6.94 A fillet with a radius of 0.15 in. is used at the junction in a stepped shaft where the diameter is reduced from 4.00 in. to 3.00 in. Determine the maximum shear stress in the fillet when the shaft is transmitting a torque of 4,000 lb-ft.
Solution 6.17b: Fillet: From Fig. 6.17b D 4.0 in. = = 1.33 d 3.0 in.
r d
=
0.15 in. 3.0 in.
=
0 .0 5
K t ≅ 2.0
∴
Section Properties:
I p
π =
32
D4
π =
32
(3.00 in in.)4
=
7.952156 in in.4
Maximum Shear Stress: TR (4,000 lb-ft)(3.00 in./2)(12 in./ft) τ = K t = ( 2.0 ) 7.952156 in.4 I p
=
18,108 psi = 18.11 ksi
Ans.
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6.95 A fillet with a radius of 12 mm is used at the junction in a stepped shaft where the diameter is reduced from 135 mm to 100 mm. Determine the maximum shear stress in the fillet when the shaft is transmitting a torque of 10 kN-m.
Solution 6.17b: Fillet: From Fig. 6.17b D 135 mm = = 1.3 5 d 100 mm
r d
=
12 mm 100 mm
=
0.12
K t ≅ 1.6
∴
Section Properties:
I p
π =
32
D4
π =
32
(100 mm)4
=
9,81 ,817,477 ,477 mm mm4
Maximum Shear Stress: TR (10 kN-m)(100 mm/2)(1,000 N/kN)(1,000 mm/m) τ = K t = (1.6) = 81.5 MPa 9,817,477 mm4 I p
Ans.
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6.96 A fillet with a radius of 1/8 in. is used at the junction in a stepped shaft where the diameter is reduced from 8.00 in. to 6.00 in. Determine the maximum torque that the shaft can transmit if the maximum shear stress in the fillet must be limited to 12 ksi.
Solution 6.17b: Fillet: From Fig. 6.17b D 8.0 in. = = 1.33 d 6.0 in.
r d
=
0.125 in. 6.0 in.
=
0.021
K t ≅ 2.6
∴
Section Properties:
I p
π =
32
D4
π =
32
(6. (6.00 in.)4
=
127 127.2345 345 in.4
Maximum Torque: TR τ = K t I p ∴
T =
I
τ p
K t R
=
(12 ksi)( ksi)(127.2345 in.4 ) (2.6)(6.00 in./2)
=
195.7 kip-in.
Ans.
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6.97 A stepped shaft has a 5-in. diameter for one-half of its length and a 4-in. diameter for the other half. If the maximum shear stress in the fillet between the two portions of the shaft must be limited to 12 ksi when the maximum shear stress in the 4-in. portion is 8 ksi, determine the minimum radius needed at the junction between the two portions of the shaft.
Solution Section Properties:
I p
π =
32
D4
π =
32
(4.0 in in.)4
=
25.13274 in in.4
6.17b: Fillet Requirement: From Fig. 6.17b τ 12 ksi K t = fillet = = 1.50 τ shaft 8 ksi D d
5.0 in.
= 1.25 4.0 in. r ∴ ≅ 0.135 d =
Minimum Fillet Radius: r r = d = (0.1 (0.135 35))(4.0 (4.0 in.) in.) = 0.54 0.54 in. in. d
Ans.
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6.98 The small portion of a stepped shaft has a diameter of 50 mm. The radius of the fillet at the junction between the large and small portions is 4.5 mm. If the maximum shear stress in the fillet between the two portions of the shaft must be limited to 40 MPa when the shaft is transmitting a torque of 614 N-m, determine the maximum diameter that can be b e used for the large portion of the shaft.
Solution Section Properties: π
I p
=
32
D4
π =
32
(50 mm)4
=
613, 59 592 mm4
Shear Stress in Shaft: TR (614 N-m)(50 N-m)(50 mm/2)(1,000 mm/m) τ =
I p
=
613, 613, 592 mm4
=
25.017 MPa
6.17b: Fillet Requirement: From Fig. 6.17b τ 40 MPa K t = fillet = = 1.60 25.017 MPa τ shaft r d
=
4.5 mm
=
50 mm D ∴ ≅ 1.20 d
0.09
Maximum Shaft Diameter: D Dmax = d = (1.20)(50 mm mm) = 60.0 0.0 mm mm d
Ans
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6.99 A 2-in.-diameter shaft contains a ½-in.-deep U-shaped groove that has a ¼-in. radius at the bottom of the groove. The shaft must transmit a torque of T = = 500 lb-in. If a factor of safety of 3.0 with respect to yield is specified, determine the minimum yield strength in shear required for the shaft material.
Solution 6.17a: Groove: From Fig. 6.17a h 0.5 in. r 0.25 in. = = 2.00 = = 0.25 r 0.25 in. d 2.0 in. − 2(0.5 in.)
K t ≅ 1.75
∴
Section Properties:
I p
π =
32
D4
π =
32
(1.00 in.)4
=
0.0 0.098175 175 in.4
Shaft Shear Stress: TR (500 lb-in.)(1.00 in./2) τ shaft = K t = (1.75) I p 0.098175 in.4
=
4, 456 psi
Minimum Required Yield Strength: τ y , min =
FS ⋅τ shaft
=
(3.0)(4, 45 456 psi) = 13, 369 psi = 13.37 ks ksi
Ans.
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6.100 A semicircular groove with a 5-mm radius is required in a 110-mm-diameter shaft. If the maximum allowable shear stress in the shaft must be limited to 60 MPa, determine the maximum torque that can be transmitted by the shaft.
Solution 6.17a: Groove: From Fig. 6.17a h 5 mm r 5 mm = = 1.00 = = 0.05 r 5 mm d 110 mm − 2(5 mm)
K t ≅ 1.85
∴
Section Properties:
I p
π =
32
D4
π =
32
(100 mm)4
=
9,81 ,817,477 ,477 mm mm4
Maximum Torque: TR τ = K t I p ∴
T =
I
τ p
K t R
=
(60 (60 N/mm N/mm2 )(9, )(9,81 817, 7,47 477 7 mm mm4 ) (1.85)(100 mm/2)
=
6,368 ,368,093 ,093 N-mm = 6.37 .37 kN kN-m
Ans.
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