1. ERT352 FARM STRUCTURES BEAM DESIGN
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2. INTRODUCTION
Steel beams of various cross-sectional shapes are commercially available.
Sections of steel beams are indicated with a combination of letters and a number.
Eg: 305 × 102 UB 25: 305 mm by 102 mm universal beam weighing 25 kg/m.

3. ANALYSIS – LOAD DISTRIBUTION
Loads from slab are normally defined in “q”kN/m2.
These loads are transferred to supporting beams in either “w”kN/m or “W” kN.
Loads from reinforced concrete solid slab:
Loads may be distributed to the supporting beams depending on the ratio of long side/short side of the slab (Ly/Lx)
Type of distribution loads from slab:
One-way spanning slab, when Ly/Lx > 2.0
Two-way spanning slab, when Ly/Lx ≤ 2.0

4. ANALYSIS – LOAD DISTRIBUTION (cont.)
Load from precast concrete slab
Loads may be distributed to the supporting beams in one direction only and not depending on the ratio of Ly/Lx.
This is because precast concrete slab are one-way spanning as they are supported by beams at the ends of the slab only.

5. ANALYSIS – LOAD DISTRIBUTION (cont.)
One-way spanning slab (Ly/Lx > 2.0)
For one way slab, the loads from the slab are distributed to supporting beams as shown in figure below.
In the following case only beams AB and CD carry the loads

6. ANALYSIS – LOAD DISTRIBUTION (cont.)
Two-way spanning slab (Ly/Lx ≤ 2.0)
For two way slab, the loads from the slab are distributed in two directions as shown in figure below.
In this case, all the four beams supporting the slab carry the loads

7. Example 1: Load distribution and beam analysis
The following figure shows s floor plan of a steel building. The floor consists of precast concrete hollow core slabs.
Load carried by the slab are follows:
Unfactored dead load from selfweight of precast slabs, selfweight of steel beams and finishing = 5.0 kN/m2
Unfactored imposed load = 4.0 kN/m2
Determine the maximum shear force and moment in beam 1/A-B.

8. Solution: Load distribution: Design load, n n = 1.35Gk + 1.5Qk
for precast slab, considered as one-way spanning and it is supported by beam AB and CD.
Design load, n
n = 1.35Gk + 1.5Qk
= 1.35 (5 kN/m2) +1.5 (4.0 kN/m2)
= kN/m2
Total design load, w
w = n x width of load transferred to beam 1/A-B
= kN/m2 x (2 m)
= 25.5 kN/m

9. Solution: (cont.)
Since beam AB is simply supported and loads are symmetry, then:
Maximum shear force, VEd = (wL)/2
= (25.5 kN/m)(5 m)/2
= kN
Maximum moment, MEd = (wL2)/8
= (25.5 kN/m)(5 m)2/8
= 79.7kNm

10. Example 2: Load distribution and beam analysis
The following figure shows the plan of a building. The slab is cast in-situ reinforced concrete. Unfactored dead and imposed loads are shown on the drawing. In addition, beam 2/A-B alo carries 4m high brickwall of 3 kN/m2. determine the maximum shear force and moment of beam 2/A-B.

11. Solution: Load distribution on reinforced concrete solid slab
Slab 1: Ly/Lx = 7m/4m = 1.75 < 2.0, two way spanning slab
Slab 2: Ly/Lx = 7m/3m = 2.3 > 2.0, one way spanning slab

12. Solution:(cont.) Load from slab 1: beam 2/A-B
Design load, n1 = 1.35Gk + 1.5Qk
= 1.35 (5 kN/m2) + 1.5(2.0kN/m2)
= 9.75kN/m2
Total design load, W1 = n1 x trapezodial area of load from slab 1
= 9.75kN/m2 x [(3m +7m)/2 x 2m]
= 97.5kN

13. Solution:(cont.)
Since beam is simply supported and loads are symmetry, then:
Maximum shear force, VEd1 = (wL)/2
= (97.5 kN/m)(5 m)/2
= kN
From the bending moment diagram; aL = 2m
a = 2m / 7m = 0.286
Maximum moment, MEd1 = [(3 – 4 a2) /24 (1 - a)] WL
= [(3 – 4 (0.2862) /24 ( )] 97.5kN (7m)
= kNm

14. Solution:(cont.) Load from slab 2: beam 2/A-B
Design load, n2 = 1.35Gk + 1.5Qk
= 1.35 (5 kN/m2) + 1.5(3.0kN/m2)
= kN/m2
Total design load, W2 = n2 x width of slab 2
= kN/m2 x 1.5 m
= 16.9 kN/m

15. Solution:(cont.) Load from beam selfweight and brick wall
Selfweight of beam = 1 kN/m
Weight of brickwall per meter length = weight of brickwall x height of wall
= 3 kN/m2 x 4 m
= 12 kN/m
Total dead load, Gk = selfweight of beam + weight of brickwall
= 1 kN/m + 12 kN/m
= 13 kN/m
Design load, w3 = 1.35Gk = 1.35(13kN/m) = kN/m
Total design load for slab 2, brickwall and self weight of beam,
w4 = w2 + w3 = 16.9 kN/m kN/m
= 34.5 kN/m

16. Solution:(cont.)
Since beam AB is simply supported and loads are symmetry, then:
Maximum shear force, VEd2 = (wL)/2
= (34.5 kN/m)(7 m)/2
= kN
Maximum moment, MEd2 = (wL2)/8
= (34.5 kN/m)(7 m)2/8
= kNm

17. Solution:(cont.) Load combination from slab 1 and slab 2
The maximum shear force and moment from beam 2/A-B is obtained by combining the loads from slab 1 and slab 2 together as follow:
Maximum shear force, VEd = VEd1 + VEd2 =
= kN
Maximum moment, MEd = MEd1 + MEd2 =
= kNm

18. Beam Design Section Classification
Any steel beam sections that are subject to compression due to bending or an axial force should be classified.
The purpose of the classification is to determine whether local buckling influences the capacity of the beam.
The occurrence of local buckling of compressed elements of a cross-section prevents the development of full section capacity.
The classification of a section is carried out by comparing the width-to-thickness ration of the elements, i.e. c/t of the flange element and c/t of the web element against the limit of c/t of the flange and c/t of the web respectively given by Table 5.2 EC3

22. Example: Section classification
Determine the classification of a 406 x 140 x 46 UB in grade S275.
Solution:
Refer Table 3.1 page 26:
Steel grade, S275  t < 40 mm  fy = 275 N/mm2
Refer Table 5.2 page 42: ε = √ (235/fy) = √ (235/275) = 0.92
Flange: c/tf = < 9ε = 9 (0.92) = 8.28  flange is Class 1
Web: c/tw = 53.0 < 72ε = 72(0.92) = 66.2  web is Class 1
Since flange and web are class 1, the section is classified as Class 1.

23. Restrained and Unrestrained Beams
Steel beams may be design as either restrained or unrestrained.
If a beam full lateral restraint to its compression flange along the span, the beam is considered fully restrained.
Cases where beams can be designed as fully restrained along the spans are:
Beams carrying in-situ reinforced concrete slab
Beams with steel decking flooring systems, with or without shear studs. Shear studs function as a simple concrete anchor.

24. Restrained and Unrestrained Beams (cont.)
Deflection of restrained beam
The restrained beam is fully prevented from moving sideways.
The deflection will only take place vertically about the major axis without any lateral deflection.
Types of restraint
There are two conditions of restraint:
Fully restrained along the beam contributed by the concrete slab
Restrained at particular points only, normally at the connections, such the connections between beam-to-column or beam-to-beam.

25. Design of Restrained Beam
The design process for fully restrained beam is follows:
Analyze the beam and determine the reaction (R) ; maximum shear force (VEd); maximum moment (MEd) and maximum deflection.
Select suitable steel UB section.
Classify the section
Check the bending moment resistance
Check shear resistance
Check shear buckling resistance of web
Check combination of shear and moment
Check deflection

26. Bending Moment, Clause 6.2.5 EC3
The design resistance for bending for classes 1 and 2 cros sections
For class 3 cross section

27. Shear, Clause EC3
For plastic design, shear resistance of the section is given as
Shear area for main beam, Av
For main beams of rolled I and H sections, the shear area of the section is given as:
Where;

28. Shear area for secondary beam, Av
Where A = gross sectional area of the section
Shear Buckling Resistance of Web, Clause (6) EC3
If , shear buckling resistance no need to be carried out.
Where, depth of web,

29. Shear and Moment, Clause 6.2.8 EC3
If the shear force is small, in which, the moment resistance remains as for class 1 section.
If the shear force is small, in which, the design strength, fy should be reduced by factor (1 – ρ).
where

30. Example:
Beam AB with span 5.0 m is simply supported at A and B. From the analysis results, the maximum external design shear, VEd = 192 kN and maximum external design moment, MEd = 240 kNm. Using grade S275, Design beam AB.

31. Solution: Trial section size based on moment resistance, Wpl
So, try section 406 x 140 x 46 UB in grade S275 (Wpl = 888cm3)

32. Solution: (cont.) Section properties 406 x 140 x 46 UB
h = mm d = 360.4mm
b = mm A = 58.6 cm2
tw = 6.8 mm tf = 11.2 mm
Iy = cm4 Iz = 538 cm4
Wpl,y = 888 cm3 Wel, y = 778 cm3
d/tw = 53 c/tf = 5,13
E = N/mm2 G = N/mm2

33. Solution: (cont.) Design strength Section classification
Table 3.1: Steel grade S275
T< 40 mm fy = 275 N/mm2, fu = 430 N/mm2
Section classification
Refer Table 5.2 page 42: ε = √ (235/fy) = √ (235/275) = 0.92
Flange: c/tf = < 9ε = 9 (0.92) = 8.28  flange is Class 1
Web: c/tw = 53.0 < 72ε = 72(0.92) = 66.2  web is Class 1
Since flange and web are class 1, the section is classified as Class 1.

34. Solution: (cont.) Shear Resistance of Section (Clause 6.2.6)
Maximum external design shear force, VEd = 192 kN
Shear resistance of section, V c, Rd = V pl, Rd
For rolled I and H section and load parallel to the web, the shear area, Av must not less than Av,web
So, shear are of web only

35. Solution: (cont.) Shear area for main beam, Av
Therefore, use the biggest Av = mm2

36. Solution: (cont.)
Design check
 section is satisfactory.

37. Solution: (cont.) Bending Moment Resistance of Section (Clause 6.2.5)
Maximum external design moment, MEd = 240 kN
Moment resistance for class 1 cross-section:
Design check  section is satisfactory.

38. Solution: (cont.) Shear Buckling Resistance of Web (Clause 6.2.6(6))
If , shear buckling resistance no need to be carried out.
Web depth ,
So,
local web buckling is likely to occur and hence shear buckling check no need to be carried out.

39. Solution: (cont.) Combined bending and shear resistance (Clause 6.2.8)
Since , consider a section where
moment is maximum
Referring to shear force and bending moment diagrams, consider mid-span section where MEd = 240 kN;
Shear force at maximum moment is V = 0 kN.
0.5 Vc,Rd = 0.5 x 473 = kN
Since V = 0 kN < 0.5 Vc,Rd, it does not affect the moment resistance, Mc,Rd.
This beam section is able to carry the most critical combination of bending and shear.

40. Unrestrained Beam
The possibility of lateral-torsional buckling must be taken into consideration when the compression flange of the beam is not fully restrained along the span.
The buckling capacity of unrestrained beam depends on the
Section type.
Unrestrained length
Restraint condition
Type of load applied

41. Design of Unrestrained Beam
The following are the steps for designing unrestrained beams:
Divide the beam into segments between lateral restraints
Check the moment buckling resistance for each segment where is given as;
Section modulus, Wy;
Wy = Wpl,y; Plastic modulus for Class 1 and Class 2 sections
Wy = Wel,y, Elastic modulus for Class 3 section

42. The reduction factor χLT, is given by
Where,
Non dimensional slenderness,
Buckling parameter,

43. The critical elastic buckling moment,
Where:
E, G are material properties
Iz, It, Iw are section properties
Lcr is the buckling length of the member
C1 is the factor that depends on the shape of bending moment
diagram

44. If the moment along the beam is not uniform, then Mcr is modified by the C1 factor
The modification factor, C1 is governed by the ratio ψ which is the ration of the smaller to larger moments of restrained ends.
The ratio of smaller moment to larger moment is given as
ψ = Msmall / Mlarge for restrained ends
The C1 factor can be obtained from the following table (refer BS5950).

46. The C1 factor is never less than 1.0.
If C1 = 1.0, its means the moment along the beam is uniform.
C1 = 1.88 – 1.40 ψ ψ2

47. Buckling length, Lcr for designing unrestrained beams
Typical example of beams without intermediate lateral restraints and their corresponding buckling lengths.

48. Simple beams with intermediate lateral restraints
Simple supported beams with intermediate lateral restraints are described as beams where
Lateral restraints are provided at beam supports and at intermediate of beams

49. For example:
Effective length for segment AB,

50. The C1 factor: ψ = Msmall / Mlarge for restrained ends

51. Example: Design of Unrestrained Beam

53. Solution: Design load, q = 1.35Gk + 1.5Qk
= 1.35(3.0kN/m2) + 1.5(5.0kN/m2)
= 11.55kN/m2
Reaction of beam CD at support C
= point load at point C of beam AE
= (q x width x length)/2
= (11.55kN/m2 x 2.5 m x 6 m)/2
= 86.6 kN
Self-weight beam AE, w = 53kg/m x 9.81 x 10-3
= 0.52 kN/m

55. Section properties 406 x 140 x 53 UKB
h = mm d = 360.4mm
b = mm A = 67.9 cm2
tw = 7.9 mm tf = 12.9 mm
Iy = 183,00 cm4 Iz = 635 cm4
Iw = dm4 It = 29 cm4
Wpl,y = 1030 cm3 Wel, y = 899 cm3
d/tw = c/tf = 4.46
E = N/mm2 G = N/mm2
Lcr = 2.5 m r = 10.2 mm

56. Section classification
Design strength
Table 3.1: Steel grade S275
T< 40 mm fy = 275 N/mm2, fu = 430 N/mm2
Section classification
Refer Table 5.2 page 42: ε = √ (235/fy) = √ (235/275) = 0.92
Flange: c/tf = < 9ε = 9 (0.92) = 8.28  flange is Class 1
Web: c/tw = 45.6 < 72ε = 72(0.92) = 66.2  web is Class 1
Since flange and web are class 1, the section is classified as Class 1.

57. Solution: (cont.) Shear Resistance of Section (Clause 6.2.6)
Maximum external design shear force, VEd = 44.6 kN
Shear resistance of section, V c, Rd = V pl, Rd
Web shear area
So, shear are of web only

58. Solution: (cont.)
More accurate calculation of shear area considering the fillet area
Therefore, use the biggest Av = 3458 mm2

59. Solution: (cont.)
Design check
 section is satisfactory.

60. Solution: (cont.) Bending Moment Resistance of Section (Clause 6.2.5)
Maximum external design moment, MEd = kN
Moment resistance for class 1 cross-section:
Design check  section is satisfactory.

61. Solution: (cont.) Lateral torsional buckling (LTB)
Maximum external design moment, MEd = kNm
Moment buckling resistance
If the moment along the beam is not uniform, then Mcr is modified by the C1 factor.

62. Determine the modification factor, C1
Msmall = 0 kNm, Mlarge = kNm
ψ = Msmall / Mlarge = 0/109.9 = 0 for restrained ends = 0
C1 = 1.88 – 1.40 ψ ψ2
= 1.88 – 1.40(0) (0)2
= 1.88
Elastic critical buckling moment

63. 1 dm = 100 mm,  hence 1 dm6 = 1012 mm6
Non-dimensional slenderness for LTB

64. Determine the buckling parameter, ΦLT
Imperfection factor, αLT for LTB curves
h/b = mm/143.3 mm = 2.84 >2
curve ‘b’ } hence, αLT = 0.34
Reduction factor, ΧLT for LTB

65. Moment buckling resistance
Design check  section is satisfactory

66. Example: Design of restrained beam

68. Solution:

69. Solution: Section properties 533 x 210 x122 UKB
h = mm d = 476.5mm
b = mm A = 155 cm2
tw = 12.7 mm tf = 21.3 mm
Iy = 76,000 cm4 Iz = 3390 cm4
Iw = 2.32 dm4 It = 178 cm4
Wpl,y = 3200 cm3 Wel, y = 2790 cm3
d/tw = c/tf = 4.08
E = N/mm2 G = N/mm2
Lcr = 4.0 m r = 12.7 mm

70. Solution: (cont.) Design strength Section classification
Table 3.1: Steel grade S275
T< 40 mm fy = 275 N/mm2, fu = 430 N/mm2
Section classification
Refer Table 5.2 page 42: ε = √ (235/fy) = √ (235/275) = 0.92
Flange: c/tf = < 9ε = 9 (0.92) = 8.28  flange is Class 1
Web: c/tw = 37.5 < 72ε = 72(0.92) = 66.2  web is Class 1
Since flange and web are class 1, the section is classified as Class 1.

71. Solution: (cont.) Bending Moment Resistance of Section (Clause 6.2.5)
Maximum external design moment, MEd = 821 kNm
Moment resistance for class 1 cross-section:
Design check  section is satisfactory.

72. Solution: (cont.) Lateral torsional buckling (LTB)
Maximum external design moment, MEd = 821 kNm
Moment buckling resistance
If the moment along the beam is not uniform, then Mcr is modified by the C1 factor.

73. Determine the modification factor, C1
Msmall = 794 kNm, Mlarge = 794 kNm
ψ = Msmall / Mlarge = 794/794 = 1 for restrained ends
C1 = 1.88 – 1.40 ψ ψ2
= 1.88 – 1.40(1) (1)2
= 1.0
Elastic critical buckling moment

74. 1 dm = 100 mm,  hence 1 dm6 = 1012 mm6
Non-dimensional slenderness for LTB

75. Determine the buckling parameter, ΦLT
Imperfection factor, αLT for LTB curves
h/b = mm/211.9 mm = 2.6 >2
curve ‘b’ } hence, αLT = 0.34
Reduction factor, ΧLT for LTB

76. Moment buckling resistance
Design check  section is not satisfactory,
hence, larger section should be used.

77. Deflection
A beam may not fail due to excessive deflection, however, it is necessary to ensure that deflections are not excessive under unfactored imposed loading to prevent:
Damage to various architectural features such as interior walls, partitions, ceilings and exterior cladding
Severe cracking in brittle finishes such as brick wall with plaster finished
Damage to non-structural element such as glass façade, ceilings, partitions and other fragile elements.
Vertical deflection limit
The following table gives suggested limits for calculated vertical deflections of certain members under the characteristic load combination due to variable loads and should not include permanent loads.

78. Design situation
Vertical Deflection limit

79. Maximum Vertical Deflection due to external load
The following table shows the formula to calculate vertical deflection at mid-span of a simply supported beam with different loading types.
The formula can be used to calculate deflection of a restrained or unrestrained beam

81. Example:
Check deflection of beam shown in figure below

82. Solution: Deflection is checked under serviceability.
Therfore, only unfactored imposed load (1.0Qk) are considered in calculating the deflection. Dead loads are not included
Beam 406 x 178 x 60 UB
Clause EC3:
Modulus of Elasticity, E = N/mm2
From table of properties:
Second moment of inertia, Iy = cm4

83. Solution:(con.t) Maximum deflection due to unfactored imposed loads
If the beam is carrying wall with plaster finish or any other brittle finish, deflection limit,
 choose larger beam section

84. THANK YOU
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