Under normal operating conditions, the electric motor exerts...

Under normal operating conditions, the electric motor exerts a torque of $2.4 \mathrm{kN} \cdot \mathrm{m}$ on shaft $A B .$ Knowing that each shaft is solid determine the maximum shearing stress in $(a)$ shaft $A B$ (b) shaft $B C,(c)$ shaft $C D$

Key Concepts

Maximum Shearing Stress

Maximum shearing stress in a circular shaft occurs at the outer surface, where the radial distance (r) is greatest. In torsion analysis, determining where and how large this stress is helps engineers design shafts that can safely transmit the required torque without material failure.

Polar Moment of Inertia

The polar moment of inertia (J) is a geometrical property of the cross-section that quantifies its resistance to torsional deformation. For a solid circular shaft, it is typically calculated using J = (?*d^4)/32, where d is the diameter of the shaft. Correct evaluation of J is critical for accurate stress analysis in torsion problems.

Torsion in Circular Shafts

This concept deals with the study of how a torque applied to a shaft produces twisting or rotational deformation. In mechanical design, it is crucial to determine how the applied load generates shear stresses throughout the material of the shaft. The analysis typically assumes that the material is homogeneous and behaves elastically, and that the cross-section remains plane and undistorted during twisting.

Torsion Formula

The torsion formula is fundamental for calculating the shearing stress in a shaft subjected to a torque. It is expressed as ? = T*r / J, where ? is the shear stress, T is the applied torque, r is the distance from the center of the shaft, and J is the polar moment of inertia. This formula is essential for ensuring that the design does not exceed the material's allowable stress limits.

Transcript

00:01 Let us first calculate the polar moment of inertia of soft ac.

00:04 So it will be jac is equal to pi divided by 2, phi divided by 2, rac to the power of 4.

00:12 Let us put the value here.

00:14 So it will be pi divided by 2.

00:16 Radius is diameter divided by 12, 2.

00:19 So 25 divided by 2 to the power of 4.

00:22 So it will be equal to 39, 38 point 349 multiplied with 3.

00:35 38 .349 multiplied with 10 to the power of 3 mm to the power of 4 so we will write it as jac is equal to 38 .349 multiplied with 10 to the power of minus 9 meter to the power of now let's calculate let's calculate the angular velocity of soft so it will be omega is equal to 2 pi n so it will be equals to 2 pi multiplied with n n is given as 50 revolution per second so it will be 100 pi radian per second let us now calculate the torque in sacs and in shaft bc so it will be torque in saff bc is equals to power divided by omega omega which is angular velocity so power is given as 3 multiplied by 10 to 2 or 3 watt and omega we have calculated as 100 pi.

01:43 So from this we get torque in section bc is 9 .549 newton meter.

01:52 Let us now calculate the maximum shear stress in the soft bc.

01:56 So it will be tau bc is equal to the torque in section bc multiplied by the radius divided by the polar moment of inertia.

02:04 So the torque in section bc is 9 .549 which we have already calculated by the radius is 12 .5 mm...

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