A rectangular area 8 in wide by 23 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 8 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 8(23) = 184 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 8(23)^{3}/12 = 8,111 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(8,111/184) = 6.640 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 8,111 + 184(8)^{2} = 19,887 in^{4}$$

Why the other options are there

• r = 44.08 in (forgot the square root)
• I' = 9,583 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

A rectangular area 13 in wide by 13 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 11 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 13(13) = 169 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 13(13)^{3}/12 = 2,380 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(2,380/169) = 3.753 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 2,380 + 169(11)^{2} = 22,829 in^{4}$$

Why the other options are there

• r = 14.08 in (forgot the square root)
• I' = 4,239 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

A rectangular area 6 in wide by 10 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 7 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 6(10) = 60 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 6(10)^{3}/12 = 500.0 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(500.0/60) = 2.887 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 500.0 + 60(7)^{2} = 3,440 in^{4}$$

Why the other options are there

• r = 8.33 in (forgot the square root)
• I' = 920.0 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

A rectangular area 9 in wide by 15 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 12 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 9(15) = 135 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 9(15)^{3}/12 = 2,531 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(2,531/135) = 4.330 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 2,531 + 135(12)^{2} = 21,971 in^{4}$$

Why the other options are there

• r = 18.75 in (forgot the square root)
• I' = 4,151 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

A rectangular area 7 in wide by 18 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 12 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 7(18) = 126 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 7(18)^{3}/12 = 3,402 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(3,402/126) = 5.196 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 3,402 + 126(12)^{2} = 21,546 in^{4}$$

Why the other options are there

• r = 27.00 in (forgot the square root)
• I' = 4,914 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

A rectangular area 9 in wide by 19 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 3 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 9(19) = 171 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 9(19)^{3}/12 = 5,144 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(5,144/171) = 5.485 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 5,144 + 171(3)^{2} = 6,683 in^{4}$$

Why the other options are there

• r = 30.08 in (forgot the square root)
• I' = 5,657 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

A rectangular area 5 in wide by 14 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 5 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 5(14) = 70 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 5(14)^{3}/12 = 1,143 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(1,143/70) = 4.041 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 1,143 + 70(5)^{2} = 2,893 in^{4}$$

Why the other options are there

• r = 16.33 in (forgot the square root)
• I' = 1,493 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

A rectangular area 13 in wide by 8 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 12 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 13(8) = 104 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 13(8)^{3}/12 = 554.7 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(554.7/104) = 2.309 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 554.7 + 104(12)^{2} = 15,531 in^{4}$$

Why the other options are there

• r = 5.33 in (forgot the square root)
• I' = 1,803 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

A rectangular area 13 in wide by 17 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 7 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 13(17) = 221 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 13(17)^{3}/12 = 5,322 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(5,322/221) = 4.907 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 5,322 + 221(7)^{2} = 16,151 in^{4}$$

Why the other options are there

• r = 24.08 in (forgot the square root)
• I' = 6,869 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

A rectangular area 13 in wide by 19 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 4 in away.

Start with the thinking

• The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
• The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

1. Area

   $$A = b h = 13(19) = 247 in^{2}$$

2. Formula

   $$I_x = b h^{3}/12$$

3. Substituting

   $$I_x = 13(19)^{3}/12 = 7,431 in^{4}$$

4. Formula

   $$r_x = \sqrt(I/A)$$

5. Substituting

   $$r_x = \sqrt(7,431/247) = 5.485 in$$

6. Formula

   $$I' = I + A d^{2}$$

7. Substituting

   $$I' = 7,431 + 247(4)^{2} = 11,383 in^{4}$$

Why the other options are there

• r = 30.08 in (forgot the square root)
• I' = 8,419 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration