Post on 03-Apr-2018
7/28/2019 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
7/28/2019 Chap. 6-Pbs(2)
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Problem 6.2-1 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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4
Problem 6.2-1 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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5
Problem 6.2-1 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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6
Problem 6.2-1 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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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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Problem 6.2-2 (Solution)
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.
width-to-thickness ratio:
Load Calculations for LRFD:
/ 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
7/28/2019 Chap. 6-Pbs(2)
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Problem 6.2-2 (Solution)
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
Problem 6.2-2 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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11
Problem 6.2-2 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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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.
7/28/2019 Chap. 6-Pbs(2)
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Problem 6.6-1 (Solution)
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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Problem 6.6-1 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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Problem 6.6-1 (Solution)
15
b. ASD
In the plane of bending
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
7/28/2019 Chap. 6-Pbs(2)
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Problem 6.6-1 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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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.
7/28/2019 Chap. 6-Pbs(2)
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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.
width-to-thickness ratio:
Load Calculations for LRFD:
/ 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
Problem 6.6-4 (Solution)
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Problem 6.6-4 (Solution)
19
a. LRFD
In the plane of bending
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
Problem 6.6-4 (Solution)
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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Problem 6.6-4 (Solution)
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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Problem 6.6-4 (Solution)
22
b. ASD
In the plane of bending
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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Problem 6.6-4 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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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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Problem 6.7-2 (Solution)
25
width-to-thickness ratio:
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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Problem 6.7-2 (Solution)
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
Problem 6.7-2 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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28
Problem 6.7-2 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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Problem 6.7-2 (Solution)
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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Problem 6.7-2 (Solution)
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
Problem 6.7-2 (Solution)
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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Problem 6.7-2 (Solution)
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
7/28/2019 Chap. 6-Pbs(2)
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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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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
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Example 6.10-modified (Solution)
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
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Example 6.10-modified (Solution)
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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Example 6.10-modified (Solution)
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.
width-to-thickness ratio:
/ 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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Example 5.15-6-modified (Solution)
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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Example 5.15-6-modified (Solution)
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