solution authoring guidelines · (2) for problems involving usage of autocad/solid works/ansys...

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Solution Authoring

Guidelines

Version 9.4

September 2016

2 Copyright@2016 chegg.com

Subject-specific Guidelines- Mechanical Engineering

Table of Contents

ME1.Content ...................................................................................................... 3

A. Text/Explanation: ......................................................................................................... 3

B. Equations ....................................................................................................................... 3

C. Diagrams ....................................................................................................................... 4

D. Graphs ........................................................................................................................... 8

E. Tables ............................................................................................................................. 8

ME2. Technology ............................................................................................... 8

ME3. Special points / others: ............................................................................ 9

List of changes made over edition 9.3

(1). Formatting changes under correct representation B. Equations 3. (i) has been

modified………………………………………………………………………………………3

(2). Second and third points added under ME3………………………………………………9


ME1. Content

A. Text/Explanation:

1. If the problem involves interpolation, then the calculation part of the interpolation

should be explained clearly instead of writing direct values.

2. Standard constant values (like acceleration due to gravity), must be taken from the

respective textbook.

B. Equations

1. The degree symbol should be used while describing the temperature in case of

centigrade, Fahrenheit and Rankine C, F, RT T T . The notation for Kelvin should

be as per the textbook notations K or K .

2. The multiplication symbol used in units should be mid-dot.

3. The representation of sigma symbol and degree symbol that are frequently used in the

expressions:

i. Correct Representation

0

sin 60 0

24.4 sin 60 0

28.2kN

y

y AB

AB

AB

F

B F

F

F


ii. Wrong Representation

sin 0

24.4 sin 0

28.2k

0

N

60

60

y AB

A

y

B

AB

B

F

F

F

F

4. While taking moments about a point:

Correct representation

Take moments about A.

0AM

Wrong representation

Taking moments

0aM

C. Diagrams

1. If parts of a diagram are needed to be referred in the solution, label them.

For example:

2. While representing a T-s diagram, please show the difference between the actual

process and saturation lines and place the state points on the T-s diagram

correctly. For example, consider a simple Rankine cycle,


Correct Representation

3. Numbering done at the precise points/positions.

4. The actual process lines are thickened when compared to other lines.

5. The thickness of the actual process lines is 2 points and for the other lines 1 point.

Wrong Representation

Numbering NOT done at the precise points/positions.

The actual process lines NOT thickened when compared to other lines.

Numbering NOT done at the precise points/positions.

Highlight only the process, as the process varies among problems, but the

curve is always constant.


6. Representation of trusses:

Example:

Variables should be used in the same style throughout. In other words, the variables should

be shown in diagrams just as they are represented in the rest of a solution.

Consider the free-body diagram at joint B.



Wrong Representation

Here, variables A and B have been represented in lowercase, but they should be in uppercase.

Variables should be used in the same style as in the textbook.

7. In the case of shear force and bending moment diagrams, their widths should

match the width of the beam. The diagrams should also be aligned with the beam.


While drawing the shear diagram and the moment diagram, the width of the diagram should

be the same as the given Load diagram. Also, all the three diagrams should be aligned

vertically.


Wrong representation

In the above figure, Shear diagram and Moment diagram crossed the boundaries of the Load

diagram.

D. Graphs

While taking any constant values from the textbook graphs, the graph name should be

specified.

E. Tables

While taking any constant values from the textbook tables, the table name should be

specified. Do not use the number or exact title of the table used in the textbook.

ME2. Technology

1. For solving the equations and programming type exercises in Thermodynamics

problems, use EES (Engineering Equation Solver) software or IT (Interactive

Thermodynamics) software.

2. For programming in heat transfer problems, use IHT (Interactive Heat Transfer)

software.

3. For diagrams, use Serif v.X.5 Software. Should you wish to use a different

program for diagrams, please consult your Territory Manager.


4. For representing equations in your solutions, use MathType or the Microsoft

Equation 3.0 (Insert > Object (Text Panel)> Microsoft equation 3.0).

5. If a solution uses results from software (or any other tool), then the step wise

procedure for obtaining those results in software should be mentioned in the

solution.

6. Snapshots of any data (figures/charts/graphs/plots etc) generated by the software

should be produced in the solution to make solution more understandable and

clear.

ME3. Special points / others:

(1) Use only the respective tables given in the textbook while solving the solution.

However, do not scan the textbook table. Rather take certain entries from the

textbook table which are needed for the solution.

(2) For problems involving usage of AutoCAD/Solid works/Ansys etc., the author should

provide steps of construction.

(3) For problems involving graphical analysis, the author should provide steps of

construction. Example: problems related to Mohr’s Circle, Graphical Analysis of

Velocity/ Acceleration diagrams in Theory of machines

Back


Example solutions – Mechanical Engineering

List of changes made over Version 9.2

Existing solution replaced with new solution in ME example 2………………Page no. 187

Delimiter added to the existing ME example4…………………..………………….…Page no. 188

Sample solution related to fill in the blank added in ME example 3…………Page no. 187

Sample solution related to True or False added in ME example 7……………Page no. 199

Sample solution related to VSAQ added in ME example 8………………………Page no. 199

List of changes made over Version 9.3

Tab spaces removed in ME example 1……………………………………………………Page no. 195

ME example 3 modified…………………….……………………………………………………Page no. 198

Tab spaces removed in ME example 6……………………………………………………Page no. 201


ME Example 1: Calculation Based

Question:

Express the following quantities in British Gravitational system (BG).

(a)15.2 km , (b)39.14 N m , (c)

31.61 kg m , (d) 0.0320 N m s , (e) 5.67 mm hr

Solution:

(a)

Express 15.2 km in British Gravitational system (BG).

The unit of length in BG system is foot.

To perform the conversion, first, convert the length from kilometers to meters.

15.2 km 15.2 1000 m

152,00 m

Now, convert the length from meter to feet. State the conversion factor.

1 m 3.281 ft

Use the conversion factor to express the length in feet.

3.281 ft15,200 m 152,00 m

1 m

152,00 3.281 ft

49871.2 ft

Therefore, the quantity in BG units is 49,871.2 ft .

(b)

Express 39.14 N/m in British Gravitational system (BG).

State the conversion factor to convert the given quantity from 3

N m to3

lb ft3 3 31 N m 6.366 10 lb ft

Use the conversion factor to express 39.14 N/m in3

lb ft . 3 3

3 3

3

3 3

3

6.366 10 lb ft9.14 N m 9.14 N m

1 N m

9.14 6.366 10 lb ft

0.0581 lb ft

Therefore, the quantity in BG units is30.0581 lb ft .


(c)

Express 31.61 kg/m in British Gravitational system (BG).

State the conversion factor to convert the given quantity from 3

kg m to3

slugs ft . 3 3 31 kg m 1.94 10 slugs ft

Use the conversion factor to express 31.61 kg/m in

3slugs ft .

3 33 3

3

3 3

3 3

1.94 10 slugs ft1.61 kg m 1.61 kg m

1 kg m

1.61 1.94 10 slugs ft

3.1234 10 slugs ft

Therefore, the quantity in BG units is 3 33.1234 10 slugs ft .

(d)

Express 0.0320 N m s in British Gravitational system (BG).

State the conversion factor to convert the given quantity from N m s to ft lb s . 11 N m s 7.373 10 ft lb s

Use the conversion factor to express 0.0320 N m s in ft lb s . 1

1

7.373 10 ft lb s0.0320 N m s 0.0320 N m s

1 N m s

0.0320 7.373 10 ft lb s

0.0236 ft lb s

Therefore, the quantity in BG units is 0.0236 ft lb s .

(e)

Express 5.67 mm hr in British Gravitational system (BG).

State the conversion factors to convert the velocity from mm hr to m s . 31 mm 10 m

1 hr 3600 s

Use the conversion factors to express 5.67 mm hr in m s .


3

3

6

mm 10 m 1 hr5.67 mm hr 5.67

hr 1 mm 3600 s

5.67 10 m s

3600

1.575 10 m s

State the conversion factor to convert the velocity from m s to ft s .

1 m s 3.281 ft s

Use the conversion factors to express 61.575 10 m s in ft s .

6 6

6

6

3.281 ft s1.575 10 m s 1.575 10 m s

1 m s

1.575 10 3.281 ft s

5.1675 10 ft s

Therefore, the quantity in BG units is 65.1675 10 ft s .

ME Example 2: Multiple Choices type

Question:

From the deaerator, oxygen and other noncondensable gases are separated from the feedwater

and directed to the atmosphere ___________.

A. by using the closed feedwater heater.

B. through the economizer.

C. by using steam traps.

D. Through the vent.

Solution:

Closed feed water heater consists of outlet valve, bypass line and valve. It provides passage

for feed water to the boiler.

Hence, option A is wrong.

An economizer uses the heat from the combustion gases to heat the feed water before it enters

the boiler.

Hence, option B is wrong.


A steam trap provides passage for gases without any loss of steam.

Hence, option C is wrong.

Oxygen and the non-condensable gases are separated from the feed water in the deaerator and

are escaped to the atmosphere through the vent provided.

Therefore, the correct option is D .

ME Example 3: Fill in the blank type

Question:

The two broad categories of display cases are ____________ and ______________.

Solution:

Display cases are used to display the item in retail stores.

These display cases may be classified into two major categories

Open type

Closed type.

Therefore, the answers to fill the blanks are open type and closed type .


ME Example 4: Software based

Question:

A double pane window with 2 m high and 4 m wide consists 5 mm thick layers of glass (k =

0.78 W/mK) seperated by stagnant air in the space with width of ‘d’. The temperature of the

room is 24o C and that of the outdoor is 0

o C . By considering the heat transfer coefficients

on the outer and inner surfaces of window as 25 and 10 2 oW/m C

and temperature of air as 25o C , plot the rate of heat transfer through the window as a

function of the width of air space ‘d’ in the range of 3 mm to 30 mm.

Solution:

The given problem can be solved using EES software by the following steps.

1. Open EES software

2. Create a new file by clicking on the ‘New’ option of the ‘File menu’

3. Now, type the following code in dailog box:

"GIVEN"

A=2*4 [m^2]

L_glass=5 [mm]

k_glass=0.78 [W/m-C]

"d=1 [mm], parameter to be varied"

T_infinity_1=24 [C]

T_infinity_2=0 [C]

h_in=10 [W/m^2-C]

h_out=25 [W/m^2-C]

"PROPERTIES"

k_air=conductivity(Air,T=25)

"ANALYSIS"

R_conv_1=1/(h_in*A)

R_glass=(L_glass*Convert(mm, m))/(k_glass*A)

R_air=(d*Convert(mm, m))/(k_air*A)

R_conv_2=1/(h_out*A)

R_total=R_conv_1+2*R_glass+R_air+R_conv_2

Q_dot=(T_infinity_1-T_infinity_2)/R_total

To solve the equation for a range of length values, use parametric table as follows:

Click on the ‘New parametric table’ of ‘Tables’ menu

A dailog box will open.

Now, select Run 10.

Select variables in equations as d and Q, press ‘Add’ and then press ‘Ok’

A table will appear; enter the values of length under ‘d ’ as 3,6....30 in runs 1 to 10

Then, click on ‘Solve table’ of ‘Calculate’ menu


The solution will appear as follows:

To plot a graph, execute the following steps:

Click on the ‘X-Y Plot’ of ‘New Plot window’ option from ‘Plots’ menu.

Select X-axis as d and Y-axis as Q . Press OK

The following graph will be plotted:

This graph indicates that by increasing the space between the glasses, the resistance towards

heat transfer increases, and thereby decreases the heat transfer. So, in order to maintain the

room at high temperature for a long time, it is recommended to provide more space between

the glasses.


ME Example 5: Conceptual

Question:

Describe what may happen to an electric motor if the drive belt is too tight.

Solution:

In a belt drive system, an electric motor is usually used to drive a shaft. The shaft is mounted

on bearings for free rotation.

If the belt is too tight, the load is applied on the shaft, and the shaft applies the load on the

bearings. So, failure of the bearings may occur impeding the rotation of the shaft to rotate.

When the shaft rotation is impeded, it may also lead to failure of the motor.

ME Example 6: Diagrammatic

Question:

The bars shown in figure have a cross-sectional area260 mmA , modulus of elasticity

200 GPaE , and coefficient of thermal expansion6 112 10 C . If a 40 kN downward

force is applied at A and the temperature of the bars is raised 30 C , what are the normal

stresses in the three bars?

Solution:

Draw the free body diagram of the joint A and mark the tensile forces , and AC AD ABP P P

acting on the members AC, AD, and AB, respectively.


Consider the forces acting along the horizontal direction.

cos45 cos60AC ABP P

0.707AC ABP P …… (1)

Consider the forces acting along the vertical direction.

1000 N

sin 45 sin 60 40 kN1 kN

AD AC ABP P P

Substitute 0.707 for AB ACP P

0.707 0.707 0.866 40,000

1.366 40,000

AD AB AB

AD AB

P P P

P P

40,000 1.366AD ABP P …… (2)

Consider the geometry of system prior and after the deformation.

Let A be the final position of the point A after the deformation. Similarly, let v and u be the

vertical and horizontal displacements of the point A respectively.

Use trigonometric relations to obtain the lengths of the members.


Length of the member AB:

0.2 m

sin 60

0.231 m

ABL

Length of the member AD:

0.2 mADL

Length of the member AC:

0.2 m

sin 45

0.283 m

ACL

Use trigonometric relations to obtain the distance between the points.

Distance BD:

0.2 m

tan 60

0.1155 m

BDL

Distance DC:

0.2 m

tan 45

0.2 m

DCL

The total change in length will be the sum of the change in length due to the axial force and

the change in length due to thermal strain.

Write the deformation relations for the members AB.

AB ABAB AB

P LTL

EA

Here, and T are the coefficient of thermal expansion and change in the temperature of

the members respectively. Similarly, E and A are the modulus of elasticity and cross sectional

area of the members.

Substitute 2 6 10.231 m for , 200 GPa for , 60 mm for , 12 10 C for ABL E A and

30 C for .T

6 26 1

2 9 22

0.231 m 10 mm 1 GPa12 10 C 30 C 0.231 m

1 m 10 N m200 GPa 60 mm

ABAB

P

8 51.925 10 8.316 10AB ABP …… (3)

Write the deformation relations for the members AC.

AC ACAC AC

P LTL

EA


Substitute 2 6 10.283 m for , 200 GPa for , 60 mm for , 12 10 C for ACL E A and

30 C for T .

6 26 1

2 9 22

8 5

0.283 m 10 mm 1 GPa12 10 C 30 C 0.283 m

1 m 10 N m200 GPa 60 mm

2.357 10 10.182 10

ACAC

AC

P

P

From the equation (1), substitute 0.707 for AB ACP P to find AC in terms of PAB

8 52.357 10 0.707 10.182 10AC ABP

8 51.667 10 10.182 10ABP …… (4)

Write the deformation relations for the members AD.

AD ADAD AD

P LTL

EA

Substitute 2 6 10.2 m, 200 GPa, 60 mm , 12 10 C and ADL E A 30 C,T

6 26 1

2 9 22

8 5

0.2 m 10 mm 1 GPa12 10 C 30 C 0.2 m

1 m 10 N m200 GPa 60 mm

1.667 10 7.2 10

ADAD

AD

P

P

From the equation (2), substitute 40,000 1.366 for AB ADP P to find AD in terms of PAB

8 51.667 10 40,000 1.366 7.2 10AD ABP

5 873.867 10 2.277 10 ABP …… (5)

Reconsider the geometry of the system:

Apply the Pythagorean Theorem to obtain the relations between the lengths.

Consider the triangle A GD


2 22

2 2 2 2 2

2 2 2

2 2

2 2

AD AD AD

AD AD AD AD AD AD

AD AD AD AD

u L v L

u L v vL L L

u v vL L

As the deformations are comparatively small in the magnitude, neglect the second order

terms of the deformations. 2 2 2 0ADu v

Substitute zero for2 2 2, and ADu v .

2 2AD AD ADvL L

ADv

From equation (5), substitute 5 873.867 10 2.277 10 ABP for AD to find v in terms of

PAB 5 873.867 10 2.277 10 ABv P …… (6)

Consider the triangle .A FC

2 22

2 2 2 2 2

0.2

2 0.04 0.4 2

DC AC AC

DC DC AC AC AC AC

L u v L

L u L u v v L L


terms of the deformations. 2 2 2 0ACu v

Substitute zero for 2 2 2, and ACu v

2 22 0.04 0.4 2DC DC AC AC ACL L u v L L

Replace with 0.2 m and with 0.283 mDC ACL L

2 2

0.2 2 0.2 0.04 0.4 0.283 2 0.283

0.04 0.4 0.04 0.4 0.08 0.5657

0.4 0.4 0.5657

AC

AC

AC

u v

u v

v u

1.414 ACv u

From equations (6) and (4), substitute 5 873.867 10 2.277 10 ABP for v and

8 51.667 10 10.182 10ABP for AC to find u in terms of PAB.

5 8 8 573.867 10 2.277 10 1.414 1.667 10 10.182 10AB ABP u P

5 859.467 10 4.6345 10 ABu P …… (7)


Consider the triangle BEG

2 2 2

2 2 2 2 2

0.2

2 0.04 0.4 2

BD AB AB

BD BD AB AB AB AB

L u v L

L u L u v v L L


terms of the deformations. 2 2 2 0ABu v

Substitute zero for 2 2 2, and ABu v

2 22 0.04 0.4 2BD BD AB AB ABL L u v L L

Replace with 0.1155 m with 0.231 mBD ABL L

2 2

0.1155 2 0.1155 0.04 0.4 0.231 2 0.231

0.0133 0.231 0.04 0.4 0.0533 0.4619

AB

AB

u v

u v

0.5 0.866 ABu v

From the equations (3), (6), and (7), substitute8 51.925 10 8.316 10 for AB ABP ,

5 873.867 10 2.277 10 for ABP v and 5 859.467 10 4.6345 10 for ABP u to find PAB

5 8

8 5

5 8

0.5 59.467 10 4.6345 101.925 10 8.316 10

0.866 73.867 10 2.277 10

AB

AB

AB

PP

P

8 56.2142 10 85.388 10ABP

13,741 N

13.74 kN

ABP

Find the deformation in the member AB, AB from the equation (3).

8 51.925 10 8.316 10AB ABP

Substitute 13,741 kN for ABP

8 5

8 3 5

1.925 10 8.316 10

1.925 10 13.74 10 8.316 10

0.3479 mm

AB ABP

Find the deformation in the member AC, AC from the equation (4).

8 51.667 10 10.182 10AC ABP

Substitute13,741 kN for ABP .


8 5

8 3 5

1.667 10 10.182 10

1.667 10 13.74 10 10.182 10

0.3309 mm

AC ABP

Find the deformation in the member AD, AD from the equation (5).

5 873.867 10 2.277 10AD ABP


5 8

5 8 3

73.867 10 2.277 10

73.867 10 2.277 10 13.74 10

0.4260 mm

AD ABP

Find the vertical displacement of the point A, v from the equation (6).

5 873.867 10 2.277 10 ABv P


5 8

5 8 3

73.867 10 2.277 10

73.867 10 2.277 10 13.74 10

0.4260 mm

ABv P

Find the horizontal displacement of the point A, u from the equation (7). 5 859.467 10 4.6345 10 ABu P


5 8

5 8 3

59.467 10 4.6345 10

59.467 10 4.6345 10 13.74 10

0.042 mm

ABv P

Write the relation to find normal stress in the member AB.

ABAB

P

A

Replace 2 with 13.74 kN and with 60 mmABP A

6 2

2 2

6 2

13.74 kN 1000 N 10 mm

1 kN60 mm 1 m

229 10 N m

229 MPa

AB


Hence, the normal stress in the member AB is 229 MPa . As we assumed the forces in the

members to be in tension, the obtained stress is also tensile.

Write the relation to find normal stress in the member AC.

ACAC

P

A

Find the force in the member AC, ACP from the equation (2).

0.707AC ABP P

Substitute13,741 kN for ABP .

3

0.707

0.707 13.74 10

9.72 kN

AC ABP P

Replace 2 with 9.72 kN and with 60 mmACP A to find the normal stress in the member AC..

6 2

2 2

6 2

9.72 kN 1000 N 10 mm

1 kN60 mm 1 m

162 10 N m

162 MPa

AC

Hence, the normal stress in the member AC is 162 MPa . As we assumed the forces in the


Write the relation to find normal stress in the member AD.

ADAD

P

A

Find the force in the member AD, ADP from the equation (1).

40,000 1.366AD ABP P


340,000 1.366 13.74 10

21.23 kN

ADP

Replace 2 with 21.23 kN and with 60 mmADP A to find the normal stress in the member AD.

6 2

2 2

6 2

21.23 kN 1000 N 10 mm

1 kN60 mm 1 m

354 10 N m

354 MPa

AD


Hence, the normal stress in the member AD is 354 MPa . As we assumed the forces in the


ME Example 7: True or False

Question:

True or False? The air bag diagnostic monitor supplies back-up power to the air bag module

in the event that the battery or cables are damaged during the accident.

Solution:

If the vehicle undergoes an accident, the battery or cables might get damaged. Then the

power to deploy the air bag is supplied by the Air bag sensing diagnostic monitor (ASDM).

The back-up power can last up to 30 minutes after the battery has been disconnected.

Therefore, the statement that the air bag module is supplied with a backup power from air bag

sensing diagnostic monitor in the situation when the vehicle undergoes an accident (battery or

cables might get damaged) is True .

ME Example 8: Very Short Answer Type

Question:

How do dermatologists remove pre-cancerous skin blemishes cryosurgically?

Solution:

The technique of cryosurgery uses the advantage of pre-cancerous skin blemishes or cells

being less dense.

Cryosurgery applies freezing temperatures on pre-cancerous skin blemishes, due to which ice

crystals are formed in these less density cells. The formation of ice crystals eventually tears

apart these cells from and blocks the blood supply to the affected tissues.

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solution authoring guidelines · (2) for problems involving usage of autocad/solid works/ansys...

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