Ends A and D of the two solid steel shafts A B and C D are fixed, while ends B and C are connect

Ends A and D of the two solid steel shafts A B and C D are...

Key Concepts

Allowable Shearing Stress

Allowable shearing stress is the maximum shear stress that a material can safely withstand without yielding or failing. It is a critical design parameter in the analysis of torsional components, ensuring that the applied loads remain within the material's safe operational limits.

Polar Moment of Inertia

The polar moment of inertia is a geometric property of a section that reflects its resistance to torsion. For circular shafts, it plays a crucial role in the relationship between the applied torque and the resulting shear stress, governing how efficiently a shaft can transmit torque without excessive stress.

Torque

Torque is the rotational equivalent of linear force and is responsible for causing an object to rotate. In the context of shafts and torsion, torque is applied to generate shear stresses, and its magnitude must be controlled to prevent exceeding the material’s allowable shear stress.

Torsion

Torsion refers to the twisting of an object due to an applied torque. In mechanics of materials, the study of torsion includes the analysis of how internal stresses, particularly shear stresses, are distributed across the cross-section of a shaft when it is subjected to twisting forces.

Shear Stress in Torsion

Shear stress in torsion is the stress component parallel to the cross-sectional plane of the shaft. It is generated by the applied torque and varies with the radial distance from the center, reaching a maximum at the outer surface of a circular shaft. This concept is key to ensuring that the induced stresses do not exceed the material's allowable limits.

Transcript

00:01 In this problem as the peripheral motion of the gear are equal as the peripheral motions of the gear are equal motions of the gears are equal gears are equal i can write the value of r b 5 b is equal to r c 5 c now you can write the value of 5b is equal to i c 5 c by r b so it which is equal to 40 by 100 multiplication this value which is equal to 0 .4 5 c let it be equation number one now equate the sum of moment to 0 for gear c summation of mc is equal to 0 or t cd is equal to r c f or f is equal to t cd by r c also summation of mb is equal to zero equate the moment of equate the sum of moment to zero for gear b now you can write next expression edge t minus t abe is equal to r b f or t minus t ab b is equal to r b is equal to r b multiplication t cd by r c so t is equal to t a b plus 100 by 40 multiplication tcd which on further simplification you will get t is equal to t a b plus 2 .5 tcd let it be equation number now sigma b is equal to this value which is equal to t a b l by j g on further simplification it will be equal to t a b l by pi by two c to the power four g which is equal to t a b multiplication point three by pi by two multiplication 0 .03 to the power 4 multiplication g so you will get the value of 5b is equal to 235 .77 multiplication 10 to the power 3 t a b by g now this value will be equal to this which is equal to tcd l by j g that is equal to tcd l by j g that is equal to t cd l by pi by 2 multiplication 0 .0 225 to the power 4 multiplication g which on further simplification you can write the value of sigma c as 1 to 4 2 multiplication 10 to the power 3 t c d by g now from equation 1 sigma b is equal to 0 .4 sigma c from equation 1 you can write the next expression as or 235 .77 multiplication 10 to the power 3 t a b by g is equal to 0 .6 multiplication 1 to 4 2 multiplication 10 to the power 3 multiplication t cd by t c d by g finally you can write the value of t cd as 0 .47462 t a b now from equation 2 you can write t is equal to t a b plus 2 .5 tcd so t will be equal to t a b plus 2 .5 multiplication 047462 t ab from the upper equation which is equal to 2 .1865 t a b now on further simplification you can calculate the value of t is equal to t a b plus 2 .5 t cd so t is equal to t cd by 047462 plus 2 .5 tcd so t will be equal to 4 .7462 plus 2 .5 t cd so t will be equal to 4 .4162 plus 2 .4262 plus 2 .2 so t cd so t will 6 to 4 .60 4 .60 68 t cd now you can write the expression for tau a b is equal to t a b c by j also i can write t a b is equal to j b is equal to j by c multiplication tau ab which is equal to pi by 2 c cube tau a b that is equal to pi by 2 multiplication 0 .03 cube multiplication 50 multiplication 10 to the power 6 so you will get the value of t a b as 212 .6 newton meter now let us finally write the value of t so t will be equal to 2 .1865 t a b that will be equal to 2 .1865 multiplication 212 .6 so it will be equal to 4 .64 kilo newton meter as the final value now let us calculate the value of tcd so tau cd is equal to tcd e c by j t cd is equal to j by c t cd which is equal to pi by two ccube tau cd that on further simplification it will be equal to pi by two multiplication point zero two to five cube multiplication 50 multiplication 10 to the power 6 that is equal to 894 0 .62 newton meter and calculating the value of t that is equal to 4 .6068 t cd that is equal to 4 .6068 multiplication 894 .62 so you will get this value as 4 .12 kilo -neuton meter as the final value...

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