Steel Beam Design

STEEL BEAM DESIGN

The flow chart below can be used as a guide to design a steel beam member.

HJ ROSLAN BIN KOLOPPUSAT PENGAJIAN DIPLOMA

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1.0 Initial Section Selection

Since bending is the critical mode of failure for most beam, therefore bending theory is used for initial section selection.

To avoid bending failure:

Design moment < Moment capacity of section M < Mc …………… (1) where Mc = pyS ……………(2)

py = design strength of steel S = plastic modulus of section

Combine (1) and (2) give:

S > M/py

Select from steel tables for S greater than the calculated value.

2.0 Classification of Section

The selected section is classified for the purpose of:

1. To identify if the section can sustain the formation of plastic hinge –thick flange and web. Thin flange and slender web usually fail by local buckling before yield stress is achieved.

2. To know the moment capacity or the section moment of resistance.


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Clause 3.5.2 : Four classes of section.

Class 1 : Plastic cross sections. Plastic hinge can be developed with significant rotation capacity(see graph below). If plastic design method is used in structural analysis, this class must be used.


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Class 2 : Compact. Full plastic moment capacity can be developed, but local buckling may prevent the development of plastic hinge with sufficient rotation capacity to permit plastic design.

Class 3 : Semi-compact. Can develop elastic moment capacity but local buckling prevent the development of full plastic moment.

Class 4: Slender. Slender element subjected to compression due to moment or axial load. Local buckling prevent full elastic moment capacity to develop.

Limiting width to thickness ration for elements for the above classes are given in Table 11, BS5950.


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Once the section has been classified, the various strength check can be carried out to assess its suitability.

CONTOH

Semak kelas keratan 305 x 102 x 33 UB,py=275 N/mm2

Dari Jadual Keluli, B=102.4mm, T=10.8mm,t=6.6mm d = (275.8 – 2x7.6) =260.6 b= 102.4/2 = 51.2mmDari jadual 7, BS5950 ε = (275/275)0.5 = 1

b/T = 51.3/10.8 = 4.8 < 9 ε d/t = 260.6/6.6 = 39.5 <80 ε

Berdasarkan jadual di atas keratan adalah plastik.

3.0 Strength Check :

Shear

Clause 4.2.3 : The shear force, Fv, should not exceed the shear capacity, Pv

Fv ≤ Pv …………….(3)

Where Pv = 0.6 pyAv ……………..(4) (Assuming that the web carries the shear force alone)

Av = shear area ( = tD for I, H and channel section)


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Clause 4.2.3: when buckling ratio (d/t) > 70ε (see equation 4.4) web should be additionally checked for shear buckling.

MOMENT CHECK

(1) Low shear and Moment Capacity

Clause 4.2.1.1 : At critical points the combination of (i) maximum moment and co-existent shear(ii) maximum shear and co-existent momentshould be checked.

If shear force Fv < 0.6Pv, this is called low shear load.

If 0.6Pv < Fv <Pv then this is called high shear load.


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For low shear load, the moment capacity of the section is calculated using clause 4.2.5.2 :

For Class 1 Plastic or Class 2 compact, the moment capacity is:

Mc = py S ≤ 1.2 py Z ……………….(5)

py - design strength of the steel S - plastic modulus Z - section modulus

Additional check i.e ( Mc ≤ 1.2pyZ) is to guard against plastic deformation under serviceability loads and is applicable to simply supported and cantilever beams. Other beam the limit is 1.5pyZ.

For class 3 Semi-compact:

Mc = py Z ………………………. (6)Or Mc = pySeff ≤ 1.2pyZ …………….(7)

Seff – effective plastic modulus (clause 3.5.6)

For class 4 Slender, Mc = pyZeff …………….(8)

Zeff - effective section modulus (clause 3.6.2)

Note: Nearly all sections in grade S275 steel are plastic, only a few sections in higher strength are semi compact. No British rolled UB section in pure bending are slender or plastic/compact flanges and semi-compact webs]


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EXAMPLE 1

A simply supported UB steel beam of 10 m span support loads as shown below. Assuming that the beam is laterally supported find the suitable UB cross section that suits the shear and bending moment requirement.

5m 30 kN (imposed load)

5kN/m (dead load) 5 kN/m (imposed load)

10m RA RB

Answer:

1. Calculate the design shear and bending moment.

Total design load = (1.6 x 30) + (1.4 x 5 + 1.6x5)10 = 48 + 15 x 10 = 198 kN

Due to symmetry, RA = RB= 198/2 = 99 kN

Mid span moment, M = Wl/4 + wl2/8 = (48x10) + (15x10 2 ) 4 8 = 120 + 187.5 = 307.5 kNm


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2. Initial cross-section

Assume py = 275 N/mm2

Sx > M/py = 307.5 x 106/275 = 1.118 x 106 mm3 = 1118 cm3

From steel table, the following cross-section can be used:

1. 356 x 171 x 67 UB : Sx = 1210 cm3

2. 406 x 178 x 60 UB : Sx= 1200 cm3

3. 457 x 152 x 60 UB: Sx = 1290 cm3

Selection is based on economy or engineering consideration. If lighter section is required that section no. 3 is chosen. If depth consideration is important than section No. 1 is chosen.

Say we select section No. 2406 x 178 x 60 UB : Sx= 1200 cm3

3. Classification

Strength classification: T = 12.8(<16mm) therefore py=275 N/mm2 (Table 9) dan ε = (275/py)1/2 = 1 (Table 11)

Section classification:b/T = 6.95 < 9 ε plastic (Table 11)d/t = 46.2 < 80 ε plastic (Table 11)

Therefore the section 406 x 178 x 60 (UB) is class 1 plastic.


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4. Shear Force

d/t = 46.2 <70 ε therefore check on shear buckling is not required.(clause 4.2.3)

The capacity of section, Pv = 0.6pyA = 0.6 pytD = 0.6 x 275 x 7.9 x 406.4 = 523 kN>99(Fv)

(Reaction), Fv (99kN) < 0.6Pv

= 0.6 x 523 = 314 kN (Low shear case) Clause 4.2.5

5. Momen lentur

Since the condition is low shear, Mc = py S = 275 x 1200 x 103

= 327 kNm < 1.2pyZ =1.2x275x1060x103 = 349.8 kNm OK

Design moment not including self weight = 307.5 kNm.

Bending moment due to self weight :

Msw = 1.4(60x9.81/103)102/8 = 10.3 kNm

Total moment (r/b) = 307.5 + 10.3 = 317.8 kNm < 327 kNm.

Therefore the section is suitable.


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(2) High Shear and Moment Capacity

Fv > 0.6Pv (high shear load), then the moment carrying capacity of the section is reduced.This is because the web cannot take the full tensile or compressive stresses associated with bending moment as well as a substantial shear stress due to shear load.

Clause 4.2.5.3: Moment capacity should be calculated using:

Class 1 plastic and compact: Mc = py (S –ρSv)

Where, ρ =[ 2(Fv/Pv – 1]2 Sv = tD2/4 (plastic modulus of shear area)

Note: The effect of ρ factor is to reduce the moment carrying capacity of the web as the shear load rises from 50 to 100% of the web shear capacity. However, the resulting moment capacity is negligible when Fv < 0.6Pv.

EXAMPLE 2

A steel cantilever beam is encased in a concrete wall and is subjected to the load shown. Choose a suitable UB that fulfill the criteria for of moment and shear. 450 kN (total dead load) 270 kN (total imposed load)

1 m A


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Solution:

Design shear (Fv) = (1.4x450) + (1.6x270) = 1062 kN

Design momen at A = Wl/2 = 1062x1/2 = 531 kNm

Initial sizing: Assume py=275 N/mm2

Sx > M/py = 531x106/275 = 1931 cm3

From Table of steel properties, the suitable size are:

1. 457x191x89 UB: Sx = 2010 cm3,.2. 533 x 210 x 82 UB: Sx = 2060 cm3,

Classification of section shown that the section is plastic.

Shear strength :

Try section No.2, Pv = 0.6pytD = 0.6 x 275 x 9.6 x 528.3 = 837 kN < 1062 (Fv) kN not OK.

Try a new section where the shear strength is more critical than the bending strength.

Try 610 x 229 x 113 UB:Sx=3290 cm3,py=265,plastic

Pv = 0.6 x 265 x 607.3 x 11.2 = 1081 kN > Fv O.K


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But,

Fv > 0.6x1081(Pv)=648.6 kN high shear Clause 4.2.6

Calculate Moment capacity from clause 4.2.5.3:

ρ =[ 2(Fv/Pv – 1]2 = {2(1062/1081) –1}2 =0.93

Mc = py(Sx -Sv ρ )

= 265[3290x103 – (11.2 x 607.32/4)0.93] =617 kNm

Moment from self weight: Mbk =1.4(113x9.81/103)12/2 = 0.8 kNm

Total moment, Mj =Mbk + M = 531 + 0.8 =532 kNm < Mc Therefore the section is suitable

Deflection

Check for maximum deflection using combination of unfactored imposed serviceability loading.

Clause 2.5.2 and Table 8 outlined recommended limits to these deflections to avoid significant damage to the structure and finishes. Standard formula or other method such as Area-moment, Macaulay etc can be used to calculate deflection.


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Standard formula:

CONTOH 3

Calculate the deflection of the beam in example 1 and 2.

From Example 1.

Δmak = 5wL4/384EI + WL3/48EI

= 5x5x10 4 + 30x10 3 384x205x106x21500x10-8 48x205x106x21500x10-8 = 0.0148 + 0.0142 = 0.029 m = 29 mm

From Table 4.5, the maximum allowable is span/360 = 10 000/360 = 27.8 mm for beam carrying plaster and brittle finishing. It can been seen that the calculated deflection is


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bigger than the allowable deflection therefore deeper cross-section is recommended.

If other than the specified beam is used, the allowable deflection is span/200.

= 10000/200 = 50 mm which is OK.

From example 2,

Δmak = wL4/8EI = 270 x 1 4 8 x 205 x106x 87400 x10-8

= 0.00019 m = 0.19mm

From Table 4.5, the allowable deflection is span/180 = 1000/180 = 5.6mm

Deflection OK


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WEB BEARING AND WEB BUCKLING

Clause 4.5. The web bearing capacity and web buckling resistance should be performed when unstiffened webs are subjected to high concentration locals loads or reactions that act through the web and flange of the beam.

Stiffeners should be provided if the strength of the unstiffened web is not sufficient.

Web Bearing Resistance

Figure below shows how concentrated load transmitted through the flange/web connection in the span, and at supports when the distance to the end of the member from the end of the stiff bearing is zero.


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Clause 4.5.2.1 gives the bearing resistance Pbw calculated at the flange/web connection for unstiffened web as:

Pbw = (b1 + nk) tpyw

Where:

b1 _ stiff bearing length

Clause 4.5.1.3 (Fig 13) shows cases of stiff bearing length that can be calculated using the formula given.


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n as shown in Fig (a): n =5 except at the end of a member and n = 2 + 0.6be/k ≤ 5 at the end of the member.be Distance to the end of the member from the end of the stiff bearing Fig (b).k = (T + r) from rolled I or H sectionsT thickness of the flanget web thicknesspyw design strength of web

The formula is a simple check to ensure the stress at critical point on flange/web connection is less than the strength of steel.

Check also the contact stresses between load or support and flange do not exceed py.


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Web buckling

Clause 4.5.3.1 : provided the distance αe from the concentrated load or reaction from the nearer end of the member is at least 0.7d, and if the flange through which the load or reaction is applied is effectively restrained against both:

(a) rotation relative to the web(b) lateral movement relative to the other flange (Fig

below)

the buckling resistance of an unstiffened web is given by:

Px = 25εtPbw /√(b1 + nk)d

Or if αe < 0.7d, the buckling resistance of unstiffened web is given by:

Px = (αe + 0.7d).25εt.Pbw/ 1.4d√(b1 + nk)d

If the flange is not restrained against rotation and/or lateral movement the buckling resistance of the web is reduced to:


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Pxr = 0.7d.Px/Le

Where Le is the effective length of the web determine using Table 22.

Example:

Check web bearing and buckling for Example 1, assuming the beam sits on stiff bearing length, b1 of 100 mm at each end.(406x178x60, d = 360.4, r =10.2, T=12.8,t=7.9)

Ans:

Bearing capacity:Pbw = (b1 + nk) tpyw

= (100 + 2 x 23) 7.8 x 275 = 313 kN > 99 kN OK

where k = T + r = 12.8 + 10.2 = 23 mm n = 2 + 0.6be/k = 2 (since be =0)

Contact Stress At Supports

Pcs = (b1 x 2(r + T))py = (100 x 46) x 275 = 1265 kN > 99 kN OK


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Web Buckling at Support

Since αe (=50mm) < 0.7d = 0.7 x 360.5 = 252 mm, buckling resistance of web is:

Px = (αe + 0.7d).25εt.Pbw/ 1.4d√(b1 + nk)d

= (50 + 252) x 25 x 1 x 7.8 x 313 1.4 x 360.5 √(100 + 2 x 23) 360.5 = 159 kN> 99 kN OK

No web stiffners are required at supports.

Example

Compute the bearing capacity, Pbw and the buckling capacity, Px of unstiffened web of a beam subjected to a concentrated load of 200 kN as shown below. Assume steel grade of S275.


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Solutions:

For 305 x 127 UB 37,t = 7.2 mmT=10.7 mmr = 8.9 mm (root radius)

For 686x 254 UB 170 sectiont = 14.5 mmT= 23.7 mmr = 15.2 mmd = 615 mm


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CHECK WEB BEARING

Bearing capacity for web, Pbw = (b1 + nk) tpyw

where b1 = t + 1.6r + 2T (FIG. 13 for secondary beam) = 7.2 + 1.6x8.9 + 2x10.7 = 42.8 mm

n = 5 (except at end of member)

k = T + r = 23.7 + 15.2 = 38.9 mmPbw = (b1 + nk) tpyw

= (42.8 + 5x38.9)x14.5 x 275 x 10-6

= 946 kN > 200 kN OK

Buckling Capacity Check

Since αe > 0.7d, use Px = 25εtPbw /√(b1 + nk)d = 25 x 1 x 14.5 x 946 / √ (42.8 + 5x38.9)615 = 898 kN > 200 kN OK

No stiffeners is needed.


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Tutorial

Q1 The beam shown below is fully restrained along its length and has a stiff bearing of 75 mm at the supports. The beam supports a secondary beam at the centre of its length. The characteristic loads acting on the beams are shown. Design the beam using S275 steel. The secondary beam is supported by a bearing plate which has 50 mm stiff bearing length on the top flange of the beam.

Q2 The figure below shows a layout plan of a floor. The floor slabs are of precast concrete. The characteristic loadings on the floor are as follows:

Self weight = 3.5 kN/m2

Floor finishing = 1.0 kN/m2

Imposed load = 3.0 kN/m2

Design beam 2/A-C using S275 steel assuming the beam is fully restrained. The beam is supported by angles 120x120x15 mm at both ends. The self-weight of the beam is assumed as 0.7 kN/m.


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(i) Assume a trial section.(ii) Classify the section(iii) Check the shear capacity at support(iv) Check moment capacity

(v) Check deflection given: ∂=

5wd L4

384 EI+W ia

2b2

3EIL

(vi) Check bearing and buckling at support.


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Steel Beam Design

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