Question

1) Use equilibrium to solve for the unknown torque at Point C of the problem in the Appendix. Sum moments about the X-axis and given that the shaft is in equilibrium, determine the torque that must be applied at Point C. You can assume that the shaft is not connected to any other features although in a real-world application there would need to be bearing that keep the shaft from ‘wandering off.’ But note that there are no ‘forces’ applied to the shaft, only moments, so summing forces along an axis will be zero=zero.

2) Draw Two Free Body Diagrams (FBD) for the shaft in the Appendix. We are working with a shaft assembly as shown in the Appendix of this document (it is a ‘chain’ of three segments, some solid and some hollow). We will want a FBD drawn for the two ends segments. Cut the shaft halfway between Points A and B to ‘release’ the end with Point A from its’ environment. Repeat the process to draw a FBD of the end Point D by cutting halfway between Points C and D. Linked to Basic Learning Objective 1.

3) Use the Equations of Equilibrium (Sum Moments about the X-Axis) to determine the internal torque occurring between Point A and Point B and the internal torque occurring between Point C and Point D. Use the equation from the lecture notes that relates torque (T) to shear stress (τ) at two points on each cross-section: Point 1 is on the outside surface of the cross-section and Point 2 is 7 mm from the center of the cross-section. Linked to Basic Learning Objectives 2 and 3.

Material has E35 GPa and G 18 GPa Power Shaft 15 kN-mm Solid cylinder AB OD 30 mm A transmission shaft is made of three different segments: Solid from A to B, a large pipe from B to C, and a smaller pipe from C to D 25 kN-mm LAB = 1.2m We know there is an applied torque at A and B and an unknown torque at C. There is no torque applied at LBC= 0.6m Pipe BC OD 30 mm ID 10 mm LCD=0.5m Pipe CD OD 20 mm ID 10 mm Note that the drawing shows a shaft that extends from Point A to Point D and is made up of three different segments of solid or circular shaft. The blue regions represent the cross-section at the four different points along the shaft (Points A, B, C, and D)

Answer 1

1. Since, the given shaft is not connected to any external features, we can simply equate the sum of all the moments about the X axis to zero to find out the unknown torque at C as:

Since, the given shaft is not connected to any external features, we can simply equate the sum of all the moments about the X axis to zero to find out the unknown torque at C as: sigma M_x = 0 i.e. -15 kN - mm + 25 kN - mm - C kN - mm = 0 or, C kN - mm = 25 kN - mm - 15 kN - mm = 10 kN - mm. Therefore, unknown torque, C to keep the shaft in equilibrium = 10 kN - mm 2. Now, for the FBD of the two end sections, we first cut the first segment, AB midway and draw it's FBD as shown (refer attached diagram) 3. The internal torque occurring between point A and B can be found out by cutting the segment AB anywhere (preferably, at mid point) and using moment equilibrium equation. To ensure the equilibrium of the cut section, the internal torque generated in the

i.e. -15 kN-mm + 25 kN-mm - C kN-mm = 0
or, C kN-mm = 25 kN-mm - 15 kN-mm = 10 kN - mm.
Therefore, unknown torque, C to keep the shaft in equilibrium = 10 kN-mm

2. Now, for the FBD of the two end sections, we first cut the first segment, AB midway and draw it's FBD as shown (refer attached diagram)

3. The internal torque occurring between point A and B can be found out by cutting the segment AB anywhere (preferably, at mid point) and using moment equilibrium equation.
To ensure the equilibrium of the cut section, the internal torque generated in the shaft AB must be equal in magnitude to the torque at A (15 kN-mm) but opposite in direction.
For the segment CD, there is no torque present anywhere on the shaft, therefore the FBD of the cut section of CD will not have any torque or force on it.
Mathematicaly, for segment AB,

Since, the given shaft is not connected to any external features, we can simply equate the sum of all the moments about the X axis to zero to find out the unknown torque at C as: sigma M_x = 0 i.e. -15 kN - mm + 25 kN - mm - C kN - mm = 0 or, C kN - mm = 25 kN - mm - 15 kN - mm = 10 kN - mm. Therefore, unknown torque, C to keep the shaft in equilibrium = 10 kN - mm 2. Now, for the FBD of the two end sections, we first cut the first segment, AB midway and draw it's FBD as shown (refer attached diagram) 3. The internal torque occurring between point A and B can be found out by cutting the segment AB anywhere (preferably, at mid point) and using moment equilibrium equation. To ensure the equilibrium of the cut section, the internal torque generated in the

i.e. - 15 kN-mm + T kN-mm = 0
or T = 15 kN-mm.
Now, the equation that relates torque to shear stress is given by:
, where = Shear stress, T = Torque generated, r = Distance of point under consideration from the center of shaft
and I_p = Polar moment of inertia.
For Shaft AB, polar moment of inertia = = = 79521.56 mm⁴
(a) At the outside surface of the cross section: distance of point from the center, r = 30 mm.
  
i.e. = (15x10³ N-mm x 30 mm)/ 79521.56 mm⁴ = 5.659 N/mm² = 5.659 MPa
(b) At 7 mm from the center of the shaft
= (15x10³ N-mm x 7 mm)/ 79521.56 mm⁴ = 1.32 N/mm² = 1.32 MPa

For shaft CD, since there is no internal torque in the shaft, there will not be any shear stress anywhere in the shaft.