Topic 6-Beam Design

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  • BEAM DESIGN-PART 1

  • LESSON OUTCOMES

    Define and explain the need of beam sizing

    Define and calculate load distribution for analysis simply-supported beam

    Illustrate the SFD and BMD

    Design typical simply supported beam

    Illustrate beam detailing

  • Contents:

    Beam sizing

    Simply-supported beam

    Distribution of slab weight to beam

  • Introduction

    Reinforced concrete beam design consists primarily of producing member details which will adequately resist the ultimate bending moments, shear forces & torsional moments.

    Serviceability requirements must be considered to ensure that the member will behave satisfactorily under working loads.

    Both ultimate limit state (ULS) and serviceability limit state (SLS) has to be taken into consideration.

    Three (3) basic design stages: Preliminary analysis & member sizing; Detailed analysis & design of reinforcement; Serviceability calculations

  • Steps in beam design 1. Determine design life 2. Determine preliminary size of beam 3. Determine nominal cover for durability, fire and bond

    requirements 4. Calculate the effective depth, d 5. Estimate actions on beam 6. Analysis structure to obtain critical moments and

    shear forces SFD & BMD 7. Design of flexural reinforcement 8. Design shear reinforcement 9. Verify deflection 10. Verify cracking 11. Produce detailing

  • Preliminary Analysis & Member Sizing

    The layout & size of members are usually controlled by architectural details & clearances for machinery and equipment.

    Role of engineer: Check the beam sizes are adequate to carry the

    loading or;

    Decide on sizes that are adequate

  • Beam dimensions required are:

    Cover to the reinforcement

    Breadth (b)

    Effective depth (d)

    Overall depth (h)

  • Nominal cover, c

    The concrete cover is necessary to provide: safe transfer of bond forces

    adequate durability (protect reinforcement against corrosion and damage)

    fire resistance

    The value of Cmin is influenced by:

    The exposure classification

    Mix characteristics

    Intended design life of the structure

  • Spacer Blocks

  • Determination of nominal cover

    Determined based on the: a) Durability requirement (EC2- 1-1)

    b) Fire resistance requirement (EC2-1-2)

  • Effective depth, d

    d = distance from the compression face to the

    center of the tension reinforcement.

    d = distance from the compression face to the

    center of the compression reinforcement.

  • Design Load Calculation

    At ultimate limit state:

    (for reinforcement design)

    Design load, wEd = 1.35 gk + 1.5 qk

    At serviceability limit state:

    Design load, wEd = 1.0 gk + 1.0 qk

  • Actions/Loads acting on beam

    Permanent actions, gk (EC1 Table A1-A5) a) Self-weight of the beam

    = density of concrete x bw x d

    = 2500 kg/m3 x bw x d

    b) Finishes/building services

    c) Self-weight of brickwall

    d) Loading from slab

    e) Loading from secondary beams

    Variable actions, qk = loads from slab

    = can be obtained from EC1 (Table 6.1 & 6.2)

  • Moment and Shear Analysis

    Compute and draw the shear force and bending moment diagram of the beam.

    The maximum values of the shear forces and bending moment will be used in the design calculation.

    There are two (2) types of beams:

    i. Simply-supported beam

    ii. Continuous beam

  • Estimation of actions from slab

    Actions that applied on a beam may consists of: - beam self-weight

    - dead & imposed loads from slabs

    - actions from secondary beams

    - other structural /non-structural members

    supported by the beam

    The distribution of slab actions on beams depends on the slab dimension, supporting system & boundary condition.

  • Estimation of actions from slab

  • Estimation of actions from slab

  • Estimation of actions from slab

    Three (3) alternative methods:

    1. Slab shear coefficient from Table 3.15 BS 8110

    2. Yield line analysis

    3. Table 6.3 Reinforced Concrete Designer Handbook by Reynold

  • (i) Simply-supported beam

  • (ii) Continuous Beam

    Consist of more than 1 span

    Usually in cast in-site structures

    Can be obtained from Table 3.5 (BS 8110)

    Will be given in the

    design appendix

  • (ii) Continuous Beam

  • Shear Force Diagram &

    Bending Moment Diagram

  • Main reinforcement design

  • Design of Main Reinforcements

    Covered in previous topics

    Type of beam: Rectangular beam

    Flanged beam (T beam & L beam)

    Type of support:

    i. Simply-supported

    ii. Continuous

  • Design of shear reinforcement

    Deflection check

    Cracking check

    Detailing

    Already covered in previous topics

  • EXAMPLE 1

    DESIGN TYPICAL SIMPLY-SUPPORTED BEAM

    A simply-supported beam has the following characteristics:-

    Breath, b = 300 mm, Length = 6000 mm

    Effective depth, d = 540 mm

    fck = 30 N/mm2

    fyk = 500 N/mm2

    Given permanent load, gk = 60 kN/m, including self weight

    Variable load, qk = 18 kN/m

    Design the simply-supported beam. Provide the main

    reinforcement and shear link. Draw the beam detailing.

  • 1. Ultimate Loading & Maximum Moment

    Ultimate load, wu = (1.35 gk + 1.5 qk) kN/m

    = (1.35 x 60 + 1.5 x 18)

    = 108 kN/m

    Hence, maximum design moment, M = wu L2/8

    = 108 x 62/8

    = 486 kNm

    2. Bending Reinforcement

    K = M/bd2fck = 486 x 106/300 x 5402 x30

    =0.185 > Kbal =0.167

    Therefore, compression reinforcement, As is required.

    d/d = 50/540 = 0.092 < 0.171 (Table 7.1),

    therefore, fsc = 0.87fyk

  • Compression steel, As = (K-Kbal)fckbd2

    fsc(d-d)

    = (0.185 0.167) x 30 x 300 x 5402

    0.87 x 500 x (540 50)

    = 222 mm2

    Hence, provide 2 H16 bars, As = 402 mm2

    Tension steel, As = 0.167 fckbd2 + As

    0.87fykzbal

    *Refer Figure 7.5, Kbal = 0.167, la= z/d = 0.82

    Thus, As = 0.167 x 30 x 300 x 5402 + 222

    0.87 x 500 x (0.82 x 540)

    = 2275 + 222 = 2497 mm2

    Hence, provide 2 H32 bars and 2 H25 bars , area = 2592 mm2

  • 3. Check for minimum and maximum reinforcement area

    As, min = 0.26 (fctm/fyk)bd > 0.0013bd

    = 0.26 x (2.9/500) x 300 x 540 > 0.0013 (300 x 540)

    = 244.3 > 210.6

    = 244 mm2

    As, max = 0.04Ac = 0.04 b x h

    = 0.04 x 300 x 590 = 7080 mm2

    Hence, As, min < As, prov < As, max

    244 mm2 < 2592 mm2 < 7080 mm2 , OK !

  • 4. Shear Reinforcement

    Ultimate load, wu = 108 kN/m

    a) Check maximum shear at face of support

    Maximum design shear = wu x effective span

    2

    = 108 kN/m x 6.0 m = 324 kN

    2

    Design shear at face of support, VEd = 324 (108 x0.15)

    = 308 kN

    Crushing strength, VRd,max of diagonal strut, assuming = 22, cot = 2.5

    VRd,max = 0.124 bwd(1-fck/250) fck

    = 0.124 x 300 x 540 (1-30/250) x 30 x 10-3

    = 530 kN ( > VEd = 308 kN)

    Therefore, angle = 22 and cot = 2.5 as assumed.

    Variable Strut Inclination Method

  • b) Shear links

    At distance d from the face of support, VEd = 308 wu.d

    = 308 (108 x 0.54) = 250 kN

    Asw = VEd = 250 x 103

    s 0.78dfykcot 0.78 x 540 x 500 2.5

    = 0.475

    *Using Table A.4 in Appendix

    Hence, provide 8 mm links at 200 mm centres, Asw/s = 0.503

    c) Minimum links

    Asw,min = 0.08fck0.5bw = 0.08 x 30 x 300

    0.5 = 0.26

    s fyk 500

    Hence, provide 8 mm links at 350 mm centres, Asw/s = 0.287

    Shear resistance of minimum links:

    Vmin = Asw x 0.78dfykcot = 0.287 x 0.78 x 540 x 500 x 2.5 x 10 -3

    s = 151 kN

  • d) Extent of Shear links

    Shear links are required at each end of the beam from the face of the support to the point where the design shear force is Vmin = 151 kN

    From the face of support,