Strength Analysis of Beams According to ACI Code

3.1 DESIGN METHODS

From the early 1900s until the early 1960s, nearly all reinforced concrete design in the United States was performed by the working-stress design method (also called allowable-stress design or straight-line design). In this method, frequently referred to as WSD, the dead and live loads to be supported, called working loads or service loads, were first estimated. Then the members of the structure were proportioned so that stresses calculated by a transformed area did not exceed certain permissible or allowable values.

After 1963 the ultimate-strength design method rapidly gained popularity because (1) it makes use of a more rational approach than does WSD; (2) a more realistic consideration of safety is used; and (3) it provides more economical designs. With this method (now called strength design) the working dead and live loads are multiplied by certain load factors (equivalent to safety factors) and the resulting values are called factored loads. The members are then selected so they will theoretically

just fail under the factored loads.

3.2 ADVANTAGES OF STRENGTH DESIGN

Among the several advantages of the strength design method as compared to the no longer permitted working-stress design method are the following:

The derivation of the strength design expressions takes into account the non- linear shape of the stress-strain diagram. When the resulting equations are applied, decidedly better estimates of load-carrying ability are obtained.

With strength design, a more consistent theory is used throughout the designs of reinforced concrete structures. For instance, with working-stress design the transformed-area or straight-line method was used for beam design, and a strength design procedure was used for columns.

A more realistic factor of safety is used in strength design. The designer can certainly estimate the magnitudes of the dead loads that a structure will have to support more accurately than estimating the live and environmental loads. With working-stress design the same safety factor was used for dead, live, and environmental loads. For this reason, use of different load or safety factors in strength design for the different types of loads is a definite improvement.

A structure designed by the strength method will have a more uniform safety factor against collapse throughout. The strength method takes considerable advantage of higher-strength steels, whereas working-stress design did only partly so. The result is better economy for strength design.

The strength method permits more flexible designs than did the WSM. For instance, the percentage of steel may be varied quite a bit. As a result, large sections may be used with small percentages of steel or small sections maybe used with large percentages of steel. Such variations were not the case in the relatively fixed working-stress method. If the same amount of steel is used in strength design for a particular beam as would have been used with WSD, a smaller section will result. If the same size section is used as required by WSD, a smaller amount of steel will be required.

3.3 STRUCTURAL SAFETY

The structural safety of a reinforced concrete structure can be calculated with two methods. The first method involves calculations of the stresses caused by the working or service loads and their comparison with certain allowable stresses. Usually the safety factor against collapse when the working-stress method was used was said to equal the smaller of f^' c/fc or fy/fs•

The second approach to structural safety is the one used in strength design in which uncertainty is considered. The working loads are multiplied by certain load factors that are larger than one. The resulting larger or factored loads are used for designing the structure. The values of the load factors vary depending on the type and combination of the loads. To accurately estimate the ultimate strength of a structure, it is necessary to take into account the uncertainties in material strengths, dimensions, and workmanship. This is done by multiplying the theoretical ultimate strength (called the nominal strength herein) of each member by the strength reduction factor which is less than one. These values generally vary from 0.90 for bending down to 0.65 for some columns.

In summary, the strength design approach to safety is to select a member who's computed ultimate load capacity multiplied by its strength reduction factor will at least equal the sum of the service loads multiplied by their respective load factors. Member capacities obtained with the strength method are appreciably more accurate than member capacities predicted with the working-stress method.

3.4 DERIVATION OF BEAM EXPRESSIONS

Tests of reinforced concrete beams confirm that strains vary in proportion to distances from the neutral axis even on the tension sides and even near ultimate loads. Compression stresses vary approximately in a straight line until the maximum stress equals about0.50f'c. This is not the case, however, after stresses go higher. When the ultimate load is reached, the strain and stress variations are approximately as shown in Figure 3.1.

The compressive stresses vary from zero at the neutral axis to a maximum value at or near the extreme fiber. The actual stress variation and the actual location of the neutral axis vary somewhat from beam to beam depending on such variables as the magnitude and history of past loadings, shrinkage and creep of the concrete, size and spacing of tension cracks, speed of loading, and so on.

If the shape of the stress diagram were the same for every beam, it would easily be possible to derive a single rational set of expressions for flexural behavior. Because of these stress variations, however, it is necessary to base the strength design on a combination of theory and test results.

Although the actual stress distribution given in Figure 3.2(b) may seem to be important, any assumed shape (rectangular, parabolic, trapezoidal, etc.) can be used practically if the resulting equations compare favorably with test results. The most common shapes proposed are the rectangle, parabola, and trapezoid, with the rectangular shape used in this text as shown in Figure 3.2(c) being the most common one.

If the concrete is assumed to crush at a strain of about 0.003 (which is a little conservative for most concretes) and the steel to yield at fy, it is possible to make a reasonable derivation of beam formulas without knowing the exact stress distribution. However, it is necessary to know the value of the total compression force and its centroid.

Whitney replaced the curved stress block with an equivalent rectangular block of intensity 0.85f'c and depth a =β_1 c, as shown in Figure 3.2(c). The area of this rectangular block should equal that of the curved stress block and the centroid of the two blocks should coincide. Sufficient test results are available for concrete beams to provide the depths of the equivalent rectangular stress blocks. The values of f'c given by the Code (10.2.7.3) are intended to give this result. Forf'c between 17 and 28 MPa, β_1 shall be taken as 0.85. For f'c above 28 MPa, β_1 shall be reduced linearly at a rate of 0.05 for each 7 MPa of strength in excess of 28 MPa, but β_1shall not be taken less than 0.65.The values of β_1 are reduced for high-strength concretes primarily because of the shapes of their stress-strain curves.

For concretes with f'c > 28 MPa,

β_1=0.85-0.008 (f^' c-28)≥0.65

Based on these assumptions regarding the stress block, statics equations can easily be written for the sum of the horizontal forces and for the resisting moment produced by the internal couple. These expressions can then be solved separately for α and for the momentMn. Mn is defined as the theoretical or nominal resistingmoment of a section. In Section 3.3 it was stated that the usable strength of a member equals its theoretical strength times the strength reduction factor, or, in this case, ∅Mn. The usable flexural strength of a member, ∅Mnmust at least be equal to the calculated factored moment, Mu caused by the factored loads ∅Mn≥Mu.

For writing the beam expressions, reference is made to Figure 3.3. Equating the horizontal forces C and T and solving for a, we obtain:

C=T

0.85*f^' c*a*b=As*fy

Then a=(As fy)/(0.85*f^' c*b )=ρdfy/(0.85*f^' c) where ρ= As/bd=percentage of tensile steel

Because the reinforcing steel is limited to an amount such that it will yield well before the concrete reaches its ultimate strength, the value of the nominal moment Mn can bewritten as Mn=T (d-a/2)=As*fy(d-a/2)

And the usable flexural strength is: ∅Mn=∅*As*fy(d-a/2)

Substituting into this expression the value previously obtained for a (it was ρdfy/0.85*f^' c ) and equate ∅Mn to Mu we obtain the following expression:

∅Mn=Mu=∅*ρ*b*fy*d^2 (1-0.59 (ρ*fy)/( f'c))

Replacing As with ρbd and letting R = Mu/∅bd^2, we can solve this expression for ρ (the percentage of steel required for a particular beam) with the following results:

ρ=(0.85 f^' c)/fy (1-√(1-2R/(0.85*f^' c )))

3.5 STRAINS IN FLEXURALMEMBERS

As previously mentioned, Section 10.2.2 of the Code states that the strains in concrete members and their reinforcement are to be assumed to vary directly with distances from their neutral axes. Furthermore, in Section 10.2.3 the Code states that the maximum usable strain in the extreme compression fibers of a flexural member is to be 0.003.Section 10.3.3 states that for Grade 420 reinforcement and for all prestressed reinforcement we may set the strain in the steel equal to 0.002 at the balanced condition. (Theoretically, for 420MPa steel it equals fy/Es=420/200000=0.0021). In Section 3.4 a value was derived for a, the depth of the equivalent stress block of a beam. It can be related to c with the factor β_1 also given in that section.

a=(As fy)/(0.85*f^' c*b )=c β_1

Then the distance c from the extreme concrete compression fibers to the neutral axis is: c=a/β_1

Ex. 1: Determine the values of a,c,and ϵ_tfor the beam shown in figure belowfy= 420MP,f^' c =21 MPa and d=530mm.

a=(As fy)/(0.85*f^' c*b )=(3*491*420)/(0.85*20*350)=99.025 mm

c=a/β_1 =99.025/0.85=116.5 mm

ε_t=(d-c)/c 0.003=(0.003 (530-116.5))/116.5=0.0106

3.6 BALANCED SECTIONS, TENSION-CONTROLLED SECTIONS,

AND COMPRESSION-CONTROLLED OR BRITTLE SECTIONS:

A beam that has a balanced steel ratio is one for which the tensile steel will theoretically yield at the time the extreme compression concrete fibers attain a strain equal to0.003. Should a flexural member be so designed that it has a balanced steel ratio or be a member whose compression side controls (that is, if its compression strain reaches 0.003 before the steel yields), the member can suddenly fail without warning. As the load on such a member is increased, its deflections will usually not be particularly noticeable, even though the concrete is highly stressed in compression and failure will probably occur without warning to users of the structure. These members are compression controlled and are referred to as brittle members. Obviously, such members must be avoided.

The Code, in Section 10.3.4, states that members whose computed tensile strains areε_(t )≥0.005 at the same time the concrete strain is ε_(c )=0.003 are to be referred to as tension-controlled sections. For such members the steel will yield before the compression side crushes and deflections will be large, giving users warning of impending failure. Furthermore, members with ε_(t )>0.005are considered to be fully ductile. The ACl-Code chose the 0.005 value for ε_(t ) to apply to all types of steel permitted by the Code, whether regular or prestressed. The Code further states that members that have net steel strains or ε_(t )values between 0.002 and 0.005 are in a transition range between compression-controlled and tension-controlled sections.

3.7 STRENGTH REDUCTION OR ∅ FACTORS

Strength reduction factors are used to take into account the uncertainties of material strengths, inaccuracies in the design equations, approximations in analysis, possible variations in dimensions of the concrete sections and placement of reinforcement, the importance of members in the structures of which they are part, and so on. The Code (9.3) prescribes ∅values orstrength reduction factors for most situations. Among these values are the following:

- 0.90 for tension-controlled beams and slabs

- 0.75 for shear and torsion in beams

- 0.65 or 0.70 for columns

- 0.65 or 0.70 to 0.9 for columns supporting very small axial loads

- 0.65 for bearing on concrete

The sizes of these factors are rather good indications of our knowledge of the subject in question. For instance, calculated nominal moment capacities in reinforced concrete members seem to be quite accurate, whereas computed bearing capacities are more questionable.

For ductile or tension-controlled beams and slabs where〖 ε〗_(t )≥0.005, the value of ∅for bending used is 0.90. Should ε_(t )be less than 0.005 it is still possible to use the sections if ε_(t )is not less than certain values. This situation is shown in Figure R.9.3.2 in the ACI Commentary to the 2011 Code.

Members subject to axial loads equal to or less than P≤0.10f'c Ag may be used when ε_(t )isas low as 0.004 (ACI Section 10.3.5). Should the members be subject to axial loads ≥0.10f'c Agthey may be used when ε_(t ) is as small as 0.002. When ε_(t )values fall between 0.002and 0.005, they are said to be in the transition range between tension-controlled and compression-controlled sections. In this range ∅ values will fall between 0.65 or 0.70 and 0.90as shown in the figure.

The procedure for determining ∅ values in the transition range is described later in this section. It must be clearly understand that the use of flexural members in this range is usually uneconomical, and it is probably better, if the situation permits, to increase member depths and/or decrease steel percentages until ε_(t )is equal or larger than 0.005. If this is done, not only will ∅ values equal 0.9 but also steel percentages will not be so large as to cause crowding of reinforcing bars. The net result will be slightly larger concrete sections, with consequent smaller deflections.

The bottom half of Figure above gives values for c/dt ratios. If c/dt for a particular flexural member is ≤ 0.375, the beam will be ductile, and if > 0.600 it will be brittle. In between is the transition range. You may prefer to compute c/dt for a particular beam to check its ductility rather than computing ρ or ε_t.

Continuing our consideration of Figure, we can see that when ε_t less than 0.005, the values of ∅ will vary along a straight line from their 0.90 value for ductile sections to 0.65 at balanced conditions where ε_t is 0.002. Later, we will learn that ∅ can equal 0.70 rather than 0.65 at this latter strain situation if spirally reinforced sections are being considered. For this range of varying ε_tvalues, the value of ∅ may be determined by proportions or with the following formula, which is also shown in the figure.

∅ = 0.48 + 83〖*ε〗_t

Or ∅=0.65+0.25(d_t/c-5/3) for two layer of tension steel area

3.8 MINIMUM PERCENTAGE OF STEEL

A brief discussion of the modes of failure that occur for various reinforced beams was presented in Section 3.6. Sometimes because of architectural or functional requirements, beam dimensions are selected that are much larger than are required for bending alone. Such members theoretically require very small amounts of reinforcing.

Actually, another mode of failure can occur in very lightly reinforced beams. If the ultimate resisting moment of the section is less than its cracking moment, the section will fail immediately when a crack occurs. This type of failure may occur without warning. To prevent such a possibility, the ACI (10.5.1) specifies a certain minimum amount of reinforcing that must be used at every section of flexural members where tensile reinforcing is required by analysis, whether for positive or negative moments. In the following equations, bw represents the web width of beams.

〖As〗_min=√(f^' c)/(4 fy)*bw*d ≥1.4/fy*bw*d

Section 10.5.3 of the Code states that the preceding minimums do not have to be met if the area of the tensile reinforcing furnished at every section is at least one-third greater than the area required by moment.

ACI Section 10.5.4 states that for slabs and footings of uniform thickness, the minimum area of tensile reinforcing in the direction of the span is that specified in ACI Section 7.12 for shrinkage and temperature steel. When slabs are overloaded in certain areas there is a tendency for those loads to be distributed laterally to other parts of the slab, thus substantially reducing the chances of sudden failure. This explains why a reduction of the minimum reinforcing percentage is permitted in slabs of uniform thickness.

ACI-Code 2011- 10.5.4 — For structural slabs and footings of uniform thickness, As-min in the direction of the span shall be the same as that required by

7.12.2.1. Maximum spacing of this reinforcement shall not exceed three times the thickness, nor 450 mm.

7.12.2.1 — Area of shrinkage and temperature reinforcement shall provide at least the following ratios of reinforcement area to gross concrete area, but not less than 0.0014:

(a) Slabs where Grade 280 or 350 deformed bars are used ..........................0.0020

(b) Slabs where Grade 420 deformed bars or welded wire reinforcement are used .......0.0018

(c) Slabs where reinforcement with yield stress exceeding 420 MPa measured at a yield strainof0.35 percent is used..........................(0.0018*420)/fy

7.12.2.2 — Shrinkage and temperature reinforcement shall be spaced not farther apart than five times the slab thickness, nor farther apart than 450 mm.

3.9 BALANCED STEEL PERCENTAGE

An expression is derived for ρ_b, the percentage of steel required for a balanced design. At ultimate load for such a beam, the concrete will theoretically fail (at a strain of 0.003), and the steel will simultaneously yield (see Figure below).The neutral axis is located by the triangular strain relationships that follow, noting that Es= 200000MPa for the reinforcing bars:

c/d=0.003/(0.003+fy/Es)= 600/(600+fy)

Then c= (600*d)/(600+fy)

And from equating C and T: a=(As fy)/(0.85*f^' c*b )=ρdfy/(0.85*f^' c)=c*β1 and c=a/β1

Two expressions are now available for c, and they are equated to each other and

solved for the percentage of steel. This is the balanced percentage〖 ρ〗_b:

Sub. For c :

c= ρdfy/(0.85*f^' c β_1 )= (600*d)/(600+fy)

Then ρ_b=0.85*β_1*(f^' c)/fy*600/(600+fy) which is the balanced value of steel percent allowed to be used in the sectionsfor under reinforced beams the ACI Code allow to be multiplied by 0.75 before 2002. ACI code 2011 used:

ρ_b=0.85*β_1*(f^' c)/fy*0.003/(0.003+ε_t )

ε_t=0.005 for tension controls and 0.004 for member with axial loads

less than 0.10f^' c Ag

The ACI Code use the maximum percent steel ratio as:

ρ_max=0.85*β_1*(f^' c)/fy*0.003/(0.003+0.004)=(51*β_1*f^' c)/(140 fy)=0.3643 (β_1*f^' c)/( fy)

Ex. 2: Determine the ACI design moment capacity ∅Mnof the beam shown in Figure below, iff’c= 27.5 MPa and fy =420MPa.

Sol. :Checking Steel Percentageρ_ac=As/bd=2580/(350*600)=0.0122

ρ_min=√(f^' c)/(4 fy)=√27.5/(4*420)=0.00312 ≥1.4/fy=1.4/420=0.0033

ρ_(b,max.)=0.85*β_1*(f^' c)/fy*0.003/(0.003+ε_t )

ρ_b=0.85*β_1*(f^' c)/fy*0.003/(0.003+0.005)=0.85*0.85*27.5/420*0.003/0.008=0.0177

ρ_( max.)=0.85*β_1*(f^' c)/fy*0.003/(0.003+0.004)=0.85*0.85*27.5/420*0.003/0.007=0.0203( govern)

ρ_(b,yeild)=0.85*β_1*(f^' c)/fy*600/(600+fy)=0.85*0.85*27.5/420*600/1020=0.0278

〖ρ_min≤ρ〗_ac≤ρ_max

a=(As fy)/(0.85*f^' c*b )=(2580*420)/(0.85*27.5*350)=132.5mm and c=132.5/0.85=155.8mm

ε_t=(600-155.8)/155.8*0.003=0.00855 >0.005 then tension control.∅=0.9

M_n=2580*420*(600-132.5/2)/(10^6 )=578.4kN.m

〖Mu= ∅M〗_n=0.9*578.4=520.6kN.m

Ex.3: A 2.4m span cantilever beam has a rectangular section of b=200mm and d= 390mm with 3 bars of 22mm diameter, carries a uniform dead load including it’s own weight of 12kN/m and a uniform distributed live load of 10.5kN/m. Check the adequacy of the section, using f’c of 28MPa and fy of 280MPa?

W_u=1.2*12+1.6*10.5=31.2kN/m

M_u= (W_u L^2)/2= (31.2*〖(2.4)〗^2)/2=89.86 kN.m

ρ_act= As/bd= (3*π*〖22〗^2)/(4*200*390)=0.0146

ρ_min= √(f'c)/(4*fy)≥1.4/fy , ρ_min= √28/(4*280)≥1.4/280 ,

ρ_min= 0.0047≥0.005then ρ_min= 0.005

ρ_(max.)=(51*β_1)/140*(f^' c)/fy= (51*β_1)/140*28/280=0.0309

then ρ_min≤ρ_act≤ρ_max the beam is under reinforced o.k.

a= (As*fy)/(0.85*f^' c*b)= (1140*280)/(0.85*28*200)=67mm

c=a/β_1 and ε_t=(0.003 (d-c))/c≥0.005

c=67/0.85=78.82mm ε_t=0.003(390-78.82)/78.82=0.0118>0.005 o.k.

Then ∅=0.9

M_u=∅ As*fy(d-a/2)

=0.9*1140*280*(390-67/2)=102.4 kN.m>89.86kN.m acceptable

Ex.4: Determine the allowable moment and the position of neutral axis of a rectangular section which has b = 200mm, d =300mm use f’c= 21MPa and fy= 280MPa, when:

Reinforced with 3ø 20 mm

Reinforced with 3ø 28mm

Solution:

ρ_act= As/bd= (3*π*〖20〗^2)/(4*200*300)=0.0157

ρ_min= √(f'c)/(4*fy)≥1.4/fy, ρ_min= √21/(4*280)≥1.4/280 ,

ρ_min= 0.00409≥0.005 then ρ_min= 0.005

ρ_(max.)=(51*β_1)/140*(f^' c)/fy=(51*0.85)/140*21/280=0.023

then ρ_min≤ρ_act≤ρ_max then beam is under reinforced o.k.

a= (As*fy)/(0.85*f^' c*b)= (942.5*280)/(0.85*21*200)=74mm

c=a/β=74/0.85=87.06mm

c=74/0.85=87.06mm ε_t=0.003(300-87.06)/87.06=0.00734>0.005 o.k.

Then ∅=0.9

Mu=∅As fy (d-a/2) or=∅ρbd^2 fy(1-0.59ρ fy/(f^' c))

Mu=0.9*942.5*280(300-74/2)/(10^6 )=62.46 kN.m

Mu=0.9*0.0157*200*300^2*280(1-0.59*0.0157 280/21)/〖10〗^6 =62.46kN.m

ρ_act= As/bd= (3*π*〖28〗^2)/(4*200*300)=0.0308

ρ_min= √(f'c)/(4*fy)≥1.4/fy, ρ_min= √21/(4*280)≥1.4/280 ,

ρ_min= 0.00409≥0.005then ρ_min= 0.005 and ρ_(max.)=0.023

then ρ_min≤ρ_act≥ρ_maxThen beam is over reinforced and fs are less than fy, then use ρ_max

〖As〗_max=0.023*200*300=1380 〖mm〗^2

a= (As*fy)/(0.85*f^' c*b)= (1380*280)/(0.85*21*200)=108.24 mm

c=a/β=108.24/0.85=127.34 mm〖 ε〗_t=0.003(300-127.34)/127.34=0.00407<0.005 o.k.

Then ∅=0.48+83*0.00407=0.82

or ∅=0.65+0.25(d_t/c-5/3)=0.65+0.25(300/127.34-5/3)=0.822

where d_t=

Mu=∅ρ_max bd^2 fy(1-0.59ρ_max fy/(f^' c))

Mu=0.82*0.023*200*300^2*280(1-0.59*0.023 280/21)/〖10〗^6 =68.72 kN.m

3.10: Design of Rectangular Beams and One-Way Slabs:

3.10.1: Load Factors

Load factors are numbers, almost always larger than 1.0, that are used to increase the estimated loads applied to structures. The loads are increased to attempt to account for the uncertainties involved in estimating their magnitudes. The load factors for dead loads are much smaller than the ones used for live and environmental loads. In this regard, you will notice that the magnitudes of loads that remain in place for long periods of time are much less variable than are those loads applied for brief periods, such as wind and snow. Section 9.2 of the ACI-Code presents the load factors and combinations that are to be used for reinforced concrete design. The required strength, U, or the load-carrying ability of a particular reinforced concrete member, must at least equal the largest value obtained by substituting into ACI Equations 9-1 to 9-7.

U=1.2*D+1.6*L

Where U = the design or ultimate load the structure needs to be able to resist

D = dead load, L = live load.

3.10.2 Design of Rectangular Beams

Before the design of an actual beam is attempted, several miscellaneous topics need to be discussed. These include the following:

Beam proportions. The most economical beam sections are usually obtained for shorter beams (up to 6.0m or 7.6 m in length), when the ratio of d to b is in the range of 1.5 to 2. For longer spans, better economy is usually obtained if deep, narrow sections are used. The depths may be as large as 3*b or 4*b. However, today’s reinforced concrete designer is often confronted with the need to keep members rather shallow to reduce ﬂoor heights. As a result, wider and shallower beams are used more frequently than in the past.

Deﬂections. The ACI Code in its Table 9.5(a) provides minimum thicknesses of beams and one-way slabs for which such deﬂection calculations are not required. The minimum thicknesses provided apply only to members that are not supporting or attached to partitions or other construction likely to be damaged by large deﬂection.

Estimated beam weight. The weight of the beam to be selected must be included in the calculation of the bending moment to be resisted, because the beam must support itself as well as the external loads. For instance, calculate the moment due to the external loads only, select a beam size, and calculate its weight. Another practical method for estimating beam sizes is to assume a minimum overall depth, h, equal to the minimum depth speciﬁed by [ACI-318-11, Table 9.5(a)] if deﬂections are not to be calculated. Then the beam width can be roughly estimated equal to about one-half of the assumed value of hand the weight of this estimated beam calculated = bh*24 times the concrete weight per cubic meter. After M is determined for all of the loads, including the estimated beam weight, the section is selected. If the dimensions of this section are signiﬁcant different from those initially assumed, it will be necessary to recalculate the weight and Mu and repeat the beam selection.

Selection of bars. Select an appropriate reinforcement ratio between ρ_min and〖 ρ〗_max . Often a ratio of about 0.60ρ_max, will be an economical and practical choice. Selection of ρ≤ρ_(0.005) ensures that ∅ will remain equal to 0.90. For ρ_(0.005)<ρ <ρ_max an iterative solution will be necessary.

After the required reinforcing area is calculated, select diameter of and numbers of bar that provide the necessary area. For the usual situations, bars of sizes ∅36 and smaller are practical. It is usually convenient to use bars of one size only in a beam, although occasionally two sizes will be used. Bars for compression steel and stirrups are usually a different size.

Cover. The reinforcing for concrete members must be protected from the surrounding environment; that are, ﬁre and corrosion protection need to be provided. To do this, the reinforcing is located at certain minimum distances from the surface of the concrete so that a protective layer of concrete, called cover is provided. In addition, the cover improves the bond between the concrete and the steel. In Section 7.7 of the ACI Code, speciﬁed cover is given for reinforcing bars under different conditions. Values are given for reinforced concrete beams, columns, and slabs; for cast-in-place members; for precast members; for pre-stressed members; for members exposed to earth and weather.

7.7 - Concrete protection for reinforcement

7.7.1 - Cast-in-place concrete (non-prestress) ,cover for reinforcement shall not be less

than the following:

No. Position Cover, mm

a Concrete cast against and permanently exposed to earth 75

b Concrete exposed to earth or weather:

bars with diameter 19mm through 57mm

Bars with diameter 16 bar, MW200 or MD200 wire, and smaller

c Concrete not exposed to weather or in contact with ground:

Slabs, walls, joists:

bars with dia. 43mm and 57mm

bars with dia. 36 mm and smaller

Beams, columns: Primary reinforcement, ties, stirrups, spirals

Shells, folded plate members:

bars with dia. 19 mm and larger

Bars with dia. 16mm, MW200 or Pl D200 wire, and smaller

Minimum spacing of bars. The code (7.6) states that the clear distance between parallel bars cannot be less than 25mm or less than the nominal bar diameter. If the bars are placed in more than one layer, those in the upper layers are required to be placed directly over the ones in the lower layers, and the clear distance between the layers must be not less than 25mm.

7.6 — Spacing limits for reinforcement

7.6.1 — The minimum clear spacing between parallel bars in a layer shall be d_bar but not less than 25 mm.

7.6.2 — Where parallel reinforcement is placed in two or more layers, bars in the upper layers shall be placed directly above bars in the bottom layer with clear distance between layers not less than 25 mm.

7.6.5 — In walls and slabs other than concrete joist construction, primary flexural reinforcement shall not be spaced farther apart than (3h) three times the wall or slab thickness, nor farther apart than 450 mm.

A major purpose of these requirements is to enable the concrete to pass between the bars. The ACI Code further relates the spacing of the bars to the maximum aggregate sizes for the same purpose. In the code Section 3.3.2, maximum permissible aggregate sizes are limited to the smallest of (a) 1/5 of the narrowest distance between side forms, (b) 1/3 of slab depths, and (c) 3/4 of the minimum clear spacing between bars.

Ex.5: Design the simply supported rectangular beam with span of 4m support service dead load of 10kN/m and service live load of 30kN/m f’c= 21MPa and fy = 400MPa.

h_min=L/16=4000/16=250mm*(0.4+400/700)=242.8mm

Try section of b = 250mm and h = 500mm

W_d=0.25*0.5*24=3 kNlm

Total dead load =10+3=13 kN/m

W_u=1.2*13+1.6*30=63.6 kN/m

M_u=〖W_u L〗^2/8=(63.6*4^2)/8=127.2kN.m

ρ_min= √(f'c)/(4*fy)≥1.4/fy, ρ_min= √21/(4*400)≥1.4/400 ,

ρ_min= 0.00286≥0.0035 then ρ_min= 0.0035

Select an appropriate reinforcement ratio between ρ_min and ρ_max . Often a ratio of about 0.60ρ_max, will be an economical and practical choice. Selection of ρ≤ρ_(0.005) ensures that ∅ will remain equal to 0.90. For ρ_0.005<ρ <ρ_max an iterative solution will be necessary.

ρ_(max.)=(51*β_1)/140*(f^' c)/fy= (51*0.85)/140*21/400=0.0163

Choose percent of steel in between of ρ_min= 0.0035 and ρ_max= 0.0142

Try ρ_assumed=0.012 then:

Cross section will be bd^2=Mu/(∅ρfy(1-0.59ρ fy/f'c)) assuming ∅=0.9

bd^2=(127.2*〖10〗^6)/(0.9*0.012*400(1-0.59*0.012 400/21) )=34.0432*10^6 〖mm〗^3

b/d d (mm) b (mm)

0.484 412.57 200

0.677 369.02 250(govern)

0.89 336.86 300

d=√((34.0432*10^6)/b) then

Use b= 250mm and d= 380 mm then h = 380 +70 = 450>242.8mm o.k.

Check weight of beam:

W_d=0.25*0.45*24=2.7 kNlm<3 kN/m o.k.

Then As can be find from one of the following methods:

First method :

Mu=∅ρbd^2 fy(1-0.59ρ fy/(f^' c))

127.2*10^6=0.9*ρ*250*〖380〗^2*400(1-0.59ρ 400/21)

0.0097876=ρ*(1-11.2381*ρ) then 11.2381*ρ^2-ρ+0.0097876=0

ρ^2-0.08898*ρ+0.000871=0 and ρ=(0.08898∓√(0.08898^2-4*0.000871))/2=0.01119

As=0.01119*380*250= 1063.05mm2

Second method:

Mu=∅As*fy(d-a/2) and a=(As fy )/(0.85 f^' cb)

Try a≤d/2=100mm

then find As=Mu/(∅ fy ( d-a/2))=(127.2*10^6)/(0.9*400*(380-100/2))=1070.70mm^2

And a=(As fy )/(0.85 f^' cb)=(1070.70*400)/(0.85*21*250)=95.97 mm

Repeat As=(127.2*10^6)/(0.9*400*(380-(95.97 )/2))=1064.21 mm^2

And a=(1064.21*400)/(0.85*21*250)=95.39 mm o.k.≈95.97 mm ∴As=1064.21mm^2

Third method:

Ru=Mu/(∅bd^2 )=(127.2*〖10〗^6)/(0.9*250*380^2 )=3.915 μ=fy/0.85f'c= 400/(0.85*21)=22.4

ρ=1/μ [1-√(1-(2*Ru*μ)/fy)]=1/22.4 [1-√(1-(2*3.915*22.4)/400)]=0.01119

Or ρ=(0.85*f'c)/fy [1-√(1-(2*R)/(0.85*f'c))]=(0.85*21)/400 [1-√(1-(2*3.915)/(0.85*21))]=0.01119

As=0.01119*380*250= 1063.05mm2

Assume using bars with 22mm diameter then number of bars required:

A_bar=(π*〖22〗^2)/4=387mm^2 and n= 1063.05/387= 2.75 use 3 bars ∅22mm

As=3*387=1161 mm^2

clear spacing between bars will be:

(250-2*40-2*10-3*22)/2=42>25mm o.k use one layer.

The cross section required

Checking sol.: As=3*387=1161 mm^2

ρ_ac=1161/(250*380)=0.0122>ρ_min and ρ_(max.)= (51*0.85)/140*21/400=0.0163

Then the section is ductile under reinforced beam.

a=(As fy)/(0.85*f^' c*b )=(1161*400)/(0.85*21*250)=104.07 mm and c=104.07/0.85=122.4 mm

ε_t=(380-122.4)/122.4*0.003=0.0063>0.005 then tension control.∅=0.9

M_u=0.9*1161*400*(380-104.07/2)/(10^6 )=137.07 kN.m>127.2kN.m o.k.

Ex.6: Simply supported beam with 6m span, support a service dead load of 15 kN/m and point live load at mid span 50kN, Design the beam using f’c = 30 MPa and fy = 420 MPa ,find the depth if the width used b = 250mm and using 0.8 of ρ_max try using bars with 25mm diameter?

h_min= L/16= 6000/16=375mm try 400mm, Assume b= 250mm

Then Wd= 0.25* 0. 4* 24 = 2.4 kN/m try 3 kN/m

Total load load = 15 + 3 = 18 kN/m

Mu=(1.2*18*6^2)/8+ (1.6*50*6)/4=97.2+120=217.2 kN.m

f^' c>28MPa then β_1=0.85-0.008 (f^' c-28)≥0.65

β_1=0.85-0.008 (30-28)=0.834 ≥0.65

ρ_(max.)=(51*β_1)/140*(f^' c)/fy= (51*0.834)/140*30/420=0.0217

For use 〖0.8*ρ〗_(,max.) =0.8*0.0217=0.01736 , assume .∅=0.9

Mu=∅*ρ*b*d^2*fy(1-0.59ρ fy/(f^' c))

d=√(Mu/(∅ρbfy(1-0.59ρ fy/f'c)))=√((217.2*〖10〗^6)/(0.9*250*0.01736*420(1-0.59*0.01736* 420/30)))=393.14 mm

try d= 400mm

then b= 250mm ,d=400mm and h= 400 +80= 480mm>h_min

Wd= 24*0.25 *0.48 = 2.88 < 3 kN/m o.k.

Ru=Mu/(∅bd^2 )=(217.2*〖10〗^6)/(0.9*250*400^2 )=6.03 μ=fy/0.85f'c= 420/(0.85*30)=16.47

ρ=1/μ [1-√(1-(2*Ru*μ)/fy)]=1/16.47 [1-√(1-(2*6.03*16.47)/420)]=0.0166

Or ρ=(0.85*f'c)/fy [1-√(1-(2*R)/(0.85*f'c))]=(0.85*30)/420 [1-√(1-(2*6.03)/(0.85*30))]=0.0166

As=0.0166*400*250= 1660mm2

∅=25mm Abar=491mm^2 then No.of bars= 1660/491=3.3 use 4∅25mm

As= 4 * 491= 1964 mm2

Check solution: a=(As fy)/(0.85*f^' c*b )=(1964 *420)/(0.85*30*250)=129.39 mm

and c=129.39/0.834=155.14 mm

ε_t=(420-155.14 )/(155.14 )*0.003=0.00512>0.005 then tension control.∅=0.9

M_u=0.9*1964*420*(420-129.39/2)/(10^6 )=263.77 kN.m>217.2 kN.m o.k.

Check steel spacing :

(250-70-70-3*25)/3=11.67mm<25mm

Then two layers :

(250-70-70-2*25)/2=30mm>25mmo.k.then total depth will be 515 say 520mm

Ex.7: Find the minimum dimension of the cross section for the beam shown with b = 300mm, Use f’c = 25MPa and fy = 400 MPa, then find area of steel for the whole beam, bars with 25 mm?

hmin=L/18.5 (0.4+fy/700)=7000/18.5 (0.4 +400/700)=367.6mm

W_u=1.2*30+1.6*18=64.8 kN/m

ρ_min= √(f'c)/(4*fy)≥1.4/fy , ρ_min= √25/(4*400)≥1.4/400

ρ_min= 0.00315≥0.0035 then ρ_min= 0.0035

ρ_( max)=51/140*(β_1*f^' c)/fy=(51*0.85*25)/(140*400)=0.0194

To keep ∅=0.9

section use ρ_( max) for minimum dimension

d=√(Mu/(∅*ρ*b*fy*(1-0.59ρ*fy/(f^' c))))

d=√((302.19*〖10〗^6)/(0.9*0.0194*300*400*(1-0.59*0.0194*400/25)))=420.5 mm say 430mm

h=d+70=430+70≈500mm >h_min

Ru=Mu/(∅bd^2 )=(302.19*〖10〗^6)/(0.9*300*430^2 )=6.05

Or ρ=(0.85*f'c)/fy [1-√(1-(2*R)/(0.85*f'c))]=(0.85*25)/400 [1-√(1-(2*6.05)/(0.85*25))]=0.01826

〖As〗_max=0.01826*300*430=2355.54 〖mm〗^2 use ∅25mm then

A_bar=491 mm^2 no.of bar req.= 2355.54/491=4.8 Use 5∅25

check one or two layers of steel at tension zone.

(300-2*40-2*10-5*25)/5=15<25mm then two layers

(300-2*40-2*10-3*25)/2=62.5>25mm

Cover 40 + 10 + 25 + 12.5 = 87.5 say 90mm

Then the depth will be corrected to h= 430 + 90 = 520mm

Check solution: a=(As fy)/(0.85*f^' c*b )=(5*491 *400)/(0.85*25*300)=154.04 mm

and c=154.04/0.85=181.2 mm

ε_t=(430-181.2 )/181.2*0.003=0.00412<0.005

∅=0.65+0.25(d_t/c-5/3)=0.65+0.25((430+25+12.5)/181.2-5/3)=0.88

Mu=0.88*5*491*400(430-154.04/2)/(10^6 )=305.03 kN.m>302.29kN.m

For negative moment = 202.5kN.m, use the same section b=300& h = 520mm

d= 430mm try ∅=0.9

Mu=∅*ρ*b*d^2*fy*(1-0.59*ρ*fy/(f^' c))

Ru=Mu/(∅bd^2 )=(202.5*〖10〗^6)/(0.9*300*430^2 )=4.056 μ=fy/0.85f'c= 400/(0.85*25)=18.824

ρ=1/μ[1-√(1-(2*Ru*μ)/fy)] =1/18.824 [1-√(1-(2*4.056*18.824)/400)]=0.01135

As = 0.01135*300*430 = 1464.15 mm2

using bars with 25mm:

A_bar=491mm^2 no.of bar req.= 1464.15/491=2.98 Use 3∅25=1473mm^2

one layer of steel at tension zone has been checked

a=(As fy)/(0.85*f^' c*b )=(1473 *400)/(0.85*25*300)=92.4 mm and c=92.4/0.85=108.7 mm

ε_t=(430-108.7 )/(108.7 )*0.003=0.00886>0.005 then tension control.∅=0.9

Tension steel reinforcement

Breaking

Breaking News

Sunday, 17 November 2019

Strength Analysis of Beams According to ACI Code