Mechanics of Materials Solutions CH06

7/25/2019 Mechanics of Materials Solutions CH06

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6.1 A solid circular steel shaft having an outside diameter of D = 1.25 in. is subjected to a pure torque oT = 1,900 lb-in. Determine the maximum shear stress in the shaft.

Solution

The polar moment of inertia for the shaft is

4 4 4(1.25 in.) 0.239684 in.32 32

p I Dπ π

= = =

The maximum shear stress in the steel shaft is found from the elastic torsion formula:

max 4

(1,900 lb-in.)(1.25 in./ 2)4,954.430 psi 4,950 psi

0.239684 in. p

TR

I τ = = = = Ans.





6.2 A hollow aluminum shaft with an outside diameter of 100 mm and a wall thickness of 5 mm has anallowable shear stress of 40 MPa. Determine the maximum torque T that may be applied to the shaft.

Solution


4 4 4 4 4(100 mm) (90 mm) 3, 376, 230 mm32 32

p I D d π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

Rearrange the elastic torsion formula to determine the maximum torque T :2 4

allow (40 N/mm )(3,376,230 mm )2,700,984 N-mm 2,701 N-m

100 mm / 2

p I T

R

τ = = = = Ans.





6.3 A hollow steel shaft with an outside diameter of 90 mm and a wall thickness of 5 mm is subjected toa pure torque of T = 3,500 N-m.

(a) Determine the maximum shear stress in the hollow shaft.

(b) Determine the minimum diameter D of a solid steel shaft for which the maximum shear stress is thesame as in part (a) for the same torque T .

Solution


4 4 4 4 4(90 mm) (80 mm) 2, 420, 008 mm32 32

p I D d π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

(a) The maximum shear stress in the hollow steel shaft is found from the elastic torsion formula:

max 4

(3,500 N-m)(90 mm/ 2)(1,000 mm/m)65.0824 MPa 65.1 MPa

2,420,008 mm p

TR

I τ = = = = Ans.

(b) The polar moment of inertia for a solid shaft can be expressed as

4

32 p I D

π =

Rearrange the elastic torsion formula to group terms with D on the left-hand side:4

32 ( / 2)

D T

D

π

τ =

and simplify to3

16

D T π

τ =

From this equation, the unknown diameter of the solid shaft can be expressed as

316T

Dπτ

=

To support a torque of T = 3,500 N-m without exceeding the maximum shear stress determined in part

(a), a solid shaft must have a diameter of

3 32

16 16(3,500 N-m)(1,000 mm/m)64.9419 mm 64.9 mm

(65.0824 N/mm )

T D

πτ π ≥ = = = Ans.





6.4 A compound shaft consists of two pipesegments. Segment (1) has an outside diameter o

220 mm and a wall thickness of 10 mm. Segment

(2) has an outside diameter of 140 mm and a wallthickness of 15 mm. The shaft is subjected to

torques T B = 40 kN-m and T C = 12 kN-m, which act

in the directions shown in Fig. P6.4. Determine the

maximum shear stress magnitude in each shaft

segment.Fig. P6.4

Solution

Equilibrium:1 140 kN-m 12 kN-m 0 28 kN-m x M T T Σ = − + − = ∴ =

2 212 kN-m 0 12 kN-m x M T T Σ = − − = ∴ = −

Section properties:

4 4 4 4 4

1 1 1 (220 mm) (200 mm) 72, 900, 658 mm32 32

p I D d π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

4 4 4 4 4

2 2 2 (140 mm) (110 mm) 23,341,052 mm32 32

p I D d π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

Shear stress magnitudes:

1 11 4

1

(28 kN-m)(220 mm/ 2)(1,000 N/kN)(1,000 mm/m)42.2 MPa

72,900,658 mm p

T R

I τ = = = Ans.

2 22 4

2

(12 kN-m)(140 mm / 2)(1,000 N/kN)(1,000 mm/m)36.0 MPa

23,341,052 mm p

T R

I τ = = = Ans.





6.5 A compound shaft consists of two pipesegments. Segment (1) has an outside diameter o

10.75 in. and a wall thickness of 0.365 in. Segment

(2) has an outside diameter of 6.625 in. and a wallthickness of 0.280 in. The shaft is subjected to

torques T B = 68 kip-ft and T C = 16 kip-ft, which act

in the directions shown in Fig. P6.5. Determine themaximum shear stress magnitude in each shaft

segment.Fig. P6.5

Solution

Equilibrium: 1 168 kip-ft 16 kip-ft 0 52 kip-ft x M T T Σ = − + − = ∴ =

2 216 kip-ft 0 16 kip-ft x M T T Σ = − − = ∴ = −

Section properties:

4 4 4 4 4

1 1 1 (10.75 in.) (10.02 in.) 321.4685 in.32 32

p I D d π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

4 4 4 4 4

2 2 2 (6.625 in.) (6.0650 in.) 56.2844 in.32 32

p I D d π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦


1 11 4

1

(52 kip-ft)(10.75 in./ 2)(12 in./ft)10.43 ksi

321.4685 in. p

T R

I τ = = = Ans.

2 22 4

2

(16 kip-ft)(6.625 in./ 2)(12 in./ft)11.30 ksi

56.2844 in. p

T R

I τ = = = Ans.





6.6 A compound shaft (Fig. P6.6) consists of brass segment (1) andaluminum segment (2). Segment (1) is a solid brass shaft with an outside

diameter of 0.875 in. and an allowable shear stress of 6,000 psi. Segment

(2) is a solid aluminum shaft with an outside diameter of 0.75 in. and anallowable shear stress of 9,000 psi. Determine the magnitude of the largest

torque T C that may be applied at C .

Fig. P6.6

Solution

Section properties:

4 4 4

1 1 (0.875 in.) 0.057548 in.32 32

p I Dπ π

= = =

4 4 4

2 2 (0.75 in.) 0.031063 in.32 32

p I Dπ π

= = =

Allowable internal torques: 4

1 1

1

1

(6,000 psi)(0.057548 in. )789.23 lb-in.

0.875 in./2

p I T

R

τ ≤ = =

42 2

2

2

(9,000 psi)(0.031063 in. )745.51 lb-in. controls

0.75 in./2

p I T

R

τ ≤ = = ←

Equilibrium:

The internal torque magnitude in each segment equals the external torque; therefore, T 1 = T 2 = T C . The

controlling internal torque is T 2 = 745.51 lb-in.; therefore, the maximum external torque T C that may be

applied to the compound shaft is745.51 lb-in. 746 lb-in.C T ≤ = Ans.





6.7 A compound shaft (Fig. P6.7) consists of brass segment (1) andaluminum segment (2). Segment (1) is a solid brass shaft with an

allowable shear stress of 40 MPa. Segment (2) is a solid aluminum shaft

with an allowable shear stress of 60 MPa. If a torque of T C = 8,500 N-m isapplied at C , determine the minimum required diameter of (a) the brass

shaft and (b) the aluminum shaft.

Fig. P6.7

Solution

The polar moment of inertia for a solid shaft can be expressed as

4

32 p I D

π =

Rearrange the elastic torsion formula to group terms with D on the left-hand side:4

32 ( / 2)

D T

D

π

τ =

and simplify to3

16

D T π

τ =


316T

Dπτ

=

Equilibrium:

For this shaft, the internal torque magnitude in each segment equals the external torque; therefore, T 1 =T 2 = T C = 8,500 N-m.

Minimum shaft diameters:

(a) Brass shaft (1)

1331 2

allow,1

16 16(8,500 N-m)(1,000 mm/m)102.7 mm

(40 N/mm )

T D

πτ π ≥ = = Ans.

(b) Aluminum shaft (2)

2332 2

allow,2

16 16(8,500 N-m)(1,000 mm/m)89.7 mm

(60 N/mm )

T D

πτ π ≥ = = Ans.





6.8 A solid 0.75-in.-diameter shaft is subjected tothe torques shown in Fig. P6.8. The bearings shown

allow the shaft to turn freely.

(a) Plot a torque diagram showing the internaltorque in segments (1), (2), and (3) of the shaft. Use

the sign convention presented in Section 6-6.

(b) Determine the maximum shear stress magnitudein the shaft.

Fig. P6.8

Solution

Equilibrium:

1 110 lb-ft 0 10 lb-ft x M T T Σ = + = ∴ = −

2 210 lb-ft 50 lb-ft 0 40 lb-ft x M T T Σ = + − = ∴ =

3 330 lb-ft 0 30 lb-ft x M T T Σ = − − = ∴ = −

Section properties:

4 4

1 2 3 (0.75 in.) 0.031063 in.32

p p p I I I π

= = = =


1 11 4

1

(10 lb-ft)(0.75 in./ 2)(12 in./ft) 1,448.7 psi0.031063 in. p

T R I

τ = = =

2 22 4

2

(40 lb-ft)(0.75 in./ 2)(12 in./ft)5,794.7 psi

0.031063 in. p

T R

I τ = = =

3 33 4

3

(30 lb-ft)(0.75 in. / 2)(12 in./ft)4,346.0 psi

0.031063 in. p

T R

I τ = = =

The maximum shear stress in the shaft occurs in segment (2):

max 5,790 psiτ = Ans.





6.9 A solid constant-diameter shaft is subjected tothe torques shown in Fig. P6.9. The bearings shown

allow the shaft to turn freely.

(a) Plot a torque diagram showing the internaltorque in segments (1), (2), and (3) of the shaft. Use


(b) If the allowable shear stress in the shaft is 80

MPa, determine the minimum acceptable diameter

for the shaft.Fig. P6.9

Solution

3 3160 N-m 0 160 N-m x M T T Σ = − = ∴ =

2 2160 N-m 380 N-m 0 220 N-m x M T T Σ = − − = ∴ = −

1 1160 N-m 380 N-m 330 N-m 0 110 N-m x M T T Σ = − + − = ∴ =

The maximum torque magnitude in the shaft occurs in segment (2): T max = 220 N-m.

The elastic torsion formula gives the relationship between shear stress and torque in a shaft.

p

TR

I τ =

In this instance, the torque and the allowable shear stress are known for the shaft. Rearrange the elastic

torsion formula, putting the known terms on the right-hand side of the equation:

p I T

R τ =

Express the left-hand side of this equation in terms of the shaft diameter D:

4

332

/ 2 16

DT

D D

π

π

τ = =

and solve for the minimum acceptable diameter:

3 3

2

16 16(220 N-m)(1,000 mm/m)14,005.635 mm

(80 N/mm )

24.1 mm

T D

D

π τ π = = =

∴ = Ans.





6.10 A solid circular steel shaft having an outside diameter of 1.75 in. is subjected to a pure torque of= 9,000 lb-in. The shear modulus of the steel is G = 12,000 ksi. Determine:

(a) the maximum shear stress in the shaft.

(b) the magnitude of the angle of twist in a 6-ft length of shaft.

Solution


4 4 4

(1.75 in.) 0.920772 in.32 32 p I D

π π

= = =

(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula:

max 4

(9,000 lb-in.)(1.75 in./ 2)8,552.6 psi 8.55 ksi

0.920772 in. p

TR

I τ = = = = Ans.

(b) The magnitude of the angle of twist in a 6-ft length of shaft is

4

(9,000 lb-in.)(6 ft)(12 in./ft)0.058646 rad 0.0586 rad 3.36

(12,000,000 psi)(0.920772 in. ) p

TL

GI φ = = = = = ° Ans.





6.11 A solid circular steel shaft having an outside diameter of 30 mm is subjected to a pure torque of T =175 N-m. The shear modulus of the steel is G = 80 GPa. Determine:


(b) the magnitude of the angle of twist in a 1.5-m length of shaft.

Solution


4 4 4

(30 mm) 79,521.56 mm32 32 p I D

π π

= = =


max 4

(175 N-m)(30 mm/ 2)(1,000 mm/m)33.001 MPa 33.0 MPa

79,521.56 mm p

TR

I τ = = = = Ans.

(b) The magnitude of the angle of twist in a 1.5-m length of shaft is2

2 4

(175 N-m)(1.5 m)(1,000 mm/m)0.041262 rad 0.0413 rad 2.36

(80,000 N/mm )(79,521.56 mm ) p

TL

GI φ = = = = = ° Ans.





6.12 A hollow steel shaft with an outside diameter of 75 mm and a wall thickness of 5 mm is subjectedto a pure torque of T = 2,200 N-m. The shear modulus of the steel is G = 80 GPa. Determine:


(b) the magnitude of the angle of twist in a 2.5-m length of shaft.

Solution


4 4 4 4 4

(75 mm) (65 mm) 1, 353,830 mm32 32 p I D d

π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦


max 4

(2,200 N-m)(75 mm / 2)(1,000 mm/m)60.9382 MPa 60.9 MPa

1,353,830 mm p

TR

I τ = = = = Ans.

(b) The magnitude of the angle of twist in a 2.5-m length of shaft is2

2 4

(2,200 N-m)(2.5 m)(1,000 mm/m)0.050782 rad 0.0508 rad 2.91

(80,000 N/mm )(1,353,830 mm ) p

TL

GI φ = = = = = ° Ans.





6.13 A solid stainless steel [G = 12,500 ksi] shaft that is 30-in. long will be subjected to a pure torque oT = 300 lb-in. Determine the minimum diameter D required if the shear stress must not exceed 8,000 psi

and the angle of twist must not exceed 0.050 rad. Report both the maximum shear stress τ and the angle

of twist φ at this minimum diameter.

Solution

Consider shear stress:


4

32 p I D

π =

The elastic torsion formula can be rearranged to gather terms with D:4 3

32 ( / 2) 16

D D T

D

π π

τ = =


316T

Dπτ

=

For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is:

3 16(300 lb-in.) 0.576 in.(8,000 psi)

Dπ

≥ =

Consider angle of twist:

Rearrange the angle of twist equation:

4

32 p

p

TL TL I D

GI G

π φ

φ = ∴ = ≥

and solve for the minimum diameter that will satisfy the angle of twist limitation:

4 432 32(300 lb-in.)(30 in.)

0.619 in.(0.050 rad)(12,500,000 psi)

TL D

Gπ φ π ≥ = =

Therefore, the minimum diameter that could be used for the shaft is

min 0.619 in. D = Ans.

The angle of twist for this shaft is φ = 0.050 rad. To compute the shear stress in a 0.619-in.-diametershaft, first compute the polar moment of inertia:

4 4 4(0.619 in.) 0.014413 in.32 32

p I Dπ π

= = =

The shear stress in the shaft is thus:

max 4

(300 lb-in.)(0.619 in./ 2)6,442.1 psi 6,440 psi0.014413 in. p

TR

I τ = = = = Ans.





6.14 A solid stainless steel [G = 86 GPa] shaft that is 750 mm long will be subjected to a pure torque oT = 40 N-m. Determine the minimum diameter D required if the shear stress must not exceed 50 MPa

and the angle of twist must not exceed 0.050 rad. Report both the maximum shear stress τ and the angle

of twist φ at this minimum diameter.

Solution

Consider shear stress:


4

32 p I D

π =


32 ( / 2) 16

D D T

D

π π

τ = =


316T

Dπτ

=

For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is:

32

16(40 N-m)(1,000 mm/m) 15.97 mm(50 N/mm )

Dπ

≥ =

Consider angle of twist:

Rearrange the angle of twist equation:

4

32 p

p

TL TL I D

GI G

π φ

φ = ∴ = ≥

and solve for the minimum diameter that will satisfy the angle of twist limitation:

4 42

32 32(40 N-m)(750 mm)(1,000 mm/m)16.33 mm

(0.050 rad)(86,000 N/mm )

TL D

Gπ φ π ≥ = =

Therefore, the minimum diameter that could be used for the shaft is

min 16.33 mm D = Ans.

The angle of twist for this shaft is φ = 0.050 rad. To compute the shear stress in a 16.33-mm-diametershaft, first compute the polar moment of inertia:

4 4 4(16.33 mm) 6,981.434 mm32 32

p I Dπ π

= = =

The shear stress in the shaft is thus:

max 4

(40 N-m)(16.33 mm / 2)(1,000 mm/m)46.8 MPa6,981.434 mm p

TR

I τ = = = Ans.





6.15 A hollow steel [G = 12,000 ksi] shaft with an outside diameter of 3.50 in. will be subjected to a pure torque of T = 3,750 lb-ft. Determine the maximum inside diameter d that can be used if the shear

stress must not exceed 8,000 psi and the angle of twist must not exceed 3° in an 8-ft length of shaft.

Report both the maximum shear stress τ and the angle of twist φ for this maximum inside diameter.

Solution

Consider shear stress: The polar moment of inertia for a hollow shaft can be expressed as

4 4

32 p I D d

π

⎡ ⎤= −⎣ ⎦


32 / 2

D d T

D

π

τ

⎡ ⎤−⎣ ⎦=

Rearrange this equation to isolate the inside diameter d term:

4 4 4 432 32

2 2

T D T D D d d D

πτ πτ − ≥ ∴ ≤ −

From this equation, the unknown inside diameter of the hollow shaft can be expressed as

4432

2

T Dd D

πτ

≤ −

For the hollow steel shaft, the maximum diameter that will satisfy the allowable shear stress is:

44

32(3,750 lb-ft)(3.50 in.)(12 in./ft)(3.50 in.) 2.6564 in.

2 (8,000 psi)d

π ≤ − =

Consider angle of twist: Rearrange the angle of twist equation:

4 4

32 p

p

TL TL I D d

GI G

π φ

φ ⎡ ⎤= ∴ = − ≥⎣ ⎦

Rearrange this equation to isolate the inside diameter d term:

4 4 32TL

d D Gπ φ ≤ −

and solve for the maximum inside diameter that will satisfy the angle of twist limitation:2

4 444

32 32(3,750 lb-ft)(8 ft)(12 in./ft)(3.50 in.) 2.991 in.

(3 )( /180 )(12,000,000 psi)

TLd D

Gπφ π π ≤ − = − =

° °

Therefore, the maximum inside diameter that could be used for the shaft is

max 2.66 in.d = Ans.

The shear stress for this shaft is τ = 8,000 psi. To compute the angle of twist in an 8-ft length of thehollow shaft, first compute the polar moment of inertia:

4 4 4 4 4(3.50 in.) (2.66 in.) 9.817318 in.32 32

p I D d π π ⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

The angle of twist in the shaft is thus:2

4

(3,750 lb-ft)(8 ft)(12 in./ft)0.0367 rad 2.10

(12,000,000 psi)(9.817318 in. ) p

TL

GI φ = = = = ° Ans.





6.16 A compound steel [G = 80 GPa] shaft (Fig.P6.16) consists of a solid 90-mm-diameter

segment (1) and a solid 70-mm-diameter

segment (2). The allowable shear stress of thesteel is 96 MPa, and the maximum rotation

angle at the free end of the compound shaft must

be limited to φ C ≤ 4°. Determine the magnitudeof the largest torque T C that may be applied at C .

Fig. P6.16

Solution

Section properties:

4 4 4

1 1 (90 mm) 6,441,247 mm32 32

p I Dπ π

= = =

4 4 4

2 2 (70 mm) 2,357,176 mm32 32

p I Dπ π

= = =

Consider shear stress: The larger shear stress will occur in the smaller diameter segment; that is,

segment (2).2 4

max

(96 N/mm )(2,357,176 mm )6,465,397 N-mm

70 mm/2

p I T

R

τ ≤ = =

Consider angle of twist: The rotation angle at C is the sum of the angles of twist in segments (1) and

(2):

1 1 2 21 2

1 1 2 2

C

p p

T L T L

G I G I φ φ φ = + = +

Since T 1 = T 2 = T C and G1 = G2, this equation can be simplified tomax 1 2

1 2

4 p p

T L L

G I I

⎡ ⎤+ ≤ °⎢ ⎥

⎢ ⎥⎣ ⎦

and solved for the maximum torque:2 2

max 6 3 6 3

4 4

(4 )( /180 )(80,000 N/mm ) 5,585.0536 N/mm

1,200 mm 1,800 mm 186.2993 10 mm 763.6256 10 mm

6,441,247 mm 2,357,176 mm

5,879,468 N-mm

T π

− − − −

° °≤ =

⎡ ⎤ ⎡ ⎤× + ×⎣ ⎦+⎢ ⎥⎣ ⎦

=

Therefore, the magnitude of the largest torque T C that may be applied at C is,max 5,879,468 N-mm 5,880 N-mC T = = Ans.





6.17 A compound shaft (Fig. P6.17) consists of brass segment (1) andaluminum segment (2). Segment (1) is a solid brass [G = 5,600 ksi]

shaft with an outside diameter of 2.75 in. and an allowable shear stress

of 9,000 psi. Segment (2) is a solid aluminum [G = 4,000 ksi] shaftwith an outside diameter of 2.25 in. and an allowable shear stress o

12,000 psi. The maximum rotation angle at the upper end of the

compound shaft must be limited to φ C ≤ 4°. Determine the magnitudeof the largest torque T C that may be applied at C .

Fig. P6.17

Solution

From equilibrium, the internal torques in segments (1) and (2) are equal to the external torque T C . The

elastic torsion formula gives the relationship between shear stress and torque in a shaft.

p

TR

I τ =

Consider shear stress: In this compound shaft, the diameters and allowable shear stresses in segments

(1) and (2) are known. The elastic torsion formula can be rearranged to solve for the unknown torque.An expression can be written for each shaft segment:

1 1 2 2

1 2

1 2

p p I I T T

R R

τ τ = =

For shaft segment (1), the polar moment of inertia is:

4 4 4

1 1 (2.75 in.) 5.61475 in.32 32

p I Dπ π

= = =

Use this value along with the 9,000 psi allowable shear stress to determine the allowable torque T 1:4

1 1

1

1

(9,000 psi)(5.61475 in. )36,751 lb-in.

(2.75 in./2)

p I T

R

τ = = = (a)

For shaft segment (2), the polar moment of inertia is:

4 4 4

2 2 (2.25 in.) 2.51611 in.32 32

p I Dπ π

= = =

Use this value along with the 12,000 psi allowable shear stress to determine the allowable torque T 2:4

2 2

2

2

(12,000 psi)(2.51611 in. )26,838 lb-in.

(2.25 in./2)

p I T

R

τ = = = (b)

Consider angle of twist: The angles of twists in segments (1) and (2) can be expressed as:

1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I

φ φ = =

The rotation angle at C is the sum of these two angles of twist:

1 1 2 21 2

1 1 2 2

C

p p

T L T L

G I G I φ φ φ = + = +

and since T 1 = T 2 = T C:





1 2

1 1 2 2

C C

p p

L LT

G I G I φ

⎡ ⎤= +⎢ ⎥

⎢ ⎥⎣ ⎦

Solving for T C gives:

1 2

1 1 2 2

4 4

rad(4 )

18023,609 lb-in.

18 in. 24 in.

(5,600,000 psi)(5.61475 in. ) (4,000,000 psi)(2.51611 in. )

C C

p p

T L L

G I G I

φ

π

=

+

⎛ ⎞° ⎜ ⎟

°⎝ ⎠= =

+

(c)

Compare the torque magnitudes in Eqs. (a), (b), and (c). The smallest torque controls; therefore, the

maximum torque that can be applied to the compound shaft at C is T C = 23,609 lb-in. = 1,967 lb-ft. Ans.





6.18 A compound steel [G = 80 GPa] shaft consistsof solid 30-mm-diameter segments (1) and (3) and

tube segment (2), which has an outside diameter o

60 mm and an inside diameter of 50 mm (Fig.P6.18). Determine:

(a) the maximum shear stress in tube segment (2).

(b) the angle of twist in tube segment (2).

(c) the rotation angle of gear D relative to gear A.

Fig. P6.18

Solution

The torques in all shaft segments are equal; therefore, T 1 = T 2 = T 3 = 240 N-m.

Polar moments of inertia in the shaft segments will be needed for this calculation. For segments (1) and(3), which are solid 30-mm-diameter shafts, the polar moment of inertia is:

4 4 4

1 1 3(30 mm) 79,521.56 mm32 32

p p I D I π π

= = = =

For segment (2), which has a tube cross section, the polar moment of inertia is:

4 4 4 4 4

2 2 2 (60 mm) (50 mm) 658, 752.71 mm32 32

p I D d π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

Shear stress in tube segment (2): Use the elastic torsion formula to calculate the shear stress caused by

an internal torque of T 2 = 240 N-m. Note that the radius term used in the elastic torsion formula for the

tube is the outside radius; that is, R2 = 60 mm/2 = 30 mm.

2 22 4

2

(240 N-m)(30 mm)(1,000 mm/m)10.93 MPa

658,752.71 mm p

T R

I τ = = = Ans.

Angle of twist in tube segment (2): Apply the angle of twist equation to segment (2).

2 22 2 4

2 2

(240 N-m)(2,000 mm)(1,000 mm/m) 0.009108 rad 0.00911 rad(80,000 N/mm )(658,752.71 mm ) p

T LG I

φ = = == = Ans.

Rotation angle of gear D relative to gear A: The angles of twist in segments (1) and (3) must be

calculated. Since both segments have the same shear modulus, polar moment of inertia, and length, they

will both have the same angle of twist:

1 11 32 4

1 1

(240 N-m)(300 mm)(1,000 mm/m)0.011318 rad

(80,000 N/mm )(79,521.56 mm ) p

T L

G I φ φ = = = =

Since gear A is the origin of the coordinate system for this problem, we will arbitrarily define the

rotation angle at gear A to be zero; that is, φ A = 0. The rotation angle of gear D relative to gear A isfound by adding the angles of twist for the three segments to φ A:

1 2 3

0 0.011318 rad 0.009108 rad 0.011318 rad

0.031743 rad 0.0317 rad

D Aφ φ φ φ φ = + + +

= + + +

= = Ans.





6.19 A compound steel [G = 80 GPa] shaft consistsof solid 40-mm-diameter segments (1) and (3) and

tube segment (2), which has an outside diameter o

75 mm (Fig. P6.19). If the rotation of gear Drelative to gear A must not exceed 0.01 rad,

determine the maximum inside diameter that may

be used for tube segment (2).

Fig. P6.19

Solution

The polar moment of inertia for shaft segments (1) and (3) is:

4 4

1 3 (40 mm) 251,327 mm32

p p I I π

= = =

The rotation of gear D relative to gear A is equal to the sum of the angles of twist in segments (1), (2),

and (3):

1 2 3 Dφ φ φ φ = + +

The angles of twist in segments (1) and (3) is

1 3 2 4

(240 N-m)(300 mm)(1,000 mm/m)0.003581 rad

(80,000 N/mm )(251,327 mm )φ φ = = =

Therefore, the angle of twist that can allowed for segment (2) is

1 2 3 20.01 rad 0.01 rad 2(0.003581 rad) 0.002838 rad Dφ φ φ φ φ = + + = ∴ ≤ − =

The minimum polar moment of inertia required for segment (2) is thus:2

42 22 2

2 2

(240 N-m)(2 m)(1,000 mm/m)0.002838 rad 2,114,165 mm

(80,000 N/mm )(0.002838 rad) p

p

T L I

G I ≤ ∴ ≥ =

Since the outside diameter of segment (2) is D2 = 75 mm, the maximum inside diameter d 2 is:

4 4 4 4 4

2 2 2

2

(75 mm) 2,114,165 mm32 32

56.4 mm

D d d

d

π π ⎡ ⎤ ⎡ ⎤− = − ≥⎣ ⎦ ⎣ ⎦

∴ ≤ Ans.





6.20 The compound shaft shown in Fig. P6.20 consists of aluminum segment (1) and steel segment (2).Aluminum segment (1) is a tube with an outside diameter of D1 = 4.00 in., a wall thickness of t 1 = 0.25

in., and a shear modulus of G1 = 4,000 ksi. Steel segment (2) is a tube with an outside diameter of D2 =

2.50 in., a wall thickness of t 2 = 0.125 in., and a shear modulus of G2 = 12,000 ksi. The compound shaftis subjected to torques applied at B and C , as shown in Fig. P6.20.

(a) Prepare a diagram that shows the internal

torque and the maximum shear stress in

segments (1), and (2) of the shaft. Use the sign

convention presented in Section 6-6.(b) Determine the rotation angle of B with

respect to the support at A.(c) Determine the rotation angle of C with

respect to the support at A.

Fig. P6.20

SolutionSection properties: Polar moments of inertia in the shaft segments will be needed for this calculation.

4 4 4 4 4

1 1 1 (4.00 in.) (3.50 in.) 10.4004 in.32 32

p I D d π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

4 4 4 4 4

2 2 2 (2.50 in.) (2.25 in.) 1.3188 in.32 32

p I D d π π

⎡ ⎤ ⎡ ⎤= − = − =⎣ ⎦ ⎣ ⎦

(a) Equilibrium

1

1

650 lb-ft 1,700 lb-ft 01,050 lb-ft

x M T T

Σ = − + − =

∴ =

2

2

650 lb-ft 0

650 lb-ft

x M T

T

Σ = − − =

∴ = −

Shear stress:

1 11 4

1

(1,050 lb-ft)(4.00 in./2)(12 in./ft)2,422.98 psi 2,420 psi

10.4004 in. p

T R

I τ = = = = Ans.

2 22 4

2

( 650 lb-ft)(2.50 in./2)(12 in./ft)7,393.08 psi 7,390 psi

1.3188 in. p

T R

I τ

−= = = − = − Ans.





(b) Rotation angle of B with respect to A:

1 0 B A B Bφ φ φ φ φ = − = − =

Therefore,2

1 11 4

1 1

(1,050 lb-ft)(9 ft)(12 in./ft)0.032710 rad 0.0327 rad

(4,000,000 psi)(10.4004 in. ) B

p

T L

G I φ φ = = = = = Ans.

(c) Rotation angle of C with respect to A:

2 2C B C Bφ φ φ φ φ φ = − ∴ = +

The angle of twist in shaft (2) is2

2 2

2 4

2 2

( 650 lb-ft)(6 ft)(12 in./ft)

0.035486 rad(12,000,000 psi)(1.3188 in. ) p

T L

G I φ

−

= = = −

and thus, the rotation angle at C is

2 0.032710 rad ( 0.035486 rad) 0.002776 rad 0.00278 radC Bφ φ φ = + = + − = − = − Ans.





6.21 A solid 1.00-in.-diameter steel [G =12,000 ksi] shaft is subjected to the torques

shown in Fig. P6.21.

(a) Prepare a diagram that shows the internaltorque and the maximum shear stress in

segments (1), (2), and (3) of the shaft. Use


(a) Determine the rotation angle of pulley C

with respect to the support at A.(b) Determine the rotation angle of pulley D

with respect to the support at A.Fig. P6.21

Solution

Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segmentswill be needed for this calculation.

4 4 4

1 1 2 3(1.00 in.) 0.098175 in.32 32

p p p I D I I π π

= = = = =

(a) Equilibrium

1

1

240 lb-ft 215 lb-ft 110 lb-ft 0

135 lb-ft

x M T

T

Σ = − + − − =

∴ = −

2

2

215 lb-ft 110 lb-ft 0

105 lb-ft

x M T

T

Σ = − − =

∴ =

3

3

110 lb-ft 0

110 lb-ft

x M T

T

Σ = − − =

∴ = −

Shear stress:

1 1

1 41

( 135 lb-ft)(1.00 in./2)(12 in./ft)

8,250.57 psi 8,250 psi0.098175 in. p

T R

I τ

−= = = − = −

2 22 4

2

(105 lb-ft)(1.00 in./2)(12 in./ft)6,417.11 psi 6,420 psi

0.098175 in. p

T R

I τ = = = =

3 33 4

3

( 110 lb-ft)(1.00 in./2)(12 in./ft)6,722.69 psi 6,720 psi

0.098175 in. p

T R

I τ

−= = = − = −

The maximum shear stress in the entire shaft is τ max = 8,250 psi. Ans.





(b) Rotation angle of C with respect to A:

The angles of twist in the three shaft segments are:

1 11 4

1 1

( 135 lb-ft)(36 in.)(12 in./ft)0.049503 rad

(12,000,000 psi)(0.098175 in. ) p

T L

G I φ

−= = = −

2 22 4

2 2

(105 lb-ft)(36 in.)(12 in./ft)0.038503 rad

(12,000,000 psi)(0.098175 in. ) p

T L

G I φ = = =

3 33 4

3 3

( 110 lb-ft)(36 in.)(12 in./ft)0.040336 rad

(12,000,000 psi)(0.098175 in. ) p

T L

G I φ

−= = = −

The rotation angle of C with respect to A is found from the sum of the angles of twist in segments (1)

and (2):

1 2

0.049503 rad 0.038503 rad 0.011000 rad 0.01100 radC

φ φ φ = + = − + = − = − Ans.

(c) Rotation angle of D with respect to A:

The rotation angle of D with respect to A is found from the sum of the angles of twist in segments (1),

(2), and (3):

1 2 3

0.049503 rad 0.038503 rad ( 0.040336 rad)

0.051336 rad 0.0513 rad

Dφ φ φ φ = + +

= − + + −

= − = − Ans.





6.22 A compound shaft supports several pulleysas shown in Fig. P6.22. Segments (1) and (4) are

solid 25-mm-diameter steel [G = 80 GPa] shafts.

Segments (2) and (3) are solid 50-mm-diametersteel shafts. The bearings shown allow the shaft

to turn freely.

(a) Prepare a diagram that shows the internal

torque and the maximum shear stress in

segments (1), (2), (3), and (4) of the shaft. Usethe sign convention presented in Section 6-6.

(b) Determine the rotation angle of pulley D withrespect to pulley B.

(c) Determine the rotation angle of pulley E with

respect to pulley A.

Fig. P6.22

Solution

Section properties: The polar moments of inertia for the shaft segments will be needed for thiscalculation.

4 4 4

1 1 4(25 mm) 38,349.52 mm32 32

p p I D I π π

= = = =

4 4 4

2 2 3(50 mm) 613,592.32 mm32 32

p p I D I π π

= = = =

(a) Equilibrium

1

1

150 N-m 0

150 N-m

x M T

T

Σ = + =

∴ = −

2

2

1,850 N-m 150 N-m 0

1,700 N-m

x M T

T

Σ = − + =

∴ =

3

3

570 N-m 190 N-m 0

380 N-m

x M T

T

Σ = − − =

∴ =





4

4

190 N-m 0

190 N-m

x M T

T

Σ = − − =

∴ = −

Shear stress:

1 11 4

1

( 150 N-m)(25 mm/2)(1,000 mm/m)48.892 MPa 48.9 MPa

38,349.52 mm p

T R

I τ

−= = = − = −

2 22 4

2

(1,700 N-m)(50 mm/2)(1,000 mm/m)69.264 MPa 69.3 MPa

613,592.32 mm p

T R

I τ = = = =

3 33 4

3

(380 N-m)(50 mm/2)(1,000 mm/m)15.48 MPa

613,592.32 mm p

T R

I τ = = =

4 44 4

4

( 190 N-m)(25 mm/2)(1,000 mm/m)61.930 MPa 61.9 MPa

38,349.52 mm p

T R

I τ

−= = = − = −

The maximum shear stress in the entire shaft is τ max = 69.3 MPa. Ans.





Angles of twist:

The angles of twist in the four shaft segments are:

1 11 2 4

1 1

( 150 N-m)(750 mm)(1,000 mm/m)0.036669 rad

(80,000 N/mm )(38,349.52 mm ) p

T L

G I φ

−= = = −

2 22 2 4

2 2

(1,700 N-m)(500 mm)(1,000 mm/m)0.017316 rad

(80,000 N/mm )(613,592.32 mm ) p

T L

G I φ = = =

3 33 2 4

3 3

(380 N-m)(625 mm)(1,000 mm/m)0.004838 rad

(80,000 N/mm )(613,592.32 mm ) p

T L

G I φ = = =

4 44 2 4

4 4

( 190 N-m)(550 mm)(1,000 mm/m)0.034062 rad

(80,000 N/mm )(38,349.52 mm ) p

T L

G I φ

−= = = −

(b) Rotation angle of pulley D with respect to pulley B: The rotation angle of D with respect to B is

found from the sum of the angles of twist in segments (2) and (3):

/ 2 3 0.017316 rad 0.004838 rad 0.022154 rad 0.0222 rad D Bφ φ φ = + = + = = Ans.

(c) Rotation angle of pulley E with respect to pulley A: The rotation angle of E with respect to A is

found from the sum of the angles of twist in all four segments:

1 2 3 4

0.036669 rad 0.017316 rad 0.004838 rad ( 0.034062 rad)

0.048577 rad 0.0486 rad

E φ φ φ φ φ = + + +

= − + + + −

= − = − Ans.





6.23 A solid steel [G = 80 GPa] shaft of variablediameter is subjected to the torques shown in Fig.

P6.23. The diameter of the shaft in segments (1) and

(3) is 50 mm, and the diameter of the shaft insegment (2) is 80 mm. The bearings shown allow

the shaft to turn freely.

(a) Plot a torque diagram showing the internal

torque in segments (1), (2), and (3) of the shaft. Use

the sign convention presented in Section 6-6.(b) Add a plot to the torque diagram that shows the

maximum shear stress magnitude in each segmentof the shaft.

(c) Add another plot to the torque diagram that

shows the rotation angles along the shaft measuredwith respect to gear A.

Fig. P6.23

Solution

Section properties: The polar moments of inertia for the shaft segments will be needed for this

calculation.

4 4 4

1 1 3(50 mm) 613,592.32 mm32 32

p p I D I π π

= = = =

4 4 4

2 2 (80 mm) 4,021,238.60 mm32 32

p I Dπ π

= = =

(a) Equilibrium

1

1

1,200 N-m 0

1,200 N-m

x M T

T

Σ = − + =

∴ =

2

2

1,200 N-m 4,500 N-m 0

3,300 N-m

x M T

T

Σ = − + + =

∴ = −

3

3

500 N-m 0

500 N-m

x M T

T

Σ = − − =

∴ = −

Shear stress:

1 11 4

1

(1, 200 N-m)(50 mm/2)(1,000 mm/m)48.892 MPa 48.9 MPa

613,592.32 mm p

T R

I τ = = = = Ans.

2 22 4

2

( 3,300 N-m)(80 mm/2)(1,000 mm/m)32.826 MPa 32.8 MPa

4,021,238.60 mm p

T R

I τ

−= = = − = − Ans.





3 33 4

3

( 500 N-m)(50 mm/2)(1,000 mm/m)20.372 MPa 20.4 MPa

613,592.32 mm p

T R

I τ

−= = = − = − Ans.

The maximum shear stress in the entire shaft is τ max = 48.9 MPa. Ans.

Angles of twist:

The angles of twist in the three shaft segments are:2

1 11 2 4

1 1

(1,200 N-m)(0.7 m)(1,000 mm/m) 0.017112 rad(80,000 N/mm )(613,592.32 mm ) p

T LG I

φ = = =

2

2 22 2 4

2 2

( 3,300 N-m)(1.8 m)(1,000 mm/m)0.018464 rad

(80,000 N/mm )(4,021,238.60 mm ) p

T L

G I φ

−= = = −

2

3 33 2 4

3 3

( 500 N-m)(0.7 m)(1,000 mm/m)0.007130 rad

(80,000 N/mm )(613,592.32 mm ) p

T L

G I φ

−= = = −

(b) Rotation angles:

1

2

3

0 rad

0 rad 0.017112 rad 0.017112 rad

0.017112 rad ( 0.018464 rad) 0.001352 rad

0.001352 rad ( 0.007130 rad) 0.008482 rad

A

B A

C B

D C

φ

φ φ φ

φ φ φ

φ φ φ

=

= + = + =

= + = + − = −

= + = − + − = −





6.24 A compound shaft drives three gears,as shown in Fig. P6.24. Segments (1) and

(2) of the compound shaft are hollow

aluminum [G = 4,000 ksi] tubes, whichhave an outside diameter of 3.00 in. and a

wall thickness of 0.25 in. Segments (3) and

(4) are solid 2.00-in.-diameter steel [G =

12,000 ksi] shafts. The bearings shown

allow the shaft to turn freely.(a) Prepare a diagram showing the internal

torque and the maximum shear stress insegments (1), (2), (3), and (4) of the shaft.

Also, add a plot of the rotation angles

along the shaft measured with respect to A.(b) Determine the rotation angle of flange

C with respect to flange A.

(c) Determine the rotation angle of gear E with respect to flange A.

Fig. P6.24

Solution

Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segmentswill be needed for this calculation.

4 4 4 4 4

1 1 1 2(3.00 in.) (2.50 in.) 4.117204 in.32 32

p p I D d I π π

⎡ ⎤ ⎡ ⎤= − = − = =⎣ ⎦ ⎣ ⎦

4 4 4

3 3 4(2.00 in.) 1.570796 in.32 32

p p I D I π π

= = = =

(a) Equilibrium

1

1

14 kip-in. 42 kip-in. 35 kip-in. 0

7 kip-in.

x M T

T

Σ = − + − =

∴ =

2

2

14 kip-in. 42 kip-in. 0

28 kip-in.

x M T

T

Σ = − − =

∴ = −

3

3

14 kip-in. 42 kip-in. 0

28 kip-in.

x M T

T

Σ = − − =

∴ = −





4

4

14 kip-in. 0

14 kip-in.

x M T

T

Σ = − =

∴ =

Shear stress:

1 11 4

1

( 7 kip-in.)(3.00 in./2)

2.55 ksi4.117204 in. p

T R

I τ

−

= = = Ans.

2 22 4

2

( 28 kip-in.)(3.00 in./2)10.20 ksi

4.117204 in. p

T R

I τ

−= = = − Ans.

3 33 4

3

( 28 kip-in.)(2.00 in./2)17.83 ksi

1.570796 in. p

T R

I τ

−= = = − Ans.

4 44 4

4

(14 kip-in.)(2.00 in./2)8.91 ksi

1.570796 in. p

T R

I τ = = = Ans.

(b) Angles of Twist: The angles of twist in the four shaft segments are:

1 11 4

1 1

(7 kip-in.)(60 in.)0.025503 rad

(4,000 ksi)(4.117204 in. ) p

T L

G I φ = = =

2 22 4

2 2

( 28 kip-in.)(6 in.)0.010201 rad

(4,000 ksi)(4.117204 in. ) p

T L

G I φ

−= = = −

3 33 4

3 3

( 28 kip-in.)(18 in.)0.026738 rad

(12,000 ksi)(1.570796 in. ) p

T L

G I φ

−= = = −

4 44 4

4 4

(14 kip-in.)(12 in.)0.008913 rad

(12,000 ksi)(1.570796 in. ) p

T L

G I φ = = =

1

2

3

4

0 rad

0 rad 0.025503 rad 0.025503 rad

0.025503 rad ( 0.010201 rad) 0.015302 rad

0.015302 rad ( 0.026738 rad) 0.011436 rad

0.011436 rad 0.008913 rad 0.002523 rad

A

B A

C B

D C

E D

φ

φ φ φ

φ φ φ

φ φ φ

φ φ φ

=

= + = + =

= + = + − =

= + = + − = −

= + = − + = −

The rotation angle of flange C with respect to A is

0.01530 radC φ = Ans.

(c) Rotation angle of E with respect to A:

0.00252 rad E φ = − Ans.









6.25 A compound shaft drives several pulleys, asshown in Fig. P6.25. Segments (1) and (2) of the

compound shaft are hollow aluminum [G = 4,000 ksi]

tubes, which have an outside diameter of 3.00 in. anda wall thickness of 0.125 in. Segments (3) and (4) are

solid 1.50-in.-diameter steel [G = 12,000 ksi] shafts.

The bearings shown allow the shaft to turn freely.

(a) Prepare a diagram showing the internal torque and

the maximum shear stress in segments (1), (2), (3),and (4) of the shaft. Also, add a plot of the rotation

angles along the shaft measured with respect to A.(b) Determine the rotation angle of flange C with

respect to pulley A.

(c) Determine the rotation angle of pulley E withrespect to pulley A.

Fig. P6.25

Solution

Equilibrium

4 470 lb-ft 0 70 lb-ft x M T T Σ = − = ∴ =

3 370 lb-ft 200 lb-ft 0 130 lb-ft x M T T Σ = − − = ∴ = −

Note: The internal torque in shaft segment (2) is the same as

in segment (3); therefore, T 2 = –130 lb-ft.

1 1480 lb-ft 0 480 lb-ft x M T T Σ = − = ∴ =

Polar moments of inertia in the shaft segments will be needed for this calculation. Segments (1) and (2)

are hollow tubes with an outside diameter of 3.00 in. and an inside diameter of 3.00 in. – 2(0.125 in.) =

2.75 in. The polar moment of inertia in segments (1) and (2) is:

4 4 4 4 4

1 1 1 2(3.00 in.) (2.75 in.) 2.33740 in.32 32

p p I D d I π π

⎡ ⎤ ⎡ ⎤= − = − = =⎣ ⎦ ⎣ ⎦

Segments (3) and (4) are solid 1.50-in.-diameter shafts, which have a polar moment of inertia of:

4 4 4

3 3 4(1.50 in.) 0.49701 in.32 32

p p I D I π π

= = = =

Shear stresses: The shear stresses in each segment can be calculated by the elastic torsion formula:





1 11 4

1

(480 lb-ft)(1.50 in.)(12 in./ft)3,696.4 psi 3,700 psi

2.33740 in. p

T R

I τ = = = = Ans.

2 22 4

2

( 130 lb-ft)(1.50 in.)(12 in./ft)1,001.1 psi 1,001 psi

2.33740 in. p

T R

I τ

−= = = − = − Ans.

3 33 4

3

( 130 lb-ft)(0.75 in.)(12 in./ft)2,354.1 psi 2,350 psi

0.49701 in. p

T R

I τ

−= = = − = − Ans.

4 44 4

4

(70 lb-ft)(0.75 in.)(12 in./ft) 1,267.6 psi 1, 268 psi0.49701 in. p

T R I

τ = = = = Ans.

Angles of twist:

1 11 2 4

1 1

(480 lb-ft)(64 in.)(12 in./ft)0.039428 rad

(4,000,000 lb/in. )(2.33740 in. ) p

T L

G I φ = = =

2 22 2 4

2 2

( 130 lb-ft)(32 in.)(12 in./ft)0.005339 rad

(4,000,000 lb/in. )(2.33740 in. ) p

T L

G I φ

−= = = −

3 3

3 2 4

3 3

( 130 lb-ft)(48 in.)(12 in./ft)

0.012555 rad(12,000,000 lb/in. )(0.49701 in. ) p

T L

G I φ

−

= = = −

4 44 2 4

4 4

(70 lb-ft)(48 in.)(12 in./ft)0.006760 rad

(12,000,000 lb/in. )(0.49701 in. ) p

T L

G I φ = = =

Rotation angles: The angles of twist can be defined in terms of the rotation angles at the ends of each

segment:

1 2 3 4 B A C B D C E Dφ φ φ φ φ φ φ φ φ φ φ φ = − = − = − = −

The origin of the coordinate system is located at pulley A. We will arbitrarily define the rotation angle

at pulley A to be zero (φ A = 0). The rotation angle at B can be calculated from the angle of twist in

segment (1):1

1 0 0.039428 rad 0.039428 rad

B A

B A

φ φ φ

φ φ φ

= −

∴ = + = + =

Similarly, the rotation angle at C is determined from the angle of twist in segment (2) and the rotationangle of pulley B:

2

2 0.039428 rad ( 0.005339 rad) 0.034089 rad

C B

C B

φ φ φ

φ φ φ

= −

∴ = + = + − =

The rotation angle at D is:

3

3 0.034089 rad ( 0.012555 rad) 0.021534 rad

D C

D C

φ φ φ

φ φ φ

= −

∴ = + = + − =

and the rotation angle at E is:

4

4 0.021534 rad 0.006760 rad 0.028294 rad

E D

E D

φ φ φ

φ φ φ

= −

∴ = + = + =



Rotation angle of flange C with respect to pulley A: Using the rotation angles determined for the

system, the rotation angle of flange C with respect to pulley A is simply:

0.034089 rad 0.0341 radC φ = = Ans.

Rotation angle of pulley E with respect to pulley A: Using the rotation angles determined for thesystem, the rotation angle of flange C with respect to pulley A is simply:

0.028294 rad 0.0283 rad E φ = = Ans.

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Mechanics of Materials Solutions CH06