CONTINUOUS BEAM
EXAMPLE 3.2
A
A
Figure above shows a plan view of a school building. It is constructed using cast in-situ method and the slab spans 3 x 8 m each. Prepare a complete design of beam 2/A-D.
Design data: Slab thickness = 125 mm Loads from finishes, partitions etc. = 15 kN/m Characteristic live load, q k = 10 kN/m Nominal cover, c = 25 mm Concrete’s characteristic strength, f ck ck = 30 N/mm
2
Steel characteristic strength (main), f yk yk = 500 N/mm Beam size, b w x h = 250 x 450 mm
2
CONTINUOUS BEAM
EXAMPLE 3.2
Solution
1. Calculate the loads acting on the beam. Characteristic permanent action on beam, g k Self-weight of beam = 25 x bw x h
= 25 x 0.25 x 0.45 = 2.81 kN/m
∴Total charac. permanent action on beam 2/A-D, gk
= self-weight of beam + finishes = 2.81 + 15 = 17.81
kN/m
∴Total charac. variable action acting on beam 2/A-D, qk = 10 kN/m
Therefore, design load acting on beam 2/A-D, w = 1.35 g k + 1.5 qk = 1.35 ( 17.81 ) + 1.5 (10) = 39.04 kN/m
(1.35 gk + 1.5 qk)
CONTINUOUS BEAM
EXAMPLE 3.2
2. Draw the shear force (SFD) and bending moment diagram (BMD). The coefficients from Table 3.5 can only be applied to continuous beam analysis when all this provisions are fulfilled. a) Qk ≤ Gk
= 10 < 17.81 ∴OK!
b) Loads should be uniformly distributed over 3 or more spans = 3 spans ∴OK! c) Variation in span length should not exceed 15% of the longest = same span ∴OK!
Beam 2/A-D w = 39.04 kN/m
8m
8m
0.45F
0.55F
+
8m
0.6F
+
+
0.6F 0.11FL
0.09FL
-
0.55F
0.45F
0.11FL
+
-
+
0.07FL
+
0.09FL
CONTINUOUS BEAM
EXAMPLE 3.2
3. Calculate the design moments and shear force values. F = wL = ( 39.04 ) x ( 8 ) = 312.32
kN
w = 39.04 kN/m
8m
8m
140.54
171.78
+
8m
187.39
+
+
-
187.39
171.78
140.54
274.84
274.84
-
-
+
224.87
-
+
+
174.9
224.87
4. Design the main reinforcements. i)
Calculate the effective depth, d.
Assume φbar = 20 mm
φlink = 8 mm d = h – c - φlink - φbar/2 = 450 – 25 – 8 - 20/2 = 407 mm
d h
CONTINUOUS BEAM
EXAMPLE 3.2
• At mid span A-B and C-D (design as flange section) M = 0.09 FL = 224.87 kNm
3000 mm
3000 mm
bw = 250 mm
bw = 250 mm
Section A-A
Lo values Mid span A-B & C-D => l o = 0.85 (8000) = 6800 mm Mid span B – C => l o = 0.7 (8000) = 5600 mm
bw = 250 mm
CONTINUOUS BEAM
EXAMPLE 3.2
Calculate the effective width of flange, bf
≤ =
+
,
1 2 ≤ 11 2 2 1 2 ≤ =
,
= ,
.
+
.
.
= (3000 250)/2 = 1375 = , = 0.2 (1375) + 0.1(6800) =
, ,
(
)=
(1375)
,
=>
!
=
+
,
,
= (955 + 955) + 250 =
+
<
= 0.567
2
∴
= 1375 + 1375 + 250 = 3000
!
= 0.567(30)(2160)(125) 407
125 2
= 1582
Compare Mf with the design moment, M.
M =224.87 kNm < M f = 1582 kNm
∴ Neutral axis lies in flange / below flange. ∴Design the beam as rectangular / flanged beam.
2 =
=
6 2
224.87 2160
10
407
30
= 0.02 <
0.167
∴ compression reinforcement is required / not required
=
0.5 +
∴ use = 0.95d
0.25
1.134
=
Flanged beam
Design the main reinforcement.
ℎ ℎ
0.2 (6800) = 1360
0.5 +
0.25
0.02
1.134
= 0.98 > 0.95
CONTINUOUS BEAM
6 =
224.87
=
0.87
0.87 500
10
0.95
407
= 1337
EXAMPLE 3.2
2
2
∴Provide: 5H20
(Asprov = 1570 mm )
Check the area of reinforcement.
≥ ,
,
≥ 2≥ ∴ 2 ∴
0.26
=
=
0.0013
0.26 (2.9)(250)(407) 500 !
0.04
= 0.04
0.0013 (250)(407) = 132
= 153
250
450 = 4500
<
<
• At support B & C (design as rectangular section) M=
0.11 FL = 274.84 kNm
2
6 2 ′ =
=
274.84
250
10
407
30
= 0.22 > 0.167
∴ compression reinforcement is required / not required
=
0.5 +
0.25
1.134
= 0.82
Calculate d’
∅ ∅ =
+
+
2
= 25 + 8 +
20 2
= 43
!
2
CONTINUOUS BEAM
=
(
2 )
0.87
(
=
)
∴Provide: 3H16
+
0.87
∴Provide:
2
0.167)(30)(250)(407)
0.87 ( 500 )(407 2
43 )
= 416
2
(As’prov = 603 mm )
2 ′ =
( 0.22
EXAMPLE 3.2
2
0.167(30)(250)( 407 ) = + 416 = 1845 0.87(500)( 0.82 407)
3H25
2
2
(Asprov = 1963 mm )
Check the area of reinforcement.
≥
≥ 2 2 ∴
0.26
,
0.0013
= 153
,
= 4500
<
<
!
• At mid span B-C (flanged section) M = 0.07 FL = 174.9 kNm
Calculate the effective width of flange, bf
≤ =
+
1 2 ≤ 11 2 2 1 2 ≤ ,
=
,
= ,
.
+
.
.
= (3000 250)/2 = 1375 = , = 0.2 (1375) + 0.1(5600) =
, ,
(
)=
(1375)
,
=>
=
,
+
,
+
!
= (835 + 835) + 250 = <
0.2 (5600) = 1120
∴
= 1375 + 1375 + 250 = 3000
!
CONTINUOUS BEAM Flanged beam
Design the main reinforcement.
= 1582
EXAMPLE 3.2
Compare Mf with the design moment, M. M = 174.9 kNm < M f = 1582 kNm
∴ Neutral axis lies in flange / below flange. ∴Design the beam as rectangular / flanged beam.
2 =
=
6 2
174.9 1920
10
407
30
= 0.02 <
0.167
∴ compression reinforcement is required / not required
6 2 =
0.5 +
=
0.25
0.87
∴Provide: 4H20
=
1.134
=
174.9
0.87 500
0.5 +
0.25
10
0.95
407
0.02
1.134
= 0.98 > 0.95
= 1040
2
(Asprov = 1271 mm )
Check the area of reinforcement.
≥ ,
,
≥ 2 2 ∴
0.26
0.0013
= 153
= 4500
<
<
!
CONTINUOUS BEAM
5. Design the shear reinforcement.
VEd = V max = 0.6 F = 187.39 kN
Calculate VRd,c
200
= 1+
� =
,
407
=
= 1.7
≤ 2.0
1 3 ≥ 3 2 1 2 1963
250
= 0.02
407
= 0.12 (100
)
= 0.12(1.7) 100(0.02)(30) (250)(407) = 81.26 kN > Vmin /
/
= 0.035
= 43.24 kN
VRd.c = 81.26 kN
Compare VEd with VRd,c VEd (187.39) > V Rd,c (81.26) => shear reinforcement is required
Calculate VRd,max @ 22°,
,
0.36 =
1
(
+
250 )
0.36 (250)(407) 1 =
(
22° +
(22°
≤≤
30 (30) 250 = 335.88 22°)
45°)
EXAMPLE 3.2
CONTINUOUS BEAM
EXAMPLE 3.2
Compare VEd with VRd,max VEd (187.39) < VRd,max (335.88) Design shear reinforcement
=
0.78
cot
3
187.39 10 = = 0.48 0.78(500)(407)(cot 22°) 2
Try H8, Asw = 2π(∅link) /4 x 2 legs = 101 mm
=
101 0.48
= 210
2
< 0.75 = 0.75 (407) = 305.25
∴Provide: H8 – 200 c/c
6. Check beam’s capacity against deflection. Check only at mid-span with maximum moment.
0 −3 −3 ′ ′ =
,
=
=
1337
250
10
407
= 0.013
= 5.48
10
ρ > ρo
=
11 + 1.5
+
1
12
From table 7.4N, K = 1.3 (end span of continuous beam) – where the moment is maximum
= (1.3) 11 + 1.5
(30)
(5.48
−3 −3
0.013
)
10
0
+
1
12
(30)
0
(5.48
10
)
= 18.8
CONTINUOUS BEAM
(i)
EXAMPLE 3.2
Calculate the modification factor
a) Modification factor of tension reinforcement,
310
=
500 ,
,
=
500 = . 1337 500 1570
b) Modification factor for flange section,
bf /bw = 2160/250 = 8.64 > 3, therefore the modification factor (flange) = 0.8
c) Modification factor for span length more than 7 m Effective span length 8 m > 7 m, therefore modification factor span length = 7/l eff = 7/8 = 0.88
d) Calculate (L/d) allowable
=
= 18.8
1.17
0.8
0.88 = 15.5
a) Calculate (L/d)actual
ℎ ℎ =
=
8000
b) Compare with (L/d)actual with (L/d)allowable
(L/d)actual > (L/d)allowable
407
= 19.66
CONTINUOUS BEAM
EXAMPLE 3.2
Therefore, the deflection check fails !
2
Try to increase from 5H20 to 6H20 (As prov = 1890 mm ) Modification factor for tension reinforcement = Asprov/As req = 1890/1337 = 1.4 Recalculate (L/d)allowable = 18.8 x 1.4 x 0.8 x 0.88 = 18.5
2
Try to increase from 6H20 to 5H25 (As prov = 2450 mm ) Modification factor for tension reinforcement = Asprov/As req = 2450/1337 = 1.83 Recalculate (L/d)allowable = 18.8 x 1.83 x 0.8 x 0.88 = 24.2
Compare with L/d actual (L/d)actual = 19.66 < (L/d)allowable = 24.2 Therefore, the deflection check passes !!
i) ii)
(L/d)actual ≤ (L/d)allowable – Beam is safe against deflection (OK!) (L/d)actual > (L/d)allowable - Beam is not safe against deflection (Fail!)
Therefore, beam is safe / not safe against deflection.
CONTINUOUS BEAM
EXAMPLE 3.2
7. Check the beam for cracking. Check only at mid span with maximum spacing.
i)
Calculate the clear horizontal distance between bars in tension, S1. 250 mm
4H20
S1
∅ ∅ =(
=
=
=
2
250
2
2( 25)
2
2(8)
)
2( 20)
= 144
+ 0.3
1.15
500
1.15
1.35
+ 1.5
17.81 + 0.3 (10) = 232 1.35(17.81) + 1.5 (10)
Taking wk = 0.3 mm Maximum allowable clear spacing =
230 mm
∴OK! ∴Crack check passed / failed !!
The value must not exceed its maximum allowable clear spacing
CONTINUOUS BEAM
EXAMPLE 3.2
Minimum bar spacing (check for the closest bar spacing) Minimum bar spacing between reinforcements = max {k1. Bar diameter, dg + k2, 20 mm} i)
1.25 = 25 mm
ii)
dg + k2 = 20 + 5 = 25 mm
iii)
20 mm
Minimum bar spacing = 25 mm Compare with actual bar spacing = 54.5 mm (5H25) > 25 mm ∴Ok!!
