ENGR 244 Lab 5
docx
keyboard_arrow_up
School
Concordia University *
*We aren’t endorsed by this school
Course
244
Subject
Mechanical Engineering
Date
Dec 6, 2023
Type
docx
Pages
10
Uploaded by AgentSnail2002
Lab 5:
Deflection of Beams
Lab Section CI – X
Winter 2022
Professor Ahmed Soliman
Concordia University
Montreal, QC, Canada
Tuesday March 8
th
, 2022
Table of Contents
Nomenclature pg 3
List of Tables pg 3
List of Figures pg 3
Objective pg 4
Introduction pg 4
Procedure pg 4-5
Results pg 5-9
Discussion and Conclusion pg 9-10
References pg 10
Original Data pg 11
Nomenclature
ρ = radius of curvature
b = width
E = modulus of elasticity
h = height
I = moment of inertia
L = length of span
P = load
x = distance along span
y = deflection
List of Tables
Table 1: Experimental and Theoretical Values of Deflection for Brass Beam at L/2
p. 5
Table 2: Experimental and Theoretical Values of Deflection for Brass Beam at L/4
p. 5
Table 3: Experimental and Theoretical Values of Deflection for Steel Beam at L/2
p. 6
Table 4: Experimental and Theoretical Values of Deflection for Steel Beam at L/4
p. 6
Table 5: Experimental and Theoretical Values of Deflection for Aluminum at L/2
p. 6
Table 6: Experimental and Theoretical Values of Deflection for Aluminum at L/4
p. 6
Table 7: Experimental and Theoretical Values of Deflection for Cantilever Brass at L/2
p. 7
Table 8: Experimental and Theoretical Values of Deflection for Cantilever Brass at L
p. 7
Table 9: Experimental and Theoretical Values of Deflection for Cantilever Steel at L/2
p. 8
Table 10: Experimental and Theoretical Values of Deflection for Cantilever Steel at L
p. 8
Table 11: Experimental and Theoretical Values of Deflection for Cantilever Aluminum at L/2
p. 8
Table 12: Experimental and Theoretical Values of Deflection for Cantilever Aluminum at L
p. 8
List of Figures
Figure 1: Theoretical vs Experimental Deflection of the Beams at L/2
p.7
Figure 2: Theoretical vs Experimental Deflection of the Cantilever Beams at L/2
p.9
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Objective:
The goal of this experiment is to determine the modulus of elasticity using the
relationship between simply supported beams and load deflection[ CITATION Con \l 4105 ].
Introduction:
Subjecting a beam to a transverse load will cause deflection in the beam.
Different types of loads can induce deflection [CITATION Des22 \l 4105 ], but in this experiment only
transverse loading will be tested. The length of the beam, the magnitude of the loads and the
positioning of the loads will make a difference in the magnitude of the deflection. Deflection is an
important aspect in engineering as many building codes require certain limits on
deflections[ CITATION Con \l 4105 ]. These limits can include things such as deflections from
earthquakes or winds [ CITATION Mov22 \l 4105 ]. The data collected in this experiment is important
as it allows us to calculate the curvature of the beam about the neutral axis through this relation
[ CITATION Con \l 4105 ]:
1/ρ = M(x)/EI
The equation above is only valid within the elastic range where ρ is the radius of curvature, M(x) is the
bending moment at distance x, E is the modulus of elasticity of the assumed material, and I is the
moment of inertia of the entire cross-section [ CITATION Con \l 4105 ]. Various beam cross-sections
give different advantages when it comes to beam deflection [ CITATION Mov22 \l 4105 ], but for this
experiment, a full rectangular cross-section beam is used, so the value of I can be expressed as:
I = (1/12)bh
3
Though the radius of curvature, ρ, is related to the deflection of the beam, it does not allow us to
calculate it directly. The curvature has the following equation that will allow us to solve for the
deflection at any point on the beam:
1/ρ = (d
2
y/dx
2
)/(1 + (dy/dx)
2
)
3/2
Since the beam is only subjected to deflection in the elastic region, it can be said that the value of the
slope, dy/dx, is negligible and therefore our new relation between the curvature and the deflection will
be [ CITATION Con \l 4105 ]:
1/ρ = (d
2
y/dx
2
)
Substituting 1/ρ with M(x)/EI, we get:
d
2
y/dx
2
= M(x)/EI
(d
2
y/dx
2
)EI = M(x)
This equation will allow us to calculate the deflection of the beam at any point once we integrate twice
[ CITATION Con \l 4105 ].
Procedure:
Point load on a simply supported beam:
Make sure to measure and record the cross section of each
beam and its span. The span should be 455mm. Place the brass, steel or aluminum sample on the
supports of the test machine and ensure that load is at the center of the sample. Place the two
deformation gauges at the center and quarter portions of the beam and ensure it measure the vertical
deformation. Add loads in increments of 200N, up to 1000N and record the deflections at the middle
and quarter portions for each load. Repeat these steps for the other two metal beams.
Point load on cantilever beams:
Measure the cross section of your steel, brass or aluminum sample
and record it. Ensure that the span of the beam is at 250mm. Add increments of 100g loads to the beam,
up until 500g, and record the deformation at half the span and the full span. Repeat steps for the other
metal beams.
Results:
Sample Calculations
Finding the expression for the deflection at any point:
(d
2
y/dx
2
)EI = M(x)
(d
2
y/dx
2
)EI = Px/2 since load P is applied at x/2. Now… (dy/dx)EI = Px
2
/4 + C
solve for C using condition y
1
(L/2) = 0
(0)EI = P(L/2)
2
/4 + C
C = -PL
2
/16
(dy/dx)EI = Px
2
/4 -
PL
2
/16. Integrate again to get… yEI = Px
3
/12 – (PL
2
/16)x +C
1
and use
y(0) = 0 to solve for C
1
0 =
0 – 0 + C
1
so C
1
= 0
yEI = Px
3
/12 – (PL
2
/16)x
Final equation: y(x) = (Px(4x
2
-3L
2
))/48EI
Formulas used:
I = (1/12)bh
3
For beam:
y(x) = (Px(4x
2
-3L
2
))/48EI
and
E = (Px(4x
2
-3L
2
))/48I(y(x))
For cantilever beam:
y(x) = (Px
2
(x-3L))/6EI
and
E = (Px
2
(x-3L))/6I(y(x))
Calculations for Brass Beam Load 1:
Beam: I
brass
= (1/12)(0.01907m)(0.01295m)
3
= 3.451x10
-9
m
4
y(L/2) = ((200N)(0.2275m)(4(0.2275m)
2
- 3(0.455m)
2
))/48(105x10
9
GPa)(3.451x10
-9
m
4
) =
-1.08mm (Negative sign since deflection is downwards, mark as positive displacement in data)
E = ((200N)(0.2275m)(4(0.2275m)
2
- 3(0.455m)
2
))/48(3.451x10
-9
m
4
)(0.00187m) = 60.819GPa
Table for brass x=L/2:
Load (N)
200
400
603
799
999
Theoretical(mm)
1.083
2.167
3.266
4.327
5.410
Experimental(mm
)
1.87
3.54
5.25
6.88
8.58
Table 1: Experimental and Theoretical Values of Deflection for Brass Beam at L/2
Table for brass x=L/4:
Load (N)
200
400
603
799
999
Theoretical(mm)
0.745
1.489
2.245
2.975
3.720
Experimental(mm
)
1.14
2.30
3.48
4.62
5.80
Table 2: Experimental and Theoretical Values of Deflection for Brass Beam at L/4
E
ave brass
=66.191GPa
vs
105GPa
I
steel
= 3.472x10
-9
m
4
Tables for steel x=L/2:
Load(N)
205
399
597
800
1008
Theoretical(mm)
0.579
1.128
1.687
2.261
2.849
Experimental(mm
)
0.66
1.29
1.93
2.60
3.29
Table 3: Experimental and Theoretical Values of Deflection for Steel Beam at L/2
Table for steel x=L/4:
Load(N)
205
399
597
800
1008
Theoretical(mm)
0.398
0.775
1.160
1.554
1.958
Experimental(mm
)
0.50
0.96
1.42
1.89
2.37
Table 4: Experimental and Theoretical Values of Deflection for Steel Beam at L/4
E
ave steel
= 168.625GPa
vs
200GPa
I
aluminum
= 3.438x10
-9
m
4
Table for Aluminum x=L/2:
Load(N)
201
400
599
800
1000
Theoretical(mm)
1.639
3.262
4.884
6.523
8.154
Experimental(mm
)
1.67
3.47
5.29
7.11
8.92
Table 5: Experimental and Theoretical Values of Deflection for Aluminum at L/2
Table for Aluminum x=L/4:
Load(N)
201
400
599
800
1000
Theoretical(mm)
1.127
2.242
3.358
4.485
5.606
Experimental(mm
)
1.43
2.69
3.96
5.23
6.49
Table 6: Experimental and Theoretical Values of Deflection for Aluminum at L/4
E
ave aluminum
= 55.546GPa
vs
70GPa
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
0
1
2
3
4
5
6
7
8
9
10
0
200
400
600
800
1000
1200
Theoretical vs Experimental Deflection of Beams at L/2
Theoretical Brass
Linear (Theoretical Brass)
Experimental Brass
Linear (Experimental Brass)
Theoretical Steel
Linear (Theoretical Steel)
Experimental Steel
Linear (Experimental Steel)
Theoretical Aluminum
Linear (Theoretical Aluminum)
Experimental Aluminum
Linear (Experimental Aluminum)
Deflection (mm)
Load (N)
Figure 1: Theoretical vs Experimental Deflection of the Beams at L/2
Calculations for Cantilever Brass Beam Load 1:
I
brass
= (1/12)(0.0194m)(0.00323m)
3
= 54.479x10
-12
m
4
y(L) = ((119.81g x 9.81m/s
2
)(0.250m)
2
(0.250m-3(0.250m)))/6(105x10
9
Pa)(54.479x10
-12
m
4
) =
-1.070mm
(Negative sign since deflection is downwards, mark as positive displacement in data)
E = ((0.11981kg x 9.81m/s
2
)(0.250m)
2
(0.250m-3(0.250m)))/6(54.479x10
-12
m
4
)(0.00127m) =
88.477GPa
Brass x=L/2:
Load(g)
119.81
218.76
319.58
422.89
521.96
Theoretical(mm)
0.334
0.611
0.892
1.180
1.457
Experimental(mm
)
0.44
0.77
1.17
1.59
1.95
Table 7: Experimental and Theoretical Values of Deflection for Cantilever Brass at L/2
Brass x=L:
Load(g)
119.81
218.76
319.58
422.89
521.96
Theoretical(mm)
1.070
1.954
2.854
3.777
4.662
Experimental(mm
)
1.27
2.33
3.53
4.77
5.91
Table 8: Experimental and Theoretical Values of Deflection for Cantilever Brass at L
E
ave brass
= 82.694GPa
vs
105GPa
I
steel
= 81.065x10
-12
m
4
Steel x=L/2:
Load(g)
119.21
219.21
320.51
421.63
523.41
Theoretical(mm)
0.117
0.216
0.316
0.415
0.515
Experimental(mm
)
0.21
0.38
0.57
0.76
0.97
Table 9: Experimental and Theoretical Values of Deflection for Cantilever Steel at L/2
Steel x=L:
Load(g)
119.21
219.21
320.51
421.63
523.41
Theoretical(mm)
0.376
0.691
1.010
1.329
1.649
Experimental(mm
)
0.65
1.16
1.72
2.32
2.89
Table 10: Experimental and Theoretical Values of Deflection for Cantilever Steel at L
E
ave steel
= 113.258GPa
vs
200GPa
I
aluminum
= 68.346x10
-12
m
4
Aluminum x=L/2:
Load(g)
119.61
218.51
319.21
422.41
521.31
Theoretical(mm)
0.399
0.729
1.065
1.410
1.740
Experimental(mm
)
0.48
0.91
1.35
1.79
2.23
Table 11: Experimental and Theoretical Values of Deflection for Cantilever Aluminum at L/2
Aluminum x=L:
Load(g)
119.61
218.51
319.21
422.41
521.31
Theoretical(mm)
1.277
2.334
3.409
4.511
5.567
Experimental(mm
)
1.47
2.79
4.13
5.50
6.83
Table 12: Experimental and Theoretical Values of Deflection for Cantilever Aluminum at L
E
ave aluminum
= 57.098GPa
vs
70GPa
0
1
2
3
0
100
200
300
400
500
600
Theoretical vs Experimental Deflection of Cantilever Beams at L/2
Cantilever Brass Theoretical
Linear (Cantilever Brass Theoretical)
Cantilever Brass Experimental
Linear (Cantilever Brass Experimental)
Cantilever Steel Theoretical
Linear (Cantilever Steel Theoretical)
Cantilever Steel Experimental
Linear (Cantilever Steel Experimental)
Cantilever Aluminum Theoretical
Linear (Cantilever Aluminum Theoretical)
Cantilever Aluminum Experimental
Linear (Cantilever Aluminum Experimen-
tal)
Deflection (mm)
Load (g)
Figure 2: Theoretical vs Experimental Deflection of the Cantilever Beams at L/2
Discussion and Conclusion:
If we compare the theoretical and experimental values of
deflection, we can see that though they are within the same orders of magnitude, they are not
comparable. This can be seen when we look at the error percentage. At its lowest, the error percentage
is 15.11% for the cantilever aluminum specimen and at its highest, and error value of 75.26% is
calculated for the cantilever steel beam. Considering that error percentage is expected to be between
5%-10%, it can be assumed that the specimen used were either compromised or alloys with different
moduli of elasticity. Since the specimen used for the experiment had already been for multiple other
experiments, it was observed that several of the beams were already subject to some plastic
deformation. It should also be noted that uniform, rectangular beams were used as specimen since
different configurations are more resistant to deflection. An example of this would be “I” beams which
are more resistant to bending and therefore buckling[ CITATION Mov22 \l 4105 ]. These
configurations are more resistant due to the flanges resisting the bending moment along with the rest of
the beam [ CITATION Mov22 \l 4105 ].
The experimental values for aluminum were the closest to its theoretical results, but this can be
due to several different reasons. It is possible that the aluminum specimen were the purest metals out of
the three, but it is also possible that the aluminum specimen had the least plastic deformation. This is
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
not an unreasonable assumption as aluminum has the lowest value for its modulus of elasticity,
meaning that it can be subject to more elastic deformation before reaching plastic deformation.
For steel and brass, it was observed that the specimen had already undergone some plastic
deformation before the experiment. This is most likely the cause for the large error percentage, but
possibly not the only cause. Both brass and steel are alloys that can have variable percentages of
different metals in them. This causes fluctuations in their respective moduli of elasticity. The
information regarding these specimens’ compositions were unavailable to us, but it is likely that the
low moduli of elasticity’s are due to these alloys having a larger part of softer metals.
Human error while using the machines and taking the measurements is always a possibility, but
for this experiment it can be considered as negligible as the magnitudes of our calculated values are
much higher than those of our measured values.
In conclusion, even if the error percentage is so high, we know the experiment was carried out
properly as our graphing shows very clear linear relations between the theoretical and experimental
values. This experiment successfully showed that we could accurately calculate the modulus of
elasticity using its relation to deflection.
Bibliography
[1] C. U. E. a. C. Science, ENGR 244 Mechanics of Materials Lab Manual, Montreal: Concordia University.
[2] Designing Buildings, "Deflection," Designing Buildings, 11 01 2022. [Online]. Available:
https://www.designingbuildings.co.uk/wiki/Deflection. [Accessed 13 03 2022].
[3] Movie Cultists, "Why beam deflection is important," MovieCultists, [Online]. Available:
https://moviecultists.com/why-beam-deflection-is-important. [Accessed 13 03 2022].
Related Documents
Related Questions
Can someone please help me to solve the following question using the FACTOR LABEL METHOD OF UNIT CONVERSION. Showing all work and formulas please and thank you!!
arrow_forward
draw an ashby chart for a knife
By using this instructured:
arrow_forward
Problem 1: (50 points)
In Avengers 2, Captain America's improved
shield is made from unobtainium, a new
material that will soon be available in a store
near you. Unobtainium has the normal stress-
strain diagram shown. The proportional limit,
the elastic limit and the yield point are
identical in this material.
o [MPa]
2001
175
150
125
100
Fig.1 Normal stress-strain
75
50
25
0
0. 0.05 0.1 0.15 0.2 0.25
Unobtainium has a Poisson's ratio of 0.3.
a. [6pts] Identify the yield point oy, the ultimate stress ou and the fracture
stress of. Include units.
c. [25pts] A bar of unobtainium has a length of
1.5 m, a width of 100mm and a height of 50
mm, as shown. The cross-sectional area is
50mm 100mm = 5 x 10-³m². The bar is
subjected to an axial force of 375 kN.
Find the normal stress in a cross-section of the bar.
b. [5pts] How do you find Young's modulus from this graph? Find its value with units.
whesta fins dus al hoss
P = 375 KN
1.5 m
0.3
50 mm
100 mm
& [mm/mm]
Is the material within its…
arrow_forward
Constants | Periodic Table
Part A
A 71 kg man's arm, including the hand, can be modeled
as a 79-cm-long uniform cylinder with a mass of 3.5 kg
In raising both his arms, from hanging down to straight up, by how much does he raise his center of gravity?
Express your answer with the appropriate units.
?
h =
Value
Units
Submit
Request Answer
arrow_forward
O Week 2- 20527 22110L
x a MasteringEngineering Mastering x
M Inbox (10,309) - usmikail@gmail x
a Product Detail Page
a Central Service Technical Manual x
O 21) YouTube
i session.masteringengineering.com/myct/itemView?assignmentProblemiD=12443395&offset=next
KHW 1
Problem 12.3
6 of 16
I Review
A particle travels along a straight line with a velocity
v = (12 – 3t2) m/s, where t is in seconds. When
t = 1 s, the particle is located 10 m to the left of the
origin.
Express your answer to three significant figures and include the appropriate units.
As =
437
Submit
Previous Answers Request Answer
X Incorrect; Try Again; 4 attempts remaining
Part C
Determine the distance the particle travels during the time period given in previous part.
Express your answer to three significant figures and include the appropriate units.
?
ST =
Value
Units
arrow_forward
Which of these statements are correct?
arrow_forward
Needs Complete typed solution with 100 % accuracy.
arrow_forward
Units: mmgs
Mass = 538.9662 grams Volume = 199617.1105 cubic millimeters Surface area = 32076.7718 square millimeters
Material: 6061 Alloy
Any direction/guidance on how to make this would be greatly appreciated. Thank you!
arrow_forward
How to get the answer 371lbs.
I need the correct explanation within 30-45 Minutes.
arrow_forward
Hi
The first diagram is the shear force and bending moment which shown on the bottom on the same graph.
The second diagram shows shear force on the top and the bottom shoes the bedding moment.
I would like to ask if you can draw the same graph but different layout with different colours but same figures thank you
arrow_forward
1) The engineering constants for an orthotropic material are found to be
Ej= 27.579 GPa
E2=20.684 GPa
E3=21.374 GPa
V12 = 0.2
G12=41.368 GPa
V23 = 0.4
G23=48.263 GPa
V13 = 0.6
G31=13.790 GPa
Find the stiffness matrix [C] and the compliance matrix [S] for the preceding orthotropic
material.
arrow_forward
As shown in figure a Pepsi can and a plastic Pepsi bottle. Describe how they were manufactured
arrow_forward
For my assigment, I was asked to design a electric motorbike that has a peformance equal to Honda CBR1000 Fireblade which has a petrol engine. A part of the the assignment is to calculate " An estimate of maximum Power your new motor will need to generate to match the Honda’s performance." I can make the assumption, apart from changing the motor, everything else is going to stay the same so the fairing,the rider and etc they're gonna be the same for the two bikes. So can you please tell me how I can calculate that which information would I need ?
arrow_forward
Could you please explain where and what is tight side and also how to find 5-9
arrow_forward
E Machine Design & CAD-II (Engr. x
O Shigley's Mechanical Engineering X
+
O File
C:/SEMMESTER%2306/Machine%20Design%20and%20CAD/Shigley's%20Mechanical%20Engineering%20Design,%2010th%20Edition.pdf
Shigley's Mechanical Engineering Design, Tenth Edition
737 / 1105
200%
+
3-17
Shaft a in the figure rotates at 600 rev/min in the direction shown. Find the speed and direction
of rotation of shaft d.
207, = 30° RH
736
8T, = 60° RH
13-17
3
737
40T
177, = 30° RH
20T
60T
Shigley's Mechanic.pdf
Show all
W
7:22 PM
5/16/2021
II
国
arrow_forward
vio-Ou X
28523020?X-Blackboard-Expi
2/3
Ö
150% + A
=651gq|4cRVazVmd0Uq2ZHfoUuaYHSBHXymA39vD1F0g%3D&X-Blackboard-
2) The ballistic pendulum is a simple device to measure projectile speed v by
observing the maximum angle to which the box of sand with the
embedded projectile swings. Calculate the pre-impact speed of the 40-
gram projectile if the 16-kg box of sand swings up to a maximum angle of
0=23° after impact.
2 m
120
arrow_forward
F
Xp
Side View
To
Xg
gauge
Top View
The gauged beam of length 0.33 (m) and cross section base 0.013 (m) and height
0.017 (m) shown is made of aluminum with a modulus of elasticity of 69 (GPa) and an
ultimate strength of 300 (MPa). A strain of 250 (microstrain) is measured from the
gauge which is 0.18 (m) from the wall in the x direction. What is the magnitude of
load at of the end of the beam in units of Newtons?
arrow_forward
Number 3
arrow_forward
Helpe me quickly please
arrow_forward
3. Please help me solve all parts to this mech. engineering question
arrow_forward
☐ please work out the stiffness matrix for
transversely isotropic elastic solids
• Two symmetric planes and an isotropic plane
Oxy is the
isotropic plane
Step 1: Rotate Oxy along z by 90
degrees
Step 2: Rotate Oxy along z by p
degrees
ZER, WE
5 Independent
Elastic Constants
arrow_forward
The purpose of this problem is to
show the relationship between material constants typically used in engineering practice. This
is useful because you may often have access to measurements of or tabulated values of some
constants (e.g., Young's modulus and Poisson's ratio) but need another constant (e.g., shear
modulus) for a calculation. Use the expression Cijkl = µ(dildjk + dikdjl) + Ad¿jdkl to derive
the following:
(a) Young's modulus, E = µ(3X+2µ)/(X+μ), from the definition 11 = Ee11 in a unconfined
(022 = 0,033 = 0) uniaxial tension test.
(b) Poisson's ratio, v = \/(2(X + μ)), from the definition v = €22/11 in the same test as
in (a).
(c) Shear modulus, μ = G = 012/(2€12) = E/(2(1 + v)).
Use the results to show that C can also be written Cijkl
Ev/((1+v)(1 − 2v))dij§kl.
=
E/(2(1 + v))(duðjk + dikdjl) +
arrow_forward
4. Need help please. Answer with 3 sig figs only.
arrow_forward
SEE MORE QUESTIONS
Recommended textbooks for you

Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press

Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON

Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education

Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY

Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning

Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Related Questions
- Can someone please help me to solve the following question using the FACTOR LABEL METHOD OF UNIT CONVERSION. Showing all work and formulas please and thank you!!arrow_forwarddraw an ashby chart for a knife By using this instructured:arrow_forwardProblem 1: (50 points) In Avengers 2, Captain America's improved shield is made from unobtainium, a new material that will soon be available in a store near you. Unobtainium has the normal stress- strain diagram shown. The proportional limit, the elastic limit and the yield point are identical in this material. o [MPa] 2001 175 150 125 100 Fig.1 Normal stress-strain 75 50 25 0 0. 0.05 0.1 0.15 0.2 0.25 Unobtainium has a Poisson's ratio of 0.3. a. [6pts] Identify the yield point oy, the ultimate stress ou and the fracture stress of. Include units. c. [25pts] A bar of unobtainium has a length of 1.5 m, a width of 100mm and a height of 50 mm, as shown. The cross-sectional area is 50mm 100mm = 5 x 10-³m². The bar is subjected to an axial force of 375 kN. Find the normal stress in a cross-section of the bar. b. [5pts] How do you find Young's modulus from this graph? Find its value with units. whesta fins dus al hoss P = 375 KN 1.5 m 0.3 50 mm 100 mm & [mm/mm] Is the material within its…arrow_forward
- Constants | Periodic Table Part A A 71 kg man's arm, including the hand, can be modeled as a 79-cm-long uniform cylinder with a mass of 3.5 kg In raising both his arms, from hanging down to straight up, by how much does he raise his center of gravity? Express your answer with the appropriate units. ? h = Value Units Submit Request Answerarrow_forwardO Week 2- 20527 22110L x a MasteringEngineering Mastering x M Inbox (10,309) - usmikail@gmail x a Product Detail Page a Central Service Technical Manual x O 21) YouTube i session.masteringengineering.com/myct/itemView?assignmentProblemiD=12443395&offset=next KHW 1 Problem 12.3 6 of 16 I Review A particle travels along a straight line with a velocity v = (12 – 3t2) m/s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Express your answer to three significant figures and include the appropriate units. As = 437 Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Part C Determine the distance the particle travels during the time period given in previous part. Express your answer to three significant figures and include the appropriate units. ? ST = Value Unitsarrow_forwardWhich of these statements are correct?arrow_forward
- Needs Complete typed solution with 100 % accuracy.arrow_forwardUnits: mmgs Mass = 538.9662 grams Volume = 199617.1105 cubic millimeters Surface area = 32076.7718 square millimeters Material: 6061 Alloy Any direction/guidance on how to make this would be greatly appreciated. Thank you!arrow_forwardHow to get the answer 371lbs. I need the correct explanation within 30-45 Minutes.arrow_forward
- Hi The first diagram is the shear force and bending moment which shown on the bottom on the same graph. The second diagram shows shear force on the top and the bottom shoes the bedding moment. I would like to ask if you can draw the same graph but different layout with different colours but same figures thank youarrow_forward1) The engineering constants for an orthotropic material are found to be Ej= 27.579 GPa E2=20.684 GPa E3=21.374 GPa V12 = 0.2 G12=41.368 GPa V23 = 0.4 G23=48.263 GPa V13 = 0.6 G31=13.790 GPa Find the stiffness matrix [C] and the compliance matrix [S] for the preceding orthotropic material.arrow_forwardAs shown in figure a Pepsi can and a plastic Pepsi bottle. Describe how they were manufacturedarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY

Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press

Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON

Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education

Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY

Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning

Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY