For the gear train shown, the diameters of the three solid...

For the gear train shown, the diameters of the three solid shafts are: \[d_{A B}=20 \mathrm{mm} \quad d_{C D}=25 \mathrm{mm} \quad d_{E F}=40 \mathrm{mm}\] Knowing that for each shaft the allowable shearing stress is $60 \mathrm{MPa}$ determine the largest torque $\mathbf{T}$ that can be applied.

For the gear train shown, the diameters of the three solid shafts are:
\[d_{A B}=20 \mathrm{mm} \quad d_{C D}=25 \mathrm{mm} \quad d_{E F}=40 \mathrm{mm}\]
Knowing that for each shaft the allowable shearing stress is $60 \mathrm{MPa}$ determine the largest torque $\mathbf{T}$ that can be applied.

Key Concepts

Torsion in Shafts

This concept deals with the twisting of a shaft under an applied torque. It involves understanding how the internal stresses (shear stresses) are developed within a shaft and how they are distributed across the cross?section when a torque is applied.

Shear Stress in Torsion

Shear stress in a shaft subjected to torsion is calculated using the relationship ? = T*r/J, where ? is the shear stress, T is the applied torque, r is the outer radius of the shaft, and J is the polar moment of inertia. This relation is fundamental in determining whether a shaft can safely transmit the applied torque without exceeding its material limits.

Polar Moment of Inertia

The polar moment of inertia, J, is a geometric property of the cross-section of a shaft that measures its resistance to torsion. For circular cross-sections, it is typically calculated using the formula J = (?/32)d^4, where d is the diameter of the shaft. This concept is crucial for linking the shaft geometry to its torsional strength.

Design Safety with Allowable Stress

Design safety involves ensuring that the induced stresses within a component do not exceed the material's allowable stress limits. In this context, comparing the calculated shear stresses under the maximum applied torque with the allowable shearing stress ensures that the design is safe and reliable.

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...

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