5.44 A 20-mm-diameter by 3.5-m-long steel rod (1) is stress free after being attached to rigid supports, as shown in Fig. P5.44. At A, A, a 25-mm-diameter bolt is used to connect the rod to the support. Determine the normal stress in steel rod (1) and the shear stress in bolt A A after the temperature drops 6 60°C. Use E Use E = = 200 GPa and α = = 11.9 × 10− /°C.
Fig. P5.44
Solution Section properties: For the 20-mm-diameter rod, the cross-sectional area is:
A1 =
π
(20 mm)2 = 314.1593 mm2
4 Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is: F L e1 = 1 1 + α 1∆T L1 A1 E 1 Since the rod is attached to rigid supports, e1 = 0. F1 L1 + α 1∆T L1 = 0 A1 E 1 Thus, the force produced in the rod by the temperature drop is: A E F1 = −α1 ∆T L1 1 1 = −α 1∆T A1 E 1 L1 mm2 )(200, 00 000 N/ N/mm2 ) = −(11.9 ×10−6 / °C)( − 60°C)(314.1593 mm
= 44,861.95 N The normal stress in the steel rod is: 44,861.95 N σ 1 = = 142.8 MPa (T) 314.1593 mm2
Ans.
The 25-mm-diameter bolt has a cross-sectional area of: A bolt =
π
(25 mm mm)2 = 490.8739 mm mm2
4 Since the bolt is loaded in double shear, the shear stress in the bolt is 44,861.95 N τ bolt = = 45.7 MPa 2(49 2(490. 0.84 8439 39 mm mm2 )
Ans.
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5.45 A 1.5-in.-diameter by 30-ft-long steel rod (1) is stress free after being attached to rigid supports. A clevis-and-bolt connection, as shown in Fig. P5.45, connects the rod with the support at A at A.. The normal stress in the steel rod must be limited to 24 ksi, and the shear stress in the bolt must be limited to 42 ksi. 6 Assume E = 29,000 ksi and α = = 6.6 × 10− /°F and determine: (a) the temperature decrease that can be safely accommodated by rod (1) based on the allowable normal stress. (b) the minimum required diameter D diameter D for for the bolt at A for A for the temperature decrease found in part (a).
Fig. P5.45
Solution Section properties: For the 1.5-in.-diameter rod, the cross-sectional area is:
A1 =
π
(1.5 in in.)2 = 1.7671 in.2
4 Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is: F L e1 = 1 1 + α 1∆T L1 A1 E 1 Since the rod is attached to rigid supports, e1 = 0. F1 L1 + α 1∆T L1 = 0 A1 E 1 which can also be expressed in terms of the rod normal stress: L1 σ1 + α 1∆T L1 = 0 E 1 Solve for ∆T corresponding corresponding to a 24 ksi normal stress in the steel rod: L 1 σ ∆T = −σ 1 1 =− 1 α 1 E 1 E1 α1 L1
=−
24 ksi −6
(6.6 × 10 / °F)(29,000 ks ksi)
= −125.4°F
Ans.
The normal force in the steel rod is: F 1 = (24 (24 ksi) ksi)((1.76 1.7671 71 in. in.2 ) = 42.4 42.410 104 4 kips kips If the allowable shear stress in the bolt is 42 ksi, the minimum diameter required for the double shear bolt is ⎡ π 2 ⎤ 42.4104 kips 2 ⎢ D bolt ≥ = 1.009771 in.2 ⎥ 42 ksi ⎣4 ⎦
∴ D bolt ≥ 0.802 in.
Ans.
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-6
5.46 A steel [ E = = 200 GPa and α = = 11.9×10 /°C] rod containing a turnbuckle has its ends attached to rigid walls. During the summer when the temperature is 28°C, the turnbuckle is tightened to produce a stress in the rod of 25 MPa. Determine the stress in the rod in the winter when the temperature is –30°C[tap3].
Solution The normal strain in the rod can be expressed as:
=
σ
+ α ∆T E Since the rod is attached to rigid walls, the rod strain after the temperature change is ε = = 0. ε
=
σ
+ α ∆T = 0 E The change in temperature between the summer and winter is ∆T = Twinter − T summer = −30° C − 28° C = −58° C ε
Solve for the stress increase created by the 58°C drop in temperature. σ = −α ∆T E
= −(11.9 × 10−6 / °C)( − 58°C)(200,000 MPa) = 138.04 MP MPa (T) In the summer, the rod had a tension normal stress of 25 MPa; therefore, the rod stress in the winter is: σ winter =
25 MP MPa + 138.04 MP MPa = 163.0 MP MPa (T)
Ans.
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5.47 A high-density polyethylene [ E E = = 120 ksi and α = 78 × −6 10 /°F] block (1) is positioned in a fixture as shown in Fig. P5.47. The block is 2-in. by 2-in. by 32-in.-long. At room temperature, a gap of 0.10 in. exists between the block and the rigid support at B at B.. Determine: (a) the normal stress in the block caused by a temperature increase of 100°F. (b) the normal strain in block (1) at the increased inc reased temperature.
Fig. P5.47
Solution If the polyethylene block were completely free to elongate, a temperature change of 100°F would cause an elongation of e1 = α 1∆T L1 = (78 ×10−6 / °F)(100° F)(32 in.) = 0.2496 in. Since this elongation is greater than the 0.10-in. gap, the temperature change will cause the polyethylene block to contact the support at B at B,, which will create normal stress in the block. Force-Temperature-Deformation Relationship The relationship between the internal force, temperature change, and the deformation of an axial member is: F L e1 = 1 1 + α 1∆T1 L1 A1 E 1
In this situation, the elongation of member (1) equals the 0.10-in. gap: F1 L1 + α 1∆T1 L1 = 0.10 in. A1 E 1 This relationship can be stated in terms of normal stress as σ 1 L1 + α 1∆T1 L1 = 0.10 in. E 1 (a) Normal stress: The normal stress in the block due to the 100°F temperature increase is: E 1 σ 1 = [ 0.10 in. − α 1∆T1 L1 ] L1
= ⎡⎣ 0.10 in. − (78 × 10−6 / ° F)(100° F)(32 in.)⎤⎦
120 ksi
= − 0.561 ksi = 561 psi (C) 32 in. (b) Normal strain: The normal strain in the polyethylene block is: e1 0.10 in. 0.0031 3125 25 in./ in./in in.. = 3,125 ,125 με ε 1 = = = 0.00 L1 32 in.
Ans.
Ans.
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5.48 The assembly shown in Fig. P5.48 consists of a brass shell (1) fully bonded bond ed to a ceramic ce ramic core (2). The 6 brass shell [ E E = = 15,000 ksi, α = = 9.8 × 10− /°F] has an outside outside diameter diameter of 2.00 in. and an inside inside diameter diameter o 1.25 in. The ceramic core [ E E = = 42,000 ksi, α = 1.7 × −6 10 /°F] has a diameter diameter of 1.25 in. At a temperatur temperaturee o 60°F, the assembly is unstressed. Determine the largest temperature increase that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 21 ksi.
Fig. P5.48
Solution Section properties: The cross-sectional areas of brass shell (1) and ceramic core (2) are:
A1 =
π
⎡⎣ (2.00 in.)2 − (1.25 in.)2 ⎤⎦ = 1.9144 in.2 4
A2 =
π
4
(1.25 in.)2 = 1.2272 in.2
Equilibrium Consider a FBD cut through the assembly. Sum forces in the horizontal direction to obtain: (a) Σ F x = − F1 − F2 = 0 ∴ F2 = − F 1 Force-Temperature-Deformation Relationships The relationship between internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2): F L FL e1 = 1 1 + α1∆T L1 e2 = 2 2 + α 2 ∆T L2 (b) A1 E1 A2 E 2 Geometry of Deformations Relationship For this configuration, the elongations of both members will be equal; therefore, e1 = e2
(c)
Compatibility Equation Substitute the force-deformation relationships (b) into the g eometry of deformation relationship (c) to derive the compatibility equation: F1 L1 FL (d) + α1∆T L1 = 2 2 + α 2 ∆T L2 A1 E1 A2 E 2 Solve the Equations Since a limiting stress is specified for brass shell (1), express Eq. (d) in terms of normal stress: L1 L σ1 + α1∆T L1 = σ 2 2 + α 2 ∆T L2 E1 E 2
(e)
Based on Eq. (a), the normal stress σ 2 can be expressed in terms of σ 1 as:
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σ2
=
F2
=−
A2
F1
=−
A2
F1 A1 A2 A1
= −σ 1
A1 A2
Substitute this expression into Eq. (e) to obtain L1 A L σ1 + α1∆T L1 = −σ 1 1 2 + α 2 ∆T L2 E1 A2 E 2 Rearrange terms
⎡ L1
σ1 ⎢
⎣ E1
+
A1 L2 ⎤
⎥ = α 2 ∆T L2 − α1∆T L1 = (α 2 L2 − α1 L1 ) ∆ T
A2 E 2 ⎦
and solve for ∆T , recognizing that both the shell and the core have the same length:
⎡ L1
σ1 ⎢
∆T =
+
⎣ E1
A1 L2 ⎤
⎥
A2 E2 ⎦ = α 2 L2 − α1 L1
⎡1
σ 1 ⎢
⎣ E1 α2
+
A1 1 ⎤
⎥
A2 E 2 ⎦
− α1
(f)
Substitute the problem data into Eq. (f) to compute ∆T that that will produce a normal stress of 21 ksi in the brass shell:
⎡ ⎤ 1 1.9144 in.2 1 (±21 ksi) ⎢ + 2 15,000 ,000 ks ksi 1.2272 in in. 42,000 ,000 ks ksi ⎥⎦ ⎣ ∆T = 1.7 × 10−6 / °F − 9.8 × 10−6 / ° F 269.13 1319 19° F = ∓269.
Since the problem asks for the largest temperature increase, increase,
∆T max = 269°F
Ans.
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5.49 At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown in Fig. −6 P5.49. Bar (1) is an aluminum alloy [ E E = = 10,000 ksi, ν = = 0.32, α = = 12.5 × 10 /°F] bar with a width of 3 6 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [ E [ E = = 28,000 ksi, ν = = 0.12, α = = 9.6 × 10− /°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A at A and and C are are rigid. Determine: (a) the lowest temperature at which the two bars contact each other. (b) the normal stress in the two bars at a temperature of 250°F. (c) the normal strain in the two bars at 250°F. (d) the change in width of the aluminum bar at a temperature of 250°F.
Fig. P5.49
Solution (a) Lowest Contact Temperature Before the gap is closed, only thermal strains strains and the associated axial elongations exist. Write expressions for the temperature-induced elongations and set this equal to the 0.04-in. gap: α1∆T L1 + α 2 ∆T L2 = 0.04 in.
∆T [α1 L1 + α 2 L2 ] = 0.04 in. ∴ ∆T =
0.04 in.
=
0.04 in.
= 48.6381°F °F)(32 in.) + (9.6 × 10-6 / °F °F)(44 in.) + α 2 L2 (12.5 × 10-6 / °F Since the initial temperature is 60°F, the temperature at which the gap is closed is 108.6°F. α1 L1
Ans.
(b) Equilibrium From the results obtained for part (a), we know that the gap will be closed at 250°F, making this a statically indeterminate axial configuration. Knowing this, consider a FBD at joint B joint B.. Assume that both internal axial forces will be tension, tension, even though we know intuitively that both F both F 1 and F and F 2 will turn out to be compression. (a) Σ F x = − F1 + F 2 = 0 Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and the deformation of an axial member can be stated for members (1) and (2): F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2
(b)
Geometry of Deformations Relationship For this configuration, the sum of the elongations of members (1) and (2) must equal the initial gap: 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
e1 + e2 = 0.04 in.
(c)
Compatibility Equation Substitute the force-temperature-deformation relationships (b) into the geometry of deformation relationship (c) to derive the compatibility equation: F1 L1 FL + α1∆T1 L1 + 2 2 + α 2 ∆T2 L2 = 0.04 in. A1 E1 A2 E 2
(d)
Solve the Equations This part is no fun, but it must be done. From Eq. (a), F (a), F 1 = F = F 2. Substituting this into Eq. (d) (d) gives: F1 L1 FL + α1∆T1 L1 + 1 2 + α 2 ∆T2 L2 = 0.04 in. A1 E1 A2 E 2
F1 L1 A1 E1
+
F1 L2 A2 E 2
= 0.04 in. − α1∆T1 L1 − α 2 ∆T2 L2
⎡ L L ⎤ F1 ⎢ 1 + 2 ⎥ = 0.04 in. − α1∆T1 L1 − α 2 ∆T2 L2 ⎣ A1 E1 A2 E 2 ⎦ F 1 =
0.04 in. − α1∆T1 L1 − α 2 ∆T2 L2
⎡ L1 L2 ⎤ + ⎢ A E A E ⎥ 2 2 ⎦ ⎣ 1 1
(e)
For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the same for both members; therefore, ∆T 1 = ∆T 2 = ∆T = = 190°F: L1 = 32 in. L2 = 44 in. A1 = (3 in.)(0.75 in.) = 2.25 in.2
A2 = (2 in.)(0.75 in.) = 1.50 in.2
E1 = 10, 000 ksi
E 2 = 28, 000 ksi
−6 = 12.5 × 10−6 / °F α 2 = 9.6 × 10 / °F Substitute these values into Eq. (e) and calculate F calculate F 1 = −47.0702 kips. Backsubstitute into Eq. (a) to to calculate F calculate F 2 = −47.0702 kips. Note that the internal forces are compression, as expected.
α1
Normal Stresses The normal stresses in each axial member can now be calculated: F 1 −47.0702 kips 20.9201 ksi ksi = 20. 20.9 ks ksi (C) σ 1 = = = −20. A1 2.25 in.2 σ 2
=
F 2 A2
=
−47.0702 kips 31.3801 ksi = 31. 31.4 ks ksi (C (C) = −31. 1.50 in.2
Ans.
(c) Normal Strains The force-temperature-deformation relationships relationships were expressed in Eq. (b). By definition, ε = = e/ L. L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths: 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
ε1
=
F1 A1 E1
+ α1∆T1
ε2
=
F 2 A2 E 2
+ α 2 ∆T 2
(f)
Substitute the appropriate values to calculate the normal strains in each member: F 1 ε1 = + α 1∆T 1 A1 E 1
=
−47.0702 kips °F)(190°F) + (12.5 × 10−6 / °F 2 (2.25 (2.25 in. in. )(10,000 )(10,000 ksi) ksi)
000283 in./in. = 283 283 με = 0.000
ε2
=
F 2 A2 E 2
=
Ans.
+ α 2 ∆T 2
−47.0702 kips °F)(190°F) + (9.6 × 10−6 / °F 2 (1.50 (1.50 in. )(28, )(28, 000 ksi)
0.000703 703 in. in.//in. = 703 με = 0.0
Ans.
(d) The change in width of the aluminum bar (1) is caused partly by the Poisson effect and partly by the temperature change. The longitudinal strain in in the aluminum bar caused by the internal internal force F 1 is: F 1 −47.0702 kips ε long,σ = ,092.01× 10−6 in./in. = = −2,09 2 A1 E 1 (2.25 (2.25 in. in. )(10,000 )(10,000 ksi) ksi) The accompanying lateral strain due to the Poisson effect is thus ε lat,σ = −νε long,σ = −(0.32)( −2, 09 092.01× 10−6 in./in.) = 669.44 × 10−6 in./in. The lateral strain caused by the temperature change is −6 °F)(190° F) = 2, 37 375× 10−6 in./in. ε lat,T = α 1∆T 1 = (12.5 × 10 / °F Therefore, the total lateral strain in aluminum bar (1) is ε lat = ε lat,σ + ε lat,T 375× 10−6 in./in. = 3, 04 044.44× 10−6 in./in. = 669.44 × 10−6 in./in. + 2, 37 The change is width of the aluminum bar is thus 044.44 × 10−6 in./in.)(3 in in.) = 0.00913 in in. ∆width = ε lat (width) = (3, 04
Ans.
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5.50 At a temperature of 5°C, a 3-mm gap exists between two polymer bars and a rigid support, as shown in Fig. P5.50. Bar (1) is 50mm wide and 20-mm thick [ E E = = 800 MPa, α = −6 140 × 10 /°C]. Bar (2) is 75mm-wide and 256 mm thick [ E E = = 2.7 GPa, α = = 67 × 10− /°C] bar. The supports at A at A and and C are are rigid. Determine: (a) the lowest temperature at which the 3-mm gap is closed. (b) the normal stress in the two bars at a temperature of 60°C. (c) the normal strain in the two bars at 60°C.
Fig. P5.50
Solution (a) Before the gap is closed, only thermal thermal strains and the associated axial elongations exist. exist. Write expressions for the temperature-induced elongations and set this equal to the 3-mm gap: α1∆T L1 + α 2 ∆T L2 = 3 mm
∆T [α1 L1 + α 2 L2 ] = 3 mm ∴ ∆T =
3 mm
=
3 mm
= 30.99°C °C)(500 mm) + (67 × 10-6 / °C °C)(400 mm mm) + α 2 L2 (140 × 10-6 / °C Since the initial temperature is 5°C, the temperature at which the gap is closed is 35.99°C 35 .99°C = 36.0°C. Ans. α1 L1
Equilibrium (b) From the results obtained for part (a), we know that the gap will be closed at 60°C, making this a statically indeterminate axial configuration. Knowing this, consider a FBD at joint B joint B.. Assume that both internal axial forces will be tension, tension, even though we know intuitively that both F both F 1 and F and F 2 will turn out to be compression. (a) Σ F x = − F1 + F 2 = 0 Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and the deformation of an axial member can be stated for members (1) and (2): F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2
(b)
Geometry of Deformations Relationship For this configuration, the sum of the elongations of members (1) and (2) must equal the initial gap: e1 + e2 = 3 mm (c) Compatibility Equation Substitute the force-temperature-deformation relationships (b) into the geometry of deformation relationship (c) to derive the compatibility equation: 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
F1 L1 A1 E1
+ α1∆T1 L1 +
F2 L2 A2 E 2
+ α 2 ∆T2 L2 = 3 mm
(d)
Solve the Equations This part is no fun, but it must be done. From Eq. (a), F (a), F 1 = F = F 2. Substituting this into Eq. (d) (d) gives: F1 L1 FL + α1∆T1 L1 + 1 2 + α 2 ∆T2 L2 = 3 mm A1 E1 A2 E 2
F1 L1 A1 E1
+
F1 L2 A2 E 2
= 3 mm − α1 ∆T1 L1 − α 2 ∆T2 L2
⎡ L L ⎤ F1 ⎢ 1 + 2 ⎥ = 3 mm − α1 ∆T1 L1 − α 2 ∆T2 L2 ⎣ A1 E1 A2 E 2 ⎦ F 1 =
3 mm − α1∆T1 L1 − α 2 ∆T2 L2
⎡ L1 L2 ⎤ ⎢ A E + A E ⎥ 2 2 ⎦ ⎣ 1 1
(e)
For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the same for both members; therefore, ∆T 1 = ∆T 2 = ∆T = = 55°C: L1 = 500 mm L2 = 400 mm A1 = (50 mm)(20 mm) = 1, 000 mm2
A2 = (75 mm)(25 mm) = 1, 875 mm2
E1 = 800 MPa
E 2 = 2, 700 MPa
−6 = 140 × 10−6 / °C α 2 = 67 × 10 / °C Substitute these values into Eq. (e) and calculate F calculate F 1 = −3.30108 kN. Backsubstitute into Eq. (a) to calculate F calculate F 2 = −3.30108 kN. Note that the internal forces are compression, as expected.
α1
Normal Stresses The normal stresses in each axial member can now be calculated: F 1 ( −3.30108 3.30108 kN)(1, kN)(1,000 000 N/kN) N/kN) σ 1 = MPa (C) = = −3.30 MPa = 3.30 MP A1 1,000 mm2 σ 2
=
F 2 A2
=
( −3.30108 3.30108 kN)(1, kN)(1,000 000 N/kN) N/kN) 1,875 mm2
1.761 MPa = 1.761 761 MPa (C) (C) = −1.7
Ans.
Normal Strains The force-temperature-deformation relationships relationships were expressed in Eq. (b). By definition, ε = = e/ L. L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths: F1 F 2 (f) ε1 = + α1∆T1 ε2 = + α 2 ∆T 2 A1 E1 A2 E 2
Substitute the appropriate values to calculate the normal strains in each member:
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ε1
=
F 1 A1 E 1
=
+ α 1∆T 1
( −3.30108 3.30108 kN)(1, kN)(1,000 000 N/kN) N/kN) 2
2
(1,00 1,000 0 mm mm )(800 800 N/m N/mm )
°C)(55°C) + (140 × 10−6 / °C
0035736 7365 mm mm/mm = 3,570 ,570 με = 0.003
ε2
=
F 2 A2 E 2
=
Ans.
+ α 2 ∆T 2
( −3.30108 3.30108 kN)(1, kN)(1,000 000 N/kN) N/kN) 2
2
(1,87 1,875 5 mm mm )(2,70 (2,700 0 N/m N/mm )
°C)(55°C) + (67 × 10−6 / °C
0.003032 03294 mm mm/mm = 3,030 ,030 με = 0.0
Ans.
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5.51 At a temperature of 5°C, a 3-mm gap exists between the ends of the two polymer bars shown in Fig. P5.51. Bar (1) is 50-mm wide and 6 20-mm thick [ E E = = 800 MPa, α = = 140 × 10− /°C]. Bar (2) is 75-mm wide and 25-mm thick [ E E = −6 2.7 GPa, α = = 67 × 10 /°C] bar. The supports at A and A and C are are rigid. Determine: (a) the temperature at which the normal stress in bar (2) will be equal to −3.0 MPa. (b) the lengths of the two polymer bars at this temperature.
Fig. P5.51
Solution Equilibrium Consider a FBD at joint B joint B.. Assume that both internal axial forces will be tension, tension, even though we know intuitively that both F both F 1 and F and F 2 will turn out to be compression. (a) Σ F x = − F1 + F 2 = 0 Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and the deformation of an axial member can be stated for members (1) and (2): F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2
(b)
Geometry of Deformations Relationship For this configuration, the sum of the elongations of members (1) and (2) must equal the initial gap: e1 + e2 = 3 mm (c) Compatibility Equation Substitute the force-temperature-deformation relationships (b) into the geometry of deformation relationship (c) to derive the compatibility equation: F1 L1 FL + α1∆T1 L1 + 2 2 + α 2 ∆T2 L2 = 3 mm A1 E1 A2 E 2 (a) Temperature that produces −3 MPa in bar (2) Since a limiting stress is specified for bar (2), express Eq. (d) in terms of normal stress: L1 L σ1 + α1∆T L1 + σ 2 2 + α 2 ∆T L2 = 3 mm E1 E 2
(d)
(e)
Based on Eq. (a), the normal stress σ 1 can be expressed in terms of σ 2 as: F1 F2 F2 A2 A σ1 = = = = σ 2 2 A1 A1 A1 A2 A1 Substitute this expression into Eq. (e) to obtain 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
σ2
A2 L1 A1 E1
+ α1∆T L1 + σ 2
L2 E 2
+ α 2 ∆T L2 = 3 mm
Rearrange terms α1∆T L1 + α 2 ∆T L2
= 3 mm − σ 2
A2 L1 A1 E1
−σ2
L2
E 2
and solve for ∆T :
∆T =
⎡ A L L ⎤ 3 mm − σ 2 ⎢ 2 1 + 2 ⎥ ⎣ A1 E1 E 2 ⎦ α1 L1
+ α 2 L2
(f)
For this structure, the lengths, areas, and elastic moduli are given below: L1 = 500 mm L2 = 400 mm A1 = (50 mm)(20 mm) = 1, 000 mm2
A2 = (75 mm)(25 mm) = 1, 875 mm2
E1 = 800 MPa
E 2 = 2, 700 MPa
α1
= 140 × 10−6 / °C
α 2
= 67 × 10−6 / °C
Substitute these values into Eq. (f) and calculate ∆T :
⎡ 1, 875 mm2 500 mm 400 mm ⎤ 3 mm − (−3 MPa) ⎢ + ⎥ 1, 00 000 mm mm2 800 MP MPa 2, 70 700 MP MPa ⎦ ⎣ 71.901 015 5°C ∆T = = 71.9 −6 −6 (140 × 10 / °C)(500 mm) + (67 × 10 / °C)(400 mm)
Since the bars were initially at a temperature of 5°C, the temperature at which the normal stress in bar (2) reaches −3 MPa is T = 5°C + 71.9015°C = 76.9° C
Ans.
(b) Lengths of bars (1) and (2) The corresponding normal stress in bar ba r (1) is: A2 1,875 mm2 σ 1 = σ 2 = ( −3 MPa) = −5.625 MPa A1 1, 000 mm2
Thus, the elongation of bar (1) is σ L ( −5.625 5.625 MPa) MPa)(5 (500 00 mm) mm) + (140 × 10−6 / °C)(71. e1 = 1 1 + α 1∆T1 L1 = 71.9015° 15°C)( C)(500 mm) = 1.5 1.5175 mm 800 MPa E 1 The elongation of bar (2) is σ L ( −3 MPa MPa)( )(40 400 0 mm) mm) e2 = 2 2 + α 2 ∆T2 L2 = /°C)(71.9015°C)(400 mm mm) = 1.4825 mm mm + (67 ×10−6 /°C) E 2 2,700 MPa The lengths of bars (1) and (2) are thus: L1 = 500 mm + 1.5175 mm = 501.52 mm
Ans.
L2 = 400 mm + 1.4825 mm = 401.48 mm
Ans.
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5.52 The axial assembly shown in Fig. P5.52 consists of a solid 1in.diameter aluminum alloy rod (1) [ E E = 10,000 ksi, ν = 0.32, α = 12.5 × −6 10 /°F] and a solid 1.5in.diameter bronze rod (2) -6 [ E E = = 15,000 ksi, ν = = 0.15, α = = 9.4 × 10− /°F] /°F].. I the supports at A at A and and C are are rigid and the assembly is stress free at 0°F, determine: (a) the normal stress in both rods at 160°F. (b) the displacement of flange B flange B.. (c) the change in diameter of the aluminum rod.
Fig. P5.52
Solution Equilibrium Consider a FBD at joint B joint B.. Assume that both internal axial forces will be tension. tension. (a) Σ F x = − F1 + F2 = 0 ∴ F1 = F 2
Force-Temperature-Deformation Relationships F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2 Geometry of Deformations Relationship e1 + e2 = 0
(b)
(c)
Compatibility Equation F1 L1 FL + α1∆T1 L1 + 2 2 + α 2 ∆T2 L2 = 0 A1 E1 A2 E 2
(d)
Solve the Equations From Eq. (a), F (a), F 1 = F = F 2. The temperature change is the same for both members; therefore, therefore, ∆T 1 = ∆T 2 = ∆T . Eq. (d) then can be written as: F1 L1 FL + α1∆T L1 + 1 2 + α 2 ∆T L2 = 0 A1 E1 A2 E 2
Solving for F for F 1: F1 L1 F1 L2 + = −α1∆T L1 − α 2 ∆T L2 A1 E1 A2 E 2
⎡ L L ⎤ F1 ⎢ 1 + 2 ⎥ = −∆T [α1 L1 + α 2 L2 ] ⎣ A1 E1 A2 E 2 ⎦ F 1 = −
∆T [α1 L1 + α 2 L2 ]
⎡ L1 L2 ⎤ ⎢ A E + A E ⎥ 2 2 ⎦ ⎣ 1 1
(e)
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For this structure, the lengths, areas, and elastic moduli are given below. L1 = 15 in. L2 = 22 in. A1 =
π
(1 in.) 2 = 0.7854 in.2
A2 =
4 E1 = 10, 000 ksi
(1.5 in.)2 = 1.7671 in.2
4 E 2 = 15, 000 ksi
= 12.5 × 10−6 / °F
α1
π
α 2
= 9.4 × 10−6 / °F
Substitute these values along with ∆T = = +160°F into Eq. (e) and calculate F calculate F 1 = −23.0263 kips. From Eq. (a), F (a), F 2 = F = F 1 = −23.0263 kips. (a) Normal Stresses The normal stresses in each rod can now be calculated: F 1 −23.0263 kips 318 ks ksi = 29.3 ksi (C) σ 1 = = = −29.318 A1 0.7854 in.2 σ 2
=
F 2 A2
=
−23.0263 kips 13.030 30 ksi ksi = 13.0 13.03 3 ksi ksi (C) = −13.0 1.7671 in.2
Ans. Ans.
(b) Displacement of Flange B The displacement of flange B flange B is is equal to the elongation (i.e., contraction in in this instance) of rod (1). (1). The elongation of rod (1) is given by: F L e1 = 1 1 + α 1∆T1 L1 A1 E 1
=
( −23.0 23.0263 263 kip kips) s)(1 (15 5 in.) in.) 2
/°F)(160°F)(15 in in.) = −0.013977 in in. + (12.5 × 10−6 /°F)
(0.7854 (0.7854 in. )(10,000 )(10,000 ksi) ksi) The displacement of flange B flange B is is thus: u B = e1 = 0.01398 in. ←
(c) Change in diameter of the aluminum rod The change in diameter of aluminum rod (1) is caused partly by the Poisson effect and partly by the temperature change. The longitudinal strain in the aluminum rod caused by the internal force F 1 is: F 1 −23.0263 kips 931.78 × 10−6 in./in. ε long,σ = = = −2, 93 2 A1 E 1 (0.785 (0.7854 4 in. )(10,000 )(10,000 ksi)
The accompanying lateral strain due to the Poisson effect is thus 931.78 × 10−6 in./in.) = 938.17 × 10−6 in./in. ε lat,σ = −νε long,σ = −(0.32)( −2, 93 The lateral strain caused by the temperature change is −6 °F)(160° F) = 2, 00 000 × 10−6 in./in. ε lat,T = α 1∆T 1 = (12.5 × 10 / °F Therefore, the total lateral strain in aluminum rod (1) is ε lat = ε lat,σ + ε lat,T 000× 10−6 in./in. = 2, 93 938.17× 10−6 in./in. = 938.17 × 10−6 in./in. + 2, 00 The change in diameter of the aluminum rod is thus (2,938 8.17 × 10−6 in./in.)(1 in.) = 0.00294 in. ∆ D1 = ε lat D1 = (2,93
Ans.
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5.53 A load of P = 170 kN is supported by a structure consisting of rigid bar ABC , two identical solid bronze [ E = E = −6 100 GPa, α = = 16.9 × 10 /°C] rods, and a solid steel [ E = E = 200 −6 GPa, α = 11.9×10 /°C] rod, as shown in Fig. P5.53. The bronze rods (1) each have a diameter of 20 mm and they are symmetrica symmetrically lly positioned positioned relative relative to the center rod (2) and the applied load P . Steel rod (2) has a diameter of 24 mm. The bars are unstressed when the structure is assembled at 40°C. When the temperature decreases to –10°C, determine: (a) the normal stresses in the bronze and steel rods. (b) the normal strains in the bronze and steel rods. (c) the downward deflection of rigid bar ABC bar ABC .
Fig. P5.53
Solution Equilibrium: By virtue of symmetry, the forces in the two bronze rods (1) are identical. Consider a FBD of the rigid bar. Sum forces in the vertical direction to obtain: (a) Σ F y = 2 F1 + F2 − P = 0 Force-Temperature-Deformation Relationships: The relationship between internal force, temperature change, and deformation can be stated for members (1) and (2): F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 (b) A1 E1 A2 E 2 Geometry of Deformations: For this configuration, the deflections deflections of joints A joints A,, B, B, and C are are equal: v A = vB = vC (c)
All pin connections are ideal; therefore, the deflection of joints A joints A and and C will will cause an identical elongation of rods (1): v A = e1
(d)
and the rigid bar deflection v B will cause an identical elongation of rod (2): v B = e2
(e)
Substitute Eqs. (d) and (e) into Eq. (c) to obtain the geometry of deformation equation: e1 = e2
(f)
Compatibility Equation: Substitute the force-deformation relationships (b) into the geometry of deformation relationship (f) to derive the compatibility c ompatibility equation: F1 L1 FL + α1∆T1 L1 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2
(g)
Solve the Equations: For this situation, ∆T 1 = ∆T 2 = ∆T . Solve Eq. (g) for F for F 2: 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
⎡ F L ⎤ A E F2 = ⎢ 1 1 + α1∆T1 L1 − α 2 ∆T2 L2 ⎥ 2 2 ⎣ A1 E1 ⎦ L2 = F1
L1 A2 E2 L2 A1 E1
+ ∆T [α1 L1 − α 2 L2 ]
A2 E 2 L2
Substitute this expression into Eq. (a) L A E A E 2 F1 + F1 1 2 2 + ∆T [α1 L1 − α 2 L2 ] 2 2 = P L2 A1 E1 L2 and derive an expression for F for F 1:
⎡
L1 A2 E2 ⎤
A2 E 2
⎣
L2 A1 E1 ⎦
L2
F1 ⎢ 2 +
⎥ = P − ∆T [α1 L1 − α 2 L2 ]
A E P − ∆T [α1 L1 − α 2 L2 ] 2 2 L2 F 1 = L1 A2 E 2 2+ L2 A1 E 1
(h)
For this structure, P structure, P = = 170 kN = 170,000 N, and the lengths, areas, and elastic moduli are given below: L1 = 2, 400 mm L2 = 1, 800 mm A1 =
π
(20 mm)2 = 314.1593 mm2
4 E1 = 100, 000 MPa α1
= 16.9 × 10−6 / °C
A2 =
π
(24 mm)2 = 452.3893 mm2
4 E 2 = 200, 000 MPa α 2
= 11.9 × 10−6 / °C
Substitute these values along with ∆T = = −50°C into Eq. (h) and calculate F calculate F 1: A E P − ∆T [α1 L1 − α 2 L2 ] 2 2 L2 F 1 = L A E 2+ 1 2 2 L2 A1 E 1
=
(170, 00 000 N) − (−50°C) ⎡⎣ (16.9 × 10−6 )(2, 40 400) − (11.9 × 10−6 )(1, 800)⎤⎦
(452 (452.3 .389 893 3 mm mm2 )(20 )(200 0, 000 000 N/mm N/mm2 ) 1,800 mm
400 mm ⎞ ⎛ 452.3893 mm2 ⎞ ⎛ 200, 00 000 MPa ⎞ ⎛ 2, 40 2+⎜ ⎜ 2 ⎟⎜ ⎟ ⎟ ⎝ 1, 800 mm ⎠ ⎝ 314.1593 mm ⎠ ⎝ 100,000 MPa ⎠
= 37,346.59 N Backsubstitute this result into Eq. (a) to calculate F calculate F 2 = 95,306.83 N. (a) Normal Stresses: The normal stresses in each rod can now be calculated: F 1 37,346.59 N σ 1 = = = 118.9 MPa A1 314.1593 mm2 σ 2
=
F 2 A2
=
95,306.83 N 452.3893 mm2
= 211 MPa
Ans.
Ans.
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(b) Normal Strains: The force-temperature-deformation force-temperature-deformation relationships were expressed in Eq. (b). By definition, ε = = e/ L. L. Therefore, the normal strain for for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths: F1 F 2 (f) ε1 = + α1∆T1 ε2 = + α 2 ∆T 2 A1 E1 A2 E 2
Substitute the appropriate values to calculate the normal strains in each member: F 1 ε1 = + α 1∆T 1 A1 E 1 37,346.59 N
=
2
2
(314. (314.15 1593 93 mm )(100 )(100,, 000 N/mm N/mm )
°C)( − 50°C) + (16.9 × 10−6 / °C
03438 mm/mm = 344 με = 0.00034
ε2
=
F 2 A2 E 2
=
Ans.
+ α 2 ∆T 2 95,306.83 N 2
2
(452 (452.3 .389 893 3 mm )(20 )(200,00 0,000 0 N/mm N/mm )
°C)( − 50°C) + (11.9 × 10−6 / °C
mm/mm = 458 με = 0.0004584 mm/
Ans.
(c) Rigid bar deflection: The downward deflection of the rigid bar can be determined from the elongation of rods (1): F L v B = e1 = 1 1 + α 1∆T1 L1 A1 E 1
=
(37,346.59 N)(2,400 mm) 2
(314.15 (314.1593 93 mm mm )(100 )(100,, 000 MPa) MPa)
= 0.825 mm ↓
°C)( − 50°C)(2, 40 400 mm) + (16.9 × 10−6 / °C Ans.
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5.54 A solid aluminum [E = 70 GPa, α = = 22.5 × −6 10 /°C] rod (1) is connected to a solid bronze 6 [ E E = = 100 GPa, α = = 16.9 × 10− /°C] rod at flange B, B, as shown in Fig. P5.54. Aluminum rod (1) has an outside diameter of 40 mm and bronze rod (2) has an outside diameter of 120 mm. The bars are unstressed when the structure is assembled at 30°C. After the 300-kN load is applied to flange B flange B,, the temperature increases to 45°C. Determine: (a) the normal stresses in rods (1) and (2). (b) the deflection of flange B flange B..
Fig. P5.54
Solution flange B.. Sum forces in Equilibrium: Consider a FBD of flange B the horizontal direction to obtain: (a) Σ F x = − F1 + F 2 + 300 kN = 0 Force-Deformation Relationships: F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2
(b)
Geometry of Deformations: e1 + e2 = 0 Compatibility Equation: Substitute Eqs. (b) into Eq. (c) to to derive the compatibility compatibility equation: F1 L1 FL + α1∆T1 L1 + 2 2 + α 2 ∆T2 L2 = 0 A1 E1 A2 E 2
(c)
(d)
Solve the Equations: For this situation, ∆T 1 = ∆T 2 = ∆T . Solve Eq. (d) for F for F 1: L A E A E F1 = − F2 2 1 1 − ∆T [α1 L1 + α 2 L2 ] 1 1 L1 A2 E2 L1
Substitute this expression into Eq. (a): L A E A E F2 2 1 1 + ∆T [α1 L1 + α 2 L2 ] 1 1 + F 2 = −300 kN L1 A2 E2 L1 and solve for F for F 2:
⎡ L A E ⎤ A E F2 ⎢ 2 1 1 + 1⎥ = −300 kN − ∆T [α1 L1 + α 2 L2 ] 1 1 L1 ⎣ L1 A2 E2 ⎦ A E 300 kN + ∆T [α1 L1 + α 2 L2 ] 1 1 L1 F 2 = − L2 A1 E 1 +1 L1 A2 E 2
(f)
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Member lengths, areas, elastic moduli, and coefficients of thermal expansion are given below: L1 = 2, 600 mm L2 = 1, 000 mm A1 =
π
(40 mm)2 = 1, 256.637 mm2
A2 =
4 E1 = 70, 000 MPa α1
π
(120 mm)2 = 11, 309.734 mm2
4 E 2 = 100, 000 MPa
= 22.5 × 10−6 / °C
α 2
= 16.9 × 10−6 / °C
Substitute these values along with ∆T = = +15°C into Eq. (f) and compute F compute F 2 = −328.4395 kN. Backsubstitute this result into Eq. (a) to find F find F 1 = −28.4395 kN. (a) Normal Stresses: The normal stresses in each rod can now be calculated: F 1 28, 439.5 439.5 N −28, σ 1 = = = 22.6 MPa (C) A1 1,256.637 mm2 σ 2
=
F 2 A2
=
−328,439.5 N = 29.0 MPa (C) 11,309.734 mm2
Ans.
Ans.
flange B can can be determined from the elongation of rod (b) Deflection of Flange B: The deflection of flange B (1): F L u B = e1 = 1 1 + α 1∆T1 L1 A1 E 1
=
(−28, 28, 439. 439.5 5 N)(2 N)(2,60 ,600 0 mm) mm) 2
(1,256. (1,256.637 637 mm )(70,000 )(70,000 MPa) MPa)
= 0.0369 mm →
+ (22 (22.5 × 10−6 / °C)( C)(15°C)(2,60 (2,600 0 mm mm) Ans.
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5.55 A steel [ E = 30,000 ksi, α = 6.6 × 6 10− /°F] pipe column (1) with a cross2 sectional area of A of A1 = 5.60 in. is connected at flange B B to an aluminum alloy [ E E = 10,000 −6 ksi, α = = 12.5 × 10 /°F] pipe (2) with a cross2 sectional area of A of A2 = 4.40 in. . The assembly (shown in Fig. P5.55) is connected to rigid supports at A at A and and C . It is initially unstressed at a temperature of 90°F. (a) At what temperature will the normal stress in steel pipe (1) be reduced to zero? (b) Determine the normal stresses in steel pipe (1) and aluminum pipe (2) when the temperature reaches –10°F.
Fig. P5.55
Solution Equilibrium: Consider a FBD of flange B flange B.. Sum forces in the horizontal direction to obtain: (a) Σ F x = − F1 + F 2 − 60 kips = 0 Force-Temperature-Deformation Relationships: F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2
(b)
Geometry of Deformations: e1 + e2 = 0
(c)
Compatibility Equation: Substitute Eqs. (b) into Eq. (c) to derive the compatibility equation: F1 L1 FL + α1∆T1 L1 + 2 2 + α 2 ∆T2 L2 = 0 A1 E1 A2 E 2
(d)
Solve the Equations: Set F Set F 1 = 0 and solve Eq (a) to find F find F 2 = 60 kips. Substitute these values for F for F 1 and F and F 2 into Eq. (d) along with the observation that the temperature change for both axial members is the same (i.e., ∆T 1 = ∆T 2 = ∆T ) and solve for ∆T : F L FL (60 kips)(144 in.) − 1 1 − 2 2 0− A1 E1 A2 E 2 (4.40 (4.40 in. in.2 )(10,000 )(10,000 ksi) ksi) 75.758°F ∆T = = = −75. −6 α1 L1 + α 2 L2 (6.6 × 10 / °F)(120 in.) + (12.5 × 10−6 / ° F)(144 in.)
Since the pipes are initially at a temperature 90°F, the temperature at which the normal stress in steel pipe (1) is reduced to zero is T = 90°F − 75.748° F = 14.24° F (b) Solve Eq. (a) for F for F 2 to obtain F2 = F 1 + 60 kips
Ans.
(e)
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When the temperature reaches −10°F, the total change in temperature is ∆T = = −100°F. Substitute this value along with Eq. (e) into the compatibility equation [Eq. (d)] and derive an expression for F for F 1: F1 L1 ( F1 + 60 kips)L2 + = −α1 ∆T L1 − α 2 ∆T L2 A1 E1 A2 E 2
⎡ L L ⎤ (60 kips)L2 F1 ⎢ 1 + 2 ⎥ + = −∆T [α1 L1 + α 2 L2 ] A E A E A E 2 2 ⎦ 2 2 ⎣ 1 1 ⎡ L L ⎤ (60 kips)L2 F1 ⎢ 1 + 2 ⎥ = −∆T [α1 L1 + α 2 L2 ] − A2 E 2 ⎣ A1 E1 A2 E2 ⎦ −∆T [α1 L1 + α 2 L2 ] − F 1 =
L1 A1 E1
+
(60 kips) L2 A2 E 2
L2 A2 E 2
and compute F compute F 1: F 1 =
−( −100°F) ⎡⎣ (6.6 ×10−6 )(120 in.) + (12.5 × 10−6 )(144 in.)⎤⎦ − 120 in. (5.6 (5.60 0 in. in.2 )(30 (30,000 ,000 ksi ksi)
=
+
144 in. (4.4 (4.40 0 in. in.2 )(10,00 10,000 0 ksi) ksi)
0.25 0.2592 9200 00 in. in.− 0.19 0.1963 6364 64 in. in. −6
714.2857 × 10
−6
in./kip + 3, 27 272.7273× 10
(60 kips)(144 in.) (4.40 (4.40 in.2 )(10,000 )(10,000 ksi) ksi)
= in./kip
0.062836 in. 3, 98 987.0130× 10−6 in./kip
= 15.7602 kips From Eq. (a), F (a), F 2 has a value of F2 = F 1 + 60 ki kips = 15.7602 ki kips + 60 ki kips = 75.7602 ki kips (a) Normal Stresses: The normal stresses in each axial member can now be calculated: F 1 15.7602 kips σ 1 = 2.81 ksi ksi (T) = = 2.8143 ksi = 2.8 A1 5.60 in.2 σ 2
=
F 2 A2
=
75.7602 kips 4.40 in.2
17.218 182 2 ksi ksi = 17.2 17.22 2 ksi ksi (T) = 17.2
Ans.
Ans.
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5.56 A load P will be supported by a structure consisting of a rigid bar ABCD, ABCD, a polymer [ E E = −6 2,300 ksi, α = 2.9 × 10 /°F] bar (1) and an aluminum alloy [ E E = 10,000 ksi, α = 12.5 × −6 10 /°F] bar (2), as shown in Fig. P5.56. Each bar 2 has a cross-sectional area of 2.00 in. . The bars are unstressed when the structure is assembled at 30°F. After a concentrated load of P of P = = 26 kips is applied and the temperature is increased to 100°F, determine: (a) the normal stresses in bars (1) and (2). (b) the vertical deflection of point D point D..
Fig. P5.56
Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members members (1) and (2). A moment equation about pin A pin A gives gives the best information for this situation: Σ M A = (30 in.)F1 + (84 in.)F2 − (66 in.)P = 0 (a) Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2): F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2 Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: v B v (c) = D 30 in. 84 in. Since there are no gaps, clearances, or other misfits at pins B pins B and and D D,, the elongation of member (1) will equal the deflection of the rigid bar at B at B and and the elongation of member (2) will equal the deflection of the rigid bar at D at D.. Therefore, Eq. (c) can be rewritten in terms of the member elongations as: e1 e = 2 30 in. 84 in.
(b)
(d)
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Compatibility Equation Substitute the force-deformation relationships (b) into the g eometry of deformation relationship (d) to derive the compatibility equation:
⎡ F1 L1 ⎤ ⎤ 1 ⎡ F2 L2 + α1∆T1 L1 ⎥ = + α 2 ∆T2 L2 ⎥ ⎢ ⎢ 30 in. ⎣ A1 E1 ⎦ 84 in. ⎣ A2 E 2 ⎦ 1
(e)
Solve the Equations Note that the temperature change for both axial members is the same (i.e., ∆T 1 = ∆T 2 = ∆T ). ). Solve Eq. (e) for F for F 1:
F1 L1 A1 E1
=
30 in. ⎡ F2 L2
⎤ + α 2 ∆T L2 ⎥ − α 1∆T L1 ⎢ 84 in. ⎣ A2 E 2 ⎦
F1 = F2
30 in. L2 A1 E1
+
84 in. L1 A2 E2
30 in. 84 in.
α 2 ∆T L2
A1 E 1 L1
− α 1∆T A1 E 1
(f)
Substitute this expression into equilibrium equation (a):
⎡
30 in. L2 A1 E1
⎣
84 in. L1 A2 E2
(30 in.) ⎢ F2
+
30 in. 84 in.
α 2 ∆T L2
A1 E 1 L1
⎤
− α 1∆T A1 E1 ⎥ + (84 in.)F2 − (66 in.)P = 0
⎦
and solve for F for F 2: (66 in.) P −
(30 in.)2
F 2 =
84 in.
α 2 ∆T L2
A1 E 1 L1
+ (30 in.)α 1∆T A1 E 1
(30 in.) 2 L2 A1 E 1
(g)
+ 84 in. 84 in. L1 A2 E 2 For this structure, P structure, P = = 26 kips, and the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed below: L1 = 72 in. L2 = 96 in. A1 = 2.00 in.2
A2 = 2.00 in.2
E1 = 2, 300 ksi
E 2 = 10, 000 ksi
α1
= 2.9 × 10−6 / °F
α 2
= 12.5 × 10−6 / °F
Substitute these values along with ∆T = = 70°F into Eq. (g) and calculate F calculate F 2 = 19.3218 kips. Backsubstitute into Eq. (f) to calculate F calculate F 1 = 3.0991 kips. Normal Stresses The normal stresses in each axial member can now be calculated: F 1 3.0991 kips σ 1 = = = 1.550 ksi (T) A1 2.00 in.2 σ 2
=
F 2 A2
=
19.3218 kips 2.00 in.2
= 9.66 ksi (T)
Ans. Ans.
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Deflections of the rigid bar Calculate the elongation of member (2): F L (19.3218 kips)(96 in.) e2 = 2 2 + α 2 ∆T2 L2 = /°F)(70°F)(96 in.) = 0.1767 in in. + (12.5 × 10−6 /°F) 2 A2 E 2 (2.00 (2.00 in. )(10,000 )(10,000 ksi) ksi)
(h)
Since there are no gaps at pin D pin D,, the rigid bar deflection at D at D is is equal to the elongation of member (2); therefore: v D = e2 = 0.1767 in. ↓
Ans.
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5.57 Rigid bar ABCD ABCD is loaded and supported as shown in Fig. P5.57. Bar (1) is made of bronze [ E E −6 = 100 GPa, α = 16.9 × 10 /°C] and has a cross2 sectional area of 400 mm . Bar (2) is made made o −6 aluminum [ E E = = 70 GPa, α = 22.5 × 10 /°C] and 2 has a cross-sectional area of 600 mm . Bars (1) and (2) are initially unstressed. After the temperature has increased by 40°C, determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point A point A..
Fig. P5.57
Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) (1) and (2). A moment equation about pin D gives D gives the best information for this situation: (a) Σ D = (3 m) F1 − (1 m)F2 = 0 ∴ F2 = 3F 1 Force-Temperature-Deformation Relationships F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2
Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: v v A v = B = C 4 m 3m 1m
(b)
(c)
There are no gaps, clearances, or other misfits at pins B pins B and C ; therefore, Eq. (c) can be rewritten in terms of the member elongations as: e −e1 (d) = 2 ∴ e1 = −3e2 3m 1m Note: To understand the negative sign associated with e1, see Section 5-5 for discussion of statically indeterminate rigid bar configurations with oppo sing members.
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Compatibility Equation
⎡ F2 L2
F1 L1 A1 E1
+ α1∆T1 L1 = −3 ⎢
⎣ A2 E 2
Solve the Equations For this situation, T 1 =
T 2 =
⎡ ( 3 F1 ) L2
F1 L1 A1 E1
⎤
+ α 2 ∆T2 L2 ⎥
+ α1∆T L1 = −3 ⎢
⎣ A2 E 2
(e)
⎦
= 40°C. T =
Substitute Eq. (a) into Eq. (e):
⎤
+ α 2 ∆T L2 ⎥
⎦
and solve for F for F 1: ∆T ( 3α 2 L2 + α 1 L1 ) F 1 = − L1 9 L2 + A1 E1 A2 E 2
=−
(40°C) ⎡⎣3(22.5 × 10−6 / °C)(920 mm) + (16.9 × 10−6 / ° C)(840 mm)⎤⎦ 840 mm 2
2
(400 mm mm )(100,000 ,000 N/ N/mm )
+
9(920 mm) (600 mm mm2 )(70,000 ,000 N/ N/mm2 )
990 N = −13.990 kN = −13, 99 Backsubstitute into Eq. (a) to find F find F 2 = −41.970 kN. (a) Normal Stresses The normal stresses in each axial member can now be calculated: F 1 −13, 13, 990 N σ 1 = = = 35.0 MPa (C) A1 400 mm2 σ 2
=
F 2 A2
=
41, 970 N −41, = 70.0 MPa (C) 600 mm2
Ans. Ans.
(b) Deflection of the rigid bar at A Calculate the elongation of one of the axial members, say member (1): F L e1 = 1 1 + α 1∆T1 L1 A1 E 1
=
( −13,990 13,990 N)( N)(84 840 0 mm) mm) 2
2
(400 (400 mm )(10 )(100,00 0,000 0 N/m N/mm m )
/°C)(40°C)(840 mm mm) + (16.9 × 10−6 /°C
= 0.27405 mm Since there are no gaps at pin B pin B,, the rigid bar deflection at B at B is is equal to the elongation e longation of member (1); therefore, v B = e1 = 0.27405 mm (upward). From similar triangles, triangles, the deflection of the rigid bar at A at A is is related to v B by: v A v = B 4m 3m The deflection of the rigid bar at A at A is is thus: 4m 4m Ans. v A = vB = (0.27 .27405 mm) = 0.3 0.365 mm ↑ 3m 3m 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
5.58 Rigid bar ABCD bar ABCD in in Fig. P5.58 is supported by a pin connection at A at A and and by two axial bars (1) and (2). Bar (1) is a 30-in.-long bronze [ E E = = 15,000 ksi, α = = −6 9.4 × 10 /°F] bar with a cross-sectional area of 1.25 2 in. . Bar (2) is a 40-in.-long aluminum a luminum alloy [ E E = = −6 10,000 ksi, α = = 12.5 × 10 /°F] bar with a cross2 sectional area of 2.00 in. . Both bars are unstressed before the load P load P is is applied. If a concentrated load of P = = 27 kips is applied to the rigid bar at D at D and and the temperature is decreased by 100°F, determine: (a) the normal stresses in bars (1) and (2). (b) the normal strains in bars (1) and (2). (c) the deflection of the rigid bar at point D point D..
Fig. P5.58
Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces forces in members (1) and (2). A moment equation about pin A pin A gives gives the best information for this situation: Σ M A = −(36 in.)F1 + (84 in.)F 2 − (98 in.)(27 kips) = 0 (a) Force-Deformation Relationships F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E 2
(b)
Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: v v B (c) = C 36 in. 84 in. There are no gaps, clearances, or other misfits at pins B pins B and and C ; therefore, Eq. (c) can be rewritten in terms of the member elongations as:
e −e1 (d) = 2 36 in. 84 in. Note: To understand the negative sign associated with e1, see Section 5-5 for discussion of statically indeterminate rigid bar configurations with opposing members.
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Compatibility Equation
−
⎡ F1 L1 ⎤ ⎤ 1 ⎡ F2 L2 T L T L + α ∆ = + α ∆ 1 1 1⎥ 2 2 2⎥ ⎢ ⎢ 36 in. ⎣ A1 E1 8 4 i n . A E ⎦ ⎣ 2 2 ⎦ 1
(e)
Solve the Equations Solve Eq. (e) for F for F 1:
⎡ 36 in. ⎛ F2 L2 ⎤ A E ⎞ + α 2 ∆T2 L2 ⎟ − α 1 ∆T1 L1 ⎥ 1 1 ⎜ ⎠ ⎣ 84 in. ⎝ 2 E2 ⎦ L1 ⎡ 3 F2 L2 3 ⎤ A E = −⎢ + α 2 ∆T2 L2 + α 1 ∆T1 L1 ⎥ 1 1 ⎣ 7 A2 E2 7 ⎦ L1
F1 = ⎢ −
Substitute this expression into equilibrium equation (a) and solve for F for F 2 using ∆T 1 = ∆T 2 = −100°F:
−(36 in.) F1 + (84 in.)F 2 = (98 in.)(27 kips)
⎡ ⎛ 3 F2 L2
−(36 in.) ⎢ − ⎜
⎣ ⎝ 7 A2 E2
+
3 7
α 2 ∆T2 L2
⎞ A1 E 1 ⎤ ⎥ + (84 in.)F2 = (98 in.)(27 kips) L ⎠ 1 ⎦
+ α 1 ∆T1 L1 ⎟
⎡ 3 L2 A1 E 1 ⎤ ( 3 6 i n . ) 8 4 i n . + ⎢ ⎥ F 2 = (98 in.)(27 kips) 7 L A E ⎣ 1 2 2 ⎦ −(36 in.)
3 7
α 2 ∆T2 L2
A1 E1 L1
− (36 in.)α 1 ∆T1 L1
A1 E 1 L1
⎡ 3 40 in. 1.25 ⎤ 1.25 in.2 15,00 15,000 0 ks ksi (36 in.) + 84 in.⎥ F 2 = (98 in.)(27 kips) ⎢ 7 30 in. 2 in. 2.00 2.00 in. in. 10,000 10,000 ksi ksi ⎣ ⎦ 3
−6
(1.25 in.2 )(15, 000 ksi)
−(36 in.) (12.5 × 10 )(−100°F)(40 in.) 7 30 in. 2 (1.25 in. in. )(15 )(15,, 000 ksi) −(36 in.)(9.4 × 10−6 )(−100° F)(30 in.) 30 in. F 2 =
2,646 2,646 kipkip-in in.. + 482. 482.14 1429 29 kipkip-in in.. + 634. 634.50 5000 00 kipkip-in in.. 19.2857 in. + 84 in.
=
3,76 3,762. 2.64 6429 29 kipkip-in in.. 103.2857 in.
= 36.4295 kips Backsubstitute into Eq. (a) to find F find F 1: (84 (84 in.)( in.)(36.4 36.4295 295 kips) kips) − (98 in.)( in.)(27 27 kips kips)) F 1 = = 11.5022 kips 36 in. Normal Stresses The normal stresses in each axial member can now be calculated: 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
F 1
σ 1
=
σ 2
=
11.5022 kips
=
1.25 in.2
A1 F 2
=
A2
36.4295 kips 2.00 in.2
= 9.20 ksi (T) = 18.21 ksi (T)
Ans. Ans.
(b) Normal Strains The force-temperature-deformation relationships were expressed in Eq. (b). By definition, ε = = e/ L. L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths: F1 F 2 ε1 = ε2 = + α1∆T1 + α 2 ∆T 2 A1 E1 A2 E 2
Substitute the appropriate values to calculate the normal strains in each member: F 1 ε1 = + α 1∆T 1 A1 E 1 11.5022 kips
=
2
(1.25 (1.25 in. in. )(15 )(15,, 000 ksi) ksi)
°F)( − 100°F) + (9.4 × 10−6 / °F
= −326.55 × 10−6 in./in. = − 327 με
ε2
=
F 2 A2 E 2
=
Ans.
+ α 2 ∆T 2
36.4295 kips 2
(2.00 (2.00 in. in. )(10,000 )(10,000 ksi)
°F)( − 100°F) + (12.5 × 10−6 / °F
= 571.48 × 10−6 in./in. = 571 με
Ans.
(c) Deflection of the rigid bar at D Calculate the elongation of one of the axial members, say member (2): F L (36.4295 kips)(40 in.) e2 = 2 2 + α 2 ∆T2 L2 = °F)( − 100°F)(40 in.) + (12.5 × 10−6 / °F 2 A2 E 2 (2.00 (2.00 in. in. )(10,000 )(10,000 ksi)
= 0.022859 in. This elongation can also a lso be determined from the strain in member (2): in.) = 0.022859 in in. e2 = ε 2 L2 = (571.48 × 10−6 in./in.)(40 in Since there are no gaps at pin C , the rigid bar deflection at C is is equal to the elongation of member (2); therefore, vC = = e2 = 0.022859 in. (downward). From similar triangles, triangles, the deflection of the rigid bar at D at D is related to vC by: vC v = D 84 in. 98 in. The deflection of the rigid bar at D at D is is thus: 98 in. 98 in. Ans. v D = vC = (0.0228 22859 in. in.) = 0.02 .0267 in. in. ↓ 84 in. 84 in. 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
5.59 The pin-connected structure shown in Fig. P5.59 consists of a rigid bar ABC , a steel bar (1), and a steel rod (2). The cross-sectional area 2 of bar (1) is 1.5 in. and the diameter of rod (2) is 0.75 in. Assume E Assume E = = 30,000 ksi and α = 6.6 −6 × 10 /°F for both axial members. The bars are unstressed when the structure is assembled at 70°F. After application of a concentrated force of P of P = = 20 kips, the temperature is decreased to 30°F. Determine: (a) the normal stresses in bar (1) and rod (2). (b) the normal strains in bar (1) and rod (2). (c) the deflection of pin C from its original position.
Fig. P5.59
Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members members (1) and (2). A moment equation about pin B pin B gives gives the best information for this situation: Σ B = −(12 in.) F1 + (20 in.)F 2
in.)(2 (20 0 kips kips)) = 0 −(15 in.)
(a)
Force-Temperature-Deformation Relationships F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 (b) A1 E1 A2 E 2 Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: v v A (c) = C 12 in. 20 in. There are no gaps, clearances, or other misfits at pins A pins A and and C ; therefore, Eq. (c) can be rewritten in terms of the member elongations as: e −e1 (d) = 2 12 in. 20 in. Note: To understand the negative sign associated with e1, see Section 5-5 for discussion of statically indeterminate rigid bar configurations with opposing members. 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
Compatibility Equation
−
⎡ F1 L1 ⎤ ⎤ 1 ⎡ F2 L2 + α ∆ = + α ∆ T L T L 1 1 1⎥ 2 2 2⎥ ⎢ ⎢ 12 in. ⎣ A1 E1 ⎦ 20 in. ⎣ A2 E 2 ⎦ 1
(e)
Solve the Equations Solve Eq. (e) for F for F 1:
⎡ 12 in. ⎛ F2 L2 ⎤ A1E 1 ⎞ T L T L α α + ∆ − ∆ ⎜ 2 2 2 ⎟ 1 1 1⎥ ⎠ ⎣ 20 in. ⎝ A2 E2 ⎦ L1
F1 = ⎢ −
⎡ 12 F2 L2
= −⎢
⎣ 20 A2 E2
+
12 20
α 2 ∆T2 L2
⎤ A1 E 1
+ α 1∆T1L1 ⎥
⎦ L1
Substitute this expression into equilibrium equation (a) and solve for F for F 2 using ∆T 1 = ∆T 2 = −40°F:
−(12 in.) F1 + (20 in.)F 2 = (15 in.)(20 kips)
⎡ ⎛ 12 F2 L2
−(12 in.) ⎢ − ⎜
+
⎣ ⎝ 20 A2 E2
12 20
α 2 ∆T2 L2
⎞ A1 E 1 ⎤ ⎥ + (20 in.)F2 = (15 in.)(20 kips) ⎠ L1 ⎦
+ α 1∆T1 L1 ⎟ 2
2
Note that the area of rod (2) is A is A2 = π/4(0.75 in.) = 0.4418 in. .
⎡ 12 L2 A1 E 1 ⎤ ( 1 2 i n . ) 2 0 i n . + ⎢ ⎥ F 2 = (15 in.)(20 kips) ⎣ 20 L1 A2 E 2 ⎦ −(12 in.)
12 20
α 2 ∆T2 L2
A1 E1 L1
A1 E 1
− (12 in.)α 1∆T1 L1
L1
⎡ 12 80 in. 1.5 in.2 30, 00 ⎤ 000 ksi ( 1 2 i n . ) 2 0 i n . + ⎢ 20 32 in. ⎥ F 2 = (15 in.)(20 kips) in. 0.44 0.4418 18 in.2 30,000 30,000 ks ksi ⎣ ⎦ −(12 in.)
12 20
−6
(6.6 × 10 )(−40°F)(80 in.)
−(12 in.)(6.6 × 10−6 )(−40°F)(32 in.)
F 2 =
300 300 kipkip-in in.. + 213. 213.84 84 kipkip-iin.+ 142. 142.56 56 kipkip-in in.. 61.113626 in. + 20 in.
=
(1.5 in.2 )(30, 000 ksi) 32 in. 2
(1.5 in. in. )(30,000 )(30,000 ksi) ksi) 32 in. 656. 656.40 40 kipkip-iin. 81.1136267 in.
= 8.092 kips Backsubstitute into Eq. (a) to find F find F 1: (20 in.)( in.)(8.0 8.092 92 kips kips)) − (15 in.) in.)(20 (20 kips) kips) F 1 = = −11.513 kips 12 in.
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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
(a) Normal Stresses The normal stresses in each axial member can now be calculated: F 1 −11.513 kips σ 1 = = = 7.68 ksi (C) A1 1.5 in.2 σ 2
=
F 2
=
A2
8.092 kips 0.4418 in.2
= 18.32 ksi (T)
Ans. Ans.
(b) Normal Strains The force-temperature-deformation relationships were expressed in Eq. (b). By definition, ε = = e/ L. L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (b) by the respective member lengths: F1 F 2 ε1 = + α1∆T1 ε2 = + α 2 ∆T 2 A1 E1 A2 E 2
Substitute the appropriate values to calculate the normal strains in each member: F 1 ε1 = + α 1∆T 1 A1 E 1
=
−11.513 kips + (6.6 × 10−6 / °F)( − 40°F) 2 (1.5 (1.5 in. in. )(30,000 )(30,000 ksi) ksi)
= −519.8 × 10−6 in./in. = −520 με
ε2
=
F 2 A2 E 2
=
Ans.
+ α 2 ∆T 2 8.092 kips 2
(0.441 (0.4418 8 in. in. )(30,000 )(30,000 ksi) ksi)
+ (6.6 ×10−6 / °F)( − 40°F)
= 346.5 × 10−6 in./in. = 347 με
Ans.
(c) Deflection of the rigid bar at C Calculate the elongation of one of the axial members, say member (2): F L (8.092 kips)(80 in.) e2 = 2 2 + α 2 ∆T2 L2 = °F)( − 40°F)(80 in.) + (6.6 ×10−6 / °F 2 A2 E 2 (0.441 (0.4418 8 in. in. )(30,000 )(30,000 ksi) ksi)
= 0.027723 in. This elongation can also be determined from the strain in member (2): in.) = 0.027723 in in. e2 = ε 2 L2 = (346.5 × 10−6 in./in.)(80 in Since there are no gaps at pin C , the rigid bar deflection at C is is equal to the elongation of member (2); therefore: vC = e2 = 0.0277 in. ↓
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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
5.60 Three rods of different materials are connected and placed between rigid supports at A A and D, D, as shown in Fig. P5.60. Properties for each of the three rods are given below. The bars are initially unstressed when the structure is assembled at 70°F. After the temperature has been increased to 250°F, determine: (a) the normal stresses in the three rods. (b) the force exerted on the rigid supports. (c) the deflections of joints B joints B and and C relative relative to rigid support A support A.. Aluminum (1) L1 = 10 in. 2 1 = 0.8 in. E 1 = 10,000 ksi -6 α 1 = 12.5×10 /°F
Fig. P5.60
Cast Iron (2) L2 = 5 in. 2 2 = 1.8 in. E 2 = 22,500 ksi -6 α 2 = 7.5×10 /°F
Bronze (3) L3 = 7 in.
2
A3 = 0.6 in. E 3 = 15,000 ksi -6 α 3 = 9.4×10 /°F
Solution Equilibrium Consider a FBD at joint B joint B.. Assume that both internal axial forces will be tension. tension. (a) Σ F x = − F1 + F2 = 0 ∴ F1 = F 2
Similarly, consider a FBD at joint C . Assume that both internal axial forces will be tension. tension. (b) Σ F x = − F2 + F3 = 0 ∴ F3 = F 2 Force-Temperature-Deformation Relationships F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E2 Geometry of Deformations Relationship e1 + e2 + e3 = 0 Compatibility Equation F L F1 L1 FL + α1∆T1 L1 + 2 2 + α 2 ∆T2 L2 + 3 3 + α 3 ∆T3 L3 = 0 A1 E1 A2 E2 A3 E 3
e3 =
F3 L3 A3 E 3
+ α 3 ∆T3 L3
(c)
(d)
(e)
Solve the Equations From Eq. (a), F (a), F 1 = F = F 2, and from Eq. (b), F (b), F 3 = F = F 2. The temperature change is the same for all members; therefore, ∆T 1 = ∆T 2 = ∆T 3 = ∆T . Eq. (e) then can be written as: 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
( F2 ) L1 A1 E1
+ α1∆T L1 +
F2 L2 A2 E2
+ α 2 ∆T L2 +
( F2 ) L3 A3 E 3
+ α 3 ∆T L3 = 0
Solving for F for F 2: F2 L1 F2 L2 F3 L3 + + = −α1∆T L1 − α 2 ∆T L2 − α 3 ∆T L3 A1 E1 A2 E2 A3 E 3
⎡ L L L ⎤ F2 ⎢ 1 + 2 + 3 ⎥ = −∆T [α1 L1 + α 2 L2 + α 3 L3 ] ⎣ A1 E1 A2 E2 A3 E 3 ⎦ F 2 = −
∆T [α1 L1 + α 2 L2 + α 3 L3 ] L1 L L + 2 + 3 A1 E1 A2 E2 A3 E 3
(f)
Substitute the problem data along with ∆T = = +180°F into Eq. (f) and calculate F calculate F 1 = −19.1025 kips. From Eq. (a), F (a), F 1 = −19.1025 kips and from Eq. (b), F (b), F 3 = −19.1025 kips. (a) Normal Stresses The normal stresses in each rod can now be calculated: F 1 −19.1025 kips 23.878 ksi = 23.9 ksi ksi (C) σ 1 = = = −23. A1 0.8 in.2 σ 2 σ 3
−19.1025 kips 10.613 13 ksi ksi = 10.6 10.61 1 ksi ksi (C) (C) = −10.6 A2 1.8 in.2 F −19.1025 kips 838 ks ksi = 31. 31.8 ks ksi (C) (C) = 3 = = −31.838 A3 0.6 in.2 =
F 2
=
Ans. Ans. Ans.
(b) Force on Rigid Supports The force exerted on the rigid supports is equal to the internal axial force:
R A = RD = 19.10 kips
Ans.
(c) Deflection of Joints B and C The deflection of joint B joint B is is equal to the elongation (i.e., contraction in in this instance) of rod (1). (1). The elongation of rod (1) is given by: F L e1 = 1 1 + α 1∆T1 L1 A1 E 1
=
( −19.10 19.1025 25 kips kips)( )(10 10 in.) in.) 2
(0.8 in. )(10,000 )(10,000 ksi) ksi) The deflection of joint B joint B is is thus:
/°F)(180°F)(10 in in.) = −0.001378 in in. + (12.5 × 10−6 /°F)
u B = e1 = 0.001378 in. ←
Ans.
The elongation of rod (2) is given by:
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e2 =
=
F2 L2 A2 E 2
+ α 2 ∆T2 L2
( −19.10 19.1025 25 kips kips)( )(5 5 in.) in.) 2
(1.8 in. in. )(22,500 )(22,500 ksi) ksi)
/°F)(180°F)(5 in.) = 0.004392 in in. + (7.5 × 10−6 /°F)
The deflection of joint C is: is: 01378 in in. + 0.0 0.004392 392 in. in. = 0.003 00301 in. → uC = uB + e2 = −0.0013
Ans.
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5.61 Three rods of different materials are connected and placed between rigid supports at A A and D, D, as shown in Fig. P5.61. Properties for each of the three rods are given below. The bars are initially unstressed when the structure is assembled at 20°C. After the temperature has been increased to 100°C, determine: (a) the normal stresses in the three rods. (b) the force exerted on the rigid supports. (c) the deflections of joints B joints B and and C relative relative to rigid support A support A.. Aluminum (1) L1 = 440 mm 2 1 = 1,200 mm E 1 = 70 GPa -6 α 1 = 22.5×10 /°C
Fig. P5.61
Cast Iron (2) L2 = 200 mm 2 2 = 2,800 mm E 2 = 155 GPa α2 = 13.5×10-6/°C
Bronze (3)
L3 = 320 mm 2 A3 = 800 mm E 3 = 100 GPa -6 α 3 = 17.0×10 /°C
Solution Equilibrium Consider a FBD at joint B joint B.. Assume that both internal axial forces will be tension. tension. (a) Σ F x = − F1 + F2 = 0 ∴ F1 = F 2
Similarly, consider a FBD at joint C . Assume that both internal axial forces will be tension. tension. (b) Σ F x = − F2 + F3 = 0 ∴ F3 = F 2 Force-Temperature-Deformation Relationships F L FL e1 = 1 1 + α1∆T1 L1 e2 = 2 2 + α 2 ∆T2 L2 A1 E1 A2 E2 Geometry of Deformations Relationship e1 + e2 + e3 = 0 Compatibility Equation F1 L1 FL F L + α1∆T1 L1 + 2 2 + α 2 ∆T2 L2 + 3 3 + α 3 ∆T3 L3 = 0 A1 E1 A2 E2 A3 E 3
e3 =
F3 L3 A3 E 3
+ α 3 ∆T3 L3
(c)
(d)
(e)
Solve the Equations From Eq. (a), F (a), F 1 = F = F 2, and from Eq. (b), F (b), F 3 = F = F 2. The temperature change is the same for all members; therefore, ∆T 1 = ∆T 2 = ∆T 3 = ∆T . Eq. (e) then can be written as: ( F2 ) L1 (F ) L F L + α1∆T L1 + 2 2 + α 2 ∆T L2 + 2 3 + α 3 ∆T L3 = 0 A1 E1 A2 E2 A3 E 3 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 United States Copyright Act without without the permission of the copyright copyright owner is unlawful.
Solving for F for F 2: F2 L1 F2 L2 F3 L3 + + = −α1∆T L1 − α 2 ∆T L2 − α 3 ∆T L3 A1 E1 A2 E2 A3 E 3
⎡ L L L ⎤ F2 ⎢ 1 + 2 + 3 ⎥ = −∆T [α1 L1 + α 2 L2 + α 3 L3 ] ⎣ A1 E1 A2 E2 A3 E 3 ⎦ F 2 = −
∆T [α1 L1 + α 2 L2 + α 3 L3 ] L L1 L + 2 + 3 A1 E1 A2 E2 A3 E 3
(f)
Substitute the problem data along with ∆T = = +80°C into Eq. (f) and calculate F calculate F 1 = −148.80 kN. From Eq. (a), F (a), F 1 = −148.80 kN and from Eq. (b), F (b), F 3 = −148.80 kN. (a) Normal Stresses The normal stresses in each rod can now be calculated: F 1 −148,800 N 4.00 MPa MPa = 124.0 4.0 MPa MPa (C) σ 1 = = = −124.0 A1 1,200 mm2 σ 2 σ 3
−148,800 N .143 MPa = 53.1 MPa (C) = −53.14 A2 2,800 mm2 F −148,800 N 186.0 MPa MPa = 186. 86.0 MPa MPa (C) = 3 = = −186 A3 800 mm2 =
F 2
=
Ans. Ans. Ans.
(b) Force on Rigid Supports The force exerted on the rigid supports is equal to the internal axial force:
R A = RD = 148.8 kN
Ans.
(c) Deflection of Joints B and C The deflection of joint B joint B is is equal to the elongation (i.e., contraction in in this instance) of rod (1). (1). The elongation of rod (1) is given by: F L e1 = 1 1 + α 1∆T1 L1 A1 E 1
=
( −148,800 148,800 N)( N)(440 440 mm) mm) 2
2
(1, 200 200 mm mm )(70 )(70,, 000 000 N/m N/mm m ) The deflection of joint B joint B is is thus:
(22.5 ×10−6 /°C) /°C)(80°C)(440 mm) = 0.01257 mm + (22
u B = e1 = 0.01257 mm →
Ans.
The elongation of rod (2) is given by: F L e2 = 2 2 + α 2 ∆T2 L2 A2 E 2
=
( −148,80 148,800 0 N)( N)(200 200 mm) mm) 2
2
(2,800 (2,800 mm )(155 155, 000 000 N/m N/mm m )
/°C)(80°C)(200 mm mm) = 0.14743 mm mm + (13.5 × 10−6 /°C)
The deflection of joint C is: is: uC = uB + e2 = 0.01257 mm mm + 0.14743 mm mm = 0.1600 mm mm →
Ans.
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