Technical report Shear capacity verification Lightened ...€¦ · 6.1. According to rules CIRSOC...

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Technical report Shear capacity verification

Lightened slab with spheres V version

02

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Index I. Purpose of this technical report 2 1. Materials 3 2. Weights 3 3. Models 4 4. Requests 5 4.1. Moment of inertia 5 4.2. Shear 7 5. Moment of inertia verification 8 6. Shear verification 8 6.1. According to rules CIRSOC 201 8 6.2. According tests 10 6.3. Mörsch’s grid with slashes 11 7. Conclusion 12

I. Object of this technical report To analyze the behavior of a slab lightened with spheres functioning as mezzanine without beams with separations between columns of 9 m. To this end, patterned cloth corner with 2 edges wardrobes and the opposite edges with an overhang of 1 m. The slab under study is 35 cm total thickness and spheres to lighten 27 cm in diameter, spaced every 30 cm center to center.

Outline (dimensions in mm):




1. Materials tf1000kgf Takes quality concrete H-38 and steel ADN-420

fc 38.MPa

τ011 0.5MPa step bottom frame

τ12 2.70.MPa

fy 420.MPa

2. Weights qpp 541·kgf·m)–2 Dead weight of slab

qDL 500·kgf·m)–2 Dead weight

qLL 500·kgf·m)–2 Beam overload

qTOT qpp qDL qLL




3. Models Solicitations to be determined without a mezzanine beams with columns each 9 m and a cantilever edge of 1 m. For verification the moment of inertia of the section has to be determined in order to analyze the viability of the established section. It is determined by the method of yield lines, fixing the relationship between the moment of inertia at the support and at the span to 1. The shear stress is determined at a distance r from the central column supporting the scheme shown (inner column for a soft corner) Calculated stresses values are matched against a finite element model. The scheme described is modeled with four-node shell elements of 25 cm side. The columns are considered specific support, and top and left edges prevents the shift to represent continuity with the adjacent tiles. The values yielded by the MEF are highlighted in yellow, they do not not differ significantly from those calculated.

Scheme:




M.E.F., supports

M.E.F., elastic deformations (x250)




4. Requests 4.1. Moment of inertia The moment of inertia is calculated at the span and at the support by the yield lines method (Kennedy, Goodchild; Practical Yield Line Design; British Cement Association 2003)

Lvol 1m Overhang length

L 9m Support distance

Lvol 1m Overhang length

L 9m Support distance

Mvol 0.5·qTOT·Lvol² Moment of inertia of the overhang (isostatic)

Mvol 0.771· tf ·m/m

i1 1 It is fixed for the plastic calculations the relationship in between the moment of

inertia at the stretch and the support.

i2 0.0742

Mtr: = qT·T·L² / 2 √1 i1 √1 i2²

Mtr 10.392tf / mAverage moment of inertia to resist at the span.

i1·Mtr 10.392 tf / m Average moment of inertia to resist at the support.

i2·Mtr 0.771 = tf ·m / m Cantilever moment of inertia (must match the calculation for the isostatic)




El M.E.F. throws for a central strip of the section:

Mtrmax 11.58 Mtrmin 8.38

So the result is accepted




4.2. Shear verification

The shear is calculated by areas of influence. An increase in the distance to span is taken into account considering it is a case of a central column in a corner.

P(r) qTOT [(L·1.03)2 π r2] load transmitted to the column

V(r): = P(r) / π·2r

Values:

V(1.50m) 12.895·tf ·m-1

V(2.50m) 6.504·tf ·m-1

V(1.75m) 10.695·tf ·m-1

V(2.75m) 5.545·tf ·m-1

V(2.00m) 8.997·tf ·m-1

V(3.00m) 4.714·tf ·m-1

V(2.25m) 7.633·tf ·m-1

The results of M.E.F. 1.5m coincide are up to 10% higher than 3 m.




5. Moment of inertia verification The value of stretch is calculated to verify that it is a reasonable number. We work with a diagram of rectangular stress.

h 35cm total thickness of the slab

d h 3cm useful height

altura útil

Mu 1.75·Mtr compression block height

a : = d - √ d² - (2/0.85 ·fc) · Mu

a 1.775·cm remains to be seen outside the lightened area

As : = a· (0.85·fc / fy)

As 13.648· (cm²/m) approximated amount is considered a reasonable value (e.g. 2ϕ12c/15)

6. Shear verification The aim of this work is to verify the shear capacity in the most required lightened section. The solid area of the slab is not analyzed.

6.1. According to CIRSOC 201 (1982) regulation This type of lightened slabs are not alleviated within the regulation laid down. It could be adopted in the case of extremely conservative results, using only the minimum width of the nerve, it will be considered for every 30 cm width 3 cm of useful section, therefore 10% capacity of a solid slab of equal thickness. The capacity of the slab will:

z : = d –(a/2) elastic arm

z 0.311m

VR_1 (z) ·τ011 (3cm/30cm)

VR_1 1.586·tf ·m-1

value of resistance without armor. Very low value, which would require a complete solid slab.

VR_2 (z) ·τ12· (3cm/30cm)

VR_2 8.566·tf ·m-1

value of resistance with shear armor as equation 25 of the Regulation A graph for the given equation is obtained by intersecting the distance according to the column (corrected linearly to better suit the MEF) and the resistance value obtained




The solicitation is less than the resistance from the 2.25 m column. La solicitación es menor que la resistencia a partir de los 2.25 m de la columna. Conclusion

Adopting this approach should increase similarly solid section that for the above (section 6.1) but not necessary armor cut in the lightened.

6.3. Mörsch’s grid with slashes If compressed rods are placed vertically on the section between fields, you can download the tension in the concrete caused by shear and reduce the solid area. A 1.50 m of the column will:

β: = atan (z/30cm) tensile rod tilt

β: 46.043·deg

α 90deg compressed rod tilt

Z: =V(1.50m)/sin(β) Z 17.913·tf ·m-1

tensile stress on the crank

D: = V(1.50m)/sin(α) D 12.895·tf ·m-1

rod effort in the compressed

AC: = π [(30cm·√2 27cm)/2]² section of a vertical cylinder tangent to the spheres

AC 186.904·cm2

D·30cm/AC 2.03·MP average stress in the compressed cylinder For each forward slash is required

Z·30cm 5.374·tf force to be taken by the slash

(Z·30·cm·1.75)/fy 2.196 cm² minimum section of the bar, take 2φ12

Conclusion

Setting the solid section in a typical value of L / 6 should be taken cutting efforts This slash is not required to increase the massif. As low shear stress can be reduced until it becomes inclined armor unnecessary according to the criterion described in 6.2. For the case studied inclined reinforcements are needed (measured from the column) from 1.50 m in solid ending to the 2.25 m.




7. Conclusion For the selected section was determined average longitudinal reinforcement and the height was obtained section is reasonable. Having done this, we analyzed three criteria for determining the shear strength. If you want to take a purely normative and very conservative, Section 6.1 shows that taking a solid of L / 4 for the case analyzed and vertical cutting armor lightened almost the entire area can meet the requirements of the regulation. If you want to take a test criterion and the theory of rods, you can choose according to 6.2 by a clump of L / 4 and not wearing armor cutting lightened area, or set the placing solid slashes solve according to 6.3 in areas that the effort cut is high (in this case between the solid and L / 4), and set aside areas that meet reinforcements the criterion of 6.2 (from L / 4). Attached at the proposed distribution of the reinforcements and details. PLANS SUBMITTED -PLAN SCHEME -REINFORCEMENT DISTRIBUTION SCHEME -DISTRIBUTION OF REINFORCEMENT DETAIL -DETAIL OF REINFORCEMENT


Technical report Shear capacity verification Lightened ...€¦ · 6.1. According to rules CIRSOC...

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