CIV 207 – Winter 2009
Friday, March 27th
Assignment #11
Complete the first three questions. Submit your work to Box #5 on the 4th floor of the MacDonald
building by 12 noon on Tuesday April 7th. No late submissions will be accepted. The second set of
three questions are for practice only. Groups of up to 3 members are permitted. Clearly print your
names and student numbers on your submission. Ensure that the submission is properly stapled.
12.22 The stresses shown in a act at a
point on the free surface of a stressed
body. Determine the normal stresses σn
and σt and the shear stress τnt at this point
if they act on the rotated stress element
shown in Fig. b.
(a)
(b)
12.29-32 Consider a point in a structural
member that is subjected to plane stress.
Normal and shear stresses acting on
horizontal and vertical planes at the point
are shown.
(a) Determine the principal stresses and the 29)
maximum in-plane shear stress acting at
the point.
(b) Show these stresses on an appropriate
sketch (e.g., see Fig. 12-15 or Fig. 12-16)
(c) Compute the absolute shear stress at the
point.
30)
31)
32)
12.42 At a point on the free surface of a stressed body, a normal stress of 64 MPa (C) and an
unknown positive shear stress exist on a horizontal plane. One principal stress at the point is 8
MPa (C). The absolute maximum shear stress at the point has a magnitude of 95 MPa. Determine
the unknown stresses on the horizontal and vertical planes and the unknown principal stress at the
point.
For practice
12.19
12.35
12.43
12.19 The stresses shown in the figure act at a point in a
stressed body. Determine the normal and shear stresses at this
point on the inclined plane shown.
Fig. P12.19
Solution
The given stress values are:
σ x = −3,800 psi, σ y = −2,500 psi, τ xy = 8, 200 psi, θ = −59.0362°
The normal stress transformation equation [Eq. (12-3)] gives σn:
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (−3,800 psi) cos 2 (−59.0362°) + (−2,500 psi)sin 2 (−59.0362°)
+2(8, 200 psi)sin(−59.0362°) cos(−59.0362°)
= −10,079.4185 psi = 10, 080 psi (C)
Ans.
The shear stress transformation equation [Eq. (12-4)] gives τnt:
τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos 2 θ − sin 2 θ )
= −[(−3,800 psi) − (−2,500 psi)]sin(−59.0362°) cos(−59.0362°)
+ (8, 200 psi)[cos 2 (−59.0362°) − sin 2 (−59.0362°)]
= −4, 432.3424 psi = −4, 430 psi
Ans.
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12.22 The stresses shown in
Fig. P12.22a act at a point on the
free surface of a stressed body.
Determine the normal stresses σn
and σt and the shear stress τnt at
this point if they act on the
rotated stress element shown in
Fig. P12.22b.
(a)
(b)
Fig. P12.22
Solution
The given stress values are:
σ x = 1, 200 psi, σ y = 700 psi, τ xy = 400 psi, θ = 20°
The normal stress transformation equation [Eq. (12-3)] gives σn:
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (1, 200 psi) cos 2 (20°) + (700 psi)sin 2 (20°) + 2(400 psi)sin(20°) cos(20°)
= 1,398.6262 psi = 1,399 psi (T)
Ans.
To find σt, add 90° to the value of θ used in Eq. (12-3):
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ
= (1, 200 psi) cos 2 (20° + 90°) + (700 psi)sin 2 (20° + 90°)
+2(400 psi) sin(20° + 90°) cos(20° + 90°)
= (1, 200 psi) cos 2 (110°) + (700 psi) sin 2 (110°) + 2(400 psi) sin(110°) cos(110°)
= 501.3738 psi = 501 psi (T)
Ans.
The shear stress transformation equation [Eq. (12-4)] gives τnt:
τ nt = −(σ x − σ y ) sin θ cos θ + τ xy (cos 2 θ − sin 2 θ )
= −[(1, 200 psi) − (700 psi)]sin(20°) cos(20°) + (400 psi)[cos 2 (20°) − sin 2 (20°)]
= 145.7209 psi = 145.7 psi
Ans.
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
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12.29 Consider a point in a structural member that is subjected to plane
stress. Normal and shear stresses acting on horizontal and vertical
planes at the point are shown.
(a) Determine the principal stresses and the maximum in-plane shear
stress acting at the point.
(b) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15 or
Fig. 12-16)
(c) Compute the absolute shear stress at the point.
Fig. P12.29
Instructors: Problems 12.29-12.32 should be assigned as a set.
Solution
The given stress values are:
σ x = −35 MPa, σ y = −85 MPa, τ xy = −30 MPa
The principal stress magnitudes can be computed from Eq. (12-12):
σ p1, p 2 =
σx +σ y
2
2
⎛ σ x −σ y ⎞
2
± ⎜
⎟ + τ xy
⎝ 2 ⎠
2
(−35 MPa) + (−85 MPa)
⎛ (−35 MPa) − (−85 MPa) ⎞
2
± ⎜
⎟ + (−30 MPa)
2
2
⎝
⎠
= −60.0000 MPa ± 39.0512 MPa
= −20.95 MPa and σ p 2 = −99.05 MPa
=
σ p1
τ max = 39.05 MPa
(maximum in-plane shear stress)
σ avg = 60.00 MPa (C)
tan 2θ p =
(normal stress on planes of maximum in-plane shear stress)
Ans.
Ans.
Ans.
τ xy
−30 MPa
=
= −1.2000
(σ x − σ y ) / 2 [(−35 MPa) − (−85 MPa)] / 2
∴ θ p = −25.10°
(clockwise from the x axis to the direction of σ p1 )
(c) For plane stress, σz = σp3 = 0. Since σp1 and σp2 are both negative,
σ p2
−99.05 MPa
=
= 49.5 MPa
τ abs max =
2
2
Ans.
Ans.
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
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12.30 Consider a point in a structural member that is subjected to
plane stress. Normal and shear stresses acting on horizontal and
vertical planes at the point are shown.
(a) Determine the principal stresses and the maximum in-plane shear
stress acting at the point.
(b) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15
or Fig. 12-16)
(c) Compute the absolute shear stress at the point.
Instructors: Problems 12.29-12.32 should be assigned as a set.
Fig. P12.30
Solution
The given stress values are:
σ x = 16 MPa, σ y = 45 MPa, τ xy = −10 MPa
The principal stress magnitudes can be computed from Eq. (12-12):
σ p1, p 2 =
σx +σ y
2
2
⎛ σ x −σ y ⎞
2
± ⎜
⎟ + τ xy
⎝ 2 ⎠
2
(16 MPa) + (45 MPa)
⎛ (16 MPa) − (45 MPa) ⎞
2
± ⎜
⎟ + (−10 MPa)
2
2
⎝
⎠
= 30.5000 MPa ± 17.6139 MPa
= 48.11 MPa and σ p 2 = 12.89 MPa
=
σ p1
τ max = 17.61 MPa
(maximum in-plane shear stress)
σ avg = 30.50 MPa (T)
tan 2θ p =
(normal stress on planes of maximum in-plane shear stress)
Ans.
Ans.
Ans.
τ xy
−10 MPa
=
= 0.6897
(σ x − σ y ) / 2 [(16 MPa) − (45 MPa)] / 2
∴ θ p = 17.30°
(counterclockwise from the x axis to the direction of σ p 2 )
(c) For plane stress, σz = σp3 = 0. Since σp1 and σp2 are both positive,
σ
48.11 MPa
= 24.06 MPa
τ abs max = p1 =
2
2
Ans.
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
12.31 Consider a point in a structural member that is subjected to
plane stress. Normal and shear stresses acting on horizontal and
vertical planes at the point are shown.
(a) Determine the principal stresses and the maximum in-plane shear
stress acting at the point.
(b) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15
or Fig. 12-16)
(c) Compute the absolute shear stress at the point.
Fig. P12.31
Instructors: Problems 12.29-12.32 should be assigned as a set.
Solution
The given stress values are:
σ x = −66 MPa, σ y = 90 MPa, τ xy = 114 MPa
The principal stress magnitudes can be computed from Eq. (12-12):
σ p1, p 2 =
σx +σ y
2
2
⎛ σ x −σ y ⎞
2
± ⎜
⎟ + τ xy
⎝ 2 ⎠
2
(−66 MPa) + (90 MPa)
⎛ ( −66 MPa) − (90 MPa) ⎞
2
± ⎜
⎟ + (114 MPa)
2
2
⎝
⎠
= 12.0000 MPa ± 138.1304 MPa
= 150.13 MPa and σ p 2 = −126.13 MPa
=
σ p1
τ max = 138.13 MPa
(maximum in-plane shear stress)
σ avg = 12.00 MPa (T)
tan 2θ p =
(normal stress on planes of maximum in-plane shear stress)
Ans.
Ans.
Ans.
τ xy
114 MPa
=
= −1.4615
(σ x − σ y ) / 2 [(−66 MPa) − (90 MPa)] / 2
∴ θ p = −27.81°
(clockwise from the x axis to the direction of σ p 2 )
(c) For plane stress, σz = σp3 = 0. Since σp1 is positive and σp2 is negative,
τ abs max = τ max = 138.13 MPa
Ans.
Ans.
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
12.32 Consider a point in a structural member that is subjected to
plane stress. Normal and shear stresses acting on horizontal and
vertical planes at the point are shown.
(a) Determine the principal stresses and the maximum in-plane shear
stress acting at the point.
(b) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15
or Fig. 12-16)
(c) Compute the absolute shear stress at the point.
Instructors: Problems 12.29-12.32 should be assigned as a set.
Fig. P12.32
Solution
The given stress values are:
σ x = 35 ksi, σ y = 15 ksi, τ xy = 14 ksi
The principal stress magnitudes can be computed from Eq. (12-12):
σ p1, p 2 =
σx +σ y
2
2
⎛ σ x −σ y ⎞
2
± ⎜
⎟ + τ xy
⎝ 2 ⎠
2
(35 ksi) + (15 ksi)
⎛ (35 ksi) − (15 ksi) ⎞
2
± ⎜
⎟ + (14 ksi)
2
2
⎝
⎠
= 25.0000 ksi ± 17.2047 ksi
= 42.20 ksi and σ p 2 = 7.80 ksi
=
σ p1
τ max = 17.20 ksi
(maximum in-plane shear stress)
σ avg = 25.00 ksi (T)
tan 2θ p =
(normal stress on planes of maximum in-plane shear stress)
Ans.
Ans.
Ans.
τ xy
14 ksi
=
= 1.4000
(σ x − σ y ) / 2 [(35 ksi) − (15 ksi)] / 2
∴ θ p = 27.23°
(counterclockwise from the x axis to the direction of σ p1 )
(c) For plane stress, σz = σp3 = 0. Since σp1 and σp2 are both positive,
σ
42.20 ksi
τ abs max = p1 =
= 21.10 ksi
2
2
Ans.
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
12.35 Consider a point in a structural member that is subjected to
plane stress. Normal and shear stresses acting on horizontal and
vertical planes at the point are shown.
(a) Determine the principal stresses and the maximum in-plane shear
stress acting at the point.
(b) Show these stresses on an appropriate sketch (e.g., see Fig. 12-15
or Fig. 12-16)
(c) Compute the absolute maximum shear stress at the point.
Fig. P12.35
Instructors: Problems 12.33-12.36 should be assigned as a set.
Solution
The given stress values are:
σ x = −50 MPa, σ y = −35 MPa, τ xy = 16 MPa
The principal stress magnitudes can be computed from Eq. (12-12):
σ p1, p 2 =
σx +σ y
2
2
⎛ σ x −σ y ⎞
2
± ⎜
⎟ + τ xy
⎝ 2 ⎠
2
(−50 MPa) + (−35 MPa)
⎛ (−50 MPa) − ( −35 MPa) ⎞
2
± ⎜
⎟ + (16 MPa)
2
2
⎝
⎠
= −42.5000 MPa ± 17.6706 MPa
= −24.83 MPa and σ p 2 = −60.17 MPa
=
σ p1
τ max = 17.67 MPa
(maximum in-plane shear stress)
σ avg = 42.50 MPa (C)
tan 2θ p =
(normal stress on planes of maximum in-plane shear stress)
Ans.
Ans.
Ans.
τ xy
16 MPa
=
= −2.1333
(σ x − σ y ) / 2 [(−50 MPa) − (−35 MPa)] / 2
∴ θ p = −32.44°
(clockwise from the x axis to the direction of σ p 2 )
(c) For plane stress, σz = σp3 = 0. Since σp1 and σp2 are both negative,
σ p2
−60.17 MPa
=
= 30.09 MPa
τ abs max =
2
2
Ans.
Ans.
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
12.42 At a point on the free surface of a stressed body, a normal stress of 64 MPa (C) and an unknown
positive shear stress exist on a horizontal plane. One principal stress at the point is 8 MPa (C). The
absolute maximum shear stress at the point has a magnitude of 95 MPa. Determine the unknown stresses
on the horizontal and vertical planes and the unknown principal stress at the point.
Solution
The absolute maximum shear stress can be found from Eq. (12-18)
σ − σ min
τ abs max = max
2
The absolute maximum shear stress at the point has a magnitude of 95
MPa. Suppose we assume that the given principal stress of −8 MPa is
σmin. If this assumption is true, then
σ max = σ min + 2τ abs max = −8 MPa + 2(95 MPa) = 182 MPa
However, this assumption cannot be true because the normal stress on the horizontal plane is σy = −64
MPa, which is more negative than the given principal stress of −8 MPa. Therefore, we now know that
the second principal stress must be negative and its magnitude must be greater than 64 MPa.
The point in question occurs on the free surface of a stressed body. From this information, we can know
that a state of plane stress exists at the point. Therefore,
σ z = σ p3 = 0
(since it is a free surface)
Since both of the in-plane principal stresses must be negative, σmax = σp3 = 0. The minimum principal
stress can now be determined from the absolute maximum shear stress:
σ min = σ max − 2τ abs max = 0 MPa − 2(95 MPa) = −190 MPa
Thus, the two in-plane principal stresses are:
σ p1 = −8 MPa and σ p 2 = −190 MPa
Ans.
Since σy is given, σx can easily be determined from the principal of stress invariance:
σ x + σ y = σ p1 + σ p 2
∴σ x = σ p1 + σ p 2 − σ y = (−8 MPa) + ( −190 MPa) − ( −64 MPa) = −134 MPa
Ans.
The maximum in-plane shear stress can be found from
σ − σ p 2 (−8 MPa) − (−190 MPa)
τ max = p1
=
= 91 MPa
2
2
Since σx, σy, and τmax are known, the magnitude of τxy can be found from the expression
2
τ max
⎛ σ x −σ y ⎞
2
= ⎜
⎟ + τ xy
⎝ 2 ⎠
2
⎛ (−134 MPa) − (−64 MPa) ⎞
2
91 MPa = ⎜
⎟ + τ xy
2
⎝
⎠
∴τ xy = ±84 MPa
The problem states that a positive shear stress exists on a horizontal plane; therefore
τ xy = 84 MPa
Ans.
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
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12.43 At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical
plane and 30 ksi (C) on a horizontal plane. An unknown negative shear stress exists on the vertical
plane. The absolute maximum shear stress at the point has a magnitude of 32 ksi. Determine the
principal stresses and the shear stress on the vertical plane at the point.
Solution
Since σx and σy have opposite signs, the absolute maximum shear stress is
equal to the maximum in-plane shear stress:
τ max = τ abs max = 32 ksi
Since σx, σy, and τmax are known, the magnitude of τxy can be found from
the expression
2
τ max
⎛ σ x −σ y ⎞
2
= ⎜
⎟ + τ xy
⎝ 2 ⎠
2
⎛ (20 ksi) − (−30 ksi) ⎞
2
32 ksi = ⎜
⎟ + τ xy
2
⎝
⎠
∴τ xy = ±19.9750 ksi
The problem states that a negative shear stress exists on the vertical plane; therefore
τ xy = −19.98 ksi
Ans.
From the principal of stress invariance:
σ x + σ y = σ p1 + σ p 2
∴σ p1 + σ p 2 = (20 ksi) + (−30 ksi) = −10 ksi
(a)
The maximum in-plane shear stress is equal to one-half of the difference between the two in-plane
principal stresses
σ − σ p2
τ max = p1
2
∴σ p1 − σ p 2 = 2τ max = 2(32 ksi) = 64 ksi
(b)
Add Eqs. (a) and (b) to find σp1:
2σ p1 = 54 ksi
∴σ p1 = 27 ksi = 27 ksi (T)
Ans.
and subtract Eq. (b) from Eq. (a) to find σp2:
2σ p 2 = −74 ksi
∴σ p2 = −37 ksi = 37 ksi (C)
Ans.
The point in question occurs on the free surface of a stressed body. From this information, we can know
that a state of plane stress exists at the point. Therefore,
σ z = σ p3 = 0
(since the point is on a free surface)
Ans.
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
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