Strain Energy 275 Since the strain energy/unit volume in the two shafts is equal, G as then 2EA 2Es o7_EA_ 67 a Es 200=(approximately) (1) 3o7=o3 (2) Now where P is the applied load area Therefore from(1) [品：- D 1 Di=3 D4=3×D5=3×(50)4 =3×625×10 D4=(1875×104)=65.8mm The required diameter of the aluminium shaft is 65.8 mm. From(2) 3a月=3 s=√/3 C Example 11.2 Two shafts are of the same material,length and weight.One is solid and 100 mm diameter, the other is hollow.If the hollow shaft is to store 25%more energy than the solid shaft when transmitting torque,what must be its internal and external diameters? Assume the same maximum shear stress applies to both shafts. Solution Let A be the solid shaft and B the hollow shaft.If they are the same weight and the same material their volume must be equal. g0xL-[o房-L D3=D3-d8=100m2=10x10-3m (1) Now for the same maximum shear stress Tr TD T= 了=2万Strain Energy 275 Since the strain energyjunit volume in the two shafts is equal, then Now 05 EA 67 - =-=-- - f (approximately) 0% Es 200 30; = a: P a=- where P is the applied load area Therefore from (1) Df 1 Dt 3 0; = 3 x Df = 3 x (50)4 DA = 4/(1875 x lo4) = 65.8 mm .. - =- .. = 3 x 625 x 104 .. The required diameter of the aluminium shaft is 65.8mm. From (2) 30: = a: .. ““43 aA Example 11.2 Two shafts are of the same material, length and weight. One is solid and 100 mm diameter, the other is hollow. If the hollow shaft is to store 25 % more energy than the solid shaft when transmitting torque, what must be its internal and external diameters? Assume the same maximum shear stress applies to both shafts. Solution Let A be the solid shaft and B the hollow shaft. If they are the same weight and the same material their volume must be equal. Now for the same maximum shear stress Tr TD J 25 T=-=-