PROBLEM 7.1
For the given state of stress, determine the normal and shearing stresses exerted on the
oblique face of the shaded triangular element shown. Use a method of analysis based on
the equilibrium of that element, as was done in the derivations of Sec. 7.2.
SOLUTION
Stresses
Areas
Forces
F = 0: σ A − 15 A sin 30° cos 30° − 15A cos 30° sin 30° + 10A cos 30 cos 30° = 0
σ = 30 sin 30° cos 30° − 10 cos 2 30°
σ = 5.49 ksi
ΣF = 0: τ A + 15 A sin 30° sin 30° − 15 A cos 30° cos 30° − 10 A cos 30° sin 30° = 0
τ = 15(cos 2 30° − sin 2 30°) + 10 cos 30° sin 30°
τ = 11.83 ksi
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PROBLEM 7.4
For the given state of stress, determine the normal and shearing stresses
exerted on the oblique face of the shaded triangular element shown. Use a
method of analysis based on the equilibrium of that element, as was done in the
derivations of Sec. 7.2.
SOLUTION
Stresses
Areas
Forces
ΣF = 0: σ A + 18 A cos 15° sin 15°
+ 45 A cos 15° cos 15° − 27 A sin 15° sin 15°
+ 18A sin 15° cos 15° = 0
σ = −18 cos 15° sin 15° − 45 cos 2 15°
+ 27sin 2 15° − 18 sin 15° cos 15°
σ = −49.2 MPa
ΣF = 0: τ A + 18 A cos 15° cos 15°
− 45 A cos 15° sin 15°
− 27 A sin 15° cos 15°
− 18 A sin 15° sin 15° = 0
τ = −18(cos 2 15° − sin 2 15°) + (45 + 27) cos 15° sin 15°
τ = 2.41 MPa
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PROBLEM 7.5
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
SOLUTION
σ x = −60 MPa σ y = −40 MPa τ xy = 35 MPa
(a)
tan 2θ p =
2τ xy
σx −σ y
=
(2) (35)
= −3.50
−60 + 40
2θ p = −74.05°
(b)
σ max, min =
=
σx +σy
2
θ p = −37.0°, 53.0°
±
−60 − 40
±
2
σx −σ y
2
2
−60 + 40
2
2
+ τ xy
2
+ (35) 2
= −50 ± 36.4 MPa
σ max = −13.60 MPa
σ min = −86.4 MPa
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 7.8
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
SOLUTION
σ x = −8 ksi σ y = 12 ksi τ xy = 5 ksi
(a)
tan 2θ p =
2τ xy
σx − σy
2(5)
= −0.5
−8 − 12
=
2θ p = −26.5651°
(b)
σ max, min =
=
σx + σy
2
±
−8 + 12
±
2
= 2 ± 11.1803
θ p = −13.3°, 76.7°
2
σx − σy
2
+τ xy
2
−8 − 12
2
2
+ (5)2
σ max = 13.18 ksi
σ min = −9.18 ksi
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PROBLEM 7.19
A steel pipe of 12-in. outer diameter is fabricated from 14 -in. -thick plate by
welding along a helix that forms an angle of 22.5° with a plane perpendicular
to the axis of the pipe. Knowing that a 40-kip axial force P and an 80-kip ⋅ in.
torque T, each directed as shown, are applied to the pipe, determine σ and τ
in directions, respectively, normal and tangential to the weld.
SOLUTION
1
d 2 = 6 in., t = 0.25 in.
2
c1 = c2 − t = 5.75 in.
d 2 = 12 in., c2 =
(
π
J = (c
2
)
π
− c ) = (6
2
A = π c22 − c12 = π (62 − 5.752 ) = 9.2284 in 2
4
2
4
1
4
− 5.754 ) = 318.67 in 4
Stresses:
σ =−
P
A
40
= −4.3344 ksi
9.2284
Tc
τ= 2
J
(80)(6)
=
= 1.5063 ksi
318.67
σ x = 0, σ y = −4.3344 ksi, τ xy = 1.5063 ksi
=−
Choose the x′ and y ′ axes, respectively, tangential and normal to the weld.
Then
σ w = σ y′ and τ w = τ x′y′ θ = 22.5°
σ y′ =
σx +σy
σx −σ y
cos 2θ − τ xy sin 2θ
2
2
(−4.3344) [ −(−4.3344)]
=
−
cos 45° − 1.5063 sin 45°
2
2
= −4.76 ksi
τ x′y′ = −
−
σ w = −4.76 ksi
σx −σy
sin 2θ + τ xy cos 2θ
2
[ −(−4.3344)]
=−
sin 45° + 1.5063 cos 45°
2
= −0.467 ksi
τ w = −0.467 ksi
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PROBLEM 7.22
Two steel plates of uniform cross section 10 × 80 mm are welded
together as shown. Knowing that centric 100-kN forces are applied to
the welded plates and that the in-plane shearing stress parallel to the
weld is 30 MPa, determine (a) the angle β, (b) the corresponding normal
stress perpendicular to the weld.
SOLUTION
Area of weld:
Aw =
=
(a)
Fs = 0: Fs − 100sin β = 0
τw =
Fs
Aw
sin β cos β =
(b)
30 × 106 =
(10 × 10−3 )(80 × 10−3 )
cos β
800 × 10−6 2
m
cos β
Fs = 100sin β kN = 100 × 103 sin β N
100 × 103 sin β
= 125 × 106 sin β cos β
−6
800 × 10 / cos β
1
30 × 106
sin 2β =
= 0.240
2
125 × 106
Fn = 0: Fn − 100cos β = 0
β = 14.34°
Fn = 100cos14.34° = 96.88 kN
Aw =
800 × 10−6
= 825.74 × 10−6 m 2
cos14.34
σ =
Fn
96.88 × 103
=
= 117.3 × 106 Pa
Aw
825.74 × 10−6
σ = 117.3 MPa
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PROBLEM 7.31
Solve Probs. 7.5 and 7.9, using Mohr’s circle.
PROBLEM 7.5 through 7.8 For the given state of stress, determine (a) the
principal planes, (b) the principal stresses.
PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in-plane shearing stress, (b) the maximum
in-plane shearing stress, (c) the corresponding normal stress.
SOLUTION
σ x = −60 MPa,
σ y = −40 MPa,
τ xy = 35 MPa
σ ave =
σx +σy
2
= −50 MPa
Plotted points for Mohr’s circle:
X : (σ x , −τ xy ) = ( −60 MPa, − 35 MPa)
Y : (σ y , τ xy ) = (−40 MPa, 35 MPa)
C : (σ ave , 0) = ( −50 MPa, 0)
(a)
GX 35
=
= 3.500
CG 10
= 74.05°
1
= − β = −37.03°
2
= 180° − β = 105.95°
1
= α = 52.97°
2
tan β =
β
θb
α
θa
2
θb = −37.0°
θ a = 53.0°
2
R = CG + GX = 102 + 352 = 36.4 MPa
(b)
(a′)
σ min = σ ave − R = −50 − 36.4
σ min = −86.4 MPa
σ max = σ ave + R = −50 + 36.4
σ max = −13.6 MPa
θ d = θ B + 45° = 7.97°
θ d = 8.0°
θe = θ A + 45° = 97.97°
θe = 98.0°
τ max = R = 36.4 MPa
(b′)
σ ′ = σ ave = −50 MPa
τ max = 36.4 MPa
σ ′ = −50.0 MPa
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PROBLEM 7.35
Solve Prob. 7.13, using Mohr’s circle.
PROBLEM 7.13 through 7.16 For the given state of stress, determine the normal and
shearing stresses after the element shown has been rotated through (a) 25° clockwise,
(b) 10° counterclockwise.
SOLUTION
σ x = 0,
σ y = 8 ksi,
τ xy = 5 ksi
σ ave =
σx +σy
2
= 4 ksi
Plotted points for Mohr’s circle:
X : (0, − 5 ksi)
Y : (8 ksi, 5 ksi)
C : (4 ksi, 0)
FX 5
= = 1.25
FC 4
2θ p = 51.34°
tan 2θ p =
2
2
R = FC + FX = 42 + 52 = 6.40 ksi
(a)
θ = 25° .
2θ = 50°
ϕ = 51.34° − 50° = 1.34°
σ x′ = σ ave − R cos ϕ
(b)
θ = 10°
.
σ x′ = −2.40 ksi
τ x′y′ = R sin ϕ
τ x′y′ = 0.15 ksi
σ y′ = σ ave + R cos ϕ
σ y′ = 10.40 ksi
2θ = 20°
ϕ = 51.34° + 20° = 71.34°
σ x′ = σ ave − R cos ϕ
σ x′ = 1.95 ksi
τ x′y′ = R sin ϕ
τ x′y′ = 6.07 ksi
σ y′ = σ ave + R cos ϕ
σ y′ = 6.05 ksi
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PROBLEM 7.39
Solve Prob. 7.17, using Mohr’s circle.
PROBLEM 7.17 The grain of a wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in-plane shearing stress
parallel to the grain, (b) the normal stress perpendicular to the grain.
SOLUTION
σ x = −4 MPa σ y = −1.6 MPa τ xy = 0
σ ave =
σx + σy
2
= −2.8 MPa
Plotted points for Mohr’s circle:
X : (σ x , − τ xy ) = (−4 MPa, 0)
Y : (σ y , τ xy ) = (−1.6 MPa, 0)
C : (σ ave , 0) = (−2.8 MPa, 0)
θ = −15°. 2θ = −30°
CX = 1.2 MPa
R = 1.2 MPa
(a)
τ x′y′ = −CX ′ sin 30° = − R sin 30° = −1.2sin 30°
(b)
σ x′ = σ ave − CX ′ cos 30° = −2.8 − 1.2 cos 30°
τ x′y′ = −0.600 MPa
σ x′ = −3.84 MPa
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PROBLEM 7.44
Solve Prob. 7.22, using Mohr’s circle.
PROBLEM 7.22 Two steel plates of uniform cross section 10 × 80 mm
are welded together as shown. Knowing that centric 100-kN forces are
applied to the welded plates and that the in-plane shearing stress parallel
to the weld is 30 MPa, determine (a) the angle β, (b) the corresponding
normal stress perpendicular to the weld.
SOLUTION
σx =
P
100 × 103
=
= 125 × 106 Pa = 125 MPa
A (10 × 10−3 )(80 × 10−3 )
σy = 0
τ xy = 0
From Mohr’s circle:
30
= 0.48
62.5
(a)
sin 2β =
(b)
σ = 62.5 + 62.5cos 2β
β = 14.3°
σ = 117.3 MPa
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PROBLEM 7.98
A spherical gas container made of steel has a 5-m outer diameter and a wall thickness of 6 mm. Knowing that
the internal pressure is 350 kPa, determine the maximum normal stress and the maximum shearing stress in
the container.
SOLUTION
d = 5 m t = 6 mm = 0.006 m, r =
σ =
d
− t = 2.494 m
2
pr
(350 × 103 Pa)(2.494 m)
=
= 72.742 × 106 Pa
2t
2(0.006 m)
σ = 72.7 MPa
σ max = 72.742 MPa
σ min ≈ 0 (Neglecting small radialstress)
τ max =
1
(σ max − σ min )
2
τ max = 36.4 MPa
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PROBLEM 7.109
The unpressurized cylindrical storage tank shown has a 163 -in. wall thickness
and is made of steel having a 60-ksi ultimate strength in tension. Determine
the maximum height h to which it can be filled with water if a factor of safety
of 4.0 is desired. (Specific weight of water = 62.4 lb/ft 3.)
SOLUTION
d 0 = (25)(12) = 300 in.
1
3
d − t = 150 − = 149.81 in.
2
16
σ
60 ksi
σ all = U =
= 15 ksi = 15 × 103 psi
F .S .
4.0
pr
σ all =
t
3
3
tσ all ( 16 ) (15 × 10 )
p=
=
= 18.77 psi = 2703 lb/ft 2
r
149.81
r=
But p = γ h,
h=
p
γ
=
2703 lb/ft 2
62.4 lb/ft 3
h = 43.3 ft
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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