5.1 A steel [ E E = 200 GPa] rod with a circular cross section is 6-m long. Determine the minimum diameter D required D required if the rod must transmit a tensile force of 30 kN without exceeding an allowable stress of 180 MPa or stretching more than 5 mm.
Solution If the normal stress in the rod cannot exceed 180 MPa, the cross-sectional area must equal or exceed P (30 kN)(1,000 N/kN) 2 A ≥ = = 166.6667 mm 2 180 N/mm σ If the elongation must not exceed 5 mm, the cross-sectional area must equal or exceed PL (30 kN)(1,000 N/kN)(6,000 mm) 2 A ≥ = = 180.0 mm 2 Ee (200,0 (200,000 00 N/m N/mm m )(5 mm) mm) 2
Therefore, the minimum cross-sectional area that may be used for the rod is A is Amin = 180 mm . The corresponding rod diameter D diameter D is is π 2 Drod ≥ 180.0 mm2 ∴ Drod ≥ 15.1388 mm = 15.14 mm Ans. 4
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5.2 An aluminum [ E E = = 10,000 ksi] control rod with a circular cross section must not stretch more than 0.25 in. when the tension in the rod is 2,200 lb. If the maximum allowable normal stress in the rod is 20 ksi, determine: (a) the smallest diameter that can be used for the rod. (b) the corresponding maximum length of the rod.
Solution (a) If the normal stress in the rod cannot exceed 20 ksi, the cross-sectional c ross-sectional area must equal or exceed P 2.20 kips A ≥ = = 0.1100 in.2 σ 20 ksi The corresponding rod diameter D diameter D is is π 2 Drod ≥ 0.1100 in.2 ∴ Drod ≥ 0.374 in. 4
Ans.
(b) If the elongation must not exceed excee d 0.25 in., the aluminum a luminum control rod having a cross-sectional area of 2 A = A = 0.1100 in. can have a length no greater than AEe (0.1100 in. in.2 )(10,000 )(10,000 ksi)( ksi)(0.25 0.25 in.) 2 L ≤ = = 125.0 in. Ans. P 2.20 kips
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5.3 A 16-mm-diameter steel [ E E = = 200 GPa] rod (2) is connected to a 50-mm-wide by 8-mm-thick rectangular aluminum [ E E = 70 GPa] bar (1), as shown in Fig. P5.3. Determine the force P force P required required to stretch the assembly 3.00 mm.
Fig. P5.3
Solution The elongations in the two axial members are expressed by F L FL e1 = 1 1 and e2 = 2 2 A1 E1 A2 E 2 The total elongation of the assembly is thus F L FL uC = e1 + e2 = 1 1 + 2 2 A1 E1 A2 E 2 Since the internal forces F forces F 1 and F and F 2 are equal to external load P load P , this expression can be simplified to
⎡ L L ⎤ uC = P ⎢ 1 + 2 ⎥ ⎣ A1 E1 A2 E 2 ⎦ For rectangular aluminum bar (1), the cross-sectional area is A1 = (50 mm)(8 mm mm) = 400 mm2 and the cross-sectional area of steel rod (2) is π A2 = (16 mm)2 = 201.0619 mm2 4 The force P force P required required to stretch the assembly 3.00 mm is thus uC P = L1 L + 2 A1 E1 A2 E 2 =
=
3.00 mm
⎡ ⎤ 450 mm 1,300 mm + ⎢ (400 mm2 )(70,000 ⎥ ,000 N/ N/mm2 ) (201 201.0619 mm2 )(200 (200,00 ,000 0 N/ N/mm2 ) ⎦ ⎣ 3.00 mm
⎡⎣1.6071× 10−5 mm/N + 3.2328× 10−5 mm/N⎤⎦
= 61,983.7574 N = 62.0 62.0 kN
Ans.
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5.4 Two polymer bars are connected to a rigid plate at B, B, as shown shown in in Fig. P5.4. P5.4. Bar Bar (1) has a 2 cross-sectional area of 2.2 in. and an elastic modulus of 2,400 ksi. Bar (2) has a cross-sectional 2 area of 1.30 in. and an elastic modulus of 4,000 ksi. Determine the total elongation of the bar.
Fig. P5.4
Solution Draw a FBD that cuts through member (1) to find that the internal axial force in member (1) is F 1 = 20 kips (T)
Similarly, draw a FBD that cuts through member (2) to find that the internal axial force in member (2) is F 2 = 30 kips (T) The elongation in bar (1) can be computed as F L (20 kips)(30 in.) = 0.113636 in. e1 = 1 1 = A1 E 1 (2.2 (2.2 in. in.2 )(2,400 )(2,400 ksi) ksi) and the elongation in bar (2) can be computed as F L (30 kips)(36 in.) e2 = 2 2 = 0.207692 in. = 0.207692 A2 E 2 (1.30 (1.30 in. in.2 )(4,000 )(4,000 ksi) The total elongation of the bar is thus e1 + e2 = 0.11363 3636 in.+ 0.20769 7692 in in. = 0.3 0.321 in.
Ans.
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5.5 An axial member consisting of two polymer bars is supported at C as shown in Fig. P5.5. Bar (1) has a 2 cross-sectional area of 720 mm and an elastic modulus of 28 GPa. Bar (2) has a cross-sectional area of 1,560 2 mm and an elastic modulus of 16.5 GPa. Determine the deflection of point A point A relative relative to support C .
Fig. P5.5
Solution Draw a FBD that cuts through member (1) to find that the internal axial force in member (1) is F 1 = 70 kN (T) Similarly, draw a FBD that cuts through member (2) and includes the free end of the axial member. From this FBD, the equilibrium equation is Σ F x = −70 kN + 90 kN + 90 kN + F 2 = 0 Therefore, the internal axial force in member (2) is F 2 = − 110 kN kN = 110 kN kN (C) The elongation in bar (1) can be computed as F L (70 kN)(1,000 N/kN)(850 mm) e1 = 1 1 = = 2.9514 mm A1 E 1 (720 720 mm mm 2 )(28 (28,000 ,000 N/m N/mm m2 ) and the elongation in bar (2) can be computed as F L (−110 kN)(1, kN)(1,000 000 N/kN) N/kN)(1, (1,150 150 mm) mm) e2 = 2 2 = = −4.9145 mm A2 E 2 (1,5 (1,560 60 mm2 )(16 )(16,500 ,500 N/m N/mm m2 ) The deflection of point A point A relative relative to the support at C is is the sum of these two elongations: e longations: u A = e1 + e2 = 2.9514 mm + (− 4.9145 mm) = − 1.963 mm = 1.963 mm →
Ans.
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5.6 The roof and second floor of a building are supported by the column shown in Fig. P5.6. The column is a structural steel W10 × 60 wide-flange section [ E = E = 29,000 2 ksi; A ksi; A = = 17.6 in ]. The roof and floor subject the column to the axial forces shown. Determine: (a) the amount that the first floor will settle. (b) the amount that the roof will settle.
Fig. P5.6
Solution (a) Draw a FBD that cuts through member (1) and includes the free end of the column. From this FBD, the sum of forces in the vertical direction can be written as Σ F y = −150 kips − 180 kips − F 1 = 0 Therefore, the internal axial force in member (1) is F 1 = − 330 kips = 330 kips (C) (C) The settlement of the first floor is determined by the elongation (i.e., contraction in this case) that occurs in member (1). F L u B = e1 = 1 1 A1 E 1 =
(−330 kips) kips)(14 (14 ft)(12 ft)(12 in./ft in./ft)) (17.6 (17.6 in.2 )(29, )(29, 000 ksi) ksi)
10862 in. in. = 0.1086 086 in. in.↓ = −0.108 (b) Draw a FBD that cuts through member (2) and includes the free end of the column. From this FBD, the equilibrium equation is Σ F x = −150 kips − F 2 = 0 Therefore, the internal axial force in member (2) is F 2 = − 150 150 ki kips = 150 kip kipss (C (C) The elongation in member (2) (which will be contraction in this instance) can be computed as e2 =
F2 L2 A2 E 2
=
(−150 kips)( kips)(12 12 ft)( ft)(12 12 in./f in./ft) t) (17.6 (17.6 in.2 )(29, )(29, 000 ksi) ksi)
= −0.04232 in.
The settlement of the roof is found from the sum of the contractions in the two members: uC = e1 + e2 = −0.10862 in. + (− 0.04232 in.) = − 0.1509 in. = 0.1509 in.↓
Ans.
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5.7 Aluminum [ E E = 70 GPa] member ABC supports a load of 28 kN, as shown in Fig. P5.7. Determine: (a) the value of load P load P such such that the deflection of joint C is zero. (b) the corresponding deflection of joint B joint B..
Fig. P5.7
Solution Cut a FBD that exposes the internal axial force in member (2): Σ F y = 28 kN − F2 = 0 ∴ F 2 = 28 kN Similarly, cut a FBD that exposes expo ses the internal axial force in member (1): Σ F y = 28 kN − P − F1 = 0 ∴ F1 = 28 kN − P The deflection at joint C , which must ultimately equal zero, can be expressed in terms of the member elongations e1 and e2: F L FL uC = e1 + e2 = 1 1 + 2 2 = 0 A1 E1 A2 E 2 or uC = =
F1 L1 A1 E1
+
F2 L2 A2 E 2
(28, 00 000 N − P )(1,000 mm) (28, 00 000 N)(1,300 mm) + =0 π π 2 2 2 2 (32 mm) (70, 00 000 N/mm ) (50 mm) (70, 00 000 N/mm ) 4 4
The only unknown in this equation is P is P , which can be computed as P = kN = 80.5841 kN = 80.6 kN The corresponding deflection at joint B joint B can be found from the elongation in member (2): F L (28 (28,000 N − 80,584 80,584 N)( N)(1, 1,30 300 0 mm mm) u B = e1 = 1 1 = = −0.497 mm = 0.497 mm ↓ π A1 E 1 2 2 (50 mm) (70 (70,000 ,000 N/m N/mm m ) 4
Ans.
Ans.
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5.8 A solid brass [ E = 100 GPa] axial member is loaded and supported as shown in Fig. P5.8. Segments (1) and (2) each have a diameter of 30 mm and segment (3) has a diameter of 16 mm. Determine: (a) the elongation of segment (2). (b) the deflection of joint D joint D with with respect to the fixed support at A at A.. (c) the maximum normal stress in the entire axial member.
Fig. P5.8
Solution (a) Draw a FBD that cuts through segment (2) and includes the free end of the axial member. From this FBD, the sum of forces in the vertical direction reveals the internal force in the segment: Σ F y = F2 − 14 kN − 28 kN = 0 ∴ F 2 = 42 kN (T) The cross-sectional area of the 30-mm-diameter segment is π A2 = (30 mm)2 = 706.858 mm2 4 The elongation in segment (2) is thus F L (42,000 N)(1,200 mm) e2 = 2 2 = A2 E 2 (706 (706.85 .858 8 mm2 )(10 )(100,000 0,000 N/mm N/mm2 ) mm = 0.71301 mm = 0.713 mm (b) The internal forces in segments (1) and (3) must be determined at the outset. From a FBD that cuts through segment (1) and includes the free end of the axial member: Σ F y = F1 − 40 kN − 14 kN − 28 kN = 0 ∴ F 1 = 82 kN (T) The elongation in segment (1) can be computed as F L (82,000 N)(1,800 mm) e1 = 1 1 = = 2.08811 mm (706.85 .858 8 mm2 )(10 )(100,000 0,000 N/mm N/mm2 ) A1 E 1 (706
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Similarly, consider a FBD that cuts through segment (3) and includes the free end of the axial member: Σ F y = F3 − 28 kN = 0 ∴ F 3 = 28 kN (T) The cross-sectional area of the 16-mm-diameter segment is π A3 = (16 mm mm)2 = 201.062 mm mm2 4 The elongation in segment (3) can be computed as F L (28,000 N)(1,600 mm) e3 = 3 3 = = 2.22817 mm A3 E 3 (201 (201.0 .062 62 mm2 )(10 )(100,000 0,000 N/mm N/mm2 ) The deflection of joint D joint D with with respect to the fixed support at A at A is is found from the sum of the three segment elongations: u D = e1 + e2 + e3 = 2.08811 mm + 0.71301 mm + 2.22817 mm mm = 5.02929 mm mm = 5.03 mm
Ans.
(c) Since segments (1) and (2) have hav e the same cross-sectional area, the maximum normal no rmal stress in these two segments occurs where the axial force is greater; that is, in segment (1): F 82,000 N σ 1 = 1 = = 116.0 MPa (T) A1 706.858 mm2 The normal stress in segment (3) is F 28, 28, 000 N = 139.3 MPa (T) σ 3 = 3 = A3 201.062 mm2 Therefore, the maximum normal stress in the axial member occurs in segment (3):
σ max = 139.3 MPa (T)
Ans.
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5.9 A hollow steel [ E E = = 30,000 ksi] tube (1) with an outside outside diameter diameter of 2.75 in. and a wall wall thickness thickness o 0.25 in. is fastened to a solid aluminum [ E = E = 10,000 ksi] rod (2) that has a 2-in.-diameter and a solid 1.375-in.-diameter aluminum rod (3). The bar is loaded as shown in Fig. P5.9. Determine: (a) the change in length of steel tube (1). (b) the deflection of joint D joint D with with respect to the fixed support at A at A.. (c) the maximum normal stress in the entire axial assembly.
Fig. P5.9
Solution Before proceeding, it is convenient to determine the internal forces in each of the three axial segments. Segment (1): Draw a FBD that cuts through segment (1) and includes the free end of the axial member. From this FBD, the sum of forces in the the horizontal direction gives the force in segment (1): Σ F x = − F 1 − 2(22 kips) + 2(13 kips) − 30 kips = 0 ∴ F 1 = − 48 kips Segment (2): Draw a FBD that cuts through segment (2) and includes the free end of the axial member. From this FBD, the sum of forces in the the horizontal direction gives the force in segment (2): Σ F x = − F 2 + 2(13 kips) − 30 kips = 0
∴ F 2 = − 4 kips Segment (3): Draw a FBD that cuts through segment (3) and includes the free end of the axial member. From this FBD, the sum of forces in the the horizontal direction gives the force in segment (3): Σ F x = − F 3 − 30 kips = 0
∴ F 3 = − 30 kips (a) The cross-sectional area of the hollow steel tube (inside diameter = 2.25 in.) is π A1 = ⎡⎣(2.75 in.)2 − (2.25 in.)2 ⎤⎦ = 1.963495 in.2 4 The elongation in segment (1) is thus F L (−48 kip kips) s)((42 in. in.)) = −0.034225 in. = − 0.0342 in. e1 = 1 1 = A1 E 1 (1.9634 (1.963495 95 in. in.2 )(30,000 )(30,000 ksi)
Ans.
(b) For segment (2), the cross-sectional area of the 2-in.-diameter aluminum rod is π A2 = (2 in.)2 = 3.141593 in.2 4 The elongation in segment (2) can be computed 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 translation of this work beyond that that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
e2 =
F2 L2 A2 E 2
=
(−4 kips kips)( )(30 30 in.) in.) (3.1415 (3.141593 93 in.2 )(10,000 )(10,000 ksi)
0.003820 in. = −0.003820
For segment (3), the cross-sectional area of the 1.375-in.-diameter aluminum rod is π A3 = (1.3 1.375 in. in.))2 = 1.48489 4893 in in.2 4 The elongation in segment (3) can be computed as F L (−30 kip kips) s)(2 (24 4 in.) in.) = −0.048488 in. e3 = 3 3 = A3 E 3 (1.484 (1.484893 893 in.2 )(10,000 )(10,000 ksi) The deflection of joint D joint D with with respect to the fixed support at A at A is is found from the sum of the three segment elongations: u D = e1 + e2 + e3 = −0.03 0.0342 4225 25 in. in. − 0.00 0.0038 3820 20 in. in.− 0.04 0.0484 8488 88 in. = −0.086533 in. = − 0.0865 in. = 0.0865 in. ←
Ans.
(c) Compute the normal stress in each of the three segments: F 48 kips σ 1 = 1 = = 24.4462 ksi (C) A1 1.963495 in.2
σ 2 = σ 3 =
F 2
=
A2 F 3 A3
=
4 kips 3.141593 in.2 30 kips 1.484893 in.2
= 1.2732 ksi (C) = 20.2035 ksi (C)
Therefore, the maximum normal stress in the axial member occurs in segment (1):
σ max = 24.4 ksi (C)
Ans.
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5.10 A solid 5/8-in. steel [ E = 29,000 ksi] rod (1) supports beam AB, AB, as shown in Fig. P5.10. If the stress in the rod must not exceed 30 ksi and the maximum elongation in the rod must not exceed 0.25 in., determine the maximum load P load P that may be supported.
Fig. P5.10
Solution The area of the 5/8-in.-diameter rod is π A1 = (0.625 in in.)2 = 0.306796 in in.2 4 If the stress in the rod must not exceed 30 ksi, then the maximum force that can be applied to rod (1) is (30 ksi) ksi)(0 (0.3 .306 0679 796 6 in. in.2 ) = 9.20 9.2038 3880 80 kips kips F1 ≤ σ allow,1 A1 = (30 The length of rod (1) is L1 = (20 ft)2 + (16 ft) 2 = 25.612497 ft = 307.35 in. If the maximum elongation in the rod must not exceed 0.25 in., the maximum internal force in the rod must be limited to e1 A1 E 1 (0.25 (0.25 in.)(0.3 in.)(0.30679 06796 6 in.2 )(29,000 )(29,000 ksi) ksi) ≤ = = 7.236932 kips F 1 L1 307.35 in. Therefore, the maximum internal force in rod (1) must not exceed 7.237 kips. Consider a FBD of the rigid rigid beam. Rod (1) is a twoforce member at an orientation θ defined defined by: 16 ft tan θ = = 0 .8 ∴θ = 38.660° 20 ft Write the equilibrium equation for the sum of moments about A about A to to find the relationship between F between F 1 and P and P : 38.660°) − (8 ft)P = 0 Σ M A = (20 ft)(F1 sin 38 ∴ P =
(20 ft)(7.236932 ft)(7.236932 kips) kips) sin 38.660° 8 ft
11.302 0223 236 6 kip kipss = 11.3 11.30 0 kip kipss = 11.3
Ans.
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5.11 A control rod in a mechanism must stretch 5 mm when a load of 5 kN is applied. The rod is is made from an aluminum alloy that has a yield strength of σ Y of E = 73 Y = 330 MPa and an elastic modulus of E GPa. If a factor of safety of 2 with respect to yield is specified, determine the diameter and length of the lightest rod that can be used.
Solution The allowable normal stress is σ 330 MPa σ allow = Y = = 165 MPa FS 2 The minimum cross-sectional required for the control rod is thus (5 kN)(1,000 N/kN) = 30.303030 mm2 A ≥ 2 165 N/mm and the diameter is π 2 D ≥ 30.303030 mm2 ∴ D ≥ 6.2115 mm = 6.21 mm 4
Ans.
Since the rod is required to be the lightest possible, the rod diameter is taken as D as D = = 6.21 mm ( A = A = 2 30.2882 mm ). The rod must stretch 5 mm when subjected to a 5 kN load; therefore, the length of the rod must be eAE (5 mm) mm)(3 (30.2 0.288 882 2 mm mm2 )(73 )(73,, 000 000 N/mm N/mm2 ) L = 211 mm = 2.21 m = = 2, 21 Ans. F (5 kN)(1,000 N/kN)
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3
5.12 A 1-in.-diameter by 16-ft-long cold-rolled bronze bar [ E E = = 15,000 ksi and γ = 0.320 lb/in ] hangs vertically while suspended from one end. Determine the change in length of the bar due to its own weight.
Solution An incremental length dy of dy of the bar has an incremental elongation given by F ( y ) de = dy AE The force in the bar can be expressed as the product of the unit density of the bronze (γ bronze) and the volume of the bar below the incremental slice dy: dy: F ( y ) = γ bronze A y Therefore, the incremental elongation can be expressed as γ γ A y y de = bronze dy = bronze dy AE E Integrate this expression over the entire length L length L of of the bar: L γ γ bronze L γ bronze 1 2 L γ bronze L2 bronze ⎡y ⎤ = e= dy = y dy = 0 E E 0 E 2 ⎣ ⎦0 2 E
∫
∫
The change in length of the bar due to its own weight is therefore: γ bronze L2 (0.3 (0.320 20 lb/ lb/in in..3 )(16 )(16 ft ft × 12 in./ in./ft ft))2 = = 0.00 e= 0.0003 0393 9321 216 6 in. in. = 0.00 0.0003 0393 93 in. in. 2 E 2(15, 000, 000 psi)
Ans.
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5.13 Determine the extension, due to its own weight, o the conical bar shown in Fig. P5.13. The bar is made o 3 aluminum alloy [ E E = 10,600 ksi and γ = 0.100 lb/in. ]. The bar has a 2-in. radius at its upper end and a length o L = L = 20 ft. Assume the taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.
Fig. P5.13
Solution An incremental length dy of dy of the bar has an incremental elongation given by F ( y ) de = dy AE The force in the bar can be expressed as the product of the unit density of the aluminum (γ alum) and the volume of the bar below the incremental slice dy. dy. The volume below the slice dy i dy iss a cone. At y At y,, the cross-sectional area of the base of the cone is:
⎛ y ⎞ A y = π ⎜ r ⎟ ⎝ L ⎠
2
and the volume of a cone is given by: 1 V = (area of base)(altitude) 3 Therefore, the internal force at y at y can can be expressed as 1 F ( y) = γ alum × A y y 3 Therefore, the incremental elongation can be expressed as γ y alum A y y de = dy = alum dy 3 A y E 3E Integrate this expression over the entire length L length L of of the bar: L γ γ alum L γ alum 1 2 L γ alum L2 alum y ⎡y ⎤ = e= dy = y dx = 0 3 E 3E 0 3E 2 ⎣ ⎦ 0 6 E
∫
∫
The change in length of the bar due to its own weight is therefore: γ alum L2 (0.1 (0.100 00 lb/ lb/in in..3 )(20 )(20 ft ft × 12 in. in./f /ft) t)2 5 6 e= = = 9.0566 × 10− in. = 90.6 × 10− in. 6 E 6(10, 600, 000 psi)
Ans.
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