Axially-loaded-compression-members.docx

AXIALLY LOADED COMPRESSION MEMBERS 1.) A column that is 9 m. long is to carry a load of 890 kN. The member will be braced about both principal axis at top and bottom and in addition will be braced about its minor axis at mid-height. Using Fy = 345.5 MPa, design a section to carry the loads safely. Kx = Ky = 1.0 Properties of Steel Sections Available Sections W 8 x 40 W 8 x 48 W 10 x 49 W 12 x 50

Area 7613 mm2 9097 mm2 9290 mm2 9484 mm2

rx 89.66 mm 91.69 mm 110.49 mm 131.57 mm

Given: L=9m Fy = 345.5 MPa Kx = Ky = 1.0 C = 890 kN

890 kN y

Solution: a) Assume Fa (Max Fa = 0.6Fy) Fa = 0.4 Fy Fa = 0.4 (345.5) Fa = 138.2 MPa b) Gross – Area Required C

Ag = Assumed Fa Ag =

4.5 m

890 x 103 138.2

Ag = 6439.942 mm2 c) Try W 80 x 40 A = 7613 mm2 rx = 89.66 mm ry = 51.82 mm Slenderness Ratio: KxL 1(9000) = 89.66 = 100.38 rx KyL 1(4500) = 51.82 = 86.64 ry 𝐾𝐿 Use 𝑟 = 100.38 < 200

4.5 m (SAFE)

Determine if Intermediate or Long Column: 2π2 E Fy

Cc = √

2π2 (200000) 345.5

Cc = √

Cc = 106.89 > 100.38 ; Therefore, Intermediate Column Determine the Actual Allowable Fa: Use: Fa = [1 -

ry 51.82 mm 52.83 mm 64.52 mm 49.78 mm

2 (KL⁄r)

2 Cc 2

Fy

] F.S.

x

3 3(KL⁄r) (KL⁄r) 8 Cc 8 Cc 3 5 3(100.38 ) (100.38 )3 = 3 + 8(106.89) - 8 (106.89)3

5

F.S. = 3 + F.S.

F.S. = 1.92 Fa = [1 -

(100.38)2 2 (106.89)2

345.5 1.92

]

Fa = 100.60 MPa Determine the Capacity C Cap. C = Fa Ag Cap. C = 100.60 (7613) Cap. C = 765866 N Cap. C = 765.87 kN < 890 kN ; Therefore, try another section. d) W 8 x 48 A = 9097 mm2 rx = 91.69 mm ry = 52.83 mm Slenderness Ratio: KxL 1(9000) = = 98.16 rx 91.69 KyL 1(4500) = = 85.18 ry 52.83 𝐾𝐿 Use = 98.16 < 200 (SAFE) 𝑟

Determine if Intermediate or Long Column: 2π2 E Fy

Cc = √

2π2 (200000) 345.5

Cc = √

Cc = 106.89 > 98.16 ; Therefore, Intermediate Column Determine the Actual Allowable Fa: 2

Use: Fa = [1 5

F.S. = 3 + F.S.

(KL⁄r) 2 Cc 2

3(KL⁄r)

-

8 Cc 5 3(98.16 ) = 3 + 8(106.89)

]

Fy F.S.

3 (KL⁄r)

8 Cc 3 (98.16)3

- 8 (106.89)3

F.S. = 1.91 Fa = [1 -

(98.16)2 2 (106.89)2

]

345.5 1.91

Fa = 104.62 MPa Determine the Capacity C Cap. C = Fa Ag Cap. C = 104.62 (9097) Cap. C = 951728 N Cap. C = 951.73 kN < 890 kN ; SAFE Therefore, use W 8 x 48

2.) Design the web members U1L1/U1’L1’ & U3M/U3’N of your riveted fink truss which are under compression. Using two unequal leg angles long legs back to back straddling on a 12 mm gusset plate. Use AISC specifications & ASTM A36 steel. K=1.0 (pin riveted connection.)

√15.62 +7.82

𝐿𝑜 𝑈1 = 4 = 4.36 m

7.8 𝜃 = tan−1 ( ) 15.6 = 26.57° 𝑈1 𝐿1 = 4.36 tan 26.57° = 2.18 m FOR MEMBERS U1L1/U1’L1’ & U3M/U3’N: C = 66.56 kN L = 2.18 m tgp = 12 mm

For AISC ASTM A36 steel: Fy = 248 MPa

Solution: a) Assume for Fa : Max. Fa = 0.6Fy = 0.6(248) Max. Fa = 148.8 MPa Assume Fa = 60 MPa b) Compute for Gross Area: 𝐶

𝐴𝑔 = 𝐴𝑠𝑠𝑢𝑚𝑒𝑑 𝐹

𝑎

=

66560 60

Ag = 1109.33 mm2 (for 2∠s) 1109.33 = 2 Ag = 554.665 mm2 c) Try Sections: Try 2∠s 75x50x5 Area = 600 mm2 rx = 24.1 mm

ry = 14.5 mm x = 11.9 mm y = 24.4 mm

d) Compute for Radius of Gyration & check for Slenderness Ratio: rx = 24.1 mm r𝑦 = √ry 2 + (𝑥 +

𝑡𝑔𝑝 2 ) 2

= √14.5 2 + (11.9 +

12 2 ) 2

ry = 23.04 mm (use) 𝐾𝐿 ≤ 200 𝑟 1(2180) ? 200 23.04

𝑆𝑅 =

94.62 < 200 (OKAY) e) Compute for allowable compressive stress, Fa: 2𝜋2 𝐸 𝐹𝑦

𝐶𝐶 = √ 𝐹. 𝑆. = =

= √

2𝜋2 (200000) 248

= 126.17 > 𝑆𝑅 ∴ 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖𝑎𝑡𝑒 𝐶𝑜𝑙𝑢𝑚𝑛

5 3(𝑆𝑅) 𝑆𝑅 3 + − 3 8𝐶𝐶 8(𝐶𝐶 )3 5 3(94.62) (94.62)3 + − 3 8(126.17) 8(126.17)3

𝐹. 𝑆. = 1.895 𝐹𝑎 = (1 −

𝑆𝑅2 𝐹𝑦 )( ) 2𝐶𝐶 2 𝐹𝑆 (94.62)2

248

= (1 − 2(126.17)2 ) [1.895] Fa = 94.07 MPa f) Compute for allowable compressive strength, Capacity C: Cap. C = AgFa = 2(600)(94.07)(10-3) Cap. C = 112.884 kN > Actual C (OKAY!!) ‫ ؞‬USE 2∠s 75x50x5

3.) Determine the maximum length of a W 250x167 section used as a hinged-end column (k=1.0) to support a load of 1600 kN. Use AISC specifications with Fy = 380 MPa. Properties of W 250x167: A = 21300 mm2 rx = 119 mm ry = 68.1 mm Given: C = 1600 kN K = 1.0 Fy = 380 Mpa Solution: a) Determine Fa: C Fa = Ag Fa =

L

1600(1000) 21300

Fa = 75.12 MPa b) Determine Cc: 2π2 E Fy

Cc = √

2π2 (200000) 380

Cc = √

Cc = 101.93 c) Assume column is intermediate KL Let x = R x2

Fy

Fa =(1 - 2Cc 2 )( Fs ) 5

3x

x3

FS = 3 + 8Cc - 8Cc3 x2

Fa =(1 - 2Cc 2 )(

Fy 5 3x x 3 + 3 8Cc 8Cc 3

)

Solving for x: 75.12 = (1 −

𝑥2 )( 2(101.93)2 5+

380

3𝑥 𝑥3 − 3 8(101.93) 8(101.93)3

)

x1 = 1.686 x2 = - 1.775 x3 = 0.089 Solving for L using the largest value of x x= L=

KL R 68.1(1.686) 1.0

L = 114.817 mm (unrealistic)

d) Assume column is long Fa=

12π2 E KL R

23( )2

75.12=

12π2 (200000) 23(

L= 7.974 m

1.0L 2 ) 68.1

4.) A W 250 x 73 is to serve as a pin-ended 12m long column. It is braced at mid-height with respect to its weak axis. Properties of W 250 x 73 A= 9280mm² 𝑟𝑥 = 110 𝑟𝑦 = 64.7 d= 253mm 𝑏𝑓 = 254mm 𝐼𝑥 = 113x106 𝑚𝑚4 𝑡𝑓 = 14.2mm 𝐼𝑦 = 38.8x106 𝑚𝑚4 a. Determine the slenderness ratio with respect to y-axis b. Determine the Euler’s Buckling stress c. Determine the allowable axial compressive load using a factor of safety of 2.5 Solution: a) Slenderness ratio with respect to y-axis 𝐾𝐿 1(6000) = 64.7 𝑟𝑦 𝑲𝑳 = 92.74 𝒓𝒚

b) Euler’s Buckling stress 𝐾𝐿 1(6000) = 𝑟𝑥 110 𝐾𝐿 = 109.9 𝑟𝑥 𝐿

use 𝑟= 109.9 𝐹𝑒 =

𝜋²𝐸 𝐿 ( )² 𝑟

=

𝜋²(200000) (109.9)²

𝑭𝒆 = 165.87MPa c) Allowable axial compressive load using a factor of safety of 2.5 𝑃𝑒 = 𝐹𝑒 𝐴 𝑃𝑒 = 165.87(9280) 𝑃𝑒 = 1539274N 𝑃𝑒 = 1539.3kN 𝑃𝑒 𝐹.𝑆. 1539.3 P= 2.5

P=

P= 615.72kN

5.) A W 12 x 79 column carries an axial load of 2463.134 kN. A rectangular base plate is required to support this column. Assume that the base plate will cover the full area of concrete with fc’ = 28 MPa. Use Fy = 245 MPa. Properties of W 12 x 79 d = 314.45 mm bf = 306.83 mm

B m

Given: C = 2463.134 kN fc’ = 28 MPa Fy = 245 MPa Solution:

C

0.95 d

a) Size of Base Plate: Allowable bearing stress of concrete: Fp = 0.35 fc’ Fp = 0.35 (28) Fp = 9.8 MPa

m

Trial area of base plate: C A= Fp 2463134 A = 9.8

n

0.80 bf

n

A = 251340 mm2 Assume m = n B = 2n + 0.80 bf B = 2n + 0.80(306.83) B = 2n + 245.46 **

Equivalent Rectangular Section

Base Plate

C = 2m + 0.95 d C = 2m + 0.95(314.45) C = 2m + 298.73 ** BC = 251340 mm2 (2m + 245.46) (2m + 298.73) = 251340 4m + 1088.38m + 73326.27 = 251340 4m2 + 1088.38m + 178013.73 = 0 m2 + 272.10m – 44503.43 = 0 m = 114.97 Determine Size: B = 2(114.97) + 245.46 B = 475.1 say 480 mm C = 2(114.97) + 298.73 C = 528.67 say 530 mm Use 480 mm x 530 mm Base Plate fp fp

P = BC 2463134 = 480 (530)

fp = 9.68 < 9.8 MPa (OK)

b) Thickness of Base Plate: B = 2n + 245.46 480 = 2n + 245.46 n = 117.27 C = 2m + 298.73 530 = 2m + 298.73 m = 115.64 mm Use x = 117.27 (bigger value) 3 Fp x 2

t = √0.75 Fy t=√

3 (9.8) (117.27)2 0.75 (245)

t = 46.9 mm say 50 mm

6.) A W 14 x 550 is used as a column to carry an axial load of 3600kN. Design a square base plate to support the column. The base plate rest on full area of a square concrete block with fc’=21MPa. Use A36 steel base plate. Properties of W 14 x 550 𝑏𝑓 = 514 d= 437

L m

Given: C = 3600 kN fc’ = 21 MPa Fy = 248 Mpa

0.95 d Solution:

L

a) Allowable bearing stress of concrete: 𝐹𝑝 = 0.35fc’ 𝐹𝑝 = 0.35(21) = 7.35MPa b) Size of base plate: 𝑃

Area of base plate = 𝐹𝑝 =

m

3600000 7.35

Area of base plate = 489795 mm² L = √𝐴 = √489795 L = 699.854 say 700mm 700= 2n + 0.8(514) n= 144.4mm 700= 2m + 0.95(437) m= 142.425mm c) Thickness of Base Plate: use x= n 3𝐹 𝑥 2

𝑝 t= √0.75𝐹

𝑦

3(7.35)(144.4)2 0.75(248)

t= √

t= 49.718mm say 50mm therefore use 700 x 700 x 50 mm base plate

n Equivalent Rectangular Section

0.80 bf

n Base Plate

7.) Determine the safe load of the column section shown, if it has a yield strength of 250 MPa. E = 200000 MPa. Use NSCP Specifications. Properties of Channel Section: A = 3929 mm2 d = 305 mm tf = 12.7 mm tw = 7.2 mm Ix = 53.7 x 106 mm4 Iy = 1.61 x 106 mm4 rx = 117 mm x = 17.7 mm Properties of W 460x74: A = 9450 mm2 d = 457 mm bf = 190 mm tf = 14.5 mm tw = 9.0 mm Ix = 333 x 106 mm4 rx = 118 mm Iy = 16.6 x 106 mm4 ry = 41.9 mm a.) When the height of column is 6 m. b.) When the height of column is 10 m. Assume K = 1.0 Solution: a) When the height of column is 6 m. A = A1 + A2 = 3929 + 9450 = 13,379 mm2 13379ӯ = 3929(17.7) + 9450(235.7) Ӯ = 171.68 mm Ix = [1.61 x 106 + 3929(153.98)2] + [333 x 106 + 9450(64.02)2] Ix = 466.5 x 106 mm4 Iy = 53.7 x 106 + 16.6 x 106 = 70.3 x 106 mm4 Use Least I 70.3 x 106 13379

r= √

r = 72.488 mm 𝑆𝑅 =

L r

6000 72.488

=

2π2 E

CC = √

Fy 5

=

5 3

2π2 (200000)

=√

3(SR)

F.S. = 3 +

= 82.77

250 SR3

- 8(C

3 8CC C) 3(82.77) (82.77)3

+ 8(125.66) − 8(125.66)3

F.S. =1.88 Fa =(1-

=125.66 > SR ∴Intermediate Column

SR2 2CC

2

Fy

)( FS)

(82.77)2

250

= [1- 2(125.66)2] [ 1.88] Fa = 104 MPa

P = FaA P = 104(13379)(10-3) P = 1391.4 kN b) when L = 10 m: L

SR= r =

10000 =138> CC ∴Long Column 72.488

Fa =

12π2 E 23(SR)2

Fa =

12π2 (200000) 23(138)2

Fa = 54.1 MPa P = FaA = 54.1(13379)(10-3) = 723.8 kN

8.) A column is made of steel pipe with an outside diameter of 280 mm. The base plate of the column rests on a circular base plate on concrete pedestal. The column is 3.6 m long and subjected to an axial load of 900 kN. The allowable compressive stress in steel pipe is 65 MPa and the allowable bearing stress in concrete pedestal is 12 MPa. a. What is the required column thickness without exceeding its allowable compressive stress? b. What is the required diameter of the steel base plate? c. If the pipe is 10 mm thick, what is the effective slenderness ratio assuming that column is hinged at both ends? (k=1.0) Given: D = 280 mm L = 3.6 m C = 900 kN

Fp = 12 MPa Fa = 65 MPa

Solution: a. Fa =

C A

𝜋(2802 − 𝑑2 ) 900(1000) = 4 65 𝑑 = 246.517 𝑚𝑚 t

Since D – d = thickness therefore, Thickness = 33.483 mm b. Fp=

280 mm

C Ap

𝜋(𝐷 2 ) 900(1000) = 4 12 𝐝 = 𝟑𝟎𝟗. 𝟎𝟏𝟗 𝐦𝐦

d c.

KL R

KL R

=

=

3600(1.0) √D2 +d2 4

3600(1.0) √2802 +246.5172 4

KL = 38.60 R

Republic of the Philippines Nueva Ecija University of Science and Technology College of Engineering Department of Civil Engineering Cabanatuan City

Comprehensive Examination 3

Submitted by: GROUP 2: Lara Mariz B. Gatbonton Justine Chris Lina Kim Eucasion Gabriel Lopez

BSCS 5-A

Submitted to: Engr. Ermino G. Enriquez