PROBLEM 7.66
For the state of plane stress shown, determine the maximum shearing stress
when (a) σ x = 0 and σ y = 60 MPa, (b) σ x = 105 MPa and σ y = 45 MPa.
(Hint: Consider both in-plane and out-of-plane shearing stresses.)
SOLUTION
(a)
σ x = 0, σ y = 60 MPa, τ xy = 40 MPa
σ ave =
1
(σ x + σ y ) = 30 MPa
2
2
R=
σx − σy 
2

 + τ xy
2


= (−30) 2 + 402 = 50 MPa
σ a = σ ave + R = 80 MPa (max)
σ b = σ ave − R = −20 MPa (min)
σc = 0
σ max = 80 MPa
(b)
σ min = −20 MPa
1
τ max = (σ max − σ min ) = 50 MPa
2
σ x = 105 MPa, σ y = 45 MPa τ xy = 40 MPa

σ ave = 75 MPa
2
R=
σx − σy 
2

 + τ xy
2


= (−30) 2 + 402 = 50 MPa
σ a = σ ave + R = 125 MPa (max)
σ b = σ ave − R = 25 MPa
σc = 0
σ max = 125 MPa, σ min = 0
τ max =
(min)
1
(σ max − σ min ) = 62.5 MPa
2

PROBLEM 7.76
For the state of stress shown, determine the value of τ xy for which the
maximum shearing stress is (a) 63 MPa, (b) 84 MPa.
SOLUTION
σ x = 105 MPa σ y = 42 MPa
1
2
σ ave = (σ x + σ y ) = 73.5 MPa
U=
σx −σy
2
= 31.5 MPa
τ (MPa)
(a)
For τ max = 63 MPa
Center of Mohr’s circle lies at point C.
Lines marked (a) show the limits on τ max .
Limit on σ max is σ max = 2τ max = 126 MPa.
For the Mohr’s circle σ a = σ max
corresponds to point Aa.
R = σ a − σ ave
= 126 − 73.5
= 52.5 MPa
2
R = U 2 + τ xy
τ xy = ± R 2 − U 2
= ± 52.52 − 31.52
= ± 42 MPa
(b)

For τ max = 84 MPa
Center of Mohr’s circle lies at point C.
R = 84 MPa
τ xy = ± R 2 − U 2 = ± 78.7 MPa
Checking

σ a = 73.5 + 84 = 157.5 MPa
σ b = 73.5 − 84 = −10.5 MPa
σc = 0
1
2
τ max = (σ max − σ min ) = 84 MPa
O.K.
PROBLEM 7.99
The maximum gage pressure is known to be 8 MPa in a spherical steel pressure vessel having a 250-mm outer
diameter and a 6-mm wall thickness. Knowing that the ultimate stress in the steel used is σ U = 400 MPa,
determine the factor of safety with respect to tensile failure.
SOLUTION
r =
d
250
−t =
− 6 = 119 mm = 119 × 10−3 m, t = 6 × 10−3 m
2
2
σ1 = σ 2 =
F .S . =
pr
(8 × 106 Pa)(119 × 10−3 m)
=
= 79.333 × 106 Pa
2t
2(6 × 10−3 m)
σU
400 × 106
=
σ max 79.333 × 106
F .S . = 5.04 
PROBLEM 7.108
A cylindrical storage tank contains liquefied propane under a pressure of 1.5 MPa at a temperature of 38°C.
Knowing that the tank has an outer diameter of 320 mm and a wall thickness of 3 mm, determine the
maximum normal stress and the maximum shearing stress in the tank.
SOLUTION
r =
d
320
−t =
− 3 = 157 mm = 157 × 10−3 m
2
2
t = 3 × 10−3 m
σ1 =
(1.5 × 106 Pa)(157 × 10−3 m)
pr
=
= 78.5 × 106 Pa
−3
t
3 × 10 m
σ max = σ 1 = 78.5 × 106 Pa
σ max = 78.5 MPa 
σ min = − p ≈ 0
τ max =
1
(σ max − σ min ) = 39.25 × 106 Pa
2
τ max = 39.3 MPa 
PROBLEM 7.115
The steel pressure tank shown has a 750-mm inner diameter and a 9-mm wall
thickness. Knowing that the butt-welded seams form an angle β = 50° with
the longitudinal axis of the tank and that the gage pressure in the tank is
1.5 MPa, determine (a) the normal stress perpendicular to the weld, (b) the
shearing stress parallel to the weld.
SOLUTION
r =
d
= 375 mm = 0.375 m
2
σ1 =
pr
(1.5 × 106 Pa × 0.375 m)
=
= 62.5 × 106 Pa = 62.5 MPa
t
0.009 m
σ2 =
1
σ1 = 31.25 MPa
2
2β = 100°
1
(σ1 + σ 2 ) = 46.875 MPa
2
σ − σ2
R= 1
= 15.625 MPa
2
σ ave =
(a)
σ w = σ ave + R cos100°
σ w = 44.2 MPa 
(b)
τ w = R sin100°
τ w = 15.39 MPa 
PROBLEM 7.120
The compressed-air tank AB has an inner diameter of 450 mm and a
uniform wall thickness of 6 mm. Knowing that the gage pressure
inside the tank is 1.2 MPa, determine the maximum normal stress
and the maximum in-plane shearing stress at point a on the top of the
tank.
SOLUTION
Internal pressure:
1
d = 225mm t = 6 mm
2
pr
(1.2) (225)
σ1 =
=
= 45 MPa
6
t
pr
σ2 =
= 22.5 MPa
2t
r =
Torsion: c1 = 225 mm, c2 = 225 + 6 = 231 mm
J =
π
(c
2
4
2
)
− c14 = 446.9 × 106 mm 4 = 446.9 × 10−6 m 4
T = (5 × 103 )(500 × 10−3 ) = 2500 N ⋅ m
τ =
Tc (2500) (231 × 10−3 )
=
= 1.292 × 106 Pa = 1.292 MPa
J
446.9 × 10−6
Transverse shear:
τ = 0 at point a.
Bending:
I =
At point a,
1
J = 223.45 × 10−6 m 4 , c = 231 × 10−3 m
2
M = (5 × 103 ) (750 × 10−3 ) = 3750 N ⋅ m
σ =
Mc (3750) (231 × 10−3 )
=
= 3.88 MPa
I
223.45 × 10−6
Total stresses (MPa).
Longitudinal:
σ x = 22.5 + 3.88 = 26.38 MPa
Circumferential:
σ y = 45 MPa
Shear:
τ xy = 1.292 MPa
PROBLEM 7.120 (Continued)
σ ave =
1
(σ x + σ y ) = 35.69 MPa
2
2
R=
σx − σy 
2

 + τ xy = 9.40 MPa
2


σ max = σ ave + R = 45.1 MPa
τ max(in-plane) = R = 9.40 MPa
σ max = 45.1 MPa 
τ max (in-plane) = 9.40 MPa 
PROBLEM 7.150
A single gage is cemented to a solid 100-mm-diameter steel shaft at an
angle β = 25° with a line parallel to the axis of the shaft. Knowing that
G = 79 GPa, determine the torque T indicated by a gage reading of
300 × 10−6 mm/mm.
SOLUTION
For torsion,
σ x = σ y = 0, τ = τ 0
1
(σ x − vσ y ) = 0
E
1
ε y = (σ y − vσ x ) = 0
E
τ0 1
τ
γ xy =
γ xy = 0
2G
G 2
εx =
Draw the Mohr’s circle for strain
R=
τ0
2G
ε x′ = R sin 2 β =
But
τ0
2G
sin 2β
τ0 =
2G ε x′
Tc 2T
= 3 =
J πc
sin 2 β
T=
π c3Gε x′
sin 2β
=
π (0.05)3 (79 × 109 )(300 × 10−6 )
= 12.2 kNm
sin 50°

PROBLEM 7.156
A centric axial force P and a horizontal force Q are both applied at point C
of the rectangular bar shown. A 45° strain rosette on the surface of the bar at
point A indicates the following strains:
ε1 = −75 × 10−6 mm/mm ε 2 = +300 × 10−6 mm/mm
ε 3 = +250 × 10−6 mm/mm
Knowing that E = 200 GPa and v = 0.30, determine the magnitudes of P
and Q.
SOLUTION
ε x = ε1 = −75 × 10−6 ε y = ε 3 = 250 × 10−6
γ xy = 2ε 2 − ε1 − ε 3 = 425 × 10−6
200 × 109
E
(ε x + vε y ) =
[−75 + (0.3)(250)](10−6 ) = 0
2
2
1− v
1 − 0.3
200 × 109
E
(ε y + vε x ) =
[250 + (0.3)( −75)](10−6 ) = 50 MPa
σy =
1 − v2
1 − (0.3) 2
P
= σ y P = Aσ y = (50)(150)(503 )
A
= 375 kN
σx =
G=

E
200 × 10
=
= 76.92 GPa
2(1 + v)
(2)(1.3)
9
τ xy = Gγ xy = (76.92 × 109 )(425)(10−6 ) = 32.69 MPa
1 3 1
bh = (50)(150)3 = 14062500 mm 4
12
12
 75 
Q = A y = (75)(50)   = 140625 mm3
t = 50 mm
 2 
VQ
τ xy =
It
Itτ xy (14062500)(50)(32.69)
V=
=
= 163.45 kN
140625
Q
I=
Qx = V = 163.5 kN
