Mechanical Engineering Case Studies and Calculations

This document contains case studies and calculations related to mechanical engineering. It covers topics such as beam, space station, barge, and tank. The calculations and solutions for various mechanical engineering problems are provided in detail.

MECHANICAL Project

Table of Contents
1. Case Study A...........................................................................................................................2
1.1 Question 1.........................................................................................................................2
1.2 Overview of the Beam......................................................................................................4
2. Case Study B............................................................................................................................7
2.1 Question 1.........................................................................................................................7
2.2 Question 2.........................................................................................................................8
2.3 Question 3.........................................................................................................................8
3. Case study C..........................................................................................................................11
3.1 Question 1.......................................................................................................................11
3.2 Question 2.......................................................................................................................13
4. Case study D..........................................................................................................................15
4.1 Question 1.......................................................................................................................15
4.2 Question 2.......................................................................................................................16
4.3 Question 3.......................................................................................................................17
5. References..............................................................................................................................18
6. Appendix................................................................................................................................19
1

1. Case Study A
1.1 Question 1
a) Reaction forces
Here, summation of horizontal forces are zero and summation of moments are zero.
Take moment about point A and point B.
Calculate RD and RA
Reaction force at point A, (RA) = 82.5 kN.
Reaction force at point D, (RD) = 77.50 kN.
Here, summation of vertical forces are zero.
Find the shear force at Point A, B, C and D (QA, QB, QC, and QD)
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After that, find the bending moment at Point A, B, C and D (MA, MB, MC, and MD)
The calculations are added in appendix.
b) Shear force diagram
It is given in below diagram.
c) Bending moment diagram
The shear force and bending moment diagrams are shown below.
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d) Max S.F and B.M
From the calculation done, the maximum shear force value and the maximum bending
moment value is found.
Maximum shear force value = 82.5 kN.
It acts at the point “A”.
Maximum Bending Moment value = 85 kN.m.
It is found in Point “C”.
1.2 Overview of the Beam
What is Beam?
Beam is a mechanical element. It is used to support the loads. Mostly it has a horizontal
orientation, in some cases we can see this at an inclined positions. In general, vertical loads and
moments are act on this element. And in some rare cases horizontal loads also may act on this
element. In other words, beam is a structural element. Its length is higher than its cross-sectional
dimensions like width and thickness. Also it undergoes to the lateral loading (Zhang et al. 2014).
Types of Beams
Beams are classified by many factors like type of support, Cross section etc. But
generally beams are classified according to the type of support used. In this section, the brief
overview of all five types of beams are shown.
Type Diagram Commercial
Application.
Cantilever Beam Aircraft wings,
& Balcony
support
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Simply
Supported Beam
Bridges
Continuous
Beam
Bridges
Fixed Beam Rigid Structural
elements
Over hanging
Beam
Balcony
Types of loadings
Type Diagram
Point load
Uniformly Distributed Load
Uniformly varying load
Bending moment diagram and Shear force diagram shape
In this section, the Bending moment diagram and Shear force diagram shapes are drawn
for various types of loads acts on the various types of Beams.
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Calculations
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2. Case Study B
2.1 Question 1
a) Finding speed for developing the gravity
In this kind of space stations, the centripetal force developed by the rotary motion acts similar to
the gravitational pull.
So
Here the r= 0.9 km or 900 m
Centripetal acceleration a = 9.81 m/s2
According to this formula, the ring must rotate at 1 rpm in counter clockwise direction to
develop the centripetal acceleration as 9.81 m/s2.
Here, w= (2*pi*N)/60.
Using this, N value is found.
The calculation for this is added in appendix.
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2.2 Question 2
Best place for constructing the zero gravity lab is its center. Because the center of the
space station has the zero centripetal force.
Because the acceleration due to centripetal force,
At center, r = 0.
Applying r = 0 to the formula. Then, the acceleration value is zero (a=0).
This calculation is added in appendix.
2.3 Question 3
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a) Advantages of having two counter rotating cylinders.
This design is proposed, because this design allows to get the sunlight constantly. In space
stations, the sun light is the one and only source for developing energy for various purposes
(Kamisaka 2001). So this design helps to reduce the complications involved in the energy
extraction from the sun. And another important thing it avoids other gyroscopic effects.
b) Required speed for develop the acceleration 9.81 m/s2
Here,
Radius r = 8 km
Length L = 32 km
And w= (2*pi*N)/60.
According to this formula, the cylinder must rotate at 0.33 rpm in counter clockwise direction to
develop the centripetal acceleration as 9.81 m/s2.
The N value is 0.33 rpm.
This calculation is added in appendix.
Calculations
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3. Case study C
3.1 Question 1
Here, the initial load is 450 kN.
i) Given, Original area (Ad)= 35 m2
Find area (Aa)?
Formula:
Buoyancy force= volume of submerged portion of barge * density of water *
gravitational force. (GROBARCIKOVA and SOSEDOVA 2016)
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Using this formula, the Area (Aa) is calculated. Area (Aa) is 32.63 m2.
This calculation is added in appendix
ii) Here, maximum weight of cargo barge can carry (t) = 2.0 m
Formula used here is Wwave= Aa* t*998*9.81
Using this formula, Wwave value is calculated.
This calculation is added in appendix.
iii) Here for float on surface of water (l) = 3 w
Answer
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3.2 Question 2
I. Answer – False
According to the first question, all the forces acts in the cylinder increases with the
increase of the water level. But it’s wrong because the increase of water level only increase the
Buoyancy force other but the weight is remains constant. So the given statement is false.
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II. Answer – True
According to the second statement the difference between the buoyancy force and tension
remains constant when the buoyancy force and tension increases. The statement is correct.
Because the difference between the Buoyancy and Tension is the Weight of barge. So it is
constant.
Fb = W
Here the W = W – T
So Fb = W – T
Fb – T = W
III. Answer – False
The third statement is Buoyancy force and tension of the barge decreases when the
weight is constant. This statement is wrong because there is no relation between the weight and
Buoyancy force. So it won’t affect the barge. So the given statement is false.
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4. Case study D
4.1 Question 1
Height of the plat form above sea level
Pressure at mean sea level
Pb = (13000*9.81*0.760) = 96.922 kPa.
Pressure at top of the platform
Pt = (13000*9.81*0.731) = 93.735 kPa.
Change in pressure = 3.188 kPa.
Hair = (3188)/(1.17*9.81)
Hair = 277.76 m above the mean sea level.
Detailed calculations are added in appendix section.
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4.2 Question 2
1) Yes the system needs ballast tanks. Because it needs to float at the depth of 10 under the
water line. Here the actual dry weight of the tank is 8000 kg. For the desired floating
condition the Buoyancy force at 10 m depth must be same with the load of the ship (Park et
al. 2016).
Force at 10 m depth
Fb = ρghA
Fb = 3*10*998*9.81
Fb = 293.711 kN.
But the actual weight is 8000*9.81 = 74.48 kN.
So the system needs ballast tanks. Detailed calculation are added in the Appendix section.
2) Volume of concrete
Formula to calculate volume is given below
Volume = Mass/ Density
Mass required for float at 10 m depth is 215.231 kN.
Density of concrete is 2500 kg/m3
So, the required volume is calculated as,
Vc = 21939.9/2500 = 8.7759 m3.
So, 8.7759 m3 of concrete required to pour into the tank.
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4.3 Question 3
In excel, we need to enter the details in the Blue color cells and we can found that the corrected
values are in yellow color cells.
The excel file is attached below.
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5. References
GROBARCIKOVA, A. and SOSEDOVA, J. (2016) "DESIGN OF AGENT-BASED MODEL
FOR BARGE CONTAINER TRANSPORT". Transport Problems 11 (4), 95-101
Kamisaka, S. (2001) "Artificial Gravity And Space Plant Science". Biological Sciences In
Space 15 (3), 153-153
Park, S., Song, Y. and Kim, Y. (2016) "Optimal Design Of Mud Flushing System In Ballast
Tank Of LNG Carrier". Transactions Of The KSME C: Industrial Technology And Innovation 4
(2), 85-92
Zhang, H., Shen, Y. and Zhang, S. (2014) "To Solve The Load Of Beam By Shearing Force
Diagram And Bending Moment Diagram - Reverse Thinking In Strength Of Materials". Applied
Mechanics And Materials 488-489, 546-549
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6. Appendix
Calculations for Case study A
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Calculations for Case Study B
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Calculations for Case Study C
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Calculations for Case Study D
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