4.Steel Tutorial solution Tension and Compression PDF

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Steel Design Tutorial - Tension and Compression Members - SolutionQuestion 1Determine the maximum design load for a tension member consisting of two 75x50x Onesteel 300 PLUS unequal angles welded back to back, connected to a loading plate by the short leg.2 x 75 x 50 x 8 Onesteel 300 Plus unequal an...

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Steel Design Tutorial - Tension and Compression Members - Solution Question 1 Determine the maximum design load for a tension member consisting of two 75x50x8 Onesteel 300 PLUS unequal angles welded back to back, connected to a loading plate by the short leg. 2 x 75 x 50 x 8 Onesteel 300 Plus unequal angles, welded back to back by the short leg N* ≤  Nt Nt = min ( Agfy and 0.85 ktAnfu ) AS4100 Cl 7.2

From Onesteel tables for 75 x 50 x 8 angles: Ag = An for no boltholes = 2 x 921 mm2 = 1842 mm2 Onesteel Table 26 fy = 320 MPa fu = 440 MPa Onesteel Table 27 kt = 0.75

AS 4100 Table 7.3.2

Nt = Agfy = 1842 x 320 = 589 kN Or Nt = 0.85 ktAnfu = 0.85 x 0.75 x 1842 x 440 = 517 kN  = 0.9 AS4100 Table 3.4 Hence N* = 0.9 x 517 = 465kN = max design load

Question 2 Determine a suitable equal angle section of Onesteel 300 PLUS to carry an ultimate tensile force N* = 420 kN. Assume that the angle is connected through one leg by a single line of 16mm bolts in 18mm holes.

N* ≤  Nt Nt = min ( Agfy and 0.85 ktAnfu ) AS4100 Cl 7.2  = 0.9 AS4100 Table 3.4 N* = 420 kN  Nt = 420/0.9 = 467kN

Need to satisfy:

and

467 kN ≤ Agfy 467 kN ≤ 0.85 kt Anfu

Unknowns are Ag, fy, An For Onesteel 300 Plus equal angles, fu = 440 MPa, fy varies between 280, 300 or 320 MPa Assume fy = 320 MPa as this will give the smallest value of Ag possible.  Ag = 467 x 103/320 = 1459 mm2 From Onesteel tables, try 100 x 100 x 8 EA (Ag = 1500 mm2) Onesteel Table 23 (Check fy = 320 MPa  assumption OK) Onesteel Table 26 Check fracture through holes: An = 1500 – 18 x 7.8 = 1360 mm2 AS 4100 Table 7.3.2 kt = 0.85  0.85 ktAnfu = 0.85 x 0.85 x 1360 x 440 = 432 kN < 467 kN no good Try 100 x 100 x 10 EA (Ag = 1810 mm2) Onesteel Table 23 Agfy = 1810 x 320 = 579 kN > 467 kN OK 0.85 ktAnfu = 0.85 x 0.85 x (1810 – 18 x 9.5) x 440 = 521 kN > 467 kN OK  Adopt 100 x 100 x 10 EA

Question 3 Determine the maximum design load for a tension member consisting of a single 65x65x8 EA of Onesteel 300 PLUS connected eccentrically through one leg by a single line of 16mm diameter bolts (hence 18 mm holes). N*t = min ( Agfy and 0.85 ktAnfu ) AS4100 Cl 7.2 Agfy = 0.9 x 957 mm2 x 320 MPa = 276 kN (Ag = 957 mm2) Onesteel Table 23 0.85 ktAnfu

= 0.9 x 0.85 x 0.85 x (957 – 18 x 7.8) x 440 = 234 kN

 Max design load N*t = 234 kN

Question 4 Design a pin ended UC section (Onesteel 300PLUS) of effective length equal to 4m to carry a concentric ultimate compressive design load N*c = 500 kN. N*c = 500 kN

lex = ley = 4m

N*c ≤ Ns and Nc N c =  cN s c (AS4100 Table 6.3.3 (3)), find b & n b for kf = 1.0, UC section  AS4100 Table 6.3.3(1)  b = 0

l  λn = e  k f  r 

fy 250

You can use the trial and error method to try different sections until you find one that works or you can use one of the methods below: Method 1 – Euler Buckling Method Pcr = 2 EIy / Ley2

→

Iy ≥ Pcr Ley2 / ( 2 E )

Set Pcr = N*c /  = 500 kN / 0.9 = 556 kN Iy ≥ 556 x 1000 x 40002 / (2 x 200 x 1000) = 4.5 x 106 mm4 Choose size with Iy MUCH greater than the required value to allow for cy Method 2 – Euler Buckling Method N*c = 500kN ≤ Ns = kfAn fy = 0.9 x 1 x An x 300 An > 1852mm2 Set An = Ag then choose size with Ag MUCH greater than the required value to allow for cy Try 150 UC 37.2, Ag = 4730 mm2, ry = 38.5 mm Onesteel Table 15 If the effective length is the same for both axes then the weak axis (or y-y axis) will be the most critical case.

300  4000 λ ny =  = 113.8  1 250  38.5  Table 6.3.3(3)  c = 0.455 Ncy

= cy kf Anfy = 0.455 x 1.0 x 4730 x 300 [no holes  An = Ag]

= 646 kN Ncy = 0.9 x 646 = 581 kN > 500 kN OK Adopt 150 UC 37 [Check 150 UC 30]

4000  320 λ ny =  = 119  1 250  38.1   cy = 0.426 Ncy = 0.426 x 1 x 3860 x 320 = 526 kN Ncy = 474 kN < 500 kN  no good

Question 5 A column ABC is 12m in length and is pinned at ends A & C. At midpoint B, lateral restraint is provided about the YY axis. The column is a Onesteel 300 PLUS 250UC72.9 section. Determine the maximum compressive load N*c the column can carry considering the possibility of buckling about either principal axis. l = 12 m = lex ley = 6 m 250 UC 72.9  fy = 300 MPa, Ag = 9320 mm2 rx = 111 mm, ry = 64.5 mm, kf = 1.0 Consider buckling about x-x axis b = 0

l  λ nx =  e  k f  r

fy 250

=

12000 300 = 118 1 111 250

Table 6.3.3 (3)  cx = 0.432 Ncx = cxNs

= 0.432 x 1 x 9320 x 300 = 1208 kN

Ncx = 0.9 x 1208 = 1087 kN Consider buckling about y-y axis

300  6000  λ ny =  = 102  1 250  64.5  cy

= 0.528

Ncy

= 0.9x0.528 x 9320 x 300 = 1329 kN  not critical

Max compressive load is governed by x-x axis buckling = 1087 kN...