Please help me with question. I need only part C (the graph question) with detailed answers please. Thanks! E8.3 A cheap column that must not buckle orcrush (Figure E.15) The best choice of materialfor a light, strong column depends on its aspectratio: the ratio of its height H to its diameter D.This is because short, fat columns fail by crushing;tall, slender columns buckle instead.Force FDerive two performance equations for the materialcost of a column of a solid circular section and ofHeightCross-sectionarea nD2/4Ha specified height H, designed to support a load F(i.e., large compared to its self-load), one usingthe constraints that the column must not crush,the other that it must not buckle. The followingtable summarizes the needs.FIGURE E.15FunctionColumnConstraintsMust not fail by compressive crushingMust not buckleHeight H and compressive load F specifiedObjectiveMinimize material cost CFree variablesDiameter DChoice of materiala. Proceed as follows1. Write down an expression for the material cost of the column-itsmass times its cost per unit mass Cm.2. Express the two constraints as equations, and use them to substitutefor the free variable to find the cost of the column that will justsupport the load without failing by either mechanism.3. Identify the material indices M1 and M2 that enter the two equations forthe mass, showing that they are[Cmp]E/2(CmPM, = ()and M2where Cm is the material cost per kg, p the material density, o, itscrushing strength, and E its modulus. c. Figure E.16 shows a material chart with the two indices as axes.Identify and plot coupling lines for selecting materials for a columnwith F= 10° N and H=3 m (the same conditions as previously noted),and for a second column with F= 10³ N and H=20 m.Tungstenalloys,TitaniumalloysGFRP1000CFRPSMCAluminaMg alloysSilica glass100Al alloysLow alloy steelCopperCarbonsteel10Cast ironBrickStoneWoodsCementConcrete110100100010,000M, =[density] * [price]/[compressive strength] ($/m³V(MPa)M2=[density] * [price]/[Young's modulus]0.5)($/m/(MP20.5)

Answered: Image /qna-images/answer/1ca90267-2e1e-4c02-b7eb-3cb7670e6fa2.jpg

c. Figure E.16 shows a material chart with the two indices as axes. Identify and plot coupling lines for selecting materials for a column with F= 10° N and H=3 m (the same conditions as previously noted), and for a second column with F= 10³ N and H=20 m. Tungsten alloys, Titanium alloys GFRP 1000 CFRP SMC Alumina Mg alloys Silica glass 100 Al alloys Low alloy steel Copper Carbon steel 10 Cast iron Brick Stone Woods Cement Concrete 1 10 100 1000 10,000 M, =[density] * [price]/[compressive strength] ($/m³/(MPa) M2=[density] * [price]/([Young's modulus]0.5) (S/m/(MPA05)

c. Figure E.16 shows a material chart with the two indices as axes. Identify and plot coupling lines for selecting materials for a column with F= 10° N and H=3 m (the same conditions as previously noted), and for a second column with F= 10³ N and H=20 m. Tungsten alloys, Titanium alloys GFRP 1000 CFRP SMC Alumina Mg alloys Silica glass 100 Al alloys Low alloy steel Copper Carbon steel 10 Cast iron Brick Stone Woods Cement Concrete 1 10 100 1000 10,000 M, =[density] * [price]/[compressive strength] ($/m³/(MPa) M2=[density] * [price]/([Young's modulus]0.5) (S/m/(MPA05)

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Project Management Techniques

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Please help me with question. I need only part C (the graph question) with detailed answers please. Thanks!

E8.3 A cheap column that must not buckle or
crush (Figure E.15) The best choice of material
for a light, strong column depends on its aspect
ratio: the ratio of its height H to its diameter D.
This is because short, fat columns fail by crushing;
tall, slender columns buckle instead.
Force F
Derive two performance equations for the material
cost of a column of a solid circular section and of
Height
Cross-section
area nD2/4
H
a specified height H, designed to support a load F
(i.e., large compared to its self-load), one using
the constraints that the column must not crush,
the other that it must not buckle. The following
table summarizes the needs.
FIGURE E.15
Function
Column
Constraints
Must not fail by compressive crushing
Must not buckle
Height H and compressive load F specified
Objective
Minimize material cost C
Free variables
Diameter D
Choice of material
a. Proceed as follows
1. Write down an expression for the material cost of the column-its
mass times its cost per unit mass Cm.
2. Express the two constraints as equations, and use them to substitute
for the free variable to find the cost of the column that will just
support the load without failing by either mechanism.
3. Identify the material indices M1 and M2 that enter the two equations for
the mass, showing that they are
[Cmp]
E/2
(CmP
M, = ()
and M2
where Cm is the material cost per kg, p the material density, o, its
crushing strength, and E its modulus.

c. Figure E.16 shows a material chart with the two indices as axes.
Identify and plot coupling lines for selecting materials for a column
with F= 10° N and H=3 m (the same conditions as previously noted),
and for a second column with F= 10³ N and H=20 m.
Tungsten
alloys,
Titanium
alloys
GFRP
1000
CFRP
SMC
Alumina
Mg alloys
Silica glass
100
Al alloys
Low alloy steel
Copper
Carbon
steel
10
Cast iron
Brick
Stone
Woods
Cement
Concrete
1
10
100
1000
10,000
M, =[density] * [price]/[compressive strength] ($/m³V(MPa)
M2=[density] * [price]/[Young's modulus]0.5)
($/m/(MP20.5)

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