A materials engineer computes the true strain of a tensile specimen at necking. Given original length (L0) = 1.3500 in; final length (Lf) = 2.9000 in, determine the true strain (epsT).

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except epsT is given, so isolate epsT 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for epsT:

   $$epsT = \ln(L_f/L_0)$$

3. Step 3

   $$List the givens: original length (L_{0}) = 1.3500 in, final length (Lf) = 2.9000 in$$

4. Step 4 — Substitute the given values:

   $$epsT = \ln(L_f/L_0)$$

5. Step 5 — Evaluate:

   $$epsT = 0.7646$$

6. Step 6 — Check: returning epsT = 0.7646 to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• 1.5292 — kept a factor of two that cancels in the correct rearrangement.
• 0.3823 — dropped that same factor in the other direction.
• 0.8411 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain

A student compares true strain to engineering strain for a stretched rod. Given original length (L0) = 2.2000 in; true strain (epsT) = 1.1610, determine the final length (Lf) in in.

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except Lf is given, so isolate Lf 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for Lf:

   $$Lf = L_0 e^{\varepsilon_T}$$

3. Step 3

   $$List the givens: original length (L_{0}) = 2.2000 in, true strain (epsT) = 1.1610$$

4. Step 4 — Substitute the given values:

   $$Lf = L_0 e^{\varepsilon_T}$$

5. Step 5 — Evaluate:

   $$Lf = 7.0249\ \text{in}$$

6. Step 6 — Check: returning Lf = 7.0249 in to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• 14.0497 — kept a factor of two that cancels in the correct rearrangement.
• 3.5124 — dropped that same factor in the other direction.
• 7.7274 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain

The true strain of a drawn wire is calculated from its length change. Given final length (Lf) = 6.0500 in; true strain (epsT) = 0.7750, determine the original length (L0) in in.

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except L0 is given, so isolate L0 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for L0:

   $$L_{0} = L_f e^{-\varepsilon_T}$$

3. Step 3

   $$List the givens: final length (Lf) = 6.0500 in, true strain (epsT) = 0.7750$$

4. Step 4 — Substitute the given values:

   $$L_{0} = L_f e^{-\varepsilon_T}$$

5. Step 5 — Evaluate:

   $$L_{0} = 2.7873\ \text{in}$$

6. Step 6 — Check: returning L0 = 2.7873 in to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• 5.5745 — kept a factor of two that cancels in the correct rearrangement.
• 1.3936 — dropped that same factor in the other direction.
• 3.0660 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain

A materials engineer computes the true strain of a tensile specimen at necking. Given original length (L0) = 4.2500 in; final length (Lf) = 6.4000 in, determine the true strain (epsT).

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except epsT is given, so isolate epsT 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for epsT:

   $$epsT = \ln(L_f/L_0)$$

3. Step 3

   $$List the givens: original length (L_{0}) = 4.2500 in, final length (Lf) = 6.4000 in$$

4. Step 4 — Substitute the given values:

   $$epsT = \ln(L_f/L_0)$$

5. Step 5 — Evaluate:

   $$epsT = 0.4094$$

6. Step 6 — Check: returning epsT = 0.4094 to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• 0.8188 — kept a factor of two that cancels in the correct rearrangement.
• 0.2047 — dropped that same factor in the other direction.
• 0.4503 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain

A student compares true strain to engineering strain for a stretched rod. Given original length (L0) = 1.8000 in; true strain (epsT) = 0.1170, determine the final length (Lf) in in.

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except Lf is given, so isolate Lf 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for Lf:

   $$Lf = L_0 e^{\varepsilon_T}$$

3. Step 3

   $$List the givens: original length (L_{0}) = 1.8000 in, true strain (epsT) = 0.1170$$

4. Step 4 — Substitute the given values:

   $$Lf = L_0 e^{\varepsilon_T}$$

5. Step 5 — Evaluate:

   $$Lf = 2.0234\ \text{in}$$

6. Step 6 — Check: returning Lf = 2.0234 in to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• 4.0468 — kept a factor of two that cancels in the correct rearrangement.
• 1.0117 — dropped that same factor in the other direction.
• 2.2258 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain

The true strain of a drawn wire is calculated from its length change. Given final length (Lf) = 2.1000 in; true strain (epsT) = 1.0640, determine the original length (L0) in in.

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except L0 is given, so isolate L0 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for L0:

   $$L_{0} = L_f e^{-\varepsilon_T}$$

3. Step 3

   $$List the givens: final length (Lf) = 2.1000 in, true strain (epsT) = 1.0640$$

4. Step 4 — Substitute the given values:

   $$L_{0} = L_f e^{-\varepsilon_T}$$

5. Step 5 — Evaluate:

   $$L_{0} = 0.7247\ \text{in}$$

6. Step 6 — Check: returning L0 = 0.7247 in to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• 1.4493 — kept a factor of two that cancels in the correct rearrangement.
• 0.3623 — dropped that same factor in the other direction.
• 0.7971 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain

A materials engineer computes the true strain of a tensile specimen at necking. Given original length (L0) = 2.4000 in; final length (Lf) = 1.5000 in, determine the true strain (epsT).

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except epsT is given, so isolate epsT 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for epsT:

   $$epsT = \ln(L_f/L_0)$$

3. Step 3

   $$List the givens: original length (L_{0}) = 2.4000 in, final length (Lf) = 1.5000 in$$

4. Step 4 — Substitute the given values:

   $$epsT = \ln(L_f/L_0)$$

5. Step 5 — Evaluate:

   $$epsT = -0.4700$$

6. Step 6 — Check: returning epsT = -0.4700 to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• -0.9400 — kept a factor of two that cancels in the correct rearrangement.
• -0.2350 — dropped that same factor in the other direction.
• -0.5170 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain

A student compares true strain to engineering strain for a stretched rod. Given original length (L0) = 4.6500 in; true strain (epsT) = 0.7850, determine the final length (Lf) in in.

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except Lf is given, so isolate Lf 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for Lf:

   $$Lf = L_0 e^{\varepsilon_T}$$

3. Step 3

   $$List the givens: original length (L_{0}) = 4.6500 in, true strain (epsT) = 0.7850$$

4. Step 4 — Substitute the given values:

   $$Lf = L_0 e^{\varepsilon_T}$$

5. Step 5 — Evaluate:

   $$Lf = 10.1947\ \text{in}$$

6. Step 6 — Check: returning Lf = 10.1947 in to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• 20.3894 — kept a factor of two that cancels in the correct rearrangement.
• 5.0973 — dropped that same factor in the other direction.
• 11.2142 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain

The true strain of a drawn wire is calculated from its length change. Given final length (Lf) = 1.5000 in; true strain (epsT) = 0.9680, determine the original length (L0) in in.

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except L0 is given, so isolate L0 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for L0:

   $$L_{0} = L_f e^{-\varepsilon_T}$$

3. Step 3

   $$List the givens: final length (Lf) = 1.5000 in, true strain (epsT) = 0.9680$$

4. Step 4 — Substitute the given values:

   $$L_{0} = L_f e^{-\varepsilon_T}$$

5. Step 5 — Evaluate:

   $$L_{0} = 0.5698\ \text{in}$$

6. Step 6 — Check: returning L0 = 0.5698 in to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• 1.1395 — kept a factor of two that cancels in the correct rearrangement.
• 0.2849 — dropped that same factor in the other direction.
• 0.6267 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain

A materials engineer computes the true strain of a tensile specimen at necking. Given original length (L0) = 1.6000 in; final length (Lf) = 2.6500 in, determine the true strain (epsT).

Start with the thinking

• The governing relation printed in this handbook section is True strain.
• Everything except epsT is given, so isolate epsT 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.
• True strain uses the natural log of the length ratio, unlike engineering strain's linear ratio.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

2. Step 2 — Rearrange symbolically for epsT:

   $$epsT = \ln(L_f/L_0)$$

3. Step 3

   $$List the givens: original length (L_{0}) = 1.6000 in, final length (Lf) = 2.6500 in$$

4. Step 4 — Substitute the given values:

   $$epsT = \ln(L_f/L_0)$$

5. Step 5 — Evaluate:

   $$epsT = 0.5046$$

6. Step 6 — Check: returning epsT = 0.5046 to

   $$\varepsilon_T = \ln\!\left(\dfrac{L_f}{L_0}\right)$$

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

Why the other options are there

• 1.0091 — kept a factor of two that cancels in the correct rearrangement.
• 0.2523 — dropped that same factor in the other direction.
• 0.5550 — rounded an intermediate value before the final step.

Reference: FE Handbook — True Strain