Problem Number (3.3)

Knowing that the internal diameter of the hollow shaft shown is d = 22 mm, determine the maximum shearing stress caused by a torque of magnitude T = 900N.m.

Solution:

τ=T CJ

, J=π2C4

τ= 900×20×10−3

3.142

(0.024−0.0114)=78.87 MPa

Problem Number (3.8)

The solid spindle AB has a diameter ds = 38 mm and is made of a steel with an allowable shearing stress of 84 MPa, while sleeve CD is made of a brass with an allowable shearing stress of 50 MPa. Determine the largest torque T that can be applied at A.

Solution:

τ=T CJ

, J=π2C4

T= τ JC

, ForholloCylindre :J=π2(Couter

4 −C inner4 )

For solid spindle AB :

T=84×106×2.046×10−7

0.019=904.565N .m

For sleeveCD :

T=50×106×1.558×10−6

0.0375=2077.5N .m

Problem Number (3.11)

Under normal operating conditions, the electric motor exerts a torque of 2.8 KN.m on shaft AB. Knowing that each shaft is solid, determine the maximum shearing stress in (a) shaft AB, (b) shaft BC, (c) shaft CD.

Solution:

TAB = 2.8 KN

A

B

τ=T CJ

, J=π2C4

τ AB=2.8×103×0.028

9.65×10−7 =81.2 MPa

TBC = 1.4 KN

B

C

τ BC=1400×0.024

5.21×10−7=64.5MPa

TCD = 0.5 KN

C

D

τCD=500×0.024

5.21×10−7=23.033 MPa

Problem Number (3.17)

The solid shaft shown is formed of a brass for which the allowable shearing stress is 55 MPa. Neglecting the effect of stress concentrations, determine the smallest diameter dAB and dBC for which the allowable shearing stress is not exceeded.

Solution:

TAB = 800N

A

B

τ=T CJ

, J=π2C4

τ AB=800×C3.14

2C4

=55×106

∴C=21mm,∴d AB=42mm

TBC = 400 N

B

C

τ BC=400×C3.14

2C4

=55×106

∴C=16.67mm ,∴dBC=33.34 mm

Problem Number (3.20)

The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa. Determine (a) the largest inner diameter of rod AB for which the factor of safety is the same for each rod.(b) the largest torque that can be applied at A.

Solution:

T= τ JC

=25×106×7.948×10−8

15×10−3 =132.5 N .m

J=T Cτ

=132.5×12.5×10−3

50×106 =3.142

(2.44×10−8−C4)

C=7.588×10−3

Then the inner diameter of rod AB = 15.18 mm

Problem Number (3.21)

A torque of magnitude T = 900 N.m is applied at D as shown. Knowing that the allowable shearing stress is 50 MPa in each shaft, determine the required diameter of (a) shaft AB, (b) shaft CD.

Solution:

τ=T CJ

, J=π2C4

T c

0.04=

T B

0.1,

9000.04

=T B

0.1,T B=2250N .m

C3=2Tπ τ

,CAB=3√ 2Tπ τ

= 3√ 2×22503.14×50×106 =0.031m

Then Diameter of shaft ABis61.2mm

CDC=3√ 2Tπ τ

=3√ 2×9003.14 ×50×106 =0.0225m

Then theDiameter of shaft DC is45mm

Problem Number (3.22)

A torque of magnitude T = 900 N.m is applied at D as shown. Knowing that the diameter of shaft AB is 60 mm and that the diameter of shaft CD is 45 mm, determine the maximum shearing stress in (a) shaft AB, (b) shaft CD.

Solution:

Problem Number (3.35)

The electric motor exerts a 500 N.m torque on the aluminum shaft ABCD when it is rotating at a constant speed. Knowing that G = 27 GPa and that the torques exerted on pulleys B and C are as shown, determine the angle of twist between (a) B and C, (b) B and D.

Solution:

ϕ=T LGJ

, J= π2C4

JBC=¿ 3.68×10−7

ϕ BC=800×1.2

27×109×3.68×10−7=0.097 rad=5.5 4o

JCD=5.21×10−7

ϕCD=500×0.9

27×109×5.21×10−7=0.032 rad=1.83 4o

ϕBD=ϕCD+ϕBC=5.54+1.834=7.37 4o

Problem Number (3.38)

The aluminum rod AB (G = 27 GPa) is bonded to the brass rod BD (G = 39 GPa). Knowing that portion CD of the brass rod is hollow and has an inner diameter of 40 mm, determine the angle of twist at A.

Solution:

For Part AB:

ϕ=T LGJ

= 800×0.4

27×109×1.65×10−7=0.072rad

For Part BC:

ϕ=T LGJ

= 2400×0.375

39×109×1.273×10−6=0.018 rad

For Part CD:

ϕ=T LGJ

= 2400×0.25

39×109×1.0205×10−6=0.0152 rad

Then ϕA=0.072+0.018+0.0152=0.1052 rad=6.0 3o