Steel Module 4

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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. Whereas tensile loads tend to hold a member straight compressive loads tend to bend them out of the plane of the loads (buckling).

2. 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:

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 c KL C r  :

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2 / 1 2 y a c F KL r F FS C    _ _   (505-1) where: 3 3 5 3( / ) ( / ) 3 8 _c 8 _c KL r KL r FS C C    c 2 2 y E C F

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 (505-1a) 505.3.2 When c KL C r  , (505-2)

For E = 200,000 MPa, this gives,

6 2 1.03 10 ( / ) a x F KL r  (505-2a) 2 2 12 23( / ) a E F KL r

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 Four angle box section channel W shape Tee Double angle

Single angle Square

Tubing Pipe Rectangular Tubing W shape with cover PLs 2-channels with lacing Built-up W and Channels Built-up

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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:

200 KL

r  all compression members Other column formulas:

1. Straight-Line Formula when 30 KL 120 r   , a 110 0.483 KL F r     _ _   MPa when KL 30 r  , Fa110 MPa 2. Rankine-Gordon Formula:

for main members: 60 KL 120 r

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for secondary members: 60 KL 200 r   2 124 ( / ) 1 18000 a F KL r   MPa when KL 60 r  , use Fa103MPa 3. Euler’s Formula

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2 2 c EI P KL   _P Pc FS  FS = factor of safety when KL 100 r  , 2 2 ( / ) c P E A KL r   when KL 100 r  , c PL P F A

Effective Lengths for Main Members Only

Buckled shape of column is shown by dashed line (a) (b) (c ) (d) (e) (f) Effective Length, KL 0.5L 0.7L 1.0L 1.0L 2.0L 2.0L L L L L L L

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Recommended design value when ideal conditions are approximated 0.65L 0.80L 1.20L 1.00L 2.10L 2.00L End Condition Code

NSCP Table 4. Values of Cc (From Equation 505-1a)

Fy (MPa) Fy (ksi) Cc Fy (MPa) Fy (ksi) Cc

227 33 131.9 317 46 111.6 241 35 128.0 345 50 107.0 248 36 126.2 379 55 102.1 269 39 121.1 414 60 97.7 276 40 119.6 448 65 93.9 290 42 116.7 620 90 79.8 310 45 112.8 689 100 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)7.0m PcP FS( )450(3) 1350 kN

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2 2 c EI P KL   :

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2 ₂ ₉ 6 2 3 2 1350(7) 10 33.516 10 200 10 ( ) c P KL I x E x     mm4 least 7000 70 100 100 KL r   mm W250x73: Iy = 38.8x106 mm4, ry = 64.6 mm : 3 1350 10 6750 200 c PL P x A F    mm2 100, KL r  least 7000 70 100 100 KL r   mm W310x97: A = 12300 mm2_{, r}_y_{= 76.9 mm}

Therefore, use W250x73 section.

2) one end fixed and the other hinged, KL0.7(7)4.9m

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2 2 c EI P KL  

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2 ₂ 6 2 3 2 1350(4.9) 16.421 10 200 10 ( ) c P KL I x E x     mm4 100, KL r  least 4900 49 100 100 KL r   mm W360x64: Iy = 18.8x106 mm4, ry = 48.1 mm C PL P F A  : 3 1350 10 6750 200 c PL P x A F    mm2 100, KL r  C PL P F A 

Rotation free and translation free

Rotation free and translation

Rotation fixed and translation

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100, KL r  least 4900 49 100 100 KL r   mm W250x58: A = 7420 mm2_{, r}_y_{= 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 at KL 0 r  , 5 3 FS : 250 150 5 / 3 y a F F FS    MPa

assume 0.80Fa = 0.80(150) = 120 MPa required 3 700 10 5833 0.80 a 120 P x A F    mm2

2) Select trial section and compute axial capacity [NSCP 505.1.1]: Try W200x46: A = 5860 mm2_{, r}_y_{= 51.2 mm}

From Table 4 (505-1a), C_c126.2 5500 107.4 51.2 c KL C r    , use (505-1) 3 3 5 3(107.4) (107.4) 1.909 3 8(126.2) 8(126.2) FS    2 2 (107.4) 248 1 82.85 2(126.2) 1.909 a F       MPa 3 82.85(5860)10 486 a PF A   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} 5500 107.8 51.0 c KL C r    3 3 5 3(107.8) (107.8) 1.909 3 8(126.2) 8(126.2) FS    1 (107.8)2₂ 248 82.49 2(126.2) 1.909 a F       MPa 3 82.49(8550)10 705 a PF A   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 6 6 2 6 45.5 10 2 2.12 10 5690(97.5) 157.92 10 x I  x  _ x  _ x mm4 6

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6

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6 15.3 10 2 67.3 10 149.90 10 y I  x  x  x mm4_{(least I)}

x

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least 6 149.90 10 93.25 17240 I x r A    mm

1.2) Allowable axial stress

From Table 4 (505-1a), Cc126.2 10000 107.2 93.25 c KL C r    use (505-1) 3 3 5 3(107.2) (107.2) 1.909 3 8(126.2) 8(126.2) FS    2 2 (107.2) 248 1 83.04 2(126.2) 1.909 a F       MPa

1.3) Allowable axial load

3

83.04(17240)10 1432 a

PF A   kN 2) Straight-line Formula

2.1) Allowable axial stress since KL r/ 107.2 120 use a 110 0.483 KL F r     _ _  MPa

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110 0.483 107.2 58.22 a F    MPa

2.2) Allowable axial load

3

58.22(17240)10 1004 a

PF A   kN 3) Rankine-Gordon Formula

3.1) Allowable axial stress since KL r/ 107.2 120 2 2 124 124 75.68 ( / ) (107.2) 1 1 18000 18000 a F L r      MPa 3.2) Allowable axial load

3 75.68(17240)10 1305 a PF A   kN 114.5 114.5 229 mm Section Properties: W200x46: C310x45: A = 5860 mm2 _{A = 5690 mm}2 d = 203 mm d = 305 mm bf = 203 mm bf = 80 mm 13.0 13.0 x y 305 mm