RC II Ch. 5 Serviceability Prof. Dr. Bayan Salim 1

5 Serviceability Although adequate strength is guaranteed, it is important that the following serviceability

concerns be taken into consideration:

- Service load deflections under full load may be excessively large.

- Long-term deflections due to sustained loads may cause damage.

- Tension cracks in beams may be wide enough to be visually disturbing, and in

some cases may reduce the durability of the structure.

- Other questions, such as vibration or fatigue, require consideration.

In this chapter, methods will be developed to ensure that:

- Cracks associated with flexure of RC beams are narrow and well distributed.

- Short and long-term deflections at loads up to the full service load are not

objectionably large.

Cracking in Flexural Members All RC beams crack, generally starting at loads well below service loads, and even prior

to loading due to restrained shrinkage.

If loads further increased, crack widths increase further. For reasons of durability and

appearance, many fine cracks are preferable to a few wide cracks.

Based on test observations, it was confirmed that crack width at service loads is

proportional to reinforcement stress fs.

Variables Affecting Width of Cracks - Beams with bars having proper surface deformation will show a larger number of

very fine cracks.

- Stress in the reinforcement.

- Concrete cover distance.

- Distribution of the reinforcement in the tension zone. It is better to use larger

number of smaller-diameter bars than two larger bars of equivalent area.

ACI Code Provisions for Crack Control, ACI 24.3.2 Crack width is controlled in the ACI Code by establishing a maximum center-to-center

spacing s for the reinforcement closest to the surface of a tension member:

s = 380 (280 / fs) – 2.5 cc ≤ 300 (280 / fs) … (1) Where

fs = bar stress under service load, MPa, calculated based on unfactored moment

fs = M /(As jd). (See ch.2, ex.2). It may be taken as (⅔ fy)

cc = clear cover from the nearest surface in tension, mm

Example 1: If fy = 420 MPa and clear cover = 50 mm, then

s = 380 (280 / fs) – 2.5 cc ≤ 300 (280 / fs);→ s = 255 ≤ 300. Use smax = 250 mm.

RC II Ch. 5 Serviceability Prof. Dr. Bayan Salim 2

Skin Reinforcement, ACI 9.7.2.3 For deep members h > 900 mm, skin reinforcement shall be uniformly distributed along

both side faces of the member, at distance h/2 from tension face, at spacing ≤ s found

from equation (1), as shown in Fig. below:

(h > 900 mm) Example 2: Fig. below shows the main reinforcement at mid span for a T beam carrying a service

load moment of 975 kNm. The clear cover on the side and bottom of the beam stem is 60

mm. Determine if the beam meets the ACI crack control criteria. Take fy = 420 MPa

Solution: Since h is not greater than 900 mm, no skin reinforcement is required.

Check bar spacing: estimate closely jd = d – hf /2

fs = M / [As (d – hf / 2)] = 975 (10)6 / (5100×725) = 263 MPa

(Alternatively, ACI Code permits fs = ⅔fy = 280 MPa)

s = 380 (280 / fs) – 2.5 cc ≤ 300 (280 / fs)

= 255 mm < 319 mm

Available s = (675 – 2×75)/4 – 25 = 106 mm < 255 mm OK, crack width is controlled.

(If the result had been unfavorable, redesign using smaller-diameter bars)

HW: Rectangular section with cover cc = 60 mm, h = 560 mm, d = 500 mm, b = 450

mm. Check crack control requirements for 2 No.36 tension reinforcement. If not satisfied,

specify alternative design. Take fy = 420 MPa

RC II Ch. 5 Serviceability Prof. Dr. Bayan Salim 3

Control of Deflections

There are two approaches to control deflections:

- Setting suitable upper limits on the span-depth ratio.

- Otherwise, calculating deflections and comparing with specific limitations

imposed by codes.

Immediate Deflections: Deflections that occur upon application of load.

Time-dependent Deflections: Deflections that take place gradually over an extended

period of time (several years). These are chiefly due to concrete creep and shrinkage, and

may be two or more times the initial elastic deflections.

ACI Code Provisions for Control of Deflections

a. Minimum Beam Depth; ACI Code 9.3: For beams not supporting or attached to partitions or other construction likely to be

damaged by large deflections, overall beam depth h shall satisfy the limits in Table

9.3.1.1, unless the calculated deflection limits of 9.3.2 are satisfied.

For fy other than 420 MPa, the expressions in Table 9.3.1.1 shall be multiplied by

(0.4 + fy / 700).

For beams made of lightweight concrete having wc in the range of 1450 – 1850 kg/m3 ,

the expressions in Table 9.3.1.1 shall be multiplied by the greater of (a) and (b):

(a) 1.65 – 0.0003 wc

(b) 1.09

Example 3: A lightweight (1700 kg/m3) SS beam with span 4.8 m, what is the

minimum depth of this beam to control deflection. Take fy = 560 MPa. The beam is not

supporting susceptible partitions.

Solution

Using Table 9.3.1.1 (The beam is not supporting susceptible partitions):

Min h = (l / 16) (0.4 + 560/700)(1.65 – 0.0003×1700)

= 300 (1.2)(1.14) = 411 mm, say 425 mm. [Check 1.14 > 1.09 OK]

RC II Ch. 5 Serviceability Prof. Dr. Bayan Salim 4

b. Calculation of Immediate Deflections Where there is need to use member depths shallower than ACI table 9.3.1.1, or when

members support construction that is likely to be damaged by large deflections,

deflections must be calculated and compared with limiting values.

Elastic midspan deflection for UD loaded simple beam:

Elastic end deflection for UD loaded cantilever beam:

Elastic midspan deflection for UD loaded continuous beam:

δm = 5 l2 [Mm + 0.1(MA + MB)] / (48EI) The moments Mm, MA, and MB, refer to midspan and ends respectively. Appropriate

signs must be included for the moments, usually (+) for Mm and (−) for MA and MB

Similar deflection equations can be used for many other loading and span

arrangements.

The particular problem in RC structures is the determination of the appropriate

flexural rigidity EI for a member.

In the regions of flexural cracks, the position of NA varies; the corresponding

moment of inertia I will be:

- The I directly at each crack, I = Icr,

- The I midway between cracks, I = Ig

RC II Ch. 5 Serviceability Prof. Dr. Bayan Salim 5

It is seen from extensive studies that a deflection, occurring in a beam after the maximum

moment Ma has reached and exceeds ⅔ of the cracking moment (⅔Mcr), can be

calculated using an effective moment of inertia Ie;

Ie was developed to provide a transition between the upper and lower bounds Ig and Icr as

a function of the ratio Mcr / Ma.

That is:

The effective moment of inertia (≤ Ig) for the calculations of deflections shall be as

follows: (ACI 24.2.3.5)

The modulus of rupture to be taken fr = 0.62 λ√fc

′

λ = 1.0 (NW Concrete), λ = 0.85 (Sand LW Concrete), λ = 0.75 (All LW

Concrete)

Continuous Spans ACI 24.2.3.6 For continuous spans, Ie shall be taken as the average of values obtained from Table

24.2.3.5 for the critical positive and negative moment sections.

For prismatic continuous spans, Ie shall be taken as the value obtained from Table

24.2.3.5 at midspan for simple and continuous spans, and at support for cantilevers.

c. Calculation of Time-Dependent Deflections

Long-term Multipliers Additional long-term deflection ∆t shall be determined by multiplying the immediate

deflection ∆i caused by sustained load considered, by the factor λ∆:

∆t = λ∆ ∆i

λ∆ = ξ / (1 + 50 ρ') Where

ρ' = ratio of compression reinforcement at midspan for simple and continuous spans

and at support of cantilevers.

ξ = time-dependent factor for sustained loads, (Fig. R24.2.4.1):

RC II Ch. 5 Serviceability Prof. Dr. Bayan Salim 6

d. Permissible Deflections Deflection computed shall not exceed limits stipulated in Table 24.2.2:

RC II Ch. 5 Serviceability Prof. Dr. Bayan Salim 7

Example 4: The beam shown below is designed to carry calculated dead load of 24 kN/m and a

service live load of 48 kN/m. Of the total live load, 20% is sustained. The moment

diagram due to full dead and live load is shown in Fig. (c). The beam will support

nonstructural partitions that would be damaged if large deflection were to occur.

f'c = 28 MPa and fy = 420 MPa

Calculate that part of the 5-year total deflection that would adversely affect the partitions.

c.c. span = 7.9 m, and clear span = 7.6 m.

Solution:

Ec = 4700√28 = 24,870 MPa, Es = 200,000 MPa, n = Es / Ec = 8

fr = 0.62√28 = 3.28 MPa.

From Fig.(b), ‾y = 193 mm, and Ig = 1.347×1010 mm4 (check this!)

From Fig.(d), find the neutral axis location kd below the top of slab, and Icr .

(Ans.: kd = 95 mm and Icr = 4.539×109 mm4) (Check them!)

RC II Ch. 5 Serviceability Prof. Dr. Bayan Salim 8

In the +ve M region,

Mcr = fr Ig / yt = 3.28×1.347×1010 / 427] ×10–6 = 104 kNm

Ma = 218 kNm > ⅔ Mcr = ⅔ (104) = 69.33 kNm

From Table 24.2.3.5, Ie =

= 4.539×109 / [1 – (69.33 / 218)2 (1 – 4.539/13.47)] = 4.865×109 mm4

(This Ie = 4.865×109 mm4 -midspan value- is permitted by ACI Code)

λ∆ = ξ / (1 + 50 ρ') = 2.00 (5 years with ρ' = 0)

Full load = D + L = 24 + 48 = 72 kN/m

∆D+L = 5 l2 [Mm + 0.1(MA + MB)]/ (48EI)

= 5(7600)2 [218 – 0.1(302 + 302)] 106 / (48×24870×4.865×109) = 7.8 mm

This ∆D+L = 7.8 mm will be considered the reference value; using it as a basis:

Time-dependent portion of D deflection:

∆D = 7.8 (24/72) ×2.00 = 5.2 mm

Sum of immediate and time-dependent portion (sustained) of L deflection:

∆0.2L = 7.8 (48/72)×0.2×(1 + 2.00) = 3.1 mm (0.2 means 20% of L is sustained)

Immediate deflection due to short-term L:

∆0.8L = 7.8 (48/72) ×0.80 = 4.2 mm (0.8 means 80% of L is not-sustained)

Total deflection that would adverse the partitions:

∆total = 5.2 + 3.1 + 4.2 = 12.5 mm

For comparison with Table 24.2.2, the limitation imposed by ACI Code is

l / 480 = 7900 / 480 = 16.5 mm > 12.5 mm OK.

HW: For Example 4, if 2 – No.25 is provided in the compression zone, what would be

the 3-year and 5-year deflections? Ans. 10.4 mm, 11.7 mm.

RC II Ch. 5 Serviceability Prof. Dr. Bayan Salim 9

Problems in Serviceability 1. A rectangular beam of width b = 375 mm effective depth d = 515 mm, and total depth

h = 575 mm spans 5.6 m between simple supports. It will carry a computed dead load of

16 kN/m including self-weight, plus a service live load of 33 kN/m. Reinforcement

consists of four evenly spaced (No. 22) bars in one row. The clear cover on the sides is 50

mm. Material strengths are fy = 420 MPa and fc’ = 28 MPa.

(a) Compute the stress in the steel at full service load.

(b) Confirm the suitability of the proposed design based on ACI Eq. for crack control.

2. An alternative design is proposed for the beam in Problem 1, using two (No. 29) Grade

520 bars to provide approximately the same steel strength as the originally proposed four

(No. 22) Grade 420 bars. Check to determine if the redesigned beam is satisfactory with

respect to cracking according to the ACI Code. What modification could you suggest that

would minimize the no. of bars to reduce cost, yet satisfy crack control requirements?

3. For the beam in Problem 1:

(a) Calculate the total deflection affecting partitions.

(b) Compare your results with the limitations imposed by the ACI Code, Table 24.2.2.

Assume that the beam supports partitions susceptible to cracking if deflections are large.

4. A beam having b = 300 mm, d = 540 mm, and h = 600 mm is reinforced with 3 (No.

36) bars. Material strengths are fy = 420 MPa and f’c = 28 MPa. It is used on a 8.5 m

simple span to carry a total service load of 35 kN/m. For this member, the sustained loads

include self-weight of the beam plus additional superimposed dead load of 7.5 kN/m,

plus 6 kN/m representing that part of the live load that acts more or less continuously.

The remaining 18 kN/m live load consists of short-duration loads,

(a) Find the increment of deflection under sustained loads due to creep.

(b) Find the additional deflection increment due to the intermittent part of the live load.

Compare with ACI Code limits from Table 24.2.2. Assume that construction details are

provided that will avoid damage to supported elements due to deflections. If ACI Code

limitations are not met, what changes would you recommend to improve the design?

5. A reinforced concrete beam is continuous over two equal 6.7 m spans, simply

supported at the two exterior supports, and fully continuous at the interior support.

Concrete dimensions are b = 250 mm, h = 550 mm and d = 490 m for both positive and

negative bending regions. Positive reinforcement in each span consists of two (No. 29)

bars, and negative reinforcement at the interior support is made up of three (No. 32) bars.

No compression steel is used. fy = 420 MPa and f’c = 28 MPa. The beam will carry a

service live load of 26 kN/m; 20 percent of this load will be sustained permanently, while

the rest is intermittent. The total service dead load is 15 kN/m including self-weight.

(a) Find the immediate deflection when the full dead load is applied.

(b) Find the long-term deflection under sustained load.

(c) Find the increment of deflection when the short-term part of the live load is applied.

Compare with ACI Code deflection limits; nonstructural elements are carried that would

be damaged by large deflections. Note that midspan deflection may be used as a close

approximation of maximum deflection.