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Mechanical Behaviour
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Ain Shams University
Faculty of Engineering
New Program
8th assignment
Presented to: Dr. Nahed Abd El-Salam
Presented by: Ahmed Hassan Ibrahim
Mostafa sherif Ibrahim
S.MANF
7.1 An engineering component is made of the silicon nitride (Si3N4) ceramic of Table 3.10. most severely stressed point is subjected to the following state of stress: Οx = 125, Οy = 15, πxy = -25, and Οz = πxz = πyz = 0 MPa. Determine the safely factor against fracture.
π1 , π2 =ππ₯ + ππ¦
πΒ± β(
ππ₯ β ππ¦
π)
2
+ ππππ =
125 + 15
2Β± β(
πππ β ππ
π)
π
+ (βππ)π = 70 Β± 60.41522
π1 = 130.4152299 πππ , π2 = 9.584770132 πππ
ππ = ππ΄π(|π1|, |π2|, |π3|) = ππ΄π(130.4152299, 9.584770132,0) = 130.4152299πππ
π =ππ’
ππ=
450 πππ
130.4152299 πππ= 3.450517
Ceramic Melting
Temp. Density Elastic
Modulus Typical Strength Uses
Tm π E ππ’, MPa (ksi)
tension compression Silicon nitride.
Si3N4 1900 3.18 310 450 3450 fibers for
composites (hot pressed) (3450) (199) (45) (65) (500) cutting tool
inserts 7.2 In an engineering component made of gray cast iron, the most severely stressed
poi subjected to the following state of stress: Οx = 50, Οy = 80, πxy = 20 and Οz = πxz = πyz = 0 MPa, Determine the safety factor against fracture. The material has a tensile strength 214 MPa and a compressive strength of 770 MPa.
π1 , π2 =ππ₯ + ππ¦
πΒ± β(
ππ₯ β ππ¦
π)
2
+ ππππ =
50 + 80
2Β± β(
ππ β ππ
π)
π
+ (ππ)π = 65 Β± 25
π1 = 90 πππ , π2 = 40πππ
ππ = ππ΄π(|π1|, |π2|, |π3|) = ππ΄π(90, 25,0) = 90 πππ
Safety factor against tension fracture :
π =ππ’
ππ=
214 πππ
90 πππ= 2.377778
Safety factor against compression fracture :
π =ππ’
ππ=
770 πππ
90 πππ= 8.5556
7.5 A pipe 10 m long has closed ends, a wall thickness of 5 mm. and an inner diameter of 3 m, and it is filled with a gas at a pressure of 2MPa. Neglecting any localized effects of the end closure, what is the safety factor against yielding if the material is 18 Ni maraging steel (250 grade)? Employ (a) the maximum shear stress criterion, and (b) the octahedral shear stress criterion.
Material Elastic
Modulus E
0.2% yield Strength
Οo
Ultimate Strength
Οu
Elongation
100πΊf
Reduction in area %RA
18 Ni maraging steel 250
GPa (103ksi)
MPa (ksi)
MPa (ksi) % %
186 (27)
1791 (260)
1860 (270) 8 56
πβ =πππ
π‘=
( 2 πππ )(3/2 π)
0.005 π= 600 πππ, πβ =
πππ
2π‘=
( 2 πππ )(3/2π)
2(0.005 π)= 300 πππ
ππ = β2 πππ πππ πππ , , ππ’π‘π πππ ππ = 0
inside
ππ = ππ΄π(|π1 β π2|, |π2 β π3|, |π3 β π1|) = ππ΄π(|600 β 300|, |300 + 2|, |β2 β 600|) = 602 πππ
ππ =ππ¦
ππ =
1791 πππ
602 πππ= 2.975
outside
ππ = ππ΄π(|π1 β π2|, |π2 β π3|, |π3 β π1|) = ππ΄π(|600 β 300|, |300 + 0|, |0 β 600|) = 600 πππ
ππ =ππ¦
ππ =
1791 πππ
600 πππ= 2.985
ππ» =1
β2β(π1 β π2)2 + (π2 β π3)2 + (π3 β π1)2
Inside
ππ» =1
β2β(600 β 300)2 + (300 + 2)2 + (β2 β 600)2 = 521.34 πππ
ππ» =ππ¦
ππ»=
1791 πππ
521.34 πππ= 3.43532
Inside
ππ» =1
β2β(600 β 300)2 + (300 + 0)2 + (0 β 600)2 = 519.615 πππ
ππ» =ππ¦
ππ»=
1791 πππ
519.615 πππ= 3.4467
7.6 In an engineering component made of AISI 1020 steel (as rolled), the most severely stressed point is subjected to the following state of stress: Οx = -100, Οy
= 40, πxy = -50 and Οz = πxz = πyz = 0 MPa. Determine the safely factor against yielding by (a) the maximum shear stress criterion, and (b) the octahedral shear stress criterion.
Material Elastic
Modulus E
0.2% yield Strength
Οo
Ultimate Strength
Οu
Elongation
100πΊf
Reduction in area %RA
AISI 1020 steel
GPa (103ksi)
MPa (ksi)
MPa (ksi) % %
203 (29.4)
260 (37.7)
441 (64) 36 61
π1, π2 =ππ₯ + ππ¦
πΒ± β(
ππ₯ β ππ¦
π)
2
+ ππππ =
β100 + 40
2Β± β(
βπππ β ππ
π)
π
+ (βππ)π = β30 Β± 86.023
π1 = 56.02325 πππ , π2 = β116.0232 πππ
ππ = ππ΄π(|π1 β π2|, |π2 β π3|, |π3 β π1|)= ππ΄π(|56.02325 + 116.0232|, |β116.0232 + 0|, |0β 56.02325|) = 172.04650 πππ
ππ =ππ¦
ππ =
260 πππ
172.0465 πππ= 1.511219
ππ» =1
β2β(ππ₯ β ππ¦)
2+ (ππ¦ β ππ§)
2+ (ππ§ β ππ₯)2 + 6(ππ₯π¦
2 + ππ¦π§2 + ππ§π₯
2)
ππ» =1
β2β(56.02325 + 116.0232 )2 + (β116.0232 β 0)2 + (0 β 56.02325)2 + 6((β50)2 + 0 + 0)
= 151.9868 πππ
ππ» =ππ¦
ππ»=
260 πππ
151.9868 πππ= 1.7106744
7.10 strain are measured on the surface of a part made from AISI 1020 steel as follows: ex = 190 x I0-6 ey = -760 x 10-6, and πΎxy = 300 x 10-6. Assume that no yielding has occurred, and also that no loading is applied directly to the surface,
so that Οz = πxz = πyz = 0. What is the safety factor against yielding?
ππ₯ =πΈ
1 β π£2(ππ₯ + π£ππ¦) =
203 Γ 103 πππ
1 β 0. 2932(0.00019 + 0.293 Γ β0.00076) = β7.257 πππ
ππ¦ =πΈ
1 β π£2(ππ¦ + π£ππ₯) =
203 Γ 103 πππ
1 β 0. 2932(0.00019 + 0.293 Γ β0.00076) = β156.4063 πππ
ππ¦π₯ = πΊπΎπ¦π₯ ==πΈ
2(1 + π£)πΎπ¦π₯ =
203 Γ 103 πππ
2(1 + 0.293)(0.0003) = 23.54988 πππ
π1, π2 =ππ₯ + ππ¦
πΒ± β(
ππ₯ β ππ¦
π)
2
+ ππππ =
β7.257 β 156.4063
2Β± β(
β7.257 + 156.4063
π)
π
+ (23.54988)π
= β80.83165 Β± 78.2047
π1 = β2.626948 πππ , π2 = β159.0363511 πππ
ππ = ππ΄π(|π1 β π2|, |π2 β π3|, |π3 β π1|) = ππ΄π(|β2.626948 β 159.0363511|, |β159.0363511 + 0|, |0 + 2.626948|)
= 161.6632991 πππ
ππ =ππ¦
ππ =
260 πππ
161.6632991 πππ= 1.60828
ππ» =1
β2β(ππ₯ β ππ¦)
2+ (ππ¦ β ππ§)
2+ (ππ§ β ππ₯)2 + 6(ππ₯π¦
2 + ππ¦π§2 + ππ§π₯
2)
ππ» =1
β2β(β7.257 β 156.4063 )2 + (β156.4063 β 0)2 + (0 β 7.257 )2 + 6((23.54988)2 + 0 + 0)
= 165.277651 πππ
ππ» =ππ¦
ππ»=
260 πππ
165.277651 πππ= 1.5731756
7.11 A strain gage rosette, as in Ex. 6.9, is applied to the surface of a component made of 7075-T6 aluminum. Assume that no yielding has occurred, and also that no loading is applied directly to the surface, so that Οz = πxz = πyz = 0. Strains are measured as follows: ex = 1200 x 10-6 , ey = -650 x 10-6, and e45 = 1900 x 10 -6. What is the safety factor against yielding?
Material Elastic
Modulus E
0.2% yield Strength
Οo
Ultimate Strength
Οu
Elongation
100πΊf
Reduction in area %RA
7075-T6 aluminum
GPa (103ksi)
MPa (ksi)
MPa (ksi) % %
71 (10.3)
469 (68)
578 (84) 11 33
Material Poisson ratio
Aluminum 0.345
π45 =ππ₯ + ππ¦
2+
πΎπ₯π¦
2 β πΈππ = 2π45 β ππ₯ β ππ¦ = 2(0.0019) β 0.0012 + 0.00065 = 0.00325
ππ₯ =πΈ
1 β π£2(ππ₯ + π£ππ¦) =
71 Γ 103 πππ
1 β 0. 3452(0.0012 + 0.345 Γ β0.00065) = 78.638 πππ
ππ¦ =πΈ
1 β π£2(ππ¦ + π£ππ₯) =
71 Γ 103 πππ
1 β 0. 3452(β0.00065 + 0.345 Γ 0.0012) = β19.0198 πππ
ππ¦π₯ = πΊπΎπ¦π₯ ==πΈ
2(1 + π£)πΎπ¦π₯ =
71 Γ 103 πππ
2(1 + 0.345)(0.00325) = 85.780669 πππ
π1, π2 =ππ₯ + ππ¦
πΒ± β(
ππ₯ β ππ¦
π)
2
+ ππππ =
78.638 β 19.0198 2
Β± β(78.638 + 19.0198
π)
π
+ (85.780669)π
= 29.8091 Β± 98.7045
π1 = 129.51363 πππ , π2 = β68.8954 πππ
ππ = ππ΄π(|π1 β π2|, |π2 β π3|, |π3 β π1|)
= ππ΄π(|129.51363 + 68.8954 |, |β68.8954 + 0|, |0 + 129.51363|) = 198.40903 πππ
ππ =ππ¦
ππ =
469 πππ
198.40903 πππ= 2.3638
ππ» =1
β2β(ππ₯ β ππ¦)
2+ (ππ¦ β ππ§)
2+ (ππ§ β ππ₯)2 + 6(ππ₯π¦
2 + ππ¦π§2 + ππ§π₯
2)
ππ» =1
β2β(78.638 + 19.0198 )2 + (β19.0198 β 0)2 + (0 β 78.638 )2 + 6(85.7806692 + 0 + 0)
= 173.54087 πππ
ππ» =ππ¦
ππ»=
469 πππ
173.54087 πππ= 2.7025
7.12 A solid circular shaft subjected to pure torsion must be designed to avoid yielding, with a factor X . Find the required diameter as a function of the torque T and the yield strength ππ using (a) the maximum shear stress criterion, and (b) the octahedral shear stress criterion. How much do these two sizes differ?
π =ππ
π½=
ππ2
π½, π½ =
ππ4
32, π =
16π
ππ3
For maximum shear stress criterion
ππ =ππ¦
ππ , ,
ππ¦
ππ = ππ = ππ΄π(|π1 β π2|, |π2 β π3|, |π3 β π1|) = 2π
ππ¦
ππ = 2 (
16π
ππ3 )
π π = (πππΏππ»
π ππ)
ππβ
the octahedral shear stress criterion
ππ» =ππ¦
ππ»
=1
β2β(π1 β π2)2 + (π2 β π3)2 + (π3 β π1)2 =
1
β2β(2π)2 + π2 + π2 = πβ3
ππ¦
ππ»= πβ3 = β3 (
16π
ππ3)
π π― = (ππβππΏπ―
π ππ)
ππβ
π π―
π π=
(ππβππΏπ―
π ππ)
ππβ
(πππΏππ»
π ππ)
ππβ
=ππβπ
ππ= π. ππππ
π π― = π. πππππ π
7.14 A pipe with closed ends has an outer diameter of 80 mm and a wall thickness of 3.0 mm. It is subjected to an internal pressure of 20MPa and a bending moment of 2.0 kN-m. Determine the safety factor against yielding if the material is 707S-T6 aluminum. Employ (a) me maximum shear stress criterion, and (b) the octahedral shear stress criterion
7.19 A circular tube must support an axial load of 60 kN tension and a torque of 1.0 kN-m. It is made of 7075-T6 aluminum and has an inside diameter of 46.0 mm. (a) What is the safety factor against yielding if the wall thickness is 2.5 mm"? (b)For the situation of (a), what adjusted value of thickness with the same inside diameter is required to obtain a safety factor against yielding of 2.0?
ππ₯ =πΉ
π΄=
πΉ
2ππ π‘=
60 kN
2π (0.046π
2 ) (0.0025π)= 166.0747 πππ
ππ₯π¦ =ππ
π½, π½ =
π(ππ4 β ππ
4)
32=
π(0. 046π4 β 0. 0435π4)
32= 8.8 Γ 10β8π4
7.21 A thin-walled tube with closed ends has an inside radius ri = 50 mm and a wall thickens t = 2 mm. It is .subjected to an internal pressure p = 24MPa and a torque T = 8.0kN m. A safety factor against yielding of 2.2 is required. Select a material from Table 4.2 that would be suitable for this application.
πβ =πππ
π‘=
( 24 πππ )(0.05 π)
0.002 π= 600 πππ, πβ =
πππ
2π‘=
( 24 πππ )(0.05π)
2(0.002 π)= 300 πππ
ππ₯π¦ =ππ
π½=
(8 πππ)(0.026π)π
64(0.0524 β 0.0504)
7.22 A piece of a ductile metal is confined on two sides by a rigid die. as shown in Fig. P7.22. A uniform compressive stress Οz is applied to the surface of the metal. Assume that there is no friction against the die, and also that the material behaves in an elastic, perfectly plastic manner with Uniaxial yield strength Οy Derive an equation for the value of Οz necessary to cause yielding in terms of Οy and the elastic constants of the material. Is the value of Οz that causes yielding affected significantly by Poisson's ratio? Employ (a) the maximum shear stress criterion, and (b) the octahedral shear stress criterion, (c) What stress Οz is expected to cause yielding if the material is A1SI 1020 steel (as rolled)?
7.23 Repeat Prob. 7.22(a). (b), and (c) for the case where the die confines the material on all four sidesβthat is, in both the x- and y-directions, as shown in Fig. P7.23.
7.26 A block of AISI 1020 steel (as rolled) is subjected to a stress Οz = -l20MPa, along with a shear stress πxy. as shown in Pig. P7.26. (a) What is the largest value of = πxy that can be applied if the safety factor against yield must be 2.0? (b)Is there a large effect of Οz on the πxy required to cause yielding? Briefly discus effect of Οz as to whether the effect is large, small, or absent, and explain why.
π1, π2 =ππ₯+ππ¦
πΒ± β(
ππ₯βππ¦
π)
2
+ ππππ = Β±πππ
, π3 = ππ§ = β120 πππ
Based on maximum shear stress criterion
ππ = ππ΄π(|π1 β π2|, |π2 β π3|, |π3 β π1|) = ππ΄π(|πππ + πππ|, |βπππ + 120|, |β120 β πππ|)
ππ = 2πππ ππ = πππ + πππ
Material Elastic
Modulus E
0.2% yield Strength
Οo
Ultimate Strength
Οu
Elongation
100πΊf
Reduction in area %RA
AISI 1020 steel
GPa (103ksi)
MPa (ksi)
MPa (ksi) % %
203 (29.4) 260
(37.7) 441 (64) 36 61
ππ =ππ¦
ππ , ππ =
ππ¦
ππ =
260
2= 130 πππ
ππ ππ = πππ + πππ, ππ = 130 πππ , πππ = 130 β 120 = 10 πππ
If ππ = 2πππ , , ππ = 130 πππ , πππ =130 πππ
2= 65 πππ
- in the first case πππ = 65 πππ the result is vary with Οz
Based on octahedral shear stress criterion
ππ» =1
β2β(π1 β π2)2 + (π2 β π3)2 + (π3 β π1)2 =
1
β2β(πππ+πππ)
2+ (β πππ + 120)
2+ (β120 β πππ)
2
=1
β2β4πππ
π + ππππ β ππππππ + πππππ + πππππ + ππππππ + πππ
π =1
β2β6πππ
π + π(ππππ) = β3ππππ + (ππππ)
ππ» =ππ¦
ππ»=
260
2= β3πππ
π + (ππππ) , 2602 = 4 (3ππππ + (ππππ)) , πππ = 28.86 πππ