Question

a) A hollow circular shaft designed to have an inner diameter of 0.7Do; where Do is an outer diameter.

i) Determine the inner and outer diameters of the shaft if the maximum allowable shearing stress is Y0 N/mm² and torque is X00 kNm. Then, calculate the polar moment of inertia.

ii) By using inner diameter calculated in (i), calculate the maximum shearing stress and a new torque, if the shaft is changed to solid circular shaft. The shaft is 1.5 m long with the twisted through an angle of X°. Take modulus of rigidity is 85 GPa.

 

b) A steel shaft consists of three-disks attached on the shaft as illustrated in Figure 3. The shaft is fixed at A against rotation at its left end but free to rotate in a bearing at its right end, E. Three-disks has 300 mm diameter is located at B, C and D. For disks B and C, the downward forces of 5X N and 2Y0 N is applied. Conversely, the upward force of 1XY N is subjected to disk at D. The forces act at the outer surfaces of the disks so that torques are applied to the rod. The modulus of rigidity, G for the steel is 80 GPa.

Note: Shows all the assumptions and diagrams.

 

i) Determine the torque that occurs at each segment of the shaft.

ii) Draw the torque diagram.

iii) Calculate the maximum shear stress of the shaft.

iv) Calculate the angle of twist of section E relative to the fixed section A.

 

 

 

 

 

 

 

 

 

Answer 1

g Given that in (Di) inner dia = 0.1 Do Duter dia = Do Imax - so N/mm² T= 500 KN-m = 500x106 K10° N- N-mm we know the, Torsion formula - C ४. . т Ip GO L Taking T o = T Ip IP = 2p TE Tx 2p

o Tmax Tmax x 2p 500x106 50 x * (bu - DiM) 16 D. 10? DOT -10.70.)" 16 DO 10? 3 Do X 0.0475 Do = 406.18 mm Di = 0.7 Do = 284.33 - Polar moment of intertia Ip = 17 1 / 2 (BY-D") Ip = P (466.184_284.324) = 32 2030 68106 many Ip = 2030.6 x 106 mary Now the Dia shaft changes into solid, inner Di = 284.33 using the O=5 L-1.5m G=85 Gpa = 85010 N/mm using T z Go ER= Di 2 at the L surface of shafe U Tmax = Go ха L for Tmax

З Inax 85х10 с” ХК Х X 2.4.33 2 1. (xjooo Iso° Стах = 703 • 02 N/mm2 — Ip = ° С я x (284-23 ) с 6Ч1.69 xio“ уч 32 2 using T - Сое L 1 p. 85x10° x r° xт x 6Ч1.69 xlos а та 180° x S X 1000 — Т- 31 73 хуб 106 T= 3173 KCN -т