Long span beam design for wooden trusses. Wood design.Full description
Full description
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A bridge deck is the portion of a bridge that acts as the roadway in the support of vehicular or pedestrian traffic. While deck parts like trusses, girders, rails, arches, posts and cantilevers assume a number of forms and types, there are relatively
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A bridge deck is the portion of a bridge that acts as the roadway in the support of vehicular or pedestrian traffic. While deck parts like trusses, girders, rails, arches, posts and cantilevers assume a number of forms and types, there are relatively
Preliminary Design of Very Long-span Suspension Bridge
DESIGN OF DECKSLAB Clear Span Thickness of slab (assumed)
= =
7.450 m 65 cm
Thickness of w.c = 75 mm Clear cover = 4.00 cm (as per clause 304.3.1 and table 10 of IRC bridge code 21-2000 section III ) Main reinf inforcement = 20 mm dia. HYSD bars conforming to IS-1786 (Deformed bars) Concrete Mix is M 20 grade Bearing(assumed) = 49 cm Effective depth
= =
65
-
4.00
-
1.00
60 cm
Effective Span As per clause 305.1.2 of IRC code 21-2000 Effective span shall be the least of the following
i). Effective span where
l1
= =
l1
+
d
clear span
= 7.45 m d = effective depth = 0.6 m Effective span = 7.45 + 0.6000 = 8.050 m ii) l = distance from centre of supports = 7.45 + 2x 0.490 2 = 7.940 m Effective span shall be least of (i) and (ii) Effective span = 7.940 m Carriage way width = 4.25 m (clause 113.1 of IRC 5-1985) Kerb width Width of slab
= 0.225 m = 4.25 + 0.225 x 2 = 4.700 m Loading IRC class 'A' 'A' As per per clau clause se 303.1 03.1 and and Tab Table le 6 of of IRC IRC code code 21-2 1-2000, 00, fo for M 20 Permissible flexural compressive stresses 66.67 c allowable = 6.667 M.Pa = 66.67 kg/cm2 Modular ratio = Es = 10.00 Ec
grad rade RCC RCC
As 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 + = 0.250 j = 1 n/3 = 1 0.25 /3 = 0.917
2000
Q
= =
1/2*c*n*j 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 Weight of w.c Total dead load
= =
0.65 0.075
=
2
=
4.70 7.94
* *
2500 2400
= =
1625 kg/m 180 kg/m 1805 kg/m
wl = 1805 x 7.940 8 8 = 14224.21225 kg-m = 1422421.225 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 Bending Moment due to dead load
Ratio
=
b/lo
= =
1.72
7.94
0.592
for simply supported slab 1.72 1.96
0.5 0.6 For b/lo
x
+
0.24 * 0.1
0.092
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
where
bef bef lo a
= = = =
b1
=
a (1 a/lo) +b1 The effective width of slab on which the load acts. The effective span as indicated in clause 305.1 The distance of the centre of gravity of the concentrated load from the nearer support The breadth of concentration area of the load, i.e the dimension
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 3.2
11.4 1.2
6.8
6.8
4.3
3
Class 'A' Train 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 A
C
11.4 3.2 D 7.940
11.4 1.2 E
B
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 0.671 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 = a = 0.671 a from B = 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
0.671 7.269
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
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 < 0.7713 here left is going beyond the slab edge, hence the left dispersion is limited to 0.48
=
Dispersion width under one wheel 0.475 + 1.543 /2 + Intensity of load under 'C' =
Dispersion Width Under D bef = a (1 = 1.941 a = 3.871 lo = 7.94 b1 = 0.65 m bef = 4.500 m Dispersion Widths overlap
= 1.246 m 1.35 (1.00 0.323 ) 1.246 = 1.433 t 11.4 t load a/lo) +b1 a from A = 3.871 a from B = 4.069
>
0
1.8 m
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 < 2.250 here left is going beyond the slab edge, hence the left dispersion is limited to 0.6250 Combined dispersion width = 0.6250 + 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 < 2.103 here left dispersion is going beyond the slab edge, hence the left dispersion is limited 0.6250 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
A
1.433 3.226 R 3.330 0.099 0.099 1.002 0.671 3.2 1.2 2.869 C D E 3.97 3.970 7.94
B
Ra
x
Ra
x
Taking moments about 'B' 7.9 = 3.330 * 1.433 * 7.9 = 33.219 Ra = 4.184 t Rb = 1.433 + Rb = 3.805 t
2.869 + 7.269
3.226
3.226 +
3.330 -
*
4.07
4.184
Maximum live load Bending Moment = ( 4.184 = = = Total B.M = = = Effective depth required
=
= Steel steel at bottom Main steel reinforcement require
=
* 3.871 ) - ( 1.433 * 16.196 4.585 11.61078 t-m 1161078 kg-cm Dead load B.M + Live Load B.M 1422421.2 + 1161077.5 2583499 kg-cm M Q*b
=
58.153 cm
M tjd
)
2583499 7.640 x100 < OK
=
###
60 cm
2583499 2000 x 0.917 x
60
2
= 23.486 cm 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 2
provide
= 7.80 cm 20 mm dia. HYSD bars at
2
cm < 23.486 130 mm c/c
2
OK Ast provided = 24.15 cm 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 1161077.5 + 0.2x 1422421.23 = 632807.4993 kg-cm Assuming the dia. Of distributio 12 mm HYSD Effective depth = 58.40 cm Ast required = M = 632807.50 tjd 2000 x 0.917 x 58.4 2
= 5.910 cm 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 =
2
7.008 cm
>
2
5.910 cm
2
Hence provide minimun distribution reinforcem 7.008 cm provide 12 mm dia. HYSD bars at 130 Ast provided
=
2
8.695 cm
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mm c/c OK
Top Reinforcement 2
Min. Reinforcement = 7.008 cm 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 C
4.3
D 3.970
A
1.590 E 3.970
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 < 1.0616 here left is going beyond the slab edge, hence the left dispersion is limited to 0.625
Combined dispersion wid = =
0.625
+ 2.123 /2 + 3.487 m
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 < 1.8009 here left is going beyond the slab edge, hence the left dispersion is limited to 0.6250 Combined dispersion width 0.6250 + 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 < 1.4991 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
= 4.325 0.85 C A
2.328 t
3.569 1.2 D 3.970
2.328 4.3
1.59 E 3.970
7.940
B
Ra
Taking moments about 'B' x 7.94 = 4.325 + 2.328
Ra
x
7.94 Ra Rb
Shear due to dead load
= = = = =
* *
7.09 1.59
55.385 6.975 t 3.246 t wl 2 1805 x
7.940
+
3.569
*
5.89
2 Total shear due to dead load & live lo
= =
7165.9 kgs 6975.45 + 14141.301
7165.850 kgs
As per clause 304.7.1 of IRC 21-2000 Design shear stress
τc
=
where
V d b
= = =
here
V b d
= = = =
V bd The design shear across the section The depth of the section The breadth of the rectangular beam or slab, or the breadth of the rib in the case of flanged beam 14141 kgs 100 cms 60 cms 14141
100x 60 = 2.36 kg/cm2 As per clause 304.7.1.1 of IRC 21-2000 Max. permissible shear 20 2 τcmax = 1.8 N/mm 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 2
0.9 N/mm 9.0 kg/cm2
Hence max. permissible stress
= =
Ast provided
=
24.154 cm
100 Ast / b d from Table 12B of IRC:21-2000
=
0.403
τc
grade concrete
2
=
0.269 N/mm2
=
2.690 kg/cm
2
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*
