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1) For a cold-drawn AISI 1050 steel component: - The yield strength is 515 MPa - The maximum and minimum in-plane normal stresses are 318.2 MPa and -318.2 MPa - The factor of safety from maximum shear stress theory is 0.809 and from distortion theory is 1.32 2) Section properties and forces acting on a pipe are calculated, resulting in bending moments of -136.8, 233.4, and -198 kip-in and shear stresses of -10.536 and 8.988 ksi - Additional stresses due to internal pressure are 2.56 and 5.12 ksi - The maximum

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48 views2 pages

X y X y Ab Xy

1) For a cold-drawn AISI 1050 steel component: - The yield strength is 515 MPa - The maximum and minimum in-plane normal stresses are 318.2 MPa and -318.2 MPa - The factor of safety from maximum shear stress theory is 0.809 and from distortion theory is 1.32 2) Section properties and forces acting on a pipe are calculated, resulting in bending moments of -136.8, 233.4, and -198 kip-in and shear stresses of -10.536 and 8.988 ksi - Additional stresses due to internal pressure are 2.56 and 5.12 ksi - The maximum

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A Musiclover
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1)

For Cold-drawn AISI 1050 steel


Yield strength is 515MPa

The maximum and minimum in-plane normal stresses are given by


2
s +s y s - s y
s a ,b = x x + t xy
2

2 2
2
-225 + 225 -225 - 225
= + 225
2

2 2
= 0 318.2
s a = -318.2 MPa
s b = 318.2 MPa

Factor safety from maximum shear stress theory


515
n=
318.2 - ( -318.2)
= 0.809

Factor safety from Distortion theory


( -225) - ( -225 225) + 2252 + 3 225
2
s a ,b =
= 390.6 MPa
515
n=
390.6
= 1.32

2)

Section Properties
p
A= 8.52 - ( 8 - 2 0.25 )
2

4
= 6.4795 in 2
p
IP = 8.54 - ( 8 - 2 0.25 )
4

32
= 110.354 in 4
p
Ix = I y = 8.54 - ( 8 - 2 0.25 )
4

64
= 55.177 in 4

1
Q= 8.53 - ( 8 - 2 0.25 )
3

32
3
=8.51 in

Vector representation of the forces


F = 2.75i + 0 j - 1.9k
Fx = 2.75 kips
Fy = 0
Fz =-1.9 kips

And
r = 3i + 6 j + 5k
Calculate the moments
r F = ( 3i + 6 j + 5k ) ( 2.75i + 0 j - 1.9k )
= -11.4i - ( -5.7 - 13.75) j - 16.5k
= -11.4i + 19.45 j - 16.5k
Equivalent moments
M x = -11.4 12
= -136.8 kip-in
M y = 19.45 12
= 233.4 kip-in
M z = -16.5 12
= -198 kip-in

Mx y
sz =
Ix
8.5
-136.8
= 2
55.177
= -10.536 ksi
M R
t yz = y
IP
8.5
233.4
= 2
110.354
= 8.988 ksi
Stresses due to internal pressure
pd 320 8
s long = =
4t 4 0.25
= 2560 psi
= 2.56 ksi
pd 320 8
s long = =
2t 2 0.25
= 5120 psi
= 5.12 ksi
Summary of stresses
s y = 2.56 ksi
s z = -10.536 + 5.120
= 5.416 ksi
t y = 8.988 ksi

The maximum and minimum in-plane normal stresses are given by


2
s +sz s - s z
s a ,b = y y + t yz
2

2 2
2
2.56 + 5.416 2.56 - 5.416
= + 8.988
2

2 2
= 3.988 9.1
s a = 13.088 ksi
s b = 5.112 ksi

Factor safety from maximum shear stress theory


29.7
n=
13.088 - 5.112
= 3.72

Factor safety from Distortion theory


s a ,b = 2.562 - ( 2.56 5.416 ) + 5.4162 + 3 8.988
= 6.998 ksi
29.7
n=
6.998
= 4.24

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