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, q k = 10 kN/m Nominal cover, c = 25 mm Concrete’s characteristic strength, f ck = 30 N/mm 2 Steel characteristic strength (main), f yk = 500 N/mm 2 Beam size, b w x h = 250 x 450 mm A A
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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
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)