A steel beam's factored demand is set by loads (ASCE 7-16) combinations of dead and live load. Given dead load (D) = 69.0000 kip; live load (L) = 25.5000 kip, determine the factored load (W_u) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• Everything except W_u is given, so isolate W_u 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for W_u:

   $$W_{u} = 1.2 D + 1.6 L$$

3. Step 3

   $$List the givens: dead load (D) = 69.0000 kip, live load (L) = 25.5000 kip$$

4. Step 4 — Substitute the given values:

   $$W_{u} = 1.2 69.0000 + 1.6 25.5000$$

5. Step 5 — Evaluate:

   $$W_{u} = 123.6\ \text{kip}$$

6. Step 6 — Check: returning W_u = 123.6 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

• 247.2 — kept a factor of two that cancels in the correct rearrangement.
• 61.8000 — dropped that same factor in the other direction.
• 136.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)

An office floor system is designed under loads (ASCE 7-16) with governing D and L combinations. Given live load (L) = 74.5000 kip; factored load (W_u) = 232.5 kip, determine the dead load (D) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for D:

   $$D = \dfrac{W_u - 1.6 L}{1.2}$$

3. Step 3

   $$List the givens: live load (L) = 74.5000 kip, factored load (W_u) = 232.5 kip$$

4. Step 4 — Substitute the given values:

   $$D = \dfrac{W_u - 1.6 74.5000}{1.2}$$

5. Step 5 — Evaluate:

   $$D = 94.4167\ \text{kip}$$

6. Step 6 — Check: returning D = 94.4167 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

• 188.8 — kept a factor of two that cancels in the correct rearrangement.
• 47.2083 — dropped that same factor in the other direction.
• 103.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)

A roof girder's factored load is computed per loads (ASCE 7-16) provisions. Given dead load (D) = 54.0000 kip; factored load (W_u) = 209.0 kip, determine the live load (L) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• Everything except L is given, so isolate L 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for L:

   $$L = \dfrac{W_u - 1.2 D}{1.6}$$

3. Step 3

   $$List the givens: dead load (D) = 54.0000 kip, factored load (W_u) = 209.0 kip$$

4. Step 4 — Substitute the given values:

   $$L = \dfrac{W_u - 1.2 54.0000}{1.6}$$

5. Step 5 — Evaluate:

   $$L = 90.1250\ \text{kip}$$

6. Step 6 — Check: returning L = 90.1250 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

• 180.2 — kept a factor of two that cancels in the correct rearrangement.
• 45.0625 — dropped that same factor in the other direction.
• 99.1375 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)

A steel beam's factored demand is set by loads (ASCE 7-16) combinations of dead and live load. Given dead load (D) = 63.0000 kip; live load (L) = 18.0000 kip, determine the factored load (W_u) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• Everything except W_u is given, so isolate W_u 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for W_u:

   $$W_{u} = 1.2 D + 1.6 L$$

3. Step 3

   $$List the givens: dead load (D) = 63.0000 kip, live load (L) = 18.0000 kip$$

4. Step 4 — Substitute the given values:

   $$W_{u} = 1.2 63.0000 + 1.6 18.0000$$

5. Step 5 — Evaluate:

   $$W_{u} = 104.4\ \text{kip}$$

6. Step 6 — Check: returning W_u = 104.4 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

• 208.8 — kept a factor of two that cancels in the correct rearrangement.
• 52.2000 — dropped that same factor in the other direction.
• 114.8 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)

An office floor system is designed under loads (ASCE 7-16) with governing D and L combinations. Given live load (L) = 37.0000 kip; factored load (W_u) = 169.0 kip, determine the dead load (D) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for D:

   $$D = \dfrac{W_u - 1.6 L}{1.2}$$

3. Step 3

   $$List the givens: live load (L) = 37.0000 kip, factored load (W_u) = 169.0 kip$$

4. Step 4 — Substitute the given values:

   $$D = \dfrac{W_u - 1.6 37.0000}{1.2}$$

5. Step 5 — Evaluate:

   $$D = 91.5000\ \text{kip}$$

6. Step 6 — Check: returning D = 91.5000 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

• 183.0 — kept a factor of two that cancels in the correct rearrangement.
• 45.7500 — dropped that same factor in the other direction.
• 100.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)

A roof girder's factored load is computed per loads (ASCE 7-16) provisions. Given dead load (D) = 57.0000 kip; factored load (W_u) = 137.5 kip, determine the live load (L) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• Everything except L is given, so isolate L 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for L:

   $$L = \dfrac{W_u - 1.2 D}{1.6}$$

3. Step 3

   $$List the givens: dead load (D) = 57.0000 kip, factored load (W_u) = 137.5 kip$$

4. Step 4 — Substitute the given values:

   $$L = \dfrac{W_u - 1.2 57.0000}{1.6}$$

5. Step 5 — Evaluate:

   $$L = 43.1875\ \text{kip}$$

6. Step 6 — Check: returning L = 43.1875 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

• 86.3750 — kept a factor of two that cancels in the correct rearrangement.
• 21.5938 — dropped that same factor in the other direction.
• 47.5063 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)

A steel beam's factored demand is set by loads (ASCE 7-16) combinations of dead and live load. Given dead load (D) = 79.0000 kip; live load (L) = 71.0000 kip, determine the factored load (W_u) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• Everything except W_u is given, so isolate W_u 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for W_u:

   $$W_{u} = 1.2 D + 1.6 L$$

3. Step 3

   $$List the givens: dead load (D) = 79.0000 kip, live load (L) = 71.0000 kip$$

4. Step 4 — Substitute the given values:

   $$W_{u} = 1.2 79.0000 + 1.6 71.0000$$

5. Step 5 — Evaluate:

   $$W_{u} = 208.4\ \text{kip}$$

6. Step 6 — Check: returning W_u = 208.4 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

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

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)

An office floor system is designed under loads (ASCE 7-16) with governing D and L combinations. Given live load (L) = 69.5000 kip; factored load (W_u) = 194.5 kip, determine the dead load (D) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for D:

   $$D = \dfrac{W_u - 1.6 L}{1.2}$$

3. Step 3

   $$List the givens: live load (L) = 69.5000 kip, factored load (W_u) = 194.5 kip$$

4. Step 4 — Substitute the given values:

   $$D = \dfrac{W_u - 1.6 69.5000}{1.2}$$

5. Step 5 — Evaluate:

   $$D = 69.4167\ \text{kip}$$

6. Step 6 — Check: returning D = 69.4167 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

• 138.8 — kept a factor of two that cancels in the correct rearrangement.
• 34.7083 — dropped that same factor in the other direction.
• 76.3583 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)

A roof girder's factored load is computed per loads (ASCE 7-16) provisions. Given dead load (D) = 97.5000 kip; factored load (W_u) = 185.0 kip, determine the live load (L) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• Everything except L is given, so isolate L 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for L:

   $$L = \dfrac{W_u - 1.2 D}{1.6}$$

3. Step 3

   $$List the givens: dead load (D) = 97.5000 kip, factored load (W_u) = 185.0 kip$$

4. Step 4 — Substitute the given values:

   $$L = \dfrac{W_u - 1.2 97.5000}{1.6}$$

5. Step 5 — Evaluate:

   $$L = 42.5000\ \text{kip}$$

6. Step 6 — Check: returning L = 42.5000 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

• 85.0000 — kept a factor of two that cancels in the correct rearrangement.
• 21.2500 — dropped that same factor in the other direction.
• 46.7500 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)

A steel beam's factored demand is set by loads (ASCE 7-16) combinations of dead and live load. Given dead load (D) = 46.0000 kip; live load (L) = 45.5000 kip, determine the factored load (W_u) in kip.

Start with the thinking

• The governing relation printed in this handbook section is LRFD load combination (ASCE 7-16).
• Everything except W_u is given, so isolate W_u 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.
• Loads (ASCE 7-16) requires combining dead and live loads with LRFD load factors to find the factored design load.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$W_u = 1.2 D + 1.6 L$$

2. Step 2 — Rearrange symbolically for W_u:

   $$W_{u} = 1.2 D + 1.6 L$$

3. Step 3

   $$List the givens: dead load (D) = 46.0000 kip, live load (L) = 45.5000 kip$$

4. Step 4 — Substitute the given values:

   $$W_{u} = 1.2 46.0000 + 1.6 45.5000$$

5. Step 5 — Evaluate:

   $$W_{u} = 128.0\ \text{kip}$$

6. Step 6 — Check: returning W_u = 128.0 kip to

   $$W_u = 1.2 D + 1.6 L$$

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

Why the other options are there

• 256.0 — kept a factor of two that cancels in the correct rearrangement.
• 64.0000 — dropped that same factor in the other direction.
• 140.8 — rounded an intermediate value before the final step.

Reference: FE Handbook — Structural Design: Loads (ASCE 7-16)