Mechanics of Materials Solutions Chapter06 Probs67 78

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6.67 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 2.00 in.and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainless

steel [G2 = 12,500 ksi] core (2) as shown in Fig. P6.67. If a torque of T = 20 kip-in. is applied to the

tube-and-core assembly, determine: (a) the torques produced in tube (1) and core (2).

(b) the maximum shear stress in the bronze tube

and in the stainless steel core.

(c) the angle of twist produced in a 10-in. length

of the assembly.

Fig. P6.67

Solution

Section Properties: For tube (1) and core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(2.00 in.) (1.25 in.) 1.331112 in.32

(1.25 in.) 0.239684 in.32

p

p

I

I

π

π

⎡ ⎤= − =⎣ ⎦

= =

Equilibrium:

1 2 20 kip-in. 0 x M T T Σ = + − = (a)

Torque-Twist Relationships:

1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I φ φ = = (b)

Geometry of Deformation Relationship: Since the tube and core are securely bonded together, theangles of twist in both members must be equal; therefore,

1 2φ φ = (c)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of

deformation relationship [Eq. (c)] to obtain the compatibility equation:

1 1 2 2

1 1 2 2 p p

T L T L

G I G I = (d)

Solve the Equations: Solve Eq. (d) for T 1:4

12 11 2 2 24

1 2 2

6,500 ksi 1.331112 in.2.887878

12,500 ksi 0.239684 in.

p

p

I L GT T T T

L G I

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

and substitute this result into Eq. (a) to compute the torque T 2 in core (2):

1 2 2 2 2

2

2.887878 3.887878 20 kip-in.

5.144 kip-in. 5.14 kip-in.

T T T T T

T

+ = + = =

∴ = = Ans.

The torque in tube (1) is therefore:

1 220 kip-in. 20 kip-in. 5.144 kip-in. 14.856 kip-in. 14.86 kip-in.T T = − = − = = Ans.





(b) Maximum Shear Stress: The maximum shear stress in tube (1) is:

1 11 4

1

(14.856 kip-in.)(2.00 in./ 2)11.16 ksi

1.331112 in. p

T R

I τ = = = Ans.

The maximum shear stress in core (2) is:

2 22 4

2

(5.144 kip-in.)(1.25 in. / 2)13.41 ksi

0.239684 in. p

T R

I τ = = = Ans.

(c) Angle of Twist produced in a 10-in. length: Since both the tube and the core twist exactly the

same amount [i.e., Eq. (c)], either torque-twist relationship can be used to compute the angle of twist of

the entire assembly.

1 11 4

1 1

(14.856 kip-in.)(10 in.)0.017170 rad 0.01717 rad

(6,500 ksi)(1.331112 in. ) p

T L

G I φ = = = = Ans.





6.68 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 2.00 in.and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainless

steel [G2 = 12,500 ksi] core (2) as shown in Fig. P6.68. The allowable shear stress of tube (1) is 27 ksi,

and the allowable shear stress of core (2) is 60 ksi. Determine: (a) the allowable torque T that can be applied to

the tube-and-core assembly.

(b) the corresponding torques produced in tube (1)

and core (2).

(c) the angle of twist produced in a 10-in. lengthof the assembly by the allowable torque T .

Fig. P6.68

Solution

Section Properties: For tube (1) and core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(2.00 in.) (1.25 in.) 1.331112 in.32

(1.25 in.) 0.239684 in.32

p

p

I

I

π

π

⎡ ⎤= − =⎣ ⎦

= =

Equilibrium:

1 2 0 x M T T T Σ = + − = (a)


1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I φ φ = = (b)

Geometry of Deformation Relationship: Since the tube and core are securely bonded together, the

angles of twist in both members must be equal; therefore,

1 2φ φ = (c)



1 1 2 2

1 1 2 2 p p

T L T L

G I G I = (d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient torewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:

p p

TR T

I I R

τ

τ = ∴ =

which allows Eq. (d) to be rewritten as:

1 1 2 2

1 1 2 2

L L

G R G R

τ τ = (e)

Solve Eq. (e) for τ 1:

2 1 11 2 2 2

1 2 2

6,500 ksi (2.00 in./2)0.832

12,500 ksi (1.25 in./2)

L G R

L G Rτ τ τ τ

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ (f)





Assume the core controls: If the shear stress in core (2) reaches its allowable value of 60 ksi (in otherwords, if the core controls), then the shear stress in tube (1) will be:

1 0.832(60 ksi) 49.92 ksi 27 ksiτ = = ≥ N.G.

which is larger than the 27-ksi allowable shear stress for the tube. Therefore, the shear stress in the tubecontrols.

Tube shear stress actually controls: Rearrange Eq. (f) to solve for the shear stress in core (2) giventhat the shear stress in tube (1) is at its 27-ksi allowable value:

2

27 ksi32.4519 ksi 60 ksi

0.832τ = = ≤ O.K.

Now that the maximum shear stresses in the tube and the core are known, the torques in each componentcan be computed:

( )

( )

41 1

1

1

42 2

2

2

(27 ksi)(1.331112 in. )35.940 kip-in.

2.00 in./2

(32.4519 ksi)(0.239684 in. )12.445 kip-in.

1.25 in./2

p

p

I T

R

I T

R

τ

τ

= = =

= = =

(a) Total Torque: From Eq. (a), the total torque acting on the assembly must not exceed:

max 1 2 35.940 kip-in. 12.445 kip-in. 48.385 kip-in.= 48.4 kip-in.T T T = + = + = Ans.

(b) Torques in each component: As computed previously, the torque in tube (1) is:

1 35.9 kip-in.T = Ans.

and the torque in core (2) is:

2 12.45 kip-in.T = Ans.

(c) Angle of Twist produced in a 10-in. length: Since both the tube and the core twist exactly the

same amount [i.e., Eq. (c)], either torque-twist relationship can be used to compute the angle of twist ofthe entire assembly.

1 11 4

1 1

(35.940 kip-in.)(10 in.)0.041538 rad 0.0415 rad

(6,500 ksi)(1.331112 in. ) p

T L

G I φ = = = = Ans.





6.69 A composite assembly consisting of an inner steel [G = 80 GPa] shaft (2) connected by rigid plates

at the ends of an aluminum [G = 28 GPa] tube (1) is shown in Fig. P6.69a. The cross-sectional

dimensions of the assembly are shown in Fig. P6.69b. If a torque of T = 1,500 N-m is applied to the

composite assembly, determine:(a) the maximum shear stress in the aluminum tube and in the steel core.

(b) the rotation angle of end B relative to end A.

Fig. P6.69a Tube-and-Core Composite Shaft Fig. P6.69b Cross-Sectional Dimensions

Solution

Section Properties: For aluminum tube (1) and steel core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(50 mm) (40 mm) 362, 265 mm32

(20 mm) 15,708.0 mm32

p

p

I

I

π

π

⎡ ⎤= − =⎣ ⎦

= =

Equilibrium:

1 2 1,500 N-m 0 x M T T Σ = + − = (a)


1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I φ φ = = (b)

Geometry of Deformation Relationship: Since the tube and core are connected by rigid plates, the


1 2φ φ = (c)



1 1 2 2

1 1 2 2 p p

T L T L

G I G I = (d)


12 11 2 2 24

1 2 2

28 GPa 362,265 mm8.071858

80 GPa 15,708.0 mm

p

p

I L GT T T T

L G I

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

and substitute this result into Eq. (a) to compute the torque T 2 in steel core (2):

1 2 2 2 2

2

8.071858 9.071858 1,500 N-m

165.346 N-m

T T T T T

T

+ = + = =

∴ =





The torque in aluminum tube (1) is therefore:

1 21,500 N-m 1,500 N-m 165.346 N-m 1,334.654 N-mT T = − = − =

(a) Maximum Shear Stress: The maximum shear stress in aluminum tube (1) is:

1 11 4

1

(1,334.654 N-m)(50 mm / 2)(1,000 mm/m)92.1 MPa

362,265 mm p

T R

I τ = = = Ans.

The maximum shear stress in steel core (2) is:

2 22 4

2

(165.346 N-m)(20 mm / 2)(1,000 mm/m) 105.3 MPa15,708.0 mm p

T R I

τ = = = Ans.

(b) Rotation Angle of B relative to A: Since both the tube and the core twist exactly the same amount[i.e., Eq. (c)], either torque-twist relationship can be used to compute the angle of twist of the entire

assembly.

1 11 2 4

1 1

(1,334.654 N-m)(450 mm)(1,000 mm/m)0.059210 rad 0.0592 rad

(28,000 N/mm )(362,265 mm ) p

T L

G I φ = = = = Ans.





6.70 A composite assembly consisting of an inner steel [G = 80 GPa] shaft (2) connected by rigid platesat the ends of an aluminum [G = 28 GPa] tube (1) is shown in Fig. P6.70a. The cross-sectional

dimensions of the assembly are shown in Fig. P6.70b. The allowable shear stress of aluminum tube (1) is

120 MPa, and the allowable shear stress of steel core (2) is 165 MPa. Determine:(a) the allowable torque T that can be applied to the composite shaft.

(b) the corresponding torques produced in tube (1) and core (2).

(c) the angle of twist produced by the allowable torque T .

Fig. P6.70a Tube-and-Core Composite Shaft Fig. P6.70b Cross-Sectional Dimensions

Solution

Section Properties: For aluminum tube (1) and steel core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(50 mm) (40 mm) 362, 265 mm32

(20 mm) 15,708.0 mm32

p

p

I

I

π

π

⎡ ⎤= − =⎣ ⎦

= =

Equilibrium:

1 2 0 x M T T T Σ = + − = (a)


1 1 2 21 2

1 1 2 2 p p

T L T LG I G I

φ φ = = (b)

Geometry of Deformation Relationship: Since the tube and core are connected by rigid plates, the


1 2φ φ = (c)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry ofdeformation relationship [Eq. (c)] to obtain the compatibility equation:

1 1 2 2

1 1 2 2 p p

T L T L

G I G I

= (d)


p p

TR T

I I R

τ τ = ∴ =






1 1 2 2

1 1 2 2

L L

G R G R

τ τ = (e)


2 1 11 2 2 2

1 2 2

28 GPa (50 mm/2)0.875

80 GPa (20 mm/2)

L G R

L G Rτ τ τ τ

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ (f)

Assume the core controls: If the shear stress in core (2) reaches its allowable value of 165 MPa (in

other words, if the core controls), then the shear stress in tube (1) will be:

1 0.875(165 MPa) 144.375 MPa 120 MPaτ = = ≥ N.G.

which is larger than the 120 MPa allowable shear stress for the tube. Therefore, the shear stress in the

tube controls.

Tube shear stress actually controls: Rearrange Eq. (f) to solve for the shear stress in core (2) given

that the shear stress in tube (1) is at its 120 MPa allowable value:

2

120 MPa137.1429 MPa 165 MPa

0.875τ = = ≤ O.K.

Now that the maximum shear stresses in the tube and the core are known, the torques in each componentcan be computed:

( )

( )

2 4

1 11

1

42 2

2

2

(120 N/mm )(362,265 mm ) 1,738,872 N-mm50 mm/2

(137.1429 MPa)(15,708.0 mm )215,424 N-mm

20 mm/2

p

p

I T R

I T

R

τ

τ

= = =

= = =


max 1 2 1,738,872 N-mm 215,424 N-mm 1,954,286 N-mm=1,954 N-mT T T = + = + = Ans.

(b) Torques in each component: As computed previously, the torque in tube (1) is:

1 1,739 N-mT =

Ans. and the torque in core (2) is:

2 215 N-mT = Ans.

(c) Angle of Twist: Since both the tube and the core twist exactly the same amount [i.e., Eq. (c)], either

torque-twist relationship can be used to compute the angle of twist of the entire assembly.

1 11 2 4

1 1

(1,738,872 N-mm)(450 mm)0.077143 rad 0.0771 rad

(28,000 N/mm )(362,265 mm ) p

T L

G I φ = = = = Ans.





6.71 The composite shaft shown in Fig. P6.71 consistsof a bronze sleeve (1) securely bonded to an inner steel

core (2). The bronze sleeve has an outside diameter o

35 mm, an inside diameter of 25 mm, and a shear modulus of G1 = 45 GPa. The solid steel core has a

diameter of 25 mm and a shear modulus of G2 = 80 GPa.

The composite shaft is subjected to a torque of T = 750 N-m. Determine:

(a) the maximum shear stresses in the bronze sleeve andthe steel core.

(b) the rotation angle of end B relative to end A.Fig. P6.71

Solution

Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(35 mm) (25 mm) 108,974 mm32

(25 mm) 38,349.5 mm32

p

p

I

I

π

π

⎡ ⎤= − =⎣ ⎦

= =

Equilibrium:

1 2 750 N-m 0 x M T T Σ = + − = (a)


1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I φ φ = = (b)

Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the


1 2φ φ = (c)


1 1 2 2

1 1 2 2 p p

T L T L

G I G I = (d)


12 11 2 2 24

1 2 2

45 GPa 108,974 mm1.598401

80 GPa 38,349.5 mm

p

p

I L GT T T T

L G I

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

and substitute this result into Eq. (a) to compute the torque T 2 in steel core (2):

1 2 2 2 2

2

1.598401 2.598401 750 N-m

288.639 N-m

T T T T T

T

+ = + = =

∴ =

The torque in bronze sleeve (1) is therefore:

1 2750 N-m 750 N-m 288.639 N-m 461.361 N-mT T = − = − =





(a) Maximum Shear Stress: The maximum shear stress in bronze sleeve (1) is:

1 11 4

1

(451.351 N-m)(35 mm / 2)(1,000 mm/m)74.1 MPa

108,974 mm p

T R

I τ = = = Ans.

The maximum shear stress in steel core (2) is:

2 22 4

2

(288.639 N-m)(25 mm/ 2)(1,000 mm/m)94.1 MPa

38,349.5 mm p

T R

I τ = = = Ans.

(b) Rotation Angle of B relative to A: Since both the sleeve and the core twist exactly the sameamount [i.e., Eq. (c)], either torque-twist relationship can be used to compute the angle of twist of the

entire assembly.

1 11 2 4

1 1

(461.361 N-m)(360 mm)(1,000 mm/m)0.033869 rad 0.0339 rad

(45,000 N/mm )(108,974 mm ) p

T L

G I φ = = = = Ans.





6.72 The composite shaft shown in Fig. P6.72 consists of a bronze sleeve (1) securely bonded to an inner steel core (2). The bronze sleeve has an outside diameter of 35 mm, an inside diameter of 25 mm, and a

shear modulus of G1 = 45 GPa. The solid steel core has a diameter of 25 mm and a shear modulus of G2

= 80 GPa. The allowable shear stress of sleeve (1) is 180 MPa, and the allowable shear stress of core (2)is 150 MPa. Determine:

(a) the allowable torque T that can be applied to the

composite shaft.(b) the corresponding torques produced in sleeve (1) and

core (2).(c) the rotation angle of end B relative to end A that is

produced by the allowable torque T .

Fig. P6.72

Solution

Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(35 mm) (25 mm) 108,974 mm32

(25 mm) 38,349.5 mm32

p

p

I

I

π

π

⎡ ⎤= − =⎣ ⎦

= =

Equilibrium:

1 2 0 x M T T T Σ = + − = (a)


1 1 2 2

1 21 1 2 2 p p

T L T L

G I G I φ φ = = (b)

Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the


1 2φ φ = (c)


1 1 2 2

1 1 2 2 p p

T L T L

G I G I = (d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to

rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:

p p

TR T

I I R

τ τ = ∴ =






1 1 2 2

1 1 2 2

L L

G R G R

τ τ = (e)


2 1 11 2 2 2

1 2 2

45 GPa (35 mm/2)0.7875

80 GPa (25 mm/2)

L G R

L G Rτ τ τ τ

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ (f)

Assume the core controls: If the shear stress in core (2) reaches its allowable value of 150 MPa (in

other words, if the core controls), then the shear stress in sleeve (1) will be:

1 0.7875(150 MPa) 118.125 MPa 180 MPaτ = = ≤ O.K.

This calculation shows that the shear stress in the core does in fact controls.

Now that the maximum shear stresses in the sleeve and the core are known, the torques in each

component can be computed:

( )

( )

2 41 1

1

1

42 2

2

2

(118.125 N/mm )(108,974 mm )735,574 N-mm

35 mm/2

(150 MPa)(38,349.5 mm )460,194 N-mm

25 mm/2

p

p

I T

R

I T

R

τ

τ

= = =

= = =


max 1 2 735,574 N-mm 460,194 N-mm 1,195,769 N-mm= 1,196 N-mT T T = + = + = Ans.

(b) Torques in each component: As computed previously, the torque in sleeve (1) is:

1 736 N-mT = Ans.

and the torque in core (2) is:

2 460 N-mT = Ans.

(c) Angle of Twist: Since both the sleeve and the core twist exactly the same amount [i.e., Eq. (c)],either torque-twist relationship can be used to compute the angle of twist of the entire assembly.

1 11 2 4

1 1

(735,574 N-mm)(360 mm)0.054000 rad 0.0540 rad

(45,000 N/mm )(108,974 mm ) p

T L

G I φ = = = = Ans.





6.73 The composite shaft shown in Fig. P6.73consists of two steel pipes that are connected at

flange B and securely attached to rigid walls at A

and C . Steel pipe (1) has an outside diameter o168 mm and a wall thickness of 7 mm. Steel pipe

(2) has an outside diameter of 114 mm and a wall

thickness of 6 mm. Both pipes are 3-m long andhave a shear modulus of 80 GPa. If a concentrated

torque of 20 kN-m is applied to flange B,determine:

(a) the maximum shear stress magnitudes in pipes(1) and (2).

(b) the rotation angle of flange B relative to

support A.

Fig. P6.73

Solution

Section Properties: The polar moments of inertia for the two steel pipes are:

4 4 4

1

4 4 4

2

(168 mm) (154 mm) 22,987,183 mm32

(114 mm) (102 mm) 5, 954, 575 mm32

p

p

I

I

π

π

⎡ ⎤= − =⎣ ⎦

⎡ ⎤= − =⎣ ⎦

Equilibrium:

1 220 kN-m 0

x M T T Σ = − + + = (a)


1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I φ φ = = (b)

Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixed

supports at A and C , the sum of the angles of twist in the two pipes must equal zero:

1 20φ φ + = (c)


1 1 2 2

1 1 2 2

0 p p

T L T L

G I G I + = (d)

Solve the Equations: Solve Eq. (d) for T 1:12 1

1 2

1 2 2

4

2 4

4

2 24

3 m 80 GPa 22, 987,183 mm

3 m 80 GPa 5,954,575 mm

22,987,183 mm3.860424

5,954,575 mm

p

p

I L GT T

L G I

T

T T

= −

⎛ ⎞⎛ ⎞⎛ ⎞= − ⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠





and substitute this result into Eq. (a) to compute the torque T 2:

( )1 2 2 2 2

2

3.860424 4.860424 20 kN-m

4.1149 kN-m

T T T T T

T

− + = − − + = = −

∴ = −

The torque in member (1) is therefore:

1 2 20 kN-m 4.1149 kN-m 20 kN-m 15.8851 kN-mT T = + = − + =

(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is:2

1 11 4

1

(15.8851 kN-m)(168 mm/ 2)(1,000)58.0 MPa

22,987,183 mm p

T R

I τ = = = Ans.

The maximum shear stress magnitude in member (2) is:2

2 22 4

2

(4.1149 kN-m)(114 mm / 2)(1,000)39.4 MPa

5,954,575 mm p

T R

I τ = = = Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be

defined by the difference in rotation angles at the two ends; hence,

1 B Aφ φ φ = −

Since joint A is restrained from rotating, φ A = 0 and thus

1 Bφ φ =

The rotation angle at B can be determined by computing the angle of twist in member (1):

1 11

1 1

2 4

(15.8851 kN-m)(3,000 mm)(1,000 mm/m)(1,000 N/kN)

(80,000 N/mm )(22,987,183 mm )

0.025914 rad 0.0259 rad

p

T L

G I φ =

=

= = Ans.





6.74 The composite shaft shown in Fig. P6.74 consists of two steel pipes that are connected at flange Band securely attached to rigid walls at A and C . Steel pipe (1) has an outside diameter of 8.625 in., a wall

thickness of 0.322 in., and a length of L1 = 15 ft. Steel pipe (2) has an outside diameter of 6.625 in., a

wall thickness of 0.280 in., and a length of L2 = 25 ft. Both pipes have a shear modulus of 12,000 ksi.If a concentrated torque of 20 kip-ft is applied to

flange B, determine:

(a) the maximum shear stress magnitudes in

pipes (1) and (2).

(b) the rotation angle of flange B relative tosupport A.

Fig. P6.74

Solution


4 4 4

1

4 4 4

2

(8.625 in.) (7.981 in.) 144.978 in.

32

(6.625 in.) (6.065 in.) 56.284 in.32

p

p

I

I

π

π

⎡ ⎤= − =

⎣ ⎦⎡ ⎤= − =⎣ ⎦

Equilibrium:

1 2 20 kip-ft 0 x M T T Σ = − + + = (a)


1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I φ φ = = (b)

Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixedsupports at A and C , the sum of the angles of twist in the two pipes must equal zero:

1 2 0φ φ + = (c)



1 1 2 2

1 1 2 2

0 p p

T L T L

G I G I + = (d)

Solve the Equations: Solve Eq. (d) for T 1:

12 11 2

1 2 2

4

2 24

25 ft 12,000 ksi 144.978 in.4.293050

15 ft 12, 000 ksi 56.284 in.

p

p

I L GT T

L G I

T T

= −

⎛ ⎞⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠






( )1 2 2 2 2

2

4.293050 5.293050 20 kip-ft

3.778540 kip-ft

T T T T T

T

− + = − − + = = −

∴ = −


1 2 20 kip-ft 3.778540 kip-ft 20 kip-ft 16.221460 kip-ftT T = + = − + =

(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is:

1 1

1 41

(16.221460 kip-ft)(8.625 in. / 2)(12 in./ft)

5.79 ksi144.978 in. p

T R

I τ = = =

Ans.

The maximum shear stress magnitude in member (2) is:

2 22 4

2

(3.778540 kip-ft)(6.625 in./ 2)(12 in./ft)2.67 ksi

56.284 in. p

T R

I τ = = = Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can bedefined by the difference in rotation angles at the two ends; hence,

1 B Aφ φ φ = −


1 Bφ φ =


1 11

1 1

2

4

(16.221460 kip-ft)(15 ft)(12 in./ft)

(12,000 ksi)(144.978 in. )

0.020140 rad 0.0201 rad

p

T L

G I φ =

=

= = Ans.





6.75 The composite shaft shown in Fig. P6.75 consists of a solid brass segment (1) and a solid aluminumsegment (2) that are connected at flange B and securely attached to rigid supports at A and C . Brass

segment (1) has a diameter of 1.00 in., a length of L1 = 6 in., a shear modulus of 5,600 ksi, and an

allowable shear stress of 15 ksi. Aluminum segment (2) has a diameter of 0.75 in., a length of L2 = 8 in.,a shear modulus of 4,000 ksi, and an allowable shear stress of 12 ksi. Determine:

(a) the allowable torque T B that can be

applied to the composite shaft at flange B.

(b) the magnitudes of the internal torques in

segments (1) and (2).(c) the rotation angle of flange B that is

produced by the allowable torque T B.

Fig. P6.75

Solution

Section Properties: The polar moments of inertia for the two shafts are:

4 4

1

4 4

2

(1.00 in.) 0.098175 in.32

(0.75 in.) 0.031063 in.32

p

p

I

I

π

π

= =

= =

Equilibrium:

1 2 0 x B M T T T Σ = − + + = (a)


1 1 2 2

1 21 1 2 2 p p

T L T L

G I G I φ φ = =

(b)

Geometry of Deformation Relationship: Since the two shafts are securely attached to fixed supportsat A and C , the sum of the angles of twist in the two members must equal zero:

1 2 0φ φ + = (c)



1 1 2 2

1 1 2 2

0 p p

T L T L

G I G I + = (d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to

rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:

p p

TR T

I I R

τ τ = ∴ =






1 1 2 2

1 1 2 2

L L

G R G R

τ τ = − (e)


2 1 11 2 2 2

1 2 2

8 in. 5,600 ksi (1.00 in./2)2.488889

6 in. 4,000 ksi (0.75 in./2)

L G R

L G Rτ τ τ τ

⎛ ⎞⎛ ⎞⎛ ⎞= − = − = −⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠ (f)

Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 12 ksi,

then the shear stress magnitude in shaft (1) will be:

1 2.488889(12 ksi) 29.866667 ksi 15 ksiτ = − = ≥ N.G.

which is larger than the 15 ksi allowable shear stress magnitude for shaft (1). Therefore, the shear stress

magnitude in shaft (1) must control.

Shaft (1) actually controls: Rearrange Eq. (f) to solve for the shear stress magnitude in shaft (2) given

that the shear stress magnitude in shaft (1) is at its 15 ksi allowable magnitude:

2

15 ksi6.026785 ksi 12 ksi

2.488889τ = = ≤

−O.K.

Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in

each component can be computed:

( )

( )

41 1

1

1

42 2

2

2

(15 ksi)(0.098175 in. )2.94525 kip-in.

1.00 in./2

(6.026785 ksi)(0.031063 in. )0.49923 kip-in.

0.75 in./2

p

p

I T

R

I T

R

τ

τ

= = =

= = =

Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated

equilibrium equation, it is apparent that T 2 must act opposite to the direction assumed in the FBD, givingit a negative value. Therefore, by inspection

2 0.49923 kip-in.T = −

(a) Total Torque: From Eq. (a), the total torque acting at flange B must not exceed:

,max 1 2 2.94525 kip-in. ( 0.49923 kip-in.) 3.44448 kip-in.= 3.44 kip-in. BT T T = − = − − = Ans.

(b) Torques Magnitudes: As computed previously, the torque magnitude in shaft (1) is:

1 2.95 kip-in.T = Ans.

and the torque magnitude in shaft (2) is:

2 0.499 kip-in.T = Ans.

(c) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be


1 B Aφ φ φ = −


1 Bφ φ =


1 11 4

1 1

(2.94525 kip-in.)(6 in.)0.032143 rad 0.0321 rad

(5,600 ksi)(0.098175 in. ) p

T L

G I φ = = = = Ans.





6.76 The composite shaft shown in Fig. P6.76consists of a solid brass segment (1) and a solid

aluminum segment (2) that are connected at flange

B and securely attached to rigid walls at A and C .Brass segment (1) has a diameter of 18 mm, a

length of L1 = 235 mm, and a shear modulus of 39

GPa. Aluminum segment (2) has a diameter of 24

mm, a length of L2 = 165 mm, and a shear modulus

of 28 GPa. If a concentrated torque of 270 N-m isapplied to flange B, determine:

(a) the maximum shear stress magnitudes insegments (1) and (2).

(b) the rotation angle of flange B relative to

support A.Fig. P6.76

Solution


4 4

1

4 4

2

(18 mm) 10,306.0 mm32

(24 mm) 32,572.0 mm32

p

p

I

I

π

π

= =

= =

Equilibrium:

1 2 270 N-m 0 x M T T Σ = − + + = (a)


1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I

φ φ = = (b)

Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixedsupports at A and C , the sum of the angles of twist in the two pipes must equal zero:

1 2 0φ φ + = (c)



1 1 2 2

1 1 2 2

0 p p

T L T L

G I G I + = (d)


12 11 2

1 2 2

4

2 24

165 mm 39 GPa 10, 306.0 mm0.309434

235 mm 28 GPa 32, 572.0 mm

p

p

I L GT T

L G I

T T

= −

⎛ ⎞⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠






( )1 2 2 2 2

2

0.309434 1.309434 270 N-m

206.1959 N-m

T T T T T

T

− + = − − + = = −

∴ = −


1 2 270 N-m 206.1959 N-m 270 N-m 63.8041 N-mT T = + = − + =


1 1

1 41

(63.8041 N-m)(18 mm / 2)(1,000 mm/m)

55.7 MPa10,306.0 mm p

T R

I τ = = =

Ans.


2 22 4

2

(206.1959 N-m)(24 mm/ 2)(1,000 mm/m)76.0 MPa

32,572.0 mm p

T R

I τ = = = Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can bedefined by the difference in rotation angles at the two ends; hence,

1 B Aφ φ φ = −


1 Bφ φ =


1 11

1 1

2 4

(63.8041 N-m)(235 mm)(1,000 mm/m)

(39,000 N/mm )(10,306.0 mm )

0.037305 rad 0.0373 rad

p

T L

G I φ =

=

= = Ans.





6.77 The composite shaft shown in Fig. P6.77consists of a stainless steel tube (1) and a brass

tube (2) that are connected at flange B and securely

attached to rigid supports at A and C . Stainlesssteel tube (1) has an outside diameter of 2.25 in., a

wall thickness of 0.250 in., a length of L1 = 40 in.,

and a shear modulus of 12,500 ksi. Brass tube (2)

has an outside diameter of 3.500 in., a wall

thickness of 0.219 in., a length of L2 = 20 in., and ashear modulus of 5,600 ksi. If a concentrated

torque of T B = 42 kip-in. is applied to flange B,determine:

(a) the maximum shear stress magnitudes in tubes

(1) and (2).(b) the rotation angle of flange B relative to

support A.Fig. P6.77

Solution

Section Properties: The polar moments of inertia for the two tubes are:

4 4 4

1

4 4 4

2

(2.250 in.) (1.750 in.) 1.595340 in.32

(3.500 in.) (3.062 in.) 6.102156 in.32

p

p

I

I

π

π

⎡ ⎤= − =⎣ ⎦

⎡ ⎤= − =⎣ ⎦

Equilibrium:

1 2 42 kip-in. 0 x M T T Σ = − + + = (a)


1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I φ φ = = (b)

Geometry of Deformation Relationship: Since the two tubes are securely attached to fixed supports at

A and C , the sum of the angles of twist in the two tubes must equal zero:

1 2 0φ φ + = (c)



1 1 2 2

1 1 2 2

0

p p

T L T L

G I G I + = (d)


12 11 2

1 2 2

4

2 24

20 in. 12,500 ksi 1.595340 in.0.291784

40 in. 5, 600 ksi 6.102156 in.

p

p

I L GT T

L G I

T T

= −

⎛ ⎞⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠






( )1 2 2 2 2

2

0.291784 1.291784 42 kip-in.

32.5132 kip-in.

T T T T T

T

− + = − − + = = −

∴ = −


1 2 42 kip-in. 32.5132 kip-in. 42 kip-in. 9.4868 kip-in.T T = + = − + =


1 11 4

1

(9.4868 kip-in.)(2.250 in./ 2) 6.69 ksi1.595340 in. p

T R

I τ = = = Ans.


2 22 4

2

(32.5132 kip-in.)(3.500 in./ 2)9.32 ksi

6.102156 in. p

T R

I τ = = = Ans.



1 B Aφ φ φ = −


1 Bφ φ =


1 11

1 1

4

(9.4868 kip-in.)(40 in.)

(12,500 ksi)(1.595340 in. )

0.019029 rad 0.01903 rad

p

T L

G I φ =

=

= = Ans.





6.78 The composite shaft shown in Fig. P6.78consists of a stainless steel tube (1) and a brass

tube (2) that are connected at flange B and securely

attached to rigid supports at A and C . Stainlesssteel tube (1) has an outside diameter of 2.25 in., a

wall thickness of 0.250 in., a length of L1 = 40 in.,

and a shear modulus of 12,500 ksi. Brass tube (2)

has an outside diameter of 3.500 in., a wall

thickness of 0.219 in., a length of L2 = 20 in., and ashear modulus of 5,600 ksi. The allowable shear

stress in the stainless steel is 50 ksi, and theallowable shear stress in the brass is 18 ksi.

Determine:

(a) the allowable torque T B that can be applied tothe composite shaft at flange B.

(b) the rotation angle of flange B that is produced

by the allowable torque T B.

Fig. P6.78

Solution

Section Properties: The polar moments of inertia for the two tubes are:

4 4 4

1

4 4 4

2

(2.250 in.) (1.750 in.) 1.595340 in.32

(3.500 in.) (3.062 in.) 6.102156 in.32

p

p

I

I

π

π

⎡ ⎤= − =⎣ ⎦

⎡ ⎤= − =⎣ ⎦

Equilibrium:

1 2 0 x B M T T T Σ = − + + = (a)


1 1 2 21 2

1 1 2 2 p p

T L T L

G I G I φ φ = = (b)

Geometry of Deformation Relationship: Since the two tubes are securely attached to fixed supports at

A and C , the sum of the angles of twist in the two tubes must equal zero:

1 2 0φ φ + = (c)



1 1 2 2

1 1 2 20

p p

T L T L

G I G I + = (d)


p p

TR T

I I R

τ τ = ∴ =




Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only

1 1 2 2

1 1 2 2

L L

G R G R

τ τ = − (e)


2 1 11 2 2 2

1 2 2

20 in. 12,500 ksi (2.250 in./2)0.717474

40 in. 5,600 ksi (3.500 in./2)

L G R

L G Rτ τ τ τ

⎛ ⎞⎛ ⎞⎛ ⎞= − = − = −⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠ (f)

Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 18 ksi,

then the shear stress magnitude in shaft (1) will be:

1 0.717474(18 ksi) 12.914541 ksi 50 ksiτ = − = ≤ O.K.

This calculation shows that the shear stress in shaft (2) does in fact controls.

Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes ineach component can be computed:

( )

( )

41 1

1

1

42 2

2

2

(12.914541 ksi)(1.595340 in. )18.3139 kip-in.

2.250 in./2

(18 ksi)(6.102156 in. )62.7650 kip-in.

3.500 in./2

p

p

I T

R

I T

R

τ

τ

= = =

= = =

Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated

equilibrium equation, it is apparent that T 2 must act opposite to the direction assumed in the FBD, giving

it a negative value. Therefore, by inspection

2 62.7650 kip-in.T = −

(a) Allowable Torque T B: From Eq. (a), the total torque acting at flange B must not exceed:

,max 1 2 18.3138 kip-in. ( 62.7650 kip-in.) 81.0788 kip-in.= 81.1 kip-in. BT T T = − = − − = Ans.


defined by the difference in rotation angles at the two ends; hence,1 B Aφ φ φ = −


1 Bφ φ =


1 11 4

1 1

(18.3139 kip-in.)(40 in.)0.036735 rad 0.0367 rad

(12,500 ksi)(1.595340 in. ) p

T L

G I φ = = = = Ans.

Mechanics of Materials Solutions Chapter06 Probs67 78

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Transcript of Mechanics of Materials Solutions Chapter06 Probs67 78