fe review - university of memphis review-steel design 2015-2.pdf · fe review structural steel...
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![Page 1: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/1.jpg)
FE Review
Structural Steel Design
![Page 2: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/2.jpg)
Overview of Reference Handbook
• LRFD only
• Form of Equations is sometimes different from AISC.
• Some tables are not in same form as they are in the Manual.
• 14th Edition Steel Construction Manual
• Pages of formulas followed by tables
![Page 3: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/3.jpg)
Example: Find Cb for 30-ft segment.
A B C
30 ft 10 ft
16 k
4 k 12 k
30 60
90 120
0.367.1)90(3)60(4)30(3)120(5.2
)120(5.12
bC
![Page 4: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/4.jpg)
Example
Given:
W18 x 55, Fy = 50 ksi, Cb = 1.67
Find bMn for
a. Lb = 5 ft
Solution:
5 ft < Lp = 5.90 ft
bMn = b Mp = 420 kip-ft.
![Page 5: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/5.jpg)
Example
b. Lb = 10 ft.
Solution
ft-kip 420 use
ft-kip 420 ft kip 606
ft-kip 60690.5109.1342067.1
)ft5.17ft10ft90.5(
ft 10
nMb
pMb
nMb
rLpL
![Page 6: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/6.jpg)
Or, from Charts,
kips-ft 420 Use
kips-ft 606)363(67.1
,67.1For
1for kips-ft 363
n
n
b
bn
M
M
C
CM
![Page 7: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/7.jpg)
![Page 8: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/8.jpg)
Example:
Lb = 13 ft
Mu = 405 ft-kips
Cb = 1
Fy = 50 ksi
Select a W shape
Solution:
Use a W21 x 62
![Page 9: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/9.jpg)
![Page 10: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/10.jpg)
![Page 11: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/11.jpg)
Example:
The beam has continuous lateral support. Fy = 50 ksi
Select a shape
30 ft
wD = 1.00 kips/ft wL = 2.00 kips/ft
![Page 12: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/12.jpg)
Solution: Try a W24 x 55
kips-ft 495)30)(4.4(8
1
kips/ft 4.4)2(6.1)1(2.1
2
u
u
M
w
(OK) kips 66 kips 251
kips 662
)30(4.4
kips-ft 503
nv
u
pbnb
V
V
MM
Use a W24 x 55
![Page 13: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/13.jpg)
13 ft
Columns
Example:
W12 x 50 Fy = 50 ksi
What is the design strength?
![Page 14: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/14.jpg)
Solution:
kips 489)6.14(46.33
ksi 46.33
)50(658.09.0
658.09.0
11367.63
11350
1.8021.802
67.6396.1
)1213(8.0
220,286
)67.63(50
220,286
)/(
2
2
gcrcnc
y
rKLF
crc
y
y
AFP
FF
F
r
KL
r
KL
y
![Page 15: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/15.jpg)
Alternate: From Table 4-22,
ksi 5.334.337.3367.07.33or
ksi 4.33
crcF
Alternate: From Table 4-1, with KL = 0.8(13) = 10.4 ft,
kips 488 471) - 0.4(499 - 499or
kips 471
ncP
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Design:
KL = 14.5 ft Pu = 300 kips Fy = 50 ksi
Select a W10 shape
Solution:
OK kips 332 45 x W10a Use
(N.G.) kips 294 kips 300
kips 2942
282305
39 x W10aTry
c
c
n
n
P
P
![Page 17: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/17.jpg)
2 3 3 2
x
+ 3.5 D = 35 k L = 105 k
L6 x 6 x ½
A36 steel
A = 5.77 in.2
x bar = 1.67 in.
¾ - in. A325 bearing-type bolts
Determine design strength for all limit states.
t = 3/8 in.
![Page 18: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/18.jpg)
kips 187)7.207(90.0
kips 7.207)77.5(36
n
gyn
T
AFT
1. Tension on Gross Area
2. Tension on Net Area
kips 168)5.224(75.0
kips 224.53.871) (58
in.871.3 )364.5(7217.0
7217.0)3(2
67.111
in. 364.5)16/14/3)(5.0(77.5
2
2
n
eun
ne
holegn
T
AFT
UAA
L
xU
AAA
![Page 19: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/19.jpg)
2 3 3 2
3.5
t = 3/8 in.
3. Block Shear
![Page 20: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/20.jpg)
2
2
2
in. 047.1)16/13(5.05.22
1
in. 984.2)16/13(5.23322
1
in. 0.43322
1
in. 16
13
16
1
4
3
16
1
nt
nv
gv
bh
A
A
A
dd
![Page 21: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/21.jpg)
kips 123
047.10.1984.26.0)58(75.0
6.075.0
ntbsnvun AUAFT
Upper Limit:
kips 110 Use
kips 110
)047.1)(58(0.1)0.4)(36(6.075.0
6.075.0
n
ntubsgvyn
T
AFUAFT
![Page 22: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/22.jpg)
Example. Staggered Bolts
Find the smallest net area. Holes are for 1-inch diameter bolts.
![Page 23: FE Review - University of Memphis Review-Steel Design 2015-2.pdf · FE Review Structural Steel Design . Overview of Reference Handbook •LRFD only •Form of Equations is sometimes](https://reader031.vdocuments.site/reader031/viewer/2022021900/5b5d218a7f8b9a68368e2d48/html5/thumbnails/23.jpg)
Hole diameter = 1 + 1/16 = 1.063 in. (?)
tg
sdbA hgn
4
2
For line abde, 2in. 41.10)75.0(063.1(216 nA
For line abcde, 22
in. 27.10 )75.0()5(4
)3(2063.1(316
nA