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Análisis de opciones de políticas

Evaluación de impacto del Programa Nacional de Apoyo Directo a los más Pobres (Juntos) en la provincia

4. Conclusiones

3.4 Análisis de opciones de políticas

In the previous discussions, beams of rectangular section (which are most common) and with tensile steel alone (‘singly reinforced’) were considered, for the sake of simplicity. The procedure of analysis is similar for other cross-sectional shapes.

Frequently, rectangular sections of beams are coupled with flanges ⎯ on top or bottom. If the flanges are located in the compression zone, they become effective (partly or wholly) in adding significantly to the area of the concrete in compression.

However, if the flanges are located in the tension zone, the concrete in the flanges becomes ineffective in cracked section analysis.

T−beams and L−beams

Beams having effectively T-sections and L-sections (called T-beams and L-beams) are commonly encountered in beam-supported slab floor systems [refer Figs. 1.10, 4.14]. In such situations, a portion of the slab acts integrally with the beam and bends in the longitudinal direction of the beam. This slab portion is called the flange of the T- or L-beam. The beam portion below the flange is often termed the web, although, technically, the web is the full rectangular portion of the beam other than the overhanging parts of the flange. Indeed, in shear calculations, the web is interpreted in this manner.

When the flange is relatively wide, the flexural compressive stress is not uniform over its width. The stress varies from a maximum in the web region to progressively lower values at points farther away from the web. In order to operate within the framework of the theory of flexure, which assumes a uniform stress distribution across the width of the section, it is necessary to define a reduced effective flange.

The term ‘shear lag’ is sometimes used to explain this behaviour. The longitudinal stresses at the junction of the web and flange are transmitted through in-plane shear to the flange regions.

The resulting shear deformations in the flange are maximum at the junction and reduce progressively at regions farther away from the web. Such ‘shear lag’ behaviour can be easily visualised in the case of a rectangular piece of sponge that is compressed in the middle.

The ‘effective width of flange’ may be defined as the width of a hypothetical flange that resists in-plane compressive stresses of uniform magnitude equal to the peak stress in the original wide flange, such that the value of the resultant longitudinal compressive force is the same (Fig. 4.14).

effective flange width bf

BEAM-SUPPORTED FLOOR SLAB SY STEM

s1 s2

bf s1/2 + bw /2

flange

w eb bw

Df

L-BEAM T-BEAM

bw

(s1 + s2)/2 bf

d bf

Equivalent flange w idth Actual distribution of

compressive stress (total force = C)

Assumed uniform distribution (total force = C)

Fig. 4.14 T-beams and L-beams in beam-supported floor slab systems The effective flange width is found to increase with increased span, increased web width and increased flange thickness. It also depends on the type of loading

(concentrated, distributed, etc.) and the support conditions (simply supported, continuous, etc.). Approximate formulae for estimating the ‘effective width of flange’ bf (Cl. 23.1.2 of Code) are given as follows:

For example, it is seen that the equivalent flange width is less when a concentrated load is applied at the midspan of a simply supported beam, compared to the case when the same load is applied as a uniformly distributed load.

is the “distance between points of zero moments in the beam” (which may be assumed as 0.7 times the effective span in continuous beams and frames).

Obviously, b l0

f cannot extend beyond the slab portion tributary to a beam, i.e., the actual width of slab available. Hence, the calculated bf shouldbe restricted to a value that does not exceed (s1+s2)/2 in the case of T−beams, and s1/2 + bw/2 in the case of L−beams, where the spans s1 and s2 of the slab are as marked in Fig. 4.14.

In some situations, isolated T−beams and L−beams are encountered, i.e., the slab is discontinuous at the sides, as in a footbridge or a ‘stringer beam’ of a staircase. In such cases, the Code [Cl. 23.1.2(c)] recommends the use of the following formula to estimate the ‘effective width of flange’ bf:

b

where b denotes the actual width of flange; evidently, the calculated value of bf

should not exceed b.

Analysis of T−beams and L−beams

The neutral axis may lie either within the flange [Fig. 4.15(b)] or in the web of the flanged beam [Fig. 4.15(c)]. In the former case (kd ≤ Df), as all the concrete on the tension side of the neutral axis is assumed ineffective in flexural computations, the flanged beam may just as well be treated as a rectangular beam having a width bfand an effective depth d. The analytical procedures described in Sections 4.6.1 and 4.6.3, therefore, are identically applicable here, the only difference being that bf is to be used in lieu of b.

In the case kd > Df , the area of concrete in compression spreads into the web region of the beam [Fig. 4.15(c)]. The exact location of the neutral axis (i.e., kd) is determined by equating moments of areas of the cracked-transformed section in tension and compression [Eq. 4.31] and solving for kd:

bf

(a) section (b) transformed section

(N.A. in flange)

Fig. 4.15 Example 4.4 — Cracked section analysis (WSM) of a T-beam

(

bfbw

)

Df

(

kdDf 2

)

+bw

( )

kd 2 2 = mAst

(

dkd

)

(4.31) This is valid only if the resulting kd exceeds Df .

With reference to the stress distribution shown in Fig. 4.15(d), the area of the concrete in compression can be conveniently obtained by considering the difference between the two rectangles bf ×kd and

(

bf bw

) (

× kdDf

)

. Accordingly, considering equilibrium of forces (C = T),

( )

c

(

f w

)(

f

)

st st Also, taking moments of forces about the centroid of tension steel,

( )( ) ( )( ) { ( )

3

}

then fcmay be determined first by solving Eq. 4.34 (after substituting Eq. 4.33) and fst can then be determined either by solving Eq. 4.32, or by considering similar triangles in the stress distribution diagram (Eq. 4.21). On the other hand, if the

problem is one of determining the allowable moment capacity (Mall) of the section, then it should first be verified whether the section is ‘under−reinforced (WSM)’ or

‘over−reinforced (WSM)’ ⎯ by comparing the neutral axis depth factor k with kb

(given by Eq. 4.23). If k < kb, the section is ‘under−reinforced (WSM),’ whereby fst

= σst. The corresponding value of fc can be calculated using the stress distribution diagram. On the other hand, if k > kb, the section is ‘over-reinforced (WSM)’, whereby fccbc. Using the appropriate value of fc in Eq. 4.34, Mall can be determined.

EXAMPLE 4.4

An isolated T-beam, having a span of 6 m and cross sectional dimensions shown in Fig. 4.15(a), is subjected to a service load moment of 200 kNm. Compute the maximum stresses in concrete and steel, assuming M 20 concrete and Fe 250 steel.

SOLUTION

• It must be verified first whether the actual flange width b = 1000 mm is fully effective or not. Applying Eq. 4.30(b) for isolated T-beams with l0 = 6000 mm

4 250

• First assuming [Fig. 4.15(b)], and equating moments of compression and (transformed) tension areas about the neutral axis,

kdDf

• Taking moments of forces about the tension steel centroid,

(which, incidentally, is less than the permissible stress σcbc = 7.0 MPa for M 20 concrete).

• Alternatively, from the stress distribution diagram [Fig. 4.15(d)]

f mf d kd (which, incidentally, is less than the permissible stress

σ

st = 130 MPa )

EXAMPLE 4.5

For the T-beam problem in Example 4.4, determine the allowable moment capacity.

SOLUTION

• From the previous Example, the neutral axis depth factor k =210.9/520 = 0.4057. substituting in Eq. 4.34,

6.66 850 210.9 (520 210.93)

The answer could have been easily obtained using the result of Example 4.4, and making use of the linear elastic assumption underlying WSM. A compressive stress fc = 5.95 MPa results from a moment M = 200 kNm.

Hence, the allowable stress fc = 6.66 MPa corresponds to a moment

⎟⎠

⎜ ⎞

×⎛

= 5.95 66 . 200 6

Mall = 223.9 kNm (as before).