3.Steel Tutorial solution Flexural members PDF

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Steel Design Tutorial - Flexural Members - SolutionQuestion 1:A simply supported universal beam is subject to bending about its x-axis due to thefollowing uniformly distributed service loads.Permanent action (including estimated beam weight) = 10 kN/mImposed action (not storage) = 5 kN/mBeam span = ...

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Steel Design Tutorial - Flexural Members - Solution Question 1: A simply supported universal beam is subject to bending about its x-axis due to the following uniformly distributed service loads. Permanent action (including estimated beam weight) = 10 kN/m Imposed action (not storage) = 5 kN/m Beam span = 12m. Assume that the beam is continuously laterally restrained over its whole length. Determine a suitable UB section (grade Onesteel-300PLUS) considering bending and an allowable maximum deflection of span/250. G = 10 kN/m, Q = 5 kN/m w* = 1.2G + 1.5Q = 1.2 * 10 + 1.5 * 5 = 19.5 kN/m simply supported beam, span L = 12 m  M* =

V* =

 * L2 8

=

19.5*12 2 = 351kNm 8

ω*L 19.5 *12 = = 117 kN 2 2

Check serviceability first wserv = G + sQ = 10 + 0.7 * 5 = 13.5 kN/m

=

5 wserv L4 span 12000 max value = = = 48mm 384 EIx 250 250

Solve for Ix =

13.5*120004 5 * = 380*10 6 mm 4 3 384 200 *10 * 48

From Onesteel handbook Table 13, try 530UB82.0 (Ix = 477 * 106 mm4) Check bending: Beam is laterally restrained for full length  Mbx = Msx  M* ≤  Msx = 0.9 fy Zex For 530 UB 82.0, fy = 300 MPa (use flange value for bending) Onesteel Handbook – Table 14 Zex = 2070 * 103 mm3   Msx

= 0.9 * 300 * 2070 * 103 = 559 kNm > 351 kNm  OK for bending

Question 2 Check if the UB section selected in problem 1 would be satisfactory if it is fully laterally restrained at the supports and at mid span only. Check 530 UB 82.0 if le = 6m, instead of fully restrained Check if full lateral restraint

i.e.

le 250  (80 + 50m ) ry fy

Sec. 5.3.2.4

le = 6000 mm, ry = 43.8 mm  le / ry = 137

m = -1 (conservative) or ratio of end moments in segment = 0/M* = 0

(80 + 50 * 0)

250 le = 73.0   not fully restrained 300 ry

 M* ≤ Mbx m: (1)

where Mb = m s Msx ≤ Msx

Could take as 1.0 – always OK, always conservative

(2)

Could use Table 5.6.1 – remember table shows segments not beams. In this case none of them actually apply

(3)

Could use the 1.7 M*m formula. Most accurate, but for parabolas, time consuming.

So take

m = 1.0

 αs = 0.6   

 M  M  2  s  + 3  −  s   M oa  Moa 

    

Ms = 300 * 2070 * 103 (from Q1) = 621 kNm

Mo =

 π 2 EI y   π 2 EI w   GJ +  2 2  l e   le

   

E = 200 * 103 MPa, G = 80 * 103 MPa – constants for steel Iy = 20.1 * 106mm4 ) J = 526 * 103mm4 ) Onesteel table 13 Iw = 1330 * 109mm6 ) le = 6000 mm

Mo = =



 π 2 * 200 *10 3 * 20.1*10 6  6000 2 

 π 2 * 200 *10 3 *1330 *10 9   80 *10 3 * 526 *10 3 +    6000 2  

   

356 kNm

 αs = 0.6   

  621   621 2   + 3 −   356 356       

= 0.428 

Mbx = 1.0 * 0.428 * 621 = 266 kNm Mbx = 0.9 * 266 = 239 kNm < 351 kNm

 530 UB 82.0 would not be adequate if only laterally restrained at ends and mid-span NB

m for ‘half’ a parabola is 1.33 Hence Mbx = 0.9 * 266 * 1.33 = 318 kNm < 351 kNm So the beam is still not sufficient with only lateral restraints at each end and midspan. m for a full parabola is 1.13.

Question 3 Determine a suitable UB section (grade Onesteel-300PLUS) for the beam shown. The P beam is laterally restrained at A, B, C, D, P P and E. The self-weight of the beam may be D B neglected initially but should be included in C E A the final checking. Check the beam for L/6 L/3 L/3 L/6 bending, shear, shear and bending and an allowable deflection of span/250. Span of the beam is 12m. Pu = 10kN, Ps = 7kN PL3 Hint :  C = 40.9*10 −3 EIx L = 12 m max = span/250 = 48 mm Pu = 10 kN Ps = 7 kN Check serviceability first for initial member selection

Δc = 40.9 *10 −3

PL3 EI x

Ps = 7 kN, L = 12 m, E = 200 * 103 MPa, c = 48 mm max  Ix =

40.9 *10-3 * 7000 * 120003 200 *103 * 48

= 51.5 * 106 mm4 Try 310 UB 32.0 (Ix = 63.2 * 106 mm4) Onesteel Handbook Table 13 [Note: 250 UB 37.3 also OK, but heavier  more cost, would use if height restrictions governed beam depth]

Check bending

M* = 50kNm ≤ Mbx and Msx Msx = fyZe

= 320 * 467 * 103 mm3 = 149 kNm

Msx = 0.9 x 149 = 134 kNm > 50 kNm OK Mbx = msMsx ≤ Msx First check if fully laterally restrained: 5.3.2.4

le 4000 = = 122 ry 32.9

m =

− 30 = − 0.60 for middle segments, 0 for end segments 50

Check middle segments since larger le and higher moments  critical

(80 + 50βm )

250 250 le = (80 +50 * − 0.6) = 44.2   not fully restrained fy 320 ry

Hence find m, s m = 1.0 or Table 5.6.1, first case with m = -0.6

 m = 1.75 + 1.05 * (-0.6) + 0.3 * (-0.6)2 = 1.23 or m =

Mo =

(35

1.7 * 50 2

+ 40 2 + 45 2

)

= 1.22

 π 2 * 200*10 3 * 4.42 *10 6    4000 2  

  π 2 * 200*10 3 * 92.9 *10 9  3 3   80 * 10 * 86.5 * 10 +  4000 2   

= 100 kNm

 149 2    0 . 6 = s  +   100    Mb

 149  3 −   = 0.48  100  

= 1.23 x 0.48 x 149 = 88.0 kNm

Mb = 0.9 x 88.0 = 79.2 kNm > 50 kNm OK  310 UB 32.0 OK for bending Check shear: V* ≤ Vv Check

dp tw

=

282 = 51.3  5.5

82 fy 250

 Vv = Vo = Vw = 0.6 fyAw

=

82

= 72.5

320 250

= 0.6 * 320 * 282 * 5.5 = 298 kN

Vv = 0.9 * 298 = 268 kN >> 15 kN OK for shear Check shear and bending Is M* ≤ 0.75  Ms = 0.75 * 134 = 101 kNm > 50 kNm  shear/bending interaction not critical  Adopt 310 UB 32.0 [Note: Self wt of beam = 32 kg/m = 0.32 kN/m, additional M* = (1.2 x 0.32) x problem, beam capacity still adequate].

12 2 = 6.9 kNm  no 8

Question 4 A 410 UB 53.7 kg/m section (grade Onesteel-300PLUS) carries an ultimate point load P = 80kN and is laterally and torsionally restrained A at A, B and C. Check bearing under the point load at mid span assuming that the load is applied by means of a 310UB 40.4 cross beam.

P B 5m

C 5m

Refer to AS4100, Section 5.13 R* = 80 kN R* ≤  Rb  = 0.9 Rb = lesser of Rby and Rbb Bearing yield capacity: Rby = 1.25 bbf twfy bbf = bs + (2.5 tf) * 2 = bs + 5*tf (interior bearing – since top beam is not at the end of the supporting beam, the force can be fully dispersed in the web) See AS4100, bearing width, bbf, shown in Figure 5.13.1.1(a) Load applied by 310 UB 40 bs = 2 tf + tw + 1.172r1

(Stiff bearing Length, bs - AS 4100, Figure 5.13.1.2. The term 1.172r1 can be geometrically determined) = 2 * 10.2 + 6.1 + 1.172 * 11.4 = 39.9 mm

bbf = 39.9 + 5 * 10.9 = 94.4 mm tw = 7.6 mm fy = 320 MPa Rby = 1.25 * 94.4 * 7.6 * 320 = 287 kN Bearing buckling capacity, Rbb

In accordance with Section 5.13.4 first paragraph for this question

b = 0.5 kf = 1.0 Area = twbb = tw [bbf + 2 * bbw]

Section 5.13.4 (a) Section 5.13.4 (b)

= 7.6 [94.4 + 2 *

381 ] 2

bb, total bearing width, AS4100, Section 5.13.4 (c), (e) and Figure 5.13.1.1 (a) 381mm = d1 in Onesteel book Table 13

= 3613 mm2 le/r

= 2.5 d1/tw = 2.5 * 50.1 = 125 Geometrical Slenderness Ratio Section 5.13.4 (d)

λn =

le r

kf

fy 250

= 125 1

320 = 141 250

Section 6.3.3, note second to last paragraph on page 84

c = 0.301 (Table 6.3.3(3)) Nc

Section 6.3.3, note second to last paragraph on page 84

Section 6.2.1. & Flexural Buckling, Section 6.3.3 = cNs = ckfAnfy = 0.301 * 1 * 3613 * 320 = 348 kN – not critical – Rby critical

 Rb = 287 kN Rb = 0.9 * 287 = 258 kN > 80 kN  Bearing capacity is adequate

Question 5 A steel cantilever beam is acted on by a concentrated imposed service load and a uniformly distributed imposed service load as shown. The beam is laterally restrained over its entire span. Determine a suitable UB section (grade Onesteel-300PLUS) considering bending, shear, shear and bending. Deflection is limited to span/125. Self-weight of the beam may be neglected.

10 kN 30 kN/m

4m

P = 10kN  = 30 kN/m Both are imposed loads Neglect self weight

Deflection of cantilever

Part load u dl

Δ=

PL3   3EIx 

ωl4  Δ =  8EIx 

0.7 *10 *103 * 40003 0.7 * 30 * 4000 4 Total Δ = + 3* 200 *103 * Ix 8 * 200*103 * Ix

max = span/125 = 32 mm Substitute and solve for Ix = 128 * 106 mm4 From Onesteel table13 try 360 UB 50.7 Bending P* = 1.5 * 10 = 15 kN * = 1.5 * 30 = 45 kN/m M* = 15 * 4 + 45 *

42 = 420 kNm 2

Ix = 142 * 106 mm4

Section capacity: M* ≤  Msx = 0.9 fyZex = 0.9 * 300 * 897 * 103 (use flange value for fy) = 242 kNm < 420

no good, need bigger beam, find the min Zex required instead

420 *10 6 = 1556 *10 3 mm 3 0.9 * 300

Need min. Zex =

Try 460 UB 74.6 (Zex = 1660 * 103 mm3) Onesteel Handbook - Table 14 Member capacity - fully laterally restrained  Mb = Ms Shear V* ≤  Vv

V* = 15 + 45 * 4 = 195 kN Vv

82 fy 250 Vw

dp

Cl 5.11.2

=

tw 82

320

= 47.1 for 460 UB 74.6 (

d1 in Onesteel table13) tw

= 72.5  47.1  Vu = Vw

250

= 0.6 fy Aw = 0.6 * 320 (web) * 428 (depth between flanges) * 9.1 (tw) = 748 kN

Vw = 0.9 * 748 = 673 kN > 195 kN OK for shear Shear and Bending Interaction method 0.75  Msx = 0.75 * 0.9 * 300 * 1660 * 103 m3 = 336 kNm M* = 420 kNm > 336 kNm



 1.6 M *   

 V* ≤  Vvm = Vv  2.2 −   φM sx 





1.6* 420*106



 = 0.9 * 748  2.2 −  3  0.9 * 300 * 1660 * 10    = 472 kN > 195 kN OK for shear/bending interaction  Adopt 460 UB 74.6...