Spring force due to Axial and Torque load?

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SUMMARY

The discussion focuses on the mechanics of a spring subjected to axial loads and torque, specifically analyzing the relationships between torque and bending moments. The torque (T) is defined as Wcos(a) for axial loads and Wsin(a) for bending moments, with the angle alpha (α) being crucial in determining the orientation of these forces. The confusion arises from the interpretation of the angle relative to the axes, clarified by the realization that the torque vector aligns with the X-axis, making cos(0) = 1 applicable. This understanding resolves the initial misconceptions regarding the calculations of torque and bending moments.

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Mechanical engineers, physics students, and anyone involved in the analysis of mechanical systems under axial and torque loads will benefit from this discussion.

Mechaman
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Homework Statement


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The first theory shows a spring under an axial load where the Torque is given as WRcos(a)

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The second theory shows a spring under axial Torque. T is given as Wsin(a)

Homework Equations


Given in links

The Attempt at a Solution


A torque T would be at full strength along the X axis, where cos(0) = 1. Where a bending moment would be at full strength along Y where sin(90) = 1.

It seems to be other way around though, what am I missing?
 
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Mechaman said:
It seems to be other way around though, what am I missing?
In Fig 12.5, alpha is the angle between T and the Y axis, not between T and the X axis.
 
haruspex said:
In Fig 12.5, alpha is the angle between T and the Y axis, not between T and the X axis.

Ok I see now how the axis has has rotated and has the same angle relative to the Y axis. So that means X axis is 0 degrees and Y is 90 degrees.

Why is the bending moment = Tcos(alpha) and axial Torque = TSin(alpha). Wouldn't measuring Torque be TCos(alpha) as Cos(0) = 1?
 
Mechaman said:
Why is the bending moment = Tcos(alpha) and axial Torque = TSin(alpha). Wouldn't measuring Torque be TCos(alpha) as Cos(0) = 1?
The torque is around the X axis, so the vector is along the X axis, i.e. where alpha is zero. So yes, it is cos because cos(0)=1.
 
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haruspex said:
The torque is around the X axis, so the vector is along the X axis, i.e. where alpha is zero. So yes, it is cos because cos(0)=1.
Thanks for response sorry for late reply. I see now.
 

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