A torque T=900 N ·m is applied to shaft A B of the gear...

A torque $T=900 \mathrm{N} \cdot \mathrm{m}$ is applied to shaft $A B$ of the gear train shown. Knowing that the allowable shearing stress is $80 \mathrm{MPa}$ $(b) \operatorname{shaft} C D$ determine the required diameter of $(a)$ shaft $A B$ $(c)$ shaft $E F$

Key Concepts

Shaft Sizing for Torsion

Determining the minimum required shaft diameter involves combining the torsional stress equation with the polar moment of inertia for a circular section and then equating the resulting shear stress to the allowable shear stress. This calculation ensures that the shaft’s diameter is sufficient to transmit the applied torque safely.

Torque Transmission in Gear Trains

In a gear train, the input torque is distributed among the various shafts according to the gear ratios. This means different shafts may experience different torques, and understanding this distribution is essential for designing each shaft to safely carry its specific load.

Polar Moment of Inertia

The polar moment of inertia is a measure of a shaft’s resistance to torsional deformation. For a circular cross-section, it is given by J = (?/32)d^4. It is a crucial parameter when relating applied torque to the resulting shear stress within the material.

Torsional Stress

Torsional stress occurs when a shaft is subjected to a twisting moment (torque), which results in shear stresses over its cross-sectional area. The maximum shear stress is typically found at the outer surface of the shaft and must be limited to prevent failure.

Allowable Shear Stress

Allowable shear stress is the maximum shear stress that a material can sustain without yielding or failing. In design, it sets the upper limit for the stress due to torsion and is used to determine the required dimensions of components to ensure safe operation.

Transcript

00:03 In this question we have given diameter of shaft ab equals to 20 m m and diameter of soft c d equals to 25 m m and diameter of soft c d equals to 25 m and diameter of soft e f is equal to 40 m and allowable searing stress tau is equal to 60 mega pascal okay so now as the peripheral motions of the gears are equal so we can write between soft ab and cd as t cd divided by r c equals to t divided by r b okay so from here t cd will be equals to t divided by r b okay so from here t cd will be equals to r c by r b multiplied by t okay so by putting the value of r c and r b r c and r b are the radius of gear c and b so torque in cd will be equals to 75 divided by 30 these values are directly returned from the given diagram so this will be equals to 2 .5 times of okay similarly for ef and cd we can write t ef divided by r f equals to t cd d okay so from here torque in ef equals to rf divided by rf divided by rd multiplied by tcd so by putting value of tcd we get t ef equals to rf from the diagram is 90 and rd from the diagram is 30 and tcd is 2 .53 from 2 .5 t from this equation okay so from here torque in ef comes out to be 7 .5 t okay so now the magnitude of torque in soft a b will be equals to t ab equals to t equals to pi by 2 r a b squared by allowable shearing stress tau so by putting values t will be equals to pi multiplied by 10 cube this is q okay multiplied by allowable stress is 60 divided by 2 so from here we get capital t equals to 94 .248 kilo kilo kilo per kilo newton mm okay so now the magnitude of torque in the soft cd will be equals to t cd equals to pi by 2 r cd q multiplied by t so now putting values we get t cd equals to pi by 12 .5 multiplied by 60 divided by 2.

04:13 So from here, tcd is 2 .5 times of t equals to pi multiplied by 12 .5 q multiplied by 60 divided by 2.

04:34 So from here we get t equals to 73 .6...

Please give Ace some feedback