|
|
 |
Concrete Grades
Beam fcu = 50 N/mm2, fci = 40 N/mm2
Slab fcu = 40 N/mm2
|
|
BS
5400 Pt. 4 |
|
Section Properties |
|
cl.7.4.1 |
Modular
ratio effect for different concrete strengths between beam
and slab may be ignored. |
|
|
| Property |
Beam |
Composite |
| Area (mm2) |
449.22 x 103 |
599.22 x 103 |
| Centroid (mm) |
456 |
623 |
| 2nd Moment of Area (mm4) |
52.905 x 109 |
103.515 x 109 |
| Modulus @ Level 1 (mm3) |
116.020 x 106 |
166.156 x 106 |
| Modulus @ Level 2 (mm3) |
89.066 x 106 |
242.424 x 106 |
| Modulus @ Level 3 (mm3) |
_____ |
179.402 x 106 |
|
|
Temperature Difference Effects |
|
Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4)
to a simplified beam section
Cl. 5.4.6 - Coefficient of thermal expansion = 12 x 10-6
per ºC.
From BS 5400 Pt4 Table 3 : Ec = 34 kN/mm2
for fcu = 50N/mm2
Hence restrained temperature stresses per °C = 34 x 103
x 12 x 10-6 = 0.408 N/mm2
a) Positive temperature difference
Force F to restrain temperature strain :
0.408 x 1000 x [ 150 x ( 3.0 + 5.25 ) ] x 10-3 +
0.408 x ( 300 x 250 x 1.5 + 750 x 200 x 1.25 ) x 10-3 =
504.9 + 122.4 = 627.3 kN
Moment M about centroid of section to restrain
curvature due to temperature strain :
0.408 x 1000 x [ 150 x ( 3.0 x 502 + 5.25 x 527 ) ] x 10-6
+
0.408 x ( 300 x 250 x 1.5 x 344 - 750 x 200 x 1.25 x 556 ) x 10-6
= 261.5 - 26.7 = 234.8 kNm
b) Reverse temperature difference
Force F to restrain temperature strain :
- 0.408 x [ 1000 x 150 x ( 3.6 + 2.3 ) + 300 x 90 x ( 0.9 + 1.35 )
] x 10-3
- 0.408 x 300 x ( 200 x 0.45 + 150 x 0.45 ) x 10-3
- 0.408 x 750 x [ 50 x ( 0.9 + 0.15 ) + 240 x ( 1.2 + 2.6 ) ] x 10-3
= - 385.9 - 19.3 - 295.1 = - 700.3 kN
Moment M about centroid of section to restrain
curvature due to temperature strain :
- 0.408 x [ 150000 x ( 3.6 x 502 + 2.3 x 527 ) + 27000 x ( 0.9 x
382 + 1.35 x 397 ) ] x 10-6
- 0.408 x 300 x ( 200 x 0.45 x 270 - 150 x 0.45 x 283 ) x 10-6
+ 0.408 x 750 x [ 50 x ( 0.9 x 358 + 0.15 x 366 ) + 240 x ( 1.2 x
503 + 2.6 x 543 ) ] x 10-6
= - 194.5 - 0.6 + 153.8 = - 41.3 kNm
|
|
Differential Shrinkage Effects |
|
BS 5400 Pt 4. cl.7.4.3.4 |
Use cl.6.7.2.4 Table 29 :
Total shrinkage of insitu concrete = 300 x 10-6
Assume that 2/3 of the total shrinkage of the precast concrete takes
place before the deck slab is cast and that the residual shrinkage
is 100 x 10-6 ,
hence the differential shrinkage is 200 x 10-6 |
|
cl.7.4.3.5 |
Force to restrain differential shrinkage : F = - ediff
x Ecf x Acf x f
F = -200 x 10-6 x 34 x 1000 x 150 x 0.43 = -439 kN
Eccentricity acent = 502mm
Restraint moment Mcs = -439 x 0.502 = -220.4 kNm 
|
|
Self weight of beam and
weight of deck slab is supported by the beam. When the deck slab
concrete has cured then any further loading (superimposed and live
loads) is supported by the composite section of the beam and slab. |
|
Dead Loading (beam and
slab) |
|
Total
load for serviceability limit state = |
(1.0
x 3.6)+(1.0 x 10.78) = 14.4kN/m |
|
Design
serviceability moment = |
14.4 x 242 / 8 = 1037 kNm |
|
|
|
Combination 1 Loading |
|
Super. & HA live load for serviceability limit state =
=
=
|
[(1.2 x 2.4)+(1.2 x 10)]udl & [(1.2 x
33)]kel
(2.88 + 12.0)udl & 39.6kel
14.9 kN/m & 39.6kN |
|
Super. & HB live
load for serviceability limit state =
=
|
2.88
& 4 wheels @ 1.1 x 62.5
2.9 kN/m & 4 wheels @ 68.75 kN
|
|
Total
load for ultimate limit state =
=
= |
[(1.15
x 3.6)+(1.15 x 10.78)+(1.75 x 2.4)+(1.5 x 10)]udl & [(1.5 x
33)]kel
(4.14
+ 12.40 + 4.20 + 15.0)udl & 49.5kel
35.7
kN/m & 49.5kN |
|
HA Design
serviceability moment =
=
25 units HB Design
serviceability moment =
= |
14.9 x 24.02 / 8 + 39.6 x 24 / 4
1310 kNm
2.9 x 24.02 / 8 + 982.3(from grillage analysis)
1191.1 kNm |
|
Design
ultimate moment =
= |
35.7
x 24.02 / 8 + 49.5 x 24 / 4
2867
kNm |
|
Combination 3 Loading |
|
|
Super. & HA live
load for serviceability limit state =
=
= |
[(1.2 x 2.4)+(1.0 x 10)]udl & [(1.0 x
33)]kel
(2.88 + 10.0)udl & 33kel
12.9 kN/m & 33kN |
|
Total
load for ultimate limit state =
=
= |
[(1.15
x 3.6)+(1.15 x 10.78)+(1.75 x 2.4)+(1.25 x 10)]udl & [(1.25 x
33)]kel
(4.14
+ 12.40 + 4.20 + 12.5)udl & 41.3kel
33.2
kN/m & 41.3kN |
|
Design
serviceability moment =
= |
12.9 x 24.02 / 8 + 33 x 24 / 4
1127 kNm |
|
Allowable stresses in precast concrete |
|
cl.6.3.2.2
|
b) At transfer :
|
Compression ( Table 23 )
0.5fci (<=0.4fcu) = 20 N/mm2 max. |
|
cl.6.3.2.4 |
b)
|
Tension = 1.0 N/mm2 |
|
cl.7.4.3.2 |
At serviceability limit state : |
Compression (1.25 x Table 22)
1.25 x 0.4fcu = 25 N/mm2
Tension = 0 N/mm2 (class 1)
= 3.2 N/mm2 (class 2 - Table 24)
|
|
|
|
|
Stresses at Level 1 due to SLS
loads (N/mm2) : |
|
|
|
Com. 1 (HA) |
Com. 1 (HB) |
Comb. 3 |
| Dead Load M / Z = (1037 x
106) / (116.020 x 106) |
- 8.94 |
- 8.94 |
- 8.94 |
| Super. & Live Load M / Z =
M / (166.156 x 106) |
- 7.88 |
-7.17 |
- 6.78 |
| Reverse Temperature = gfL
x -1.69 = 0.8 x -1.69 |
- |
- |
-1.35 |
| Differential shrinkage |
- 0.60 |
- 0.60 |
- 0.60 |
|
Total Stress at Level 1 = |
-17.42 |
-16.71 |
-17.67 * |
|
|
|
|
Hence Combination 3 is critical |
|
Prestressing Force
and Eccentricity |
|
|
Using
straight, fully bonded tendons (constant force and eccentricity).
Allow for 20% loss of prestress after transfer.
Initial prestress at Level 1 to satisfy class 2 requirement for SLS
(Comb. 3). |
|
|
Stress at transfer = ( 17.67 -
3.2 ) / 0.8 =
|
18.1 N/mm2 (use
allowable stress of 20 N/mm2) |
|
|
The critical
section at transfer occurs at the end of the transmission zone. The
moment due to the self weight at this section is near zero and
initial stress conditions are: |
|
|
P/A + Pe/Zlevel 1
= 20 |
.....................(eqn. 1) |
|
|
P/A
- Pe/Zlevel 2 >= - 1.0 |
.....................(eqn. 2)
|
|
|
(eqn. 1) x Zlevel 1 + (eqn. 2) x Zlevel 2 gives
|
:
|
|
|
P >= A x (20 x Zlevel 1 - 1.0 x Zlevel 2) / (Zlevel
1 + Zlevel 2)
|
|
|
P = 449.22 x 103
x ( 20 x 116.02 - 89.066) / ( 116.02 + 89.066) x 10-3 =
4888 kN |
|
|
Allow 10% for loss
of force before and during transfer, then the initial force Po
= 4888 / 0.9 = 5431kN |
|
|
Using 15.2mm class 2 relaxation standard strand at
maximum initial force of 174kN (0.75 x Pu)
Area of tendon = 139mm2
Nominal tensile strength = fpu =1670 N/mm2
Hence 32 tendons required.
Initial force Po = 32 x 174 = 5568 kN
P = 0.9 x 5568 = 5011 kN |
|
|
Substituting P = 5011 kN in (eqn. 2) |
|
|
e <= Zlevel
2 / A + Zlevel 2 / P = (89.066 x 106 /
449.22 x 103) + (89.066 x 106 / 5011 x 103) |
|
|
e = 198 + 18 = 216 mm |
|
|
|
Arrange 32 tendons
symmetrically about the Y-Y axis to achieve an eccentricity of about
216mm. |
|
|
|
|
Taking moments about bottom of beam : |
| 2 at |
1000 = |
2000 |
| 2 at |
900 = |
1800 |
| 4 at |
260 = |
1040 |
| 8 at |
160 = |
1280 |
| 10 at |
110 = |
1100 |
| 6 at |
60 = |
360 |
| 32 |
|
7580 |
|
|
|
|
e = 456 - 7580 / 32 = 456 - 237 = 219mm |
|
|
|
|
|
Allowing for 1% relaxation loss in steel before transfer and elastic
deformation of concrete at transfer :
|
|
cl. 6.7.2.3 |
P = 0.99 Po / [ 1 + Es x (Aps
/ A) x (1 + A x e2 / I)
/ Eci ] |
|
|
P = 0.99 x Po / [ 1
+ 196 x ( 32 x 139 / 449220) x (1 + 449220 x 2192 /
52.905 x 109) / 31 ] |
|
|
P
= 0.91 Po = 0.91 x 5568 = 5067 kN |
|
|
Initial stresses
due to prestress at end of transmission zone : |
|
|
Level 1 : P / A x ( 1 + A x e / Zlevel
1 ) = |
11.3
x ( 1 + 219 / 258 ) = 20.89 N/mm2 |
|
|
Level 2 : P / A x ( 1 - A x e / Zlevel
2 ) = |
11.3 x ( 1 - 219 / 198 ) = - 1.20 N/mm2
|
|
|
Moment due to self
weight of beam at mid span = 10.78 x 242 / 8 =
776.2 kNm |
|
|
Stress due to self weight of beam at mid span :
@ Level 1 = - 776.2 / 116.02 = - 6.69 N/mm2
@ Level 2 = 776.2 / 89.066 = 8.71 N/mm2
Initial stresses at mid span :
|
|
|
 |
|
cl. 6.7.2.5 |
Allowing for 2% relaxation loss in steel after transfer, concrete
shrinkage ecs = 300 x 10-6
and concrete specific creep ct = 1.03 x 48 x 10-6
per N/mm2
Loss of force after transfer due to :
|
|
cl. 6.7.2.2 |
Steel relaxation = 0.02 x 5568 = 111 |
|
|
cl. 6.7.2.4 |
Concrete shrinkage = (ecs
x Es x Aps ) = 300 x 10-6 x 196 x
32 x 139 = 262 |
|
cl. 6.7.2.5 |
Concrete creep = ( ct x fco
x Es x Aps ) = 1.03 x 48 x 10-6 x
12.76 x 196 x 32 x 139 = 550
|
|
|
Total
Loss = 111 + 262 + 550 = 923 kN |
|
|
Final force after all loss of prestress = Pe
= 5067 - 923 = 4144 kN (Pe/P = 0.82) |
|
|
Final stresses due to prestress after all loss of
prestress at : |
|
|
Level 1 f1,0.82P = 0.82 x 20.89 =
17.08 N/mm2 |
|
|
Level 2 f2,0.82P = 0.82 x - 1.20 = - 0.98
N/mm2 |
|
|
Combined stresses in final condition for worst
effects of design loads, differential shrinkage and temperature
difference : |
|
|
Level 1, combination 1 HB : f = 17.08 - 16.71 = 0.37
N/mm2 (> 0 hence O.K.) |
|
|
Level
1, combination 3 : f = 17.08 - 17.67 = - 0.59 N/mm2 (>
- 3.2 hence O.K.) |
|
|
Level 2, combination 1 : f = - 0.98 + 1037 / 89.066 +
1310 / 242.424 + 1.64 = 17.71 (< 25 O.K.)
|
|
|
Level 3. combination 3 : f = (1127 / 179.402) + (0.8
x 3.15) = 8.8 N/mm2 (< 25 O.K.) |
|
Ultimate Capacity of Beam and Deck Slab (Composite Section)
|
|
|
Ultimate
Design Moment = gf3 x M = 1.1
x 2867 = 3154 kNm |
|
cl. 6.3.3 |
Only
steel in the tension zone is to be considered :
Centroid of tendons in tension zone = (6x60 + 10x110 + 8x160 +
4x260) / 28 = 135mm
Effective depth from Level 3 = 1200 - 135 = 1065mm |
|
|
Assume that the
maximum design stress is developed in the tendons, then :
Tensile force in tendons Fp = 0.87 x 28 x 139 x 1670 x 10-3 = 5655
kN |
|
|
Compressive
force in concrete flange :
Ff = 0.4 x 40 x 1000 x 150 x 10-3 = 2400 kN |
|
|
Let X = depth to
neutral axis.
Compressive force in
concrete web :
Fw = 0.4 x 50 x [393 - (393 - 200) x (X - 150) / (671 x
2)] x (X - 150) x 10-3
Fw = ( -2.876X2 + 8722.84X - 1243717) x 10-3
Equating forces to obtain X :
5655 = 2400 + ( -2.876X2 + 8722.84X - 1243717) x 10-3
X = 659 mm
|
|
|
Stress in tendon
after losses = fpe = 4144 x 103 / (32 x 139) =
932 N/mm2
Prestrain epe = fpe
/ Es = 932 / 200 x 103 = 0.0047
 |
|
|
Determine depth to
neutral axis by an iterative strain compatibility analysis
Try X = 659 mm as an initial estimate
Width of web at this depth = 247mm
| epb6
= |
e6
+ epe = |
-459 * 0.0035 / 659 + 0.0047 |
= 0.0022 |
| epb5
= |
e5
+ epe = |
-359 * 0.0035 / 659 + 0.0047 |
= 0.0028 |
| epb4
= |
e4
+ epe = |
281 * 0.0035 / 659 + 0.0047 |
= 0.0062 |
| epb3
= |
e3
+ epe = |
381 * 0.0035 / 659 + 0.0047 |
= 0.0067 |
| epb2
= |
e2
+ epe = |
431 * 0.0035 / 659 + 0.0047 |
= 0.0069 |
| epb1
= |
e1
+ epe = |
481 * 0.0035 / 659 + 0.0047 |
= 0.0072 |
|
|
|
| fpb6
= |
0.0022 x 200 x 103 |
= 444 N/mm2 |
| fpb5
= |
0.0028 x 200 x 103 |
= 551 N/mm2 |
| fpb4
= |
1162 + 290 x (0.0062 - 0.0058)
/ 0.0065 |
= 1178 N/mm2 |
| fpb3
= |
1162 + 290 x (0.0067 - 0.0058)
/ 0.0065 |
= 1201 N/mm2 |
| fpb2
= |
1162 + 290 x (0.0069 - 0.0058)
/ 0.0065 |
= 1213 N/mm2 |
| fpb1
= |
1162 + 290 x (0.0072 - 0.0058)
/ 0.0065 |
= 1225 N/mm2 |
|
|
|
Tensile force in
tendons :
| Fp6
= |
2 x 139 x 444 x 10-3 |
= 124 |
| Fp5
= |
2 x 139 x 551 x 10-3 |
= 153 |
| Fp4
= |
4 x 139 x 1178 x 10-3 |
= 655 |
| Fp3
= |
8 x 139 x 1201 x 10-3 |
= 1336 |
| Fp2
= |
10 x 139 x 1213 x 10-3 |
= 1686 |
| Fp1
= |
6 x 139 x 1225 x 10-3 |
= 1022 |
|
Ft = |
4976 kN |
|
|
|
Compressive force in
concrete :
| Ff
= |
0.4 x 40 x 1000 x 150 x 10-3 |
= 2400 |
| Fw
= |
0.4 x 50 x 0.5 x (393 + 247) x
(659 - 150) x 10-3 |
= 3258 |
|
Fc = |
5658 kN |
|
|
|
Fc > Ft therefore
reduce depth to neutral axis and repeat the calculations.
Using a depth of 565mm will achieve equilibrium.
The following forces are obtained :
| Fp6
= |
134 |
Ff
= |
2400 |
| Fp5
= |
168 |
Fw
= |
2765 |
| Fp4
= |
675 |
Fc
=
|
5165 |
| Fp3
= |
1382 |
|
|
| Fp2
= |
1746 |
|
|
| Fp1
= |
1060 |
|
|
| Ft = |
5165 |
|
|
|
|
|
Taking Moments about
the neutral axis :
| Fp6
= |
134 x -0.365 |
= -49 |
|
| Fp5
= |
168 x -0.265 |
= -45 |
|
| Fp4
= |
675 x 0.375 |
= 253
|
|
| Fp3
= |
1382 x 0.475 |
= 656 |
|
| Fp2
= |
1746 x 0.525 |
= 917 |
|
| Fp1
= |
1060 x 0.575 |
= 610 |
|
| Ff = |
2400 x 0.49 |
= 1176 |
|
| Fw = |
3258 x 0.207 |
= 674 |
|
|
Mu
|
= 4192 kNm > 3154 kNm hence
O.K. |
|
|
|
|
Mu / M = 4192 / 3154
= 1.33 ( > 1.15 ) |
