C
Clleeaar r SSppaann == 7.4507.450 mm T
Thhiicckknneesss s oof f ssllaabb == 6565 cmcm (assumed)
(assumed) T
Thhiicckknneesss s oof f ww..cc == 7575 mmmm C
Clleeaar r ccoovveerr == 4.004.00 cmcm (as per clause 304.3.1 and table
(as per clause 304.3.1 and table 10 of IRC bridge code 21-200010 of IRC bridge code 21-2000 section III )
section III ) M
Maaiin n rreeininffoorrcceemmeenntt == 2020 mm dia. HYSD bars conforming to IS-1786mm dia. HYSD bars conforming to IS-1786 (Deformed bars)
(Deformed bars) Concrete Mix is M
Concrete Mix is M 2020 grade grade B
Beeaarriinngg((aassssuummeedd)) == 4949 cmcm E Effffeeccttiivve e ddeepptthh == 6655 -- 44..0000 -- 11..0000 = = 6600 ccmm Effective Span Effective Span
As per clause 305.1.2 of IRC code 21-2000 As per clause 305.1.2 of IRC code 21-2000 Effective span shall be the least of the following Effective span shall be the least of the following ii)). . EEffffeeccttiivve e ssppaann == ll11 ++ d d where
where ll11 == cclleeaar r ssppaann = = 77..4455 mm d d == eeffffeeccttiivve e ddeepptthh = = 00..66 mm E Effffeeccttiivve e ssppaann == 7..47 45 +5 + 00..66000000 = = 88..005500 mm
iiii) l) l == ddiissttaanncce e ffrroom m cceennttrre e oof f ssuuppppoorrttss = = 77..4455 ++ 22xx 00..449900 2 2 = = 77..994400 mm Effective span shall be least of (i) and (ii) Effective span shall be least of (i) and (ii) E
Effffeeccttiivve e ssppaann == 77..994400 mm C
Caarrrriiaagge e wwaay y wwiiddtthh == 44..2255 mm (clause 113.1 of IRC (clause 113.1 of IRC 5-1985)5-1985) K Keerrb b wwiiddtthh == 00..222255 mm W Wiiddtth h oof f ssllaabb == 4..24 25 5 + + 00..22225 5 x x 22 = = 44..770000 mm Loading
Loading IRC IRC class 'Aclass 'A'' A
As ps per er clclauause se 330303.1 .1 anand Td Tabable le 6 o6 of If IRC RC cocode de 221-1-22000000, f, foor Mr M 220 0 ggraradde Re RCCCC P
Peerrmmiissssiibblle e fflleexxuurraal l ccoommpprreessssiivve e ssttrreesssseess 6666..6677 =
= 6.6676.667 M.PaM.Pa =
= 6666..6677 kkgg//ccmm22 M
Moodduullaar r rraattiioo == EEss == 1100..0000 Ec Ec
D
D E
E S
S I
I G
G N
N O
O F
F D
D E
E C
C K
K S
S L
L A
A B
B
c allowable c allowableAs per clause 303.2.1 of IRC code 21-2000
Permissible tensile stress in steel for combined bending For steel S 415 t = 200 M.Pa
m = 10.00 n = mc / (mc+t) n = 10.00 x 66.67 66.67 x 10.00 + 2000 = 0.250 j = 1 - n/3 = 1 - 0.25 /3 Q = 1/2*c*n*j = 0.917 = 7.640
As per clause 211.2 of IRC code 6 Impact factor fractio = 4.5
6+L
where L is Effective span= 7.940 m Impact factor fractio = 4.5
6 + 7.940
= 0.323
Dead Load Bending Moment
Weight of slab = 0.65 * 2500 = 1625 kg/m
Weight of w.c = 0.075 * 2400 = 180 kg/m
Total dead load 1805 kg/m
Bending Moment = wl2 = 1805 x 7.940 x 7.94
duetodeadload 8 8
= kg-m
= kg-cm
Live Load Bending Moment
As per clause 305.13.2.1 of IRC code 21-2000 Ratio b / lo
where b = width of slab = 4.70 m
lo = Effective span = 7.940 m
Ratio = 4.70 = 0.592
7.94
for simply supported slab
0.5 1.72
0.6 1.96
For b/lo = 1.72 + 0.24 * 0.092
0.1
= 1.941
As per clause 305.1.3.2(1) IRC 21-2000 Solid slabs spanning in one direction
For a single concentrated load, the effective width may be calculated in accordance with the following equation
bef = a (1 - a/lo) +b1
where bef = The effective width of slab on which the load acts. lo = The effective span as indicated in clause 305.1
a = The distance of the centre of gravity of the concentrated load from the nearer support
b1 = The breadth of concentration area of the load, i.e the dimension b/lo
14224.21225 1422421.225
of the tyre or track contact area over the road surface of the slab in a direction at right angles to the span plus twice the thickness of the wearing coat or surface finish above the structural slab. = a constant depending upon the ratio
b/lo where b is the width of the slab
2.7 11.4 11.4 6.8 6.8
3.2 1.2 4.3 3
For maximum bending moment, the two loads of 11.40 t should be kept such that the resultant o the load system and the load under consideration should be equidistant from the centre of span.
Position of Loads
2.7 11.4 11.4
3.2 1.2
A C D E B
7.940 CG of the load sytem from C
= (11.4 x 3.2)+(11.4 x 4.4) = 86.64 = (2.7 + 11.4 + 11.4) 25.5 CG from D = 0.198 2.7 11.4 R 11.4 0.099 0.099 1.002 3.2 1.2 2.869 A C D E B 3.970 3.970 7.940
Dispersion Width Under 'C' 2.7 t load
bef = a ( 1 - a/lo) + b1 = 1.941 a from A = 0.671 a = 0.671 a from B = 7.269 lo = 7.94 b1 = 0.35 m bef = 1.543 m < 1.8 m
Dispersion Widths do not overlap
0.225 0.2 0.1 1.8
0.2 0.2
4.70
Left Dispersion calculated (bef/2) = 0.7713
left dispersion means the possible dispersion on the left side from the centre Class 'A' Train
3.398
of the left wheel of the vehicle = kerb width + kerb to wheel gap + tyre width / 2 = = 0.225 + 0.15 + 0.2/2 0.475 < here left is going beyond the slab edge, hence the left dispersion is limited to 0.48
Dispersion width under one wheel
= 0.475 + /2 + 0 = 1.246 m
Intensity of load under 'C' = 1.35 (1.00 0.323 ) 1.246
= 1.433 t
Dispersion Width Under D 11.4 t load
bef = a (1 - a/lo) +b1 = 1.941 a from A = 3.871 a = 3.871 a from B = 4.069 lo = 7.94 b1 = 0.65 m bef = 4.500 m > 1.8 m
Dispersion Widths overlap 1.543
0.225 0.2 0.25 1.8
0.5 0.5
4.70
Left Dispersion calculated (bef/2) = 2.250
here left dispersion means the possible dispersion on the left side from the centre
of the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 = = 0.225 + 0.15 + 0.5/2 0.625 <
here left is going beyond the slab edge, hence the left dispersion is limited to Combined dispersion width = + 4.500 /2 + 1.8
= 4.675 m
Intensity of load under 'D' = 11.4 ( 1.00 0.323 )
4.675
= 3.226 t
Dispersion Width Under 'E' 11.4 t load
bef = a (1 - a/lo) +b1 = 1.941 a from A = 5.071 a = 2.869 a from B = 2.869 lo = 7.940 b1 = 0.65 m bef = 4.206 m > 1.8 m
Dispersion Widths overlap
0.225 0.2 0.25 1.8
0.5 0.5
4.70
Left Dispersion calculated (bef/2) = 2.103
left dispersion means the possible dispersion on the left side from the centre
of the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 = 0.225 + 0.15 + 0.5/2 0.625 <
here left dispersion is going beyond the slab edge, hence the left dispersion is limited Combined dispersion width = 0.625 + 4.206 /2 + 1.8
= 4.528 m
Intensity of load under 'E' = 11.4 ( 1.00 0.323 )
4.528 = 3.330 t 1.433 3.226 R 3.330 0.099 0.099 1.002 3.2 1.2 2.869 C D E 3.97 3.970 A 7.94 B 0.6250 2.250 0.6250 0.6250 2.103 0.671
Taking moments about 'B' Ra x 7.9 = 3.330 * 2.869 + 3.226 * 1.433 * 7.269 Ra x 7.9 = Ra = 4.184 t Rb = 1.433 + 3.226 + 3.330 -Rb = 3.805 t 4.07 33.219 4.184
Maximum live load Bending Moment
= ( 4.184 * 3.871 ) - ( 1.433 * ### )
= 16.196 - 4.585
= t-m
= kg-cm
Total B.M = Dead load B.M + Live Load B.M
= + = kg-cm Effective depth = M = required Q*b 7.640 x100 = 58.153 cm < 60 cm Steel OK steel at bottom
Main steel reinforcement require = M =
t j d 2000 x 0.917 x 60 = 23.486 cm2
As per clause 305.19 of IRC code 21-2000 Minimum area of tension
reinforcement = 0.12 % of total cross sectional area
On each face = 0.12 * 100 * 65
* 100
= 7.80 cm2 < cm2
provide 20 mm dia. HYSD bars at 130 mm c/c
Ast provided = 24.15 cm2 OK
Distribution Steel
As per clause 305.18.1 of IRC code 21-2000 Resisting moment
= 0.3 times the moment due to concentrated live loads plus 0.2 times the moment due to other loads such as dead load, shrinkage, temperature etc. = 0.3 * L.L B. Mom. + 0.2*D.L B. Mom.
= 0.3x + 0.2x
= kg-cm
Assuming the dia. Of distributio 12 mm HYSD Effective depth = 58.40 cm
Ast required = M =
t j d 2000 x 0.917 x 58.4
= 5.910 cm2 As per clause 305.19 of IRC code 21-2000
Minimum distribution reinforce = 0.12 % of total cross section area
= 0.12 x 100 x 58.40 100 = 7.008 cm2 > 5.910 cm2 2583499 2583499 2583499 632807.50 11.61078 1161078 1161077.5 1422421.2 23.486 1161077.5 1422421.23 632807.4993
Hence provide minimun distribution reinforcem 7.008 cm2
provide 12 mm dia. HYSD bars at 130 mm c/c
Ast provided = 8.695 cm2 OK
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Top Reinforcement
Min. Reinforcement = 7.008 cm2
Provide 12 mm dia. HYSD bars at 130 mm c/c
Also provide distribution steel o 12 mm dia. HYSD bars at 130 mm c/c Check for shear
As per clause 305.13.3 of IRC code 21-2000 Dispersion of loads along the span
Longitudinal dispersion =
The effect of contact of wheel or track load in the direction of span length shall be taken as equal to the dimension of the tyre contact area over the wearing surface of the slab in the direction of the span plus twice the overall depth of the slab inclusive of the thickness of
the wearing surface.
Longitudinal dispersion = 0.25 + 2 ( 0.65 + 0.075 ) = 1.70 m
so 11 t load may be 0.85 m from the support to get max. shear
11.4 11.4 6.8 t
0.85 1.2 4.3 1.590
C D E
3.970 3.970
A 7.940 B
The load of 11.40 t may be kept at 1.70 /2 = 0.85 m from the support to Dispersion Width Under 'C' 11.4 t get max. shear
bef = a (1 - a/lo) +b1 = 1.941 a from A = 0.850 a = 0.850 a from B = 7.090 lo = 7.940 b1 = 0.65 m bef = 2.123 m > 1.8 m
Dispersion Widths overlap
0.225 0.15 0.25 1.8
0.5 0.5
4.70
Left Dispersion (bef/2) = 1.0616
left dispersion means the possible dispersion on the left side from the centre
of the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 = = 0.225 + 0.15 + 0.5/2 = 0.625 <
here left is going beyond the slab edge, hence the left dispersion is limited to 0.625 1.0616
Combined dispersion wid = 0.625 + 2.123 /2 + 1.8
Intensity of load under 'C' = 11.4 ( 1.00 0.323 ) 3.487
= 4.325 t
Dispersion Width Under 'D' 11.4 t
bef = a (1 - a/lo) +b1 = 1.94 a from A = 2.05 a = 2.050 a from B = 5.890 lo = 7.94 b1 = 0.65 m bef = 3.602 m > 1.8 m
Dispersion Widths overlap
Left calculated Dispersion (bef/2) = 1.8009
left dispersion means the possible dispersion on the left side from the centre
of the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 = = 0.225 + 0.15 + 0.5/ 0.625 <
here left is going beyond the slab edge, hence the left dispersion is limited to
Combined dispersion width + 3.602 /2 + 1.80
= 4.226 m
Intensity of load under 'D' = 11.4 ( 1.0 + 0.323 )
4.226
= 3.569 t
Dispersion Width Under 'E' 6.8 t
bef = a (1 - a/lo) +b1 = 1.941 a from A = 6.350 a = 1.590 a from B = 1.590 lo = 7.940 b1 = 0.53 m bef = 2.998 m > 1.8 m
Dispersion Widths overlap
0.225 0.2 0.19 1.8
0.38 0.5
4.70
Left Dispersion calculated (bef/2) = 1.4991
left dispersion means the possible dispersion on the left side from the centre
of the left wheel of the vehicle. = kerb width + kerb to wheel gap + tyre width / 2 = 0.225 + 0.15 + 0.38/2 = 0.565 <
here left is going beyond the slab edge, hence the left dispersion is limited to 0.565 Combined dispersion width = 0.565 + 2.998 /2 + 1.8
= 3.864 m
Intensity of load under 'E' = 6.8 ( 1.00 0.323 ) 3.864
1.8009
0.6250
1.4991 0.6250
= 2.328 t 4.325 3.569 2.328 0.85 1.2 4.3 1.59 C D E 3.970 3.970 A 7.940 B
Taking moments about 'B' Ra x 7.94 = 4.325 * 7.09 + 3.569 * 5.89 + 2.328 * 1.59 Ra x 7.94 = 55.385 Ra = 6.975 t Rb = 3.246 t
Shear due to dead load = wl 2
= 1805 x 7.940
2 = 7165.9 kgs
Total shear due to dead load & live lo = +
kgs As per clause 304.7.1 of IRC 21-2000
Design shear stress τc = V
bd
where V = The design shear across the section d = The depth of the section
b = The breadth of the rectangular beam or slab, or the breadth of the rib in the case of flanged beam
here V = 14141 kgs b = 100 cms d = 60 cms = 14141 100x 60 = 2.36 kg/cm2
As per clause 304.7.1.1 of IRC 21-2000 Max. permissible shear 20 grade concrete
τcmax = 1.8 N/mm2
for solid slabs the permissible shear stress shall not exceed half the value of
τcmax given in Table 12 A
Hence τcmax = 0.5 x 1.8
= 0.9 N/mm2 Hence max. permissible stress = 9.0 kg/cm2
Ast provided = 24.154 cm2
100 Ast / b d = 0.403
from Table 12B of IRC:21-2000
τc = 0.269 N/mm2
= 2.690 kg/cm2 For solid slabs the permissible shear in concrete k * τc
(vide cl. 304.7.1.3.2 of IRC:21-2000)
k for solid slab of 65 cm thick = 1.00 (Table 12C of IRC:21-2000)
k * τc= 1.0 x 2.690 = 2.690 Kg/cm2 > 2.36 kg/cm2
14141.301
