5.14 Rigid bar ABCD bar ABCD is is loaded and supported as shown in Fig. P5.14. Steel [ E E = = 30,000 ksi] bars (1) and (2) are unstressed before the load P is applied. applied. Bar (1) has a cross-secti cross-sectional onal area o 2 0.625 in. and bar (2) has a cross-sectional area 2 of 1.25 in. . After load P load P is is applied, the strain in bar (2) is found to be 900 µε. Determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point D point D.. (c) the load P load P .
Fig. P5.14
Solution From the strain in bar (2), the elongation in bar (2) is e2 = ε 2 L2 = (900 × 10−6 in./in.)(75 in.) = 0.067500 in in. Since the joint at C is is a perfect connection, the rigid bar deflection at C must must equal the elongation of bar (2): vC = e2 = 0.067500 in. ↓ From a deformation diagram of the rigid bar, the vertical deflection of joint B joint B is is related to C by similar triangles: v B vC 36 in.
=
72 in. 36 in. 1 ∴ v B = vC = vC 72 in. 2 1 (0.067 675 5 in.) in.) = 0.03 0.0337 3750 50 in. in.↓ = (0.0 2 The joint at B at B is is also a perfect connection; therefore, the downward displacement of B of B also also causes an equal elongation in bar (1): e1 = v B = 0.033750 in. (a) Now that the elongations in both bars are known, the normal stresses in each can be computed. The normal stress in bar (1) is σ L F L e E (0.033750 in.)(30,000 ksi) Ans. e1 = 1 1 = 1 1 ∴σ 1 = 1 1 = = 20.25 ksi (T) A1 E1 E1 L1 50 in. and the normal stress in bar (2) is σ L F L e E ∴ σ 2 = 2 2 e2 = 2 2 = 2 2 A2 E2 E2 L2
=
(0.067500 in.)(30,000 ksi) 75 in.
= 27.0 ksi (T)
Ans.
Alternatively, this stress could be calculated from Hooke’s Law: 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
,000 ksi)(900 × 10−6 in./in.) = 27.0 ksi (T) = E 2ε 2 = (30,000
(b) From a deformation diagram of the rigid bar, the vertical deflection of o f joint D joint D is is related to B to B and and C by similar triangles: v B vC vD 36 in.
=
72 in.
=
∴ v D = vC
96 in. 96 in. 4
4
72 in.
3
0.0675 in.) = 0.090 0900 in in. ↓ = vC = (0.0 3
Ans.
(c) The force in each bar can be determined from the stresses computed previously: F1 = σ 1 A1 = (20. (20.25 25 ksi ksi))(0.6 (0.625 25 in. in.2 ) = 12.6 12.652 5250 50 kip kipss F2
(27.0 0 ksi) ksi)((1.25 1.25 in. in.2 ) = 33.7 33.750 5000 00 ki kips = σ 2 A2 = (27.
Consider a FBD of the rigid bar. The equilibrium equation for the sum of moments about A about A can can be used to determine the load P load P : Σ M A = (36 in.)F1 + (72 in.)F2 − (96 in.)P = 0
∴ P =
(36 in.) in.)(12 (12.652 .65250 50 kips kips)) + (72 in.)( in.)(33.7 33.750 50 kips kips)) 96 in.
057 ki kips = 30. 30.1 kip kips = 30.057
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.
5.15 Rigid bar ABCD ABCD is loaded and supported as shown in Fig. P5.15. Bars (1) and (2) are unstressed before the load P load P is is applied. Bar (1) is made of bronze [ E E = 100 GPa] and has a cross2 sectional area of 400 mm . Bar Bar (2) (2) is made made o aluminum [ E E = = 70 GPa] and has a cross-sectional 2 area of 600 mm . After the load P is is applied, the force in bar (1) is found to be 60 kN (in compression). Determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point A point A.. (c) the load P load P .
Fig. P5.15
Solution Given that the axial force in bar (1) is 60 kN (in compression), the elongation (i.e., contraction in this case) can be computed as: F L ( −60,000 60,000 N)( N)(84 840 0 mm mm) e1 = 1 1 = = −1.260 mm A1 E 1 (400 (400 mm 2 )(10 )(100,00 0,000 0 N/m N/mm m2 ) Since the pin at B at B is is a perfect connection, the deflection of the rigid bar at B at B is is equal to the contraction of bar (1): v B = e1 = 1.260 mm ↓ From a deformation diagram of the rigid bar, the vertical deflection of joint B joint B is is related to C by similar triangles: v B vC 3m
=
1m 1m 1 ∴ vC = vB = vB 3m 3 1 mm) = 0.420 420 mm mm ↓ = (1.260 mm 3 The joint at C is is also a perfect connection; therefore, the downward displacement of C also also causes an equal elongation in bar (2): e2 = vC = 0.420 mm Note that a downward displacement at C causes causes elongation (and hence, tension) in bar (2). (a) The normal stress in bronze bar (1) can be computed from the known force in the bar: F 1 −60,000 N σ 1 = = = −150.0 0.0 MP MPa = 150 150.0 MPa MPa (C) A1 400 mm 2
Ans.
The normal stress in aluminum bar (2) can be computed from its elongation:
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e2
=
F2 L2 A2 E2
∴σ 2 =
=
σ 2 L2
E 2
e2 E 2 L2
=
(0.420 (0.420 mm)( mm)(70,0 70,000 00 N/mm N/mm2 ) 920 mm
31.95652 6522 MPa MPa = 32.0 MP MPa (T) (T) = 31.
Ans.
(b) From a deformation diagram of the rigid bar, the vertical deflection of joint A joint A is is related to joint B joint B by by similar triangles: v A vB 4m
=
3m
∴ v A = vB
4m 3m
4
4
3
3
1.680 mm mm ↓ = vB = (1.260 mm) = 1.6
Ans.
(c) The force in bar (1) is given as F as F 1 = −60 kN. The force in bar (2) can be determined determined from the stress computed previously: N/mm2 )(600 mm mm2 ) = 19,174 N = 19.174 kN kN F2 = σ 2 A2 = (31.956522 N/
Consider a FBD of the rigid bar. The equilibrium equation for the sum of moments about D about D can can be used to determine the load P load P : Σ M D = (3 m)F1 − (1 m)F2 + (4 m)P = 0
∴ P =
(1 m)(19.1 9.174 kN) kN) − (3 m)( − 60 kN kN) 4m
= 49.7935 kN = 49.8 kN
Ans.
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5.16 In Fig. P5.16, bronze [ E E = = 100 GPa] links (1) and (2) support rigid beam ABC . Link (1) has a 2 cross-sectional area of 300 mm and link (2) has a 2 cross-sectional area of 450 mm . For an applied load of P = 70 kN, determine the rigid beam deflection at point B point B..
Fig. P5.16
Solution From a FBD of the rigid beam, write two equilibrium equations: Σ F y = F1 + F 2 − 70 kN = 0
(a)
460 mm)F 2 − (540 mm)(70 kN) = 0 Σ M A = (1, 46
(b)
Solve Eq. (b) for F for F 2: (540 mm)(70 kN) F 2 = = 25.8904 kN 1, 460 mm and backsubstitute into Eq. (a) to obtain F obtain F 1: F1 = 70 kN − F 2 = 70 kN − 25.8904 kN = 44.1096 kN Next, determine the elongations in links (1) and (2): F L (44,109.6 N)(2,000 mm) e1 = 1 1 = = 2.9406 mm A1 E 1 (300 (300 mm 2 )(10 )(100,00 0,000 0 N/m N/mm m2 ) e2
=
F2 L2 A2 E 2
=
(25,890.4 N)(3,000 mm) (450 (450 mm2 )(100,00 100,000 0 N/m N/mm m2 )
= 1.7260 mm
Since the connections at A at A and and C are are perfect, the rigid beam deflections at these joints are equal to the elongations of links (1) and (2), respectively: v A = e1 = 2.9406 mm and vC = e2 = 1.7260 mm
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The deflection at B at B can can be determined from similar triangles: v A − vC vB − vC 1, 46 460 mm
=
920 mm
Solve this expression for v B: 920 mm (vA − vC ) + vC v B = 1,460 mm
=
920 mm 1,460 mm
(2. (2.9406 mm mm − 1.7260 260 mm) mm) + 1.7 1.7260 mm mm
(0.630137)( 37)(1.21 1.2146 46 mm) mm) + 1.7260 1.7260 mm = (0.6301 = 2.49 2.49 mm ↓
Ans.
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5.17 Rigid bar ABC bar ABC is is supported by bronze rod (1) and aluminum rod (2), as shown in Fig P5-17. A concentrated load P load P is is applied to the free end of aluminum rod (3). Bronze rod (1) has an elastic modulus of E 1 = 15,000 ksi and a diameter of D1 = 0.375 in. Aluminum rod (2) has an elastic modulus of E of E 2 = 10,000 ksi and a diameter of D of D2 = 0.625 in. Aluminum Aluminum rod rod (3) has a diameter diameter o D3 = 1.0 in. The yield strength of the bronze is 50 ksi and the yield strength of the aluminum is 36 ksi. (a) Determine the magnitude of load P load P that that can safely be applied to the structure if a minimum factor of safety of 1 .5 is required. (b) Determine the deflection of point D for the load determined in part (a). (c) The pin used at B at B has has an ultimate shear strength of 70 ksi. If a factor of safety of 3.0 is required for this double shear pin connection, determine the minimum pin diameter that can be used at B at B..
Fig. P5.17
Solution Before beginning, the cross-sectional areas of the three rods can be calculated from the specified diameters: A1 = 0.110447 in.2 A2 = 0.306796 in.2 A3 = 0.785398 in.2 The allowable stress of the bronze is σ Y ,bronze 50 ksi σ allow,1 = = = 33.3333 ksi FS 1. 5 and the allowable stress of the aluminum is σ Y ,alum 36 ksi σ allow,2 = σ allow,3 = = = 24 ksi FS 1 .5 (a) From a FBD cut through throug h rod (3), equilibrium requires that the internal force in rod (3) is F is F 3 = P = P . From a FBD of the rigid bar, write two equilibrium equations: Σ F y = F1 + F2 − F3 = F1 + F2 − P = 0
Σ M A = (4 ft)F2 − (2.5 ft)F3 = (4 ft)F2 − (2.5 ft)P = 0
(a) (b)
Solve Eq. (b) for F for F 2 in terms of the unknown load P load P : 2.5 ft F2 = P = 0.6250 P (c) 4 ft Backsubstitute this expression into Eq. (a) to obtain F obtain F 1 in terms of the unknown load P load P : F1 = P − F2 = P − 0.6250 P = 0.3750P (d)
Rearrange Eqs. (c) and (d) to express P express P in in terms of F of F 2 and F and F 1, respectively: 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.
P = 1.6000 F 2
(e)
P = 2.6667 F 1
(f)
Based on the allowable stress for the bronze, the maximum allowable force that can be supported by rod (1) is: F1 ≤ σ allow,1 A1 = (33. (33.33 3333 33 ksi ksi)( )(0. 0.11 1104 0447 47 in. in.2 ) = 3.68 3.6816 16 kip kipss Thus, from Eq. (f), the corresponding maximum load P load P is: is: P ≤ 2.6667 F 1 = 2.6667(3.6816 kips) = 9.8177 kips
(g)
Next, the maximum allowable force that can be supported by rod (2) based on the allowable stress for the aluminum is: F2 ≤ σ allow,2 A2 = (24 (24 ksi ksi))(0.3 (0.306 0679 796 6 in. in.2 ) = 7.36 7.3631 31 kips kips which leads to a corresponding maximum load P load P from from Eq. (e): P ≤ 1.6000F 2 = 1.6000(7.3631 kips) = 11.7810 kips
(h)
Finally, we should also check the capacity of rod (3): F3 ≤ σ allow,3 A3 = (24 (24 ksi) ksi)(0 (0.7 .785 8539 398 8 in. in.2 ) = 18.8 18.849 496 6 kips kips which means that P ≤ ≤ 18.8496 kips
(i)
From a comparison of Eqs. (g), (h), and (i), the maximum load that may be applied to the structure is P max
kips = 9.82 ki kips = 9.8177 ki
Ans.
Based on this maximum load P load P , the internal forces in the three rods are: F1 = 0.3750P = 0.3750(9.8177 kip kips) = 3.6816 kip kips
= 0.6250P = 0.6250(9.8177 kips) = 6.1361 kips F3 = P = 9.8177 kips F2
(b) Next, determine the elongations in rods (1), (2), and (3): F L (3.6816 kips)(6 ft)(12 in./ft) e1 = 1 1 = = 0.1600 in. A1 E 1 (0.110 (0.110447 447 in.2 )(15 )(15,, 000 ksi) e2
=
e3
=
F2 L2 A2 E 2 F3 L3 A3 E 3
=
(6.1361 kips)(8 ft)(12 in./ft)
=
(9.8177 kips)(3 ft)(12 in./ft)
(0.3067 (0.306796 96 in.2 )(10,000 )(10,000 ksi) (0.785 (0.785398 398 in. in.2 )(10,000 )(10,000 ksi)
= 0.1920 in. = 0.0450 in.
Since the connections at A at A and and C are are perfect, the rigid bar deflections at these joints are equal to the elongations of rods (1) and (2), respectively: v A = e1 = 0.1600 in. and vC = e2 = 0.1920 in.
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The rigid bar deflection at B at B can can be determined from similar triangles: vC − v A v B − vA 4 ft
=
2.5 ft
Solve this expression for v B: 2.5 ft v B = (vC − vA ) + vA 4 ft 2.5 ft (0.19 .1920 in. in. − 0.16 .1600 in. in.) + 0.160 1600 in. in. = 4 ft (0.6250 50)( )(0. 0.03 0320 20 in. in.)) + 0.16 0.1600 00 in. in. = (0.62
= 0.1800 in. The deflection of joint D joint D is is equal to the deflection of the rigid bar at B at B plus plus the elongation in rod (3): v D
in. + 0.0450 in. = 0.2250 in. = 0.225 in.↓ = vB + e3 = 0.1800 in
Ans.
(c) The pin at B at B has has an allowable shear stress of τ ult 70 ksi τ allow = = = 23.3333 ksi FS 3 .0 The force tending to shear this pin is equal to the load P load P = = 9.8177 kips. The shear area A area AV required required for this pin is thus V 9.8177 kips AV ≥ = = 0.4208 in.2 23.3333 ksi τ allow Since the pin is used in a double shear connection, the shear area is equal to twice the cross-sectional area of the pin: AV
π
= 2 A in = 2 × Dp2in
4 and so the minimum pin diameter is π
2 2 × D pin 4
≥ 0.4208 in.2
∴ Dpin ≥ 0.5176 in. = 0.518 in.
Ans.
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5.18 Solve problem 5.14 when there is a clearance of 0.05 in. in the pin connection at C . 5.14 (Repeated here for convenience). Rigid bar ABCD ABCD is loaded and supported as shown in Fig. P5.14. Steel [ E E = = 30,000 ksi] bars (1) and (2) are unstressed before the load P is applied. Bar (1) 2 has a cross-sectional area of 0.625 in. and bar (2) 2 has a cross-sectional area of 1.25 in. . After load P load P is applied, the strain in bar (2) is found to be 900 µε. Determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point D point D.. (c) the load P load P .
Fig. P5.14 (repeated)
Solution From the strain in bar (2), the elongation in bar (2) is e2 = ε 2 L2 = (900 × 10−6 in./in.)(75 in.) = 0.067500 in in. There is a 0.05-in. clearance in the connection at C ; therefore, vC = e2 + 0.05 0.05 in. in. = 0.11 0.1175 7500 00 in.↓ From a deformation diagram of the rigid bar, the vertical deflection of joint B joint B is is related to C by by similar triangles: v B vC 36 in.
=
72 in. 3 6 i n. 1 ∴ v B = vC = vC 72 in. 2 1 (0.117 175 5 in.) in.) = 0.05 0.0587 875 5 in. in. ↓ = (0.1 2 The joint at B at B is is a perfect connection; therefore, the downward displacement of B of B also also causes an equal elongation in bar (1): e1 = v B = 0.05875 in. (a) Now that the elongations in both bars are known, the normal stresses in each can be computed. The normal stress in bar (1) is σ L F L e E (0.05875 in.)(30,000 ksi) e1 = 1 1 = 1 1 ∴σ 1 = 1 1 = = 35.25 ksi (T) Ans. A1 E1 E1 L1 50 in. and the normal stress in bar (2) is σ L F L e E e2 = 2 2 = 2 2 ∴ σ 2 = 2 2 A2 E2 E2 L2
=
(0.067500 in.)(30,000 ksi) 75 in.
= 27.0 ksi (T)
Ans.
Alternatively, this stress could be calculated from Hooke’s Law: σ2
,000 ksi)(900 × 10−6 in./in.) = 27.0 ksi (T) = E 2ε 2 = (30,000
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(b) From a deformation diagram of the rigid bar, the vertical deflection of joint D joint D is is related to B to B and and C by similar triangles: v B vC vD 36 in.
=
72 in.
=
∴ v D = vC
96 in. 96 in. 4
4
72 in.
3
0.1175 in.) = 0.156 1567 in in. ↓ = vC = (0.1 3
Ans.
(c) The force in each bar can be determined from the stresses computed previously: F1 = σ 1 A1 = (35. (35.25 25 ksi) ksi)(0 (0.6 .625 25 in. in.2 ) = 22.0 22.031 3125 25 kips kips F2
(27.0 0 ksi) ksi)((1.25 1.25 in. in.2 ) = 33.7 33.750 5000 00 kip kipss = σ 2 A2 = (27.
Consider a FBD of the rigid bar. The equilibrium equation for the sum of moments about A about A can can be used to determine the load P load P : Σ M A = (36 in.)F1 + (72 in.)F2 − (96 in.)P = 0
∴ P =
(36 in.) in.)(22 (22.031 .03125 25 kips kips)) + (72 in.)( in.)(33. 33.750 750 kips) kips) 96 in.
5742 kips = 33.6 kip kips = 33.574
Ans.
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5.19 The rigid beam in Fig. P5.19 is supported by links (1) and (2), which are made from a polymer material [ E E = 16 GPa]. Link (1) has a cross2 sectional area of 400 mm and link (2) has a cross2 sectional area of 800 mm . Determine the maximum load P that may by applied if the deflection of the rigid beam is not to exceed 20 mm at point C .
Fig. P5.19
Solution Equilibrium: Consider a FBD of the rigid beam and assume tension in each link. (a) Σ F y = − F1 + F2 − P = 0
Σ M A = (600 mm)F2 − (900 mm)P = 0 From Eq. (b): 900 mm F2 = P = 1.5 P 600 mm Backsubstituting into Eq. (a): F1 = F2 − P = 1.5 P − P = 0.5P
(b)
(c) (d)
Force-deformation relationships: The relationship between internal force and member elongation for links (1) and (2) can be expressed as: F L F L e1 = 1 1 and e2 = 2 2 (e) A1 E 1 A2 E 2 Geometry of deformations: Consider a deformation diagram of the rigid beam. Using similar similar triangles, one way to express the relationships between v A, v B, and vC is: v A + vC v −v (f) = C B 900 mm 300 mm Note: Here, v A, v B, and vC are treated as unsigned magnitudes in the directions shown on the deformation diagram.
The rigid beam deflection at A at A will will equal the elongation that occurs in link (1): 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.
v A
= e1
and similarly at B at B:: v B = e2 Equation (f) can now be rewritten in terms of e1 and e2 as: e1 + vC vC − e2 900 mm
=
300 mm
or e1 + vC
=
900 mm
300 mm Solving for vC gives: 2vC = e1 + 3e2
(vC
− e2 ) = 3vC − 3e2 ∴ vC = 0.5e1 + 1.5e2
Substitute the force-deformation relationships from Eq. (e) to ob tain: 0.5 F1 L1 1.5F2 L2 vC = 0.5e1 + 1.5e2 = + A1 E1 A2 E 2 and then substitute Eqs. (c) and (d) to derive an expression for vC in terms of the unknown load P load P : vC =
0.5(0.5 P ) L1 A1 E1
+
1.5(1.5P ) L2 A2 E2
⎡ 0.25L1
= P ⎢
⎣ A1 E1
+
2.25L2 ⎤
⎥
A2 E 2 ⎦ The unknown load P load P is thus related to the rigid beam deflection at C by: by: vC P = 0.25 L1 2.25L2
+
A1 E1 A2 E 2 Substituting the appropriate values into this relationship gives the maximum load P load P that that may be applied to the rigid beam at C without without causing more than 20 mm deflection: 20 mm P = = (0.25)(1,000 mm) (2.25)(1, 25 250 mm) (400 mm mm2 )(16,000 ,000 N/ N/mm2 ) 283 N = 77.3 kN = 77, 28
+
(800 mm mm2 )(16,000 ,000 N/ N/mm2 ) Ans.
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5.20 Three aluminum [ E E = 10,000 ksi] bars are used to support the loads shown in Fig. P5.20. The elongation in each bar must be limited to 0.25 in. Determine the minimum cross-sectional area required for each bar.
Fig. P5.20
Solution Determine the inclination angles for bars (1), (2), and (3): 7 ft tan θ1 = ∴θ 1 = 34.992° 10 ft 3 ft tan θ 2 = ∴θ 2 = 18.435° 9 ft 8 ft tan θ3 = ∴θ 1 = 57.995° 5 ft The bar lengths are: L1
ft) 2 + (10 ft ft ) 2 = 12.206556 ft ft = 146.4787 in in. = (7 ft
L2
= (3 ft)2 + (9 ft ) 2 = 9.486833 ft = 113.8420 in.
L3
= (8 ft)2 + (5 ft ) 2 = 9.433981 ft = 113.2078 in.
joint B and and write two equilibrium Equilibrium: Consider a FBD of joint B equations: Σ F x = − F1 cos 34.992° + F 2 cos18.435° = 0 34.992° + F 2 sin 18.435° − 31 kips = 0 Σ F y = F1 sin 34 Solve these two equations simultaneously to obtain F obtain F 1 = 36.61967 kips and F and F 2 = 31.62278 kips. Next, consider a FBD of joint C and and write two additional equilibrium equations: Σ F x = − F2 cos18.435° + F 3 cos 57.995° = 0
Σ F y = − F2 sin 18.435° + F 3 sin 57.995° − 38 kips = 0 From this, the internal force in bar (3) is F is F 3 = 56.60389 kips. The minimum cross-sectional area required for eac h bar is: F L (36.61967 kips)(146.4787 in.) A1 ≥ 1 1 = = 2.15 in.2 e1 E 1 (0.25 in.)(10,000 ksi)
Ans.
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A2
≥
A3
≥
F2 L2 e2 E 2 F3 L3 e3 E 3
=
(31.62278 kips)(113.8420 in.)
=
(56.60389 kips)(113.2078 in.)
(0.25 in.)(10,000 ksi) (0.25 in.)(10,000 ksi)
= 1.440 in.2
Ans.
= 2.56 in.2
Ans.
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5.21 A tie rod (1) and a pipe strut (2) are used to support an 80-kip load as shown in Fig. P5.21. Pipe strut (2) has an outside diameter of 6.625 in. and a wall thickness of 0.280 in. Both the tie rod and the pipe strut are made of structural steel with a modulus of elasticity of E of E = = 29,000 ksi and a yield strengt strength ho tie rod, the minimum minimum factor factor o σ Y = 36 ksi. For the tie safety with respect to yield is 1.5 and the maximum allowable axial elongation is 0.20 in. (a) Determine the minimum diameter D required to satisfy both constraints for tie rod (1). (b) Draw a deformation diagram showing the final position of joint B joint B..
Fig. P5.21
Solution The angles of inclination for members (1) and (2) are: 12 ft ∴θ 1 = 26.565° tan θ1 = 24 ft 30 ft tan θ 2 = ∴θ 2 = 51.340° 24 ft The member lengths are: L1
= (12 ft)2 + (24 ft )2 = 26.8328 ft = 321.9938 in.
L2
= (30 ft)2 + (24 ftft ) 2 = 38.4187 ftft = 461.0249 in.
joint B and and write two equilibrium equ ilibrium (a) Equilibrium: Consider a FBD of joint B equations: Σ F x = − F1 cos 26.565° − F 2 cos 51.340° = 0 26.565° − F 2 sin 51.340° − 80 kips = 0 Σ F y = F1 sin 26 Solve these two equations simultaneously to obtain F obtain F 1 = 51.1103 kips and F 2 = −73.1786 kips. The allowable normal stress for tie rod (1) is σ Y 36 ksi σ allow = = = 24 ksi FS 1 .5 The minimum cross-sectional area required to satisfy the normal stress requirement is F 1 51.1103 kips A ≥ = = 2.1296 in.2 σ allow 24 ksi The maximum allowable axial elongation for the tie tie rod is 0.20 in. The minimum cross-sectional area required to satisfy the deformation requirement is F L (51.1103 kips)(321.9938 in.) A ≥ 1 1 = = 2.8374 in.2 e1 E 1 (0.20 in.)(29,000 ksi)
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To satisfy both the stress and deformation requirements, the tie rod must have a minimum cross2 sectional area of A of A = = 2.8374 in. . The corresponding rod diameter is π
4
Dr2od
≥ 2.8374 in.2
Drod
≥ 1.901 in.
Ans. 2
(b) The pipe has a cross-sectional area of A of A2 = 5.5814 in. . The pipe elongation is F L ( −73.1786 73.1786 kips kips)( )(461 461.024 .0249 9 in.) in.) e2 = 2 2 = = −0.2084 in. 2 A2 E 2 (5.581 (5.5814 4 in. in. )(29,000 )(29,000 ksi) ksi) Rod (1) elongates 0.20 in. and pipe (2) contracts 0.2084 in. Therefore, the deformation diagram showing the final position of joint B joint B is is shown.
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5.22 Two axial members are used to support a load of P of P = 72 kips as shown in Fig. P5.22. Member (1) is 12-ft long, it has a 2 cross-sectional area of A of A1 = 1.75 in. , and it is made of structural steel [ E E = = 29,000 ksi]. Member (2) is 16-ft long, it has a cross2 sectional area of A of A2 = 4.50 in. , and it is made of an aluminum alloy [ E E = = 10,000 ksi]. (a) Compute the normal stress in each axial member. (b) Compute the elongation of each axial member. (c) Draw a deformation deformation diagram diagram showing showing the final position position o joint B joint B.. (d) Compute the horizontal and vertical displacements of joint B joint B..
Fig. P5.22
Solution (a) Consider a FBD of joint B joint B and and write two equilibrium equations: Σ F x = − F1 + F 2 cos 55° = 0
Σ F y = F 2 sin 55° − 72 kips = 0 Solve these two equations simultaneously to obtain F obtain F 1 = 50.4149 kips and F and F 2 = 87.8958 kips. The normal stress in member (1) is: F 1 50.4149 kips σ 1 = = = 28.8 ksi A1 1.75 in.2
Ans.
and the normal stress in member (2) is: F 2 87.8958 kips σ 2 = = = 19.53 ksi A2 4.50 in.2 (b) The member elongations are F L (50.4149 kips)(12 ft)(12 in./ft) e1 = 1 1 = A1 E 1 (1.7 (1.75 5 in.2 )(29,000 )(29,000 ksi) ksi) e2
=
F2 L2 A2 E 2
=
(87.8958 kips)(16 ft)(12 in./ft) (4.50 (4.50 in. in.2 )(10,000 )(10,000 ksi)
Ans.
= 0.1430 in.
Ans.
= 0.375 in.
Ans.
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(c) Member (1) elongates 0.1430 in. in. and member (2) elongates 0.375 in. Therefore, the deformation diagram showing the final position of joint B joint B is is shown.
(d) The horizontal displacement of B is B is
∆ x = 0.1430 in.
Ans.
The vertical displacement is found from 0.37 0.375 5 in. in. + 0.14 0.1430c 30cos5 os55 5° sin55° = ∆ y
∴ ∆ y =
0.375 in. + 0.1430 co cos 55 55° sin 55°
=
0.375 in.+ 0.0820 in in. sin 55°
= 0.558 in. ↓
Ans.
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