Mechanics of Materials Chapter06 Probs79 93

6.79 The torsional assembly of Fig. P6.79 consists of a cold-rolled stainless steel tube connected to a solid cold-rolled brass segment at flange C. The assembly is securely fastened to rigid supports at A and D. Stainless steel tube (1) and (2) has an outside diameter of 3.50 in., a wall thickness of 0.120 in., and a shear modulus of G = 12,500 ksi. The solid brass segment (3) has a diameter of 2.00 in. and a shear modulus of G = 5,600 ksi. A concentrated torque of TB = 7 kip-ft is applied to the stainless steel pipe at B. Determine: (a) the maximum shear stress magnitude in the stainless steel tube. (b) the maximum shear stress magnitude in brass segment (3). (c) the rotation angle of flange C.

Fig. P6.79

Solution Section Properties: The polar moments of inertia for the stainless steel tube and the solid brass segment are:

I p1 = I p3 =

π

⎡ (3.500 in.) 4 − (3.260 in.) 4 ⎤⎦ = 3.643915 in.4 = I p 2 32 ⎣

π 32

(2.0 in.) 4 = 1.570796 in.4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: ΣM x = −T1 + T2 + TB = 0 (a)

and also consider a FBD cut around joint C. Although there is not an external torque applied at joint C, the section properties of the torsion structure change at C. ΣM x = −T2 + T3 = 0 ∴T2 = T3 (b) Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

φ3 =

T3 L3 G3 I p 3

(c)

Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero: φ1 + φ2 + φ3 = 0 (d) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (d)] to obtain the compatibility equation: TL T1 L1 TL + 2 2 + 3 3 =0 G1 I p1 G2 I p 2 G3 I p 3

(e)

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 States Copyright Act without the permission of the copyright owner is unlawful.

Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1 and T3 with equivalent expressions involving T2. From Eq. (a): T1 = T2 + TB (f) Substitute Eqs. (b) and (f) into Eq. (e) and simplify to derive an expression for T2: TL (T2 + TB ) L1 T2 L2 + + 2 3 =0 G1 I p1 G2 I p 2 G3 I p 3 ⎡ L L ⎤ L T L T2 ⎢ 1 + 2 + 3 ⎥ = − B 1 G1 I p1 ⎢⎣ G1 I p1 G2 I p 2 G3 I p 3 ⎥⎦

TB L1 G1 I p1 T2 = − L L1 L + 2 + 3 G1 I p1 G2 I p 2 G3 I p 3 Note that G1 = G2 and Ip1 = Ip2. Compute T2: (7 kip-ft)(30 in.)(12 in./ft) (12,500 ksi)(3.643915 in.4 ) T2 = − 30 in. + 22 in. 18 in. + 4 (12,500 ksi)(3.643915 in. ) (5,600 ksi)(1.570796 in.4 ) = −17.3547 kip-in. Substitute this result into Eq. (f) to compute internal torque T1: T1 = T2 + TB = −17.3547 kip-in. + (7 kip-ft)(12 in./ft) = 66.6453 kip-in. and from Eq. (b): T3 = T2 = −17.3547 kip-in.

(e)

Maximum Shear Stress: The maximum shear stress magnitudes in the three members are: T R (66.6453 kip-in.)(3.500 in./ 2) τ1 = 1 1 = = 32.0 ksi I p1 3.643915 in.4 TR (17.3547 kip-in.)(3.500 in./ 2) τ2 = 2 2 = = 8.34 ksi I p2 3.643915 in.4 T R (17.3547 kip-in.)(2.000 in./ 2) τ3 = 3 3 = = 11.05 ksi I p3 1.570796 in.4 (a) Maximum Shear Stress Magnitude in Stainless Steel Tube: τ 1 = 32.0 ksi

Ans.

(b) Maximum Shear Stress Magnitude in Brass Shaft: τ 3 = 11.05 ksi

Ans.

(c) Rotation Angle of Flange C Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends: φ1 = φB − φ A ∴φB = φ A + φ1 Similarly, the angle of twist in member (2) can be defined by: φ2 = φC − φB ∴φC = φB + φ2

(f) (g)


To derive an expression for φC, substitute Eq. (f) into Eq. (g), and note that joint A is restrained from rotating; therefore, φA = 0. φC = φ1 + φ2 The angle of twist in member (1) is: TL (66.6453 kip-in.)(30 in.) φ1 = 1 1 = = 0.043895 rad G1 I p1 (12,500 ksi)(3.643915 in.4 ) The angle of twist in member (2) is: TL (−17.3547 kip-in.)(22 in.) = −0.008382 rad φ2 = 2 2 = G2 I p 2 (12,500 ksi)(3.643915 in.4 ) The rotation angle of flange C is thus: φC = φ1 + φ2 = 0.043895 rad + (−0.008382 rad) = 0.035513 rad = 0.0355 rad

Ans.


6.80 The torsional assembly of Fig. P6.80 consists of a cold-rolled stainless steel tube connected to a solid cold-rolled brass segment at flange C. The assembly is securely fastened to rigid supports at A and D. Stainless steel tubes (1) and (2) have an outside diameter of 3.50 in., a wall thickness of 0.120 in., a shear modulus of G = 12,500 ksi, and an allowable shear stress of 50 ksi. The solid brass segment (3) has a diameter of 2.00 in., a shear modulus of G = 5,600 ksi, and an allowable shear stress of 18 ksi. Determine the maximum permissible magnitude for the concentrated torque TB.

Fig. P6.80

Solution Section Properties: The polar moments of inertia for the stainless steel tube and the solid brass segment are:

I p1 = I p3 =

π

⎡ (3.500 in.) 4 − (3.260 in.) 4 ⎤⎦ = 3.643915 in.4 = I p 2 32 ⎣

π 32

(2.0 in.) 4 = 1.570796 in.4


and also consider a FBD cut around joint C. Although there is not an external torque applied at joint C, the section properties of the torsion structure change at C. ΣM x = −T2 + T3 = 0 ∴T2 = T3 (b) Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

φ3 =

T3 L3 G3 I p 3

(c)


(e)

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


TR T τ ∴ = Ip Ip R which allows Eq. (e) to be rewritten as: τ 1 L1 τ 2 L2 τ 3 L3 + + =0 (f) G1 R1 G2 R2 G3 R3 Also using the elastic torsion formula, Eq. (b) can be expressed in terms of stress: τ 2 I p 2 τ 3 I p3 τ I τ = ∴ 3 = 2 p2 (g) R2 R3 R3 R2 I p 3 Substitute Eq. (g) into Eq. (f) to obtain τ 1 L1 τ 2 L2 L3 τ 2 I p 2 + + =0 G1 R1 G2 R2 G3 R2 I p 3 Simplify: ⎡ L2 L I ⎤ + 3 p2 ⎥ ⎢ ⎡ L G2 R2 G3 R2 I p 3 ⎥ L I ⎤ L τ 1 1 = −τ 2 ⎢ 2 + 3 p 2 ⎥ ∴τ 1 = −τ 2 ⎢ L1 ⎢ ⎥ G1 R1 ⎢⎣ G2 R2 G3 R2 I p 3 ⎥⎦ ⎢ ⎥ G1 R1 ⎣ ⎦ Substitute values: ⎡ ⎛ 3.643915 in.4 ⎞ ⎤ 22 in. 18 in. + ⎢ ⎜ ⎟⎥ (12,500 ksi)(3.50 in./2) (5, 600 ksi)(3.50 in./2) ⎝ 1.570796 in.4 ⎠ ⎥ ⎢ τ 1 = −τ 2 30 in. ⎢ ⎥ ⎢ ⎥ (12,500 ksi)(3.50 in./2) ⎣ ⎦ = −3.840193τ 2 This calculation demonstrates that the shear stress in segment (1) of the stainless steel tube is much larger than the shear stress in segment (2). If the shear stress magnitude in segment (1) is 50 ksi, then the shear stress magnitude in segment (2) will be:

τ=

τ2 = −

τ1

= 13.020179 ksi 3.840193 Next, we need to check the corresponding shear stress in brass shaft (3). From Eq. (g): R I τ 3 = τ 2 3 p2 R2 I p 3

(h)

Substitute the magnitude obtained for τ2 in Eq. (h) into this expression and calculate the corresponding shear stress magnitude in shaft (3): 4 ⎛ 2.00 in./2 ⎞ ⎛ 3.643915 in. ⎞ τ 3 = (13.020179 ksi) ⎜ = 17.259466 ksi ≤ 18 ksi O.K. ⎟⎜ 4 ⎟ ⎝ 3.50 in./2 ⎠ ⎝ 1.570796 in. ⎠ Now that the maximum shear stresses in the three shaft segments are known, the torques in each component can be computed:


T1 = T2 = T3 =

τ 1 I p1 R1

τ 2 I p2 R2

τ 3 I p3 R3

=

(50 ksi)(3.643915 in.4 ) = 104.1119 kip-in. ( 3.50 in./2 )

=

(−13.020179 ksi)(3.643915 in.4 ) = −27.1111 kip-in. ( 3.50 in./2 )

(−17.259466 ksi)(1.570796 in.4 ) = = −27.1111 kip-in. ( 2.00 in./2 )

Maximum Permissible Torque TB: From Eq. (a), the torque TB acting on the assembly must not exceed: TB ,max = T1 − T2 = 104.1119 kip-in. − (−27.1111 kip-in.) = 131.2230 kip-in.= 131.2 kip-in. Ans.


6.81 The torsional assembly of Fig. P6.81a consists of a solid 75-mm-diameter bronze [G = 45 GPa] segment (1) securely connected at flange B to solid 75-mm-diameter stainless steel [G = 86 GPa] segments (2) and (3). The flange at B is secured by four 14-mm-diameter bolts, which are each located on a 120-mm-diameter bolt circle (Fig. P6.81b). The allowable shear stress of the bolts is 90 MPa, and friction effects in the flange can be neglected for this analysis. Determine: (a) the allowable torque TC that can be applied to the assembly at C without exceeding the capacity of the bolted flange connection. (b) the maximum shear stress magnitude in bronze segment (1). (c) the maximum shear stress magnitude in stainless steel segments (2) and (3).

Fig. P6.81a

Fig. P6.81b Flange B Bolts

Solution Section Properties: For bronze segment (1) and stainless steel segments (2) and (3), the polar moments of inertia are identical:

I p1 = I p 2 = I p 3 =

π

32

(75 mm) 4 = 3,106,311 mm 4

Equilibrium: Consider a free-body diagram cut around flange B: ΣM x = −T1 + T2 = 0 ∴T1 = T2 (a)

and also consider a FBD cut around joint C, where the external torque TC is applied: ΣM x = −T2 + T3 + TC = 0 (b) Capacity of Flange B: The flange at B is secured by four 14-mm-diameter bolts, which are each located on a 120-mm-diameter bolt circle. The allowable shear stress of the bolts is 90 MPa, and each bolt acts in single shear. The cross-sectional area of a single bolt is:

Abolt =

π

(14 mm) 2 = 153.9380 mm 2

4 The shear force that can be provided by each bolt is: Vbolt = (90 N/mm 2 )(153.9380 mm 2 ) = 13,854.4236 N The total torque that can resisted by the bolts is thus: TB ,max = (4 bolts)(13,954.4236 N/bolt)(120 mm/2) = 3,325, 062 N-mm 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 States Copyright Act without the permission of the copyright owner is unlawful.

The internal torques in segments (1) and (2) cannot exceed this magnitude; therefore, based on Eq. (a): T1 = T2 = 3,325, 062 N-mm (c) From Eq. (b), we observe that the external torque applied to the shaft is related to internal torques T2 and T3. We must determine the relationship between these two internal torques by developing a compatibility equation, which is based on the geometry of deformations for this configuration. Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

φ3 =

T3 L3 G3 I p 3

(d)

Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three segments must equal zero: φ1 + φ2 + φ3 = 0 (e) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of deformation relationship [Eq. (e)] to obtain the compatibility equation: TL T1 L1 TL + 2 2 + 3 3 =0 G1 I p1 G2 I p 2 G3 I p 3 Solve the Equations: Solve Eq. (f) for T3, noting that Ip1 = Ip2 = Ip3, G2 = G3, and T1 = T2: ⎡ TL ⎡L G L ⎤ T L ⎤G I T3 = − ⎢ 1 1 + 2 2 ⎥ 3 p 3 = −T1 ⎢ 1 3 + 2 ⎥ ⎣ L3 G1 L3 ⎦ ⎣⎢ G1 I p1 G2 I p 2 ⎦⎥ L3 Substitute the values determined for T1 and T2 in Eq. (c) into Eq. (g) to compute T3: ⎡⎛ 400 mm ⎞⎛ 86 GPa ⎞ 750 mm ⎤ T3 = −(3,325, 062 N-mm) ⎢⎜ ⎟⎜ ⎟+ ⎥ ⎣⎝ 400 mm ⎠⎝ 45 GPa ⎠ 400 mm ⎦

(f)

(g)

= −(3,325, 062 N-mm)(3.786111) = −12,589, 054 N-mm (a) Allowable Torque TC: From equilibrium equation (b), TC = T2 − T3 = 3,325, 062 N-mm − (−12,589, 054 N-mm) = 15,914,116 N-mm = 15.91 kN-m

Ans.

(b) Shear Stress Magnitude in Bronze Segment (1): T R (3,325, 062 N-mm)(75 mm / 2) τ1 = 1 1 = = 40.1 MPa I p1 3,106,311 mm 4

Ans.

(c) Shear Stress Magnitude in Stainless Steel Segments (2) and (3): T R (12,589, 054 N-mm)(75 mm / 2) τ3 = 3 3 = = 152.0 MPa 3,106,311 mm 4 I p3

Ans.


6.82 The torsional assembly shown in Fig. P6.82 consists of solid 2.50-in.-diameter aluminum [G = 4,000 ksi] segments (1) and (3) and a central solid 3.00-in.-diameter bronze [G = 6,500 ksi] segment (2). Concentrated torques of TB = T0 and TC = 2T0 are applied to the assembly at B and C, respectively. If T0 = 20 kip-in., determine: (a) the maximum shear stress magnitude in aluminum segments (1) and (3). (b) the maximum shear stress magnitude in bronze segment (2). (c) the rotation angle of joint C.

Fig. P6.82

Solution Section Properties: The polar moments of inertia for the aluminum and bronze segments are:

I p1 = I p 3 = I p2 =

π 32

π

32

(2.50 in.) 4 = 3.834952 in.4

(3.00 in.)4 = 7.952156 in.4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: ΣM x = −T1 + T2 + T0 = 0 (a)

and also consider a FBD cut around joint C, where the external torque TC is applied: ΣM x = −T2 + T3 + 2T0 = 0 Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

φ3 =

(b)

T3 L3 G3 I p 3

(c)


(e)

Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1 and T3 with equivalent expressions involving T2. From Eq. (a): T1 = T2 + T0 (f) and from Eq. (b): 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 States Copyright Act without the permission of the copyright owner is unlawful.

T3 = T2 − 2T0 Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2: (T2 + T0 ) L1 T2 L2 (T2 − 2T0 ) L3 + + =0 G1 I p1 G2 I p 2 G3 I p 3 ⎡ L L ⎤ 2T L TL L T2 ⎢ 1 + 2 + 3 ⎥ = 0 3 − 0 1 ⎢⎣ G1 I p1 G2 I p 2 G3 I p 3 ⎥⎦ G3 I p 3 G1 I p1 2T0 L3 T0 L1 − G3 I p 3 G1 I p1 T2 = L L1 L + 2 + 3 G1 I p1 G2 I p 2 G3 I p 3 Since G1 = G3 and Ip1 = Ip3, Eq. (h) can be further simplified and T2 can be computed as: T ⎡ ( 2 L3 − L1 ) ⎤ T2 = 0 ⎢ ⎥ G1 I p1 ⎢ L1 + L3 + L2 ⎥ ⎣⎢ G1 I p1 G2 I p 2 ⎦⎥

(g)

(h)

20 kip-in. 2(12 in.) − 12 in. ⎡ ⎤ 4 ⎢ ⎥ 12 in. + 12 in. 24 in. (4,000 ksi)(3.834952 in. ) + ⎢ 4 4 ⎥ ⎣ (4,000 ksi)(3.834952 in. ) (6,500 ksi)(7.952156 in. ) ⎦ = 7.711461 kip-in. Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3: T1 = 7.711461 kip-in. + 20 kip-in. = 27.711461 kip-in. T3 = 7.711461 kip-in. − 2(20 kip-in.) = −32.288539 kip-in. =

Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are: T R (27.711461 kip-in.)(2.500 in./ 2) τ1 = 1 1 = = 9.03 ksi I p1 3.834952 in.4 T2 R2 (7.711461 kip-in.)(3.000 in./ 2) = = 1.455 ksi I p2 7.952156 in.4 T R (32.288539 kip-in.)(2.500 in./ 2) τ3 = 3 3 = = 10.52 ksi 3.834952 in.4 I p3

τ2 =

(a) Maximum Shear Stress Magnitude in Aluminum Segments: τ 3 = 10.52 ksi

Ans.

(b) Maximum Shear Stress Magnitude in Bronze Segment: τ 2 = 1.455 ksi

Ans.

(c) Rotation Angle of Flange C Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends: φ1 = φB − φ A ∴φB = φ A + φ1 Similarly, the angle of twist in member (2) can be defined by: φ2 = φC − φB ∴φC = φB + φ2

(i) (j)


To derive an expression for φC, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from rotating; therefore, φA = 0. φC = φ1 + φ2 The angle of twist in member (1) is: TL (27.711461 kip-in.)(12 in.) φ1 = 1 1 = = 0.021678 rad G1 I p1 (4, 000 ksi)(3.834952 in.4 ) The angle of twist in member (2) is: TL (7.711461 kip-in.)(24 in.) φ2 = 2 2 = = 0.003581 rad G2 I p 2 (6,500 ksi)(7.952156 in.4 ) The rotation angle of flange C is thus: φC = φ1 + φ2 = 0.021678 rad + 0.003581 rad = 0.025259 rad = 0.0253 rad

Ans.


6.83 The torsional assembly shown in Fig. P6.83 consists of solid 2.50-in.-diameter aluminum [G = 4,000 ksi] segments (1) and (3) and a central solid 3.00-in.-diameter bronze [G = 6,500 ksi] segment (2). Concentrated torques of TB = T0 and TC = 2T0 are applied to the assembly at B and C, respectively. If the rotation angle at joint C must not exceed 3°, determine: (a) the maximum magnitude of T0 that may be applied to the assembly. (b) the maximum shear stress magnitude in aluminum segments (1) and (3). (c) the maximum shear stress magnitude in bronze segment (2).

Fig. P6.83


I p1 = I p 3 = I p2 =

π 32

π

32

(2.50 in.) 4 = 3.834952 in.4

(3.00 in.)4 = 7.952156 in.4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: ΣM x = −T1 + T2 + T0 = 0 (a)

and also consider a FBD cut around joint C, where the external torque TC is applied: ΣM x = −T2 + T3 + 2T0 = 0 Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

φ3 =

(b)

T3 L3 G3 I p 3

Torque in Segment (3): From the problem statement, the rotation angle at C must not exceed 3°; consequently, the angle of twist in segment (3) must be the opposite of this value: φ3 = −3° = −0.052360 rad From the torque-twist relationships, the internal torque in segment (3) can be computed: φGI TL (−0.052360 rad)(4,000 ksi)(3.834952 in.4 ) φ3 = 3 3 ∴T3 = 3 3 p 3 = = −66.9327 kip-in. 12 in. G3 I p 3 L3 Consider Rotation Angle of Flange C Relative to Support A: The rotation angle at C is given by the sum of the angles of twist in segments (1) and (2). This rotation angle must not exceed 3°. TL TL (c) φC = φ1 + φ2 = 1 1 + 2 2 ≤ 3° = 0.052360 rad G1 I p1 G2 I p 2 from Eq. (b): 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 States Copyright Act without the permission of the copyright owner is unlawful.

T2 = T3 + 2T0 (d) and from Eq. (a): T1 = T2 + T0 = (T3 + 2T0 ) + T0 = T3 + 3T0 (e) Substitute the expressions derived for T1 and T2 in Eqs. (d) and (e) into Eq. (c): (T3 + 3T0 ) L1 (T3 + 2T0 ) L2 + ≤ 0.052360 rad G1 I p1 G2 I p 2 Simplify ⎡ L ⎡ 3L L ⎤ 2 L2 ⎤ T3 ⎢ 1 + 2 ⎥ + T0 ⎢ 1 + ⎥ ≤ 0.052360 rad ⎣⎢ G1I p1 G2 I p 2 ⎦⎥ ⎣⎢ G1I p1 G2 I p 2 ⎦⎥ Solve for T0: ⎡ L L ⎤ 0.052360 rad − T3 ⎢ 1 + 2 ⎥ ⎢⎣ G1 I p1 G2 I p 2 ⎥⎦ T0 ≤ 3L1 2 L2 + G1 I p1 G2 I p 2 and compute: ⎡ ⎤ 12 in. 24 in. + 0.052360 rad − (−66.9327 kip-in.) ⎢ 4 4 ⎥ ⎣ (4,000 ksi)(3.834952 in. ) (6,500 ksi)(7.952156 in. ) ⎦ T0 ≤ 3(12 in.) 2(24 in.) + 4 (4,000 ksi)(3.834952 in. ) (6,500 ksi)(7.952156 in.4 ) ≤ 41.4591 kip-in. Backsubstitute this value into Eq. (d) to compute T2: T2 = T3 + 2T0 = −66.9327 kip-in. + 2(41.4591 kip-in.) = 15.9855 kip-in. and backsubstitute this value into Eq. (e) to compute T1: T1 = T3 + 3T0 = −66.9327 kip-in. + 3(41.4591 kip-in.) = 57.4446 kip-in. (a) Maximum magnitude of T0: T0,max = 41.5 kip-in.

Ans.

Maximum Shear Stress Magnitudes: The maximum shear stress magnitudes are: T R (57.4446 kip-in.)(2.500 in./ 2) τ1 = 1 1 = = 18.72 ksi I p1 3.834952 in.4 T2 R2 (15.9855 kip-in.)(3.000 in./ 2) = = 3.02 ksi I p2 7.952156 in.4 T R (66.9327 kip-in.)(2.500 in./ 2) τ3 = 3 3 = = 21.8 ksi I p3 3.834952 in.4

τ2 =

(b) Maximum Shear Stress Magnitude in Aluminum Segments: τ 3 = 21.8 ksi

Ans.

(c) Maximum Shear Stress Magnitude in Bronze Segment: τ 2 = 3.02 ksi

Ans.


6.84 The torsional assembly shown in Fig. P6.84 consists of a solid 60-mm-diameter aluminum [G = 28 GPa] segment (2) and two bronze [G = 45 GPa] tube segments (1) and (3), which have an outside diameter of 75 mm and a wall thickness of 5 mm. If concentrated torques of TB = 9 kN-m and TC = 9 kN-m are applied in the directions shown, determine: (a) the maximum shear stress magnitude in bronze tube segments (1) and (3). (b) the maximum shear stress magnitude in aluminum segment (2). (c) the rotation angle of joint C.

Fig. P6.84


I p1 = I p 3 = I p2 =

π 32

π

⎡⎣(75 mm) 4 − (65 mm) 4 ⎤⎦ = 1,353,830 mm 4 32

(60 mm) 4 = 1, 272,345 mm 4


and also consider a FBD cut around joint C, where the external torque TC is applied: ΣM x = −T2 + T3 − TC = 0 Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

φ3 =

(b)

T3 L3 G3 I p 3

(c)


(e)

Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1 and T3 with equivalent expressions involving T2. From Eq. (a): T1 = T2 + TB (f) 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 States Copyright Act without the permission of the copyright owner is unlawful.

and from Eq. (b): T3 = T2 + TC Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2: (T2 + TB ) L1 T2 L2 (T2 + TC ) L3 + + =0 G1 I p1 G2 I p 2 G3 I p 3

(g)

⎡ L L ⎤ T L L T L T2 ⎢ 1 + 2 + 3 ⎥ = − B 1 − C 3 G1 I p1 G3 I p 3 ⎣⎢ G1 I p1 G2 I p 2 G3 I p 3 ⎦⎥

TB L1 TC L3 + G1 I p1 G3 I p 3 T2 = − L L1 L + 2 + 3 G1 I p1 G2 I p 2 G3 I p 3 Since G1 = G3 and Ip1 = Ip3, Eq. (h) can be simplified further and T2 can be computed as: 1 ⎡ (TB L1 + TC L3 ) ⎤ T2 = − ⎢ ⎥ G1 I p1 ⎢ L1 + L3 + L2 ⎥ ⎢⎣ G1 I p1 G2 I p 2 ⎥⎦

(h)

1 ⎡ (9, 000, 000 N-mm)(325 mm) + (9, 000, 000 N-mm)(325 mm) ⎤ 4 ⎢ ⎥ 325 mm + 325 mm 400 mm (45,000 N/mm )(1,353,830 mm ) + ⎢ ⎥ (45,000)(1,353,830) (28,000)(1,272,345) ⎣ ⎦ = −4,385, 216 N-mm Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3: T1 = −4,385, 216 N-mm + 9, 000, 000 N-mm = 4, 614, 784 N-mm T3 = −4,385, 216 N-mm + 9, 000, 000 N-mm = 4, 614, 784 N-mm =−

2

Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are: T R (4, 614, 784 N-mm)(75 mm / 2) τ1 = 1 1 = = 127.8 MPa 1,353,830 mm 4 I p1 T2 R2 (4,385, 216 N-mm)(60 mm / 2) = = 103.4 MPa 1, 272,345 mm 4 I p2 T R (4, 614, 784 N-mm)(75 mm / 2) τ3 = 3 3 = = 127.8 MPa 1,353,830 mm 4 I p3

τ2 =

(a) Maximum Shear Stress Magnitude in Bronze Segments: τ 1 = 127.8 MPa

Ans.

(b) Maximum Shear Stress Magnitude in Aluminum Segment: τ 2 = 103.4 MPa

Ans.

(c) Rotation Angle of Joint C Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends: φ1 = φB − φ A ∴φB = φ A + φ1 Similarly, the angle of twist in member (2) can be defined by:

(i)


φ2 = φC − φB

∴φC = φB + φ2 To derive an expression for φC, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from rotating; therefore, φA = 0. φC = φ1 + φ2 The angle of twist in member (1) is: TL (4, 614, 784 N-mm)(325 mm) φ1 = 1 1 = = 0.024618 rad G1 I p1 (45, 000 N/mm 2 )(1,353,830 mm 4 ) The angle of twist in member (2) is: TL (−4,385, 216 N-mm)(400 mm) = −0.049237 rad φ2 = 2 2 = G2 I p 2 (28, 000 N/mm 2 )(1, 272,345 mm 4 ) The rotation angle of flange C is thus: φC = φ1 + φ2 = 0.024618 rad + (−0.049237 rad) = −0.024618 rad = −0.0246 rad

(j)

Ans.


6.85 The torsional assembly shown in Fig. P6.85 consists of a solid 60-mm-diameter aluminum [G = 28 GPa] segment (2) and two bronze [G = 45 GPa] tube segments (1) and (3), which have an outside diameter of 75 mm and a wall thickness of 5 mm. If concentrated torques of TB = 6 kN-m and TC = 10 kN-m are applied in the directions shown, determine: (a) the maximum shear stress magnitude in bronze tube segments (1) and (3). (b) the maximum shear stress magnitude in aluminum segment (2). (c) the rotation angle of joint C.

Fig. P6.85


I p1 = I p 3 = I p2 =

π 32

π

⎡⎣(75 mm) 4 − (65 mm) 4 ⎤⎦ = 1,353,830 mm 4 32

(60 mm) 4 = 1, 272,345 mm 4


and also consider a FBD cut around joint C, where the external torque TC is applied: ΣM x = −T2 + T3 − TC = 0 Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

φ3 =

(b)

T3 L3 G3 I p 3

(c)


(e)

Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1 and T3 with equivalent expressions involving T2. From Eq. (a): T1 = T2 + TB (f) 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 States Copyright Act without the permission of the copyright owner is unlawful.

and from Eq. (b): T3 = T2 + TC Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2: (T2 + TB ) L1 T2 L2 (T2 + TC ) L3 + + =0 G1 I p1 G2 I p 2 G3 I p 3

(g)

⎡ L L ⎤ T L L T L T2 ⎢ 1 + 2 + 3 ⎥ = − B 1 − C 3 G1 I p1 G3 I p 3 ⎣⎢ G1 I p1 G2 I p 2 G3 I p 3 ⎦⎥

TB L1 TC L3 + G1 I p1 G3 I p 3 T2 = − L L1 L + 2 + 3 G1 I p1 G2 I p 2 G3 I p 3 Since G1 = G3 and Ip1 = Ip3, Eq. (h) can be simplified further and T2 can be computed as: 1 ⎡ (TB L1 + TC L3 ) ⎤ T2 = − ⎢ ⎥ G1 I p1 ⎢ L1 + L3 + L2 ⎥ ⎢⎣ G1 I p1 G2 I p 2 ⎥⎦

(h)

1 ⎡ (6, 000, 000 N-mm)(325 mm) + (10, 000, 000 N-mm)(325 mm) ⎤ 4 ⎢ ⎥ 325 mm + 325 mm 400 mm (45,000 N/mm )(1,353,830 mm ) + ⎢ ⎥ (45,000)(1,353,830) (28,000)(1,272,345) ⎣ ⎦ = −3,897,970 N-mm =−

2

Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3: T1 = −3,897,970 N-mm + 6, 000, 000 N-mm = 2,102, 030 N-mm T3 = −3,897,970 N-mm + 10, 000, 000 N-mm = 6,102, 030 N-mm Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are: T R (2,102, 030 N-mm)(75 mm / 2) τ1 = 1 1 = = 58.2 MPa 1,353,830 mm 4 I p1 T2 R2 (3,897,970 N-mm)(60 mm / 2) = = 91.9 MPa 1, 272,345 mm 4 I p2 T R (6,102, 030 N-mm)(75 mm / 2) τ3 = 3 3 = = 169.0 MPa 1,353,830 mm 4 I p3

τ2 =

(a) Maximum Shear Stress Magnitude in Bronze Segments: τ 3 = 169.0 MPa

Ans.

(b) Maximum Shear Stress Magnitude in Aluminum Segment: τ 2 = 91.9 MPa

Ans.

(c) Rotation Angle of Joint C Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends: φ1 = φB − φ A ∴φB = φ A + φ1

(i)


Similarly, the angle of twist in member (2) can be defined by: φ2 = φC − φB ∴φC = φB + φ2 To derive an expression for φC, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from rotating; therefore, φA = 0. φC = φ1 + φ2 The angle of twist in member (1) is: TL (2,102, 030 N-mm)(325 mm) φ1 = 1 1 = = 0.011214 rad G1 I p1 (45, 000 N/mm 2 )(1,353,830 mm 4 ) The angle of twist in member (2) is: TL (−3,897,970 N-mm)(400 mm) = −0.043766 rad φ2 = 2 2 = G2 I p 2 (28, 000 N/mm 2 )(1, 272,345 mm 4 ) The rotation angle of flange C is thus: φC = φ1 + φ2 = 0.011214 rad + (−0.043766 rad) = −0.032552 rad = −0.0326 rad

(j)

Ans.


6.86 A solid 1.50-in.-diameter brass [G = 5,600 ksi] shaft [segments (1), (2), and (3)] has been stiffened between B and C by the addition of a cold-rolled stainless steel tube (4) (Fig. P6.86a). The tube (Fig. P6.86b) has an outside diameter of 3.50 in., a wall thickness of 0.12 in., and a shear modulus of G = 12,500 ksi. The tube is attached to the brass shaft by means of rigid flanges welded to the tube and to the shaft. (The thickness of the flanges can be neglected for this analysis.) If a torque of 400 lb-ft is applied to the shaft as shown in Fig. P6.86a, determine: (a) the maximum shear stress magnitude in segment (1) of the brass shaft. (b) the maximum shear stress magnitude in segment (2) of the brass shaft (i.e., between flanges B and C) (c) the maximum shear stress magnitude in the stainless steel tube (4). (d) the rotation angle of end D relative to end A.

Fig. P6.86a

Fig. P6.86b Cross Section Through Tube

Solution Section Properties: The polar moments of inertia for the brass shaft and the stainless steel tube are:

I p1 = I p 2 = I p 3 = I p4 =

π

32

(1.50 in.) 4 = 0.497010 in.4

π

⎡⎣(3.50 in.) 4 − (3.26 in.)4 ⎤⎦ = 3.643915 in.4 32

(a) Shear Stress Magnitude in Brass Segment (1): T R (400 lb-ft)(1.500 in./ 2)(12 in./ft) τ1 = 1 1 = = 7, 243.3 psi = 7, 240 psi 0.497010 in.4 I p1

Ans.

Statically Indeterminate Portion of the Shaft: The portion of the shaft between B and C is statically indeterminate. Equilibrium: Consider a free-body diagram that cuts through the shaft between B and C and includes the free end of the shaft at A: ΣM x = T2 + T4 + 400 lb-ft = 0 Torque-Twist Relationships: TL TL φ2 = 2 2 φ4 = 4 4 G2 I p 2 G4 I p 4

(b)

Geometry of Deformation Relationship: Since the tube and shaft are securely connected by the rigid flanges, the angles of twist in both members must be equal; therefore, 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 States Copyright Act without the permission of the copyright owner is unlawful.

φ2 = φ4 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T2 L2 TL = 4 4 G2 I p 2 G4 I p 4

(c)

(d)

Solve the Equations: Solve Eq. (d) for T2: 4 L G I p2 ⎛ 20 in. ⎞ ⎛ 5, 600 ksi ⎞ ⎛ 0.497010 in. ⎞ T2 = T4 4 2 T = 4⎜ = 0.0611047 T4 ⎟⎜ ⎟⎜ 4 ⎟ L2 G4 I p 4 ⎝ 20 in. ⎠ ⎝ 12,500 ksi ⎠ ⎝ 3.643915 in. ⎠ and substitute this result into Eq. (a) to compute the torque T4 in the stainless steel tube (4): T2 + T4 = 0.0611047 T4 + T4 = 1.0611047 T4 = −400 lb-ft

∴T4 = −376.9656 lb-ft The torque in brass shaft segment (2) is therefore: T2 = −400 lb-ft − T4 = −400 lb-ft − (−376.9656 lb-ft) = −23.0344 lb-ft = −23.0 lb-ft (b) Maximum Shear Stress in Brass Segment (2): The maximum shear stress in segment (2) is: TR (23.0344 lb-ft)(1.50 in./ 2)(12 in./ft) Ans. τ2 = 2 2 = = 417.1 psi = 417 psi 0.497010 in.4 I p2 (c) Maximum Shear Stress in Stainless Steel Tube (4): The maximum shear stress in tube (4) is: TR (376.9656 lb-ft)(3.50 in./ 2)(12 in./ft) Ans. τ4 = 4 4 = = 2,172.5 psi = 2,170 psi 3.643915 in.4 I p4 (d) Rotation Angle of End D Relative to End A: The angle of twist of segment (1) is: TL (−400 lb-ft)(12 in.)(12 in./ft) = −0.020695 rad φ1 = 1 1 = G1 I p1 (5, 600, 000 psi)(0.497010 in.4 ) The angle of twist in brass segment (3) has the same value: φ3 = −0.020695 rad The angle of twist in brass segment (2) is: TL (−23.0344 lb-ft)(20 in.)(12 in./ft) = −0.0019863 rad φ2 = 2 2 = (5, 600, 000 psi)(0.497010 in.4 ) G2 I p 2

The rotation angle of end D relative to end A is the sum of these three angles of twist: φD = φ1 + φ2 + φ3 = −0.020695 rad + (−0.0019863 rad) + ( −0.020695 rad) = −0.0433763 rad = −0.0434 rad

Ans.


6.87 A solid 60-mm-diameter cold-rolled brass [G = 39 GPa] shaft that is 1.25-m long extends through and is completely bonded to a hollow aluminum [G = 28 GPa] tube, as shown in Fig. P6.87. Aluminum tube (1) has an outside diameter of 90 mm, an inside diameter of 60 mm, and a length of 0.75 m. Both the brass shaft and the aluminum tube are securely attached to the wall support at A. When the two torques shown are applied to the composite shaft, determine: (a) the maximum shear stress magnitude in aluminum tube (1). (b) the maximum shear stress magnitude in brass shaft segment (2). (c) the maximum shear stress magnitude in brass shaft segment (3). (d) the rotation angle of joint B. (e) the rotation angle of joint C.

Fig. P6.87

Solution Section Properties: The polar moments of inertia for tube (1) and brass shafts (2) and (3) are:

I p1 =

π

⎡⎣ (90 mm)4 − (60 mm) 4 ⎤⎦ = 5,168,902 mm 4 32

I p 2 = I p3 =

π 32

(60 mm) 4 = 1, 272,345 mm 4

Equilibrium: Consider a free-body diagram cut around joint C through segment (3): ΣM x = −T3 − 8 kN-m = 0 ∴T3 = −8 kN-m (a)

and also consider a FBD cut around joint B: ΣM x = −T1 − T2 + T3 + 20 kN-m = 0 ∴T1 + T2 = T3 + 20 kN-m = 12 kN-m

(b)

Eq. (b) reveals that the portion of the shaft between A and B is statically indeterminate. The five-step solution procedure will be used to determine T1 and T2 in this portion of the shaft. Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(c)

Geometry of Deformation Relationship: Since the aluminum tube and the brass shaft are securely bonded together, the angles of twist in both members must be equal; therefore, 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 States Copyright Act without the permission of the copyright owner is unlawful.

φ1 = φ2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (d)] to obtain the compatibility equation: T1 L1 TL = 2 2 G1 I p1 G2 I p 2

(d)

(e)

Solve the Equations: Solve Eq. (e) for T1: 4 L G I p1 ⎛ 28 GPa ⎞ ⎛ 5,168,902 mm ⎞ T1 = T2 2 1 T = 2⎜ = 2.916667 T2 ⎟⎜ 4 ⎟ L1 G2 I p 2 ⎝ 39 GPa ⎠ ⎝ 1, 272,345 mm ⎠ and substitute this result into Eq. (b) to compute the torque T2 in brass shaft (2): T1 + T2 = 2.916667 T2 + T2 = 3.916667 T2 = 12, 000 N-m

∴T2 = 3, 063.830 N-m The torque in aluminum tube (1) is therefore: T1 = 12, 000 N-m − T2 = 12, 000 N-m − 3, 063.830 N-m = 8,936.170 N-m (a) Maximum Shear Stress in Aluminum Tube: The maximum shear stress in aluminum tube (1) is: T R (8,936.170 N-m)(90 mm / 2)(1, 000 mm/m) Ans. τ1 = 1 1 = = 77.8 MPa 5,168,902 mm 4 I p1 (b) Maximum Shear Stress Magnitude in Brass Shaft Segment (2): TR (3, 063.830 N-m)(60 mm / 2)(1, 000 mm/m) τ2 = 2 2 = = 72.2 MPa 1, 272,345 mm 4 I p2

Ans.

(c) Maximum Shear Stress Magnitude in Brass Shaft Segment (3): T R (8, 000 N-m)(60 mm / 2)(1, 000 mm/m) τ3 = 3 3 = = 188.6 MPa 1, 272,345 mm 4 I p3

Ans.

(d) Rotation Angle of Joint B: TL (8,936.170 N-m)(0.75 m)(1, 000 mm/m) 2 φ1 = 1 1 = = 0.046308 rad G1 I p1 (28, 000 N/mm 2 )(5,168,902 mm 4 ) or

φ2 =

T2 L2 (3, 063.830 N-m)(0.75 m)(1, 000 mm/m) 2 = = 0.046308 rad G2 I p 2 (39, 000 N/mm 2 )(1,272,345 mm 4 )

∴φB = 0.0463 rad

Ans.

(e) Rotation Angle of Joint C: TL (−8, 000 N-m)(0.5 m)(1, 000 mm/m) 2 = −0.080610 rad φ3 = 3 3 = (39, 000 N/mm 2 )(1,272,345 mm 4 ) G3 I p 3

∴φC = φB + φ3 = 0.046308 rad + (−0.080610 rad) = −0.0343 rad

Ans.


6.88 The gear assembly shown in Fig. P6.88 is subjected to a torque of TC = 100 N-m. Shafts (1) and (2) are solid 20-mm-diameter steel shafts, and shaft (3) is a solid 25-mm-diameter steel shaft. Assume L = 400 mm and G = 80 GPa. Determine: (a) the maximum shear stress magnitude in shaft (1). (b) the maximum shear stress magnitude in shaft segment (3). (c) the rotation angle of gear E. (d) the rotation angle of gear C.

Fig. P6.88

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

I p1 = I p 2 = I p3 =

π 32

π

32

(20 mm) 4 = 15, 708.0 mm 4

(25 mm) 4 = 38,349.5 mm 4

Equilibrium: Consider a free-body diagram cut through shaft (2) and around gear C: ΣM x = −T2 + TC = 0 ∴T2 = TC = 100 N-m (a)

Next, consider a FBD cut around gear B through shafts (1) and (2). The teeth of gear C exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by RB for now (even though the gear radius is not given explicitly). ΣM x = T2 − T1 − F ⋅ RB = 0 (b) Finally, consider a FBD cut around gear E through shaft (3). The teeth of gear B exert an equal magnitude force F on the teeth of gear C, acting opposite to the direction assumed in the previous FBD. The radius of gear E will be denoted by RE for now. T ΣM x′ = −T3 − F ⋅ RE = 0 ∴F = − 3 (c) RE The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give R T1 = 100 N-m + T3 B RE The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth NB and NE: N 24 teeth T1 = 100 N-m + T3 B = 100 N-m + T3 = 100 N-m + 0.4T3 (d) 60 teeth NE 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 States Copyright Act without the permission of the copyright owner is unlawful.

Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the problem, an additional equation must be developed. This second equation will be derived from the relationship between the angles of twist in shafts (1) and (3). Torque-Twist Relationships: TL TL φ1 = 1 1 φ3 = 3 3 G1 I p1 G3 I p 3

(e)

Geometry of Deformation Relationship: The rotation of gear B is equal to the angle of twist in shaft (1): φB = φ 1

and the rotation of gear E is equal to the angle of twist in shaft (3): φE = φ3 However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The arclengths associated with the respective rotations must be equal, but the gears turn in opposite directions. The relationship between gear rotations can be stated as: RBφB = − REφE where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to the shaft angles of twist, this relationship can be expressed as: RBφ1 = − REφ3 (f) Compatibility Equation: Substitute the torque-twist relationships [Eq. (e)] into the geometry of deformation relationship [Eq. (f)] to obtain: TL TL RB 1 1 = − RE 3 3 G1 I p1 G3 I p 3 which can be rearranged and expressed in terms of the gear ratio NB /NE: TL N B T1 L1 (g) =− 3 3 N E G1 I p1 G3 I p 3 Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3). Solve the Equations: Solve for internal torque T3 in Eq. (g): N ⎛ L ⎞⎛ G ⎞⎛ I ⎞ T3 = −T1 B ⎜ 1 ⎟ ⎜ 3 ⎟ ⎜ p 3 ⎟ N E ⎝ L3 ⎠ ⎝ G1 ⎠ ⎜⎝ I p1 ⎟⎠

and substitute this result into Eq. (d): T1 = 100 N-m + 0.4T3

⎡ N ⎛ L ⎞⎛ G ⎞ ⎛ I ⎞ ⎤ = 100 N-m + 0.4 ⎢ −T1 B ⎜ 1 ⎟⎜ 3 ⎟ ⎜ p 3 ⎟ ⎥ N E ⎝ L3 ⎠ ⎝ G1 ⎠ ⎜⎝ I p1 ⎟⎠ ⎦⎥ ⎣⎢ ⎡ ⎛ 24 teeth ⎞ ⎛ 38,349.5 mm 4 ⎞ ⎤ = 100 N-m − 0.4 ⎢T1 ⎜ ⎟⎜ 4 ⎟⎥ ⎣ ⎝ 60 teeth ⎠ ⎝ 15, 708.0 mm ⎠ ⎦ = 100 N-m − 0.390624 T1 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 States Copyright Act without the permission of the copyright owner is unlawful.

Group the T1 terms to obtain: 100 N-m T1 = = 71.9102 N-m 1.390624 Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3): T1 = 100 N-m + 0.4T3 T1 − 100 N-m 71.9102 N-m − 100 N-m = = −70.2246 N-m 0.4 0.4 (a) Maximum Shear Stress Magnitude in Shaft (1): T R (71.9102 N-m)(20 mm / 2)(1, 000 mm/m) τ1 = 1 1 = = 45.8 MPa 15, 708.0 mm 4 I p1 ∴T3 =

(b) Maximum Shear Stress Magnitude in Shaft (3): T R (70.2246 N-m)(25 mm / 2)(1, 000 mm/m) τ3 = 3 3 = = 22.9 MPa I p3 38,349.5 mm 4

Ans.

Ans.

(c) Rotation Angle of Gear E: TL (−70.2246 N-m)(1.25)(400 mm)(1, 000 mm/m) φ3 = 3 3 = = −0.011445 rad G3 I p 3 (80, 000 N/mm 2 )(38,349.5 mm 4 )

∴φE = φ3 = −0.01145 rad

Ans.

(d) Rotation Angle of Gear C: TL (71.9102 N-m)(1.25)(400 mm)(1, 000 mm/m) φ1 = 1 1 = = 0.028612 rad (80, 000 N/mm 2 )(15,708.0 mm 4 ) G1 I p1

φ2 =

T2 L2 (100 N-m)(400 mm)(1, 000 mm/m) = = 0.031831 rad (80, 000 N/mm 2 )(15,708.0 mm 4 ) G2 I p 2

∴φC = φB + φ2 = φ1 + φ2 = 0.028612 rad + 0.031831 rad = 0.0604 rad

Ans.


6.89 The gear assembly shown in Fig. P6.89 is subjected to a torque of TC = 800 lb-ft. Shafts (1) and (2) are solid 1.625-in.-diameter aluminum shafts and shaft (3) is a solid 2.00-in.-diameter aluminum shaft. Assume L = 20 in. and G = 4,000 ksi. Determine: (a) the maximum shear stress magnitude in shaft (1). (b) the maximum shear stress magnitude in shaft segment (3). (c) the rotation angle of gear E. (d) the rotation angle of gear C.

Fig. P6.89


I p1 = I p 2 = I p3 =

π 32

π

32

(1.625 in.) 4 = 0.684563 in.4

(2.00 in.) 4 = 1.570796 in.4

Equilibrium: Consider a free-body diagram cut through shaft (2) and around gear C: ΣM x = −T2 + TC = 0 ∴ T2 = TC = 800 lb-ft (a)

Next, consider a FBD cut around gear B through shafts (1) and (2). The teeth of gear C exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by RB for now (even though the gear radius is not given explicitly). ΣM x = T2 − T1 − F ⋅ RB = 0 (b) Finally, consider a FBD cut around gear E through shaft (3). The teeth of gear B exert an equal magnitude force F on the teeth of gear C, acting opposite to the direction assumed in the previous FBD. The radius of gear E will be denoted by RE for now. T ΣM x′ = −T3 − F ⋅ RE = 0 ∴F = − 3 (c) RE The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give R T1 = 800 lb-ft + T3 B RE The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth NB and NE: N 24 teeth T1 = 800 lb-ft + T3 B = 800 lb-ft + T3 = 800 lb-ft + 0.4T3 (d) 60 teeth NE 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 States Copyright Act without the permission of the copyright owner is unlawful.


(e)


and the rotation of gear E is equal to the angle of twist in shaft (3): φE = φ3 However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The arclengths associated with the respective rotations must be equal, but the gears turn in opposite directions. The relationship between gear rotations can be stated as: RBφB = − REφE where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to the shaft angles of twist, this relationship can be expressed as: RBφ1 = − REφ3 (f) Compatibility Equation: Substitute the torque-twist relationships [Eq. (e)] into the geometry of deformation relationship [Eq. (f)] to obtain: TL TL RB 1 1 = − RE 3 3 G1 I p1 G3 I p 3 which can be rearranged and expressed in terms of the gear ratio NB /NE: TL N B T1 L1 (g) =− 3 3 N E G1 I p1 G3 I p 3 Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3). Solve the Equations: Solve for internal torque T3 in Eq. (g): N ⎛ L ⎞⎛ G ⎞ ⎛ I ⎞ T3 = −T1 B ⎜ 1 ⎟⎜ 3 ⎟ ⎜ p 3 ⎟ N E ⎝ L3 ⎠ ⎝ G1 ⎠ ⎜⎝ I p1 ⎟⎠

and substitute this result into Eq. (d): T1 = 800 lb-ft + 0.4T3

⎡ N ⎛ L ⎞⎛ G ⎞ ⎛ I ⎞ ⎤ = 800 lb-ft + 0.4 ⎢ −T1 B ⎜ 1 ⎟⎜ 3 ⎟ ⎜ p 3 ⎟ ⎥ N E ⎝ L3 ⎠ ⎝ G1 ⎠ ⎜⎝ I p1 ⎟⎠ ⎥⎦ ⎢⎣ ⎡ ⎛ 24 teeth ⎞ ⎛ 1.570796 in.4 ⎞ ⎤ = 800 lb-ft − 0.4 ⎢T1 ⎜ ⎟⎜ 4 ⎟⎥ ⎣ ⎝ 60 teeth ⎠ ⎝ 0.684563 in. ⎠ ⎦ = 800 lb-ft − 0.367135 T1 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 States Copyright Act without the permission of the copyright owner is unlawful.

Group the T1 terms to obtain: 800 lb-ft T1 = = 585.165 lb-ft 1.367135 Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3): T1 = 800 lb-ft + 0.4T3 T1 − 800 lb-ft 585.165 lb-ft − 800 lb-ft = = −537.087 lb-ft 0.4 0.4 (a) Maximum Shear Stress Magnitude in Shaft (1): T R (585.165 lb-ft)(1.625 in./ 2)(12 in./ft) τ1 = 1 1 = = 8,334.3 psi = 8.33 ksi 0.684563 in.4 I p1 ∴T3 =

(b) Maximum Shear Stress Magnitude in Shaft (3): T R (537.087 lb-ft)(2.00 in./ 2)(12 in./ft) τ3 = 3 3 = = 4,103.0 psi = 4.10 ksi I p3 1.570796 in.4

Ans.

Ans.

(c) Rotation Angle of Gear E: TL (−537.087 lb-ft)(1.25)(20 in.)(12 in./ft) φ3 = 3 3 = = −0.025644 rad (4, 000, 000 psi)(1.570796 in.4 ) G3 I p 3

∴φE = φ3 = −0.0256 rad

Ans.

(d) Rotation Angle of Gear C: TL (585.165 lb-ft)(1.25)(20 in.)(12 in./ft) φ1 = 1 1 = = 0.064110 rad (4, 000, 000 psi)(0.684563 in.4 ) G1 I p1

φ2 =

T2 L2 (800 lb-ft)(20 in.)(12 in./ft) = = 0.070118 rad G2 I p 2 (4, 000, 000 psi)(0.684563 in.4 )


Ans.


6.90 A torque of TC = 380 N-m acts on gear C of the assembly shown in Fig. P6.90. Shafts (1) and (2) are solid 35-mm-diameter aluminum shafts and shaft (3) is a solid 25-mm-diameter aluminum shaft. Assume L = 200 mm and G = 28 GPa. Determine: (a) the maximum shear stress magnitude in shaft (1). (b) the maximum shear stress magnitude in shaft segment (3). (c) the rotation angle of gear E. (d) the rotation angle of gear C.

Fig. P6.90


I p1 = I p 2 = I p3 =

π 32

π

32

(35 mm) 4 = 147,323.5 mm 4

(25 mm) 4 = 38,349.5 mm 4

Equilibrium: Consider a free-body diagram cut through shaft (2) and around gear C: ΣM x = −T2 + TC = 0 ∴T2 = TC = 380 N-m (a)

Next, consider a FBD cut around gear B through shafts (1) and (2). The teeth of gear E exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by RB for now (even though the gear radius is not given explicitly). ΣM x = T2 − T1 − F ⋅ RB = 0 (b) Finally, consider a FBD cut around gear E through shaft (3). The teeth of gear B exert an equal magnitude force F on the teeth of gear C, acting opposite to the direction assumed in the previous FBD. The radius of gear E will be denoted by RE for now. T ΣM x′ = −T3 − F ⋅ RE = 0 ∴F = − 3 (c) RE The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give R T1 = 380 N-m + T3 B RE The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth NB and NE: N 54 teeth T1 = 380 N-m + T3 B = 380 N-m + T3 = 380 N-m + 1.285714 T3 (d) NE 42 teeth 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 States Copyright Act without the permission of the copyright owner is unlawful.


(e)



and substitute this result into Eq. (d): T1 = 380 N-m + 1.285714 T3

⎡ N ⎛ L ⎞⎛ G ⎞ ⎛ I ⎞ ⎤ = 380 N-m + 1.285714 ⎢ −T1 B ⎜ 1 ⎟⎜ 3 ⎟ ⎜ p 3 ⎟ ⎥ N E ⎝ L3 ⎠ ⎝ G1 ⎠ ⎜⎝ I p1 ⎟⎠ ⎥⎦ ⎢⎣ ⎡ ⎛ 54 teeth ⎞ ⎛ 38,349.5 mm 4 ⎞ ⎤ = 380 N-m − 1.285714 ⎢T1 ⎜ ⎟⎜ 4 ⎟⎥ ⎣ ⎝ 42 teeth ⎠ ⎝ 147,323.5 mm ⎠ ⎦ = 380 N-m − 0.430305 T1 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 States Copyright Act without the permission of the copyright owner is unlawful.

Group the T1 terms to obtain: 380 N-m T1 = = 265.6776 N-m 1.430305 Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3): T1 = 380 N-m + 1.285714 T3 ∴T3 =

T1 − 380 N-m 265.6776 N-m − 380 N-m = = −88.9174 N-m 1.285714 1.285714

(a) Maximum Shear Stress Magnitude in Shaft (1): T R (265.6776 N-m)(35 mm / 2)(1, 000 mm/m) τ1 = 1 1 = = 31.6 MPa I p1 147,323.5 mm 4

Ans.

(b) Maximum Shear Stress Magnitude in Shaft (3): T R (88.9174 N-m)(25 mm / 2)(1, 000 mm/m) τ3 = 3 3 = = 29.0 MPa I p3 38,349.5 mm 4

Ans.

(c) Rotation Angle of Gear E: TL (−88.9174 N-m)(2)(200 mm)(1, 000 mm/m) φ3 = 3 3 = = −0.033123 rad (28, 000 N/mm 2 )(38,349.5 mm 4 ) G3 I p 3

∴φE = φ3 = −0.0331 rad

Ans.

(d) Rotation Angle of Gear C: TL (265.6776 N-m)(2)(200 mm)(1, 000 mm/m) φ1 = 1 1 = = 0.025762 rad G1 I p1 (28, 000 N/mm 2 )(147,323.5 mm 4 )

φ2 =

T2 L2 (380 N-m)(200 mm)(1, 000 mm/m) = = 0.018424 rad G2 I p 2 (28, 000 N/mm 2 )(147,323.5 mm 4 )


Ans.


6.91 A torque of TC = 30 kip-in. acts on gear C of the assembly shown in Fig. P6.91. Shafts (1) and (2) are solid 2.00-in.-diameter stainless steel shafts and shaft (3) is a solid 1.75-in.diameter stainless steel shaft. Assume L = 8 in. and G = 12,500 ksi. Determine: (a) the maximum shear stress magnitude in shaft (1). (b) the maximum shear stress magnitude in shaft segment (3). (c) the rotation angle of gear E. (d) the rotation angle of gear C.

Fig. P6.91


I p1 = I p 2 = I p3 =

π 32

π

32

(2.00 in.) 4 = 1.570796 in.4

(1.75 in.) 4 = 0.920772 in.4

Equilibrium: Consider a free-body diagram cut through shaft (2) and around gear C: ΣM x = −T2 + TC = 0 ∴T2 = TC = 30 kip-in. (a)

Next, consider a FBD cut around gear B through shafts (1) and (2). The teeth of gear E exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by RB for now (even though the gear radius is not given explicitly). ΣM x = T2 − T1 − F ⋅ RB = 0 (b) Finally, consider a FBD cut around gear E through shaft (3). The teeth of gear B exert an equal magnitude force F on the teeth of gear C, acting opposite to the direction assumed in the previous FBD. The radius of gear E will be denoted by RE for now. T ΣM x′ = −T3 − F ⋅ RE = 0 ∴F = − 3 (c) RE The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give R T1 = 30 kip-in. + T3 B RE The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth NB and NE: N 54 teeth T1 = 30 kip-in. + T3 B = 30 kip-in. + T3 = 30 kip-in. + 1.285714 T3 (d) NE 42 teeth 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 States Copyright Act without the permission of the copyright owner is unlawful.


(e)



and substitute this result into Eq. (d): T1 = 30 kip-in. + 1.285714 T3

⎡ N ⎛ L ⎞⎛ G ⎞ ⎛ I ⎞ ⎤ = 30 kip-in. + 1.285714 ⎢ −T1 B ⎜ 1 ⎟⎜ 3 ⎟ ⎜ p 3 ⎟ ⎥ N E ⎝ L3 ⎠ ⎝ G1 ⎠ ⎜⎝ I p1 ⎟⎠ ⎥⎦ ⎢⎣ ⎡ ⎛ 54 teeth ⎞ ⎛ 0.920772 in.4 ⎞ ⎤ = 30 kip-in. − 1.285714 ⎢T1 ⎜ ⎟⎜ 4 ⎟⎥ ⎣ ⎝ 42 teeth ⎠ ⎝ 1.570796 in. ⎠ ⎦ = 30 kip-in. − 0.968994 T1 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 States Copyright Act without the permission of the copyright owner is unlawful.

Group the T1 terms to obtain: 30 kip-in. T1 = = 15.2362 kip-in. 1.968994 Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3): T1 = 30 kip-in. + 1.285714 T3 ∴T3 =

T1 − 30 kip-in. 15.2362 kip-in. − 30 kip-in. = = −11.4830 kip-in. 1.285714 1.285714

(a) Maximum Shear Stress Magnitude in Shaft (1): T R (15.2362 kip-in.)(2.00 in./ 2) τ1 = 1 1 = = 9.70 ksi I p1 1.570796 in.4

Ans.

(b) Maximum Shear Stress Magnitude in Shaft (3): T R (11.4830 kip-in.)(1.75 in./ 2) τ3 = 3 3 = = 10.91 ksi I p3 0.920772 in.4

Ans.

(c) Rotation Angle of Gear E: TL (−11.4830 kip-in.)(2)(8 in.) φ3 = 3 3 = = −0.015963 rad G3 I p 3 (12,500 ksi)(0.920772 in.4 )

∴φE = φ3 = −0.01596 rad

Ans.

(d) Rotation Angle of Gear C: TL (15.2362 kip-in.)(2)(8 in.) φ1 = 1 1 = = 0.012416 rad G1 I p1 (12,500 ksi)(1.570796 in.4 )

φ2 =

T2 L2 (30 kip-in.)(8 in.) = = 0.012223 rad G2 I p 2 (12,500 ksi)(1.570796 in.4 )


Ans.


6.92 The steel [G = 12,000 ksi] pipe shown in Fig. P6.92 is fixed to the wall support at C. The bolt holes in the flange at A were supposed to align with mating holes in the wall support; however, an angular misalignment of 2° was found to exist. To connect the pipe to its supports, a temporary installation torque T′B must be applied at B to align flange A with the mating holes in the wall support. The outside diameter of the pipe is 3.50 in. and its wall thickness is 0.216 in. (a) Determine the temporary installation torque T′B that must be applied at B to align the bolt holes at A. (b) Determine the maximum shear stress τinitial in the pipe after the bolts are connected and the temporary installation torque at B is removed. (c) If the maximum shear stress in the pipe shaft must not exceed 12 ksi, determine the maximum external torque TB that can be applied at B after the bolts are connected.

Fig. P6.92

Solution Section Properties: The polar moment of inertia for the pipe is:

I p1 = I p 2 =

π

⎡⎣(3.50 in.) 4 − (3.068 in.) 4 ⎤⎦ = 6.034313 in.4 32

(a) Temporary Installation Torque T′B: The shaft, which is fixed to the support at C, must be rotated 2° so that the connection at A can be completed. Consequently, segment (2) must be twisted 2° (or 0.034907 rad) by the temporary torque T′B. From the torque-twist relationship for shaft (2) TL φ2 = 2 2 G2 I p 2 the temporary torque T′B is: φGI (0.034907 rad)(12,000 ksi)(6.034313 in.4 ) Ans. = 17.55 kip-in. TB′ = 2 2 p 2 = (12 ft)(12 in./ft) L2 (b) Maximum Shear Stress after Temporary Torque is Removed: After T′B is removed, the 2° angle of twist now applies to the total pipe length, i.e., 20 ft. The internal torque magnitude in the pipe due to the 2° misfit is thus: φ GI (0.034907 rad)(12,000 ksi)(6.034313 in.4 ) = 10.5320 kip-in. Tinitial = misfit p = ( L1 + L2 ) (20 ft)(12 in./ft) and the maximum shear stress magnitude in the pipe due to this internal torque is: T R (10.5320 kip-in.)(3.50 in./ 2) Ans. τ initial = initial = = 3.05 ksi 6.034313 in.4 Ip


After Bolts Installed: After the bolts are installed, the pipe is analyzed as a statically indeterminate torsion structure. Equilibrium: ΣM x = −T1 + T2 + TB = 0 (a) Note: By inspection, the internal torque T2 will end up having a negative value. Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

Geometry of Deformation Relationship: The two shafts are securely attached to fixed supports at A and C; however, in this instance, the sum of the angles of twist in the two members must equal the misfit angle. Note: If the rotation angle at A is a positive 2° (as can be inferred from the problem sketch), then the pipe must twist in a negative sense to reach a zero rotation angle at C. φ1 + φ2 = −0.034907 rad (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL + 2 2 = −0.034907 rad G1 I p1 G2 I p 2

(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: TR T τ τ= ∴ = Ip Ip R which allows Eq. (d) to be rewritten as: τ 1 L1 τ L = −0.034907 rad − 2 2 (e) G1 R1 G2 R2 Solve Eq. (e) for τ 1: GR L G R τ 1 = (−0.034907 rad) 1 1 − τ 2 2 1 1 L1 L1 G2 R2 (12, 000 ksi)(3.50 in./2) ⎛ 12 ft ⎞ −τ 2 ⎜ ⎟ (8 ft)(12 in./ft) ⎝ 8 ft ⎠ = −7.635906 ksi − 1.5τ 2 = (−0.034907 rad)

(f)

Assume the shaft (2) controls: By inspection of the FBD, the internal torque T2 should have a negative value. Therefore, we can assume that the maximum shear stress in shaft (2) will be −12 ksi. If the shear stress in shaft (2) reaches this value, then the shear stress in shaft (1) will be: τ 1 = −7.635906 ksi − 1.5(−12 ksi) = 10.3641 ksi ≤ 12 ksi O.K. This calculation demonstrates that the shear stress magnitude in shaft (2) must control.

Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in each component can be computed: 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 States Copyright Act without the permission of the copyright owner is unlawful.

T1 = T2 =

τ 1 I p1 R1

τ 2 I p2 R2

(10.3641 ksi)(6.034313 in.4 ) = = 35.7373 kip-in. ( 3.50 in./2 ) =

(−12 ksi)(6.034313 in.4 ) = −41.3781 kip-in. ( 3.50 in./2 )

(c) Maximum Allowable Torque: From Eq. (a), the total torque acting at B must not exceed: TB ,max = T1 − T2 = 35.7373 kip-in. − (−41.3781 kip-in.) = 77.1154 kip-in.= 77.1 kip-in.

Ans.


6.93 The steel [G = 12,000 ksi] pipe shown in Fig. P6.93 is fixed to the wall support at C. The bolt holes in the flange at A were supposed to align with mating holes in the wall support; however, an angular misalignment of 2° was found to exist. To connect the pipe to its supports, a temporary installation torque T′B must be applied at B to align flange A with the mating holes in the wall support. The outside diameter of the pipe is 2.875 in. and its wall thickness is 0.203 in. (a) Determine the temporary installation torque T′B that must be applied at B to align the bolt holes at A. (b) Determine the maximum shear stress τinitial in the pipe after the bolts are connected and the temporary installation torque at B is removed. (c) Determine the magnitude of the maximum shear stress in segments (1) and (2) if an external torque of TB = 80 kip-in. is applied at B after the bolts are connected.

Fig. P6.93

Solution Section Properties: The polar moment of inertia for the pipe is:

I p1 = I p 2 =

π

⎡⎣(2.875 in.) 4 − (2.469 in.) 4 ⎤⎦ = 3.059108 in.4 32

(a) Temporary Installation Torque T′B: The shaft, which is fixed to the support at C, must be rotated 2° so that the connection at A can be completed. Consequently, segment (2) must be twisted 2° (or 0.034907 rad) by the temporary torque T′B. From the torque-twist relationship for shaft (2) TL φ2 = 2 2 G2 I p 2 the temporary torque T′B is: φGI (0.034907 rad)(12,000 ksi)(3.059108 in.4 ) Ans. = 8.90 kip-in. TB′ = 2 2 p 2 = (12 ft)(12 in./ft) L2 (b) Maximum Shear Stress after Temporary Torque is Removed: After T′B is removed, the 2° angle of twist now applies to the total pipe length, i.e., 20 ft. The internal torque magnitude in the pipe due to the 2° misfit is thus: φ GI (0.034907 rad)(12,000 ksi)(3.059108 in.4 ) = 5.3392 kip-in. Tinitial = misfit p = ( L1 + L2 ) (20 ft)(12 in./ft) and the maximum shear stress magnitude in the pipe due to this internal torque is: T R (5.3392 kip-in.)(2.875 in./ 2) Ans. τ initial = initial = = 2.51 ksi 3.059108 in.4 Ip


After Bolts Installed: After the bolts are installed, the pipe is analyzed as a statically indeterminate torsion structure. Equilibrium: ΣM x = −T1 + T2 + TB = 0 (a) Note: By inspection, the internal torque T2 will end up having a negative value. Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

Geometry of Deformation Relationship: The two shafts are securely attached to fixed supports at A and C; however, in this instance, the sum of the angles of twist in the two members must equal the misfit angle. Note: If the rotation angle at A is a positive 2° (as can be inferred from the problem sketch), then the pipe must twist in a negative sense to reach a zero rotation angle at C. φ1 + φ2 = −0.034907 rad (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL + 2 2 = −0.034907 rad G1 I p1 G2 I p 2

(d)

Solve the Equations: Solve Eq. (d) for T1: G1 I p1 L G I p1 T1 = (−0.034907 rad) − T2 2 1 L1 L1 G2 I p 2 (12, 000 ksi)(3.059108 in.4 ) ⎛ 12 ft ⎞ − T2 ⎜ ⎟ (8 ft)(12 in./ft) ⎝ 8 ft ⎠ = −13.348035 kip-in. − 1.5 T2 and substitute this result into Eq. (a) to compute the torque T2: −T1 + T2 = − ( −13.348035 kip-in. − 1.5 T2 ) + T2 = 13.348035 kip-in. + 2.5 T2 = −80 kip-in. = (−0.034907 rad)

−93.348035 kip-in. = −37.3392 kip-in. 2.5 The torque in member (1) is therefore: T1 = T2 + 80 kip-in. = −37.3392 kip-in. + 80 kip-in. = 42.6608 kip-in. ∴T2 =

(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is: T R (42.6608 kip-in.)(2.875 in./ 2) τ1 = 1 1 = = 20.0 ksi I p1 3.059108 in.4 The maximum shear stress magnitude in member (2) is: TR (37.3392 kip-in.)(2.875 in./ 2) τ2 = 2 2 = = 17.55 ksi I p2 3.059108 in.4

Ans.

Ans.


Mechanics of Materials Chapter06 Probs79 93

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