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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, A’s is required.
d’/d = 50/540 = 0.092 < 0.171 (Table 7.1),
therefore, fsc = 0.87fyk
Compression steel, A’s = (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, A’s = 402 mm2
Tension steel, As = 0.167 fckbd2 + A’s
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 θ = 22°, cot θ = 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 θ = 22° and cot θ = 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 θ 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 3000.5 = 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 θ = 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,
Distance, x = VEd - Vmin = 308 -151 = 1.45 m
wu 108
Therefore, the number of H8 links at 200 mm centres required at each end of the beam is :
1 + (x/s) = 1 + (1450/200) = 9
spaced over a distance of (9-1) 200 = 1600 mm.
e) Additional longitudinal reinforcement
∆Ftd = 0.5VEdcot θ
= 0.5 x 308 x 2.5
= 385 kN
Additional longitudinal reinforcement,
As = ∆Ftd /0.87fyk
= 385 x 103 /(0.87 x 500)
= 885 mm2
Hence, provide 2H25 bars = 982 mm2
5. Deflection
ρ = 100As,req/bd
= (100 x 2497)/ (300 x 540)
= 1.54%.
Refer Table 6.10 or Fig. 6.3, basic span-effective depth ratio =14
Modification for steel area provided:
(l/d)allowable = Modified ratio = 14 x (2592/2497) = 14.5
(l/d)actual = Span-effective depth ratio provided
= 6000/540 = 11.1 < (l/d)allowable , OK!
6. Cracking (Table 6.7)
Limiting crack width, wmax = 0.3 mm
Steel stress, fs = fyk x Gk + 0.3Qk . 1
1.15 (1.35Gk+1.5Qk) δ
= (500/1.15) x (60 +(0.3x18) . 1
108
= 263 N/mm2
Maximum allowable bar spacing = 150 mm
Bar spacing,
s = [300 – 2(50) -2(8) – 32]/2
= 76 mm < 15mm, OK!
7. Draw Beam Detailing !
EXERCISE 1
8.0 m
20 kN/m
1. Estimate Beam Size
Overall depth, h = L/13 = 8000/13 = 615 mm
Width, b = 0.4h = 0.4 x 615 = 246 mm
Hence, provide b x h = 250 x 650 mm
2. Cover, Cnom = Cmin + ∆Cdev
With regard to bond, Cmin,b = 20 mm
With regard to durability, C,dur = 15 mm (Refer to Table)
Cnom = Cmin, + ∆Cdev = 22 mm + 10 mm = 32mm, Hence use 35 mm
(Table 5.5 -Appendix) asd = a +10
3. Loading Analysis
Beam self-weight = (0.25 x 0.65) x 25 = 4.06 kN/m Permanent load (excluding self-weight) = 20 kN/m Characteristic permanent action, gk = 24.06 kN/m Characteristic variable action, qk = 10 kN/m Design action, wd = 1.35 gk + 1.5qk
= (1.35 x 24.06 + 1.5 x 10) = 47.48 kN/m
Hence maximum shear force, V = wd L/2 = 190 kN L =8m
Hence, maximum design moment, M = wd L2/8
= 47.48 x 82/8 = 379.8 kNm
4. Main Reinforcement
Effective depth, d = h – Cnom – ølink –øbar = 587 mm d’ = Cnom + ølink + øbar/2 = 53 mm K = M/bd2fck = 379.8 x 106/250 x 5872 x 20 =0.22 > Kbal =0.167
Therefore, compression reinforcement, A’s is required. d’/d = 53/587 = 0.09 < 0.171 (Table 7.1), OR
therefore, fsc = 0.87fyk
Calculate z
Compression steel, As’ = (K-Kbal)fckbd2
fsc(d-d’)
= (0.22 – 0.167) x 20 x 250 x 5872
0.87 x 500 x (587 – 53)
As’ = 393 mm2
Hence, provide 2 H16 bars, A’s = 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 20 x 250 x 5872 + 393
0.87 x 500 x (0.82 x 587)
= 1374 + 393 = 1767 mm2
Hence, provide 6 H20 bars , area As = 1890 mm2
5. Check for minimum and maximum reinforcement area
As, min = 0.26 (fctm/fyk)bd > 0.0013bd
= 0.26 x (2.2/500) x 250 x 587 > 0.0013 (250 x 587)
= 0.001bd < 0.0013bd
= 191 mm2
As, max = 0.04Ac = 0.04 b x h
= 0.04 x 250 x 650 = 6500 mm2
Hence, As, min < As, prov < As, max
191 mm2 < 1890 mm2 < 6500 mm2 , OK !
6. Shear Reinforcement
Design shear force, VEd = 190 kN
Crushing strength, VRd,max of diagonal strut, assuming θ = 22°, cot
θ = 2.5
VRd,max = 0.124 bwd(1-fck/250) fck
= 0.124 x 250 x 587 (1-20/250) x 20 x 10-3
= 335 kN ( > VEd = 190 kN)
Therefore, angle θ = 22° and cot θ = 2.5 as assumed.
b) Shear links
Asw = VEd = 190 x 103
s 0.78dfykcot θ 0.78 x 587 x 500 x 2.5
= 0.332
*Using Table A.4 in Appendix Hence, provide 8 mm links at 300 mm centres, Asw/s = 0.335
OR
Try H8 , Asw = 2 x П x (82/4) = 101 mm2
Spacing, s = 101/0.332 = 304 mm
Max spacing, smax = 0.75d = 0.75 x 587 = 440 mm
c) Minimum links
Asw,min = 0.08fck0.5bw = 0.08 x 200.5 x 250 = 0.179
s fyk 500
Try H8 , Asw = 2 x П x (82/4) = 101 mm2
Spacing, s = 101/0.179 = 564 mm
Max spacing, smax = 0.75d = 0.75 x 587 = 440 mm; s > smax, Hence, Use H8-425
Shear resistance of minimum links:
Vmin = Asw x 0.78dfykcot θ = (101/425) x 0.78 x 587 x 500 x 2.5 x 10 -3
s
= 136 kN
190kN
136kN x = [(190 -136) / 47.48] = 1.14 m
x
x
136kN
190 kN
H8 – 300
1.14 m
H8 – 425
5.72 m
H8 – 300
1.14 m
Minimum links
e) Additional longitudinal reinforcement
∆Ftd = 0.5VEdcot θ
= 0.5 x 190 x 2.5
= 237.5 kN
Additional longitudinal reinforcement,
As = ∆Ftd /0.87fyk
= 238 x 103 /(0.87 x 500)
= 547 mm2
Hence, provide 2H20 bars = 628 mm2
7. Deflection
ρ = 100As,req/bd
= (100 x 1890)/ (250 x 587)
= 1.29%.
(l/d)actual = Span-effective depth ratio provided
= 8000/587 = 13.6 < (l/d)allowable , OK!
Texbook Section
6.2
8. Cracking (Table 6.7)
Limiting crack width, wmax = 0.3 mm
Steel stress, fs = fyk x Gk + 0.3Qk . 1
1.15 (1.35Gk+1.5Qk) δ
= (500/1.15) x (24.06 +(0.3x10) . 1
47.48
= 248 N/mm2
Maximum allowable bar spacing = 150 mm
Bar spacing,
s = [250 – 2(35) -2(8) – 20]/2
= 72 mm < 150 mm, OK!
9. Provide beam detailing
EXERCISE 2 This figure shows part of the first floor plan of a reinforced concrete office building. During construction, slab and beam are cast together. The overall thickness of the slab is 125 mm and the dimensions of the beams are as given in the diagram. The finishes, ceiling and services form a characteristic permanent action of 1.5 kN/m2. The characteristic variable action is 3.0 kN/m2. Three metre high brickwall weighing 2.6 kN/m2 is placed over the entire span of all beams. The construction materials consist of Grade C25 concrete and grade 500 steel reinforcement. For durability consideration, a nominal cover of 30 mm is required. Based on the information provided, (a) Calculate the design action carried by beam 2/B-C, (b) Sketch the bending moment and shear force diagrams of beam 2/B-C, (c) design the beam for ultimate and serviceability limit state.
