A metallurgist evaluates the hardenability of a steel alloy using Jominy end-quench data. Given surface hardness (HRC0) = 53.5000 HRC; hardness gradient (k) = 11.5000 HRC/in; Jominy distance (d) = 2.4000 in, determine the hardness at distance d (HRC) in HRC.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except HRC is given, so isolate HRC symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for HRC:

   $$HRC = HRC_0 - kd$$

3. Step 3 — List the givens: surface hardness (HRC0) = 53.5000 HRC, hardness gradient (k) = 11.5000 HRC/in, Jominy distance (d) = 2.4000 in.

4. Step 4 — Substitute the given values:

   $$HRC = HRC_0 - k2.4000$$

5. Step 5 — Evaluate:

   $$HRC = 25.9000\ \text{HRC}$$

6. Step 6 — Check: returning HRC = 25.9000 HRC to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 51.8000 — kept a factor of two that cancels in the correct rearrangement.
• 12.9500 — dropped that same factor in the other direction.
• 28.4900 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability

An engineer selects a steel grade based on its hardenability requirement. Given hardness gradient (k) = 6.5000 HRC/in; Jominy distance (d) = 0.6000 in; hardness at distance d (HRC) = 10.2000 HRC, determine the surface hardness (HRC0) in HRC.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except HRC0 is given, so isolate HRC0 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for HRC0:

   $$HRC_{0} = HRC + kd$$

3. Step 3 — List the givens: hardness gradient (k) = 6.5000 HRC/in, Jominy distance (d) = 0.6000 in, hardness at distance d (HRC) = 10.2000 HRC.

4. Step 4 — Substitute the given values:

   $$HRC_{0} = 10.2000 + k0.6000$$

5. Step 5 — Evaluate:

   $$HRC_{0} = 14.1000\ \text{HRC}$$

6. Step 6 — Check: returning HRC0 = 14.1000 HRC to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 28.2000 — kept a factor of two that cancels in the correct rearrangement.
• 7.0500 — dropped that same factor in the other direction.
• 15.5100 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability

The hardenability curve shows hardness dropping with distance from the quenched face. Given surface hardness (HRC0) = 54.0000 HRC; hardness gradient (k) = 7.0000 HRC/in; hardness at distance d (HRC) = 15.3000 HRC, determine the Jominy distance (d) in in.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except d is given, so isolate d symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for d:

   $$d = \dfrac{HRC_0 - HRC}{k}$$

3. Step 3 — List the givens: surface hardness (HRC0) = 54.0000 HRC, hardness gradient (k) = 7.0000 HRC/in, hardness at distance d (HRC) = 15.3000 HRC.

4. Step 4 — Substitute the given values:

   $$d = \dfrac{HRC_0 - 15.3000}{7.0000}$$

5. Step 5 — Evaluate:

   $$d = 5.5286\ \text{in}$$

6. Step 6 — Check: returning d = 5.5286 in to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 11.0571 — kept a factor of two that cancels in the correct rearrangement.
• 2.7643 — dropped that same factor in the other direction.
• 6.0814 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability

A metallurgist evaluates the hardenability of a steel alloy using Jominy end-quench data. Given surface hardness (HRC0) = 52.0000 HRC; hardness gradient (k) = 5.0000 HRC/in; Jominy distance (d) = 0.8000 in, determine the hardness at distance d (HRC) in HRC.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except HRC is given, so isolate HRC symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for HRC:

   $$HRC = HRC_0 - kd$$

3. Step 3 — List the givens: surface hardness (HRC0) = 52.0000 HRC, hardness gradient (k) = 5.0000 HRC/in, Jominy distance (d) = 0.8000 in.

4. Step 4 — Substitute the given values:

   $$HRC = HRC_0 - k0.8000$$

5. Step 5 — Evaluate:

   $$HRC = 48.0000\ \text{HRC}$$

6. Step 6 — Check: returning HRC = 48.0000 HRC to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 96.0000 — kept a factor of two that cancels in the correct rearrangement.
• 24.0000 — dropped that same factor in the other direction.
• 52.8000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability

An engineer selects a steel grade based on its hardenability requirement. Given hardness gradient (k) = 5.5000 HRC/in; Jominy distance (d) = 2.6000 in; hardness at distance d (HRC) = 23.0000 HRC, determine the surface hardness (HRC0) in HRC.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except HRC0 is given, so isolate HRC0 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for HRC0:

   $$HRC_{0} = HRC + kd$$

3. Step 3 — List the givens: hardness gradient (k) = 5.5000 HRC/in, Jominy distance (d) = 2.6000 in, hardness at distance d (HRC) = 23.0000 HRC.

4. Step 4 — Substitute the given values:

   $$HRC_{0} = 23.0000 + k2.6000$$

5. Step 5 — Evaluate:

   $$HRC_{0} = 37.3000\ \text{HRC}$$

6. Step 6 — Check: returning HRC0 = 37.3000 HRC to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 74.6000 — kept a factor of two that cancels in the correct rearrangement.
• 18.6500 — dropped that same factor in the other direction.
• 41.0300 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability

The hardenability curve shows hardness dropping with distance from the quenched face. Given surface hardness (HRC0) = 52.0000 HRC; hardness gradient (k) = 12.0000 HRC/in; hardness at distance d (HRC) = 34.4000 HRC, determine the Jominy distance (d) in in.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except d is given, so isolate d symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for d:

   $$d = \dfrac{HRC_0 - HRC}{k}$$

3. Step 3 — List the givens: surface hardness (HRC0) = 52.0000 HRC, hardness gradient (k) = 12.0000 HRC/in, hardness at distance d (HRC) = 34.4000 HRC.

4. Step 4 — Substitute the given values:

   $$d = \dfrac{HRC_0 - 34.4000}{12.0000}$$

5. Step 5 — Evaluate:

   $$d = 1.4667\ \text{in}$$

6. Step 6 — Check: returning d = 1.4667 in to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 2.9333 — kept a factor of two that cancels in the correct rearrangement.
• 0.7333 — dropped that same factor in the other direction.
• 1.6133 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability

A metallurgist evaluates the hardenability of a steel alloy using Jominy end-quench data. Given surface hardness (HRC0) = 55.5000 HRC; hardness gradient (k) = 4.5000 HRC/in; Jominy distance (d) = 0.6000 in, determine the hardness at distance d (HRC) in HRC.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except HRC is given, so isolate HRC symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for HRC:

   $$HRC = HRC_0 - kd$$

3. Step 3 — List the givens: surface hardness (HRC0) = 55.5000 HRC, hardness gradient (k) = 4.5000 HRC/in, Jominy distance (d) = 0.6000 in.

4. Step 4 — Substitute the given values:

   $$HRC = HRC_0 - k0.6000$$

5. Step 5 — Evaluate:

   $$HRC = 52.8000\ \text{HRC}$$

6. Step 6 — Check: returning HRC = 52.8000 HRC to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 105.6 — kept a factor of two that cancels in the correct rearrangement.
• 26.4000 — dropped that same factor in the other direction.
• 58.0800 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability

An engineer selects a steel grade based on its hardenability requirement. Given hardness gradient (k) = 5.0000 HRC/in; Jominy distance (d) = 1.8000 in; hardness at distance d (HRC) = 40.4000 HRC, determine the surface hardness (HRC0) in HRC.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except HRC0 is given, so isolate HRC0 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for HRC0:

   $$HRC_{0} = HRC + kd$$

3. Step 3 — List the givens: hardness gradient (k) = 5.0000 HRC/in, Jominy distance (d) = 1.8000 in, hardness at distance d (HRC) = 40.4000 HRC.

4. Step 4 — Substitute the given values:

   $$HRC_{0} = 40.4000 + k1.8000$$

5. Step 5 — Evaluate:

   $$HRC_{0} = 49.4000\ \text{HRC}$$

6. Step 6 — Check: returning HRC0 = 49.4000 HRC to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 98.8000 — kept a factor of two that cancels in the correct rearrangement.
• 24.7000 — dropped that same factor in the other direction.
• 54.3400 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability

The hardenability curve shows hardness dropping with distance from the quenched face. Given surface hardness (HRC0) = 61.5000 HRC; hardness gradient (k) = 5.0000 HRC/in; hardness at distance d (HRC) = 30.6000 HRC, determine the Jominy distance (d) in in.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except d is given, so isolate d symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for d:

   $$d = \dfrac{HRC_0 - HRC}{k}$$

3. Step 3 — List the givens: surface hardness (HRC0) = 61.5000 HRC, hardness gradient (k) = 5.0000 HRC/in, hardness at distance d (HRC) = 30.6000 HRC.

4. Step 4 — Substitute the given values:

   $$d = \dfrac{HRC_0 - 30.6000}{5.0000}$$

5. Step 5 — Evaluate:

   $$d = 6.1800\ \text{in}$$

6. Step 6 — Check: returning d = 6.1800 in to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 12.3600 — kept a factor of two that cancels in the correct rearrangement.
• 3.0900 — dropped that same factor in the other direction.
• 6.7980 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability

A metallurgist evaluates the hardenability of a steel alloy using Jominy end-quench data. Given surface hardness (HRC0) = 60.5000 HRC; hardness gradient (k) = 8.5000 HRC/in; Jominy distance (d) = 2.2000 in, determine the hardness at distance d (HRC) in HRC.

Start with the thinking

• The governing relation printed in this handbook section is Hardenability — Jominy distance to hardness ratio.
• Everything except HRC is given, so isolate HRC symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hardenability describes how hardness decreases with distance from the quenched end in a Jominy test.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$HRC = HRC_0 - k \, d$$

2. Step 2 — Rearrange symbolically for HRC:

   $$HRC = HRC_0 - kd$$

3. Step 3 — List the givens: surface hardness (HRC0) = 60.5000 HRC, hardness gradient (k) = 8.5000 HRC/in, Jominy distance (d) = 2.2000 in.

4. Step 4 — Substitute the given values:

   $$HRC = HRC_0 - k2.2000$$

5. Step 5 — Evaluate:

   $$HRC = 41.8000\ \text{HRC}$$

6. Step 6 — Check: returning HRC = 41.8000 HRC to

   $$HRC = HRC_0 - k \, d$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 83.6000 — kept a factor of two that cancels in the correct rearrangement.
• 20.9000 — dropped that same factor in the other direction.
• 45.9800 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hardenability