ERT 352 FARM STRUCTURES BEAM DESIGN It doesnt

ERT 352 FARM STRUCTURES BEAM DESIGN It doesn’t matter what the subject is; once you’ve learnt how to study, you can do anything you want.

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.

ANALYSIS – LOAD DISTRIBUTION • Loads from slab are normally defined in “q”k. N/m 2. • These loads are transferred to supporting beams in either “w”k. N/m or “W” k. N. • Loads from reinforced concrete solid slab: o Loads may be distributed to the supporting beams depending on the ratio of long side/short side of the slab (Ly/Lx) o Type of distribution loads from slab: • One-way spanning slab, when Ly/Lx > 2. 0 • Two-way spanning slab, when Ly/Lx ≤ 2. 0

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

ANALYSIS – LOAD DISTRIBUTION (cont. ) 1. One-way spanning slab (Ly/Lx > 2. 0) o o 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

ANALYSIS – LOAD DISTRIBUTION (cont. ) 2. Two-way spanning slab (Ly/Lx ≤ 2. 0) o o 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

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: o Unfactored dead load from selfweight of precast slabs, selfweight of steel beams and finishing o Unfactored imposed load = 5. 0 k. N/m 2 = 4. 0 k. N/m 2 • Determine the maximum shear force and moment in beam 1/A-B.

Solution: • Load distribution: o for precast slab, considered as one-way spanning and it is supported by beam AB and CD. • Design load, n n = 1. 35 Gk + 1. 5 Qk = 1. 35 (5 k. N/m 2) +1. 5 (4. 0 k. N/m 2) = 12. 75 k. N/m 2 • Total design load, w w = n x width of load transferred to beam 1/A-B = 12. 75 k. N/m 2 x (2 m) = 25. 5 k. N/m

Solution: (cont. ) • Since beam AB is simply supported and loads are symmetry, then: o Maximum shear force, VEd = (w. L)/2 = (25. 5 k. N/m)(5 m)/2 = 63. 75 k. N o Maximum moment, MEd = (w. L 2)/8 = (25. 5 k. N/m)(5 m)2/8 = 79. 7 k. Nm

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 4 m high brickwall of 3 k. N/m 2. determine the maximum shear force and moment of beam 2/A-B.

Solution: • Load distribution on reinforced concrete solid slab o Slab 1: Ly/Lx = 7 m/4 m = 1. 75 < 2. 0, two way spanning slab o Slab 2: Ly/Lx = 7 m/3 m = 2. 3 > 2. 0, one way spanning slab

Solution: (cont. ) • Load from slab 1: beam 2/A-B o Design load, n 1 = 1. 35 Gk + 1. 5 Qk = 1. 35 (5 k. N/m 2) + 1. 5(2. 0 k. N/m 2) = 9. 75 k. N/m 2 o Total design load, W 1 = n 1 x trapezodial area of load from slab 1 = 9. 75 k. N/m 2 x [(3 m +7 m)/2 x 2 m] = 97. 5 k. N

Solution: (cont. ) • Since beam is simply supported and loads are symmetry, then: o Maximum shear force, VEd 1 = (w. L)/2 = (97. 5 k. N/m)(5 m)/2 = 48. 75 k. N o From the bending moment diagram; a. L = 2 m a = 2 m / 7 m = 0. 286 o Maximum moment, MEd 1 = [(3 – 4 a 2) /24 (1 - a)] WL = [(3 – 4 (0. 2862) /24 (1 - 0. 286)] 97. 5 k. N (7 m) = 106. 5 k. Nm

Solution: (cont. ) • Load from slab 2: beam 2/A-B o Design load, n 2 = 1. 35 Gk + 1. 5 Qk = 1. 35 (5 k. N/m 2) + 1. 5(3. 0 k. N/m 2) = 11. 25 k. N/m 2 o Total design load, W 2 = n 2 x width of slab 2 = 11. 25 k. N/m 2 x 1. 5 m = 16. 9 k. N/m

Solution: (cont. ) • Load from beam selfweight and brick wall o Selfweight of beam = 1 k. N/m o Weight of brickwall per meter length = weight of brickwall x height of wall = 3 k. N/m 2 x 4 m = 12 k. N/m o Total dead load, Gk = selfweight of beam + weight of brickwall = 1 k. N/m + 12 k. N/m = 13 k. N/m o Design load, w 3 = 1. 35 Gk = 1. 35(13 k. N/m) = 17. 55 k. N/m o Total design load for slab 2, brickwall and self weight of beam, w 4 = w 2 + w 3 = 16. 9 k. N/m + 17. 55 k. N/m = 34. 5 k. N/m

Solution: (cont. ) • Since beam AB is simply supported and loads are symmetry, then: o Maximum shear force, VEd 2 = (w. L)/2 = (34. 5 k. N/m)(7 m)/2 = 120. 75 k. N o Maximum moment, MEd 2 = (w. L 2)/8 = (34. 5 k. N/m)(7 m)2/8 = 211. 3 k. Nm

Solution: (cont. ) • Load combination from slab 1 and slab 2 o 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: o Maximum shear force, VEd = VEd 1 + VEd 2 = 48. 75 + 120. 75 = 169. 5 k. N o Maximum moment, MEd = MEd 1 + MEd 2 = 106. 5 + 211. 3 = 317. 8 k. Nm

Beam Design • Section Classification o Any steel beam sections that are subject to compression due to bending or an axial force should be classified. o The purpose of the classification is to determine whether local buckling influences the capacity of the beam. o The occurrence of local buckling of compressed elements of a cross-section prevents the development of full section capacity. o 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 EC 3

Example: Section classification • Determine the classification of a 406 x 140 x 46 UB in grade S 275. • Solution: o Refer Table 3. 1 page 26: • Steel grade, S 275 t < 40 mm fy = 275 N/mm 2 o Refer Table 5. 2 page 42: ε = √ (235/fy) = √ (235/275) = 0. 92 o Flange: c/tf = 5. 13 < 9ε = 9 (0. 92) = 8. 28 flange is Class 1 o Web: c/tw = 53. 0 < 72ε = 72(0. 92) = 66. 2 web is Class 1 o Since flange and web are class 1, the section is classified as Class 1.

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: o Beams carrying in-situ reinforced concrete slab o Beams with steel decking flooring systems, with or without shear studs. Shear studs function as a simple concrete anchor.

Restrained and Unrestrained Beams (cont. ) • Deflection of restrained beam o The restrained beam is fully prevented from moving sideways. o The deflection will only take place vertically about the major axis without any lateral deflection. • Types of restraint o 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.

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

Bending Moment, Clause 6. 2. 5 EC 3 • The design resistance for bending for classes 1 and 2 cros sections • For class 3 cross section

Shear, Clause 6. 2. 6 EC 3 • For plastic design, shear resistance of the section is given as • Shear area for main beam, Av o For main beams of rolled I and H sections, the shear area of the section is given as: o Where;

• Shear area for secondary beam, Av o Where A = gross sectional area of the section • Shear Buckling Resistance of Web, Clause 6. 2. 6 (6) EC 3 o If , shear buckling resistance no need to be carried out. o Where, depth of web,

• Shear and Moment, Clause 6. 2. 8 EC 3 o If the shear force is small, in which, the moment resistance remains as for class 1 section. o If the shear force is small, in which, the design strength, fy should be reduced by factor (1 – ρ). where

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 k. N and maximum external design moment, MEd = 240 k. Nm. Using grade S 275, Design beam AB.

Solution: • Trial section size based on moment resistance, Wpl • So, try section 406 x 140 x 46 UB in grade S 275 (Wpl = 888 cm 3)

Solution: (cont. ) • Section properties 406 x 140 x 46 UB h = 403. 2 mm d = 360. 4 mm b = 142. 2 mm A = 58. 6 cm 2 tw = 6. 8 mm tf = 11. 2 mm Iy = 15700 cm 4 Iz = 538 cm 4 Wpl, y = 888 cm 3 Wel, y = 778 cm 3 c/tw = 53 c/tf = 5, 13 E = 210000 N/mm 2 G = 81000 N/mm 2

Solution: (cont. ) • Design strength o Table 3. 1: Steel grade S 275 o T< 40 mm fy = 275 N/mm 2, fu = 430 N/mm 2 • Section classification o Refer Table 5. 2 page 42: ε = √ (235/fy) = √ (235/275) = 0. 92 o Flange: c/tf = 5. 13 < 9ε = 9 (0. 92) = 8. 28 flange is Class 1 o Web: c/tw = 53. 0 < 72ε = 72(0. 92) = 66. 2 web is Class 1 o Since flange and web are class 1, the section is classified as Class 1.

Solution: (cont. ) • Shear Resistance of Section (Clause 6. 2. 6) o Maximum external design shear force, VEd = 192 k. N o Shear resistance of section, V c, Rd = V pl, Rd o For rolled I and H section and load parallel to the web, the shear area, Av must not less than Av, web o So, shear are of web only

Solution: (cont. ) • Shear area for main beam, Av • Therefore, use the biggest Av = 2979. 3 mm 2

Solution: (cont. ) • Design check section is satisfactory.

Solution: (cont. ) • Bending Moment Resistance of Section (Clause 6. 2. 5) o Maximum external design moment, MEd = 240 k. N o Moment resistance for class 1 cross-section: o Design check section is satisfactory.

Solution: (cont. ) • Shear Buckling Resistance of Web (Clause 6. 2. 6(6)) o If , shear buckling resistance no need to be carried out. o Web depth , o So, local web buckling is likely to occur and hence shear buckling check no need to be carried out.

Solution: (cont. ) • Combined bending and shear resistance (Clause 6. 2. 8) o Since , consider a section where moment is maximum o Referring to shear force and bending moment diagrams, consider mid-span section where MEd = 240 k. N; • Shear force at maximum moment is V = 0 k. N. • 0. 5 Vc, Rd = 0. 5 x 473 = 236. 5 k. N • Since V = 0 k. N < 0. 5 Vc, Rd, it does not affect the moment resistance, Mc, Rd. o This beam section is able to carry the most critical combination of bending and shear.

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 o Section type. o Unrestrained length o Restraint condition o Type of load applied

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

• The reduction factor χLT, is given by • Where, o Non dimensional slenderness, o Buckling parameter,

• The critical elastic buckling moment, • Where: o E, G o Iz, It, Iw o Lcr o C 1 are material properties are section properties is the buckling length of the member is the factor that depends on the shape of bending moment diagram

• If the moment along the beam is not uniform, then Mcr is modified by the C 1 factor • The modification factor, C 1 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 C 1 factor can be obtained from the following table (refer BS 5950).

• The C 1 factor is never less than 1. 0. • If C 1 = 1. 0, its means the moment along the beam is uniform. C 1 = 1. 88 – 1. 40 ψ + 0. 52 ψ2

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

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

• For example: o Effective length for segment AB, o Effective length for segment CD,

• The C 1 factor: o ψ = Msmall / Mlarge for restrained ends o C 1 = 1. 88 – 1. 40 ψ + 0. 52 ψ2

Example: Design of Unrestrained Beam

Solution: • Design load, q = 1. 35 Gk + 1. 5 Qk = 1. 35(3. 0 k. N/m 2) + 1. 5(5. 0 k. N/m 2) = 11. 55 k. N/m 2 • Reaction of beam CD at support C = point load at point C of beam AE = (q x width x length)/2 = (11. 55 k. N/m 2 x 2. 5 m x 6 m)/2 = 86. 6 k. N • Self-weight beam AE, w = 53 kg/m x 9. 81 x 10 -3 = 0. 52 k. N/m

• Section properties 406 x 140 x 53 UKB h = 406. 6 mm d = 360. 4 mm b = 143. 3 mm A = 67. 9 cm 2 tw = 7. 9 mm tf = 12. 9 mm Iy = 183, 00 cm 4 Iz = 635 cm 4 Iw = 0. 246 dm 4 It = 29 cm 4 Wpl, y = 1030 cm 3 Wel, y = 899 cm 3 c/tw = 45. 6 c/tf = 4. 46 E = 210000 N/mm 2 G = 81000 N/mm 2 Lcr = 2. 5 m r = 10. 2 mm

• Design strength o Table 3. 1: Steel grade S 275 o T< 40 mm fy = 275 N/mm 2, fu = 430 N/mm 2 • Section classification o Refer Table 5. 2 page 42: ε = √ (235/fy) = √ (235/275) = 0. 92 o Flange: c/tf = 4. 46 < 9ε = 9 (0. 92) = 8. 28 flange is Class 1 o Web: c/tw = 45. 6 < 72ε = 72(0. 92) = 66. 2 web is Class 1 o Since flange and web are class 1, the section is classified as Class 1.

Solution: (cont. ) • Shear Resistance of Section (Clause 6. 2. 6) o Maximum external design shear force, VEd = 44. 6 k. N o Shear resistance of section, V c, Rd = V pl, Rd o Web shear area o So, shear are of web only

Solution: (cont. ) • More accurate calculation of shear area considering the fillet area • Therefore, use the biggest Av = 3458 mm 2

Solution: (cont. ) • Design check section is satisfactory.

Solution: (cont. ) • Bending Moment Resistance of Section (Clause 6. 2. 5) o Maximum external design moment, MEd = 109. 9 k. N o Moment resistance for class 1 cross-section: o Design check section is satisfactory.

Solution: (cont. ) • Lateral torsional buckling (LTB) o Maximum external design moment, MEd = 109. 9 k. Nm o Moment buckling resistance o If the moment along the beam is not uniform, then Mcr is modified by the C 1 factor.

o Determine the modification factor, C 1 Msmall = 0 k. Nm, Mlarge = 109. 9 k. Nm ψ = Msmall / Mlarge = 0/109. 9 = 0 for restrained ends = 0 C 1 = 1. 88 – 1. 40 ψ + 0. 52 ψ2 = 1. 88 – 1. 40(0) + 0. 52(0)2 = 1. 88 o Elastic critical buckling moment

1 dm = 100 mm, hence 1 dm 6 = 1012 mm 6 • Non-dimensional slenderness for LTB

o Determine the buckling parameter, ΦLT Imperfection factor, αLT for LTB curves h/b = 406. 6 mm/143. 3 mm = 2. 84 >2 curve ‘b’ } hence, αLT = 0. 34 o Reduction factor, ΧLT for LTB

• Moment buckling resistance • Design check satisfactory section is

Example: Design of restrained beam

Solution:

Solution: • Section properties 533 x 210 x 122 UKB h = 544. 5 mm d = 476. 5 mm b = 211. 9 mm A = 155 cm 2 tw = 12. 7 mm tf = 21. 3 mm Iy = 76, 000 cm 4 Iz = 3390 cm 4 Iw = 2. 32 dm 4 It = 178 cm 4 Wpl, y = 3200 cm 3 Wel, y = 2790 cm 3 c/tw = 37. 5 c/tf = 4. 08 E = 210000 N/mm 2 G = 81000 N/mm 2 Lcr = 4. 0 m r = 12. 7 mm

Solution: (cont. ) • Design strength o Table 3. 1: Steel grade S 275 o T< 40 mm fy = 275 N/mm 2, fu = 430 N/mm 2 • Section classification o Refer Table 5. 2 page 42: ε = √ (235/fy) = √ (235/275) = 0. 92 o Flange: c/tf = 4. 08 < 9ε = 9 (0. 92) = 8. 28 flange is Class 1 o Web: c/tw = 37. 5 < 72ε = 72(0. 92) = 66. 2 web is Class 1 o Since flange and web are class 1, the section is classified as Class 1.

Solution: (cont. ) • Bending Moment Resistance of Section (Clause 6. 2. 5) o Maximum external design moment, MEd = 821 k. Nm o Moment resistance for class 1 cross-section: o Design check section is satisfactory.

Solution: (cont. ) • Lateral torsional buckling (LTB) o Maximum external design moment, MEd = 821 k. Nm o Moment buckling resistance o If the moment along the beam is not uniform, then Mcr is modified by the C 1 factor.

o Determine the modification factor, C 1 Msmall = 794 k. Nm, Mlarge = 794 k. Nm ψ = Msmall / Mlarge = 794/794 = 1 for restrained ends C 1 = 1. 88 – 1. 40 ψ + 0. 52 ψ2 = 1. 88 – 1. 40(1) + 0. 52(1)2 = 1. 0 o Elastic critical buckling moment

1 dm = 100 mm, hence 1 dm 6 = 1012 mm 6 • Non-dimensional slenderness for LTB

o Determine the buckling parameter, ΦLT Imperfection factor, αLT for LTB curves h/b = 544. 5 mm/211. 9 mm = 2. 6 >2 curve ‘b’ } hence, αLT = 0. 34 o Reduction factor, ΧLT for LTB

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

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: o Damage to various architectural features such as interior walls, partitions, ceilings and exterior cladding o Severe cracking in brittle finishes such as brick wall with plaster finished o Damage to non-structural element such as glass façade, ceilings, partitions and other fragile elements. • Vertical deflection limit o 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.

Design situation Vertical Deflection limit

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

Example: • Check deflection of beam shown in figure below

Solution: • Deflection is checked under serviceability. • Therfore, only unfactored imposed load (1. 0 Qk) are considered in calculating the deflection. Dead loads are not included • Beam 406 x 178 x 60 UB • Clause 3. 2. 6 EC 3: o Modulus of Elasticity, E = 210 000 N/mm 2 • From table of properties: o Second moment of inertia, Iy = 21 600 cm 4

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

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