mechanics of materials solutions chapter06 probs94-100

7/21/2019 Mechanics of Materials Solutions Chapter06 Probs94-100

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6.94 A fillet with a radius of 0.15 in. is used at the junction in a stepped shaft where the diameter isreduced 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

Fillet: From Fig. 6.17b:

4.0 in. 0.15 in.1.33 0.05 2.0

3.0 in. 3.0 in.t

D r K

d d = = = = ∴ ≅

Section Properties:

4 4 4(3.00 in.) 7.952156 in.32 32

p I Dπ π

= = =

Maximum Shear Stress:

4

(4,000 lb-ft)(3.00 in./2)(12 in./ft)(2.0) 18,108 psi 18.11 ksi

7.952156 in.t

p

TR K

I τ = = = = Ans.





6.95 A fillet with a radius of 12 mm is used at the junction in a stepped shaft where the diameter isreduced 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


135 mm 12 mm1.35 0.12 1.6

100 mm 100 mmt

D r K

d d = = = = ∴ ≅

Section Properties:

4 4 4(100 mm) 9,817,477 mm32 32

p I Dπ π

= = =

Maximum Shear Stress:

4

(10 kN-m)(100 mm/2)(1,000 N/kN)(1,000 mm/m)(1.6) 81.5 MPa

9,817,477 mmt

p

TR K

I τ = = = Ans.





6.96 A fillet with a radius of 1/8 in. is used at the junction in a stepped shaft where the diameter isreduced 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


8.0 in. 0.125 in.1.33 0.021 2.6

6.0 in. 6.0 in.t

D r K

d d = = = = ∴ ≅

Section Properties:

4 4 4(6.00 in.) 127.2345 in.32 32

p I Dπ π

= = =

Maximum Torque:

4(12 ksi)(127.2345 in. )195.7 kip-in.

(2.6)(6.00 in./2)

t

p

p

t

TR K

I

I T

K R

τ

τ

=

∴ = = = Ans.





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:

4 4 4

(4.0 in.) 25.13274 in.32 32 p I D

π π

= = =

Fillet Requirement: From Fig. 6.17b:

fillet

shaft

12 ksi1.50

8 ksi

5.0 in.1.25

4.0 in.

0.135

t K

D

d

r

d

τ

τ

= = =

= =

∴ ≅

Minimum Fillet Radius:

(0.135)(4.0 in.) 0.54 in.r

r d d

= = = Ans.





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 used for the large portion of the shaft.

Solution

Section Properties:

4 4 4

(50 mm) 613,592 mm32 32 p I D

π π

= = =

Shear Stress in Shaft:

4

(614 N-m)(50 mm/2)(1,000 mm/m)25.017 MPa

613,592 mm p

TR

I τ = = =

Fillet Requirement: From Fig. 6.17b:

fillet

shaft

40 MPa1.60

25.017 MPa

4.5 mm 0.0950 mm

1.20

t K

r d

D

d

τ

τ

= = =

= =

∴ ≅

Maximum Shaft Diameter:

max (1.20)(50 mm) 60.0 mm D

D d d

= = = Ans





6.99 A 2-in.-diameter shaft contains a ½-in.-deep U-shaped groove that has a ¼-in. radius at the bottomof 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

Groove: From Fig. 6.17a:

0.5 in. 0.25 in.2.00 0.25 1.75

0.25 in. 2.0 in. 2(0.5 in.)t

h r K

r d = = = = ∴ ≅

−

Section Properties:

4 4 4(1.00 in.) 0.098175 in.32 32

p I Dπ π

= = =

Shaft Shear Stress:

shaft 4

(500 lb-in.)(1.00 in./2)(1.75) 4, 456 psi

0.098175 in.t

p

TR K

I τ = = =

Minimum Required Yield Strength:

,min shaftFS (3.0)(4, 456 psi) 13,369 psi 13.37 ksi yτ τ = ⋅ = = = Ans.





6.100 A semicircular groove with a 5-mm radius is required in a 110-mm-diameter shaft. If themaximum allowable shear stress in the shaft must be limited to 60 MPa, determine the maximum torque

that can be transmitted by the shaft.

Solution

Groove: From Fig. 6.17a:

5 mm 5 mm1.00 0.05 1.85

5 mm 110 mm 2(5 mm)t

h r K

r d = = = = ∴ ≅

−

Section Properties:

4 4 4(100 mm) 9,817,477 mm32 32

p I Dπ π

= = =

Maximum Torque:

2 4(60 N/mm )(9,817,477 mm )6,368,093 N-mm 6.37 kN-m

(1.85)(100 mm/2)

t

p

p

t

TR K

I

I T

K R

τ

τ

=

∴ = = = = Ans.

mechanics of materials solutions chapter06 probs94-100

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