Determine the design capacity using ASD and NSCP specifications for the connection shown. Use A36 steel and D20 diameter bolt on an oversized hole diameter. Show complete solution and round your final answer to the nearest thousandths.
Determine the design capacity using ASD and NSCP specifications for the connection shown. Use A36 steel and D20 diameter bolt on an oversized hole diameter. Show complete solution and round your final answer to the nearest thousandths.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![Determine the design
capacity using ASD and NSCP
specifications for the connection
shown. Use A36 steel and D20
diameter bolt on an oversized hole
117
75
75
diameter. Show complete solution
and round your final answer to the
Pu
OR
nearest thousandths.
PA
125
Given formula:
Nominal hole diameter
Due GSY:
Bolt
Hole diameter
ØPn = ØA,F, ;ø= 0.90
dnet
diameter
Standard
Oversized
Due TRS
M16
18
20
+ 2mm
ØP, = ØUA„Fu ;ø= 0.75
M20
22
24
+ 2mm
M22
24
28
+ 2mm
U = 1-
M24
27
30
+ 2mm
Due to BSS
ØRn = Ø[0. 60A,nyFu + U psAntFu]
< ø[0.60A9„F, + UbgAntFu] ; ¢ =
y
-X = 14.35 mm
0.75
Slendemess Ratio
< 300 ;
C8x18.75
I'min
For section properties
I = Iox + A(y²)
ly = Ioy + A(x²)
A, = 3,555 mm?
bf = 64.2 mm
tf = 9.9 mm
tw = 12.4 mm
d = 203.2 mm
ly = 18.31 x106 mm*
Ix = 0.82 x106 mm*
r =
008
75 75](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb606620c-8dad-4305-84ad-a62406e6d584%2Ff9fb96bb-692e-4d96-bf4b-9c33698c1dba%2Fngznl7_processed.png&w=3840&q=75)
Transcribed Image Text:Determine the design
capacity using ASD and NSCP
specifications for the connection
shown. Use A36 steel and D20
diameter bolt on an oversized hole
117
75
75
diameter. Show complete solution
and round your final answer to the
Pu
OR
nearest thousandths.
PA
125
Given formula:
Nominal hole diameter
Due GSY:
Bolt
Hole diameter
ØPn = ØA,F, ;ø= 0.90
dnet
diameter
Standard
Oversized
Due TRS
M16
18
20
+ 2mm
ØP, = ØUA„Fu ;ø= 0.75
M20
22
24
+ 2mm
M22
24
28
+ 2mm
U = 1-
M24
27
30
+ 2mm
Due to BSS
ØRn = Ø[0. 60A,nyFu + U psAntFu]
< ø[0.60A9„F, + UbgAntFu] ; ¢ =
y
-X = 14.35 mm
0.75
Slendemess Ratio
< 300 ;
C8x18.75
I'min
For section properties
I = Iox + A(y²)
ly = Ioy + A(x²)
A, = 3,555 mm?
bf = 64.2 mm
tf = 9.9 mm
tw = 12.4 mm
d = 203.2 mm
ly = 18.31 x106 mm*
Ix = 0.82 x106 mm*
r =
008
75 75
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