4.Steel Tutorial solution Beam-Columns PDF

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Steel Design Tutorial - Beam Columns - SolutionQuestion 1: Determine whether a 200 UC 59 of grade 300PLUS steel of 3 effective height can carry a ultimate compressive load of 420kN acting at an eccentricity of 100mm along the member x axis at both ends. Ie Bending about weak (YY) Axis.N* = 420kN M*y...

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Steel Design Tutorial - Beam Columns - Solution Question 1: Determine whether a 200 UC 59.5 of grade 300PLUS steel of 3.1m effective height can carry a ultimate compressive load of 420kN acting at an eccentricity of 100mm along the member x axis at both ends. Ie Bending about weak (YY) Axis. N* = 420kN M*y = 420 * 0.1m = 42kNm lex = ley = 3100 mm 200 UC 59.5 Check first for compression by itself: AS4100 Cl 6.2.1 N s = kf A n f y = 1 * 7620 * 300 = 2286kN Ncx: n =

3100 1 89.7

b = 0 c = 0.913

300 = 37.9 Cl 6.3.3 250

Table 6.3.3 (1) Table 6.3.3 (3)

 Ncx = 0.913 * 2286 = 2088 kN Ncy: n =

3100 1 51.7

300 = 65.7 Cl 6.3.3 250

b = 0  c = 0.775

Table 6.3.3 (3)

 Ncy = 0.775 * 2286 = 1772 kN

Hence Ncy is critical

øNcy = 0.9 * 1772 = 1595kN > N* = 420kN = OK for compression Then check for bending by itself: Section:

M*y  øMsy

Capacity:

Msy = fy Zey = 300 * 299 * 103 mm3 = 89.7 kNm

AS4100 Cl 5.1

Onesteel Table 14

øMsy = 0.9 * 89.7 = 80.7 kNm > 42 kNm OK For y.axis bending Msy = Mby

No member CANT buckle about yy or weak axis

Then check for compression and bending combined: Section capacity:

N* M* + ≤1 M sy N s This is a re-arranged way to write the formulae for øMry in Cl 8.3.3 Similar to Cl 8.3.4 42 420 + = 0.72 < 1 OK 0.9 * 89.7 0.9 * 2286

Member capacity – in-plane capacity

Cl 8.4.2.2 Re-arranged for øMiy

M* N* + ≤1 N cy M sy



42 420 + = 0.78 < 1 OK 0.9 * 89.7 0.9 *1772

[Out-of-plane capacity doesn’t apply to weak axis bending]

Question 2 Determine the ultimate compressive load that can be carried by a 410 UB 59.7 in grade 300 PLUS steel over an effective height of 5.6m, assuming that it is braced against out-ofplane failure and that the maximum ultimate moment about its major axis is 72 kNm. 410UB 59.7 lex = 5600 mm ley = 0 as the beam is prevented from buckling about the Y axis lez = 5600mm effective length for bending M*x = 72 kNm Find N* max Bending: Section capacity Msx = fyZex AS4100 Cl 5.2.1 = 300 MPa * 1200 * 103 mm3 Onesteel Table 14 = 360 kNm First check for compression by itself: Section capacity Ns = kf An fy = 0.938 * 7640 * 300 = 2150 kN

Cl 6.2.1 Onesteel Table 13 & 14

Member capacity: Buckling about y.axis (out-of-plane) is prevented ie Ncy = Ns = 2150kN Ncx:- Ncx = c Ns l  nx =  e   rx 

=

5600 168

b = 0 c = 0.923

0.938

kf

Cl 6.3.3

 fy    250   300 = 35.4 250

Table 6.3.3(2) Table 6.3.3(3)

Ncx = 0.923 * 2150 = 1984 kN Check for bending by itself: Member capacity: Mbx = m s Msx m = 1 (no other information)

 π 2 * 200 *10 3 *12.1*10 6 s: Mo = √  5600 2  = 207 kNm

 π 2 * 200 *103 * 467 *109  80 *10 3 *337 *10 3 +  5600 2  

   

 360  2   360  s = 0.6 [   + 3 -    207    207  = 0.429



Cl 5.6.1.1 (2)

øMbx = 0.9 * 1 * 0.429 * 360 = 139 kNm M* = 72 < øMbx = 139 kNm

OK for bending

Now check for combined compression and bending actions: Section capacity: Rearrange formulae from Cl 8.3.2 M* N* + ≤1 N s M sx 72 N* = 1  N* = 1505 kN + 0.9 * 360 0.9 * 2150

Member capacity: In-plane failure only Rearrange formulae from Cl 8.4.2.2 M* N* + ≤1  N cx M sx 72 N* = 1  N* = 1389 kN + 0.9 * 360 0.9 *1984

Member capacity: Out-of-plane failure only Rearrange formulae from Cl 8.4.4.1

𝑀* 𝜑𝑀𝑏𝑥 72 139

+

+

𝑁* ≤1 𝜑𝑁𝑐𝑦 𝑁*

0.9*2150

= 1  N* = 933 kN

Take minimum of Cl 8.3.2, 8.4.2.2, 8.4.4.1  N*max = 933 kN

Question 3 Determine a suitable UB section in grade 300PLUS steel to carry end ultimate moments of 10 and 3.4 kNm about the section's major axis and an ultimate compressive load of 205 kN over an effective height of 4.8m. Assume that the column is bent in single curvature and that there is no buckling restraint except at the column ends. N* = 205 kN M*x = 10 kNm lex = ley = lez = 4800 mm As lex = ley – yy axis for buckling will be critical

Try 250UB 31.4 Compression: Ns = kf A n f y = 1.0 * 4010 * 320 = 1283 kN

Onesteel Table 13 and 14

Ncy = c Ns n =

le ry

kf

fy 250

=

4800 33.4

1

320 = 163 250

b = 0 Table 6.3.3.(1) c = 0.255 Ncy = 0.255 * 1283 = 327 kN ØNcy = 294kN > 205 kN OK

nx = 51.7 c = 0.853 Ncx = 0.85 * 1283 = 1094 kN ØNcx = 984 kN

Note: Even though y axis direction is critical for compression alone, it is still necessary to determine the value of ØNcx for the combined capacity in-plane calculations Bending Msx = fyZex = 320 * 395 * 103 = 126 kNm ØMsx = 0.9 * 126 = 114 kNm > 10 OK

Onesteel Table 14

Member capacity: No restraint over 4.8m  not fully laterally restrained by inspection

Mb = m s Ms −3.4

m: Table 5.6.1 first case m = = -0.34 (m is negative as the bending moment causes 10 single curvature and tension is only on the one side of the column.)  m = 1.43 or

1.7x10

= 1.44

(5.05) + (6.7) 2 + (8.35) 2 2

s: Mo =

(  2 * 200 *103 * 4.47 *106 ) 4800 2

  2 * 200 * 10 3 * 65.9 * 10 9 )  3 3 80 * 10 * 89.3 * 10 + (  48002  

= 70 kNm

  126 2  126     s = 0.6    + 3 − 70   70      Mbx

= 1.44 * 0.419 * 126 = 76 kNm ØMbx = 68 kNm > 10 kNm OK for bending

Combined actions: Section capacity:

N* M * +  1 ØNs ØM s

Rearrange formulae from Cl 8.3.2

205 10 + = 0.27  1 0.9 *1283 114

Member capacity – In-plane

N* M * +  1 Rearrange formulae from Cl 8.4.2.2 ØN cx ØM s

10 205 + = 0.296 < 1 OK 984 114

Member capacity – Out-of-plane

M* ØM bx

10 205 + = 0.84 < 1 OK 68 294  Adopt 250 UB 31 column

+

N*  1Rearrange formulae from Cl 8.4.4.1 ØN cy

Question 4 Check the ability of a 250 UC 72.9 of effective height 3.4m in grade 300PLUS steel to carry the following load combination:

N* = 750 kN (compressive) M*x = 52 kNm (top) 0.0 kNm (bottom) M*y = 14 kNm (top) 11.0 kNm (bottom) Lex = Ley = Lez = 3400mm

Compression N* = 750 kN N s = kf A n f y = 1 * 9320 * 300 = 2796 kN Nc = cNs Ncy:

Ncx 3400 64.5

300 250

nx =

3400 111

300 250

ny

=

c

= 0.822

= 33.6

Ncy

= 0.822 * 2796 = 2298 kN

Ncx = 0.93 * 2796 = 2600 kN

1

= 57.7

1

 c = 0.93

ØNcy = 2068 kN > 750 OK

ØNcx = 2340 kN > 750 kN OK

Bending: x-x axis Msx = fy Zex = 300 * 986 * 103 = 296 kNm ØMsx = 266 kNm > 52 kNm

y-y axis Msy = fy Zey = 300 * 454 * 103 = 136 kNm ØMsy = 122 kNm > 14 kNm OK = Mby

Mbx = m s Msx m = 1.75 (Table 5.6.1, 1st case, m = 0)   2 * 200 * 103 * 38.8 * 106   3400 2  = 970 kNm

Mo =

  2 * 200 *10 3 * 557 *10 9    80 * 10 3 * 586 * 10 3 +   34002  

2    296   296  s = 0.6    = 0.87  +3 −   970   970   

ms  1  1.75 * 0.87 > 1 ms = 1 Mb = Ms = 296 kNm  OK for bending Biaxial Bending: AS4100 Cl 8.4.5 for doubly I symmetric with kf = 1.0 Member capacity  M x*   M  cx

 M y*   +    M  cy  1 .4

1. 4

  1  

M*x = 52 kNm, M*y = 14 kNm ØMcx = lesser of in-plane Mix and out of plane Mox Mix

 N*  = MsX 1 −   N cx  750  = 296 1 −  = 201 kNm  2340 

 N*   Mox = Mbx 1 −  N  cy   750   = 296 1 −  = 189 kNm = Mcx  2068   N *  = 136 Miy = Msy 1 −  N  cy    52      0.9 *189 

1 .4

750   1 −  = 86.7 kNm  2068  1 .4

14   +  = 0.28 < 1 OK  0.9 * 86.7 

Section capacity (Cl 8.3.4) Applies if kf < 1.0

M *y M x* N* + + N s M sx M sy 

1

52 14 750 + = 0.61 < 1 OK + 0.9 * 296 0.9 *136 0.9 * 2796...