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: σx = 125, σy = 15, 𝜏xy = -25, and σz = 𝜏xz = 𝜏yz = 0 MPa. Determine the safely factor against fracture.

𝝈1 , 𝝈2 =𝝈𝑥 + 𝝈𝑦

𝟐± √(

𝝈𝑥 − 𝝈𝑦

𝟐)

2

+ 𝝉𝒙𝒚𝟐 =

125 + 15

2± √(

𝟏𝟐𝟓 − 𝟏𝟓

𝟐)

𝟐

+ (−𝟐𝟓)𝟐 = 70 ± 60.41522

𝝈1 = 130.4152299 𝑀𝑃𝑎 , 𝝈2 = 9.584770132 𝑀𝑃𝑎

𝜎𝑁 = 𝑀𝐴𝑋(|𝝈1|, |𝝈2|, |𝝈3|) = 𝑀𝐴𝑋(130.4152299, 9.584770132,0) = 130.4152299𝑀𝑃𝑎

𝑋 =𝝈𝑢

𝜎𝑁=

450 𝑀𝑃𝑎

130.4152299 𝑀𝑃𝑎= 3.450517

Ceramic Melting

Temp. Density Elastic

Modulus Typical Strength Uses

Tm 𝝆 E 𝝈𝑢, 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: σx = 50, σy = 80, 𝜏xy = 20 and σz = 𝜏xz = 𝜏yz = 0 MPa, Determine the safety factor against fracture. The material has a tensile strength 214 MPa and a compressive strength of 770 MPa.

𝝈1 , 𝝈2 =𝝈𝑥 + 𝝈𝑦

𝟐± √(

𝝈𝑥 − 𝝈𝑦

𝟐)

2

+ 𝝉𝒙𝒚𝟐 =

50 + 80

2± √(

𝟓𝟎 − 𝟖𝟎

𝟐)

𝟐

+ (𝟐𝟎)𝟐 = 65 ± 25

𝝈1 = 90 𝑀𝑃𝑎 , 𝝈2 = 40𝑀𝑃𝑎

𝜎𝑁 = 𝑀𝐴𝑋(|𝝈1|, |𝝈2|, |𝝈3|) = 𝑀𝐴𝑋(90, 25,0) = 90 𝑀𝑃𝑎

Safety factor against tension fracture :

𝑋 =𝝈𝑢

𝜎𝑁=

214 𝑀𝑃𝑎

90 𝑀𝑃𝑎= 2.377778

Safety factor against compression fracture :

𝑋 =𝝈𝑢

𝜎𝑁=

770 𝑀𝑃𝑎

90 𝑀𝑃𝑎= 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

σo

Ultimate Strength

σu

Elongation

100𝜺f

Reduction in area %RA

18 Ni maraging steel 250

GPa (103ksi)

MPa (ksi)

MPa (ksi) % %

186 (27)

1791 (260)

1860 (270) 8 56

𝜎ℎ =𝑝𝑟𝑖

𝑡=

( 2 𝑀𝑃𝑎 )(3/2 𝑚)

0.005 𝑚= 600 𝑀𝑃𝑎, 𝜎ℎ =

𝑝𝑟𝑖

2𝑡=

( 2 𝑀𝑃𝑎 )(3/2𝑚)

2(0.005 𝑚)= 300 𝑀𝑃𝑎

𝜎𝑟 = −2 𝑀𝑃𝑎 𝑖𝑛𝑠𝑖𝑑𝑒 , , 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝜎𝑟 = 0

inside

𝜎𝑠 = 𝑀𝐴𝑋(|𝝈1 − 𝝈2|, |𝝈2 − 𝝈3|, |𝝈3 − 𝝈1|) = 𝑀𝐴𝑋(|600 − 300|, |300 + 2|, |−2 − 600|) = 602 𝑀𝑃𝑎

𝑋𝑠 =𝝈𝑦

𝜎𝑠=

1791 𝑀𝑃𝑎

602 𝑀𝑃𝑎= 2.975

outside

𝜎𝑠 = 𝑀𝐴𝑋(|𝝈1 − 𝝈2|, |𝝈2 − 𝝈3|, |𝝈3 − 𝝈1|) = 𝑀𝐴𝑋(|600 − 300|, |300 + 0|, |0 − 600|) = 600 𝑀𝑃𝑎

𝑋𝑠 =𝝈𝑦

𝜎𝑠=

1791 𝑀𝑃𝑎

600 𝑀𝑃𝑎= 2.985

𝜎𝐻 =1

√2√(𝝈1 − 𝝈2)2 + (𝝈2 − 𝝈3)2 + (𝝈3 − 𝝈1)2

Inside

𝜎𝐻 =1

√2√(600 − 300)2 + (300 + 2)2 + (−2 − 600)2 = 521.34 𝑀𝑃𝑎

𝑋𝐻 =𝝈𝑦

𝜎𝐻=

1791 𝑀𝑃𝑎

521.34 𝑀𝑃𝑎= 3.43532

Inside

𝜎𝐻 =1

√2√(600 − 300)2 + (300 + 0)2 + (0 − 600)2 = 519.615 𝑀𝑃𝑎

𝑋𝐻 =𝝈𝑦

𝜎𝐻=

1791 𝑀𝑃𝑎

519.615 𝑀𝑃𝑎= 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: σx = -100, σy

= 40, 𝜏xy = -50 and σz = 𝜏xz = 𝜏yz = 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

σo

Ultimate Strength

σu

Elongation

100𝜺f

Reduction in area %RA

AISI 1020 steel

GPa (103ksi)

MPa (ksi)

MPa (ksi) % %

203 (29.4)

260 (37.7)

441 (64) 36 61

𝝈1, 𝝈2 =𝝈𝑥 + 𝝈𝑦

𝟐± √(

𝝈𝑥 − 𝝈𝑦

𝟐)

2

+ 𝝉𝒙𝒚𝟐 =

−100 + 40

2± √(

−𝟏𝟎𝟎 − 𝟒𝟎

𝟐)

𝟐

+ (−𝟓𝟎)𝟐 = −30 ± 86.023

𝝈1 = 56.02325 𝑀𝑃𝑎 , 𝝈2 = −116.0232 𝑀𝑃𝑎

𝜎𝑠 = 𝑀𝐴𝑋(|𝝈1 − 𝝈2|, |𝝈2 − 𝝈3|, |𝝈3 − 𝝈1|)= 𝑀𝐴𝑋(|56.02325 + 116.0232|, |−116.0232 + 0|, |0− 56.02325|) = 172.04650 𝑀𝑃𝑎

𝑋𝑠 =𝝈𝑦

𝜎𝑠=

260 𝑀𝑃𝑎

172.0465 𝑀𝑃𝑎= 1.511219

𝜎𝐻 =1

√2√(𝝈𝑥 − 𝝈𝑦)

2+ (𝝈𝑦 − 𝝈𝑧)

2+ (𝝈𝑧 − 𝝈𝑥)2 + 6(𝜏𝑥𝑦

2 + 𝜏𝑦𝑧2 + 𝜏𝑧𝑥

2)

𝜎𝐻 =1

√2√(56.02325 + 116.0232 )2 + (−116.0232 − 0)2 + (0 − 56.02325)2 + 6((−50)2 + 0 + 0)

= 151.9868 𝑀𝑃𝑎

𝑋𝐻 =𝝈𝑦

𝜎𝐻=

260 𝑀𝑃𝑎

151.9868 𝑀𝑃𝑎= 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 𝛾xy = 300 x 10-6. Assume that no yielding has occurred, and also that no loading is applied directly to the surface,

so that σz = 𝜏xz = 𝜏yz = 0. What is the safety factor against yielding?

𝜎𝑥 =𝐸

1 − 𝑣2(𝑒𝑥 + 𝑣𝑒𝑦) =

203 × 103 𝑀𝑃𝑎

1 − 0. 2932(0.00019 + 0.293 × −0.00076) = −7.257 𝑀𝑃𝑎

𝜎𝑦 =𝐸

1 − 𝑣2(𝑒𝑦 + 𝑣𝑒𝑥) =

203 × 103 𝑀𝑃𝑎

1 − 0. 2932(0.00019 + 0.293 × −0.00076) = −156.4063 𝑀𝑃𝑎

𝜏𝑦𝑥 = 𝐺𝛾𝑦𝑥 ==𝐸

2(1 + 𝑣)𝛾𝑦𝑥 =

203 × 103 𝑀𝑃𝑎

2(1 + 0.293)(0.0003) = 23.54988 𝑀𝑃𝑎

𝝈1, 𝝈2 =𝝈𝑥 + 𝝈𝑦

𝟐± √(

𝝈𝑥 − 𝝈𝑦

𝟐)

2

+ 𝝉𝒙𝒚𝟐 =

−7.257 − 156.4063

2± √(

−7.257 + 156.4063

𝟐)

𝟐

+ (23.54988)𝟐

= −80.83165 ± 78.2047

𝝈1 = −2.626948 𝑀𝑃𝑎 , 𝝈2 = −159.0363511 𝑀𝑃𝑎

𝜎𝑠 = 𝑀𝐴𝑋(|𝝈1 − 𝝈2|, |𝝈2 − 𝝈3|, |𝝈3 − 𝝈1|) = 𝑀𝐴𝑋(|−2.626948 − 159.0363511|, |−159.0363511 + 0|, |0 + 2.626948|)

= 161.6632991 𝑀𝑃𝑎

𝑋𝑠 =𝝈𝑦

𝜎𝑠=

260 𝑀𝑃𝑎

161.6632991 𝑀𝑃𝑎= 1.60828

𝜎𝐻 =1

√2√(𝝈𝑥 − 𝝈𝑦)

2+ (𝝈𝑦 − 𝝈𝑧)

2+ (𝝈𝑧 − 𝝈𝑥)2 + 6(𝜏𝑥𝑦

2 + 𝜏𝑦𝑧2 + 𝜏𝑧𝑥

2)

𝜎𝐻 =1

√2√(−7.257 − 156.4063 )2 + (−156.4063 − 0)2 + (0 − 7.257 )2 + 6((23.54988)2 + 0 + 0)

= 165.277651 𝑀𝑃𝑎

𝑋𝐻 =𝝈𝑦

𝜎𝐻=

260 𝑀𝑃𝑎

165.277651 𝑀𝑃𝑎= 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 σz = 𝜏xz = 𝜏yz = 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

σo

Ultimate Strength

σu

Elongation

100𝜺f

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

𝑒45 =𝑒𝑥 + 𝑒𝑦

2+

𝛾𝑥𝑦

2 → 𝜸𝒙𝒚 = 2𝑒45 − 𝑒𝑥 − 𝑒𝑦 = 2(0.0019) − 0.0012 + 0.00065 = 0.00325

𝜎𝑥 =𝐸

1 − 𝑣2(𝑒𝑥 + 𝑣𝑒𝑦) =

71 × 103 𝑀𝑃𝑎

1 − 0. 3452(0.0012 + 0.345 × −0.00065) = 78.638 𝑀𝑃𝑎

𝜎𝑦 =𝐸

1 − 𝑣2(𝑒𝑦 + 𝑣𝑒𝑥) =

71 × 103 𝑀𝑃𝑎

1 − 0. 3452(−0.00065 + 0.345 × 0.0012) = −19.0198 𝑀𝑃𝑎

𝜏𝑦𝑥 = 𝐺𝛾𝑦𝑥 ==𝐸

2(1 + 𝑣)𝛾𝑦𝑥 =

71 × 103 𝑀𝑃𝑎

2(1 + 0.345)(0.00325) = 85.780669 𝑀𝑃𝑎

𝝈1, 𝝈2 =𝝈𝑥 + 𝝈𝑦

𝟐± √(

𝝈𝑥 − 𝝈𝑦

𝟐)

2

+ 𝝉𝒙𝒚𝟐 =

78.638 − 19.0198 2

± √(78.638 + 19.0198

𝟐)

𝟐

+ (85.780669)𝟐

= 29.8091 ± 98.7045

𝝈1 = 129.51363 𝑀𝑃𝑎 , 𝝈2 = −68.8954 𝑀𝑃𝑎

𝜎𝑠 = 𝑀𝐴𝑋(|𝝈1 − 𝝈2|, |𝝈2 − 𝝈3|, |𝝈3 − 𝝈1|)

= 𝑀𝐴𝑋(|129.51363 + 68.8954 |, |−68.8954 + 0|, |0 + 129.51363|) = 198.40903 𝑀𝑃𝑎

𝑋𝑠 =𝝈𝑦

𝜎𝑠=

469 𝑀𝑃𝑎

198.40903 𝑀𝑃𝑎= 2.3638

𝜎𝐻 =1

√2√(𝝈𝑥 − 𝝈𝑦)

2+ (𝝈𝑦 − 𝝈𝑧)

2+ (𝝈𝑧 − 𝝈𝑥)2 + 6(𝜏𝑥𝑦

2 + 𝜏𝑦𝑧2 + 𝜏𝑧𝑥

2)

𝜎𝐻 =1

√2√(78.638 + 19.0198 )2 + (−19.0198 − 0)2 + (0 − 78.638 )2 + 6(85.7806692 + 0 + 0)

= 173.54087 𝑀𝑃𝑎

𝑋𝐻 =𝝈𝑦

𝜎𝐻=

469 𝑀𝑃𝑎

173.54087 𝑀𝑃𝑎= 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 𝝈𝑜 using (a) the maximum shear stress criterion, and (b) the octahedral shear stress criterion. How much do these two sizes differ?

𝜏 =𝑇𝑟

𝐽=

𝑇𝑑2

𝐽, 𝐽 =

𝜋𝑑4

32, 𝜏 =

16𝑇

𝜋𝑑3

For maximum shear stress criterion

𝑋𝑠 =𝝈𝑦

𝜎𝑠, ,

𝝈𝑦

𝑋𝑠= 𝜎𝑠 = 𝑀𝐴𝑋(|𝝈1 − 𝝈2|, |𝝈2 − 𝝈3|, |𝝈3 − 𝝈1|) = 2𝜏

𝝈𝑦

𝑋𝑠= 2 (

16𝑇

𝜋𝑑3 )

𝒅𝒔 = (𝟑𝟐𝑿𝒔𝑻

𝝅𝜎𝒚)

𝟏𝟑⁄

the octahedral shear stress criterion

𝜎𝐻 =𝝈𝑦

𝑋𝐻

=1

√2√(𝝈1 − 𝝈2)2 + (𝝈2 − 𝝈3)2 + (𝝈3 − 𝝈1)2 =

1

√2√(2𝜏)2 + 𝜏2 + 𝜏2 = 𝜏√3

𝝈𝑦

𝑋𝐻= 𝜏√3 = √3 (

16𝑇

𝜋𝑑3)

𝒅𝑯 = (𝟏𝟔√𝟑𝑿𝑯

𝝅𝜎𝒚)

𝟏𝟑⁄

𝒅𝑯

𝒅𝒔=

(𝟏𝟔√𝟑𝑿𝑯

𝝅𝜎𝒚)

𝟏𝟑⁄

(𝟑𝟐𝑿𝒔𝑻

𝝅𝜎𝒚)

𝟏𝟑⁄

=𝟏𝟔√𝟑

𝟑𝟐= 𝟎. 𝟖𝟔𝟔𝟔

𝒅𝑯 = 𝟎. 𝟖𝟔𝟔𝟔𝒅𝒔

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?

𝜎𝑥 =𝐹

𝐴=

𝐹

2𝜋𝑟 𝑡=

60 kN

2𝜋 (0.046𝑚

2 ) (0.0025𝑚)= 166.0747 𝑀𝑃𝑎

𝜏𝑥𝑦 =𝑇𝑟

𝐽, 𝐽 =

𝜋(𝑑𝑜4 − 𝑑𝑖

4)

32=

𝜋(0. 046𝑚4 − 0. 0435𝑚4)

32= 8.8 × 10−8𝑚4

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.

𝜎ℎ =𝑝𝑟𝑖

𝑡=

( 24 𝑀𝑃𝑎 )(0.05 𝑚)

0.002 𝑚= 600 𝑀𝑃𝑎, 𝜎ℎ =

𝑝𝑟𝑖

2𝑡=

( 24 𝑀𝑃𝑎 )(0.05𝑚)

2(0.002 𝑚)= 300 𝑀𝑃𝑎

𝜏𝑥𝑦 =𝑇𝑟

𝐽=

(8 𝑘𝑁𝑚)(0.026𝑚)𝜋

64(0.0524 − 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 σz 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 σy Derive an equation for the value of σz necessary to cause yielding in terms of σy and the elastic constants of the material. Is the value of σz 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 σz 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 sides—that 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 σz = -l20MPa, along with a shear stress 𝜏xy. as shown in Pig. P7.26. (a) What is the largest value of = 𝜏xy that can be applied if the safety factor against yield must be 2.0? (b)Is there a large effect of σz on the 𝜏xy required to cause yielding? Briefly discus effect of σz as to whether the effect is large, small, or absent, and explain why.

𝝈1, 𝝈2 =𝝈𝑥+𝝈𝑦

𝟐± √(

𝝈𝑥−𝝈𝑦

𝟐)

2

+ 𝝉𝒙𝒚𝟐 = ±𝝉𝒙𝒚

, 𝝈3 = 𝝈𝑧 = −120 𝑀𝑃𝑎

Based on maximum shear stress criterion

𝜎𝑠 = 𝑀𝐴𝑋(|𝝈1 − 𝝈2|, |𝝈2 − 𝝈3|, |𝝈3 − 𝝈1|) = 𝑀𝐴𝑋(|𝝉𝒙𝒚 + 𝝉𝒙𝒚|, |−𝝉𝒙𝒚 + 120|, |−120 − 𝝉𝒙𝒚|)

𝜎𝑠 = 2𝝉𝒙𝒚 𝒐𝒓 = 𝝉𝒙𝒚 + 𝟏𝟐𝟎

Material Elastic

Modulus E

0.2% yield Strength

σo

Ultimate Strength

σu

Elongation

100𝜺f

Reduction in area %RA

AISI 1020 steel

GPa (103ksi)

MPa (ksi)

MPa (ksi) % %

203 (29.4) 260

(37.7) 441 (64) 36 61

𝑋𝑠 =𝝈𝑦

𝜎𝑠 , 𝜎𝑠 =

𝝈𝑦

𝑋𝑠=

260

2= 130 𝑀𝑃𝑎

𝑖𝑓 𝜎𝑠 = 𝝉𝒙𝒚 + 𝟏𝟐𝟎, 𝜎𝑠 = 130 𝑀𝑃𝑎 , 𝝉𝒙𝒚 = 130 − 120 = 10 𝑀𝑃𝑎

If 𝜎𝑠 = 2𝝉𝒙𝒚 , , 𝜎𝑠 = 130 𝑀𝑃𝑎 , 𝝉𝒙𝒚 =130 𝑀𝑃𝑎

2= 65 𝑀𝑃𝑎

- in the first case 𝝉𝒙𝒚 = 65 𝑀𝑃𝑎 the result is vary with σz

Based on octahedral shear stress criterion

𝜎𝐻 =1

√2√(𝝈1 − 𝝈2)2 + (𝝈2 − 𝝈3)2 + (𝝈3 − 𝝈1)2 =

1

√2√(𝝉𝒙𝒚+𝝉𝒙𝒚)

2+ (– 𝝉𝒙𝒚 + 120)

2+ (−120 − 𝝉𝒙𝒚)

2

=1

√2√4𝝉𝒙𝒚

𝟐 + 𝝉𝒙𝒚𝟐 − 𝟐𝟒𝟎𝝉𝒙𝒚 + 𝟏𝟒𝟒𝟎𝟎 + 𝟏𝟒𝟒𝟎𝟎 + 𝟐𝟒𝟎𝝉𝒙𝒚 + 𝝉𝒙𝒚

𝟐 =1

√2√6𝝉𝒙𝒚

𝟐 + 𝟐(𝟏𝟐𝟎𝟐) = √3𝝉𝒙𝒚𝟐 + (𝟏𝟐𝟎𝟐)

𝜎𝐻 =𝝈𝑦

𝑋𝐻=

260

2= √3𝝉𝒙𝒚

𝟐 + (𝟏𝟐𝟎𝟐) , 2602 = 4 (3𝝉𝒙𝒚𝟐 + (𝟏𝟐𝟎𝟐)) , 𝝉𝒙𝒚 = 28.86 𝑀𝑃𝑎