A simply supported steel beam spans 21 ft and carries a uniform service load of 1.2 kip/ft. With E = 29,000 ksi and I = 819 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 1.2(21)²/8 = 66.15 kip·ft

3. Unit conversion

   $$w = 1.2/12 = 0.1000 kip/in, L = 252 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.1000)(252)^{4} / [384(29,000)(819)]$$

6. Evaluate

   $$\Delta = 0.2211 in$$

7. Limit

   $$L/360 = 252/360 = 0.700 in \to acceptable$$

Why the other options are there

• 0.00013 in (units not converted)
• 0.0442 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

A simply supported steel beam spans 27 ft and carries a uniform service load of 2.4 kip/ft. With E = 29,000 ksi and I = 219 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 2.4(27)²/8 = 218.7 kip·ft

3. Unit conversion

   $$w = 2.4/12 = 0.2000 kip/in, L = 324 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.2000)(324)^{4} / [384(29,000)(219)]$$

6. Evaluate

   $$\Delta = 4.5186 in$$

7. Limit

   $$L/360 = 324/360 = 0.900 in \to too flexible$$

Why the other options are there

• 0.00261 in (units not converted)
• 0.9037 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

A simply supported steel beam spans 33 ft and carries a uniform service load of 1.0 kip/ft. With E = 29,000 ksi and I = 932 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 1.0(33)²/8 = 136.1 kip·ft

3. Unit conversion

   $$w = 1.0/12 = 0.0833 kip/in, L = 396 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.0833)(396)^{4} / [384(29,000)(932)]$$

6. Evaluate

   $$\Delta = 0.9872 in$$

7. Limit

   $$L/360 = 396/360 = 1.100 in \to acceptable$$

Why the other options are there

• 0.00057 in (units not converted)
• 0.1974 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

A simply supported steel beam spans 20 ft and carries a uniform service load of 2.8 kip/ft. With E = 29,000 ksi and I = 228 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 2.8(20)²/8 = 140.0 kip·ft

3. Unit conversion

   $$w = 2.8/12 = 0.2333 kip/in, L = 240 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.2333)(240)^{4} / [384(29,000)(228)]$$

6. Evaluate

   $$\Delta = 1.5245 in$$

7. Limit

   $$L/360 = 240/360 = 0.667 in \to too flexible$$

Why the other options are there

• 0.00088 in (units not converted)
• 0.3049 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

A simply supported steel beam spans 39 ft and carries a uniform service load of 3.4 kip/ft. With E = 29,000 ksi and I = 1231 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 3.4(39)²/8 = 646.4 kip·ft

3. Unit conversion

   $$w = 3.4/12 = 0.2833 kip/in, L = 468 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.2833)(468)^{4} / [384(29,000)(1231)]$$

6. Evaluate

   $$\Delta = 4.9575 in$$

7. Limit

   $$L/360 = 468/360 = 1.300 in \to too flexible$$

Why the other options are there

• 0.00287 in (units not converted)
• 0.9915 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

A simply supported steel beam spans 23 ft and carries a uniform service load of 3.0 kip/ft. With E = 29,000 ksi and I = 1196 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 3.0(23)²/8 = 198.4 kip·ft

3. Unit conversion

   $$w = 3.0/12 = 0.2500 kip/in, L = 276 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.2500)(276)^{4} / [384(29,000)(1196)]$$

6. Evaluate

   $$\Delta = 0.5446 in$$

7. Limit

   $$L/360 = 276/360 = 0.767 in \to acceptable$$

Why the other options are there

• 0.00032 in (units not converted)
• 0.1089 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

A simply supported steel beam spans 38 ft and carries a uniform service load of 3.2 kip/ft. With E = 29,000 ksi and I = 1995 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 3.2(38)²/8 = 577.6 kip·ft

3. Unit conversion

   $$w = 3.2/12 = 0.2667 kip/in, L = 456 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.2667)(456)^{4} / [384(29,000)(1995)]$$

6. Evaluate

   $$\Delta = 2.5949 in$$

7. Limit

   $$L/360 = 456/360 = 1.267 in \to too flexible$$

Why the other options are there

• 0.00150 in (units not converted)
• 0.5190 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

A simply supported steel beam spans 29 ft and carries a uniform service load of 2.8 kip/ft. With E = 29,000 ksi and I = 1219 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 2.8(29)²/8 = 294.3 kip·ft

3. Unit conversion

   $$w = 2.8/12 = 0.2333 kip/in, L = 348 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.2333)(348)^{4} / [384(29,000)(1219)]$$

6. Evaluate

   $$\Delta = 1.2605 in$$

7. Limit

   $$L/360 = 348/360 = 0.967 in \to too flexible$$

Why the other options are there

• 0.00073 in (units not converted)
• 0.2521 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

A simply supported steel beam spans 18 ft and carries a uniform service load of 4.0 kip/ft. With E = 29,000 ksi and I = 1077 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 4.0(18)²/8 = 162.0 kip·ft

3. Unit conversion

   $$w = 4.0/12 = 0.3333 kip/in, L = 216 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.3333)(216)^{4} / [384(29,000)(1077)]$$

6. Evaluate

   $$\Delta = 0.3025 in$$

7. Limit

   $$L/360 = 216/360 = 0.600 in \to acceptable$$

Why the other options are there

• 0.00018 in (units not converted)
• 0.0605 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

A simply supported steel beam spans 36 ft and carries a uniform service load of 1.7 kip/ft. With E = 29,000 ksi and I = 223 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Start with the thinking

• Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
• Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

1. Formula

   $$M_max = w L^{2}/8$$

2. Substituting — M = 1.7(36)²/8 = 275.4 kip·ft

3. Unit conversion

   $$w = 1.7/12 = 0.1417 kip/in, L = 432 in$$

4. Formula

   $$\Delta = 5 w L^{4} / (384 E I)$$

5. Substituting

   $$\Delta = 5(0.1417)(432)^{4} / [384(29,000)(223)]$$

6. Evaluate

   $$\Delta = 9.9343 in$$

7. Limit

   $$L/360 = 432/360 = 1.200 in \to too flexible$$

Why the other options are there

• 0.00575 in (units not converted)
• 1.9869 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams