I have a question from solid mechanics with its answer. in part a, it ask me abo

Question

I have a question from solid mechanics with its answer. in part a, it ask me about maximum T (torque). (in the red part that I highlited ) why I should not use C = 0.02 ? because it gives me bigger T to me it makes more sense to use C = 0.02 instead of C = 0.03.

Can you explain why I should use C= 0.03 in this question?

Thanks

Concept Application 3.1 A hollow cylindrical steel shaft is 1.5 m long and has inner and outer the largest torque that can be applied to the shaft if the shearing stress value of the shearing stress in the shaft? 60 mm diameters respectively equal to 40 and 60 mm (Fig. 3.15). (a) What is 40 mm is not to exceed 120 MPa? (b) What is the corresponding minimunm 1.5 m Fig. 3.15 Hollow, fixed-end shaft having torque T applied at end. The largest torque T that can be applied to the shaft is the torque for which Tax 120 MPa. Since this is less than the yield strength for any steel, use Eq. (3.9). Solving this equation for T, Recalling that the polar moment of inertia J of the cross section is given by Eq. (3.11), where 40 mm) 0.02 m and c2 = (60 mm) = 0.03 m, write J =?(4-d) 1m(0.034-0.02) = 1.021 × 10-6m2 if I put C 0.02 T will be a bigger number Substituting for J and m into Eq. (1) and letting c 2 0.03 m, Jrmax (1.021 X 10 m)(120 x 10 Pa) 4.08 k-m 0.03 m The minimum shearing stress occurs on the inner surface of the shaft. Equation (3.7) expresses that Tmin and Tmax are respectively pro- portional to ci and cz 00m(120 MPa)80 MPa max 0.03 m

Explanation / Answer

In the equation for T , tmax is taken. tmax is possible at the outer fibre. So c = 0.03 is taken which is the outer diameter.

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