It is necessary to select a ceramic material to be stressed using a three-point loading scheme as shown in the figure. The specimen must have a circular cross section, a radius of 3.8 mm, and must not experience fracture or a deflection of more than 0.021 mm at its center when a load of 445 N is applied. If the distance between support points is 50.8 mm, which of the materials in the accompanying Table are candidates? The magnitude of the center point deflection may be computed using the equation: Ay=FL³/(48 EI) where E is the modulus of elasticity, and I is the cross-sectional moment of inertia. Support k½ F Circular σ = stress = c where M = maximum bending moment c = distance from center of specimen to outer fibers M 드 Rectangular 뜸 를 FL Possible cross sections I = moment of inertia of cross section F = applied load R 5+ D 3FL 2bd2 bk FL =R³ R d Rectangular Circular Tabulation of Flexural Strength (Modulus of Rupture) and Modulus of Elasticity of Ten Common Ceramic Materials: Material Silicon nitride (Si3N4) Zirconia (ZrO₂) Silicon carbide (SIC) Aluminum oxide (Al₂O3) Glass-ceramic (Pyroceram) Mullite (3Al2O3-2SiO₂) Spinel (MgAl₂O4) Magnesium oxide (MgO) Fused silica (SiO₂) Soda-lime glass Flexural Strength MPa 250-1000 800-1500 100-820 275-700 247 185 110-245 105b 110 69 "Partially stabilized with 3 mol % Y₂03. b Sintered and containing approximately 5% porosity. ksi 35-145 115-215 15-120 40-100 36 27 16-35.5 15b 16 10 Modulus of Elasticity GPa 304 205 345 393 120 145 260 225 73 69 10º psi 44 30 50 57 17 21 38 33 11 10
It is necessary to select a ceramic material to be stressed using a three-point loading scheme as shown in the figure. The specimen must have a circular cross section, a radius of 3.8 mm, and must not experience fracture or a deflection of more than 0.021 mm at its center when a load of 445 N is applied. If the distance between support points is 50.8 mm, which of the materials in the accompanying Table are candidates? The magnitude of the center point deflection may be computed using the equation: Ay=FL³/(48 EI) where E is the modulus of elasticity, and I is the cross-sectional moment of inertia. Support k½ F Circular σ = stress = c where M = maximum bending moment c = distance from center of specimen to outer fibers M 드 Rectangular 뜸 를 FL Possible cross sections I = moment of inertia of cross section F = applied load R 5+ D 3FL 2bd2 bk FL =R³ R d Rectangular Circular Tabulation of Flexural Strength (Modulus of Rupture) and Modulus of Elasticity of Ten Common Ceramic Materials: Material Silicon nitride (Si3N4) Zirconia (ZrO₂) Silicon carbide (SIC) Aluminum oxide (Al₂O3) Glass-ceramic (Pyroceram) Mullite (3Al2O3-2SiO₂) Spinel (MgAl₂O4) Magnesium oxide (MgO) Fused silica (SiO₂) Soda-lime glass Flexural Strength MPa 250-1000 800-1500 100-820 275-700 247 185 110-245 105b 110 69 "Partially stabilized with 3 mol % Y₂03. b Sintered and containing approximately 5% porosity. ksi 35-145 115-215 15-120 40-100 36 27 16-35.5 15b 16 10 Modulus of Elasticity GPa 304 205 345 393 120 145 260 225 73 69 10º psi 44 30 50 57 17 21 38 33 11 10
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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Question
It is necessary to select a ceramic material to be stressed using a three-point loading scheme as
shown in the figure. The specimen must have a circular cross section, a radius of 3.8 mm, and
must not experience fracture or a deflection of more than 0.021 mm at its center when a load of
445 N is applied. If the distance between support points is 50.8 mm, which of the materials in the
accompanying Table are candidates? The magnitude of the center point deflection may be
computed using the equation:
Δy =FL3/(48 EI)
where E is the modulus of elasticity, and I is the cross-sectional moment of inertia.
shown in the figure. The specimen must have a circular cross section, a radius of 3.8 mm, and
must not experience fracture or a deflection of more than 0.021 mm at its center when a load of
445 N is applied. If the distance between support points is 50.8 mm, which of the materials in the
accompanying Table are candidates? The magnitude of the center point deflection may be
computed using the equation:
Δy =FL3/(48 EI)
where E is the modulus of elasticity, and I is the cross-sectional moment of inertia.
Transcribed Image Text:It is necessary to select a ceramic material to be stressed using a three-point loading scheme as
shown in the figure. The specimen must have a circular cross section, a radius of 3.8 mm, and
must not experience fracture or a deflection of more than 0.021 mm at its center when a load of
445 N is applied. If the distance between support points is 50.8 mm, which of the materials in the
accompanying Table are candidates? The magnitude of the center point deflection may be
computed using the equation:
Ay=FL³²/(48 EI)
where E is the modulus of elasticity, and I is the cross-sectional moment of inertia.
Support
σ = stress =
Circular
where M = maximum bending moment
Rectangular
Mc
c = distance from center of specimen
to outer fibers
I = moment of inertia of cross section
F = applied load
M 4
FL
4
C
FL d
2
Possible cross sections
R
喔喔~
σ
3FL
2bd²
FL
TR³
R
Rectangular
Circular
Tabulation of Flexural Strength (Modulus of Rupture) and Modulus of Elasticity of Ten
Common Ceramic Materials:
Material
Silicon nitride (Si3N4)
Zirconia (ZrO₂)
Silicon carbide (SIC)
Aluminum oxide (Al₂O3)
Glass-ceramic (Pyroceram)
Mullite (3Al2O3-2SiO2)
Spinel (MgAl₂O4)
Magnesium oxide (MgO)
Fused silica (SiO₂)
Soda-lime glass
Flexural Strength
MPa
250-1000
800-1500
100-820
275-700
247
185
110-245
105b
110
69
a Partially stabilized with 3 mol% Y₂O3.
b Sintered and containing approximately 5% porosity.
ksi
35-145
115-215
15-120
40-100
36
27
16-35.5
15b
16
10
Modulus of
Elasticity
GPa
304
205
345
393
120
145
260
225
73
69
106 psi
44
30
50
57
17
21
38
33
11
10
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