A T-section has a 200 mm × 40 mm flange on top of a 40 mm × 260 mm web. Find the centroid from the bottom and I about the centroidal axis.

Start with the thinking

• Take areas and their own centroids first, then use the parallel-axis theorem.
• Measure all distances from a single datum — the bottom fibre.

Step-by-step solution

1. Areas

   $$A_web = 40(260) = 10,400 mm^{2} and A_flange = 200(40) = 8,000 mm^{2}$$

2. Local centroids

   $$y_web = 130 mm and y_flange = 260 + 20 = 280 mm$$

3. Centroid

   $$ȳ = [10,400(130) + 8,000(280)]/18,400 = (1,352,000 + 2,240,000)/18,400 = 195.2 mm$$

4. Web term

   $$I = 40(260)^{3}/12 + 10,400(195.2 - 130)^{2} = 58.6\times10^{6} + 44.2\times10^{6} = 102.8\times10^{6} mm^{4}$$

5. Flange term

   $$I = 200(40)^{3}/12 + 8,000(280 - 195.2)^{2} = 1.07\times10^{6} + 57.5\times10^{6} = 58.6\times10^{6} mm^{4}$$

6. Total

   $$I_x = 102.8\times10^{6} + 58.6\times10^{6} = 161\times10^{6} mm^{4}$$

Why the other options are there

• I = 59.7 × 10⁶ mm⁴ (parallel-axis terms omitted)
• ȳ = 150 mm (areas averaged, not weighted)

Reference: FE Reference Handbook — Statics — Centroids and moments of inertia

A statics problem uses Moment of a force. Given force (F) = 1,232 lb; moment arm (d) = 5.0000 ft, determine the moment (M) in lb·ft.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force.
• Everything except M is given, so isolate M 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.
• Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M = F \times d$$

2. Step 2 — Rearrange the relation so that M stands alone on the left-hand side.

3. Step 3

   $$List the givens: force (F) = 1,232 lb, moment arm (d) = 5.0000 ft$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$M = 6160\ \text{lb·ft}$$

6. Step 6 — Check: returning M = 6,160 lb·ft to

   $$M = F \times d$$

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

Why the other options are there

• 12,320 — kept a factor of two that cancels in the correct rearrangement.
• 3,080 — dropped that same factor in the other direction.
• 6,776 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

a cantilevered sign arm Given force (F) = 3,441 N; perpendicular distance (d) = 0.7000 m, determine the moment (M) in N\cdot m.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force about a point.
• Everything except M is given, so isolate M 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.
• A force acting at a perpendicular distance d produces a moment.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M = F d$$

2. Step 2 — Rearrange symbolically for M:

   $$M = M = F d$$

3. Step 3

   $$List the givens: force (F) = 3,441 N, perpendicular distance (d) = 0.7000 m$$

4. Step 4 — Substitute the given values:

   $$M = M = 3441 0.7000$$

5. Step 5 — Evaluate:

   $$M = 2409\ \text{N\cdot m}$$

6. Step 6 — Check: returning M = 2,409 N\cdot m to

   $$M = F d$$

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

Why the other options are there

• 4,817 — kept a factor of two that cancels in the correct rearrangement.
• 1,204 — dropped that same factor in the other direction.
• 2,649 — rounded an intermediate value before the final step.

Reference: FE Handbook — Statics (Moments)

A rectangular concrete section's moment of inertia is used in a stiffness calculation. Given width (b) = 8.5000 in; height (h) = 15.4000 in, determine the moment of inertia (I) in in^4.

Start with the thinking

• The governing relation printed in this handbook section is Rectangular moment of inertia.
• Everything except I is given, so isolate I 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.
• The moment of inertia of a rectangular cross section about its centroidal axis governs bending stiffness.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$I = \dfrac{b h^3}{12}$$

2. Step 2 — Rearrange symbolically for I:

   $$I = \dfrac{b h^3}{12}$$

3. Step 3

   $$List the givens: width (b) = 8.5000 in, height (h) = 15.4000 in$$

4. Step 4 — Substitute the given values:

   $$I = \dfrac{8.5000 15.4000^3}{12}$$

5. Step 5 — Evaluate:

   I = 2587\ \text{in^4}

6. Step 6 — Check: returning I = 2,587 in^4 to

   $$I = \dfrac{b h^3}{12}$$

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

Why the other options are there

• 5,174 — kept a factor of two that cancels in the correct rearrangement.
• 1,294 — dropped that same factor in the other direction.
• 2,846 — rounded an intermediate value before the final step.

Reference: FE Handbook — Statics: Moment of Inertia

A statics problem uses Moment of a force. Given moment arm (d) = 7.5000 ft; moment (M) = 16,502 lb·ft, determine the force (F) in lb.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force.
• Everything except F is given, so isolate F 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.
• Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M = F \times d$$

2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

3. Step 3 — List the givens: moment arm (d) = 7.5000 ft, moment (M) = 16,502 lb·ft.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F = 2200\ \text{lb}$$

6. Step 6 — Check: returning F = 2,200 lb to

   $$M = F \times d$$

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

Why the other options are there

• 4,401 — kept a factor of two that cancels in the correct rearrangement.
• 1,100 — dropped that same factor in the other direction.
• 2,420 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

a wrench applied to an anchor nut Given moment (M) = 4,687 N\cdot m; perpendicular distance (d) = 1.9000 m, determine the force (F) in N.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force about a point.
• Everything except F is given, so isolate F 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.
• A force acting at a perpendicular distance d produces a moment.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M = F d$$

2. Step 2 — Rearrange symbolically for F:

   $$F = F = \dfrac{M}{d}$$

3. Step 3

   $$List the givens: moment (M) = 4,687 N\cdot m, perpendicular distance (d) = 1.9000 m$$

4. Step 4 — Substitute the given values:

   $$F = F = \dfrac{4687}{1.9000}$$

5. Step 5 — Evaluate:

   $$F = 2467\ \text{N}$$

6. Step 6 — Check: returning F = 2,467 N to

   $$M = F d$$

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

Why the other options are there

• 4,933 — kept a factor of two that cancels in the correct rearrangement.
• 1,233 — dropped that same factor in the other direction.
• 2,713 — rounded an intermediate value before the final step.

Reference: FE Handbook — Statics (Moments)

A rectangular wood beam's moment of inertia is needed for a deflection check. Given height (h) = 11.0000 in; moment of inertia (I) = 11,078 in^4, determine the width (b) in in.

Start with the thinking

• The governing relation printed in this handbook section is Rectangular moment of inertia.
• Everything except b is given, so isolate b 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.
• The moment of inertia of a rectangular cross section about its centroidal axis governs bending stiffness.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$I = \dfrac{b h^3}{12}$$

2. Step 2 — Rearrange symbolically for b:

   $$b = \dfrac{12 I}{h^3}$$

3. Step 3

   $$List the givens: height (h) = 11.0000 in, moment of inertia (I) = 11,078 in^4$$

4. Step 4 — Substitute the given values:

   $$b = \dfrac{12 11078}{11.0000^3}$$

5. Step 5 — Evaluate:

   $$b = 99.8768\ \text{in}$$

6. Step 6 — Check: returning b = 99.8768 in to

   $$I = \dfrac{b h^3}{12}$$

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

Why the other options are there

• 199.8 — kept a factor of two that cancels in the correct rearrangement.
• 49.9384 — dropped that same factor in the other direction.
• 109.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Statics: Moment of Inertia

A statics problem uses Moment of a force. Given force (F) = 819.0 lb; moment (M) = 8,453 lb·ft, determine the moment arm (d) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force.
• 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.
• Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M = F \times d$$

2. Step 2 — Rearrange the relation so that d stands alone on the left-hand side.

3. Step 3 — List the givens: force (F) = 819.0 lb, moment (M) = 8,453 lb·ft.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$d = 10.3211\ \text{ft}$$

6. Step 6 — Check: returning d = 10.3211 ft to

   $$M = F \times d$$

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

Why the other options are there

• 20.6422 — kept a factor of two that cancels in the correct rearrangement.
• 5.1606 — dropped that same factor in the other direction.
• 11.3532 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

a bracket bolted to a column flange Given moment (M) = 7,286 N\cdot m; force (F) = 1,936 N, determine the perpendicular distance (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force about a point.
• 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.
• A force acting at a perpendicular distance d produces a moment.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M = F d$$

2. Step 2 — Rearrange symbolically for d:

   $$d = d = \dfrac{M}{F}$$

3. Step 3

   $$List the givens: moment (M) = 7,286 N\cdot m, force (F) = 1,936 N$$

4. Step 4 — Substitute the given values:

   $$d = d = \dfrac{7286}{1936}$$

5. Step 5 — Evaluate:

   $$d = 3.7627\ \text{m}$$

6. Step 6 — Check: returning d = 3.7627 m to

   $$M = F d$$

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

Why the other options are there

• 7.5255 — kept a factor of two that cancels in the correct rearrangement.
• 1.8814 — dropped that same factor in the other direction.
• 4.1390 — rounded an intermediate value before the final step.

Reference: FE Handbook — Statics (Moments)

A steel plate's moment of inertia about its centroidal axis is computed for buckling. Given width (b) = 10.9000 in; moment of inertia (I) = 13,423 in^4, determine the height (h) in in.

Start with the thinking

• The governing relation printed in this handbook section is Rectangular moment of inertia.
• Everything except h is given, so isolate h 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.
• The moment of inertia of a rectangular cross section about its centroidal axis governs bending stiffness.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$I = \dfrac{b h^3}{12}$$

2. Step 2 — Rearrange symbolically for h:

   $$h = \sqrt[3]{\dfrac{12 I}{b}}$$

3. Step 3

   $$List the givens: width (b) = 10.9000 in, moment of inertia (I) = 13,423 in^4$$

4. Step 4 — Substitute the given values:

   $$h = \sqrt[3]{\dfrac{12 13423}{10.9000}}$$

5. Step 5 — Evaluate:

   $$h = 24.5396\ \text{in}$$

6. Step 6 — Check: returning h = 24.5396 in to

   $$I = \dfrac{b h^3}{12}$$

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

Why the other options are there

• 49.0793 — kept a factor of two that cancels in the correct rearrangement.
• 12.2698 — dropped that same factor in the other direction.
• 26.9936 — rounded an intermediate value before the final step.

Reference: FE Handbook — Statics: Moment of Inertia