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Calculate reactions at B1 and B2 and draw a Bending Moment Diagram. Show clearly the values on the BM Diagram. What type of simple direct stress will be calculated using the information on the diagram?

Category: Engineering Paper Type: Online Exam | Quiz | Test Reference: APA Words: 1150


Bending moment diagram


(i)                 What type of simple shear stress is present due to the loading? Where on the shaft is the shear stress at maximum value?

If you do not complete these steps, assume values of Mmax = 62Nm and T = 71.5 Nm to allow you to proceed.


(ii)              Hence identify the location of maximum overall stress on the shaft and calculate values for both of the simple stresses acting there.

 

There are three types of simple stress which shown as below;


(iii)            Calculate the relevant principal stresses. Explain why these stress values are useful to the engineer. Use the result of the calculation to find the real shaft safety factor.

There are two types of principal stresses; 2-D and 3-D. The equation of 2-D principal stress is calculated by the angle 


(b) 30 marks

For the Shaft System above in part (a)

(i)                 Review the design parameter values used in the full calculation process from part (a). Based on these, identify four different suggested changes to the shaft design parameters to improve the stress behavior of the shaft. Assume the overall shaft length (B1 – B2) cannot be changed. You should explain how  the changes are of benefit to the shaft.

The design parameters for the shaft are follows

Parameter of shaft

Symbol

value

Unit

Diameter of shaft

D

18

mm

Length of shaft

L

0.58

mm

 

Four changes

·         A shaft rotates at a constant speed about its longitudinal axis.

·         A shaft has a uniform, circular cross section.

·         The shaft is perfectly balanced

·         All damping as well as nonlinear effects are excluded.

(ii)              What do you think would be the effects on the shaft if the rotor was keyed to the shaft? i.e. the rotor no longer idling, it rotates with the shaft. Refer to two separate effects.  Assume the rotor center of gravity is not in line with the shaft axis.

·         The effect of shaft is occurring when the acceleration is being applied to a rear wheel and it also makes the reactive force on drive shaft.

·         The shaft drive also has the rigid connection to a hub by the reactionary force which turns the shaft backwards regarding to the rear wheel.

(iii)            For the following situations A & B shown below, clearly identify the point of maximum stress in each situation. Sketch an appropriate stress element for that position. Clearly define each item on the stress element diagram.

NOTE: No actual calculation is required.


 

(c) 10 marks

A specimen of material is loaded in shear as shown. Calculate the simple shear stress present along the y direction. Assuming shear is the only loading, draw a stress element diagram at the location of loading.

Taking into account the grain direction of timber, comment on a potential failure issue arising from this loading situation that may not be obvious on first inspection.


Q.2 (a) 60 marks

A manufacturer of hydraulic cylinders is developing a new type of cylinder. Their engineering team have decided to first conduct a failure test and then design a prototype.

Before prototyping, they have developed a smaller closed-end hydraulic test cylinder for pressure testing. They have then conducted a cylinder pressure test (Test Sample 1) on a proposed cylinder material. The internal pressure was increased slowly until yielding was detected at the inside radius of the cylinder wall (this is considered as a failure). The Failure Pressure below shows the internal pressure at which the cylinder wall began to yield.

A subsequent inspection of the test piece failure site observed that the failure was actually caused by excessive maximum shear stress.

The test data is shown below:

Table Q.1.1 Test Data for SAMPLE 1

Inner Diameter (m)

0.1

Wall Thickness (m)

0.02

Material

Mild Steel

Internal Pressure at Failure (MPa)

130

 

(i)                 Calculate the values of the three usual cylinder stresses at the internal radius in SAMPLE 1 where failure is detected.

Answer:


The engineering team learned from the test observation and used this to design a full scale prototype cylinder (PROTOTYPE 1). The parameters they used are shown:

 

Table Q. 1.2 Design Data for PROTOTYPE 1

Internal Diameter (m)

0.17

Material Shear Strength (MPa)

195

Safety Factor

4

In-Service Internal Pressure (MPa)

14

External Pressure (MPa)

0

Material Shear Strength is the Yield Strength in Shear for the cylinder material.

 

(ii)              Calculate the necessary outside diameter for the PROTOTYPE cylinder.


(b) 30 marks

(i)        What factors in the design specification/design parameters should be controlled and carefully considered to minimise the chances of cylinder failure. You should briefly explain three different options.

There are the following factors;

·         Flexibility

·         Use of space

·         Installation during durable material

·         To minimize the chances of cylinder failure the proper inspection and regular training decrease the chances.

·         To minimize the chances of cylinder failure the preventative maintenance plan  decrease the chances .

(iii)            This analysis used Thick cylinder theory. Would Thin cylinder theory be suitable for this case?

Yes the Thin cylinder is suitable for this case because the wall thickness of thin cylinder is lesser than 1/20 times of its internal diameter


In service, the pressurised cylinder body (not the ram) will be subject to a torque T acting around its main longitudinal axis as shown.

Take a 2D section y-y through the cylinder wall.  Draw a stress element at one point on the inner wall, including the cylinder stresses that act in the section plane.

Show generally the effect of the extra stress from this additional torque T on the stress element. No calculation is needed here.

Answer:

The effect of the extra stress causes the compression and tension when the force is applied to the area. A torsion twist is also induced when the torque is applied and shaft causes the distribution stress over the shaft cross section area.


 

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