chap. 6-pbs(2)

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Problem 6.2-1

1

6.2-1 Determine whether the given member satisfies the appropriate AISC

interaction equation. Do not consider amplification. The loads are 50% dead load

and 50% live load. Bending is about thex-axis,Fy = 355 MPa.

a. Use LRFD.

b. Use ASD.

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Problem 6.2-1 (Solution)

2

HE 300 B: d= 300 mm, bf= 300 mm, tw = 11 mm, tf= 19 mm, h = 208 mm,Iy

= 8563 cm4 , ry = 7.58 cm , Sx = 1678 cm3 ,Zx = 1869 cm

3 ,J= 189 cm4, Cw =

1690000 cm6.

width-to-thickness ratio:

Load Calculations for LRFD:

/ 2 300 / 2 /19 7.895 0.56 / 0.56 200000 / 355 13.292

/ 208 /11 18.909 1.49 / 1.49 200000 / 355 35.366

No Local Buckling

f f y

y

b t E F

h t E F

1.2(0.5 1110) 1.6(0.5 1110) 1554 kN

1.2(0.5 325) 1.6(0.5 325) 455 kN.m

u

u

P

M

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3

(4.2 m) (4.00 m) [ (4.25 m) (4.00 m)](4.20 4.00) / (4.25 4.00)

(4.2 m) 3860 [3757 3860](0.20) / 0.25 3777.6 kN

/ (4.2 m) 3777.6 / 1.5 2518.4 kN

c n c n c n c n

c n

n c

P P P P

P

P

cPncalculation:

SinceKxL =KyL, KyL/ry Controls, Column Load Tables give, by interpolation

bMnxcalculation:

3 6355(1869 10 ) 10 663.495 kN.mpx y x

M F Z

/ 2 300 / 2 /19 7.895 0.38 / 0.38 200000 / 355 9.020

/ 208 /11 18.909 3.76 / 3.76 200000 / 355 89.246

Shape is compact

f f y

y

b t E F

h t E F

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4


52 10

1.76 1.76(7.58) 316.7 cm355

p y

y

EL r

F

62 28563(1.69 10 ) 71.691 cm 8.47 cm

1678

y w

ts ts

x

I Cr rS

0 (300 19) /10 28.1 cmfh d t

2

0

0

25

5

0.71.95 1 1 6.76

0.7

2 10 189(1.0) 0.7 355 1678 28.11.95(8.47) 1 1 6.76

0.7(355) 1678(28.1) 2 10 189 1.0

1272 cm

y x

r ts

y x

r

r

F S hE JcL r

F S h EJc

L

L

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5


3 6

316.7 cm 420 cm 1272 cm

( 0.7 )

420 316.71.0 663.495 (663.495 0.7(355)(1678 10 ) 10 )

1272 316.7

636.839 kN.m 0.9(636.839) 57

p b r

b p

nx b p p y x p

r p

nx

nx b nx

L L L

L LM C M M F S M

L L

M

M M

3.155 kN.m

/ 0.6 0.6(636.839) 382.103 kN.mnx b nxM M

1554 0.411 0.23777.6

u

c n

P

P

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6


8

9

8 4550.411 0

9 573.1550.411 0.706 1.117 1.0, N.G.

uyu ux

c n b nx b ny

MP M

P M M

15540.411 0.2

3777.6

u

c n

P

P

11100.441 0.2

/ 2518.4

u

n c

P

P

8

/ 9 / /

8 3250.441 09 382.103

0.441 0.756 1.197 1.0, N.G.

aya ax

n c nx b ny b

MP M

P M M

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Problem 6.2-2

7

6.2-2 How much service live load, in kN/m, can be supported? The member

weight is the only dead load. The axial compression load consists of a service dead

load of 45 kN and a service live load of 90 kN. Do not consider amplification

Bending is about thex-axis,Fy = 355 MPa.

a. Use LRFD.b. Use ASD.

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8

HE 450 A: d= 440 mm, bf= 300 mm, tw = 11.5 mm, tf= 21 mm, h = 2344

mm,Iy = 9465 cm4 , ry = 7.29 cm , Sx = 2896 cm

3 ,Zx = 3216 cm3 ,J= 250

cm4, Cw = 4150000 cm6, m = 139.8 kg/m.



/ 2 300 / 2 / 21 7.143 0.56 / 0.56 200000/ 355 13.292

/ 344 /11.5 29.913 1.49 / 1.49 200000/355 35.366

No Local Buckling

f f y

y

b t E F

h t E F

2 2

,max

1.2(45) 1.6(90) 198 kN

1.2(139.8 10/1000) 1.6( ) 1.678 1.6 kN/m

/8 (1.678 1.6 )(6) /8 7.549 7.2 kN.m

u

u L L

u u L L

P

w w w

M w L w w

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9

(6 m) 3415 kN

/ (6 m) 2272 kN

c n

n c

P

P

cPncalculation:

SinceKxL =KyL, KyL/ry Controls, Column Load Tables give

bMnxcalculation:

/ 2 300 / 2 / 21 7.143 0.38 / 0.38 200000 /355 9.020

/ 344/11.5 29.913 3.76 / 3.76 200000 /355 89.246

Shape is compact

f f y

y

b t E F

h t E F

Load Calculations for LRFD:`

2 2

,max

45 90 135 kN

139.8 10/1000 1.398 kN/m

/8 (1.398 )(6) /8 6.291 4.5 kN.m

a

a L L

a a L L

P

w w w

M w L w w

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10


1.14 for uniform load, lateral support at the supports only.

Beam Design Tables give

for 1.00; 865 kN.m 961.111 kN.m

(961.111) 1.14(961.111) 1095.667 kN.m 1142 kN.m

0.9(1

b

b b nx nx

nx b px

b nx

C

C M M

M C M

M

095.667) 986.1 kN.m

/ 0.6 0.6(1095.667) 657.4 kN.mnx b nxM M

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11


2

(7.549 7.2 )0.0580

2 986.1

0.029 0.0077 0.0073 1.0

131.959 kN/m

uyu ux

c n b nx b ny

L

L

L

MP M

P M M

w

w

w

1980.058 0.2

3415

u

c n

P

P

1350.059 0.2

/ 2272

u

n c

P

P

2 / / /

6.291 4.50.0590

2 657.4

0.0295 0.0096 0.0068 1.0

141.309 kN/m

aya ax

n c nx b ny b

L

L

L

MP M

P M M

w

w

w

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Problem 6.6-1

12

6.6-1 Compute the moment amplification factorB1 for the member of Problem

6.2-1. The frame analysis was performed using the requirements for approximate

second-order analysis method of AISC Appendix 8. This means that a reduced

stiffness, EI*, was used in the analysis, and an effective length factor ofKx = 1.0

can be used.

a. Use LRFD.

b. Use ASD.

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13

a. LRFD

In the plane of bending

2 21554 0 1554 kN ( 0 for braced frame)r u nt lt P P P B P B

(4.2 m) 3777.6 kN

/ (4.2 m) 2518.4 kN

c n

n c

P

P

*

3

* 4 13 2

0.8

1.00(1554)0.294 0.5 1.0

14900(355)(10 )

0.8(1.0)(200000)(25166 10 ) 4.027 10 N.mm

b x

r rb

y g y

EI EI

P P

P A F

EI

2 * 2 13

3

1 2 2

1

(4.027 10 )10 22529 kN

( ) (1.0 4200)e

EIP

K L

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14

1

2

1

1

4550.6 0.4 0.6 0.4 1.0

455

1.01.074

1 ( / ) 1 (1.00)(1554) / 22529

m

m

r e

MC

M

CB

P P

1 2 1.074(455) 0 488.71 kN.mrx ux nt lt M M B M B M

8

9

8 488.710.411 09 573.155

0.411 0.758 1.169 1.0, N.G.

uyu ux

c n b nx b ny

MP M

P M M

15540.411 0.2

3777.6

u

c n

P

P

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15

b. ASD


2 21110 0 1110 kN ( 0 for braced frame)r a nt lt P P P B P B

(4.2 m) 3777.6 kN

/ (4.2 m) 2518.4 kN

c n

n c

P

P

*

3

* 4 13 2

0.8

1.60(1110)0.336 0.5 1.0

14900(355)(10 )

0.8(1.0)(200000)(25166 10 ) 4.027 10 N.mm

b x

r rb

y g y

EI EI

P P

P A F

EI

2 * 2 13

3

1 2 2

1

(4.027 10 )10 22529 kN

( ) (1.0 4200)e

EIP

K L

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16

1

2

1

1

3250.6 0.4 0.6 0.4 1.0

325

1.01.086

1 ( / ) 1 (1.60)(1110) / 22529

m

m

r e

MC

M

CB

P P

1 2 1.086(325) 0 352.813 kN.mrx ax nt lt M M B M B M

11100.441 0.2

/ 2518.4

u

n c

P

P

8

/ 9 / /

8 352.8130.441 09 382.103

0.441 0.821 1.262 1.0, N.G.

aya ax

n c nx b ny b

MP M

P M M

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Problem 6.6-4

17

6.6-4 The member shown is part of a braced frame. The load and moments are

computed from service loads, and bending is about the x-axis (the end shears are

not shown). The frame analysis was performed consistent with the effective length

method, so the flexural rigidity, EI, was unreduced. Use Kx = 0.9. the load and

moments are 30% dead load and 70% live load. Determine whether this member

satisfies the appropriate AISC interaction equation.Fy = 355 MPa.

a. Use LRFD.

b. Use ASD.

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18

HE 300 A: d= 290 mm, bf= 300 mm, tw = 8.5 mm, tf= 14 mm, h = 208 mm,Ix

= 18263 cm4 ,Iy = 6310 cm4 , ry = 7.49 cm , Sx = 1260 cm

3 ,Zx = 1383 cm3 ,J

= 87.8 cm4, Cw = 1200000 cm6.



/ 2 300 / 2 /14 10.714 0.56 / 0.56 200000 / 355 13.292

/ 208 / 8.5 24.471 1.49 / 1.49 200000 / 355 35.366

No Local Buckling

f f y

y

b t E F

h t E F

1

1

1.2(0.3 535) 1.6(0.7 535) 1.48 535 791.8 kN

1.2(0.3 90) 1.6(0.7 90) 1.48 90 133.2 kN.m

1.2(0.3 180) 1.6(0.7 180) 1.48 180 266.4 kN.m

u

u

u

P

M

M

18


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19

a. LRFD


2 2791.8 0 791.8 kN ( 0 for braced frame)r u nt lt P P P B P B

(4.8 m) 2667 (2581 2667)(4.8 4.75) / (5 4.75) 2649.6 kN

/ (4.8 m) 1766.5 kN

c n

n c

P

P

*

* 4 13 2(200000)(18263 10 ) 3.653 10 N.mm

xEI EI

EI

2 * 2 13

31 2 2

1

(3.653 10 ) 10 19317 kN( ) (0.9 4800)

e EIPK L

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20


20

1 2 1

max

( ) / 4.8 1.48[90 ( 180 90) / 4.8] 1.48[90 56.25 ]

( 4.8/ 4 1.2) 1.48 22.5 kN.m

( 4.8/ 2 2.4) 1.48 ( 45) kN.m

( 3 4.8/ 4 3.6) 1.48 ( 112.5) kN.m( 4.8) 1.48 ( 180) kN.m

u u u

A

B

B

M M M M x x x

M M x

M M x

M M xM M x

max max12.5 /(2.5 3 4 3 )

12.5(180) /[2.5(180) 3(22.5) 4(45) 3(112.5)]

2.174

b A B C

b

b

C M M M M M

C

C

(4.8 m) 2.174[407 (402 407)(4.8 4.75) /(5 4.75)]

(4.8 m) 882.61 kN.m 427 kN.m, use 427 kN.m

/ (4.8 m) 284 kN.m

b nx

b nx b px b nx b px

nx b

M

M M M M

M

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21

1

2

1 1

1

900.6 0.4 0.6 0.4 0.4

180

0.40.417 1.0, use 1.0

1 ( / ) 1 (1.00)(791.8) /19317

m

m

r e

MC

M

CB B

P P

1 2 1.0(266.4) 0 266.4 kN.mrx ux nt lt M M B M B M

8

9

8 266.40.299 09 427

0.299 0.555 0.854 1.0, O.K.

uyu ux

c n b nx b ny

MP M

P M M

791.80.299 0.2

2649.6

u

c n

P

P

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22

b. ASD


2 2535 0 535 kN ( 0 for braced frame)r a nt lt P P P B P B

/ (4.2 m) 1766.5 kNn cP

*

* 4 13 2(200000)(18263 10 ) 3.653 10 N.mm

xEI EI

EI

2 * 2 13

31 2 2

1

(3.653 10 ) 10 19317 kN( ) (0.9 4800)

e EIPK L

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23

1

2

1 1

1

900.6 0.4 0.6 0.4 0.4

180

0.40.419 1.0, use 1.0

1 ( / ) 1 (1.60)(535) /19317

m

m

r e

MC

M

CB B

P P

1 2 1.0(180) 0 180 kN.mrx ux nt lt M M B M B M

8

/ 9 / /

8 1800.303 09 271

0.303 0.590 0.893 1.0, O.K.

aya ax

n c nx b ny b

MP M

P M M

5350.303 0.2

/ 1766.4

a

n c

P

P

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Problem 6.7-2

24

6.7-2 An HE 360 A ofFy = 355 MPa, 4.8 m long, is used as a column in an

unbraced frame. The results of a first-order analysis for service D, L, and Ware

shown in the figure. Bending is about the x-axis. The effective length factors are

Kx = 0.85 for the braced case, Kx = 1.2 for the unbraced case, and Ky = 1.0.

Determine whether this member is in compliance with the AISC Specification.

a. Use LRFD.

b. Use ASD.

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25


LRFD:

Load Combination 2: 1.2D + 1.6L

/ 2 300 / 2 /17.5 8.57 0.56 / 0.56 200000 / 355 13.29

/ 261/10 26.1 1.49 / 1.49 200000 / 355 35.37

No Local Buckling

f f y

y

b t E F

h t E F

,Bot

,Top

,Bot

1.2(535) 1.6(1070) 2354 kN

1.2(25) 1.6(65) 134 kN.m, cw1.2(20) 1.6(55) 112 kN.m, cw

2354 kN, 134 kN.m, cw, 0 (because of symmetry)

u

u

u

nt nt u lt

P

MM

P M M M

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26

1

2

2 *2 * 2

1 2 2 2

1

1 1

1

1120.6 0.4 0.6 0.4 0.27

134

(200000)(330800)39226 kN

( ) ( ) [(0.85)(4800)]

0.270.28 1.0, use 1.0

1 ( / ) 1 (1.00)(2354) / 39226

m

xe

x

m

r e

MC

M

EIEIP

K L K L

CB B

P P

1 2 1.0(134) 0 134 kN.mrx ux nt lt M M B M B M

2 2354 0 2354 kNr u nt lt P P P B P

(4.8 m) [613 (605 613)(4.8 4.75) / (5 4.75)]

(4.8 m) 611.4 kN.m, for 1.0

b nx

b nx b

M

M C

(4.8 m) [3359 (3249 3359)(4.8 4.75) / (5 4.75)]

(4.8 m) 3337 kN

c n

c n

P

P

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27


1 2 1

max

( ) / 4.8 [134 ( 112 134) / 4.8] 134 51.25

( 4.8 / 4 1.2) 72.5 kN.m

( 4.8 / 2 2.4) 11 kN.m

( 3 4.8 / 4 3.6) 50.5 kN.m( 0) 134 kN.m

u u u

A

B

C

M M M M x x x

M M x

M M x

M M xM M x

max max12.5 / (2.5 3 4 3 )

12.5(134) / [2.5(134) 3(72.5) 4(11) 3(50.5)]

2.24

b A B C

b

b

C M M M M M

C

C

For 2.24, (4.8 m) 2.24(611.4) 1369.11 kN.m 667 kN.m

Use 667 kN.m

b b nx b px

b nx b px

C M M

M M

134 kN.m

112 kN.m

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28


2354 80.71 0.2

3337 9

8 1340.71 0

9 6670.71 0.20 0.91 1.0, OK

uyu u ux

c n c n b nx b ny

MP P M

P P M M

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29

LRFD: Load Combination 4: 1.2D + 1.0W+ 0.5L

,Bot

,Top

1.2(535) 0.5(1070) 1.0(135) 1312 kN

[1.2(25) 0.5(65)] 1.0(175) [62.5] 175 237.5 kN.m, cw

[1.2(20) 0.5(55)] 1.0(175) [51.5] 175 226.5 kN.m, cw

1177 kN, 62.5 kN.m, cw, 0, 175

u

u

u

nt nt lt lt

P

M

M

P M P M

2

2

1

2

1

1 1

1

kN.m, cw

For the braced condition, 0 and

1177 0 1177 kN

51.50.6 0.4 0.6 0.4 0.27

62.5

0.27

1 ( / ) 1 [( ) / ] 1 (1.00)(1177 135) /39226

0.28 1.0, Us

r u nt lt

m

m m

r e nt lt e

B

P P P B P

MC

M

C CB

P P P P P

B

1e 1.0B

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30

2

story

2story story

2story

2 *2 *

2 2 2

2

For the unbraced condition, must be computed

1 11 since we do not dispose of

11

1312 kN

, where is taken for the unbraced case( ) ( )

u e

ee

u

xe x

x

B

PB

P P P

PP

P

EIEIP K

K L K L

P

2

2 2(200000)(330800) 19681 kN[(1.2)(4800)]

e

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31


1 2 1

max

( ) / 4.8 [237.5 ( 226.5 237.5) / 4.8] 237.5 96.67

( 4.8/ 4 1.2) 121.5 kN.m

( 4.8/ 2 2.4) 5.5 kN.m

( 3 4.8/ 4 3.6) 110.5 kN.m

( 0) 237.5 kN.m

u u u

A

B

C

M M M M x x x

M M x

M M x

M M x

M M x

max max12.5 /(2.5 3 4 3 )

12.5(237.5) /[2.5(237.5) 3(121.5) 4(5.5) 3(110.5)]

2.26

b A B C

b

b

C M M M M M

C

C

For 2.26, (4.8 m) 2.24(611.4) 1383.72 kN.m 667 kN.m

Use 667 kN.m

b b nx b px

b nx b px

C M M

M M

237.5 kN.m

226.5 kN.m

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32

2

2

2

1 2

1 11.071 1

1.0(1312)11

19681

The amplified axial load is

1177 1.071(135) 1322 kNThe total amplified moment load is

1.0(62.5) 1.071(175) 250 kN.m

1312

3337

u

e

r u nt lt

r u nt lt

r

c n

BP

P

P P P B P

M M B M B M

P

P

8

0.39 0.2

9

8 2500.39 0

9 667

0.39 0.33 0.73 1.0, OK

uyu ux

c n b nx b ny

MP M

P M M

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Example 6.10-modified

33

A multistory structure composed of ten three-bay stories must be stabilized by

diagonal X bracing in one of the three bays. The braced frame is shown in the

figure. The loading is typical. Use a steel ofFy = 235 MPa to determine the

required cross-sectional area of the bracing. Use LRFD.

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34/41

Example 6.10-modified (Solution)

34

The total vertical load to stabilized by the bracing is

1

{[1.2(32) 1.6(10)](6) 3} 10 [1.2(20) 0.5(4)](6) 3

{54.4(6) 3} 10 [26](6) 3 979.2 10 468 10260 kN

0.004 0.004(10260) 41.04 kN

41.0445.88 kN

cos cos[tan (3 / 6)] cos[26.57]

Based on

r u

r

rb r

rb rb

P P

P

P P

P PF

2 2

the limit state of tension yielding, the required area below

the first floor is

45880217 mm 2.17 cm

0.9 0.9(235)g

y

FA

F

7/28/2019 Chap. 6-Pbs(2)

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35

The required lateral stiffness is

1 2 1 2(10260)9120 kN/m

0.75 3

The axial stiffness of the brace is given by / ,

where is the axial elongation of the brace, which is related to the horizontal

displacement by

cos

rbr

b

P

L

F

2

2

The axial stiffness of the brace can be written as

/ cos 1 1 2cos

cos cos

rb rb rb r

b

P P PF F P

L

7/28/2019 Chap. 6-Pbs(2)

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36

2

2 6 3

2 2 st

min

Knowing that

21 6

9120000 0.000382 mcos 200000 10 cos (26.57)

3.82 cm Controls, use L50 50 4 3.89 cm , for the 1 story.

/ 600 / cos(26.57) / 0.979 670.82 / 0.979 685

g

r

gb

g

A EF

L

P L

A L E

A

L r

min

2

min

.21 300, N.G.

/300 670.82/300 2.24 cm,

use L120 120 8 18.7 cm , 2.37 cm, for the entire building.

r L

r

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37

The required cross-sectional areas for all stories are

Story No.

Pr @

Story No.

(kN)

Prb @

Story No.

(kN)

Ag-

Yielding

@ Story

No.

(cm2)

Lateral

Stiffness

br @Story No.

(kN/m)

Ag-Axial

Stiffness @

Story No.

(cm2)

Ag @

Story No.

(cm2)

Equal Leg

Angle, No

Slenderness

Limit

1 10260 41.04 1.94 9120 3.82 3.82 L 50x50x4

2 9280.8 37.12 1.76 8249.6 3.46 3.46 L 50x50x4

3 8301.6 33.21 1.57 7379.2 3.09 3.09 L 50x50x4

4 7322.4 29.29 1.38 6508.8 2.73 2.73 L 50x50x3

5 6343.2 25.37 1.20 5638.4 2.36 2.36 L 50x50x3

6 5364 21.46 1.01 4768 2.00 2.00 L 50x50x3

7 4384.8 17.54 0.83 3897.6 1.63 1.63 L 30x30x3

8 3405.6 13.62 0.64 3027.2 1.27 1.27 L 30x30x3

9 2426.4 9.71 0.46 2156.8 0.90 0.90 L 30x30x3

10 1447.2 5.79 0.27 1286.4 0.54 0.54 L 30x30x3

Roof 468 1.87 0.09 416 0.17 0.17 L 30x30x3

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Example 5.15-6-modified

38

An IPE 160 of a steel ofFy = 355 MPa is used as purlins supported by

trusses spaced at 3 m on centers. The total gravity load of 1.92 kN/m2 of

roof surface, half dead load half snow load. The inclination of the roof is 1

vertical to 4 horizontal. The tributary area of a purlin is 1.855 m. Assume

that wind load is not a factor and investigate the adequacy of an IPE 160

for use as a purlin. No sag rods are used, so lateral support is at the ends

only.

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Example 5.15-6-modified (Solution)

39

IPE 160: d= 160 mm, bf= 82 mm, tw = 5 mm, tf= 7.4 mm, h = 127.2

mm,Ix = 869 cm4 ,Iy = 68.3 cm

4 , ry = 7.49 cm , Sx = 109 cm3 ,Zx =

124 cm3 , Sy = 16.7 cm3 ,Zy = 26.1 cm

3.


/ 2 82 / 2 / 7.4 5.54 0.56 / 0.56 200000/ 355 13.292/ 127.2 /5 25.44 1.49 / 1.49 200000 /355 35.366

Shape is Compact

f f y

y

b t E F h t E F

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40

Load Calculations for LRFD:2

1

Roof

Roof

1.2(0.5 1.92) 1.6(0.5 1.92) 1.48 535 2.688 kN/m

1.855(2.688) 4.986 kN/m

weight of the beam;

16.8 10/1000 0.168 kN/m

roof slope:

1tan 14.04

4

( ) ( 1.2 )cos( ) [4.986 1.2(

u

u

u u

u

q

w

w

w w ww

Roof

Roof

0.168)]cos14.04 5.033 kN/m

( ) ( 1.2 )sin

( ) [4.986 1.2(0.168)]sin14.04 1.258 kN/m

u u

u

w w w

w

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41

2 2

Roof

2 2Roof

( ) 5.033(3)5.66 kN.m

8 8

( ) 1.258(3)1.42 kN.m

8 8

uux

u

ux

wM

wM

6

6

1.14 ( 1.0) 1.14(21) 23.94 kN.m 39.6 kN.m

0.9 0.9(355)(26100) 10 8.34 kN.m

(1.6 ) 0.9[1.6(355)(16700) 10 8.54 kN.m

(1.6 ) 8.34 kN.m

nx nx b px

ny y y

y y

ny y y ny

M M C M

M F Z

F S

M F S M

5.66 1.420.768 1.0

0.5 23.94 0.5(5.34)

The shape is adequate.

uyux

b nx b ny

MM

M M

chap. 6-pbs(2)

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