## Related questions with answers

The A-36 steel bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed by the equation

$t = (kx^{2/3}) N · m/m,$

where x is in meters. If a torque of T = 50 N · m is applied to the bolt head, determine the constant k and the amount of twist in the 50-mm length of the shank. Assume the shank has a constant radius of 4 mm.

Solutions

Verified**Given:**

- Shaft subjected to the torque effects
- Diameter of the shaft

**Required:**

- The angle of the twist
- Constant k

Since we have equilibrium the sum of the external Torques on the bolt must be 0. Meaning T and the sum of the total reactive Torque must be 0 (difference in this case since they have different signs ). Since t gives us the reactive Torque at any point then the sum of all points gives us the total reactive Torque so :

$\begin{align*} T &- \int_{0}^{0.05} t dx = 0 \\ \\ T &= \int_{0}^{0.05} kx ^{\frac{2}{3}} dx=k \int_{0}^{0.05} x ^{\frac{2}{3}} dx \\ \\ &= \dfrac{kx^{\frac{2}{3}+1}}{{\frac{2}{3}}+1} \Big|_{0}^{0.05} = \dfrac{3kx^{\frac{5}{3}}}{5} \Big|_{0}^{0.05} \\ \\ &= \dfrac{3k0.05^{\frac{5}{3}}}{5} \\ \\ \\ k &= \dfrac{5 T}{3 \cdot 0.05^{\frac{5}{3}}}= \dfrac{5 \cdot 50 }{3 \cdot 0.05^{\frac{5}{3}}}\\ \\ &= 12280 \dfrac{\text{ N}}{\text{ m}^2}\\ \\ &= \approx 12.3 \dfrac{\text{ KN}}{\text{ m}^2} \end{align*}$

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