columnas a1085.pdf

5
DESIGN EXAMPLE SQUARE HSS COMPRESSION MEMBER WITHOUT SLENDER ELEMENTS A Comparison of ASTM A500 and ASTM A1085 Given: Select an ASTM A1085 square HSS compression member, with a length of 20 ft, to support a dead load of 150 kips and live load of 440 kips in axial compression. The base is fixed and the top is pinned. Also, compare what size would be required for an ASTM A500 Grade B square HSS. Solution: The material properties are as follows: ASTM A1085 F y = 50 ksi F u = 65 ksi From Chapter 2 of ASCE/SEI 7, the required compressive strength is: LRFD ASD 1.2(150 kips) 1.6(440 kips) 884 kips u P 150 kips 440 kips 590 kips a P From AISC Specification Commentary Table C-A-7.1, for a fixed-pinned condition, K = 0.8. ( ) ( ) 0.80(20.0 ft) 16.0 ft x y KL KL Try an ASTM A1085 HSS12122. The geometric properties are as follows: HSS12122 A g = 22.4 in. 2 r x = r y = 4.66 in. t des = 0.500 in. Slenderness Check Note: According to ASTM A1085, the lower limit for corner radius is equal to 1.8t (for t > 0.4 in.) and can be conservatively used to calculate wall slenderness. 12.0in. 1.8 2 (0.500in.) 0.500 in. 20.4 h t Determine the wall limiting slenderness ratio, r , from AISC Specification Table B4.1a Case 6.

Upload: jose-rojas-guzman

Post on 27-Oct-2015

56 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Columnas A1085.pdf

DESIGN EXAMPLE SQUARE HSS COMPRESSION MEMBER WITHOUT SLENDER ELEMENTS A Comparison of ASTM A500 and ASTM A1085 Given: Select an ASTM A1085 square HSS compression member, with a length of 20 ft, to support a dead load of 150 kips and live load of 440 kips in axial compression. The base is fixed and the top is pinned. Also, compare what size would be required for an ASTM A500 Grade B square HSS. Solution: The material properties are as follows:

ASTM A1085 Fy = 50 ksi Fu = 65 ksi

From Chapter 2 of ASCE/SEI 7, the required compressive strength is:

LRFD ASD 1.2(150 kips) 1.6(440 kips)

884 kipsuP

150 kips 440 kips

590 kipsaP

From AISC Specification Commentary Table C-A-7.1, for a fixed-pinned condition, K = 0.8.

( ) ( ) 0.80(20.0 ft) 16.0 ftx yKL KL

Try an ASTM A1085 HSS12122. The geometric properties are as follows:

HSS12122 Ag = 22.4 in.2 rx = ry = 4.66 in.

tdes = 0.500 in. Slenderness Check Note: According to ASTM A1085, the lower limit for corner radius is equal to 1.8t (for t > 0.4 in.) and can be conservatively used to calculate wall slenderness.

12.0in. 1.8 2 (0.500in.)

0.500 in.20.4

h

t

Determine the wall limiting slenderness ratio, r, from AISC Specification Table B4.1a Case 6.

Page 2: Columnas A1085.pdf

1.40

29,000 ksi1.40

50 ksi33.7

ry

E

F

r ; therefore, the section does not contain slender elements.

Because x yr r and ( ) ( )x yKL KL , the available compressive strength will be the same about both axes.

Determine the applicable equation.

0.8(20.0 ft.)(12 in./ft)

4.66 in.41.2

KL

r

29,000 ksi4.71 4.71

50 ksi

113 41.2, therefore, use AISC Equation E3 2

y

E

F

Specification

2

2eE

FKL

r

(Spec. Eq. E3-4)

2

2

(29,000 ksi)

(41.2)

169 ksi

0.658y

e

F

Fcr yF F

(Spec. Eq. E3-2)

50 ksi

169 ksi0.658 (50 ksi)

44.2 ksi

n cr gP F A (Spec. Eq. E3-1)

244.2 ksi(22.4 in. )

990 kips

From AISC Specification Section E1, the available compressive strength is:

LRFD ASD c = 0.90 c = 1.67

0.90(990 kips)

891 kips 884 kips c nP

o.k.

990 kips

1.67

593 kips > 590 kips

n

c

P

o.k.

Page 3: Columnas A1085.pdf

For comparison purposes, the above design example can be re-calculated as follows, using an ASTM A500 Grade B square HSS compression member.

From AISC Manual Table 2-4, the material properties are as follows:

ASTM A500 Grade B Fy = 46 ksi Fu = 58 ksi

Table Solution Enter AISC Manual Table 4-4 for square sections. Try an ASTM A500 HSS1212s. From AISC Manual Table 4-4, the available strength in axial compression is:

LRFD ASD 947 kips 884 kips c nP o.k.

630 kips 590 kips n

c

P

o.k.

The available strength can be easily determined by using the tables of the AISC Manual. Available strength values can be verified by hand calculations, as follows. Calculation Solution From AISC Manual Table 1-12, the geometric properties are as follows:

HSS1212s Ag = 25.7 in.2 rx = ry = 4.62 in. tdes = 0.581 in.

Slenderness Check

12.0 in. 3(0.581 in.)

0.581 in.17.7

h

t

Determine the wall limiting slenderness ratio, r, from AISC Specification Table B4.1a Case 6.

1.40

29,000 ksi1.40

46 ksi35.2

ry

E

F

r ; therefore, the section does not contain slender elements.

Page 4: Columnas A1085.pdf

Because x yr r and ( ) ( )x yKL KL , the available compressive strength will be the same about both axes.

Determine the applicable equation.

0.8(20.0 ft.)(12 in./ft)

4.62 in.41.6

KL

r

29,000 ksi4.71 4.71

46 ksi

118 41.6, therefore, use AISC Equation E3 2

y

E

F

Specification

2

2eE

FKL

r

(Spec. Eq. E3-4)

2

2

(29,000 ksi)

(41.6)

165 ksi

0.658y

e

F

Fcr yF F

(Spec. Eq. E3-2)

46 ksi

165 ksi0.658 (46 ksi)

40.9 ksi

n cr gP F A (Spec. Eq. E3-1)

240.9 ksi(25.7 in. )

1051 kips

From AISC Specification Section E1, the available compressive strength is:

LRFD ASD c = 0.90 c = 1.67

0.90(1051 kips)

946 kips 884 kips c nP

o.k.

1051 kips

1.67

629 kips > 590 kips

n

c

P

o.k.

Page 5: Columnas A1085.pdf

Summary

HSS Standard

Selected Section

Area (in.2) Weight (lb / ft) Design Strength, ϕPn (kips)

Allowable Strength, Pn/Ω

(kips) ASTM A1085 HSS12122 22.4 76.1 891 593

ASTM A500 HSS12s 25.7 93.3 946 629

First, the member is designed using ASTM A1085 to carry the loads shown in the problem statement. The selected section is a HSS12122, which has a design strength of 891 kips and allowable strength of 593 kips. Then, the member is re-designed using ASTM A500 with the same nominal width. A similarly sized A500 section, HSS12122, has a design and allowable strength of 773 and 514 kips, respectively, and has insufficient axial strength capacity. Thus, the wall thickness is bumped up to s”. The resulting A500 section, HSS12s, has a design and allowable strength of 946 and 629 kips, respectively. The results indicate that for the loads given, the required A1085 section has smaller cross-sectional area and lighter weight than the section size required for A500. Designers should assess the options available to them and speak with their fabricator to determine what the best solution is for a given project.