Question 1. The cylindrical portion of a cylindrical air tank is fabricated of steel plate that is welded
along a helix that makes an angle of α=70o with respect to the longitudinal axis of the tank. Before
inflation, the length, the inside diameter, and the wall thickness of the cylinder portion of the vessel are
120 in., 48 in., 0.5 in., respectively. Determine the following quantities for the cylindrical portion of the
tank assuming an internal pressure of 240 psi.:
a. The axial stresses, σa, and hoop stresses, σh.
b. The normal stresses and shear stresses on planes parallel and perpendicular to the weld.
c. The maximum in-plane shear stress and the absolute maximum shear stress.
d. The change in length and in diameter of the cylindrical portion after the working internal pressure
of 240 psi is applied. E=30x106 psi, ν=0.3
e. Assuming that the steel has the yield strength of 30x106 psi., what is the coefficient of safety
using the maximum shear stress theory?
c
b
o
45
a
α
SOLUTION
The axial stresses, σa, and hoop stresses, σh.
pr 240 × 24
=
= 5760 psi
2t
2 × 0.5
pr 240 × 24
σh =
=
= 11520 psi
0.5
t
σa =
The normal stresses and shear stresses on planes parallel and perpendicular to the weld.
σ c = σ avg =
R=
σa +σh
σa −σh
2
2
= 8640 psi
= 2880 psi
σ n = σ avg − R cos(2θ )
= 8640 − 2880 × cos(40) = 6433 psi
σ t = σ avg + R cos(2θ )
= 8640 + 2880 × cos(40) = 10846 psi
τ nt = − R cos(2θ )
= −2880 × cos(40) = −1851 psi
2
The maximum in-plane shear stress and the absolute maximum shear stress.
(τ )
= R = 2880 psi
(τ )
=
max inplane
max absolute
σh
2
= 5760 psi
The change in length and in diameter of the cylindrical portion after the working internal pressure of 240.
psi is applied.
1
(σ a − ν × σ h ) = 1 6 (5760 − 0.3 × 11520) = 0.000077 in.
30 × 10
E
in.
1
1
(11520 − 0.3 × 5760) = 0.00033 in.
ε h = (σ h − ν × σ a ) =
6
E
30 × 10
in.
εa =
ΔL = L × ε a = 120 × 0.000077 = 0.0092in
P = 2πr ⇒ ΔP = 2πΔr
ΔP 2πdr dr
=
=
= εh
P
2πr
r
Δr = r × ε h = 24 × 0.00033 = 0.0079in
Assuming that the steel has the yield strength of 30x106 psi., what is the coefficient of safety using the
maximum shear stress theory?
n=
h
σ yield / 2 60000 / 2
=
= 5.21
τ abs max
5760
τ (ksi)
N(σn,τnt)
t
H(11.52, 0)
A(5.76, 0)
σ (ksi)
2θ=40o
a(axial)
n
τmax
τabsmax
T(σt,-τnt)
S
3
Question 2. A Torsion Load cell (to measure torque T) is constructed by mounting two strain gages
on a tubular shaft, with gages oriented at 45o to the axis of the tube, as indicated if the following
Figure. The gages are wired so that the measurement circuit gives an output εt= εb - εa. Determine the
relationship between the applied torque T and the measured strain difference, εt, if the tube has the
following properties: ro = outer radius; ri = inner radius; E = modulus of Elasticity; and ν = Poisson’s
ratio.
b
T
a
o
45
45o
T
Please note that from the orientation of the gages, gage “a” should read compression due to
a positive torque, while gage “b” should read tension.
SOLUTION:
Ip =
G=
π
2
(r
− ri 4 )
4
o
E
2(1 + ν )
ε b = −ε a =
γ max
=
2
τ
2G
ε t = ε b − ε a = 2ε b =
=
τ
2G
Tro
2GI p
=
Tro
GI p
⎛ GI ⎞
E ⎛ π ⎞ 4
⎜⎜
⎟⎟(ro − ri 4 )ε t
T = ⎜⎜ p ⎟⎟ε t =
2(1 + ν ) ⎝ 2ro ⎠
⎝ ro ⎠
hoop
σ2
τ
⎡πE (r − ri )⎤
T =⎢
⎥ε t
⎣ 4(1 + ν )ro ⎦
4
o
σ1
4
45o
τ
τ
axial
σ1
τ
σ2
4
Question 3. A circular tube of outer diameter do and inner diameter di is made of an elastic
–plastic material with yield point σy and modulus of elasticity E. Determine expressions for
the yield moment My, plastic moment Mp, and the shape factor f.
y
SOLUTION:
I=
c=
π (d o4 − d i4 )
do
2
My =
σ yI
c
=
σ yπ
32d o
x
z
64
(d
4
o
− d i4 )
Because the cross-section is symmetric, the neutral axis does not move. Therfore,
2
2
π
π
2
2
d
d
d
d
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
o
i
i
0
−
⎟−⎜
⎟
⎜
⎟⎜
⎟⎜
y c Ac ⎝ 3π ⎠⎝ 8 ⎠ ⎝ 3π ⎠⎝ 8 ⎠ 2
∑
=
dt = dc =
=
3π
⎛ πd 02 ⎞ ⎛ πd i2 ⎞
⎜
⎟−⎜
⎟
∑ Ac
⎝ 8 ⎠ ⎝ 8 ⎠
σyA
(d c + d t )
2
σ yπ (d o2 − d i2 ) / 4 ⎡ 2
=
⎢
2
⎣ 3π
Mp =
=
σy
6
f =
⎛ d o3 − d i3 ⎞
⎜⎜ 2
⎟
2 ⎟
d
d
−
i ⎠
⎝ o
(d
3
o
y
⎛ d o3 − d i3 ⎞⎤
⎜⎜ 2
⎟ ×2
2 ⎟⎥
⎝ d o − d i ⎠⎦
dt
− d i3 )
z
M p 16d o ⎛ d − d ⎞
⎜
⎟
=
My
3π ⎜⎝ d − d ⎟⎠
3
o
4
o
3
i
4
i
x
dc
5
Question 4. A long, thin, steel plate of thickness to, width 2h, and length 2a is subjected to loads
which produce uniform stresses σo at the ends as shown in Figure 1. The plate is confined along
the edges (y = +h, and y = -h) by two rigid walls. Assuming E and n to be the modulus of
elasticity and Poisson’s ratio, respectively,
a. Calculate the stress field
b. Calculate the strain field
c. Show that the displacement field is expressed by:
(1 − υ )σ
u=−
a
2
E
v = 0;
w=
υ (1 + υ )
E
o
y
a
y
x;
σo
(z)
σoz
σo z
x
to
SOLUTION
a. Stress field
σ
σ
x
= −σ
z
= 0
o
1
[σ y − ν (σ x + σ z )] = 0 ⇒ σ
E
= τ xz = τ yz = 0
εy =
τ xy
y
= ν (σ
x
+ σ z ) = −νσ
o
b. Strain field
σ
1
1
[σ −ν (σ + σ )] = [−σ +ν 2σ 0 ] = − o (1 −ν 2 )
x E x
y
z
o
E
E
1
ε = [σ −ν (σ + σ )] = 0
y E y
x
z
1
1
νσ
ε z = [σ −ν (σ + σ )] = [0 −ν (−σ −νσ )] = o (1 +ν )
x
y
o
o
E z
E
E
γ xy = γ xz = γ yz = 0
ε =
b. Displacements
σ
σ
∂u
= − o (1 − ν 2 ) ⇒ u = − o (1 − ν
∂x
E
E
∂v
ε
=
= 0 ⇒ v = g ( x, z)
y
∂y
ε
x
=
2
)x + f ( y, z)
νσ o
νσ o
∂w
=
=
(1 + ν ) ⇒ w =
(1 + ν ) z + h ( x , y )
z
∂z
E
E
Because of symmetry
u ( 0 ,0 ,0 ) = v ( 0 ,0 ,0 ) = w ( 0 ,0 ,0 ) = 0 ⇒ f ( y , z ) = g ( x , z ) = h ( x , y ) = 0
ε
and
σ
u = − o (1 − ν 2 ) x
E
v=0
w=
νσ o
E
(1 + ν ) z
h
h
6
Question 5. Using the Castigliano’s theorem, find the tension T in the wire in terms of F, axial
stiffness EA of the wire, and bending stiffness EI of the frame shown. Neglect the effects of
axial loading and shear in the members of the frame. T = f(F, EA, EI)
SOLUTION:
s
C
A to B→ M1=(T-F)s
D
EI
B to C→ M2=(T-F)L+Fy
Wire
EA
EI
D to C→ M3=Ts
L
A to D→ N=T
EI
B
L
y
A
F
U=
L
L
⎡L
⎤ L N2
1
dy
= ⎢ ∫ M 12 ds + ∫ M 22 dy + ∫ M 32 ds + ⎥ + ∫
2 EI ⎣ 0
0
0
⎦ 0 2 EA
∂U
= 0 (no gap appears at A)
∂N
L
L
L
L
I
0 = ∫ (T − F ) s ds + ∫ [(T − F ) L + FLy ]dy + ∫ Ts ds + ∫ Tdy
A0
0
0
0
2
2
F T ⎤ IL
⎡T F
0 = L3 ⎢ − + T − F + + ⎥ + T
2 3⎦ A
⎣3 3
IL
5L3
5L3
T+ T−
F
0=
A
3
6
or
5F
F
6 =
T=
I
5
6I
2+
+
2
3 AL
5 AL2
2
7
Question 6. A curved bar with circular cross-section of radius r is fixed at one end as shown.
The bar in the form of a split circular ring of radius R is loaded by a force P at the free end
applied in a diametric plane perpendicular to the plane of the ring. Using Castigliano’s theorem,
determine the deflection at the free end.
R(1-cosθ)
y
R
θ
P
θ
x
Rsinθ
2r
SOLUTION
M b = − PR sin θ
(bending moment)
M t = PR(1 − cos θ ) (torque)
1 L 2
1 L 2
U = Ub + Ut =
∫ M b dx + 2GJ ∫0 M t dx
2 EI 0
πr 4
where : J = 2 I =
2
2π
1
1 2π
∂U
δ=
=
∫ (− PR sin θ )(− R sin θ ) Rdθ + GJ ∫0 PR(1 − cos θ ) R(1 − cos θ ) Rdθ
∂P EI 0
∂M p
∂M t
Note that :
and
= − R sin θ
= R (1 − cos θ )
∂P
∂P
PR 3 2π
PR 3 2π
2
(sin θ ) dθ +
then δ =
(1 − cos θ ) 2 dθ
∫
∫
GJ 0
EI 0
After integration
1
3
)
δ = PR π ( +
EI GJ
3
where
J = 2I =
πr 4
2
8
Question 7. For the wire shown in Figure determine the vertical deflection at point A using the
Castigliano’s theorem and considering bending only. The bending stiffness of the cross-section is
EI=constant.
C
P
A
B
SOLUTION
M = − Px
R
θ
L
⇒
∂M
= −x
∂P
M = P[ L + R (1 − cosθ )]
for 0⟨ x⟨ L
x
∂M
= L + R (1 − cosθ )
∂P
⇒
π
2
∂U
1 L
δ=
= {∫ (− Px)(− x)dx + ∫ P[ L + R(1 − cosθ )]2 Rdθ }
∂P EI 0
0
3
1 L
R2 π
2 π
δ = { + R[( L + R ) − 2 R ( L + R) +
}
2
2 2
EI 3
P
δ=
{L3 + 4.71L2 R + 3.43LR 2 + 1.07 R 3 }
3EI
for 0⟨θ ⟨π
2
9
Question 8. A gear of inner and outer radii 0.1m and 0.15 m, respectively is shrunk onto a hollow shaft
of inner radius 0.05. The length of the gear wheel parallel to the shaft axis is 0.1 m, and the maximum
tangential stress induced in the gear wheel by the shrinking process is 0.21 MPa. Assuming a coefficient
of friction of 0.2 at the common surface determined the following:
a. The internal pressure developed at the contact surface between the gear and the hollow
shaft.
b. The maximum torque that may be transmitted by the gear without slip.
y
Gear wheel
b
a
c
x
Hollow
Shaft
a=0.05 m; b=0.1 m; c=1.5 m
SOLUTION:
Using the theory of thick-walled cylinders, the maximum tangential stress occurs at
r=0.1 m in the gear wheel. Then,
σmax=p(a2+b2)/(c2-b2)
In which p is the internal pressure developed at the contact surfaces. Upon substitution
of given numerical values into the above equation leads to:
0.21=p[(0.15)2+(0.1)2]/[(0.15)2-(0.1)2]
Or,
P=0.081 MPa
This internal pressure at the contact surface controls the maximum torque. The area of
contact is:
2πbl=2π (0.1)(0.1)=0.02π
For a coefficient of friction of 0.2, the torque transmitted is:
Mt=0.2(81000 x 0.02π)(0.1)=1017.36 N.m
10
Question 9. A three-element rectangular rosette (see Figure) is applied at a critical location on a
steel structure. Under a given loading condition the following readings are recorded:
in
in
in
ε A = 0μ ; ε B = 500μ ; ε C = 0μ
in
in
in
Assuming the material properties for steel, i.e., modulus of elasticity and Poisson’s ratio, to be E
= 29 x 106 lb/in2 and ν = 0.29, respectively, determine:
a. The principal strains.
y
b. The principal stresses.
c. The orientation of the principal axis
C
relative to the strain-gage rosette.
θC
Solution
ε A = ε x cox 2θ A + ε y sin 2 θ A + γ xy sin θ A cos θ A
B
θB
ε B = ε x cox θ B + ε y sin θ B + γ xy sin θ B cos θ B
2
2
x
A
ε c = ε x cox θ C + ε y sin θ C + γ xy sin θ B cos θ C
2
θΑ = 0o
θΒ = 45o
θC = 90o
2
and you can solve for ε x , ε y , and γ xy and then for ε 1 and ε 2 ,
or if you recognize that it is a state of pure shear, then
εA −εB
[
1
(ε A − ε C ) + (2ε B − ε A − ε C )2
2
2
1
(2ε B )2 = ε B = 500μ in
ε1 =
in
2
1
(2ε B )2 = −ε B = −500μ in
ε2 = −
in
2
in
Note that γ xy = 2ε B − ε A − ε C = −500 μ
in
⎛ 2ε − ε A − ε C ⎞
⎟⎟ = 90 °
2θ = tan −1 ⎜⎜ B
ε
ε
−
A
C
⎠
⎝
ε 1, 2 =
±
]
1
2
γ/2 C
B
ε2
2θ
CCW
θ = 45 ° CCW from Gage A
1
(σ 1 − υσ 2 ) ⇒ σ 1 = 11,240 psi
E
1
ε 2 = (σ 2 − υσ 1 ) ⇒ σ 2 = −11,240 psi
E
ε1 =
A
ε1
ε
11
Question 10. A thin-walled tube having internal and external diameters of 0.25 m and 0.26 m is subject to an
internal pressure of pi=2.8 MPa, a twisting moment of 31.36 kN.m, and an axial end thrust (tension) P=45
kN. The ultimate strengths in tension and in compression are 210 MPa and 500 MPa, respectively.
Determine if failure has occurred using the following theories of failure.
a) Maximum-shear-stress theory
b) Coulomb-Mohr theory
c) Maximum-principal-stress theory
d) Tresca theory
SOLUTION
Let x be the axial direction,
y
pr
P
σx =
+
=
2
2t π (ro − ri 2 )
x
2800(0.13)
45
+
= 47.635MPa
2(0.005)
π (0.132 + 0.125 2 )
pr 2.8(0.13)
σy =
=
= 72.8MPa
0.005
t
31.36(0.13)2
τ xy =
= 62.615MPa
π (0.134 − 0.125 4 )
Principal Stresses are :
=
σ 1, 2 = 60.218 ± [(158.319) + (3920.638)] 2 ⇒ σ 1 = 124.085MPa
1
a ) Maximun − Shear − Stress
σ 1 − σ 2 = σ ut'
127.734⟨ 210 ⇒ NO FAILURE
b)Coulomb − Mohr Theory
124.085 − 3.649
−
⟨⟩ = ? = 1
210
500
0.591 + 0.007⟨1 ⇒ NO FAILURE
c) Maximun − Pr inciple − Stress − Theory
σ 1 ? = σ ut' ⇒ 124.085⟨ 210
⇒
NO FAILURE
σ 2 ? = σ ut" ⇒ 3.649⟨500
⇒
NO FAILURE
c) Same as in part a. No Failure.
and
σ 2 = −3.649 MPa