Steel Module 4

Western Philippines University College of Engineering and Technology Civil Engineering Department STRUCTURAL STEEL DESIGN By Dr. Cesario A. Bacosa , Jr

Module 4. COMPRESSION MEMBERS Compression members are prismatic members subject to loads which tend to squeeze or shorten the member. Among the types of compression members are the columns, the top chords of trusses, bracing members, the compression flanges of beams. There are two significant differences between tension and compression members. These are: 1. 2.

Whereas tensile loads tend to hold a member straight compressive loads tend to bend them out of the plane of the loads (buckling). The presence of rivet/bolt holes in tension members reduces the area available for resisting the loads; but in compression members the rivets/bolts are assumed to fill the holes and the entire gross area is available for resisting load.

Types of Compression Members:

Single angle

Double angle

Tee

Rectangular Tubing

Four angle box section

2-channels with lacing

channel

W shape

W shape

W and Channels

Square Tubing

Pipe

Built-up

Built-up

with cover PLs

NSCP Specifications 505.1.1. This section applies to prismatic members with compact and non-compact sections subject to axial compression through the centroidal axis. For members with slender elements, see Section 502.6.2. From members subject to combined axial compression and flexure, see Section 508. For tapered members, see Section 506.8. 505.3.1 When

KL  Cc : r

  KL / r   Fy Fa  1   2Cc2  FS 

where: FS  505.3.2

When

KL  Cc , r

(505-1)

5 3( KL / r ) ( KL / r )3   3 8Cc 8Cc3

Fa 

Cc 

12 2 E 23( KL / r ) 2

For E = 200,000 MPa, this gives,

2 2 E Fy

(505-1a) (505-2)

Fa 

1.03x106 ( KL / r ) 2

(505-2a)

where: Fa = allowable compressive stress Fy = yield strength of steel K = effective length factor L = unsupported length of member r = radius of gyration about axis of buckling KL/r = slenderness ratio E = modulus of elasticity of steel (200,000 MPa) 502.8.1 Maximum slenderness ratio: KL  200 all compression members r Other column formulas: 1. Straight-Line Formula KL  KL   120 , Fa  110  0.483  when 30   MPa r  r  KL  30 , Fa  110 MPa when r 2. Rankine-Gordon Formula: KL  120 for main members: 60  r KL  200 for secondary members: 60  r 124 MPa Fa  ( KL / r ) 2 1 18000 KL  60 , use Fa  103 MPa when r 3. Euler’s Formula P EI  2 P c FS = factor of safety Pc  2 FS  KL  when

KL  100 , r

Pc E 2  A ( KL / r )2

when

KL  100 , r

Pc  FPL A

Effective Lengths for Main Members Only (a)

Buckled shape of column is shown by dashed line

L

Effective Length, KL

0.5L

(b)

L

0.7L

(c )

(d)

L

L

1.0L

1.0L

(e)

L

2.0L

(f)

L

2.0L

Recommended design value when ideal conditions are approximated

0.65L

0.80L

1.20L

1.00L

2.10L

2.00L

Rotation fixed and translation Rotation free and translation Rotation fixed and translation Rotation free and translation free

End Condition Code

NSCP Table 4. Values of Cc (From Equation 505-1a) Fy (MPa) 227 241 248 269 276 290 310

Fy (ksi) 33 35 36 39 40 42 45

Cc 131.9 128.0 126.2 121.1 119.6 116.7 112.8

Fy (MPa) 317 345 379 414 448 620 689

Fy (ksi) 46 50 55 60 65 90 100

Cc 111.6 107.0 102.1 97.7 93.9 79.8 75.7

Problem 201. Select the lightest W shape that can be used as a column 7 meters long to support an axial load of 450 kN with a factor of safety of 3. Assume 1) both ends hinged and 2) one end fixed and the other hinged. Use FPL= 200 MPa, E = 200 GPa and Euler’s Formula. Solution: 1) both ends hinged, KL  1.0(7)  7.0 m Pc  P(FS )  450(3)  1350 kN

Pc 

EI  2

 KL 

2

:

I

Pc  KL  E

2



2

1350(7)2109  33.516 x106 mm4 200 x103 ( 2 )

KL KL 7000  100,   70 mm least r  r 100 100 W250x73: Iy = 38.8x106 mm4, ry = 64.6 mm P P 1350 x103 FPL  C : A c   6750 mm2 A FPL 200 KL KL 7000  100,   70 mm least r  r 100 100 W310x97: A = 12300 mm2, ry = 76.9 mm Therefore, use W250x73 section. 2) one end fixed and the other hinged, KL  0.7(7)  4.9 m

Pc 

EI  2

 KL 

2

KL  100, r W360x64: P FPL  C : A

I

Pc  KL  E

2

2



1350(4.9) 2  16.421x106 mm4 200 x103 ( 2 )

KL 4900   49 mm 100 100 Iy = 18.8x106 mm4, ry = 48.1 mm P 1350 x103 A c   6750 mm2 FPL 200 least r 

KL KL 4900  100,   49 mm least r  r 100 100 W250x58: A = 7420 mm2, ry = 50.4 mm Therefore, use W250x58 section. Problem 202. Select a W shape section that can be used as a column to support an axial load of 700 kN on an effective length of 5.5 meters. Use Fy = 248 MPa and NSCP formulas. Solution: 1) trial calculations Fy 250 KL 5  0 , FS  : Fa    150 MPa at FS 5 / 3 r 3 assume 0.80Fa = 0.80(150) = 120 MPa P 700 x103 required A    5833 mm2 0.80Fa 120 2) Select trial section and compute axial capacity [NSCP 505.1.1]: Try W200x46: A = 5860 mm2, ry = 51.2 mm From Table 4 (505-1a), Cc  126.2

x

KL 5500   107.4  Cc , use (505-1) r 51.2 5 3(107.4) (107.4)3 FS     1.909 3 8(126.2) 8(126.2)3  (107.4)2  248 Fa  1   82.85 MPa 2  2(126.2)  1.909

P  Fa A  82.85(5860)103  486 kN < 700 kN NO 3) Try a larger section and compute axial capacity [NSCP 505.1.1]: Try W250x67: A = 8550 mm2, r = 51.0 mm KL 5500   107.8  Cc r 51.0 5 3(107.8) (107.8)3 FS     1.909 3 8(126.2) 8(126.2)3  (107.8)2  248 Fa  1   82.49 MPa 2  2(126.2)  1.909

P  Fa A  82.49(8550)103  705 kN > 700 kN

OK

Therefore, use W250x67. Problem 203. A hinged-end column 10 m long is fabricated from a W200x46 section and two C310x45 channels arranged as shown in figure. Using Fy = 248 MPa, determine the safe axial load 1) using NSCP Formulas, 2) using Straight-line Formula and 3) using Rankine-Gordon Formula. Solution: 1) NSCP Formulas [NSCP 505.1.1] 1.1) Properties of built-up section: A  5860  2(5690)  17240 mm2

I x  45.5 x106  2  2.12 x106  5690(97.5) 2   157.92 x106 mm4

I y  15.3x106  2  67.3x106   149.90 x106 mm4 (least I)

I 149.90 x106   93.25 mm A 17240 1.2) Allowable axial stress From Table 4 (505-1a), Cc  126.2 least r 

KL 10000   107.2  Cc use (505-1) r 93.25 5 3(107.2) (107.2)3 FS     1.909 3 8(126.2) 8(126.2)3

y

13.0 114.5 x 114.5

 (107.2)  248 Fa  1   83.04 MPa 2 2(126.2)   1.909 1.3) Allowable axial load P  Fa A  83.04(17240)103  1432 kN 2

2) Straight-line Formula 2.1) Allowable axial stress since KL / r  107.2  120  KL  use Fa  110  0.483   MPa  r 

Fa  110  0.483107.2  58.22 MPa 2.2) Allowable axial load P  Fa A  58.22(17240)103  1004 kN 3) Rankine-Gordon Formula 3.1) Allowable axial stress since KL / r  107.2  120 124 124 Fa    75.68 MPa ( L / r )2 (107.2)2 1 1 18000 18000 3.2) Allowable axial load P  Fa A  75.68(17240)103  1305 kN

229 mm

13.0 305 mm

Section Properties: W200x46:

C310x45:

A = 5860 mm2

A = 5690 mm2

d = 203 mm

d = 305 mm

bf = 203 mm

bf = 80 mm