cracks are caused by tensile stresses due to loads moments, shears, etc..

4 Crack Width Control Cracks are caused by tensile stresses due to loads moments, shears, etc..

5 Crack Width Control Heat of hydration cracking

6 Crack Width Control Bar crack development.

7 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) Reasons for crack width control?

8 Crack Width Control 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. Corrosion more apt to occur if (steel oxidizes rust )

9 Limits on Crack Width 0.40 mm for interior exposure 0.33 mm for exterior exposure max.. crack width = ACI Code’s Basis Prior to 1999 Now ACI handles crack width indirectly by limiting the bar spacings and bar cover for beams and one way slabs ACI Bar spacings must also satisfy ACI (3t or 450mm)

10 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.

11 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 and This limit is about three times that specified by ACI code which is based on empirical results.

12 3. Web face reinforcement:

13 Deflection Control Visual Appearance ( 7.5m. span 30mm ) Damage to Non-structural Elements - cracking of partitions - malfunction of doors /windows (1.) (2.) Reasons to Limit Deflection (Table 9-3)

14 Deflection Control Disruption of function - sensitive machinery, equipment - ponding of rain water on roofs Damage to Structural Elements - large  ’s than serviceability problem - (contact w/ other members modify load paths) (3.) (4.)

15 Allowable Deflections ACI Table 9.5(a) = min. thickness unless  ’s are computed

16 Allowable Deflections ACI Table 9.5(b) = max. permissible computed deflection

17 Deflection Response of RC Beams (Flexure) The maximum moments for distributed load acting on an indeterminate beam are given.

18 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 ends & midspan Note: Stiffness (slope) decreases as cracking progresses

19 Moment Vs curvature plot

20 “Moment Vs Slope” Plot The cracked beam starts to lose strength as the amount of cracking increases

21 To avoid complexity in calculations, an overall average effective moment of inertia

22 Moment of Inertia for Deflection Calculation For (intermediate values of EI) Branson derived Cracking Moment = Gross moment of inertia of rc cross-section Modulus of rupture = M cr = I g = f r = If M a / M cr > 3, the cracking will be extensive, I e = I cr If M a / M cr < 1, no cracking is likely and I e =I g

23 Moment of Inertia for Deflection Calculation Distance from centroid to extreme tension fiber maximum moment in member at loading stage for which I e (  ) is being computed or at any previous loading stage y t = M a =

24 Deflection Response of RC Beams (Flexure) ACI Com. 435 Weight Average ACI code

25 Definition of I g ACI code: I g is the moment of inertia of the gross concrete section neglecting area of tension steel. I g might be more accurate if it includes the transformed area of the reinforcement. I g 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 = E s / E c : E s = 200GPa, E c =4700  f c ).

26 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 I e 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.

27 Uncracked Transformed Section Note: (n-1) is to remove area of concrete

28 Cracked Transformed Section Finding the centroid of singly Reinforced Rectangular Section Solve for the quadratic for

29 Cracked Transformed Section Note: Singly Reinforced Rectangular Section

30 Cracked Transformed Section Note: Doubly Reinforced Rectangular Section

31 Uncracked Transformed Section Note: Moment of inertia (uncracked doubly reinforced beam)

32 Example 2 (9-1) For the shown beam of 28MPa concrete, Find: 1.Moment of inertia of uncracked section. 2.Moment of inertia of cracked section.

33 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

34 Cracked Transformed Section Finding the centroid of doubly reinforced T-Section

35 Cracked Transformed Section Finding the moment of inertia for a doubly reinforced T-Section

36 Calculate the Deflections (1) Instantaneous (immediate) deflections (2) Sustained load deflection Instantaneous Deflections due to dead loads( unfactored), live, etc.

37 Calculate the Deflections Instantaneous Deflections Equations for calculating  inst for common cases

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39 Sustained Load Deflections Sustained Load Deflections Creep causes an increase in concrete strain Curvature increases Compression steel present Increase in compressive strains cause increase in stress in compression reinforcement (reduces creep strain in concrete) Helps limit this effect.

40 Sustained Load Deflections Sustained load deflection =  i Instantaneous deflection ACI at midspan for simple and continuous beams at support for cantilever beams

41 Sustained Load Deflections  = time dependent factor for sustained load 5 years or more 12 months 6 months 3 months Also see Figure from ACI code

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43 The total long time deflection where δ L = immediate live load deflection δ D = immediate dead load deflection δ SL = sustained live load deflection (a percentage of the immediate δ L determined by expected duration of sustained load) λ  = time dependant multiplier for infinite duration of sustained load (assumed to occur after partitions are installed) λ t0,  = time dependant multiplier for infinite duration minus that at the time t0 when partitions are installed To calculate δ L (or δ SL ) due to the live loads, the following procedure has been found to be generally satisfactory:

44 Calculation of long time deflection 1. Calculate the deflection δ D+L due to dead and live loads acting simultaneously. For this calculation I e is found using Eq. 9.8 and the moment M a is the one produced when both dead and live loads are acting simultaneously. 2. Calculate the deflection δ D due to the dead load acting alone. For this calculation I e is found using Eq. 9.8 and the moment M a is the one produced when the dead load acts alone. 3. Subtract the deflection δ D from the deflection δ D+L to obtain the desired deflection δ L.

45 Handling long term deflections Surveys of partition damage have shown that damage to brittle partitions can occur with deflections as small as L/1000. A frequent limit on deflections that cause damage is L/480 after attachment of nonstructural elements. The value L/480 shall be compared with the value of the total long time deflection. 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.

46 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.

47 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?

48 Problem 1 ( ) 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 1.the immediate dead load deflection. 2.the immediate dead-plus-live load deflection 3.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.

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50 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 (a)the immediate dead-load deflection. (b)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.

51 Problem 9.10


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