CONTINUOUS BEAM EXAMPLE 3.2

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, qk = 10 kN/m Nominal cover, c = 25 mm Concrete’s characteristic strength, fck = 30 N/mm2 Steel characteristic strength (main), fyk = 500 N/mm2 Beam size, bw x h = 250 x 450 mm

A

A

CONTINUOUS BEAM EXAMPLE 3.2

1. Calculate the loads acting on the beam.

Characteristic permanent action on beam, gk

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 gk + 1.5 qk

= 1.35 ( 17.81 ) + 1.5 (10) = 39.04 kN/m

Solution

(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

8 m 8 m 8 m

w = 39.04 kN/m

0.45F

0.6F

0.55F 0.6F

0.55F 0.45F

0.09FL

0.11FL

0.07FL

0.11FL

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

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

8 m 8 m 8 m

w = 39.04 kN/m

140.54

187.39

171.78 187.39

171.78 140.54

224.87

274.84

174.9

274.84

224.87

+ + +

- - -

+ + +

- -

h d

CONTINUOUS BEAM EXAMPLE 3.2

• At mid span A-B and C-D (design as flange section)

M = 0.09 FL = 224.87 kNm

Section A-A

Lo values

Mid span A-B & C-D => lo = 0.85 (8000) = 6800 mm

Mid span B – C => lo = 0.7 (8000) = 5600 mm

3000 mm 3000 mm

bw = 250 mm bw = 250 mm bw = 250 mm

CONTINUOUS BEAM EXAMPLE 3.2

Calculate the effective width of flange, bf

𝒃𝒆𝒇𝒇 = �𝒃𝒆𝒇𝒇,𝒊 + 𝒃𝒘 ≤ 𝒃

𝒃𝒆𝒇𝒇,𝒊 = 𝟎.𝟐 𝒃𝒊 + 𝟎.𝟏𝒍𝒐 ≤ 𝟎.𝟐 𝒍𝒐 𝑏1 = 𝑏2 = (3000 − 250)/2 = 1375 𝑚𝑚 𝑏𝑒𝑓𝑓,1 = 𝑏𝑒𝑓𝑓,2 = 0.2 (1375) + 0.1(6800) = 𝟗𝟓𝟓 𝒎𝒎 ≤ 0.2 (6800) = 1360 𝑚𝑚 𝑏𝑒𝑓𝑓,1,2(𝟗𝟓𝟓) = 𝑏1,2(1375) => 𝑂𝐾! 𝒃𝒆𝒇𝒇 = �𝒃𝒆𝒇𝒇,𝟏 + 𝒃𝒆𝒇𝒇,𝟐� + 𝒃𝒘 = (955 + 955) + 250 = 𝟐𝟏𝟔𝟎 𝒎𝒎 < 𝑏 = 1375 + 1375 + 250 = 3000 𝑚𝑚 ∴ 𝑶𝑲!

Design the main reinforcement.

𝑀𝑓 = 0.567𝑓𝑐𝑘𝑏ℎ𝑓 �𝑑 −ℎ𝑓2� = 0.567(30)(2160)(125) � 407 −

1252� = 1582 𝑘𝑁𝑚

Compare Mf with the design moment, M.

M =224.87 kNm < Mf = 1582 kNm

∴ Neutral axis lies in flange / below flange.

∴Design the beam as rectangular / flanged beam.

𝐾 =𝑀

𝑏𝑑2𝑓𝑐𝑘=

224.87 𝑥 106

2160 𝑥 4072 𝑥 30= 0.02 < 0.167

∴ compression reinforcement is required / not required

𝑧 = 𝑑 �0.5 + �0.25 −𝐾

1.134� = �0.5 + �0.25 −

0.021.134

� = 0.98𝑑 > 0.95𝑑

∴ use = 0.95d

Flanged beam

CONTINUOUS BEAM EXAMPLE 3.2

𝐴𝑠𝑟𝑒𝑞 =𝑀

0.87𝑓𝑦𝑘𝑧=

224.87 𝑥 106

0.87𝑥 500 𝑥 0.95 𝑥 407= 1337 𝑚𝑚2

∴Provide: 5H20 (Asprov = 1570 mm2)

Check the area of reinforcement.

𝐴𝑠,𝑚𝑖𝑛 ≥ 0.26 𝑓𝑐𝑡𝑚𝑏𝑡𝑑

𝑓𝑦𝑘 ≥ 0.0013 𝑏𝑡 𝑑

𝐴𝑠,𝑚𝑖𝑛 = 0.26 (2.9)(250)(407)

500= 153 𝑚𝑚2 ≥ 0.0013 (250)(407) = 132 𝑚𝑚2

∴ 𝑂𝐾!

𝐴𝑠𝑚𝑎𝑥 = 0.04𝐴𝑐 = 0.04 𝑥 250 𝑥 450 = 4500 𝑚𝑚2

𝐴𝑠𝑚𝑖𝑛 < 𝐴𝑠𝑝𝑟𝑜𝑣 < 𝐴𝑠𝑚𝑎𝑥 ∴ 𝑂𝐾!

• At support B & C (design as rectangular section)

M = 0.11 FL = 274.84 kNm

𝐾 =𝑀

𝑏𝑑2𝑓𝑐𝑘=

274.84 𝑥 106

250 𝑥 4072 𝑥 30= 0.22 > 0.167

∴ compression reinforcement is required / not required

𝑧 = 𝑑 �0.5 + �0.25 −𝐾′

1.134� = 0.82𝑑

Calculate d’

𝑑′ = 𝑐 + ∅𝑙𝑖𝑛𝑘 +∅𝑏𝑎𝑟

2= 25 + 8 +

202

= 43 𝑚𝑚

CONTINUOUS BEAM EXAMPLE 3.2

𝐴𝑠′𝑟𝑒𝑞 =(𝐾 − 𝐾′)𝑓𝑐𝑘𝑏𝑑2

0.87𝑓𝑦𝑘(𝑑 − 𝑑′) =

( 0.22 − 0.167)(30)(250)(407)2

0.87 ( 500 )(407 − 43 ) = 416 𝑚𝑚2

∴Provide: 3H16 (As’prov = 603 mm2)

𝐴𝑠𝑟𝑒𝑞 = 𝐾′𝑓𝑐𝑘𝑏𝑑2

0.87𝑓𝑦𝑘𝑧+ 𝐴𝑠′𝑟𝑒𝑞 =

0.167(30)(250)( 407 )2

0.87(500)( 0.82 𝑥 407)+ 416 = 1845 𝑚𝑚2

∴Provide: 3H25 (Asprov = 1963 mm2)

Check the area of reinforcement.

𝐴𝑠,𝑚𝑖𝑛 ≥ 0.26 𝑓𝑐𝑡𝑚𝑏𝑡𝑑

𝑓𝑦𝑘 ≥ 0.0013 𝑏𝑡 𝑑

𝐴𝑠,𝑚𝑖𝑛 = 153 𝑚𝑚2

𝐴𝑠𝑚𝑎𝑥 = 4500 𝑚𝑚2

𝐴𝑠𝑚𝑖𝑛 < 𝐴𝑠𝑝𝑟𝑜𝑣 < 𝐴𝑠𝑚𝑎𝑥 ∴ 𝑂𝐾!

• At mid span B-C (flanged section)

M = 0.07 FL = 174.9 kNm

Calculate the effective width of flange, bf

𝒃𝒆𝒇𝒇 = �𝒃𝒆𝒇𝒇,𝒊 + 𝒃𝒘 ≤ 𝒃

𝒃𝒆𝒇𝒇,𝒊 = 𝟎.𝟐 𝒃𝒊 + 𝟎.𝟏𝒍𝒐 ≤ 𝟎.𝟐 𝒍𝒐 𝑏1 = 𝑏2 = (3000 − 250)/2 = 1375 𝑚𝑚 𝑏𝑒𝑓𝑓,1 = 𝑏𝑒𝑓𝑓,2 = 0.2 (1375) + 0.1(5600) = 𝟖𝟑𝟓 𝒎𝒎 ≤ 0.2 (5600) = 1120 𝑚𝑚 𝑏𝑒𝑓𝑓,1,2(𝟖𝟑𝟓) = 𝑏1,2(1375) => 𝑂𝐾! 𝒃𝒆𝒇𝒇 = �𝒃𝒆𝒇𝒇,𝟏 + 𝒃𝒆𝒇𝒇,𝟐� + 𝒃𝒘 = (835 + 835) + 250 = 𝟏𝟗𝟐𝟎 𝒎𝒎 < 𝑏 = 1375 + 1375 + 250 = 3000 𝑚𝑚 ∴ 𝑶𝑲!

CONTINUOUS BEAM EXAMPLE 3.2

Design the main reinforcement.

𝑀𝑓 = 1582 𝑘𝑁𝑚

Compare Mf with the design moment, M.

M = 174.9 kNm < Mf = 1582 kNm

∴ Neutral axis lies in flange / below flange.

∴Design the beam as rectangular / flanged beam.

𝐾 =𝑀

𝑏𝑑2𝑓𝑐𝑢=

174.9 𝑥 106

1920 𝑥 4072 𝑥 30 = 0.02 < 0.167

∴ compression reinforcement is required / not required

𝑧 = 𝑑 �0.5 + �0.25 −𝐾

1.134� = �0.5 + �0.25 −

0.021.134

� = 0.98𝑑 > 0.95𝑑

𝐴𝑠𝑟𝑒𝑞 =𝑀

0.87𝑓𝑦𝑘𝑧=

174.9 𝑥 106

0.87𝑥 500 𝑥 0.95 𝑥 407= 1040 𝑚𝑚2

∴Provide: 4H20 (Asprov = 1271 mm2)

Check the area of reinforcement.

𝐴𝑠,𝑚𝑖𝑛 ≥ 0.26 𝑓𝑐𝑡𝑚𝑏𝑡𝑑

𝑓𝑦𝑘 ≥ 0.0013 𝑏𝑡 𝑑

𝐴𝑠,𝑚𝑖𝑛 = 153 𝑚𝑚2

𝐴𝑠𝑚𝑎𝑥 = 4500 𝑚𝑚2

𝐴𝑠𝑚𝑖𝑛 < 𝐴𝑠𝑝𝑟𝑜𝑣 < 𝐴𝑠𝑚𝑎𝑥 ∴ 𝑂𝐾!

Flanged beam

CONTINUOUS BEAM EXAMPLE 3.2

5. Design the shear reinforcement.

VEd = V max = 0.6 F = 187.39 kN

Calculate VRd,c

𝑘 = 1 + �200407

= 1.7 ≤ 2.0 𝑑 𝑖𝑛 𝑚𝑚

𝜌𝑙 =𝐴𝑠𝑙𝑏𝑤𝑑

= 1963

250 𝑥 407= 0.02

𝑉𝑅𝑑,𝑐 = 0.12𝑘(100𝜌𝑙𝑓𝑐𝑘)13𝑏𝑤𝑑 ≥ 𝑉𝑚𝑖𝑛

= 0.12(1.7)�100(0.02)(30)�13(250)(407) = 81.26 kN > Vmin

𝑉𝑚𝑖𝑛 = �0.035 𝑘3/2𝑓𝑐𝑘1/2�𝑏𝑤 𝑑 = 43.24 kN

∴VRd.c = 81.26 kN

Compare VEd with VRd,c

VEd (187.39) > VRd,c (81.26) => shear reinforcement is required

Calculate VRd,max

@ 22°,

𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.36 𝑏𝑤𝑑 �1 − 𝑓𝑐𝑘250�𝑓𝑐𝑘

(𝑐𝑜𝑡 𝜃 + 𝑡𝑎𝑛 𝜃) (22° ≤ 𝜃 ≤ 45°)

= 0.36 (250)(407) �1− 30

250� (30)(𝑐𝑜𝑡 22° + 𝑡𝑎𝑛 22°)

= 335.88 𝑘𝑁

CONTINUOUS BEAM EXAMPLE 3.2

Compare VEd with VRd,max

VEd (187.39) < VRd,max (335.88)

Design shear reinforcement

𝐴𝑠𝑤𝑠

= 𝑉𝐸𝑑

0.78𝑓𝑦𝑘𝑑 cot 𝜃

= 187.39 𝑥 103

0.78(500)(407)(cot 22°)= 0.48

Try H8, Asw = 2π(∅link)2/4 x 2 legs = 101 mm2

𝑠 = 1010.48

= 210 𝑚𝑚 < 0.75𝑑 = 0.75 (407) = 305.25 𝑚𝑚

∴Provide: H8 – 200 c/c

6. Check beam’s capacity against deflection.

Check only at mid-span with maximum moment.

𝜌 =𝐴𝑠,𝑟𝑒𝑞

𝑏𝑤𝑑=

1337250 𝑥 407

= 0.013

𝜌0 = �𝑓𝑐𝑘𝑥 10−3 = 5.48 𝑥 10−3

ρ > ρo

𝑙𝑑

= 𝐾 �11 + 1.5 �𝑓𝑐𝑘𝜌𝑜

𝜌 − 𝜌′+

112�

𝑓𝑐𝑘�𝜌′𝜌𝑜�

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 𝑥 10−3)

0.013 − 0+

112

�(30)�0

(5.48 𝑥 10−3)� = 18.8

CONTINUOUS BEAM EXAMPLE 3.2

(i) Calculate the modification factor

a) Modification factor of tension reinforcement,

310𝜎𝑠

=500

𝑓𝑦𝑘 �𝐴𝑠,𝑟𝑒𝑞𝐴𝑠,𝑝𝑟𝑜𝑣

�=

500

500 �13371570�

= 𝟏.𝟏𝟕

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

�𝐿𝑑�𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

= �𝐿𝑑�𝑏𝑎𝑠𝑖𝑐

𝑥 𝑚𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟

�𝐿𝑑�𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

= 18.8 𝑥 1.17 𝑥 0.8 𝑥 0.88 = 15.5

a) Calculate (L/d)actual

�𝐿𝑑�𝑎𝑐𝑡𝑢𝑎𝑙

=𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑠𝑝𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ

𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ =

8000407

= 19.66

b) Compare with (L/d)actual with (L/d)allowable

(L/d)actual > (L/d)allowable

CONTINUOUS BEAM EXAMPLE 3.2

Therefore, the deflection check fails !

Try to increase from 5H20 to 6H20 (As prov = 1890 mm2)

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

Try to increase from 6H20 to 5H25 (As prov = 2450 mm2)

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 !!

Therefore, beam is safe / not safe against deflection.

i) (L/d)actual ≤ (L/d)allowable – Beam is safe against deflection (OK!) ii) (L/d)actual > (L/d)allowable - Beam is not safe against deflection (Fail!)

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.

𝑠 = (𝑏 − 2𝑐𝑜𝑣𝑒𝑟 − 2∅𝑙𝑖𝑛𝑘 − 2∅𝑏𝑎𝑟)

𝑠 = 250 − 2( 25) − 2(8) − 2( 20)

= 144 𝑚𝑚

𝑓𝑠 = 𝑓𝑦𝑘

1.15 𝑥

𝐺𝑘 + 0.3 𝑄𝑘1.35𝐺𝑘 + 1.5 𝑄𝑘

= 5001.15

𝑥 17.81 + 0.3 (10)

1.35(17.81) + 1.5 (10) = 232 𝑀𝑃𝑎

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

4H20

250 mm

S1

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!!