CHAPTER 1 Introduction
Concrete Concrete is a mixture of water, cement, sand, gravel crushed rock, or other aggregates. The aggregates (sand, gravel, crushed rock) are held together in a rocklike mass with a paste of cement and water.
RE INFO INFORCE RCE D CONCRETE As with most rocklike mass, concrete has very high compressive strength but have a very low tensile strength. As a structural member, concrete can be made to carry tensile stresses (as in beam in flexure). In this regard, it is necessary to provide steel bars to provide the tensile strength lacking in concrete. The composite member is called reinforced concrete .
A G G R E G A TE S Aggregates used in concrete may be fine aggregates (usually sand) and coarse aggregates (usually gravel or crushed stone). Fine aggregates are those that passes through a No. 4 sieve (about 6 mm in size). Materials retained are coarse aggregates. The nominal maximum sizes of coarse aggregate are specified in Section 5.3.3 of NSCP. These are follows: 1/5 the narrowest dimension between sides of forms, 1/3 the depth of slabs, or 3/4 the minimum clear spacing between individual reinforcing bars or wires, bundles of bars, or prestressing tendons or ducts. These limitations may not be applied if, in the judgment the Engineer, workability and methods of consolidation are such that concrete can be placed without honeycomb or voids.
WATER According to Section 5.3.4, water used in mixing concrete shall be clean and free from injurious of oils, acids, alkalis, salts organic materials or other substances that may be deleterious to concrete or reinforcement. Mixing water for prestressed concrete or for concrete that will contain aluminum embedment’s, including that portion of mixing water contributed in the form of free moisture on aggregates, shall not be used in concrete unless the following are satisfied: ( a) Selection of concrete proportions shall be based on concrete mixes using water from the same source and ( b ) mortar test cubes made with non-portable mixing water shall have 7-days and 28 day strengths equal to at least 90
Unlike steel and other materials, concrete has no definite modulus of elasticity. Its value is dependent on the characteristics of cement and aggregates used, age of concrete and strengths. According to NSCP (Section 5.8.5), modulus of elasticity E c for concrete for values of w c,c, between 1500 and 2500 kg/ m 3 may be taken as.
Eq. 1-1
′ /
= .0.043
=4700′.
Where is the day 28-day compressive strength of concrete in MPa is the unit weight on concrete on concrete in . For normal weight concrete, Modulus of elasticity E s Modulus for nonprestressed reinforced may be taken as 200,000 MPa.
DETAILS OF REINFORCEMENT STANDARD HOOKS
4 12
Standard hooks refer to one of the following: 1. 180-degree bend plus extension but not less than 60 mm at free end of bar. 2. 90-degree bed plus extension at free end of bar. 3. For stirrups and tie tie hooks: a) 61 mm diameter diameter bar and smaller, 90-degree bend plus extension at free end bar, or b) 20 and 25 mm diameter diameter bar, 90-degree bend, plus extension at free end of bar, or c) 25mm diameter diameter bar and and smaller, 135-degree bend d plus extension at free end of bar.
6 12 6
MIN IMUM IMU M BE B E ND D IA ME TE R S (S E C TION TI ON 407.3) 407. 3) Diameter of bend measured on the inside of the bar, other than for stirrups and ties in sizes 10mm through 15 mm, shall not be less than the values in Table 1.1.
4
Inside diameter of bend for stirrups and ties shall not be less than 16 mm bar and smaller. For bars larger than 16 mm, diameter of bend shall be in accordance with Table 1.1
4
2 4
Inside diameter of bend in welded wire fabric /9plain or deformed) for stirrups and ties shall not be less than for deformed wire larger than D56 and for all other wires. Bends with inside diameter of less than 8db shall not be less than from nearest welded intersection.
Table 1.1 - Minimum Diameters of Bend Bar Size
Minimum Diameter 10 mm to25 mm 28 mm, 32 mm, and 36 mm
68
PLA IN R E INFORC E ME NT (407 (407.6 .6)) Reinforcement, prestressing tendons, and ducts shall not be accurately placed and adequately before concrete is placed, and shall be secured against displacement within tolerance permitted. Unless otherwise specified by the Engineer, reinforcement prestressing tendons, and prestressing ducts shall be placed within the following tolerances: Tolerance for depth d, and minimum concrete over a flexural members walls and compression members shall be as follows:
E ffective ffectiv e depth, depth, d
≤ 200200 > 200200
d d
Tolerance on d
± 10 ± 12
Tolerance on minimum mini mum concrete conc rete cover -10 mm -12 mm
Except that tolerance for the clear distance to formed soffits shall be minus 6 mm and tolerance for cover shall not exceed minus 1/3 the minimum concrete cover required in the design drawings or specifications.
± ±12
Tolerance for longitudinal location of bends and ends of reinforcement shall be mm except at discontinuous ends of members where tolerance shall be mm.
50
S PAC ING LIMITS FOR R E IN FOR C E ME NT According for Section 5.7.6 of NSCP, the minimum clear spacing between parallel bars in a layer should be d b but not less than 25 mm. Where parallel reinforcement is placed in two or more layers, bars in the upper layers should be placed directly above bars in the bottom layer with clear distance between layers not less than 25mm. In spirally reinforced or tied reinforced compression members, clear distance between longitudinal bars shall be not less than 1.5 d b nor 40mm. In walls and slabs other than concrete joist construction, primary flexural reinforced shall be spaced not for farther apart than three times the wall or slab thickness, nor 450 mm.
BUNDLED BARS Groups of parallel reinforcing bars bundled in contact to act as unit shall be limited to four in any one bundle. Bundled bars shall be enclosed within stirrups or ties and bars larger than 32 mm shall not be bundle in beams. The individual bars within the span of flexural members should terminate at different points with at least 40 d b stagger. Since spacing limitations and minimum concrete cover of most members are based on a single diameter d b, bundled bars shall be treated as a single bar of a diameter derived from the equivalent total area.
Fig ure 1.1 Bundled-bar arrangement
Diameter of single bar equivalent to bundled bars according to NSCP to be used for spacing limitation and concrete cover.
=
( 25) 2 x 3
3-25mm
Equivalent diameter, D
D 2
CONCRETE PROTECTION FOR REINFORCEMENT (SECTION 407.8.1)
Cast-in – place C oncrete (nonpres tres s ed). The following minimum concrete cover shall be provided for reinforcement: Minimum cover, mm (a)
Concrete cast against permanently exposed to earth
75
(b)
Concrete exposed to earth or weather: 20 mm through 36 mm bars 16 mm bar, W31 or D31 wire, and smaller
50 40
(C) Concrete not exposed to weather or in contact with ground: slabs, walls, joists: 32 mm bar and smaller Beams, columns Primary reinforcement, ties, stirrups, spirals Shells, folded place members: 20 mm bar and larger 16 mm, Wr1 or D31 wire, and smaller
20 40 20 15
Precast concrete (Manufactured Under Plant Conditions). The Following minimum concrete shall be provided for reinforcement
Minimum cover, mm (a)
(b)
Concrete exposed to earth or weather: Wall panels: 20 32 mm bar and smaller Other members: 40 20 mm through 32 mm bars 30 16 mm bar, W31 wire, and smaller Concrete not exposed to weather or in contact with ground: 15 slabs, walls, joists: d b but not less 15, & 32 mm bar and smaller need not exceed 40 Beams, columns 10 Primary reinforcement Ties, stirrups, spirals Shells, folded plate members: 20 mm bar and larger 16 mm, Wr1 or D31 wire, and smaller
15 10
Prestressed Concrete The following minimum concrete cover shall be provided for prestressed and nonprestressed reinforcement, ducts and end fittings.
Minimum cover, mm (a)
Concrete cast against permanently exposed to earth
75
(b)
Concrete exposed to earth or weather: Wall panels, slabs joists other members
25 40
(C)
Concrete not exposed to weather or in contact with ground: slabs, walls, joists: Beams, columns: Primary reinforcement,
20 40 25
Ties, stirrups, spirals Shells, folded plate members: 16 mm, Wr1 or D31 wire, and smaller Other Reinforcement
10 d b but not less than 20
B undled Bars For bundled bars, the minimum concrete cover shall be equal to the equivalent diameter of the bundle, but need to be greater than 50 mm, except for concrete cast against and permanently exposed to earth, the minimum cover shall be 75 mm.
SHRINKAGE AND TEMPERATURE REINFORCEMENT (2010 NSCP)
Shrinkage and temperature reinforcement is required at right angles to the principles reinforcement to minimize cracking and to tie the structure together to ensure its acting as assumed in the design. The provisions of this section are intended for structural slabs only; they are not intended for soil-supported slabs on grade. Reinforcement for shrinkage and temperature stresses normal to flexural reinforcement shall be provided in structural slabs where the flexural reinforcement extends in one direction only.
Shrinkage and temperature reinforcement shall be provided in accordance with either of the following: a) Where shrinkage and temperature movements are significantly restrained, the requirements of 408.3.4 and 408.3.3 shall be considered. b) Deformed reinforcement conforming to 43.6.3 used for shrinkage and temperature reinforcement shall be provided in accordance with the following: Areas of shrinkage and temperature reinforced shall be provided at least the following rations of reinforcement area to gross concrete area, but no less than 0.014: 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 stress exceeding 420 MPa measured at a yield strain of 0.35 percent is used…………………………………………………….....…..0.0018x415/
Shrinkage and temperature reinforcement shall be spaced not farther apart than five times the slab thickness, nor farther apart than 450 mm.
LOADS
The most important and most critical task of an engineer is the determination of the loads that can be applied to a structure during its life, and the worst possible combination of these loads that might occur simultaneously. Loads on structure may be classified as dead loads or live loads.
DEAD LOAD Dead loads are loads of constant magnitude that remain in one position. This consists mainly of the weight of the structure and other permanent attachments to the frame .
LIVE LOAD Live loads are loads that may change in magnitude and position. Live loads that move under their own power called moving loads. Other Live loads are those caused by wind, rain, earthquakes, soils, and temperature changes. Wind and earthquake loads are called lateral loads.
A R R E NG ME NTS OF LIV E LOA D Live loads may be applied only to the floor or roof under consideration, and the far ends of columns built integrally with the structure may be considered fixed. It is permitted by the code to assume the following arrangement of live loads: (a) Factored dead load on all spans with full factored live load on two adjacent spans, and (b) Factored dead load on all spans with full factored live load on alternative spans.
R EQUIRED S TRE NGHT (FACTIRE D LOAD), U Required strength U to resist dead load (D) and live load (L) shall be at least equal to: Eq. 1-2 U=1.4D + 1.7L If resistances to structural effects of a specified wind load W are included in design, the following combination of D, L, and W shall be investigated to determine the greatest required strength U: Eq. 1-3
U=0.75(1.4D + 1.7L + 1.7W)
Where load combinations shall be include both full value and zero value of L to determine the more severe condition, and Eq. 1-4
U=0.9D + 1.3W
But for any combination of D, L, and W, required strength U shall not be less than Eq.12 If resistance to specified earthquake loads of forces E is included in design, the following combinations of D, L and E shall be investigated to determine the greatest required strength U: Eq. 1-5
U=1.1D + 1.3L + 1.1E
Where load combinations shall included both full value and zero value of L to determine the more severe condition, and Eq. 1-6
U=0.9D + 1.1E
But for any combination of D, L, and E, required strength U shall not be less than Eq. 12 If resistance to earth pressure H is included in design, required strength U shall be at least equal to: Eq. 1-7
U=1.4D + 1.7L + 1.7 H
Except where D or L reduces the effect of H, 0.9D shall be substituted for 1.4D and zero value of L shall be used to determine the greatest required strength U. For any combination of D, L and H, required strength U shall not be less than. If resistance to loadings due to weight and pressure of fluids with well defined densities and controllable maximum heights F is included in design, such loading shall have a factor of 1.4 and to be added to all loading combinations that include live load. If resistance to impact effects is taken into account in design, such effects shall be included with live load L.
Where structural effects T of differential settlement, creep, and shrinkage expansion of shrinkage-compensating concrete or temperature change may be significant in design, required strength U shall be equal to
Eq. 1-8
U=1.75(1.4D +1.4T + 1.7L)
But required strength U shall not be less than Eq. 1-9
U=1.4(D + T)
Estimations of differential settlement, creep, and shrinkage expansion of shrinkage compensating concrete or temperature change shall be based on a realistic assessment of such effects occurring in service.
STRENGTH REDUCTIONS FACTORS, φ (PHI)
The design strength provided by a concrete member, its connections to other members, and its cross sections, in terms of flexure, axial load, shear, and torsion shall be taken as the nominal strength multiplied by a strength reduction factor φ having following values. (a) Flexure without axial load………………………………… 0.90 (b) Axial tension, and axial tension with flexure…………… 0.90 (c)Axial tension and axial tension with flexure: 1. Spiral reinforcement…………………………………….. 0.75 2. The reinforcement & other reinforced members……... 0.75 (d) Shear and torsion…………………………………………….. 0.85 (e) Bearing on concrete………………………………………….. 0.70 ( f) Post-tensioned anchorage zones…………………………… 0.85
ACI-318-05 (NSCP C101-10-210)
= = =
,
Notations gross of concrete sections . For a hollow section, concrete only and does not include the area of the void(s) area of shear reinforcement spacing,
is the area of the
web width, or diameter of circular section, mm
D = dead loads, or related internal moments and forces d = distance from extreme compression fiber to centroid of longitudinal tension reinforcement, mm E
= load effects of earthquake, or related internal moments and forces
=
specified yield strength
of transverse reinforcement, MPa
F = loads due to weight and pressures of fluids with well-defined densities and controllable maximum heights, or related internal moments and forces. h
= overall thickness or height of member, mm
H = loads due to weight and pressure of soil water in soil, or other materials, or related internal moments and forces. L = live loads or related internal moments and forces.
= = =
roof live loads or related internal moments and forces. factored moment at section, N-mm
factored axial force normal to cross section occurring simultaneously with to be taken as positive for compression and negative for tension, N R = rain load, or related internal moments and forces.
or
;
T = cumulative effect of temperature, creep, shrinkage , differential settlement, and shrinkage-compensating concrete. U
W
= required strength to resist factored loads or related internal moments and forces, = nominal shear strength provided by concrete, N = nominal shear strength. = nominal shear strength provided by shear reinforcement N = factored shear force at section, N = wind load, related internal moments and forces
= net tensile strain in extreme layer of longitudinal tension steel at nominal strength, excluding strains due to effective prestress, creep, shrinkage, and temperature = strength reduction factor = ratio of
to
CHAPTER 9 – STRE NGTH AND SER VVICEAB ILITY RE QUIREMENTS 9.1- GE NER AL 9.1.1 Structures and structural members shall be designed to have design strengths at all sections at least equal to the required strengths calculated for the factored loads and forces in such combinations as are stipulated in this code.
9.1.2 Members also shall meet all other requirements of this code to ensure adequate performance at service load levels.
9.1.3 Design of structures and structural members using the load factor combinations and strength reduction factors of Appendix C shall be permitted. Use of load factor combinations from this chapter in conjunction with strength reduction factors of appendix C shall be permitted.
9.2 R equired s trength 9.2.1 Required strength U shall be at least to the effects of factored loads in Eq. (9-1) through (9-7). The effect of one or more loads not acting simultaneously shall be investigated.
U = 1.4 (D+F) U = 1.2(D+F+T) + 1.6(L+H) + 0.5( or R) U = 1.2D + 1.6( or R) + (1.0L or 0.8W) U = 1.2D + 1.6W + 1.0L + 0.5( or R) U = 1.2D + 1.0E+ 1.0L U = 0.9D + 1.6W+ 1.6H U = 0.9D + 1.0E+ 1.6H
(9-1) (9-2) (9-3) (9-4) (9-5) (9-6) (9-8)
Except as follows: a) The load factor on the live load L in Eq. (9-3) to (9-5) shall be permitted to be reduced to 0.5 except for garages, areas occupied as places of public assembly, and all where L is greater than 4.8N/ . b) Where wind load W has not been reduced by a directionality factor, it shall be permitted to use 1.3 W in Eq. (9-4) and (9-6). c) Where E, the load effects of earthquake, is based on service-level seismic forces, 1.4E shall be used in place of 1.0E Eq. (9-5) and (9-7). d) The load factor on H, loads due to weight and pressure of soil, water in soil or other materials, shall be set equal to zero in Eq. (9-6) and (9-7) if the structural action due to H counteracts that due to W or E. Where lateral earth pressure provides resistance to structural actions from other forces, it shall be not be included in H but shall be included in the design resistance.
9.2.2 If resistance to impact effects is taken into account id design, such effects shall be included with L.
9.2.3 Estimations of differential settlement, creep, shrinkage, expansion of shrinkage-compensating concrete. or temperature change shall be based on a realistic assessment of such effects occurring in service.
9.2.4 If structure is in a flood zone, or is subjected to forces from atmospheric ice loads, the flood or ice loads and the appropriate load combinations of SEI/ASCE7 shall be used.
9.2.5 For post-tensioned anchorage zone design, a load factor of 1.2 shall be applied to the maximum prestressing steel jacking force.
9.3 Desi g n s treng th 9.3.1 Design strength provided by a member, its connections to other members, and its cross sections, in terms of flexure, axial load, shear and torsion, shall be taken as the nominal strength calculated in accordance with requirements and assumptions of this code, multiplied by the strength reduction factors in 9.3.2,9.3.4, and 9.3.5.
9.3.2 Strength reduction factor shall be as given in 9.3.2.1 through 9.3.2.7: 9.3.2.1 Tension-controlled sections as defined in 10.3.4………….0.90 (See also 9.3.2.7)
9.3.2.2 Compression-controlled sections, as defined 10.3.3: a) Members with spiral reinforcement conforming to 10.9.3…..0.70 b) Other reinforced members……………………………………..0.65
For sections in which the net tensile strain in the extreme tension steel at nominal strength is between the limits for compression-controlled and tension-controlled sections, shall be permitted to be linearly increase from that for compression-limit to 0.005.
′ ′
Alternatively, when Appendix B is used, for members in which does not exceed 415 MPa, with symmetric reinforcement, and with (d- d’)/h not less than 0.70, shall be permitted to be increased linearly to 0.90 as decreases from 0.10 to zero. For other reinforced members, shall be permitted to be increased from 0.10 or , whichever is smaller, to zero.
9.3.2.3 –Shear and torsion…………………………………………0.75 9 .3.2.4 – Bearing on concrete (except for post-tensioned and anchorage zones and struct-and-tie models)…………………….……………0.65
CHAPTER 1 A nalys is and Des ig n of B eam
′ ′
NOTAIONS AND SY MB OLS USE D = depth of equivalent stress block, mm = area of tension reinforcement, mm 2 / m = area of skin reinforcement per unit height in one side face, mm 2 = width of compression face of member, mm = distance from extreme compression fiber to neutral axis, mm = distance from extreme compression fiber to centroid of tension reinforcement, mm = distance from extreme compression fiber to centroid of compression reinforcement, mm = thickness of concrete cover measured from extreme tension fiber to center of bar or wire, mm = modulus of elasticity of concrete, MPa = modulus of elasticity of steel 200,000 MPa = specified compressive stress of concrete, MPa = calculated stress in reinforcement at service loads, MPa = specified yield strength of steel, MPa = overall thickness of member, mm =moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement = moment of inertia of reinforcement about centroidal axis of member crosssection
= nominal moment, N-mm = factored moment at section, N-mm = factor defined in Section 410.4 in Page 16 = strain in concrete (maximum = 0.003) = strain in steel below yield point = = strain in steel at yield point = ration of tension reinforcement = balance steel ratio = strength reduction factor
/ /
Ø
A S S UMP TION IN S TR E NG TH D E S IG N IN F LE XUR E (CODE S E CTION 5.10.2 ) 1. Strain in reinforcement and concrete shall be based assumed directly proportional to the distance from the neutral axis. Expect for deep flexural members with overall depth to clear span to ratio, h/L> 2/5 for continuous spans and h/L >4/5 for simple spans, a nonlinear distribution of strain shall be considered (See Sec. 5.10.7). 2. Maximum usable strain at extreme concrete compression fiber, equal to 0.003
3. For
.
below shall
,
be taken as
x
for
>
,
=
shall be assumed
4. Tensile strength of concrete shall be neglected in axial and flexural calculations. 5. Relationships between compressive stress distribution and concrete strain may be assumed rectangular, trapezoidal, parabolic, or any other from that result in prediction of strength in substantial agreement with results of comprehensive tests. 6. For rectangular distribution of stress:
a) Concrete stress of 0.85 ′ shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross-section and a straight line located parallel to the maximum compressive strain. b) Distance c from fiber of maximum strain to the neutral axis hall is measured in the direction perpendicular to N.A.
c) Factor shall be taken as 0.85 for ′ ≤ 30 MPa and shall be reduced continuously at rate of 0.008 for each 1 MPa of strength in excess of 30 MPa, but shall not be taken less than 0.65. i.e
i. ii.
For For
≤ 30 MPa, = 0.85 ′ > 30 MPa, =0.85-0.008( ′ -30) but not shall be less than 0.65 ′
RE CTANGULAR BE AM RE INFORC ED FOR TENSION ONLY (SINGLY R EINFORC ED)
′
b
0.85
c
a
0.003 c
As
/
Fig ure 2.1: Stress and strain diagram for singly reinforced and rectangular beam
=
Eq. 2-1
′′ ≤30 , =0. 8 5 ≤30 , =0.85 . ′ 30 ∑ = ′ = For For
(
but shall not be less than 0.65
C=T 0.85
Eq. 2-2
=
. ′
= . ′ =
Multiplying Eq. 2-2 by d/d:
. ′
The term
Eq. 2-3
=
and
Eq. 2-4 Let
is called the ratio of steel reinforcement and is denoted as .
= ′
Eq. 2-5
Nominal Moment Capacity:
=/2 =0. 8 5′ /2 =0.85 0.85 12 0.85
From the stress diagram in Figure 2.1:
= .′ = .
= 10.59
Eq.2-6
= ℎ =0.90 = 10.59 =10. 5 9 =
Ultima Ultimate te Moment Moment C apa apacity ci ty (Des ig n S trength): treng th): Eq.2-7
C oefficient oefficient of Resi R esi s tance Eq.2-8 Eq.2-9
Solving for an steel ratio : Eq.2-10
= . [1 [1 1 . ]
in Eq. 2-8 and replacing it with,
BALANCE DESIGN
, yields the following formula the
Balance design refers to a design so proportioned that the maximum stresses in concrete (with strain of 0.003) and steel and (with strain of
) are ar e reached reach ed simultaneously once he
ultimate load is reached, causing them to fail simultaneously.
UNDERR EINFORC EINFORC ED DES DES IGN Underreinforced design is a design in which the steel reinforced is lesser than what is required for balance condition . If the ultimate load is approached, the steel will begin to yield although the compression concrete is still understressed. If the load is further increased, the steel will continue to elongate, resulting in appreciable deflections and large visible crack in the tensile concrete. Failure under this condition is ductile and will give warning to the user of the structure to decrease the load.
O VER RE INFO INFORCE RCE D DES IGN Overreinforced design is a design in which the steel reinforcement is more than what is required for balanced condition. If the beam is overreinforced, the steel will not before failure. As the load is increased, deflections are not noticeable although the compression concrete is highly stressed, and failure occurs suddenly without warning to the user of the structure. Overreinforced as well as balanced design should be avoided in concrete because of its brittle property, that is why the Code limits the tensile steel percentage (P max =0.75 =0.75 pb ) to ensure underreinforced beam with ductile type of failure to give occupants warning before occurs.
BA LANCED STEEL RA TIO TIO
: In balanced condition, the concrete and steel yield simultaneously, In this condition, the strain in concrete reached is maximum usable value of and the strain in steel is
=200,000 MPa.
By ratio and proportion in the triangle shown in Figure2.2:
= . .+
Note:
= 0.0030.003200, 00
=200,00
where
= +
Eq.2-11
But a =
= c=c
. .
= = = . +
Eq. 2-12
= = . ′
Note: Eq. 2-12 is for singly reinforced rectangular sections only. Eq. 2-11 is applicable to nay shape.
MA XIMUM XI MUM STE S TE E L R E IN F OR C E ME NT S ection ecti on 410.4.3: 410.4.3 : For flexural and for subject to combined flexure and compressive axial
0.10′
load when the design axial load strength is less than the smaller of or , the ratio of reinforcement that would produce balance strain condition for the section under flexure without axial; load. For members with compression reinforcement, the portion of equalized by compression reinforcement need not be reduced by the0.75 factor. Eq. 2-13
and Eq. 2-14
=0.75 =0.75
This limitation is to ensure that the steel reinforcement will yield first to ensure ductile failure.
MIN IMUM IMU M RE R E INF IN F OR C E ME NT OF O F F L E XUR A L ME MB E R S
410.61 At very section of flexural members where tensile reinforcement is required by analysis, the area
provided shall not be less than that given by:
= ′
Eq. 2-15 Eq.2-16
and not less than
.
410.62 For statically determinate T-section with flange in tension, the area be equal to or greater than the smaller value given either by:
shall
= ′
Eq. 2-17 or Eq. 2-15 with
set equal to the width of the flange.
410.6.3 The requirements of Sections 410.6.1 and 410.6.2 need to be applied if at every section the area of the tensile reinforcement is at least one-third greater than that required by analysis.
410.6.4 For structural slabs and footings of uniform thickness, the minimum area of tensile reinforcement in the direction of span shall be the same as that required by Section 407.13 ( ). Maximum spacing of S hrin hr inkk ing in g and Temperature R einfor ein forcement cement this reinforcement shall not exceed three times the thickness and 450 mm. The provision for minimum amount of reinforcement applies to beams, which for architectural and other reasons are much larger in cross-section than required by strength consideration. With a very small amount of tensile reinforcement, the computed moment strength as a reinforced concrete section computed from its modulus of rapture. Failure in such a case can be quite sudden.
S TE P S IN D E S IG NING NI NG A S IN G L Y R E IN F OR C E D RE CTANGULAR CTANGULAR B EA M FOR FLEXURE FLEXURE :
Note: The assumptions made in steps II, V,and VIII are the author’s recommendation based on his experience. I. II. III. IV. V.
Identify the values of the dead load and live load to be carried by the beam. (DL & LL) Approximate the weight of beam (DL) between 20% to 25% of (DL+LL).This weight is added to the de load. Compute the factored load and factored moment: Ex: factored Load =1.4 DL+1.7L Compute the factored moment to be resisted by the beam, Try a value of steel ratio from 0.5 but must not be less than . This value will provided enough alloance in the actual value of due to rounding-off of the number of bars to be used, for it not to exceed the maximum of 0.05ρb. ′
5 600 = 0.8600 =0. 8 5 ≤30 =0.8500.008( 30) >30 = 0. 4 , = = 10.59 =_____________________ / ′
′
VI.
Compute the value of
VII.
Solve for
′
′
:
′
VIII.
IX.
Try ratio ( from d=15b to d=2b), and solve for d, (round-off this value to reasonable dimension). Check also the minimum thickness of beam required by the Code a given in Table 2.1 in page 36. After solving for d, substitute its value to Step VII, and solve for b. Compute the weight of the beam and compare it to the assumption made in Step II.
=
Solve for the required steel area and number of bars.
Number of bars(diameter = D) x number of bars =
=0.75 600 =0.75 0. 85600 =____________ = == = =______________ = = 0.8 5 1 1 0.285 =___________ ==____________
S TE PS IN C OMP UTING THE R E QUIR E D TE NS ION S TE E L AR E A WITH K NOWN MOME NT NT A ND OTHE R PR OP E R TIE S : I.
OF A BEM
Solve for
′
if if
II.
(1-0.59 design as singly reinforced (Step II) design as doubly reinforced (Step III)
Solve for :
′
′
III. Compression reinforcement is necessary. (See Chapter 3)
S TE PS IN C OP UTING OF A BEAM WITH KNOWN TENSION STEEL AREA A ND OTHE R B E A M PR OP E R TIE S : I. II.
=
Solve for :
Check if steel yields by computing
= .+ ′
III.
`
≤=/ = 10.59 ≤ ≤≤ ′
if ,steel yields, proceed to III if ,steel does not yield, proceed to step IV. Note: if ,the given is not adequate for the beam dimension.
IV.
>
Solve for
from the strain diagram: [Note:
=200,000MPa]
= 0.003
=600 −
Eq. 2-18
=0 =0.85
[
T=C
600 600 =0.85 =0.85 =600 ;= = 2= 2 = 2=0.85′ 2 but a=
Solve c by quadratic formula and solve for
or
and a:
MINIMUM THIC K NE S S OF FLE XUR A L ME MB E R S According to Section 5.9.5 of NACP, minimum thickness stipulated in Table 2.1 shall apply for one-way construction not supporting are attached to portions or other construction likely to be damaged by large deflections, unless computation of deflection indicates a lesser thickness can be used without adverse effects.
Table 2.1 MINIMUM THICKNESS OF NON-PRESTRESSED BEAMS OR ONE-WAY SLABS UNLESS DEFLECTIONS ARE COMPUTED *
Minimum thickness, h Simply One end supported continuous
Both ends continuous
Cantilever
Members not supporting or attached to partitions or other construction likely to be damaged by large deflections
Member
Solid one-way L/20 slabs
L/24
L/28
L/10
Beams or L/16 ribbed one-way slabs
L/18.5
L/21
L/8
Span length L is in millimeters Values given shall be used directly for members with normal density concrete ( ) and grade 415 reinforcement. For other conditions, the values shall be modified as follows:
=2300 /
(a)
(b)
/ /700.
For structural lightweight concrete having weights in the range 1500-2000 values shall be multiplied by (1.65-0.005 ) but not less than 1.09, where the unit mass in . For other than 415 MPa, the values shall be multiplied by (0.4 +
/
B E AM DEFLE CTION (S E CTION 5.9.5
is
S ect. 5.9.5.2.2 Where deflections are to be computed, deflections that occur immediately on application of load shall be computed by usual methods or formulas for elastic deflections, considering effects of cracking and reinforcement on member stiffness.
S ect. 5.9.5.2.3 Unless stiffness values are obtained by a more comprehensive analysis, immediate deflection shall be computed with the modulus of elasticity and with the effective moment of inertia as follows, but not greater than
= 1 3 ]
Eq.2-19
Where
=
.
for concrete
=0.7
= modulus of rapture of concrete, MPa, for normal weight
Concrete
′
= maximum moment in member at stage deflections is computed. = moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement. = moment of inertia of cracked section transformed to concrete = distance from centroidal axis of gross section, neglecting reinforcement, to extreme fiber in tension. When Lightweight aggregate is used, one of the following modifications shall apply: (a)
When
is specified and concrete is proportioned in accordance with Sec. 5.5.2,
shall be modified by substituting 1.8
exceed
′ .
for
′ but the value of 1.8
shall not
(b) When is not specified, shall not be multiplied by 0.75 for “all lightweight” concrete, and 0.85 for “ sand -lightweight” concrete. Linear interpolation is permitted if partial sand replacement is used.
S ect. 5.9.5.2.4: For continuous members, effective moment of inertia may be taken as the average of values obtained from Eq. 2-19 for the critical positive and negative moment sections. For prismatic members, effective moment of inertia may be taken as the value obtained from Eq. 2-19 at midspan for simple and continuous spans, and at the support cantilevers.
S ect.5.9.5.2.5: Unless values are obtained by a more comprehensive analysis, additional long-term deflection resulting from creep and shrinkage of flexural members (normal weight or lightweight concrete) shall be determined by multiplying the immediate caused by the sustained load considered, by the factor. Eq. 2-10
= +
Where ′ shall be taken the value of reinforcement ratio for non-prestressed compression reinforcement at midspan for simple and continuous spans,a nd at support
for cantilevers. It is permitted to assume the time-dependent factor for sustained loads to be equal to: 5 years or more……………………2.0 12 months…………………………...1.4 6 months……………………………..1.2 3 months………………………………1.0
Deflection computed in accordance with Sec. 5.9.5.2.2 through Sec.5.9.5.2.5 shall not exceed limits stipulated in Table 2.2.
Table Table 2.2: Maximum Permissible Computed Deflections Type of member
Deflection to be considered
Deflection limitation
Flat roofs not supporting or Immediate deflection due to L/180* attached to nonstructural live load LL elements likely to be damage by large deflections Floors not supporting or Immediate deflection due to L/360* attached to nonstructural live load LL elements likely to be damaged by large deflections Roof or floor construction That part of the total L/480** supporting, or attached to deflection occurring after nonstructural elements not likely attachment of non structural to be damaged by large elements (sum of the longdeflections time deflection due to all Roof or floor construction sustained loads and the L/20**** supporting, or attached to immediate deflection due to nonstructural elements not likely any additional live load)**** to be damaged by large deflections Limit not intended to safeguard against ponding. Ponding should be cheated by suitable calculations of deflections, including added deflections due to ponded
water and considering long-term effects of all sustained loads, camber, construction tolerances, and reliability of provisions for damage. Limit may be exceeded if adequate measures are taken to prevent damage to supported or attached elements. Long=time deflections shall be determined in accordance with Sec.5.9.5.2.5 or Sec. Attachment of nonstructural elements. This amount shall be determined on basis of accepted engineering, data relating to time-deflection characteristics of members similar to those being considered. But not greater than tolerance provided for nonstructural elements. Limit may be exceeded if camber is provided so that deflection minus camber does not exceeded limit.
NSCP COEFFICICIENTS COEFFICICIENTS FOR C ONTI ONTINU NUOU OUS S BE AMS AMS AND SLAS BS Section 5.8.3.3 of NSCP states that in lieu of frame analysis, the following approximate moment and shears are permitted for design of continuous beams and one-way slabs (slabs reinforced to resist flexural stresses in only one direction), provided: a) There are two or more more spans, b) Spans are approximately equal, with the larger of two adjacent spans not greater than uniformly than the shorter by more than 20 percent, c) Loads are uniformly distributed, d) Unit live does not exceeded three times unit dead load, and e) Members are prismatic. Positive moment End spans Discontinuous end unrestrained…………………… Discontinuous end integral with support………….. Interior spans…………………………………
/14/11 /16
/9/10
Negative moment at exterior face of first interior suppo rt Two spans …………………………………………………………..... More than two spans……………………………………………….....
Negative moment at other faces of interior supports………….…… Negative moment at face of all supports for: Slabs with spans not exceeding 3 m; and beams Where ratio of sum of column stiffness to beams Stiffness exceeds eight at each end of the span……………………
/11 /12
Negative moment at interior face of exterior Support members built integrally with Where support is a spand rel rel beam…………………………… When support is a column…………………………….……..….
/24/16 1.5/2 /2
Shear in end members at face of First interior inte rior support……………………………......................... support……………………………..................................... ............
Shear at face of all other supports………………………………………..
When =clear span positive moment or shear and average of adjacent clear spans for negative moment.
Figure 2.3: Shear and moment for continuous beam or slab with spans and discontinuous end integral with support
Fig ure 2.5 Shear and moment for continuous beam or slab with more than two spans and discontinuous end unrestrained
A C I-318-05 (NS C P C 101-10-2010) 10.2 Des ig n as s umptions (410.3) 10.2.1 Strength design of members for flexure and axial loads shall be based on assumptions given in 10.2.2 through 10.2.7, and on satisfaction of applicable conditions of equilibrium and compatibility of strains.
10.2.2 Strain in reinforcement and concrete shall be assumed directly proportional to the distance from the neutral axis, except that, for deep beams as defined in 10.7.1, an analysis that considers a nonlinear distribution of strain shall be used alternatively, it shall be permitted to use a struct-and tie model. See 10.7,118, and Appendix A.
10.2.3 Maximum usable strain at extreme concrete compression fiber shall be assumed equal to 0.003.
10.2.4
Stress in reinforcement below shall be taken as times steel strain. For strains greater than that corresponding to , stress in reinforcement shall be considered independent of strain and equal to .
10.2.5 Tensile strength of concrete shall be neglected in axial and flexural calculations of reinforced concrete, except when meeting requirements of 18.4.
10.2.6 The relationship between concrete compressive stress distribution and concrete strain shall be assumed to be rectangular, trapezoidal, parabolic, or any other shape that results in prediction of strength in substantial agreement with results of comprehensive tests.
10.2.7 Requirements of 10.2.6 are satisfied by an equivalent rectangular concrete stress distribution defined by the following:
′
10.2.7.1 Concrete stress of 0.85
shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross section and a straight line located parallel to the neutral axis at distance a= form the fiber of maximum compressive strain.
10.2.7.2 Distance from the fiber of maximum strain to the neutral axis, c , shall be measured in direction perpendicular to the neutral axis.
10.2.7.3 For ′ between 17 and 18 MPa,
′ above 28 MPa,
shall not be taken less than 0.65
shall be taken as 0.85. For
10.3 G eneral principles and requirements (410.4) 10.3.1 Design of cross sections subject to flexure or axial loads, or to combined flexure and axial loads, shall be based on stress and strain compatibility using assumptions in10.2.
10.3.2 Balanced strain conditions exist at a cross section when tension reinforcement reaches the strain corresponding to reaches its assumed ultimate strain of 0.003.
just as concrete in compression
10.3.3 Sections are compression-controlled if the next tensile strain in the extreme tension steel, , is equal to or less than the compression-controlled strain limit when the concrete in reaches its assumed strain limit of 0.003. The compressioncontrolled strain limit is the net tensile strain in the reinforcement at balanced strain conditions. For Grade 415 reinforcement, and for all prestressed reinforcement, it shall be permitted to set the compression-controlled strain limit equal to 0.002.
10.3.4 Sections are tension-controlled if the net tensile strain in the extreme
tension steel is equal to greater than 0.005 when the concrete in compression reaches its assumed strain limit of 0.003. Sections with between the compressioncontrolled strain limit and 0.005 constitute a transition region between compressioncontrolled and tension-controlled sections. Derivation: for E = 200 GPa The beam is tension-controlled When = 0.005 (or =1000MPa)
0.003 = 0.008 = == 38 = ℎ =0.90 =0.90 0.85 2 3 8 =0.90 0.85 3/8 2 = 1 Eq. 2-21
For rectangular beam:
Eq. 2-22
10.3.5 For nonprestressed flexural members and nonprestressed members with factored axial compressive load less than 0.10 shall not be less than 0.004.
steel strain
at nominal strength
10.3.5.1 Use of compression reinforcement shall be permitted in conjunction with additional tension reinforcement to increase the strength of flexural members.
Derivation: for E =200 GPa Maximum steel area and
when beam is singly reinforced:
=0.004 = =800
0.003 = 0.007 ;= 37
=
Eq. 2-23
For rectangular section:
=0.85′ ℎ = =0.85 .
T=
) b
Eq. 2-14
= 2 =0.85 2 =0.85 37 2 = 1 − =0.650.25 − Eq. 2-25
)
Eq. 2-26
10.3.6 Design axial strength than
of compression members shall not be taken greater computed by Eq. (10-1) or (10-2).
10.3.6.1 For nonprestressed members with spiral reinforcement conforming to
7.10.4 or composite members conforming to 10.16:
=0.850.85 =0.850.85
(10-1) 10.3.6.2 For non nonprestressed members with spiral reinforcement conforming to 7.10.5: (10-2)
10.3.6.3 For prestressed members, design axial
shall not be taken greater than 0.85 (for members with spiral reinforcement) or 0.80 (for members with tie reinforcement) of the design axial strength at zero capacity .
10.3.7 Members subject to compressive axial load shall be designed for the maximum moment that can accompany the axial load. The factored axial force at given eccentricity shall not exceed that given in 10.3.6. The maximum factored moment shall be magnified for slenderness effects in accordance with 10.10
10.4 Di s tance between lateral supports of flexural members
10.4.1 Spacing of lateral supports for a beam shall not exceed 50 times b, the least width of compression flange or face.
10.4.2 Effects of lateral eccentricity of load shall be taken into account in determining spacing of lateral supports.
10.5.1 Minimum reinforc ement of flexural members 10.5.1 At every section of flexural members where tensile reinforcement is required by analysis, except as provided in 10.5.2, 10.5.3, and 10.5.4, as provided shall not be less than that given by
=
(10-3)
and not less than
= . .
(10-3)
10.5.2 –For statically determinate members with a flange in tension, less than the value given by eq. (10-3), except that width of the flange, whichever is smaller.
2
shall not be is replaced by either or the
S TE PS IN THE DE S IG N OF S ING LY R E IN FOR C E D RE CTANGULAR B EA M FOR FLEXURE Note: The assumption made in steps II, V, and VIII are the authors recommendation based on his experience. I. Determine the values of loads, Dl, LL and other loads II. Approximate the weight of beam (DL) as follows: Small beams: 2kN/m Medium-sized beams: 3.5kN/m Large-sixed beams: 7kN/m or Weight of beam in kN/m=24kN/ x beam area in III. Compute the factored load on different load combinations Example: Factored Load =1.2 DL + 1.6 LL IV. Compute the factored moment to be resisted by the beam,
V.
Try a value of steel ratio from 0.7 to 0.8 but must not be less than . This value of will provided enough allowance in the actual value of due to rounding-off the numbers bars to be used so that it will not exceed the maximum .
0. 8 5 600) = (600 =0.0.0585 ≤28 =0.85 7 ( 28) >28 = 37 0.8 5′ = ′4 ′>31. 3 6 , ℎ = 1. 4 , = : ℎ =
VI.
Compute the value of
VII.
Solve for the reduction factor Solve for c: Note: For singly reinforced rectangular beam, proportional to c:
is directly
c=(assumed factor) x
The assumed factor may range from 0.7 to 0.8 as suggested in step V.
if if
=600 90− ≥1000 , ,=0. <1000 , , =0.650.25 −
VIII.
IX.
Solve for
==____________ 10.59
:
Try a ratio d/b (from d= 1.5b to d=2b), and solve for d. (round-off this value to reasonable dimension) Check also the minimum thickness of beam required by the code as given in Table 2.1 in Page 26. After solving for d, substitute its value to Step VII, and solve for b. Compute the weight of the beam and it to the assumption made in Step II.
X.
=
Solve for the required steel area and number of bars.
Number of bars (diameter=D) x number of bars =
′ 51 = 37 0.85′ 3 = 140 800 1 14 ==0. 6 50. 2 5 1000 =________________ ≤
S TE PS IN FINDING THE R E QUIR E D TE NS ION S TE E L A R E A OF A B E AM WITH K NOW R EQUIRE MOMENT A ND OTHE R B E A M PR OP E R TIE S Given b, d ,
I.
if
Solve for
and
:
and
.
design as Singly Reinforced (Step II)
>
if
II.
design as Doubly Reinforced (Step V)
= 1 < , , =0. 9 0 >, =______ = 0. 8 5 = 1 1 0.285=_______ = ′4 ′ >31.36 ,ℎ = 1. 4 ==_________
Determine if the section in tension-controlled or transition From Eq. 2-11: if if
III.
IV.
,proceed to step III region, proceed to step IV
= ′4 ′ >31.36 ,ℎ = 1. 4 ==_________
: _ =_ /2 ℎ =600 =0.650.25 1000 = =0.6008 5 =0. 8 5 0.85 2 =0.650.25 1000 ____________= =_________ =0. 8 5′ =_____________________________ = ′ >31.36 ,ℎ = .
Solve for c and
if
V. Compression reinforcement is necessary.(See chapter 2)
′ = 0.85 600 = (600) ≤ , > , ≤, = ′4 ′ >31.36 ,ℎ = 1. 4 ≤
S TE PS IN FINDING OF A BEAM WITH KNOWN TENSION STEEL AREA A ND OTHE R B E A M PR OP E R TIE S : Given: b, d, I. II.
,
,
:
Solve for
Check if steel yields by computing
if steel yields, proceed to step III if steel dos not yield, proceed to step IV. Note: if the given is not adequate for the beam dimension.
III.
=0. 8 5 ′ =________ / =600 ,=0.90− ≥1000 , <1000 , , =0.650.25 − = 0.85 ′ 2 > =0. 65 ′ =600 − 0.85 ′ Solve for
:
;c=
=_________
if if
IV.
Compression-controlled
b
0.85
a
.
c=
d d-a/2
=0.85 ′ 600 =0.85 ′ − =600 =0.=___________ 65 = 2= 2 = 2=′ 2 T=C
but a=
c=__________ ; a=
or
=__________
T=
ILLUSTRATIVE PROBLE MS DESIGN PR OBLEMS PR OBL E M 2.1
′ =21 =345
A reinforced concrete rectangular beam 300 mm wide has an effective depth of 460 mm and is reinforced for tension only. Assuming and , determine the balance steel area in sq.mm.
S OL UTI ON
= .+ =0.85 ′0.<30 5210.85600 = 8345600345 =0. 0 2792 = PR OBL E M 2.2
′ =27.6
=276
: . = + =0.75 = =10.59
.. ==0..0+ =0.495 7 50. 0 495 =0.= 0.371 . =0.371 =27.60.37110.590.37
A rectangular beam has b = 300 mm and d =490 mm. Concrete compressive strength and steel yield strength . Calculate the required tension steel area if the factored moment is (a) 20 kN-m,(b)140 kN-m,(c) 485 kN-m, and (d)620 kN-m.
S OL UTI ON Solve for
=8.001 490 = =576.=8.207910 01300 =576. 2 79 = =0.=518.90 65576. 279 = = =20< 10 =0. =0.90309300490 = 0.8 5′ 1 1 0.0285′ 6 20. 3 09 = 0.8527. 1 1 276=0. 00113< 0.8527.6 = ′4 ′ >31.1.346,ℎ = 1. 4 = =0.005072 = =0. =746 00572300 490 =140< = 0 =0.90=2.13006 490 = 0.8 5′ 1 1 0.285′ a)
20 x
b)
(singly reinforced)
140 x 1
6 22. 1 6 = 0.8527. 1 1 276 0.8527.6 = =0. 00822> =0. 0=,0822300490 c)
=485 < = 10 =0.90 =7.30048490 = 0.8 5′ 1 1 0.285′ 6 27. 4 8 = 0.8527. 1 1 276=0.03384> 0.8527.6 = =0.=,03384300490 =600 > (singly reinforced)
485 x
d)
The beam will be doubly reinforced. See Chapter 3.
PR OB LE M 2.3 (CE MAY 2012)
=415
A reinforced concrete beam has a width of 300 mm and an overall depth of 480 mm. The beam is simply supported over span of 5 m. Steel strength MPa and concrete . Concrete cover is 70 mm from the centroid of the steel area. Unit weight concrete is 23.5kN/ .Other than the weight of the beam , the beam carries a superimposed dead of 18 kN/m and a live load of 14 kN/m. Use the strength design method. a) Determine the maximum factored moment on the beam. b) If the design ultimate moment capacity of the beam is 280 kN-m, determine the required number of 20 mm tension bars. c) If the beam will carry a factored load of 240 kN at midsoan, determine the required number of 20 mm tension bars.
′ =28
′=0.=30085 =415 = . =0.00337
S OL UTI ON Given:
b=300m d=480-70=410 mm
Bar diameter ,
=20 = =23.50.3 0.48 =3.384 =1.=53.43.7383/ 8418.7 14
Weight of beam,
a) Maximum factored moment on the beam. Factored load, Factored load, Maximum factored moment:
b)
= . ==. =280 Solve for
to determine whether compression steel is needed
. = .+ ==0..+ 0 2881 =0.75 =0.021261 = =0.03203 == 10. 5 9 =7. 2 74 =330.14 =280 = 280 =6. 10169 =.90 300410 = 0.8 5′ [1 0.285′ ] 619=0. 01755> = 0.8415528 1 1 26.0.8528 ==2159 =0.01755300410 = 4 2159= 4 20 Required
<
(singly reinforced)
N=6.9 say 7 bars
3.
_ =240 =3.384 / 1. 4 = 4 8 =314.805 < = = .. (weight of beam)
=6.936 = 0.8 5′ 1 1 0.285′ = 0.8415528 1 1 0.285′ =002031> = =0.=2498 02031300 410 =7. = 95 20 2498 =
N
PR OB LE M 2.4 (CE MAY 1993) A reinforced concrete beam has a width of 300 mm and an effective depth to tension bars of 600 mm. compression reinforcement if needed will be placed at a depth of 60 mm below the top. If and , determine the tension steel area if the beam is to resist an ultimate moment of 650 kN-m.
′ =30 =414
5′600 =0.85 ′ <10 = 0.8600 5300.85600 = 0.8414600414 =0. 0 31 =0.15 =0.=0.70523230.031 S OL UTION
Solve for
and
:
= = 0.02323414 30 = 0.3209 10.59 =′ =0.90300.3209300600 =758.1 > < =6.687 =6.0687 =0.90300600 = 0.8 5′ 1 1 0.285′ 687=0. 0 191> = 0.8414530 1 1 26.0.8530 = 1. 4 =0.00338 = =0.=34420191300600 [1-0.59(0.309)
Since
, the beam may be designed as singly reinforced. 650 x 1
Solve for :
PR OB LE M 2.5 (CE November 2000) A rectangular concrete beam has a width of 300 mm and an effective depth of 550 mm. The beam is simply supported over a span 6 m and is used to carry a uniform dead load of 25 kN/m and a uniform live load of 40 kN/m. Assume and . Compression reinforcement if necessary shall be placed at a depth 80 mm from the outermost compression concrete. a) Determine 80 mm from the outermost compression concrete. b) Determine the required tension steel area.
312
′ =21 =
c) Determine the required number of 25-mm tension bars.
S OL UTI ON a) Maximum steel area:
5 ′600 = 0.8600
=0.0319 =0.75 = 9
=0.85 ℎ 30 5210.85600 = 0.8312312600 =0.=0.7050.239903199 =, =0.02399300550
b) Required tension steel area:
=1.4 1.7 Factored load:
Required strength:
=1.=103425/ 1.740 1036 = 8
= 8 = ′ ==0.0.03299312 2156 10.59 = =0.90300.356300550 10.590.356 =463.5kN-m
Solve for
=536.5 > = 463. =0.5 1039=0.3009 550 300550 =5.67 = 0.8 5′ 1 1 0.825 , = 0.8312521 1 1 0.25.852167 =0.02269 = =0.=3743002269300550 c) Number of 25 mm bars: Number of 25-mm bars= Number of 25-mm bars=
.
=7.63 8
PR OB LE M 2.6 (CE MAY 2009) A reinforced concrete beam has a width of 300 mm and total depth of 600 mm. The beam will be design to carry a factored moment of 540kN-m. Concrete strength and steel yield strength . Solve using the strength design method. a) Determine the balanced steel ratio in percent. b) Determine the minimum effective depth of the beam using a steel ratio equal to 0.5 of balanced steel ratio. c) Determine the minimum effective depth of the beam using the maximum allowable steel ratio.
′ =28
=248
S OL UTI ON Given: b=300 mm h=600 mm
=540 =0.85
′=248 =28
a) Balanced steel ratio:
0. 8 5 = (600600)
8 5600 = 0.24885280. 600248 =0. 0 577=. % =0. 5 =0.50.0577 =0.0289 0.0289248 = ′ = 28 =0.2556 = 10.59 =280.255610.590.2556 =6.0776 = = 0 =0.908.307300 =
b) Effective depth using
540 x 1
PR OBL E M 2.7
=275
′ =21
A concrete one-way slab has a total thickness of 120 mm. The slab will be reinforced with 12-mm-diameter bars with .Concrete strength . Determine the required spacing 12 mm main bar if the total factored moment acting on 1-m width of slab is 23 kN-m width of slab is 23 kN-m. Clear concrete cover is 20 mm.
S OL UTI ON Note: Slabs are practically singly reinforced because of its small depths.
.
12mm bars
d
.
s
s
cover=20 mm
b = 1000 mm
Effective depth, d= 120 -20-1/2(12)=94 mm Width, b = 1000 mm
= = 0.8 5′ 1 1 0.825′ 0.85′600 =0.0284 = 0.75 600
h=120 mm
=0.90 100094 10 =2.892 982 = 0.8275521 1 1 22.0.8521
23 x
= 1. 4 =0.00509 = =0.=10851154100094 Spacing of bars (for walls and slabs using unit width):
=
= =
= 1000
Eq. 2-17
1000 12 4 = 1085 =
PR OBL E M 2.8
′ =21 =275
A 2.8 m square column fooring has a total thickness of 47 mm. The factored moment at critical section for moment is 640 kN-m. Assume and . Clear concrete cover is 75 mm. Determine the required number of 20 mm tension bars.
S OL UTI ON Effective depth, d=470-75-1/2(20)=385 mm Width, b =2800 mm Design strength,
=640
Maximum and minimum requirements:
=0.75 .+ =0.0284 =2528 = . =5488 = 640 =1. 10713=0.90 2800385 = 0.8 5′ 1 1 0.285′ = 0.8275521 1 1 0.21.8521713 =0.00656 = =0.=7074 006562800 >385 = = 47074 20 =22.5
(Procedure is not shown anymore see Problem 2.2)
Singly reinforced:
Number of 20 mm bars:
PR OBL E M 2.9
′ =
Design a rectangular beam reinforced for tension only to carry a dead load moment of 60 kN-m (including its own weight) and a live load moment of 48 kN- m. Use and
20.7 =276 . S OL UTI ON Required strength:
=1.4 1.7
(Note: this already includes the weight of beam)
=1.=165.460 1. 7 48 6
20.70.85600 5′600 = 0.8600 = 0.85276600276 =0.0371 = 1. 4 =0.00507 =60% =0. 60.0371 =0.02226 = ′ = 0.02226276 20.7 =0.2968 0. = 10.59 =20. 7 2 968 1 0. 5 90. 2 968 =5.068 = 10 =0. 9 05. 0 68 =36.296 10 Try
Note: this is the author’s suggestion
165.6 x
Try d = 1.75 b
b=228 mm say 230 mm d=399 say 30 mm
= S ummary: b = 230 mm d = 400 mm
=0.=2.02226 230 2400 049
=,
PR OB LE M 2.10 Design a singly reinforced rectangular beam for a 6-m simple span to support a superimposed dead load of 29 kN/m and a live load of 44 kN/m. Assume normal weigth oncrete with
= , ′ =34 , =345 . . Use
S OL UTI ON
_=24 0.3 0.6=4.32/ =1.=121.4 294. 3 2 1 4 48 / . = =546.516
Weight of beam: (this is the author’s assumption) Assuming a 300 mm x 600 mm,
=1.4 1.7 = 0.05 =0.821 =0.85 73430 5′600 = 0.8600 = =0.75 0.04369 .
.7(44)
5340.821600 = 0.8345600345 =0.04369 =0.03277>
= ′4 =0.00423 = ′ = 10.56 Assume d = 1.75 b
=
′ >31.36 = 0.03277345 34 =0.332 =34=9.00.8733210.590.332
(this is the author’s assumption)
Use b = 280 mm, d = 490 mm
=0.909.0871.75 10 =279.4 & =489
546.516 x
Minimum beam the thickness (Section 409.6.2.1)
ℎ = 16 0.4 700 ℎ = 600016 0.4 345700 ℎ =335 = =4496 =0.03277280 490 = = 44496 32 =5.6 6 Using 32 mm bars (#100):
280 mm
. .
h
≥ 25
ℎ=49025 3220 ℎ=554.5 > ℎ
Beam weight = 24 (0.28)(0.5545) Beam weight = 3.73 kN/m < 4.32(OK)
PR OB LE M 2.11 A propped cantilever beam shown in Figure 2.6 is made of reinforced concrete having a width of 290 mm overall depth of 490 mm. The beam is loaded with uniform dead load of 35 kN/m (including its own weight), and a uniform live load of 55 kN/m. Given Concrete cover is 60 mm from the centroid of the bars. Determine the required tension steel area for maximum positive moment. Assume EI=constant.
′ =
24 , =415 .
290mm
490 mm A
6m
B
2m
C
Fig ure 2.6 S OL UTI ON
′=415 =24 =290 =275 =490 =60 =35=55/ / =49060=430 =1.4 1.7 1.7 55 =1.=142.4 535/ Given:
O
=6
A
B
Moment Diagram
Solve for moment reactions using the three-moment equation:
= 142.5 21=285 2 ̅ ̅ =0 . 06 28560 0 =2498. 75 =0 Mo Lo +
D
2m C
x R
= = Maximum positive moment:
=0
= + 5′600 = 0.8600 =0.75 = ′ Solve for
:
== 10. 5 9
-489.75 = R(6)- 142.5(8)(4) R=676.875 kN
=142. 5 8 676. 8 75 =463.125 2 =0
142.5(2 + x) - 676.875 = 0 x = 2.75 m
=676.8752.75 142.5 .+ =253.828 85240.85600 = 0.415600415 =0.0247 =0.=0.7501852 0.0247 = 0.01852415 24 =0.3203 =4150. 3 203 1 0. 5 90. 3 203 =6. 2 35 290 430 =334.316
=300. =0.90334. 3 16 884 =253.828 < =.90 290430 = 10 =5.26
At a point of maximum positive moment:
(Singly reinforced)
253.828 x
= 0.8 5′ 1 1 0.85′ = 0.8415524 1 1 0.25.852426 =0.01495 =
=0.=1,081495 290 430 64
=
A NA LY S IS OF R E C TA NG UL A R S TE E L Y IE LDS ( PR OB LE M 2.12(CE MAY 1999)
B E A MS
WHE R E
=21 =
A reinforced concrete rectangular beam with b = 400 mm and d= 720 mm is reinforced for tension only with 6-25 mm diameter bars. If and
400 , ℎ :
a) The coefficient of resistance of the beam. b) The ultimate moment capacity of the beam.
S OL UTI ON
85′ = 0.600 =6 4 25 =2945 = = ′ = ′ 10.56 =
5210.85600 = 0.8400600400 =0.02276 2945 =0.01023< = 400720 = 0.01023400 21 =0.195 =210.19510.590.195 =. =0.903.62400720 =. Answer
Answer
PR OB LE M 2.13 A rectangular beam reinforced for tension only has b= 300 m, d = 490 mm. The tension steel area provided is 4,500 sq. mm. Determine the ultimate moment capcity of the beam in kN-m. Assume ,
′ =27 =275 .
S OL UTI ON
= .+ = = ′ = ′ 10.59 =
. = .+ =0.02276 , = = 0.0361275 27 =0.3118 =270.311810.590.3118 =6.87 =0.=. 906.8 7300490
PR OB LE M 2.14
=414 .
=325 ,′. =34.2 ,
A rectangular beam has b = 300 mm, d = 500 mm, grade 60 reinforcement ( Calculate the design moment
S OL UTI ON
=0.85 0.705 34.230 =0.82 = .+ = 4 25 3=1473 =
.. = .+ =0.03407
==0. 00982< = ′4 =0.00353 = ′ = 0.00982414 34.2 = 10.59 =34. 2 0. 1 188 1 0. 5 9 0. 1 188 =3.779 = =. =0.903.7 79300500 ′ =21 =275 Check if the beam satisfies the minimum requirement:
PR OB LE M 2.15
A 130-mm-thick-one-way slab is reinforced with 12-mm-diameter tension bars spaced at 110 on centers. Concrete cover is 20 mm, concrete strength MPa and steel yield strength . Unit weight of concrete is 23.5 kN/ . a) What is the ultimate moment capacity of the slab? b) If the slab is simply supported over a span of 4 m, what safe uniform live load pressure can the slab carry?
S OL UTI ON
a) Consider 1 m width of slab, b = 1000 mm Effective depth: d = h – cover- 1/2 d = 130-20-1/2(12)=104 mm
5210.85600 = 0.8275600275 =0.0378 = = 12 ==1028 1028 = = 1000104 =0.00989 = 1. 4 =0.00509 = ′ ==0.0.010989275 21 29 =′ 10.59 =21 0. 1 29 10. 5 9 0. 1 29 =2.511 = =0.=. 902.5111000104 5′600 = 0.8600
Check if the beam satisfies the minimum steel requirement on flexures:
b)
=
24.4=12. 43= 222/ =23. = 5 0.13=3.055 =1.4 1.7 12.222=1.4=1.3.0455 1 1.1.771 =.
Dead load pressure, Dead load pressure,
PROBLEM 2.16
x thickness of concrete.
A rectangular beam with b = 250 mm and d = 460 m is reinforced for tension only with 3-25 mm bars. The beam is simply supported over a span of 6 m and carries a uniform dead load of 680 N/m including its own weight. Calculate the uniform live load that the beam can carry. Assume and .
=276.5 ′ =20.7 =3 4 25 =1479 5′600 70.856005 = 0.8600 = 0.276.8520.5600276. =0.03703
S OL UTI ON
1.473 ==0. 01281< = 250460 )
Check if the beam satisfies the minimum steel requirement on flexure:
= 1. 4 =0.00506 5 = ′ = 0.01281276. 20.7 =0.171 =′ 10.59 =27 0. 1 71 10. 5 9 0. 1 71 =3.183 = =0.=151.903.561 83250460 6 = 8 151.56= 8 =33.68 / =1.4 1.7 33.6=. 8=1.40. 68 1.7 PR OBL E M 2.17 (CE J ANUAR Y 2008)
A reinforced concrete rectangular beam has a width of 300 mm and an effective depth of 55 mm. The beam is reinforced with six 25-mm-diameter tension bars. Steel yield is 415 MPa and concrete strength is 28 MPa. a) What is the balanced steel ratio? b) What is the maximum steel area for singly reinforced? c) What is the nominal moment capacity of the beam?
′
S OL UTI ON
a) Balanced steel ratio:
5′600 = 0.8600 =0.85
b) Maximum steel area
=
c) Nominal moment capacity Using 6-25 mm bars:
85280.85600 = 0.415600415 =0.028816 =.% ==0.0.7755 0. 0288163005 =,
= 4 25 6=2,945 2,300500 945 = = =0.01963< 0.01963415 = = ′ 10.59 = =280. =0. 2 91 28 5 9 0. 2 91 =6.749429110. = =6.7494300500 =.
PR OB LE M 2.18 2.18 A 350 mm x 500 mm rectangular is reinforced for tension only with 5-28 mm bars. The beam has an effective depth of 446 mm. The beam carries a uniform dead load of 4.5 kN/m (including its own weight), a uniform live load of 3 kN/m, and concentrated dead load of P and 2P as shown in Figure 2.7. Assume . Calculate the following: a) The ultimate moment capacity of the section in kN-m, and b) The maximum value of P in kN.
=
414 , ′′ = 3434..5
2P
2m
Fig ure 2.7 2.7 S OL UTI ON
=0.85 0.705 34.530 30 =0.818
P
2m
2m
5′600 = 0.8600 = 4 28 5 = 3079 =
34.50.0.81818600 = 0.855414600414 =0.03428 3079 == 300446 0.01972 <
Check if the beam satisfies the minimum requirement:
= ′ = ′ 10.59 =
= ′4 =0.00355 = 0.01972414 34.5 =0.2367 =7.=34.025 50.236710.590.2367 ==0.900.7.025 25300 300 446
1.4(2P)
1.4P
=1.44.5 1.73 =11.4/
A
B 2m
C 2m
D 2m
Figure 2.8 – 2.8 – Beam Beam with factored loads
For the given loads, the maximum moment can occur at B or C:
=1.42 11.421 421 = 149
=
At point C: Set
440.18 = 1.4P(2) + 11.4(2)(1)
4=171..1402.7 =2. 82 11.461 61
At point B: (First solve for
= 0 =
Set
8=17.17.1710..1 70.7 11.22411. 21421 440. 1=8= = 306.27
Thus the maximum value of P such that
will not exceed 440.18 kN-m is 149 kN .
≠
A N A L Y S I S O F R E C TA N G UL A R B E A MS S TE E L D O E S N O T Y I E L D S ( ) ) PR OB LE M 2.19 2.19
=414 .
WHE R E
= 6 32 , ′ =27.6 ,
A rectangular beam has b = 300 mm, d = 500 mm, grade 60 reinforcement ( Calculate Calculate the ultimate moment capacity of the beam.
S OL UTI ON
5′600 = 0.8600
= 4 32 6 = 4825 =
27.60.85600 = 0.85414600414 =0.0285 4825 = 300500 = 0.03217 >
From Eq. 2-18
=600 = =0 =600
=600 500 =0.85 ′ , = =0.85 4825600 − =0.8527.60.85300 =306. =484241, 9 64=0 2 =600 500306 306 =379.65 = =0. 8 5306. 2 =260.3 = 2 =0.904825379.65500 260.2 3 =.
PR OB LE M 2.20
=415 . ′ =21
A rectangular beam reinforced for tension only has b=300 mm, d = 490 mm. The tension steel area provided is 7-25 mm diameter bars with . Calculate the ultimate moment capacity of the beam.
S OL UTI ON
5′600 = 0.8600
= 4 25 7=3436 =
5210.85600 = 0.8415600415 =0.02161 3436 = 300490 =0.02337>
From Eq.2-18:
=600 − =0 = =600 =
=600 − =0.85 ′ , = =0.85 3436600 − =0.852210.85300 =296.24 2 4 =600 490296. 296.24 =392.43 < = =0. 8 5392. 4 3 =251.81 = . =0. 9 03436392. 4 3490 =.
A NA LY S IS & DE S IG N OF S ING LY R E INFOR C E D NON-R E CTANGULAR B E AMS PR OB LE M 2.21
345 ′ =21
Compute the ultimate moment capacity of the beam shown in Figure 2.9. Assume and .
=
S OL UTI ON
,
Note: This is not a rectangular beam. Some formulas derived above (such as , ) may not be applicable. The moment can be computed using the assumptions in the Code and the conditions of equilibrium.
= 4 32 4 =3217 =:
Solve for the balanced From Eq. 2-11
= + =
=
= + =396.825 =0.85396.825
=337. 3 =337.3375125125=110,863 345 =0.=0.885′21 110,863 =5,736
= Since
to determine whether the given steel yield or not.
<
, tension steel yields.
125= 0. 8 5 0.8521 =207. 3751255 =3,217345
= =0.90567.03 =.
= 2 1252 =0.8521207.5375625 207.2 5 =567.03
PR OB LE M 2.22
345 ′ =21
Compute the ultimate moment capacity of the beam shown in Figure 2.10. Assume and .
=
==1,412240 3 : 600 = 600 600375 = 600345 =238 = =202. =0.85238 4 = 375450 = 56 =168.7 =1/2 = = =17,066 = 345=0. =0.85′852117, 066 =883 < > < = ′ =
S OL UTI ON
Solve for
Since solve for c:
, tension steel does not yield (
0.85
=600 0.8521 125 =1140 600 = 7.4370.85 =684,00 − =250.92 = =0. 85250.92 =213.3 = 23 =0.85 23 =0.8521 213.29 375 213.3 =78.77 =0.90 78.77=. ′ =28 =345 =800 =1200 PR OB LE M 2.23
A hallow beam is shown in Figure 2.11. Assume and . a) Calculate the required tension steel area when . b) What is the balanced moment capacity of the beam? c) What is the maximum steel area under singly reinforced condition? d) What is the maximum design moment strength under singly reinforced condition? e) Calculate the required tension steel area when .
Fig ure 2.11-Hallow beam S OL UTI ON To guide us whether “a: will exceed 150 mm or not, let us solve the design moment when a=150 mm. d = 800 – 75 = 725 mm
= =0.=1044. 90 20.258528150725 =800 =800 <1044.25 , <150 . = = 2 =0.85′ 800 10 =0.90 0.85285007250.5 =111.6<150 =600 ℎ = =131.3 2 =600 725131. 131.3 =2,712 > = =0.85′ a)
Since the required
Assuming tension steel yields:
Check is steel yields:
345 =0.8528111.6500 =, b) Balanced condition (See Figure 2.12)
600 600725 =460.32 = 600 = 600345 = =0.8 5460.32 =391.3 =150=241. 2 7 =500150 =75,000 =725 21501 =650 =125241.27 =30,159 =725150 2241.1 27 =454.37 ==0. 285 2 =0.=1812.852852 75,000 6502 30,159 454.37 =0.90 1812.52=. )
Fig ure 2.12
= =0.85 2 345 =0.852875,002 30,159 =9,335 =0.75 =0.=, 759,335 : = 0.852875,0002 =7,001245 =13,244 =125 13,244=125 =105.95 =725 2105.150195 =522.03 ==0.8522 =0.=1489.852834 75,00 6502 13,244 522.03 =0.90 1189.34=. =1200<
c) Maximum steel area,
d) Maximum moment , Refer to Figure 2.12:
e)
Refer to Figure 2.12
=75, 0 00 =650 =125 =5750.5 =1200 10 =0.90 0.85 2 =0.90 0.852875,000650 21255750.5 1200 10 =53.04 = =75,0002 12553.04 =88,259.2 = =0.85′ 345 =0.=,852888,259.2 BE AM DEFLECTION PR OBLEM PR OB LE M 2.24 A reinforced concrete beam is 350 mm wide and 600 mm deep. The beam is simply supported over a span of 8 m and carries a uniform dead load of 11 kN/m including its own weight and a uniform live load of 15 kN/m. The beam is reinforced tension bars of 530 mm. Modulus of elasticity of concrete and a) Calculate the maximum instantaneous deflection due to service loads. b) Calculate the deflection for the same loads after five years assuming that 40% of the live load is sustained.
=21, ′ =20.6507 , =344. 8 , =2. 8 32 . =200 .
S OL UTI ON
Figure 2.13
: = 1 ≤ = ℎ 350600 = 12 = 12 =6300 10 = ℎ =1/2600=300 2. 8 32600 10 = 600/2 =59.472 = = 8 = =1115=26 / 268 = 8 =208 Effective moment of inertia,
Eq. 2-19
=
Moment of inertia of cracked section with steel transformed to concrete From Figure 2.13: Modular ratio,
= =9.238 =9.328 6 4 25 =27,208
Solve for c: Moment of area above N.A. = Moment of area below N.A. 350 x c x c/2 = 27,208(350-c) c = 219.7 mm
= = 3 350219. 7 = 3 27,208530219.7 =3,857 10 = 1 59. 4 72 59. 4 72 = 208 600 10 1 208 3,857 10 =3,914 10 < a) Instantaneous Deflection:
5 = 384
2 26 8000 = 38421,6503,914 10 =.
b) Long-term Deflection Since only 40% of the live load was sustained: w = 11 + 0.4(15) = 17 kN/m
= 5 17 8 1000 = 38421,6503,914 10 =10.7
Instantaneous deflection
Note: Since deflections are directly proportional to the load, the instantaneous deflection due to sustained load can be found by ratio and proportion using the result in Part”a”.
Long-term deflection =
17 = 16.2636 =10.7
= 150 ′ =2 =0 5ℎ 2 =2 = 1500 Long-term deflection = 16.36 + 2(10.7) Long-term deflection = 37.76 mm
PR OBL E M 2.25 (CE NOVEMBE R 2002) The continuous reinforced concrete beam shown in Figure 2.14 is subjected to a uniform service dead load of 16 k/m and a service live load of 32 kN/m,resulting in the bending moment diagram shown. Twenty percent of the live load will be sustained in nature, while 80% will be applied only intermittently. The concrete strength The modulus of elasticity of concrete is given by the expression and the modulus of rapture is given by the expression . Determine the following: a) The effective moment of inertia at the supports (maximum negative moment). b) The effective moment of inertia for the continuous member. c) The additional deflection (in addition to the initial deflection) after 5 years, under the sustained loading if the instantaneous deflection due to the combined service dead and live load is 5 mm.
17. =4700 2. =0.7
Fig ure 2.14
=
S OL UTI ON
=0.=4700′ =4700 1 7. 2 =19, 4 92 √ 70′ =0.7 √ 17.2 =2.903 =202
a) Effective moment of inertia at the supports
Maximum moment, Distance from NA of gross section to extreme tension fiber,
=310
=0. =0.0071500573 2. 9 030. 0 0715 1000 = = 10 =66.959 = 1 66. 9 59 66. 9 59 = 202 0.007151 202 0.00573 =. = 3 .409.6.2.4 Moment of inertia of gross section, Moment of inertia of cracked section,
b) Effective moment of inertia for the continuous member
At maximum negative moment (at support)
=0.0057817
=0.=620194=246 0138 =0.00513 Solving for
at maximum positive moment (at midspan)
=
2. 9 030. 0 0715 1000 = 310 =66.959 = 1 = 2 = 0.00578170.2 007932 =. = 150′ =0 ℎ =2 5 = 102 =2
c) Additional long term deflection= long term deflection x
Solving for the instantaneous deflection under sustained loading: Instantaneous deflection = 5mm (given) Instantaneous loading = 16 kN/m + 32 kN/m Instantaneous loading = 48 kN/m Sustained loading = 16 + 20%(32)
Sustained loading = 22.4 kN/m Sine deflection is directly proportional to the load:
22.4 = 485 =2.33
Additional long term deflection = 2.333 x =2.333 x 2 Additional long term deflection = 4.67 mm
ONE-WAY S LAB Reinforced concrete design slabs are large flat plates that are supported at its sides by reinforced concrete beams, walls, columns, steel beams, or by the ground. If a slab is supported on two opposite sides only, they are referred to a one-way slabs since the bending occurs in one direction only. If the slab is supported on all four sides, it is called two-way slab since the bending occurs in both direction. If a rectangular slab is supported in all four sides but the long is two or more times the short side, the slab will, for all practical purposes, act as one way slab, with bending occurring in the short direction.
A one-way slab is considered as a wide, swallow, rectangular beam. The reinforcing steel is usually spaced uniformly over its width. One way-way slabs are analyzed by considering one-meter strip, which is assumed independent of the adjacent strips. This method of analysis is somewhat conservative because we neglect the lateral restraint provided by the adjacent strips.
MA XIMUM SPA CIN G OF R E INFOR C E ME NT According to Section 407.7.5, the flexural reinforcement shall not be spaced farther apart than 3 times the slab thickness, nor 450 mm.
S HR INK A G E A ND TE MPE R ATUR E R E IN FOR C E ME NT,
Concrete shrinks as it hardens. In addition, temperature changes occur that causes expansion and construction of concrete. In this effect, the code (407.13) requires that one-way slab, where flexural reinforcement extends in one direction only, should be reinforced for shrinkage and temperature stresses perpendicular to flexural reinforcement. According to Section 407.132.2.1, the area of shrinkage reinforcement shall provide at least the following ratios of gross concrete area bh, (where h is the slab thickness) but not less than 0.0014. a) Where Grades 230 & 275 deformed bars are used………………..0.0020 b) Where Grade 415 deformed bars or welded wire fabric (plain or deformed ) are used…………………………………..0.0018 c) Where reinforcement with measured at
>415
yield strain of 0.35% are used………………………………………….
.
Shrinkage and temperature reinforcement may not be spaced not farther apart than 5 times the slab thickness, nor 450 mm (Section 407.13.2.2).
STEPS IN THE DESIGN OF ONE-WAY SLABS (FLEXURE) I. Identify the uniform floor pressure (Pa) to be carried by the slab. This load may consist of: 1) Live load pressure 2) Dead load pressure 3) Ceiling load and other attachments below the slab II. Determine the minimum slab thickness “h” from Table 2.1. If necessary adjust this value depending on your judgment. III. Compute the weight of slab (Pa) Weight = IV. Calculate the factored moment ( to be carried by the slab. Uniform load, . V. Compute the effective depth, d: d=h-covering (usually 20 mm)-1/2 (main bar diameter) VI. Compute the required steel ratio : Solve for from
VII.
ℎ = 1 = ℎ =1000 = 0.8 5′ 1 1 0.285′ , = = = 1000 ≥ ℎ = 1000 3 ℎ
Solve for If is less than and greater than , use If is greater than , increase the depth of slab to ensure ductile failure If is less than Compute the required main bar spacing.
Spacing,
Use the smallest of the following for the main bar spacing: a) b) c) 450 mm
VIII.
Temperature bars: See Page 81 for the required steel ratio,
= ℎ = 1000 5ℎ
Use the smallest of the following for temperature bar spacing: a) b)
c) 450 mm
ILLUS TR ATIVE PR OBLE MS Problem 2.36
′ =27.6 =276
Design a one-way slab having a simple span 3 m. The slab is to carry a uniform live load of 7,500 Pa. Assume and for main and temperature bars. The slab is not exposed to earth or weather. Use unit weight of concrete .
23.5 /
S OL UTI ON
=7.5 1=7.5 /
Consider 1 m strip of slab, b= 1000 m Uniform live load,
=
Minimum slab thickness from Table 2.1:
ℎ = 0.4 ℎ = 0.4 ℎ =119 120
Effective depth:
d = 120-20 mm (covering)-1/2 bar diameter (12mm) d=94 mm
= ℎ =23. 5 10. 1 2 2.82 / =1.4 1.7 =1.42.821.77.5
Weight of slab:
Factored floor pressure load:
= = 0. 8 5 = 1 1 0.85
=16.698 / = . =18.785 18.7=2.85 31062=0.90 100094 = 0.8527627.6 1 1 0.2852.327.626 =0.009039
= 1. 4 =0.00507 85′600 0.8527.60.85600 = 0.7 50.600 = 0.75 276600276 =0.037>0.009309 = =0.=850009039100094 1000 = 1000 =138 135 Check for
and
:
per meter width of slab
Using 12-mm main bars: Spacing s =
3ℎ=3120=360 450
Maximum spacing required by the Code: a) b)
Thus, us e 12 mm main bars at 135 mm o.c. Temperature bars: (Grade 275)
=0.002ℎ Spacing =
1000
=0.=2400021000120
= 1000 =327 325
5ℎ=5120=600
Maximum spacing required by the Code: a) b) 450 mm OK
Thus, us e 10 mm temperature bars at 325 mm o.c.
PR OB LE M 2.27
′ =21 =415
Design a one-way slab to carry a service live load of 4000 Pa. The slab has a length of 4m with both ends continuous. Assume and for main bars and for temperature bars. Steel cover is 20 mm. Unit weight of concrete is 23.5 kN/ .
=276
S OL UTI ON
=4 1=4 /
Consider 1 m strip, b = 1000 mm Uniform live load,
Minimum slab thickness from Table 2.1:
ℎ = 28
= ℎ =1.4 1.4
Weight of beam (DL):
ℎ = 400028 ℎ =143 150 =3.=23.525510.15 =1.=11.473.35525/ 1.74
Maximum factored moment, Section 408.4 (See Page 29) LL < 3 DL
Effective depth, d = 1.50 – 20 – 1/2 (12) Effective depth, d = 124 mm At midspan:
= 16 = = 0.8 5′ 1 1 0.285′ = 1. 4 =0.00337>0.0021
11. 7 35 4 = 16 =11.735 11.735 10 =0.90 1000124 =0.848 848 = 0.8415521 1 1 20.0.8521 =0.0021
= =0.00337 = 1000 Use
Spacing, s =
=0.003371000124 =418 = 1000 =271 270
3 ℎ=3150=450
Maximum spacing required by the Code: a) b) 450 mm
Thus , us e 12 mm bottom bars at 270 mm o.c. at mids pan At support:
= 10 = = 0.8 5′ 1 12 0.85′
11. 7 354 = 10 =18.776 18.776 =1.10357=0.901000124 357 = 0.8415521 1 1 21.0.8521
=0.0034> 5′600 =0.75 0.415600415 85210.85600 =0.75 0.8600 =0.0162>0.0034 =0.034 = =0.00341000124 =422 = 1000 = 1000 =268 265 Use
Spacing,
Thus , us e 12 mm top bars @ 265 mm o.c. at support
=0.002
Temperature bars (10 mm): (
=0.002ℎ Spacing, s =
1000
=0.=3000021000150 = 1000
5ℎ=5150=750
=261 260
Maximum spacing required by the Code: a) b) 450 mm
Thus , us e 10 mm temperature bars @ 260 mm o.c.
PROBLEM 2.28
=275
A one-way slab having a simple span of 3 m is 160 mm thick. The slab is reinforced with 12 mm tension bars spaced at 140 mm o.c. Steel covering is 20 mm. Calculate the uniform live load pressure that a slab can carry. Use ′ . Unit weight of concrete is 23.5 kN/ . SOLUTION
Consider 1 m strip of slab, b = 1000 m
=20.7
Dead load:
Effective depth:
Steel area,
= =23. ℎ 510.16 =3.76 d = 160 – 20 – 1/2(12) d = 134 mm
=
= 5′600 = 0.8600 = ′ =′ 10.59 = = 8
= 12 =807.8 807.8 = 1000134 =0.006028 0. 7 8 5600 = 0.8520. 275600275 =0.037> = 0.006028275 20.7 =20.70.080110.590.0801 =1.58 0. =20. 7 0 80110. 5 90. 0 801 =25.5334 3 25.5334= 8
=1.4 1.7 =
= 22.696 kN/m
22.696=1.43.761.7 =10.25 /
10.25 = Uniform pressure x 1 Uniform live load pressure = 10.25 kPa
S olved Pr oblems Us ing 2010 NS C P PR OB LE M 2.29
′ =30 =415
A reinforced concrete beam has width of 310 mm and an effective depth of 490 mm. , . Determine the following: a) The balanced steel area b) The maximum steel area for singly reinforced condition c) The maximum design strength if the beam is singly reinforced d) The required steel area if the beam is subjected to dead load moment of 120 kNm and live load moment of 170 kN-m.
′ >28 ; =0.85 0.705 ( 28)
S OL UTI ON Since
a) Balanced steel area:
0. 8 5 600) = (600
=0.85 0.705 3028 =0.836 5300.836600 = 0.8415600415
=
=0.03036 =0.= 0, 3036 310 490
b) Maximum steel area when when beam is singly reinforced:
= .... = + =0.0221 = =0.0221310490 = , : = = 0.004, = 800800 = ′ 1 1 = 14051 0.836 3630303103104901 1 143 0.836 = 558.05 − =0.650.25 − − =0.650.25 − =0.8145 From Eq. 2-24:
c) Maximum design strength, From Eq. 2-25 :
From Eq. 2-26:
d)
=0.= 8145558. 0 5 . =1.2 1.6 Thus, the beam is singly reinforced.
=1.= 4521.120 12045 1.6170
Determine if the beam is tension-controlled:
= ′ 1 1 = 451.45 ℎ , =0. 90 = = 0.8585 ′ 416 10 = 0.90 0.8530 530310 310490 490 = 139.06
From Eq. 2-22:
Since the required
Check if it is really tension-controlled:
the section is tension controlled.
= = 139.0.83606 = 166. 166.4 4 =600 =600 490166. > 1,000000 166.4 = 1,167167 PR OB LE M 2.30 2.30
=320 = 520 ′==16727 .. 5 ℎℎ = 1 8 0 , =34345 . =0.85 = 1.2 1.6 =1.2180 180 1.6167 = 483.2 = 14051 ′ 1 1 143 = 14051 0.8552727320320520 1 1 143 0.85 = 591.64 − =0.650.25 − = 487.31 > 459 ′ 1 = 1600 1 163 = 478.9 > < 0.90 = = 0.8585 ′ /2/2 Given the following data for a rectangular beam: width , , . Dead load moment moment
, effective depth Live load Live
S OL UTI ON
Solve for
to determine if compression steel area is required.
=0.8237
Solve for
Since
to determine if the section is tension-controlled.
, the section is within “transition region’, i.e 0.65 <
520 600 345 =0.650.25 1000 =0.650.25 1000345 = 119.084 0.2893 =0.85 483.2 10 = 119.084 0.2893 893 0.8527 5270.0.85320 5320525200 1/2 1/2 0.855 = 208.8 = = 177.45 = =0.85′ 34.5 =0.8505027 27177. 177.455320320 = 3,777777
PR OB LE M 2.31 2.31
′ = 21 = 414155
Given the following properties of a rectangular concrete beam: b = 280 mm, d = 480 mm, , . The beam is reinforced for tension only. Determine the design strength under the following conditions. a) When the beam is reinforced with three 25 mm diameter bars. b) When the beam is reinforced with four 25 mm diameter bars. c) When the beam is reinforced with seven 25 mm diameter bars.
S OL UTI ON
=0.85 0. 8 5 600) = (600
′ ℎ 28 8 5600 = 0.84155210. 600415 =0.0216
= 4 25 =490.87 =3 =1473 = a)
=
1473 = 280480 =0.01096< 0.85 = 0.8521280=1473415 =122.28
= =143.86 8 6 =600 =600 480143. 143.86 =1,402 >1,000 =0. 90 The section is tension-controlled,
=/2
=0.85′ /2 =0.8521122.28280480122.28/2 =255.87
=0.90255.87 =. =4 =1963 1963 = = 280480 =0.014961< = 0.85′ = 0.8521280 =1963415 =163.04 = =191.81 8 1 =600 =600 480191. 191.81 =901.5 <1,000 <0.90 5 415 =0.650.25 1000 =0.650.25 901. 1000415 =0.858 b)
)
The section within” transition region”, i. e 0.65 <
=/2 =0.=0.885′521163./2 0 4 280 480163. 0 4/2 =324. 5 04 =0.858324.504 =. =7 =3436 3436 = = 280480 =0.02557> =0.65 = =0.85′ 3436 600 − =0.85210.85280 =297.56 = =252. 9 2 =/2 =0.85 /2 =0.8521252.92280480252.92/2 =0.65446.91 =. c)
The section is compression-controlled,
PR OB LE M 2.32
′ =28 =345 =800
A hallow beam is shown in Figure 2.16. Assume and . a) Calculate the required tension steel area when b) What is the balanced moment capacity of the beam? c) What is the maximum steel area under singly reinforced condition? d) What is the maximum design moment strength under singly reinforced condition? e) Calculate the required tension steel area when .
=1200
S OL UTI ON This problem is the same as Problem 2.23.
=80075=725
To guide us whether “a” will exceed 150 mm or not, let us solve the design moment when a =150 mm.
= =176.47 =600 =1,865 >1000 , =0.90 =/2 =0.90 0.8528150500725150/2 =1044.225 a)
=800 =800 <1044.225 ,<150 . = =/2 =0.85′ /2 800 10 =0.90 0.85285007250.5 =111.6 <150 =600 ℎ = =131.3 2 =600 725131. 131.3 =2,712 > = =0.85′ Since the required
Stress in steel
345 =0.8528111.6500 =, =0.65 600 600725 =460.32 = 600 = 600345 = =0. 85460.32 =391.3
b) Balanced condition:
=150=241.27 =500150 =75,000 =7251/2150 =650 =125241.27 =30,159 =7251501/2 = 2 =0.85 2 =0.852875,000 6502 30,159 454.37 =1812.52 =0.65 1812.52 =.
241.27) = 454.37
2.17 = 37 =310.71 = =264.11 =150=114.11 =500150=75,000 =7251/2150 =650 =125114.11=14,263 =7251501/2 = =0.85 2 345 =0.852875,000 6502 14,263 =,
c) Maximum steel area,
(114.11)=517.95
= 2 =0.85′ 2 =0.852875,000 6502 14,263 517.95 =1511.9 800 =0.824 =0.650.25 1000
d) Maximum moment,
:
e)
=0.824 1511.9 =. =1200 < =0.85 2 =0.650.25 1000 =600 725 600 345 =0.650.25 1000345 = 166. 03 0.2893 =150=0. 8 5150 =125=106.2518,750 =7251501/2 =6500.425 Refer to Figure 2.17
=575-1/2(0.85c-150)
=0.85 2 1200 10 =166. 03 0.28930. 8 5 28 7 5, 0 00 650 2106.2518,7506500.425 =398.7
= 166.398.073 0.2893=0.706 =106.25398.718,750=23,615 = 345 =0. =0.858528 2 75, 0 002 23, 6 15 =8,432 PR OB LE M 2.33
=0.65
Design a singly reinforced rectangular beam to carry dead load moment of 110 kN-m (including self weight) and live load moment of180 kN-m. Use steel ratio and take Assume and
=1.9.
=276 ′ =21 .
S OL UTI ON
1.6180 =1.2 1.6 =1.=4202110 5′600 =0.03765 : =0.85 ′ <28 = 0.8600 =0.65 =0.02447 = ′ =0.322 =10.59 =5.473 600 = 600 =0.685
Note: For singly reinforced rectangular beam, is directly proportional to c.
=0. 6 5 =0.445 =600 =600 0.0.445445 =747.7 <1000 "" =0.650.25 747. 7 276 =0.650.25 1000 1000276 =0.813 = 420 10 =0.8135.4731.9 = =1.9= = =0.=,02447297564 =0. 5 =420 =0. =0.503765 =0.01883 = 10.59 =4.438 600 = 600 =0.685 Thus,
PR OB LE M 2.34
Repeat Problem 2.33 using a steel ratio
S OL UTI ON
=0. 5 =0.34247 34247 =600 =600 0.0.324247 =1152 >1000 , =0.90 = 420= 10=0.905.4731.9 =1.9= = =0.=, 01883308585 S UP PLE ME NTA R Y P R OB LE MS PR OB LE M 2.35
=250 ,=350 , =414 , ′ =20.7
A rectangular beam has . Determine (a) the maximum design moment if the beam is singly reinforced and (b) the required steel area if the beam is required to carry a dead load moment of 50 kN-m and a live load moment of 30 kN-m. Use the 2001 NSCP.
PR OB LE M 2.36 Repeat Problem 2.35 using the 2010 NSCP.
: =148.3 =1075
: =130.8 =1056 PR OB LE M 2.37
=0. 6 =276 ′ =28 :=250 ,=436 , =3,273
Design a rectangular beam reinforced for tension only carry dead load moment of 85 kN-m (including its estimated weight) and a live load of 102 kN-m. Use and use d= 1.75b. Assume and . Use the 2001 NSCP
PR OB LE M 2.38
Repeat Problem 2.37 using the 2010 NSCP.
PR OB LE M 2.39
∶=246 ,=430 , =3182 =320 =640 =400 ′ =25 350= =200,000
A reinforced concrete beam has the following properties: Use 2001 NSCP) beam with, effective depth, concrete strength, reinforcing steel, reinforcing steel modulus, service dead load moment a) If the beam is to be designed for a balanced condition, find the required area of steel area reinforcement, design balanced moment, and the corresponding service live load moment. b) Find the maximum steel area, the maximum design moment, and the
corresponding service live load moment if the beam is to be designed as singly reinforced.
: =5,5=4,49 162, ,=952. 4 4, =272 =775.46, =168 PR OB LE M 2.40
Repeat Problem 2.39 using the 2010 NSCP.
: 5,549 , =687.87 , =167.42 =3,963 , =677,7 =161 PR OB LE M 2.41
450 =525
=350 = ′ =24 =345 : =366.2
Calculate the ultimate moment capacity of a rectangular beam with , , . Assume . . Use 2001 NSCP
PR OB LE M 2.42
Repeat Problem 2.41 using the 2010 NSCP.
PR OB LE M 2.43
540 =728
: =366.2 =350 = ′ =24 =345 : =582.9
Calculate the ultimate moment capacity of a rectangular beam with , , . Assume , . Use 2010 NSCP.
PR OB LE M 2.44
Repeat Problem 2.43 using the 2010 NSCP.
PR OB LE M 2.45
500 =928
: =514.3 =300 = ′ =34 =414 : =729.6
Calculate the ultimate moment capacity of a rectangular beam with , , . Assume , . Use 2010 NSCP
PR OB LE M 2.46
Repeat Problem 2.45 using the 2010 NSCP.
: =522.5
CHAPTER 3
A nalys is and Des ig n of T-B eams and Doubly R einforced B eams
T-Beams Reinforced concrete floors usually consist of slab and beams, which are placed or poured monolithically. In this effect, the beam will have extra width on top (which is usually under compression) called flangers, and the resulting section is called a Tbeam. The beam may also be L-shaped if it is located at the end of slab.
A NA LY S IS A ND DE S IG N OF T-B E AMS WITH FLANGE IN COMPR ES SION Because of the huge amount of compression concrete when the flange of a T-beams is compression, the section is usually tension-controlled (extreme tension yields).
The compression block of T-beam may fall within the flange only or partly in the web. If it falls within the flange as shown in Figure 3.1 (a), the rectangular beam formulas in Chapter 2 applies since the concrete below neutral axis is assumed to be cracked and its shape has no effect on the flexure calculations. If however it cover part of the web as shown in Figure 3.1 (b), the compression concrete no longer consist of a single rectangle and thus the rectangular formulas do not apply.
B ALA NCE D AND MAXIMUM S TEE L AR E A A ND MOME NT The balanced value of “c” for any beam shape, as discussed in Chapter 2 is given by:
600 = 600
= 5′600 = 0.8600 = =0.85 2 =0.=0.7755 =0.85 2 and
If ”a” is less than the slab thickness, the formulas for rectangular beam may be used, o r
However, if “a” is greater than the slab thickness, the following formula will be used.
= Eq. 3-1 Eq. 3-2
=0.=85 = .+ − =0.75
DES IGN OF T-B EA MS WITH NEG ATIVE MOMENTS
When T-beams are resisting negative moments so that far their flangers are in tension and the bottom of their stems in compression, the formulas for rectangular beams can be applied. The following code requirements shall be applied for this case: 410.7.6: Where flangers of T-beam construction are in tension, part of the flexural tension reinforcement shall be distributed over an effective flange width as defined in Sec. 408.11, or width equal to 1/10 the span, whichever is smaller. If the effective flange width exceeds 1/10 the span, some longitudinal reinforcement shall be provided in the outer portions of the flange. The intention of this section is to minimize the possibilities of flexural cracks that will occur at the top face of the flange due to negative moments.
MINIMUM STE E L R A TIO For statically determinate T-section with flange in tension, the minimum steel area is equal to or greater than the smaller value of Eq. 3-3 and Eq. 3-4:
Eq. 3-3
Eq. 3-4
= =
CODE R E QUIR E ME NTS ON T-B E AMS (S E CTION 408.11) NOTE: THESE REQUIREMENTS ARE THE SAME WITH 2010 NSCP 1.
In T-beam construction, the flange and web shall be built integrally or otherwise effectively bonded together.
2.
The width of slab effective as T-beam shall not exceed 1/4 of the span of the beam, and the effective overhanging flange on each side of th web shall not exceed: a) 8 times the slab thickness, and b) 1/2 the clear distance to the next web.
3.
For beams with slab on one side only, the effective overhanging flange shall not exceed: a) 1/12 the span length of the beam, b) 6 times the slab thickness, and c) 1/2 the clear distance to the next web.
For Interior B eam
is the smallest of: 1. L/4 2. 3.
16 2 2
For E nd B eam
′
′ 6′ /2′
is the smallest of : 1. L/12 + 2. 3.
= =
For s ymmetrical interior beam (
is the smallest of: 1. L/4 2. 3. center-to-center spacing of beams 4. Isolated beams in which T-shape are used to provide a flange for additional compression area shall have a flange thickness not less than 1/2 the width of the web and an effective flange width not more than four times the width of the web.
16
≥ /2 ≤4
5.
Where primary flexural reinforcement in a slab that is considered as a T-beam flange is parallel to the beam, reinforcement perpendicular to the beam shall be provided in the top of the slab in accordance with the following: a) Transverse reinforcement shall be designed to carry the factored load on the overhanging slab with assumed to act as a cantilever. For isolated beam, the full width of the overhanging flange shall be considered. For other T-beams, only the effective overhanging slab needs to be considered. b) Transverse reinforcement shall be spaced not further apart than five times the slab thickness, nor 450 mm.
S TE PS IN F INDING THE TE NS ION S TE E L A R E A OF SINGLY RE INFORC ED T-BE AMS WITH GIVEN A ND OTHE R B E A M PR OP E R TIE S : I.
≤ > : . =0.85′
Solve for
max to determine of compression steel is nece ssary.
Follow the procedure in Page 105. If
If
II.
Solve for
, the beam is singly reinforced, proceed to Step II. , the beam is doubly reinforced
when a = t
Compressive force in concrete,
⁄ ==0. 2 85 ( 2)=______ >>,ℎ ,ℎ>, <, < if if
III.
Solve for a:
= = ∁ 2=0.85 ′ ⁄2 =______________ = =__ =0._ _8_5≥′ = = is the smaller value of:
IV.
= ==____________ : =_________________ = =0.85 ′ = ==0.85 =___________ = = ≤ is the smallest value of:
S TE PS IN FINDING OF SINGLY REINFORCED T-BEAMS WITH GIVEN A ND OTHE R B E A M PR OP E R TIE S : I.
Solve for balanced steel area to determine if tension steel yiel. Follow the procedure in Page 105. If
,tension steel yields. Proceed to step II
> : ℎ ℎ = = = 0. 85=________ ′ = = <> ,,<, >, < =_____ = ⁄ =2 = 2 > If
, tension does not yield
.
II.
Tension steel yields,
Compare If If
III.
Solve for a:
IV.
:
. Compute the area of compression concrete,
with the area of compression flange, proceed to Step III
=___________ = == =. ′ Solve for :
ILLUSTRATIVE PROBLE MS S OL VE D PR OB LE MS IN T-B E A MS US ING 2001 NS C P PR OBL E M 3.1 Determine the effective flange with for symmetrical T-beam with a span of 6 m. The beam width of web is 250 mm, the slab thickness is 120 mm, and the clear distance to adjacent beams is 3m.
S OL UTI ON 1. 2. 3.
For symmetrical T-beam, the effective flange width is the smallest of: 1/4 span = 6000/4 = 1500 mm
16 =16120250=2170 =3000250=3250 =
clear spacing of beams +
Therefore
PR OBL E M 3.2
=1200 ℎ ′ =30 =130 ℎ , =345 =470 =290 =
Given the following elements of a T-beam: Flange width, Flange thickness, Width of web, Effective depth, If the beam is reinforced for tension only, determine the ultimate moment capacity when the depth of compression concrete flange equals the flange thickness or .
ℎ =0.=8 ′ 2 =0.85 2⁄ =0.8530120130470 1802 =1611 =0.90 1611=
S OL UTI ON
Eq. 3-5
PR OBL E M 3.3
=900 ℎ ′ =20. 7 =110 ℎ, =414 =460 =310
Given the following elements of a T-beam: Flange width, Flange thickness, Width of web, Effective depth, If the beam is reinforced for tension only, determine the following: a) The balanced steel area b) The nominal and ultimate balanced moment capacity c) The maximum steel area d) The nominal and ultimate maximum moment capacity
=0.85 ′ ℎ 30
S OL UTI ON
a) Balanced condition
= 600600 =
600460 = 600414 =272.2 =0. 8 5272. 2 =231.4 >
==121. 4 == =900 110 =99, 0 00 =310 121. 4 =37, 6 22 = =136,622 = 414 =0. =0.85 8520.799,00037,622 =, → ⁄ =2 =405 =2⁄ =289.3 = =0.=0.885520.79 9, 000405 37,622289.3 = → Fig ure 3.3
=0.=.90897 → b) Maximum steel area and moment. Refer to Fi g ure 3.3.
=0.75 =0.=,755806 → =0.75 =102, =0.75136, 6 22 466 > ,ℎ > = =11.2 = =102,46699, 0 00=3, 4 66 2 =344.41 = =0. 85=0.20.78599,000 405 3, 466 289.3 =. → =0.=.90726. 5 → =820 =250 =470 , = 100 . ′ =20.7 =150, =120 =175 , =190 =0. ↑ 85 ℎ = =0.85′ 2=545.375 102,466=99,000 + 310(z)
PR OBLE M 3.4
A T-beam has the following properties:
Concrete compressive strength load conditions: a) b)
S OL UTION
,
, and steel area for the following
: 600 =278.11 = 600 < =236.39 > Balanced condition:
==136. 3 9 ===82, 0 00 =34,098 = =116,098 > ==0.782, 5 0=87, 0 73 00=5, 0 73 =20.29 = = 2 =359.85 =574. = 28 = 0,85′ FIGUR E 3.4
Maximum condition:
a)
=1.=1504 , =120 1.7 =414 < , ℎ , Since
"a" is less than t.
⁄414 10 =0.900.8520.78204702⁄ =0.85 2 =73.6 = =2,=0.58655 ′ 2 =646 =1059 =, ′
Minimum
Thus,
is the smaller of:
b)
Since
=175, =190 =1.4 1.7 =568 < , ,"" is more than
is more than t.
568 = 100 250470 10 =545. 3 750. 9 00. 8 520. 7 2 =15.78 = =3,946 = 414=0. =0.858520. 782, 0003946 =, =300 =550 = 450 ℎ, =350 .′ =27 , =415 . =0.=1.845 1.7 1.7350 =1.=12254450 PR OBL E M 3.5
Design a T-beam for a floor system for which and . The beams are 4.5 m long and spaced at 3 mo.c. The slab thickness is 100 mm.
S OL UTI ON
ℎ : 16 =16100 300=1,900 =1,125 ℎ ==100,=0. 9 0 =1161. =0.85′844 2 = + =325.123 = =276.355 > Solve for bf:
1. 2. 3.
L/4 = 1.125 m
center-to center spacing of beams = 3 m
Thus,
Solve for
Solve for
to determine if compression steel is needed.
1225 = 10 =1161. 844 10 0.90 0.8527300450 2 = =6975.02 = 415 =0. =0.858527 112, 5006,975.02 =6,607 =1033 =1937 z=23.25 mm
Minimum
is the smaller value of:
Thus,
=,
PR OBL E M 3.6
′=20.=2507 =
Determine the ultimate moment capacity of reinforced concrete T-beam with the following properties: Flange width b = 1500 mm, web width , effective depth d = 600 mm, slab thickness t = 100 mm. Assume and . The beam is reinforced with six 28 mm bars.
345
S OL UTI ON
: 600 =380.95 = 600 = =323. 8 1> ==22381 ===150, 0 00 =55,952 = 345=0. =0.85 8520. 7150, 0055,952 =10,503 =6 28 =3,695 > steel yields = = ′ = =3, 695345 =72,441< ℎ Solve for balanced
Steel area provided, Therefore,
0.85 0.85(20.7)
"a" is less than t
= =48.29 =0.85′ =0.28 520.748.291500600 48.29 2 =733. 9 9 =0.=.90733. 9 9 72,441 = a (1500)
PROBLEM 3.7 Given the following properties of T-beam: Flange width, Flange thickness, t=1200 Width of web, Effective depth, d = 580 mm Service deal load , Determine the safe service live load if the beam is reinforced for tension only with twelve (12) 28-mm-diameter bars.
=900 =345 ′ =21 =400 =410
SOLUTION
=0.=1285; 28=.90 =7,389 4
= =108,000 600 = 600 =368.25 ===193. =1 =313.0022 > = =77,206 = 345=0. =0.858521 108, 00077,206 =9,582 = 0.0.88521 5′ ==7, 389345 =142,813 > a> = =34,813 = =87. 0 3 = =520 2 = 2 =416.48 = =0. 85 =0.=1,82561.213 108,000520 34,813416.48 =0.=1135. 901,123861. 3 Solve for balance
:
Steel area provided is less than the balanced steel area. Steel yields.
142,813=108,000+
34,813 = 400z
=
1,135.=1.138=1. 4 1.44107 1.7 =.
PROBLEM 3.8 The section of a reinforced concrete T-beam is shown in Figure 3.5. The beam is reinforced with 10 32-mm-diameter tension bars with . Concrete strength . If the total service dead load moment on the beam is 330 kN-m, determine the safe service live load moment.
′ =32
4 32 =10 =8, 0 42 ==60,000 =0.=0.88365 0.705 3230 : 600 = 600 = 3 =313. =261. 8 3 > ==141.83 SOLUTION
Solve for balance
=415
= =45,385.5 = 345=0. =0.85852160, 0045,385.5 =6,907 > Since
, tension steel does not yield
=60, 0 00 = = =600 = = 8,0 42 =0. 60085− =0. 8 5 32 60, 0 00320 0. 8 36120 =327.95
= =261. 8 3 ==141. 8 3 =2⁄ =470 =
= = =49,23=332. 03 97 =0.=0.88553260,000470 49,303332.97 mm
=1,213.56
=1,=0.9001,92.22 13.56 = 1,0=1.92=1.441.33071. 7 =.
SOLVED PROBLEMS IN T-BEAMS USING 2010 NSCP PROBLEM 3.9 Repeat Problem 3.3 using the 2010 NSCP.
SOLUTION
=110 ′ =414 =20.7 =900 =0.=3210 =460 85 =0. 65 28 600 600460 = 600 = 600414 =272.2 Given:
ℎ
a) Balanced condition,
=
=0. 8 5272. 2 =231.4 >
Figure 3.6
==121. 4 == =900 110 =99, 0 00 =37, =310121. 4 6 22 = =136,622 = 414 =0. =0.858520. 799, 00037,622 =, → ⁄ =2 =405 =2⁄ =289.3 = =0.=0.885520.799,00040537, 622289.3 = → =0.=65897 → b) Maximum steel area and moment. Refer to Figure 3.6.
800 =0.815 = 37 =197.14 ; =0.650.25 1000 = =0. 8 5197. 1 4 =167.6 ==57. 5 71 =17, =2 = =31057. 6 8 47 ⁄ =321.21 = 414=0. 8=0.58599,00017,847 = → = =0.85 =0.=.8520. 799,00041517, 8 47321. 2 → =0.=.815806. 3 4 →
PROBLEM 3.10 Repeat Problem 3.2 using the 2010 NSCP.
=130 =1200 ′ =290 =30 =470 =345 =0.=0.885′5301200130470 2 130 2 =1611 SOLUTION Given:
Solving for :
=130 = =0. 85=155.0.7055(6 28)=0.836 =600 =1213 >1000 =0901611 =
",=0.90"
PROBLEM 3.11 Repeat Problem 3.4 using the 2010 NSCP. Additional questions: c) Find the required steel area if and . d) Find the maximum design moment so that section is tension-controlled if it is reinforced for tension only.
=195 =210
=820 ′ =20. 7 =250 =414 =470 =100 =0.85 ′ <28< =: = =0. 8=117. 5′ 6 52⁄ =605.97 =600 =1797>1000,=0.90 =545.375 : SOLUTION Given:
Solve for
Solve for
when
= 37 =201.43 800 =0.815 =0.650.25 1000 = =171.21 >
==71. 2 1 =2 = =17,⁄ 803.=334.6 39 = 0. 8 5′ =710. 7 2 =579 =1.=1502, =120 1.6 =372 < , a)
Since
is less than
,”a” is less than t.
=0. 90 =0.85′ 2 372=65. 1052 =0.900.8520.7820470 2 = =77. 0 8 = =600 =3, =0.05885′>1000 , , =0. 9 0 345=2,2=0.83 8520.765.52820 2′ =646 ′4 =1059 =, =1.=1752 , =190 1.6 =514 < , ℎ ,"" =0. 90 =0.85′ 2
Assume
Minimum
is the smaller value of:
Thus, b)
Since
Assume
is less than t.
514=93. 1053 =0.900.8520.78204702⁄ = =110.03
=600 =1,963 >1000, ,=0.90 = 345=0. =0.85′85 20. 793.53820 =, =195, =210 =1.2 1.6 =570 < , ℎ ,"" c)
Since
is more than t.
=0. 90 570 = 250470100 10 =545. 3 750. 9 00. 8 520. 7 2 =17.05 ==117.05 ;=⁄ =137.7 =1448>1000, ,=0.90 =600 = =3, 9 08 = =0. 8 5 = =3, =176. 666 25 ,=0.90 ==149.81 >
Assume
d)
==49. 8 1 ==12,453
= 2 =3450.9 = 0. 8 5 ′ =681.59 =. Note: If
is less than or equal to
, the beam is tension-controlled.
PROBLEM 3.12 Repeat Problem 3.6 using the 2010 NSCP.
=1500 ′ =20. 7 =250 =345 =600 =0. 8 5 =628 =3,694 = = + =323. =380.95 8 1> ===150, 0 00 =55,952 SOLUTION Given:
Solve for balanced
:
= 345=0. =0.858520. 7150, 00055,952 =10,503 =6 28 =3,695 < " " = Steel area provided, therefore ,
=
0.0.8855′20. 7 = =3, 695345 =72,441< ℎ ""
is less than t
= 72,=48. 441=1500 29 = =56.82 =600 =5,736>1000 =0.90 ⁄ =0.85′ =0. 82 48. 2 9 5 20. 7 48. 2 9 1500 600 2 =733.99 Solve for :
therefore
=0.=.90733. 9 9 PROBLEM 3.13 Repeat Problem 3.7 using 2010 NSCP.
SOLUTION
=900 ′ =21 =120 =345 =580 =400 =7,389 =410 =0.=1285; 28=0.9 0=7,389 = =108,000 600 =368.25 = 600 = =313. 0 2 > ==193. 0 2 = =77,206 = 345=0. =0.858521 108, 00077,206 =9,582 Given the following properties of a T-beam: Flange width, Flange thickness, Width of web, Effective depth, Service deal load,
Solve for balance
:
Steel area provided is less than the balanced steel area. Steel yields.
=
0.8521 ′ = =7, 389345 =142,813 >
0.85
= 142, 8 13=108, 0 00 =34, 8 13 813=400 = 34,=87. 03 = =520 2 = 2 =416.48 ==0.85 =0.=1,82561.216 108,000520 34,813416.48 ==203.03 = =243.57 Solve for
:
“a” >t
=600 =828.76 <1000 < <1000 , =0.8346 =0.650.25 1000 =0.=1,803461, 2 61. 3 52.703 = 1,052.=1.7203=1. .264101. 7 =. Since
Transition region
PROBLEM 3.14 Repeat Problem 3.8 using 2010 NSCP.
=10 4 32 =8,=0=60, 42 000 =0.=0.88215 0.705 3228 600 = 600 =313. 3 = =257. 3 5 > ==137. 3 5 ==43,953 = 345=0. =0.858521 60,00043,953 =6,813 >, SOLUTION
Solve for balanced
Since
:
=0.65 =60, 0 00 = == =600 = 530 =0.85 8,=329. 042 60027 =0.853260,0003200.8211200 = =270. 4 7 = =48, 1 51 ==150. 4 7 = =334. 7 6 =2⁄ =470 = =0.=0.885532 6 0, 0 0047048, 1 51334. 7 6 =1,205.48 =0.=783.651,562 05.48 = 78.5=1.6=1.2 23301.61. 6 =. DOUBLY REINFORCED BEAM
Occasionally, beams are restricted in small sizes by space or aesthetic requirements to such extent that the compression concrete should be reinforced with steel to carry compression. Compression reinforcement is needed to increase the moment capacity of a beam beyond that of a tensilely reinforced makes beams tough and ductile and reduces long-time deflection of beams. Compression steel also helps the beam withstand stress reversals that might occur during earthquakes. Continuous compression bars are also helpful for positioning stirrups and keeping them in place during concrete placement and vibration. Various tests show that compression reinforcement also prevents the beam to collapse even if the compression concrete crushes especially if it is enclosed by stirrups. According to Section 407.12 of NSCP, compression steel in beams must be enclosed by lateral ties, at least 10 mm in size for longitudinal bars 32 mm or smaller, and at least 12 mm in size for 36 mm and bundled bars. Deformed wire or welded wire fabric of equivalent area is allowed. The spacing of these ties shall not exceed 16 longitudinal bar diameters, 48 tie bar or wire diameters, or least dimension of the compression member.
ANALYSIS OF DOUBLY REINFORCED BEAM Doubly reinforced beam is analyzed by dividing the beam into two couples and as shown in Figure 3.8. is the couple due to compression concrete and the part of the tension steel is the couple due to the compression steel and the other part of the tension steel area
. .
′
Figure 3.8
′ = /′ ′
Compression reinforcement is provided to ensure ductile failure (i.e. tension steel must yield). For the reason, therefore, the stress in tension ( is always equal for . On the other hand, stress of compression steel ( may either be or below . This stress must always be checked.
′ ′ = ′ =600 −
If the compression steel yields, then , otherwise , where the stress of compression steel is given and is given by the following equation. (see derivation in page 137) Eq. 3-6
is
According to Section 410.4.3 of NSCP, for members with compression reinforcement, the portion of equalized by compression reinforcement need not be reduced by the 0.75 factor. Thus, the maximum permissible is:
=0.75′ = ′ =0.750. 85600 =0.75 600 = = =__________ =′ 10.59
Eq. 3-7
The expression 0.75
.
STEPS TO DETERMINE AND RECTANGULAR BEAM, GIVEN
I.
Solve for
and
OF ADOUBLY REINFORCED AND OTHER BEAM PROPERTIES
≤ > > If If
II.
design Singly Reinforced (See Chapter 2) design as Doubly Reinforced(proceed to step II)
= = == =______ =
Figure 3.9
Solve for Solve for
III.
,
and
Solve for the stress of compression steel Solve for a and c:
= =____ ′ = = =_____
0.85
′ ′/ = 0.003
′ =600 −
Eq. 3-8
′′ <≥ ′′ ≥= ′ = ′ <, ′
If If
proceed to IV proceed to V
IV.
then use
(compression steel yields)
V.
then use
(compression steel will not yield)
′ =
′,
STEPS IN FINDING OF A DOUBLY REINFORCED RECTANGULAR BEAM WITH GIVEN , AND OTHER BEAM PROPERTIES
There are three possible cases in doubly reinforced beams. Case 1: Both tension and compression yields (
=′ =
Case 2: Tension steel yields and compression steel does not (
=, ′ <
=, ′ <
Case 3: Tension steel does not yield compression steel yields. (
Note: For doubly reinforced beams with effective depth d=250 mm or more, it is not possible for both steels not to yield.
I.
II.
′ =
Assume compression steel yields (
=′ =______ = =___ = =___ ′ = = =___ =600 −<, ≥,
Solve for a and c (assuming tension steel yields):
0.85
Check:
If
If
III.
)
tension steel; yields, proceed to step III
tension steel does not yield, proceed to step IV
Solve for the stress in compression steel
′ =600 −
′′ ≥<,, ′ ≥, == = ′ <, ′ =600 − If If
IV.
Since
V.
If
proceed to step IV proceed to step V
compression steel does not yield.
From the stress diagram:
0.85′ =′ ′ = 0.85′ ′ 600 ′ = ′,′ =600 − =_____ =____ : = = 2 Solve for by equation formula. Solve for
Solve for a,a= Solve for
VI.
=.′ < ′ = = =0.6008−5′=0. ′ 8 5 ′ =___ _ _ ; = = = = 2 =.′ ( ) ______
DEEP BEAMS
According to Section 410.8 of the Code, beams with overall depth to clear span ratios greater than 2/3 for continuous spans, or 4/5 for simple spans, shall be designed as deep flexural members taking into account nonlinear distribution of strain and lateral buckling.
Beams with web depth that exceed 900 mm have a tendency to develop excessive wide cracks in the upper parts of their tension zones. According to Section 410..7 of NSCP, if the depth of web exceeds 900 mm, longitudinal skin reinforcement shall be uniformly distributed along both side faces of the member for distance d/2 nearest the flexural tension. The area of skin reinforcement per meter of height on each side face shall be Eq. 3-9
≥1.0750
The maximum spacing of the skin reinforcement shall not exceed the lesser of d/b and 300 mm. Such reinforcement may be included in strength computations if a strain compatibility analysis is made to determine stresses in the individual bars or wires. The total area of longitudinal skin reinforcement in both faces need not exceed one-half of the required flexural tensile reinforcement.
ILLUSATIVE PROBLEMS DESIGN PROBLEMS PROBLEM 3.15 a .305-mm wide rectangular beam has an overall depth of 560 mm. The beam is reinforced with four 25-mm-diameter compression bars. The centroid fiber. Assume and . Determine the following: a) The balanced tension steel area and the nominal and ultimate balanced moment. b) The maximum tension steel area and the nominal and ultimate maximum moment.
=
415 ′ =29 SOLUTION
=0.85 600 = 600 = =600 − = =
a) Balanced condition
600490 = 600415 =289.66 =0. 8 5289. 6 6 =246.21 =600 ..− =455 > 415=0. =0.85′8529246. 21305 yield
=4,460 =′ =1, =′964 = =, =2⁄ =0. 8 5′ 2 490 246.212 =0.8529=,. 246.21305 =0.901,021.4=. Balanced steel area,
b) Maximum tension steel area:
=0.75 =0.=3,754,345 460 =1,964 = =3,=, 3451,964 = 0.0.885′529 = 305 =3, 3 45415 =184.7 =/ =184. 7 /0. 8 5 =217.2 .− ′ =600 − ′′ =406. =600 7.<
According to Section 410.4.3, for members with compression reinforcement, the portion of equalized by compression reinforcement need not be reduced by the 0.75 factor.
=(2)=0.85′ ( 305490 ) 2 184. 7 =0.8529184. 7 2 49070 1964406. 7 =. =0.=. 90887. 45
PROBLEM 3.16 (CE NOVEMBER 2009)
A reinforced concrete beam has width of 300 mm and effective depth of 460 mm. The beam is reinforced with 2-28 mm compression bars placed 70 mm from extreme concrete. Concrete strength and steel strength . a) What is the balanced steel area considering the contribution of the compression steel? b) What is the maximum tension steel area allowed by the code?
′ =35
SOLUTION
=0.85 0.705 3530 =0.814 ′ = 4 28 2=1,232
=345
a) Balanced condition considering compression steel:
600 = 600345 600460 = 600 ′ =600 ′ ′ =600 29270 292 ′ =456 > ,ℎ ′ = =345 = 0.0.8855′35 0.′ ′ = 8 14 292 300 1232 345 = 345 =, b) Maximum steel area: For rectangular beams:
=0.=219.7505 =0.75292 =′ =178.37 219.0570 ′ =600 ′ =600 219.05 ′ =408 >,ℎ ′ = =345 = 0.0.88535178. 5′ ′37300 ′ = 1232 345 = 345 =, ′ =27.6 =276 PROBLEM 3.17
A rectangular beam has b=300 mm and d= 490 mm. Concrete compressive strength and steel yield strength . Compressive steel if required shall have its centroid 60 mm from extreme concrete fiber. Calculate the required tension steel area if the factored moment is 620 kN-m. SOLUTION This is the same problem in Chapter 2.
600 =335.616 = 600 = =285.27 2 =0.=0.885′527. 6 ⁄ 213. 9 6 300 490213. 9 62 =576. 7 6 =0.90576.76=519 =620 > , Solve for
Since
:
the beam must be doubly reinforced.
= =576.76 620 = ==112. 0. 9 0576. 7 6 13 =213. 9 6 =0.85′ 276=5456 =0.8527.6213.96300 = ′ : = =251.71 251. 7 160 ′ =600 ′ ′ =600 ′ =457 > 251.71
Note:
Solve for
Compression steel yields
′ = = Tension steel area, Compression steel:
′ =
112. 1=945 3 10= 27649060 = = ′ ′ = = ′ =
PROBLEM 3.18 A rectangular beam has b=310 mm and d=460 mm. The beam will be designed to carry a service dead load of 230 kN-m and service live load of190 kn-m. Compression reinforcement if necessary will have its centtoid 70 mm from extreme concrete fiber. Determine the required steel area. Use and
′ =30 =415 .
SOLUTION
=0. =1.854 1.7 =1.=64542301. 7 190 =0.75 600 =203.94 =0.75 600 Solve for : Note : For rectangular beams,
= =173.35 ⁄ =0.85′ 2 =0. 8 530173. 3 5310460173. 3 5/2 =511.58 =0.=460.90511. 5 8 42 =645 > , Since
= =511.58 = = 0.64590 511.58 =205.088 = =203. 9 4 =173.35 = 415=0. =0.85 ′8530173. 35310 =3,302 = 41546070 = 10 =1,267 Tension Steel:
205.088 x
= Compression steel:
′ =600 ′
=3, =, 3021,267 ′ =600 203.203.9470 94 ′ =394.06 ′ 394.′ =06 =1, 267415 ′ =
compression steel does nt yield,
′ =
PROBLEM 3.19 A floor system consists of a 100-mm concrete slab supported by continuous T beam with 9 m span, 1.2 m on centers as shown in Figure 3.10. Web dimensions, as determined by negative-moment requirements, are and Concrete cover is 70 mm from the centroid of the bars. The beam is subjected to a maximum positive factored moment of 1080 kN-m. Use . Unit weight of concrete is 23.5 kN/ a) Calculate the required tension steel area at the point of maximum positive moment. b) Using the tributary area method, what is the uniform service dead load acting on the beam? c) Calculate the uniform service live load acting on the beam.
.
=280 , =500. ′ =21 , =415
′=414 =21 =0.85
=500 =280 =70=1080 : =16100280=1. 8 8 =1.2 =1.2 600 600500 = 600 = 600415 = =0. 8 5295. 5 7 =251.23 SOLUTION
Maximum factored moment, Effective flange width, 1. L/4=9/4=2.25 m 2. 16t+ 3.
Use
Solve for to determine if compression steel is required. Balanced condition:
=
=251. 2 3100 =151.23 =1200 100=120, 0 00 =280151.23=42,345 = =162,345 =0.75 =0.=121,75162, 3 45 759 > Maximum condition:
= = =/2 =/2 = =0.90976.36
==1, =121,759759120, 0 00 1,=6. 759=280 28 =500100/2 =450 =396. =5001006. 2 8/2 86 =0.=0.885521120, 000450 1, 7 59 396. 8 6 =976.36
=878.72 =1,080 > , Since
= =/ ′ =600 ′
the compression reinforcement must be provided.
=1006. 2 8 =106.28 =106. 2 8/0. 8 5 =125.04 ′ =600 125.125.0470 04 ′ =264.1 < = =976.36 =
= 415=0. 0001,759 =0.85 8521120, =5,237 = == 976. 36 =223. . 64 = 223. 6=4 10 41550070 = =1,253 = =, ′ = ′ 264. ′ =1=1, 253415 ′ =1,969 Tension steel area,
Compression steel area:
b)
==23. 50.2516 =. ⁄ →
Dead load=weight of concrete: Area=1.2(0.1)+0.28(0.47)=0.2516
c) Uniform live load
Maximum positive moment (at midspan)
= 24 =1.4 1.7
9 1, =320 080= /24 5. 320=1. 4 9 1261. 7 =. ⁄ →
INVESTIGATION (ANALYSIS) PROBLEMS PROBLEM 3.20 The beam shown in Figure 3.11 is subjected to a maximum service dead load moment of 230 kN-m. Determine the service live load that the beam can carry. Use
20.7 =345 .
′ =
′ == 28 362=1, 4=4,232072
SOLUTION
Assume all steel yield:
=′=′ = =1, 232 = =2,840 = 0.0.8855′20. 7= 350 =2, 8 40345 =159. 1 = =187.18 600187. 1 8 =600 =1.=6003 23> 187. 18
=600 =407. =6007 >187. 187.1860 18
compression steel yields
Assumption is correct, all steel yield.
= = 2 = 2 159.1 =2,=739.8404 345 600 2 1,23234560060 =0.=90=1.739.44 =665. 4 3 2301. 1.7 665. 4 3=1. 4 7 =.
PROBLEM 3.21 A rectangular beam has the following properties: Width, b=400 mm Effective depth, d=620 mm Tension bars, 3 pcs 25-mm-diameter d’=70 mm Determine the design strength of the beam and the safe service live load if the service dead load is 320 kN-m.
′=415 =22
=10 4 28 =6,158 ′ =3 4 25 =1,473 =′ =1,473 SOLUTION
Assume all steel yields:
= =4,685
0.85 = 0.=260 8522400=4, 6 85415 = =305. 8 =600 ′ =616.5 > ′ =600 =463 > = ==( (2) ) 2 ⁄ 1, =4,=1288.6854156202602 4 7341562070 9 =0.=90=1.1288.4 91.=1,7160 1160=1.43201.7 =
All steel yields. Assumption is correct
PROBLEM 3.22 A 12-m long rectangular reinforced concrete beam is simply supported at its ends. The beam is provided with an addition support at the mid span. Width of beam is 300 mm and the overall depth is 450 mm. The beam is reinforced with 25-mm-diameter bars, four bars at the tension side and 2 bars at the compression side .Concrete protective
=415
=0.023.
′ =30
coverings is 70 mm form the centroid of the bars. Concrete strength and steel yield . Use 0.75 a) Determine the depth of the compression block. b) Determine the nominal moment capacity of the beam. c) Determine the factored uniform load, including its own weight, the beam can carry.
′=0.=30 85 =415 =4 4 25 ′=1963 =2 25 4 ′ =982 SOLUTION
=′ =982 = =982 = 0.0.885′53030=982415 = =53. 2 6 = =62.66 <70 = ′ 1963 =0.41585′=0. 85300. ′ 85300 ′ 982 600 − =98.87
Assuming all steel yields:
compression steel does not yield
Assuming tension steel yields and compression steel does not.
′ =600 98.98.8670 86 =175.17 < =600 =1, 7 06> ==. → ==0.8(5′ 2) =. → =0.=243.9053 c) Maximum factored uniform load:
By there-moment equation:
2 6̅ 6̅ =0 6̅= =0 6̅ = 4 =64 6 0260 =0 4 4 =. /
PROBLEM 3.23 (CE NOVEMBER 2010) A 6 meter long simply supported reinforced concrete beam has a width of 350mm and
an overall depth of 470 mm. The beam is reinforced with 2-28 mm compression bars on top and 4-28 tension bars at the bottom, each located 70 mm from the extreme concrete fiber. Concrete strength , and steel yield strength Determine the following: a) Depth of compression blocks assuming both tension and compression steel yields. b) What is the ultimate moment capacity of the beam in kN-m? c) Determine the additional concentrated live load that can be applied at midspan if the dead load including the weight of the beam is 20 kN/m.
415 .
′ =20.7
=6 ′ =20. 7 =350 =415 =400 =70 =28 = 28 4=2463 28 2=132 ′ = =0.85 =0. 90 =′ =1232 = =1232 = 0.0.885′520. 7= 350=1232415 = → =/ =83/0. 8 5 =97.64 97.97.6470 ′ =600 ′ ′′ =170 =600< 64 SOLUTION Given :
Tension steel area
Compression steel area,
Assuming tension & compression steel yields:
Thus, compression steel does not yield.
=
Since compression steel does not yield,
′ =600 −
0.8=5′ ′ ′ = 0.=130. 8520.700.8 853501232 600 70 =2463415 = =110.6 0 8 =600 =600 400130. 130.08 =1245> ′ =600 ′ ′ =600 130.130.0870 08 ′ =277.11 < = 2 =0.=0.885′520. 7110.62350400110. 6⁄21232277.1140070 =347.33 =0.=312.90347. 3 3 6 →
Assuming tension steel yields:
Ultimate moment capacity= Ultimate moment capacity=
= =312.6 =1.4 1.7 =1.4 1.7 1.7 312. 6 =1. 4 =. PROBLEM 3.24 A beam section is shown in Figure 3.12. The beam will be subjected to a maximum service dead load of 215 kN-m. What is the safe service live load moment for this beam? Use
′ =21 =415 .
=0.85 =8 25 =3,927 ′ =5 25 ′ =2,454 SOLUTION
Tension steel,
Compression steel,
=30 =42.5 =65030 2125 =607.5 =65030251/225 =582.5 ′ =21 215 =415 =′ =2, 4 54 = =1,473 = 0.0.8855′21 = 360 =1. 4 73415 =95.1 =/ =95. 1 /0. 8 5 =111.9 ′ =600 − ′ =600 ..−. ′ =372 < Effective depth to extreme tension bar:
Effective depth (to centroid of tension bar)
Assume all steel yields:
Compression steel does not yield.
: ℎ ℎ ℎ . ℎ, ℎ ℎ ℎ ℎ . = ′ =0.=0.885′5 ′ ′ − ′ 600 3,2,942754415 600 =0.−.85210.85360 =122.38 .−. ′ =600 − ′=391. =60064.< = =104. =0.85122. 3 8 03 = 2 =0.85 ′ 2′ 104.03 2, 4=0.54391. 852164582. 104.03542.3605582. 5 2 =873. 6 8 =0. 9 0873. 6 8 =786.31 = 786. =1.31=1.4 42151. 1.7 7 =. PROBLEM 3.25
′ =
A beam section is shown in Figure 3.13. The beam will be subjected to a maximum service dead load of 360 kN-m. What is the safe service live load moment for this beam? Use
21 =415 .
=0.85 =10 28 =6,158 ′ =2 25 ′ =982 ⁄ =42.5 =301225 =65030281/228 =578 ′ =21 =360 =415 =′ =982 = =5,176 = 0.0.8855′21 = 320 =5, 1 76415 =376.04 =/ =376. 0 4/0. 8 5 =442.4 ′ =600 ′ ′ =600 442.442.442.4 5 =222 < Tension steel,
Compression steel,
Effective depth (to centroid of tension bars)
Assume all steel yields:
Tension steel does not yield.
:ℎ ℎ ℎ ℎ ℎ . ℎ, ℎ ℎ . =6503014=606 =65030282814=550 = =5 4 28 =3,079 = ′ =0.85 ′ ′ 600 600 =0.85′ ′ 550 3,=0.07985216000.606 3, 0 79 600 85320 982415 =363. 9 ==309.29 9 =600 =600 606363. 363.9 =399. 2 5 < 550363. 9 =600 =600 363. 9 =306. 9 < 363.363.942.9 5 ′ =600 ′ ′′ =530 =600 > = =0. 8 5345. 4 =301.2 = =3,=1,0279399. 2 5 29.2 Solve for d:
= =3,=944.0799306.9 2,174.=1 =1, 229.2 606 944.9550 =581.66 = 2 =0.=0.8855′21 309.292′ ⁄ 320 578309. 2 92 982415581. 6 642. 5 =974.07 =0.=876.90947. 0 7 65 = 876. =1.65=1.4 43601. 1.7 7 =. 27 =350.
PROBLEM 3.26 Calculate the design flexural strength of the T-beam shown in Figure 3.14. Use
.
′ =
=0.=1085 25 =4,909 4 ′ =3 4 22 =600 =1,140110 =66, 000 =11039020102512 =251012⁄ 22=46 ⁄ 25 =432.5 =′ =1, 1 40 = =3,768 = 0.0.885′527 ==3, 768350 < ℎ < =57, 4 68 = 57,=95. 469=600 8 < = ⁄ =95. 8 /0. 8 5 =112.7 112.112.7467 ′ =600 ′ ′′ =600 =355 > SOLUTION
Flange area,
Assume all steel yields:
Area of compression concrete:
=600
432.112.5112.7 7 =600 =1, 7 03> ⁄ 12925 ⁄ =407.5 =1225 =600 =1,567 > = 2 2 =0.=0.885′52795. 8600432. 5 46 1,=661. 140350432. 5 46 5 =0.=.90661. 5 Verify if the upper layer of tension steel yields
All steel yields, assumption is correct:
PROBLEM 3.27 Calculate the design flexural strength of the T-beam shown in Figure 3.15. Use
25 =345 .
′ =
