School

New Jersey Institute Of Technology **We aren't endorsed by this school

Course

ME 315

Subject

Mechanical Engineering

Date

Dec 18, 2023

Pages

4

Uploaded by MateNightingale992 on coursehero.com

1 Prof. K.A. Narh ME-315-101: Fall 2022 Solution of Problems on Stress Concentration (H6) (1) (a) Stress at section AB: (i) ( )(
)
4300
9556 psi
3
0.15
9556 psi is below proportional limit,
55,000 psi, therefore, the stress is in elastic zone AB
AB
AB
P
BC
P
A
=
=
=
=
=
(ii) Stress at section BC: ( )(
)
4300
28,667 psi
1
0.15
28,667 psi < ; 55,000 psi.
Therefore, the stress is in elastic zone BC
BC
BC
P
P
BC
P
A
=
=
=
=
=
(b) Stress concentration factor K
= 2.9 (given) ( )
(
)(
)
(
)
max
max
max
where 28,667 psi
i 2.9
28,667
83,133 psi
ii Since 83,133 psi > 55,000 psi,the material in the region of the fillet becomes plastic.
ave
ave
BC
ave
K
K
=
=
=
=
=
=
=
55000 psi 1
(2)
1
1
1
(2)
1
1
(2)
(25)
Final grade

2 Problem 2 For this problem, we use the graph below, also given at the end of the notes on stress concentration. Solution To use the above graph, the two ratios w/d and r/d must first be computed; w
is the width of the bar, and r
is the radius of the groove. The net width, d
, equals the width minus twice the groove radius. So (
)
2.6
2 0.3
2.
d
=
−
=
Therefore, 2.6
0.3
1.3 and 0.15 2
2
w
r
d
d
=
=
=
=
To determine K
from the graph we find the curve corresponding to w/d
= 1.3, and find r/d
= 0.15 on the horizontal axis. However, since there is no curve for w/d
= 1.3, we use interpolation procedure as follows: Given: (
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2
2
2
1
2
1
2
1
2
1
2
3
1
1
2
3
1
1
3
1
3
1
3
1
3
1
To find and :
x =
; y
=
or 1.3
1.2
2.15
2.18
1.5
1.2
2.25
2.15
−
−
−
−
−
+
−
+
=
−
−
−
−
−
−
=
−
−
x
y
y
y
x
x
y
y
x
x
x
x
x
y
y
y
y
y
x
x
y
y
x
x
K
K
D/d K
t x
1
(1.2) y
1
(2.15) x
2
(1.3) y
2
(K
) x
3
(1.5) y
3
(2.25) (4)

3 (3) (
)( )
(
)(
)
2
max
1.5
0.375 4
Implies that =2.27 from the graph 1 in Fig. 3,
0.04
0.015
0.015
0.000375 m
5000
13.333 MPa.
0.000375
Then, =K
2.27
13.333 MPa
Maximum stress at large hole :
=
=
=
=
−
=
−
=
=
=
=
=
ave
n
n
ave
ave
d
w
K
P
A
A
w
d
t
(
)( )
(
)
2
30.27 MPa
2 / 3
0.2 (10 / 3)
Implies that =2.48 from the graph 1 in Fig. 3. where = net area
10
2
0.015
0.0004 m
3
3
5000 N
5000
12.5 MPa.
0.0004
Maximum stress at small hole :
=
=
=
=
−
=
−
=
=
=
=
ave
n
n
n
ave
d
w
K
P
A
A
A
w
d
t
P
(
)(
)
a
t
m x
Then, =K
2.48
12.5 MPa
31 MPa
5000
5000
=10 MPa
10 / 3
0.015
0.000
Stress concentration factor K
from Figure 4)
r = 1/3 cm d = 10/3 cm 5
(
D
Maximum stress at the fillet :
=
=
=
=
=
ave
ave
P
d
t
t
= 4 cm To determine K from the graph we find the curve corresponding to D/d = 1.2, and find r/d = on the horizontal axis.
However, we don't have a curve for D/d = 1.2, s
4 (10 / 3)
1.2
0.1
0.1
=
=
=
D d
r d
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2
2
2
1
2
1
2
1
2
1
2
3
1
1
2
3
1
1
3
1
3
1
3
1
3
1
To find and by interpolation we use one of the following Equations:
x =
; y
=
or Using the
o we use an interpolation procedue as follows:
−
−
−
−
−
+
−
+
=
−
−
−
−
x
y
y
y
x
x
y
y
x
x
x
x
x
y
y
y
y
y
x
x
y
y
x
x
(
)
(
)
(
)
(
)
6
max
0
e
above Equation:
1.2
1.05
1.5
0.099
1.5
1.5
0.2475
1.7
i
475
1.75
1
Max str
ss d
1
ue at f lleted .45
1.05
2.16
1.5
0.4
(1.75)(10
0 )
17.50 M
area: P
The maximum stress occurs at the side of −
−
=
=
=
−
=
=
+
=
−
−
=
=
t
t
t
t
t
K
K
K
K
K
the small hole.
5000 N 5000 N 4 cm 10/3 cm 2
2
(4)
2
2
(4)
2
2
(4)
(3)

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