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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,