A torque of magnitude T=120 N ·m is applied to shaft A B of the gear train shown. Knowing that t

A torque of magnitude T=120 N ·m is applied to shaft A B of...

A torque of magnitude $T=120 \mathrm{N} \cdot \mathrm{m}$ is applied to shaft $A B$ of the gear train shown. Knowing that the allowable shearing stress is $75 \mathrm{MPa}$ in each of the three solid shafts, determine the required diameter of $(a)$ shaft $A B$ (b) shaft $C D$ $(c)$ shaft $E F.$

Key Concepts

Shear Stress Analysis in Torsion

Shear stress analysis involves determining the maximum shear stress induced in a shaft when a torque is applied. The maximum shear stress at the outer surface of the shaft is given by ? = T*c/J, where T is the torque, c is the radius, and J is the polar moment of inertia. This analysis is essential to ensure that the induced stresses remain within the material's allowable limits to avoid failure.

Material Allowable Stress

The concept of allowable or permissible stress is a design criterion that ensures the material will not yield or fail under applied loads. In the context of torsion, the calculated maximum shear stress must not exceed the material’s allowable shearing stress. This safety check is crucial in engineering design to ensure reliability and longevity of components.

Torsion in Circular Shafts

This concept deals with the behavior of circular shafts under applied torques, where the twisting action induces shear stresses across the cross-section. Understanding how torque creates these internal stresses is fundamental for assessing the performance and safety of mechanical shafts in various applications.

Polar Moment of Inertia

The polar moment of inertia is a key geometrical property that measures a shaft’s resistance to torsion. For solid circular cross-sections, it is calculated as J = (?d?)/32. It underpins the relationship between the applied torque and the resulting shear stress, playing a critical role in the design and analysis of shafts subject to twisting loads.

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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