3.2 Reinforced Concrete Slabs - vsvu.sk · Figure 3.2-1: One –way slab, two-way slab, ribbed...

Sabah Shawkat Cabinet of Structural Engineering 2017

3.2 Reinforced Concrete Slabs

Slabs are divided into suspended slabs. Suspended slabs may be divided into two

groups:

(1) slabs supported on edges of beams and walls

(2) slabs supported directly on columns without beams and known as flat slabs. Supported

slabs may be one-way slabs (slabs supported on two sides and with main reinforcement in one

direction only) and two-way slabs (slabs supported on four sides and reinforced in two

directions). In one-way slabs the main reinforcement is provided along the shorter span. In order

to distribute the load, a distribution steel is necessary and it is placed on the longer side. One-

way slabs generally consist of a series of shallow beams of unit width and depth equal to the

slab thickness, placed side by side. Such simple slabs can be supported on brick walls and can

be supported on reinforced concrete beams in which case laced bars are used to connect slabs

to beams.

Figure 3.2-1: One –way slab, two-way slab, ribbed slab, flat slab, solid flat slab with drop

panel, waffle slab

In R.C. Building construction, every floor generally has a beam/slab arrangement and

consists of fixed or continuous one-way slabs supported by main and secondary beams.


Figure 3.2-2: Solid flat slab, solid flat slab with drop panels

The usual arrangement of a slab and beam floor consists of slabs supported on cross-

beams or secondary beams parallel to the longer side and with main reinforcement parallel to

the shorter side. The secondary beams in turn are supported on main beams or girders extending

from column to column. Part of the reinforcement in the continuous is bent up over the support,

or straight bars with bond lengths are placed over the support to give negative bending

moments.

Figure 3.2-3: Types of the reinforced concrete slab systems


3.2.1 Flat Slabs

Flat plate is defined as a two-way slab of uniform thickness supported by any

combination of columns, without any beams, drop panels, and column capitals. Flat plates are

most economical for spans from 4,5 to 7,5m, for relatively light loads, as experienced in

apartments or similar building.

-A flat slab is a reinforced concrete slab supported directly on and built monolithically with the

columns, the flat slab is divided into middle strips and column strips. The size of each strip is

defined using specific rules. The slab may be in uniform thickness supported on simple

columns. These flat slabs may be designed as continuous frames. However, they are normally

designed using an empirical method governed by specified coefficients for bending moments

and other requirements which include the following:

1. There should be not less than three rectangular bays in both longitudinal and transverse

directions.

2. The length of the adjacent bays should not vary by more than 10 %.

Figure 3.2.1-1: Post punching behaviour of slab- critical section

The general layout of the reinforcement is based on the both bending moments (in spans) and

bending moments in addition to direct loads (on columns).


Figure 3.2.1-2: Combined punching shear and transfer of moments

Figure 3.2.1-3


Figure 3.2.1-4


3.2.1-1 Analysis and Design of Flat Plate

To obtain the load effects on the elements of the floor system and its supporting

members using an elastic analysis, the structure may be considered as a series of equivalent

plane frames, each consisting of vertical members – columns, horizontal members - slab.

Such plane frames must be taken both longitudinally (in x-direction) and transversely (in y

direction) in the building, to assure load transfer in both directions.

For gravity load effects, these equivalent plane frames can be further simplified into

continuous beams or partial frames consisting of each floor may be analysed separately together

with the columns immediately above and below, the columns being assumed fixed at their far

ends. Such a procedure is described in the “Equivalent Frame Method”. When frame geometry

and loadings meet certain limitations, the positive and negative factored moments at critical

sections of the slab may be calculated using moment coefficients, termed “Direct Design

Method”. These two methods differ essentially in the manner of determining the longitudinal

distribution of bending moments in the horizontal member between the negative and positive

moment sections. However, the procedure for the lateral distribution of the moments is the same

for both design methods.

Figure 3.2.1.1-1: Steel shear –heads, steel plats joined by welding


Since the outer portions of horizontal members (slab) are less stiff than the part along the

support lines, the lateral distribution of the moment along the width of the member is not

Uniform. The procedure generally adopted is to divide the slab into column strips (along the

column lines) and middle strips and then apportion the moment between these strips and the

distribution of the moment within the width of each strip being assumed uniform.

Figure 3.2.1.1-2: Moments and frames


Figure: 3.2.1.1-3

Example: 3.2-1 Design and calculation of Flat Plate

Geometric Shapes

Slab thickness

The geometry of the building floor plans:

Construction height of object:

Dimensions columns:

The peripheral dimensions of the beam:

Figure: 3.2.1-1

Load calculation

Load per area

Reinforced concrete slab thickness of 300 mm

hd 300 mm

l1 7.7 m lk 2.3 m l2 3.6 m ly 7.7 m

kv 2.850 m

bs 400 mm hs bs

ho 0.5 m bo 0.30 m

qdo hd 25kN

m3

1.35 qdo 10.125kN

m2


floor layer:

Live load (apartments):

Total load on 1 m 2 of slab:

Force load Peripheral masonry thickness of 400 mm YTONG:

Total load acting on the console:

Investigation replacement frame in the X-axis Frame 1:

Calculation model

Figure: 3.2.1-2

load calculation

q1d 3kN

m2

1.4 q 1d 4.2kN

m2

vd 2.0kN

m2

1.5 vd 3kN

m2

qd qdo q1d vd qd 17.325kN

m2

F1 10kN

m3

kv ly 400 mm 1.35 F1 118.503kN

F1d F1 F1d 118.503kN


Load width in a direction perpendicular to the x:

Load in the x-direction:

Calculation of internal forces

Moment on a console:

Moment of inertia:

Transverse replacement frame:

Central girders replacement of frame:

column:

Bending stiffness:

Transverse replacement frame:

Central girders replacement frame:

Column

zsx ly

qdx qd zsx qdx 133.403kN

m

Mk F1d lk qdxlk

2

2

Mk 625.407kN m

Iply hd

312

Ip 0.017m4

Ist Ip

Is1

12bs hs

3 Is 5.208 10 3 m4

KpIp

l11000

kN

rad m2

Kp 2.25 kNm

rad

KstIst

l2 1000kN

rad m2

Kst 4.813 kNm

rad

KsIs

kv1000

kN

rad m2

Ks 1.827 kNm

rad


Figure: 3.2.1-3

Primary moments in node 9:

Primary moments in node 10:

Equilibrium conditions:

Node 9

Node 10:

9 1 10 1 M910 1 M108 1 M109 1 M1012 1M1010' 1 M10'10 1 M911 1 M97 1

M97o 0 kN m M910o1

12 qdx l1

2 M911o 0 kN m

M109o1

12qdx l1

2 M108o 0 kN m M1012o 0 kN m

M1010ò1

12 qdx l2

2 M10'10o M1010ò

GivenM97 kN m M97o Ks 3 9 rad M911 kN m M911o Ks 2 9 rad M910 kN m M910o Kp 2 9 rad 10 rad M108 kN m M108o Ks 3 10 rad M109 kN m M109o Kp 2 10 rad 9 rad M1012 kN m M1012o Ks 2 10 rad M1010' kN m M1010ò Kst 10 rad M10'10 kN m M10'10o Kst 10 rad

Mk M97 kN m M910 kN m M911 kN m 0 kN m


The calculated moments of individual members of equilibrium conditions:

The computation of shear forces in the individual members:

Maximum moment between 9-10 Mmax

Maximum moment between 10-10 Mstr

M109 kN m M1012 kN m M1010' kN m M108 kN m 0 kN mv Find M97 M910 M911 M109 M1012 M1010' M108 9 10 M10'10

M910 v 1.0( ) kN m M910 691.408 kN mM10'10 v 9 0( ) kN m M10'10 282.659kN mM1010' v 5.0( ) kN m M1010' 282.659 kN mM911 v 2.0( ) kN m M911 26.401 kN mM109 v 3.0( ) kN m M109 545.787kN m

V910o qdxl12

V109o V910o

V910 V910oM910 M109

l1

V910 532.511kN

V109 V109oM910 M109

l1

V109 494.688 kN

V1010ò qdxl14

V1010' V1010òV1010' 256.8 kN

v

0

01

2

3

4

5

6

7

8

9

39.601-691.408

26.401

545.787

-105.251

-282.659

-157.877

7.223

-28.797

282.659

a V910l1

V910 V109 a 3.992m

Mmax V910 a M910 qdxa

2

2 Mmax 371.423kN m

Mstr V1010'l14 M1010' qdx

l22

2

2 Mstr 4.431 kN m


Figure: 3.2.1-4

Transformation moments for the part columned strip and between the columns

Moments over support:

Positively support moments:

Ma M910 Mb M109 Ma 691.408 m kN Mb 545.787m kNMc' Mmax 1.25 Mc Mstr 1.25 Mc' 464.278kN m Mc 5.539 kN m

p 0.75M1a p Ma M1a 518.556 kN m M2a 1 p MaM2a 172.852 kN m M1b p Mb M1b 409.34 kN mM2b 1 p Mb M2b 136.447kN m

m 0.60M3c Mc' m M3c 278.567kN m M4c Mc' 1 m M4c 185.711kN m


Figure: 3.2.1-5

Dimensioning of the reinforcement:

Material characteristic of concrete f ckcyl and steel fyk

The top reinforcement for moments:

effective height:

width, which act the moment

Column strip M 1a:

fyd 375 MPa fcd 12 MPa

d hd 3 cm

bly2

b 3.85m

M1a 518.556 kN m M1a 0.518 MN m

d 0.27m fcd 12 MPa b 3.85m

M1a

b d2 fcd

0.0475 0.154

Ast b d fcd Ast b d fcd

MN100 cm

2 Ast 59.252cm2

Ast

3.8515.39 cm

2


Among the columned strip M 2a:

Column strip M 1b:

Among the columned strip M 2b:

The lower reinforcement for moments:

M2a 172.852 kN m M2a 0.172 MN m

d 0.27m fcd 12 MPa b 3.85m

M2a

b d2 fcd

0.01439 0.051

Ast b d fcd

MN100 cm

2 Ast 17.95 cm2

Ast

3.854.662 cm

2

M1b 409.34 kN m M1b 0.409 MN m

d 0.27m fcd 12 MPa b 3.85m

M1b

b d2 fcd

0.03596 0.121

Ast b d fcd

MN100 cm

2 Ast 44.857cm2

Ast

3.8511.651 cm

2

M2b 136.447kN m M2b 0.136 MN m

fcd 12 MPa b 3.85m d 0.27m

M2b

b d2 fcd

0.01145 0.04

Ast b d fcd

MN100 cm

2 Ast 14.283cm2

Ast

3.853.71 cm

2


Column strip M 3c:

Among the columned strip M 4c:

Investigation replacement frame in y Frame 2

Calculation Model

M3c 278.567kN m M3c 0.278 MN m

b 3.85m d 0.27m fcd 12 MPa

M3c

b d2 fcd

0.0249 0.083

Ast b d fcd

MN100 cm

2 Ast 31.06cm2

Ast

3.858.068 cm

2

M4c 185.711kN m M4c 0.185 MN md 0.27m fcd 12 MPa b 3.85m

M4c

b d2 fcd

0.055 0.01588

Ast b d fcd

MN100 cm

2 Ast 19.809 cm2


Figure: 3.2.1-6

Load calculation

Calculation internal forces

Support part:

Among the supports:

a magnification between support:

Transformation moments for the part columned strip and among columned

support t of Ma2

Between the support of M c

q2d qd1

2 l1 l2 q2d 97.886

kN

m

Ma1

12

q2d ly

2 Ma 483.64 kN m

Mc'1

16q2d ly

2 Mc' 362.73 kN m

Mc Mc' 1.25 Mc 453.412kN m

p 0.75 Ma1 p Ma Ma1 362.73 kN m

Ma2 1 p Ma Ma2 120.91 kN m

m 0.6 Mc1 m Mc Mc1 272.047kN m


Dimensioning of reinforcement

Upper reinforcement of moment:

Effective depth:

Column strip M 1a:

The width on which acting the moment:

Column strip M 1a:

Between the column strip M 2a:

Mc2 1 m Mc Mc2 181.365kN m

d hd 3 cm

bl14

l24

b 2.825m

M1a 0.518 MN m d 0.27m fcd 12 MPa b 2.825m

M1a

b d2 fcd

0.06687 0.21

Ast b d fcd

MN100 cm

2 Ast 61.206cm2

M2a 0.172MN m d 0.27m fcd 12 MPa b 2.825m

M2a

b d2 fcd

0.02037 0.07

Ast b d fcd

MN100 cm

2 Ast 18.645cm2


Column strip M 1a:

Between column strip M 2a:

Investigation extreme frame replacement

Calculation Model:

Mc1 0.272MN m d 0.27m fcd 12 MPa b 2.825m

Mc1

b d2 fcd

0.03277 0.11

Ast b d fcd

MN100 cm

2 Ast 29.994cm2

Mc2 0.181MN m d 0.27m fcd 12 MPa b 2.825m

Mc2

b d2 fcd

0.0216 0.073

Ast b d fcd

MN100 cm

2 Ast 19.77cm2


Figure: 3.2.1-7

Calculation of load:

From the slab:

Peripheral masonry thickness of 400 mm YTONG:

Total load replacement frame:

Calculation of internal forces

Moment of the end strip:

q3do qd lkl12

q3do 106.549

kN

m

F1 10kN

m3

kv 400 mm 1.4 F1 15.96kN

m

qkd q3do F1 q kd 122.509kN

m


Support bending moment:

Between the column bending moment:

Transformation moments for the part columned bands and among columned

columned strip width:

Moments over support:

Between the column moments:

Design the reinforcement to the reinforced concrete slab

The top reinforcement for moments:

Mka1

12 qkd ly

2 Mka 605.295 kN m

Mkc1

16qkd ly

2 Mkc 453.971kN m

bp3 lkl12

bp3 6.15m

MextaMka

41 2

lk

bp3

Mexta 264.509 kN m

Minta Mka Mexta Mk4a p Minta Mk4a 255.59 kN m

Mk3a 1 p Minta Mk3a 85.197 kN m

MextcMkc

41 2

lk

bp3

Mextc 198.382kN m

Mintc Mkc Mextc Mk4c m Mintc Mk4c 153.354kN m

Mk3c 1 m Mintc Mk3c 102.236kN m


effective height:

Column extreme strip Mexta:

width which act moment

Column extreme strip Mexta: see diagram B3-B3.3

Column strip inside M k4a:

width, which acts moment, see diagram B3-B3.3

Among the columned strip M k3a:

width, which acts moment

d d 3 cm

b lk b 2.3m

Mexta 0.265 MN m d 0.24m fcd 12 MPa b 2.3m

Mexta

b d2 fcd

0.0509 0.166

Ast b d fcd

MN100 cm

2 Ast 33.716cm2

bl14

b 1.925m

Mk4a 0.256 MN m d 0.24m fcd 12 MPa b 1.925m

Mk4a

b d2 fcd

0.05965 0.192

Ast b d fcd

MN100 cm

2 Ast 33.07cm2

Ast

1.92517.179 cm

2


The lower reinforcement for moments:

Column extreme strip Mextc:

width, which acts moment, see diagram B3-B3.3

Column strip inside M k4c:


bl14

b 1.925m

Mk3a 0.085 MN m d 0.24m fcd 12 MPa b 1.925m

Mk3a

b d2 fcd

0.01739 0.064

Ast b d fcd

MN100 cm

2 Ast 9.641cm2

Ast

1.9255.008 cm

2

b lk b 2.3m

Mextc 0.198MN m d 0.24m fcd 12 MPa b 2.3m

Mextc

b d2 fcd

0.03758 0.125

Ast b d fcd

MN100 cm

2 Ast 24.893cm2

Ast

2.310.823 cm

2

bl14

b 1.925m

Mk4c 0.153MN m d 0.24m fcd 12 MPa b 1.925m


Among the columned strip M k3c:


See diagram B3-B3.3

Mk4c

b d2 fcd

0.03436 0.115

Ast b d fcd

MN100 cm

2 Ast 19.049cm2

Ast

1.9259.896 cm

2

bl14

b 1.925m

Mk3c 0.102MN m d 0.24m fcd 12 MPa b 1.925m

Mk3c

b d2 fcd

0.02189 0.077

Ast b d fcd

MN100 cm

2 Ast 12.136cm2

Ast

1.9256.304 cm

2


Example 3.2-2: In the example we are considering reinforced concrete slab flat, floor slab

thickness is hd = 0.3m, Column diameter (round column) d =0.50 m, the maximum force applied

one column at Nd= 1800 kN.

Material characteristics:

Figure: 2.3.2‐1

Coefficient of shear strength

in both directions

On 1m plate

d 0.5 m hd 0.3 m b 1 m Nd 1800 kN

fcd 17 MPa fctm 1.2 MPa fyd 375 MPa

1 18 mm As1 1

2

4 n 5 Ast n As1 Ast 0.00127 m

2

stw

Ast

b hd stw 0.004241 m

stmin1

3

fctm

fyd

stmin 0.001067

b 1 s 1.159 n 1.0 f 1.25 b 1

h 1.42

3

hd

m h 1.2

g s n h f g 1.74


Carrying capacity of the concrete section

Assess the resistance of the concrete section

Maximum force per columns

Basic critical perimeter

Shear force on the critical perimeter

Shear resistance of concrete

We suggest shear reinforcement

Incorrect design, head to be designed so that they apply condition:

correct proposal

Proposal of hidden head

Maximum critical perimeter with hidden head

qbu 0.42 hd g b fctm qbu 262.86 m1

kN

Vcd Nd Vcd 1800 kN

ucr 2.51m

qd

Vcd

ucr

qd 716.2 m1

kN

qbu 262.86m1

kN qd 716.2m1

kN qd qbu

qd 2 qbu 2 qbu 525.72 m1

kN

qd 2 qbu

Ucrmax 1.9 ucr Ucrmax 4.78m

qda

Vcd

Ucrmax

qda 376.95m1

kN qbu 262.86m1

kN

qda qbu 2 qbu 525.72m1

kN


If we want to make a proposal without head, subject to the following parameters:

Carrying capacity of the concrete section

Proposal visible head

Geometry head

Figure: 2.3.2‐2

We suggest shear reinforcement

Proposal shear reinforcement - reinforced by bins

n is the number of bins reinforced, Ass area of reinforcement to a bin

Given

d 1.0 m fctm40 1.40 MPa ucr40 d 2hd

2

ucr40 4.08m

qbu40 0.42 hd g b fctm40 ucr40 qbu40 1252.47kN Vcd 1800 kN

2 qbu40 2504.94 kN 2qbu40 Vcd

45 deg sin ( ) 0.71 cos ( ) 0.71

hh 0.6 m dh 2.0 m Ucr2 dh Ucr2 6.28m

qd2

Vcd

Ucr2

qd2 286.48m1

kN qbu 262.86m1

kN qd2 qbu

qd2 2 qbu

qd2

qbu

qsu qsu n A ss ss s fyd qsu qd2 qbu qsu 23.62 m1

kN

fyd 190 MPa ss 1 s 1 Ass 1 m2

n 1

qsu n Ass ss s fyd m1

Ass Find Ass Ass 0.000124 m2


number of bars in one bin / m 'ss = 0.25m

diameter of one profile

Assessment of the punching according to EC 2

design value of shear resistance of plate without shear reinforcement (per unit length of critical perimeter)

shear resistance

average width tension section

concrete area

The maximum degree of reinforcement

The average degree of reinforcement

n1 5

1 8 mm

Asssku n1 1

2

4 Asssku 0.00025133 m

2 Asssku Ass

vRd1 Rd k 1.2 40 1 d

Rd 0.3MN

m2

hd 0.3 m k 1.6hd

m k 1.3

bt 1 m

fyk 410MN

m2

min1 0.6 bthd

fyk

kN

m4

min1 0.00000044

min2 0.0015 bthd

m2

min2 0.00045

min

min1

min2

min max min min 0.00045

Ac hd bt Ac 0.3m2

max 0.04Ac

m2

max 0.01

1

min max

2 1 0.01


The maximum design value of shear resistance of plate with shear reinforcement (per unit

length of critical perimeter)

Design value of shear resistance of plate with shear reinforcement (per unit length of critical

perimeter)

Column diameter

Diameter of critical perimeter

Critical perimeter

concrete shear area

Assumption degree of shear reinforcement

carrying capacity

vRd1 Rd k 1.2 40 1 hd vRd1 169.53m1

kN

vRd2 1.6 vRd1 vRd2 271.25m1

kN

vRd3 vRd1i

As fyd sin ( ) u

Ps 0.5 m

Pu 2 1.5 hd Ps Pu 1.4m

u Pu u 4.4m

Acw Pu

2

4

Ps2

4 Acw 1.34m

2

´w 0.0013 0.6

As ´w Acw As 0 m2

2 fyd 360

MN

m2

vRd3 vRd1

As fyd sin ( ) u

vRd3 255.28m1

kN


The load effects

Computing shear force internal columns

Figure: 2.3.2‐3

incorrect design, design head

as being applicable condition

Geometry head

Critical perimeter with head

Concrete shear area

The expected level of reinforcement by shear reinforcement

Slab with shear reinforcement to a void punching.

Vsd 1800 kN 1.15

vsd

Vsd

u vsd 470.64m

1kN vsd vRd3

vsd vRd3

lh 0.9 m hh 0.6 m d2crit 3.11 m

ucrit d2crit ucrit 9.77m

Acwh d2crit

2

4

Ps2

4 Acwh 7.4m

2

Ash ´w Acwh Ash 0.01m2

vsd

Vsd

ucrit

vsd 211.87m1

kN

vRd3 vRd1

Ash fyd sin ( ) ucrit

vRd3 382.21m1

kN vsd vRd3


Space inside the critical perimeter less the contact surface

For dimensioning elements requiring shear reinforcement

Smallest section width in the range of effective height

Height of the floor slab

Maximum distance of stirrups

Maximum diameter of reinforcement stirrups with a smooth surface

Sectional area of shear reinforcement in the length range

wi

Asw sin ( ) Ax

Ax d2crit

2

4

Ps2

4 Ax 7.4m

2

VRd2 0.5 fcd bw 0.9 d 1 cotg ( ) fck 25 MPa fcd 13.3 MPa

0.7fck

200 MPa 0.58 if 0.5 0.5 0.58 0.5

bw 1.0 m

hd 0.3 m

cot ( ) 0 VRd2 0.5 fcd bw 0.9 hd 1 cot ( ) VRd2 1032.41 kN

smax 0.3 hd smax 0.09m smax if smax 0.2 m smax 0.2 m smax 0.09m

Vsd 1800 kN2

3VRd2 688.27 kN

s 0.012 m


Necessary degree of reinforcement EC2:

Minimum design values of moments on columns in contact with the plate at the eccentric load

Internal Column, top moment

Internal Column, top moment acting shear

force

Figure: 2.3.2‐4

Asw s

2

4 Asw 0.00011 m

2

2

w

Asw sin ( )

Ax

w 0.00001528 wtab 0.0013 wmin 0.6 wtab

wmin 0.00078

w if w wmin wmin w w 0.00078

x 0.125

y 0.125 Vsd 1800 kN

msdx x Vsd msdx 225 kN msdy y Vsd msdy 225 kN


Example 3.2-3: In the middle columns of dimensions as x bs from adjacent reinforced flat slab

of thickness hs at a critical cross-section carries a full load slab, shear force Vcd = 400 kN,

shear force from accidental load Vcd = 325 kN and the bending moment Mcd = 20 kNm (moment

transmitted from slab to reinforced column).

Figure: 2.3.3-1

Material characteristics:

where fctk is the characteristic tensile strength of concrete (5-percent fractile), fctm is the mean

tensile strength and fck is the characteristic compressive strength of concrete measured on

cylinders.

The depth of reinforced concrete slabs

Dimension columns:

fckcub 20 MPa fckcyl 0.8 fckcub fckcyl 16 MPa

fcd 0.85fckcyl

1.5 fcd 9.067 MPa

fctm 1.4fckcyl

10 MPa

2

3

MPa fctm 1.915MPa

fyk 345 MPa fydfyk

1.15 fyd 300 MPa

hs 0.2 m

as 0.40 m bs 0.40 m


Bending moment and shearing forces:

If Mkontr less than the Mcd, should be respected Mcd

Figure: 2.3.3-2

Calculation of Qbu

Vcd1 400 kN Vcd 325 kN Mcd 20 kN m

Uc1 ashs

22 Uc1 0.6m Uc2 bs

hs

22 Uc2 0.6m

Ucr 2 Uc1 Uc2 Ucr 2.4m

qdmaxVcd1

Ucr qdmax 166.667m

1kN

Mkontr 0.2 Vcd hs Mkontr 13m kN

IcrU c1

3

6

U c2 U c12

2 Icr 0.144 m

3

n1

13

2

U c2

U c1

n 0.4

dmaxVcd

Ucr

Mcd n 0.5 Uc1

Icr dmax 152.08m

1kN


\

The reliability condition

The cross-section without shear reinforcement does not comply

I suggest shear reinforcement in the form of welded of mesh

d 18 mm As1 d

2

4 As1 0.00025447m

2 n 6

Astd n As1 Astd 0.00152681m2

c1 as c2 bs stxAstd

c2 4 hs hs stx 0.00636173

stmin1

3

fctm

fyd stmin 0.00212797

styAstd

c2 4 hs hs sty 0.00636173

stm stx sty stm 0.00636173

s 1 50 b stm stmin s 1.212 h 1.4 m 2hs

3 h 1.267m

qbu 0.42 s h n fctmhs

m qbur 0.42 fctm

hs

m

qbu 246.91096132m1

kN qbur 160.87453706m2

kN

qdmax 166.667m1

kN qbu 246.911m1

kN

qbur 160.87453706m2

kN qbueur 160.87453706m2

kN

ss 1 s 8 mm Ass1 s

2

4 Ass1 0.00005027m

2

ssss Ass1 s fyd

qdmax 0.5 qbu ss 0.34897549m


Perimeter displaced the critical cross section

It is less than qbu, that is, the cross section satisfies without shear reinforcement.

Alternative we suggest shear reinforcement consisting of a flexible conduit at an angle

=60.deg.

The proposal

Ucrp 2 c1 hs 2 ss c2 hs 2 ss Ucrp 5.19180391m

qdmaxpVcd1

Ucrp qdmaxp 77.04451232m

1kN

qbu 246.911m1

kN qdmax 166.667m1

kN 60 deg

Asbqdmax 0.5 qbu Ucr

sin ( ) s fyd Asb 0.00039917m

2

oh 14 mm Asoh oh

2

4 Asoh 0.00015394m

2

Asb

Asoh2.59304223


Figure: 2.3.3-3

Internal column of 500 x 500

step 1:

Step 2

hd 25 cm fctm 1.2 MPa bs 50 cm hs 50 cm

fyd 375 MPa P1 856 kN

ucr1 bs hd hs hd 2 Qbu1 0.42 hd fctm ucr1 Qbu1 378kN

P P1 0.5 Qbu1 P 667kN

AsbP

0.86 fyd Asb 0.00206822m

2 25 mm

As1

2

4 As1 0.00049087m

2 nAsb

As1 n 4.21333718

V1 0.42 hd fctm ucr1 V1 378kN P 667kN

ucr2 bs 3 hd hs 3 hd 2 ucr2 5m

Qbu2 0.42 hd fctm ucr2 Qbu2 630kN


Column of 400x 500 extreme

Step 1:

When applied to the plate even bending moment, then we take 0.5 qbu

For P we calculate the required shear reinforcement.

does not comply

P P1 0.5 Qbu2 P 541kN AsbP

0.86 fyd Asb 0.00167752m

2

nAsb

As1 n 3.42 25 mm As1

2

4 As1 0.00049087m

2

V2 0.42 hd fctm ucr2 V2 630kN

ucrP

0.42 hd fctm ucr 4.2937m

ucr

41.0734m

hd 25 cm fctm 1.2 MPa bs 40 cm hs 50 cm fyd 375 MPa

ucr1 bs 0.5 hd 2 hs hd Qbu1 0.42 hd fctm ucr1 Qbu1 226.8kN

0.5 Qbu1 113.4kN

P1 577 kN P P1 0.5 Qbu1 P 463.6kN

AsbP

0.86 fyd Asb 0.00143752m

2 20 mm

As1

2

4 As1 0.00031416m

2 nAsb

As1 n 4.575766

V1 0.42 hd fctm ucr1 V1 226.8kN P 463.6kN V1 P


Step 2

We expand the circumference in order to prevent the creation of a new shear crack

does not comply

Figure 2.3.3-4: Shear reinforcement at slab-column connection

Step 3:

ucr2 hs 3 hd bs 1.5 hd 2 ucr2 2.8m

Qbu2 0.42 hd fctm ucr2 Qbu2 352.8kN

P P1 0.5 Qbu2 P 400.6kN AsbP

0.86 fyd Asb 0.00124217m

2

20 mm As1

2

4 As1 0.00031416m

2

nAsb

As1 n 3.95395164 V2 0.42 hd fctm ucr2 V2 352.8kN

P 400.6kN V2 P

ucr3 hs 5 hd bs 2.5 hd 2 ucr3 3.8m

Qbu3 0.42 hd fctm ucr3 Qbu3 478.8kN P P1 0.5 Qbu3 P 337.6kN


OK

V3 is greater than P, thus the determination of the reinforcement to avoid the punching in

reinforced concrete slab flat over the column is o

AsbP

0.86 fyd Asb 0.00104682m

2

25 mm As1

2

4 As1 0.00049087m

2 nAsb

As1 n 2.13

V3 0.42 hd fctm ucr3 V3 478.8kN P 337.6kN V3 P

3.2 Reinforced Concrete Slabs - vsvu.sk · Figure 3.2-1: One –way slab, two-way slab, ribbed...

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Transcript of 3.2 Reinforced Concrete Slabs - vsvu.sk · Figure 3.2-1: One –way slab, two-way slab, ribbed...