
Solution :
Given,
diameter od rod, d = 2D
Each side of square section ,a = 2D
Now, we know that
shear stress, Ï„ = Torque(T)/Section Modulus(z)
Now, Section modulus of rod is
Now, Section modulus of rod is
Z_r= π d^3/16 = π (2D)^3/16 = 1.57D^3
Now, Section modulus of square section is
Z_s = a^3/6 = (2D)^3/6 = 1.33D^3
Since, the section modulus of rod is more than the section modulus of square section so, for same torque applied the shear stress induced in the rod will be less than that of square section.
Hence, rod geometry will be more appropriate.
Since, the section modulus of rod is more than the section modulus of square section so, for same torque applied the shear stress induced in the rod will be less than that of square section.
Hence, rod geometry will be more appropriate.
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