ME 437 – Strength of Materials Solutions
Problem #1
 x  10500 psi, Tensile
 y  5500 psi
 xy  4000 psi
3  0
Principal stresses:
1, 2 
x  y
2
 x  y
 
2

2

   xy2

Substitute values from above yields:
 1  11444 psi
 2  6444 psi
The maximum shear stress is determined by these two principal stresses as:
Max ( max, 12 , max, 13 , max, 23 )
 max, 12 
 max
1   2
 max, 1,3 
2
11444  6444

 8944
2
1   3
2
 max, 23 
2 3
2
psi
Note that the other maximum shear stresses are less than this value.
Problem #2
The total strain is:
 total   t 
L 0.05

 0.0005
L
100
This total strain is equal to:
t   M  T 
Substituting:
F
 T
EA
 t  0.0005
E  30 *10 6
A 1
  6.5 *10 6
T  100
and solving for F we get:
F=-4500 lbs
The stress is 4500 psi compressive.
Problem #3
2000 lb
6 ft
Q
20000 lb
4.5 ft
4.5 ft
The 2000 lbs creates a bending stress at Q that is tensile and is equal to:
x 
Mzy
Iz
Where
M z  2000 * 4.5 * 12
d
2
2
d 4
Iz 
64
y
Substituting into the bending formula, we get
 x  17188 psi
The stress due to the axial load is compressive and is equal to:
 x' 
F 20000

 1591 psi
A  (2) 2
The total stress is:
  17188  1591  15600
Problem #4
a
D
c
7/8
b
y
3/8
1.5
Location of centeroid:
Ay  2 Aa y a  Ab yb
1.125 y  2(.375)(.75)  .375(.5)
y  0.667
The area moment of inertia:
I z  2I a  I b
1
(.25)(1.5) 3  .375(.75  .667) 2
12
1
I b  (1.5)(.25) 3  .375(.667  .375  .125) 2
12
Ia 
The answer is:
I z  0.158 in 4
The stress at D is:

M z c 10000 * (0.833)

 57200
Iz
0.158
note that c=1.5-0.667=0.833
Problem #5: Curved beam
The neutral axis radius is:
rn 
ro  ri
42

 2.885
ln( ro / ri ) ln( 4 )
2
psi
The stress at inner radius (critical point) is:
M (rn  ri ) 30000(2.885  2)

 57717
eAri
(3  2.885)( 2)( 2)
i 
psi
There is also an axial stress of 5000 psi acting on the cross-section making
the total stress become
 total  62717 psi
Problem #6: Torsional stresses
The maximum torsional shear stress is:

Tr 16T 16(2000 * 6 *12)
 3 
 11460
3
J d
 (4)
psi
Form Problem # 3, the normal stress on the surface is 15600 psi. THe state
of stress is shown below:
xy
x
The principal stresses are calculated as before using:
x
 
1, 2 
  x    xy2  21662 and
2
 2 
2
The maximum shear stress at point Q is:
 max 
Problem #7
21662  6062
 13860
2
psi
 6062
psi
30”
20”
The forces in the upper portion (Fu) and lower portion (FL) are:
FU 
KU
F
KU  K L
FL 
KL
F
KU  K L
Where
KU 
EA
and
30
KL 
EA
20
Substituting into the force expressions:
FU 
1
30
F
1
1

30 20
3
FL  F  480
5
2
(800)  320 lbs
5
The maximum stress is (480/0.5)=960 psi
Problem #8
4 ft
6 ft
The torque is divided according to torsional stiffnesses. In this case the left
supports picks us (4/10)=0.4 of the torque and the right support takes 0.6 of
the torque.
Problem #9
P
y-bar
yc
Cross-section
The stress is

VQ
IZb
Finding the centroid is as before:
y
2 * 4 * (2  4  1)
 3.5
2(2 * 4)
The area moment of inertia is:
I
Q is
1
1
(2)( 4) 3  (2)( 4)(3.5  2) 2  (4)( 2) 3  (2)( 4)(5  3.5) 2  49.3
12
12
Q  (2)( 4)(5  3.5)  12
and

VQ
V (12)

 11  V  90.4 lbs
I Z b 49.3(2)
Problem #10
Y
Z

VQ 250 * (1 * 6)( 4.5)

 11.8
IZb
285.6(2)
psi
Problem #11
For this thin-walled tube:

T
200 *1000

 34.6 Mpa
2 At 2(38 * 38)2
The angle of rotation is:
TSL
200 *1000 * (4 * 38) * 50 *10

 0.011 rad
2
4 A Gt
4(38 * 38) 2 * 79 *103 * 2
  0.66 deg .

Problem #12
The critical point is the inner radius. The tangential stress is:
P P 
Pi ri 2  Po ro2  ri 2 ro2  o 2 i 
 r

t 
ro2  ri 2
Setting r=ri and Po=0 we get
2ro2
2(0.875) 2
 t   Po 2 2  11200 *
 45733 psi
ro  ri
0.8752  0.6252
The state of stress is simple – just this tangential stress which is also the
principal stress. From theory, we know that there are no shear stresses on
these surfaces when the stress element in oriented with radial edges.
x
The factor of safety is:
SF 
57000
 1.25
45733
Problem #S13
The critical point is the inner radius. Using the formula:
ri 2 ro2 1  3 2
3  2
2
 t   (
)( ri  ro  2 
r )
8
r
3 
3  .24
1  3(.24)
 t  3320(216.7) 2 (
)(12.52  752  752 
(12.5) 2 )
8
3  .24
 t  0.715 Mpa
2
[This number is off by a factor of 4 compared to the posted answer – Error is
in using radii as diameters]
Problem #S14
The interface pressure is:
E
P r
R
 (ro2  R 2 )( R 2  ri 2 ) 


2
2
2
 2 R (ro  ri ) 
The radial interference is 0.013 mm. Substituting all the numbers:
E  207 *10 3
R  20 ri  0 ro  40
The answer is P=50.4 Mpa.
Problem #S15 and #S16
6”
12”
6”
1000
# S15
1000
# S16
Problem #S15: Using the impact formula and simplifying for h>>:
 2hk 
2 * 1 * 2.5 * 10 6


Fe  
W
(1000)  70.7 * 103

1000
 W 
AE 1 * 30 * 10 6
K

 2.5 * 10 6
L
12
lbs
The stress is 70.7 ksi.
Problem #S16: The two bars form a pair of two springs in series. The
equivalent spring is:
Ke 
K1 K 2
AA
E
1(2) 30 *10 6
 1 2 ( )
(
)  33.3 *10 6
K1  K 2 A1  A2 L1
1 2
6
 2hk
Fe  
 W
 max  81.6

W 


ksi
lb / in
2 * 1 * 3.33 * 10 6
(1000)  81.6 * 10 3
1000
Problem #17
The area, moment of inertia, and radius of gyration:
A  r 2  1.767 and
and
k
I
d 4
64

 (1.5) 4
64
 .2485
I
 .375
A
The slenderness ratio is:
l
60

 160
k .375
The limit for the use of Euler versus Johnson formula is:
lbs
l
  
 k 1
2 2 CE

Sy
2 2 (1)(30 *10 6 )
 93.3
68000
Since the slenderness ratio is larger than the limit, the Euler formula applies:
Pcr  C
 2 EI
L2

 2 (30 *106 )(.2485)
60 2
 20438
The factor of Safety is:
SF=20438/5655 = 3.6
b) For this case:
A  0.60 I  0.0288 k  0.219
The slenderness ratio is:
l

k
18
 82
0.0288
.60
We have to use the Johnson Formula:
2

 SyL  1 

1
68000 *18 2 

  0.668000 
Pcr  A S y  
(
)   24975 lbs
6
2

k
CE
2

(.
219
)
30
*
10







24975
SF 
 4.42
5655
Problem #S18
Based on DET:
Tr 16T 16 * 20000


 12732.4
J d 3
 (2) 3
.58S y .58 * 50000
FS 

 2.27

12732.4

psi
Based on MST:
FS 
.5S y

2
Problem #S19
For this cast iron :
S ut  30000   0.211   .26 lbs / in 3
The critical point is the inner radius
3   2 2 ri 2 ro2 1  3 2
 t   (
)( ri  ro  2 
r )
8
r
3 
.26
3  .211
1  3(.211) 2
t 
( ) 2 (
)( 2 * 52  32 
(3) )
386
8
3  .211
 t  .0147 2
2
Since this is a principal stress and the other principal stress is zero (radial
stress is zero on the inner radius), we equate this stress to Sut.
0.0147 2  30000
   1428 rad / sec  13600 rpm
Problem #S20 - Challenge
The main factors influencing the speed are the yield strength and the density
of the material. The higher this ratio, the higher the speed can be. Some
typical materials:
Material
Yield strength
Density
Ratio
Al 7075-T6
Steel 4140
Titanium Ti-3A
Steel 6150
73 ksi
250
175
315
0.098 lbs/in3
0.282
0.163
0.282
745
886
1080
1120
Since 4140 is most available and least expensive, that is chosen. A
reasonable inner radius is 1.25 inches. With these parameters, the speed can
be as high as 120000 rpm leading to a tip velocity of more than 1000 mph.
Using the 6150 steel with an inner radius of 0.3 inches leads to a speed of
145000 rpm or a tip velocity of 1240 mph.