Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 6, Problem 6.18P

(a)

To determine

The toughness of 1045 steel specimen.

(a)

Expert Solution
Check Mark

Answer to Problem 6.18P

The toughness of 1045 steel specimen is 80.13×106J/m3 .

Explanation of Solution

Formula Used:

Write the expression for the first area.

  A1=σy22E …… (I)

Here, σy is the yield stress, A1 is the first area and E is the Young’s modulus of the specimen.

Write the expression for the second area.

  A2=(0.01εy)(σy+σy22) …… (II)

Here, A2 is the second area, εy is the strain, σy2 is the yield stresses for second area.

Write the expression for the third area.

  A3=(0.01)(σy2+σy32) …… (III)

Here, A3 is the third area, σy3 is the yield stresses for third area.

Write the expression for the fourth area.

  A4=(0.01)(σy3+σy42) …… (IV)

Here, A4 is the fourth area, σy4 is the yield stresses for fourth area.

Write the expression for the fifth area.

  A5=(0.01)(σy4+σy52) …… (V)

Here, A5 is the fifth area, σy5 is the yield stresses for fifth area.

Write the expression for the sixth area.

  A6=(0.01)(σy5+σy62) …… (VI)

Here, A6 is the sixth area, σy6 is the yield stresses for sixth area.

Write the expression for the seventh area.

  A7=(0.01)(σy6+σy72) …… (VII)

Here, A7 is the seventh area, σy7 is the yield stresses for seventh area.

Write the expression for the eighth area.

  A8=(0.01)(σy7+σy82) …… (VIII)

Here, A8 is the eighth area, σy8 is the yield stresses for eighth area.

Write the expression for the ninth area.

  A9=(0.01)(σy8+σy92) …… (IX)

Here, A9 is the ninth area, σy9 is the yield stresses for ninth area.

Write the expression for the tenth area.

  A10=(0.01)(σy9+σy102) …… (X)

Here, A10 is the tenth area, σy10 are the yield stresses for tenth area.

Write the expression for the eleventh area.

  A11=(0.008)(σy10+σy112) …… (XI)

Here, A11 is the eleventh area, σy11 is the yield stresses for eleventh area.

Write the expression for the toughness of specimen.

  Toughness=A1+A2+A3+A4+A5+A6+A7+A8+A9+A10+A11 …… (XII)

Write the expression for strain at yield point.

  εy=σyE …… (XIII)

Calculation:

Refer Figure 6.22 “A true stress-strain diagram for 1045 steel in mega Pascals MPa .” of the book “Material Science and Engineering Properties”.

The stress strain diagram for 1045 steel with sub-divided area is shown below,

  Materials Science And Engineering Properties, Chapter 6, Problem 6.18P

Figure (1)

Substitute 600MPa for σy and 211GPa for E in equation (I).

  A1=((600MPa)(106Pa1MPa))22((211Gpa)(109Pa1MPa))=(600×106Pa)2422×109Pa=((0.85×106Pa)(1J/m31Pa))=0.85×106J/m3

Substitute 600MPa for σy and 211GPa for E in equation (XIII).

  εy=((600MPa)(106Pa1MPa))((211Gpa)(109Pa1MPa))=(600×106Pa)211×109Pa0.003

Substitute 600MPa for σy and 620MPa for σy2

  0.003 for εy in equation (II).

  A2=(0.010.003)(((600MPa)(106Pa1MPa))+((620MPa)(106Pa1MPa))2)=(0.007)((1220×106Pa)2)=((4.27×106Pa)(1J/m31Pa))=4.27×106J/m3

Substitute 720MPa for σy3 and 620MPa for σy2 in equation (III).

  A3=(0.01)(((620MPa)(106Pa1MPa))+((720MPa)(106Pa1MPa))2)=(0.01)((1340×106Pa)2)=((6.7×106Pa)(1J/m31Pa))=6.7×106J/m3

Substitute 720MPa for σy3 and 800MPa for σy4 in equation (IV).

  A4=(0.01)(((720MPa)(106Pa1MPa))+((800MPa)(106Pa1MPa))2)=(0.01)((1520×106Pa)2)=((7.6×106Pa)(1J/m31Pa))=7.6×106J/m3

Substitute 860MPa for σy5 and 800MPa for σy4 in equation (V).

  A5=(0.01)(((800MPa)(106Pa1MPa))+((860MPa)(106Pa1MPa))2)=(0.01)((1600×106Pa)2)=((8.3×106Pa)(1J/m31Pa))=8.3×106J/m3

Substitute 860MPa for σy5 and 880MPa for σy6 in equation (VI).

  A6=(0.01)(((860MPa)(106Pa1MPa))+((880MPa)(106Pa1MPa))2)=(0.01)((1740×106Pa)2)=((8.7×106Pa)(1J/m31Pa))=8.7×106J/m3

Substitute 910MPa for σy7 and 880MPa for σy6 in equation (VII).

  A7=(0.01)(((880MPa)(106Pa1MPa))+((910MPa)(106Pa1MPa))2)=(0.01)((1790×106Pa)2)=((8.95×106Pa)(1J/m31Pa))=8.95×106J/m3

Substitute 910MPa for σy7 and 920MPa for σy8 in equation (VIII).

  A8=(0.01)(((910MPa)(106Pa1MPa))+((920MPa)(106Pa1MPa))2)=(0.01)((1830×106Pa)2)=((9.15×106Pa)(1J/m31Pa))=9.15×106J/m3

Substitute 910MPa for σy9 and 920MPa for σy8 in equation (IX).

  A9=(0.01)(((920MPa)(106Pa1MPa))+((920MPa)(106Pa1MPa))2)=(0.01)((1840×106Pa)2)=((9.20×106Pa)(1J/m31Pa))=9.20×106J/m3

Substitute 920MPa for σy9 and 910MPa for σy10 in equation (X).

  A10=(0.01)(((920MPa)(106Pa1MPa))+((910MPa)(106Pa1MPa))2)=(0.01)((1830×106Pa)2)=((9.15×106Pa)(1J/m31Pa))=9.15×106J/m3

Substitute 905MPa for σy11 and 910MPa for σy10 in equation (XI).

  A11=(0.008)(((910MPa)(106Pa1MPa))+((905MPa)(106Pa1MPa))2)=(0.008)((1815×106Pa)2)=((7.26×106Pa)(1J/m31Pa))=7.26×106J/m3

Substitute 0.85×106J/m3 for A1 , 4.27×106J/m3 for A2 , 6.7×106J/m3 for A3 , 7.6×106J/m3 for A4 , 8.3×106J/m3 for A5 , 8.7×106J/m3 for A6 , 8.95×106J/m3 for A7 , 9.15×106J/m3 for A8 , 9.20×106J/m3 for A9 , 9.15×106J/m3 for A10 , 7.26×106J/m3 for A11 in equation (XII).

  Toughness=[(0.85×106J/m3)+(4.27×106J/m3)+(6.7×106J/m3)+(7.6×106J/m3)+(8.3×106J/m3)+(8.7×106J/m3)+(8.95×106J/m3)+(9.15×106J/m3)+(9.20×106J/m3)+(9.15×106J/m3)+(7.26×106J/m3)]=(0.85+4.27+6.7+7.6+8.3+8.7+8.95+9.15+9.20+9.15+7.26)×106J/m3=80.13×106J/m3

Conclusion:

Thus, the toughness of 1045 steel specimen is 80.13×106J/m3 .

(b)

To determine

The comparison of magnitude of 1020 steel and 1045 steel.

(b)

Expert Solution
Check Mark

Explanation of Solution

Introduction:

The toughness of a specimen is defined as the strain energy per unit volume, required to fracture the specimen. Toughness is obtained when the stress-strain diagram to fracture is integrated. Toughness of a specimen plays a major role for its availability as an engineering material.

Refer Example Problem 6.13 in the textbook “Material Science and Engineering Properties”.

The toughness of 1020 steel specimen is 92.2×106J/m3 , while the toughness of 1045 steel specimen found in part (a) is 80.13×106J/m3 . So, it can be said that 1020 steel specimen is tougher as compared to 1045 steel specimen.

For 1020 steel, fracture occurs at strain of around 0.19 while in case of 1045 steel it occurs at strain of 0.098 . So, for higher value of fracture strain, toughness is higher, that is why the toughness of 1020 steel specimen is more as compare to 1045 steel specimen.

Conclusion:

Thus, the 1020 steel specimen is tougher as compared to 1045 steel specimen.

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Materials Science And Engineering Properties

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