Skip to content

Percent Reduction in Area (RA)

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials
1 formulas
10 exam-style examples
~47 min
All Mechanics of Materials lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA)

A steel tensile coupon starts with a 0.435 in diameter and necks to 0.270 in at fracture. The 2 in gauge length becomes 2.280 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.435ind_{0} = 0.435 in
  • df=0.270ind_f = 0.270 in
  • L0=2in,Lf=2.280inL_{0} = 2 in, L_f = 2.280 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.435)2=0.14862in2,Af=(π/4)(0.270)2=0.05713in2A_{0} = (\pi/4)(0.435)^{2} = 0.14862 in^{2}, A_f = (\pi/4)(0.270)^{2} = 0.05713 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.14862 - 0.05713)/0.14862 \times 100 = 61.56%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.280 - 2)/2 \times 100 = 14.00%
Answer:
%RA = 61.6%, %elongation = 14.0%

Why the other options are there

  • 38.0% (diameters instead of areas)
  • 38.4% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 2
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (2)

A steel tensile coupon starts with a 0.445 in diameter and necks to 0.303 in at fracture. The 2 in gauge length becomes 2.500 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.445ind_{0} = 0.445 in
  • df=0.303ind_f = 0.303 in
  • L0=2in,Lf=2.500inL_{0} = 2 in, L_f = 2.500 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.445)2=0.15553in2,Af=(π/4)(0.303)2=0.07192in2A_{0} = (\pi/4)(0.445)^{2} = 0.15553 in^{2}, A_f = (\pi/4)(0.303)^{2} = 0.07192 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.15553 - 0.07192)/0.15553 \times 100 = 53.76%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.500 - 2)/2 \times 100 = 25.00%
Answer:
%RA = 53.8%, %elongation = 25.0%

Why the other options are there

  • 32.0% (diameters instead of areas)
  • 46.2% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 3
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (3)

A steel tensile coupon starts with a 0.475 in diameter and necks to 0.371 in at fracture. The 2 in gauge length becomes 2.460 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.475ind_{0} = 0.475 in
  • df=0.371ind_f = 0.371 in
  • L0=2in,Lf=2.460inL_{0} = 2 in, L_f = 2.460 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.475)2=0.17721in2,Af=(π/4)(0.371)2=0.10781in2A_{0} = (\pi/4)(0.475)^{2} = 0.17721 in^{2}, A_f = (\pi/4)(0.371)^{2} = 0.10781 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.17721 - 0.10781)/0.17721 \times 100 = 39.16%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.460 - 2)/2 \times 100 = 23.00%
Answer:
%RA = 39.2%, %elongation = 23.0%

Why the other options are there

  • 22.0% (diameters instead of areas)
  • 60.8% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 4
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (4)

A steel tensile coupon starts with a 0.525 in diameter and necks to 0.315 in at fracture. The 2 in gauge length becomes 2.300 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.525ind_{0} = 0.525 in
  • df=0.315ind_f = 0.315 in
  • L0=2in,Lf=2.300inL_{0} = 2 in, L_f = 2.300 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.525)2=0.21648in2,Af=(π/4)(0.315)2=0.07793in2A_{0} = (\pi/4)(0.525)^{2} = 0.21648 in^{2}, A_f = (\pi/4)(0.315)^{2} = 0.07793 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.21648 - 0.07793)/0.21648 \times 100 = 64.00%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.300 - 2)/2 \times 100 = 15.00%
Answer:
%RA = 64.0%, %elongation = 15.0%

Why the other options are there

  • 40.0% (diameters instead of areas)
  • 36.0% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 5
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (5)

A steel tensile coupon starts with a 0.405 in diameter and necks to 0.227 in at fracture. The 2 in gauge length becomes 2.600 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.405ind_{0} = 0.405 in
  • df=0.227ind_f = 0.227 in
  • L0=2in,Lf=2.600inL_{0} = 2 in, L_f = 2.600 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.405)2=0.12882in2,Af=(π/4)(0.227)2=0.04040in2A_{0} = (\pi/4)(0.405)^{2} = 0.12882 in^{2}, A_f = (\pi/4)(0.227)^{2} = 0.04040 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.12882 - 0.04040)/0.12882 \times 100 = 68.64%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.600 - 2)/2 \times 100 = 30.00%
Answer:
%RA = 68.6%, %elongation = 30.0%

Why the other options are there

  • 44.0% (diameters instead of areas)
  • 31.4% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 6
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (6)

A steel tensile coupon starts with a 0.590 in diameter and necks to 0.413 in at fracture. The 2 in gauge length becomes 2.340 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.590ind_{0} = 0.590 in
  • df=0.413ind_f = 0.413 in
  • L0=2in,Lf=2.340inL_{0} = 2 in, L_f = 2.340 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.590)2=0.27340in2,Af=(π/4)(0.413)2=0.13396in2A_{0} = (\pi/4)(0.590)^{2} = 0.27340 in^{2}, A_f = (\pi/4)(0.413)^{2} = 0.13396 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.27340 - 0.13396)/0.27340 \times 100 = 51.00%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.340 - 2)/2 \times 100 = 17.00%
Answer:
%RA = 51.0%, %elongation = 17.0%

Why the other options are there

  • 30.0% (diameters instead of areas)
  • 49.0% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 7
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (7)

A steel tensile coupon starts with a 0.530 in diameter and necks to 0.419 in at fracture. The 2 in gauge length becomes 2.380 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.530ind_{0} = 0.530 in
  • df=0.419ind_f = 0.419 in
  • L0=2in,Lf=2.380inL_{0} = 2 in, L_f = 2.380 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.530)2=0.22062in2,Af=(π/4)(0.419)2=0.13769in2A_{0} = (\pi/4)(0.530)^{2} = 0.22062 in^{2}, A_f = (\pi/4)(0.419)^{2} = 0.13769 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.22062 - 0.13769)/0.22062 \times 100 = 37.59%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.380 - 2)/2 \times 100 = 19.00%
Answer:
%RA = 37.6%, %elongation = 19.0%

Why the other options are there

  • 21.0% (diameters instead of areas)
  • 62.4% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 8
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (8)

A steel tensile coupon starts with a 0.490 in diameter and necks to 0.402 in at fracture. The 2 in gauge length becomes 2.380 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.490ind_{0} = 0.490 in
  • df=0.402ind_f = 0.402 in
  • L0=2in,Lf=2.380inL_{0} = 2 in, L_f = 2.380 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.490)2=0.18857in2,Af=(π/4)(0.402)2=0.12680in2A_{0} = (\pi/4)(0.490)^{2} = 0.18857 in^{2}, A_f = (\pi/4)(0.402)^{2} = 0.12680 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.18857 - 0.12680)/0.18857 \times 100 = 32.76%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.380 - 2)/2 \times 100 = 19.00%
Answer:
%RA = 32.8%, %elongation = 19.0%

Why the other options are there

  • 18.0% (diameters instead of areas)
  • 67.2% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 9
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (9)

A steel tensile coupon starts with a 0.595 in diameter and necks to 0.405 in at fracture. The 2 in gauge length becomes 2.680 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.595ind_{0} = 0.595 in
  • df=0.405ind_f = 0.405 in
  • L0=2in,Lf=2.680inL_{0} = 2 in, L_f = 2.680 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.595)2=0.27805in2,Af=(π/4)(0.405)2=0.12857in2A_{0} = (\pi/4)(0.595)^{2} = 0.27805 in^{2}, A_f = (\pi/4)(0.405)^{2} = 0.12857 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.27805 - 0.12857)/0.27805 \times 100 = 53.76%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.680 - 2)/2 \times 100 = 34.00%
Answer:
%RA = 53.8%, %elongation = 34.0%

Why the other options are there

  • 32.0% (diameters instead of areas)
  • 46.2% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

Example 10
Percent reduction in area and elongation from a tension test — Percent Reduction in Area (RA) (10)

A steel tensile coupon starts with a 0.505 in diameter and necks to 0.389 in at fracture. The 2 in gauge length becomes 2.640 in. Compute the percent reduction in area and the percent elongation.

Given

  • d0=0.505ind_{0} = 0.505 in
  • df=0.389ind_f = 0.389 in
  • L0=2in,Lf=2.640inL_{0} = 2 in, L_f = 2.640 in

Find

%RA and %elongation

Start with the thinking

  • Percent reduction in area uses areas, not diameters — the ratio is squared.
  • Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

  1. Areas

    A0=(π/4)(0.505)2=0.20030in2,Af=(π/4)(0.389)2=0.11876in2A_{0} = (\pi/4)(0.505)^{2} = 0.20030 in^{2}, A_f = (\pi/4)(0.389)^{2} = 0.11876 in^{2}

  2. Formula

    %RA = (A_{0} - A_f)/A_{0} \times 100

  3. Substituting

    %RA = (0.20030 - 0.11876)/0.20030 \times 100 = 40.71%

  4. Formula

    %EL = (L_f - L_{0})/L_{0} \times 100

  5. Substituting

    %EL = (2.640 - 2)/2 \times 100 = 32.00%
Answer:
%RA = 40.7%, %elongation = 32.0%

Why the other options are there

  • 23.0% (diameters instead of areas)
  • 59.3% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Reduction in Area (RA)

© 2026 Dr. Steve Efe. Civil Engineering Capstone Studio. All rights reserved.