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Equilibrium Requirements

Statics · FE Reference Handbook section

Statics
2 formulas
10 exam-style examples
~49 min
All Statics 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
Reactions on a simply supported beam

A 10 m simply supported beam carries a 40 kN point load 3 m from A and a 12 kN/m uniform load over the full span. Find both reactions.

Given

  • L=10mL = 10 m
  • P=40kNat3mfromAP = 40 kN at 3 m from A
  • w=12kN/mover10mw = 12 kN/m over 10 m

Find

R_A and R_B

Start with the thinking

  • Replace the distributed load with its resultant at midspan.
  • Sum moments about A so R_A drops out of that equation.
12 kN/m40 kNPinRollerL = 10 units

Figure 1 — schematic for Reactions on a simply supported beam

Step-by-step solution

  1. UDL resultant

    W=wL=12(10)=120kNat5.00mW = wL = 12(10) = 120 kN at 5.00 m
  2. Moments about A — ΣM_A = 0: R_B(10) = 40(3) + 120(5)

  3. Right side — 120 + 600 = 720 kN·m

  4. Reaction

    RB=720/10=72.0kNR_B = 720/10 = 72.0 kN
  5. Vertical equilibrium

    RA=40+120−72.0=88.0kNR_A = 40 + 120 - 72.0 = 88.0 kN
Answer:
RA=88.0kN,RB=72.0kNR_A = 88.0 kN, R_B = 72.0 kN

Why the other options are there

  • 80/80 kN (point load position ignored)
  • R_A = 72, R_B = 88 (reactions swapped)

Reference: FE Reference Handbook — Statics — Equilibrium of rigid bodies

Example 2
Simply supported beam reaction — solve for far reaction — Equilibrium Requirements

A statics problem uses Simply supported beam reaction. Given point load (P) = 35.5000 kip; load location (a) = 17.5000 ft; span (L) = 14.0000 ft, determine the far reaction (Rb) in kip.

Given

  • pointload(P)=35.5000kippoint load (P) = 35.5000 kip
  • loadlocation(a)=17.5000ftload location (a) = 17.5000 ft
  • span(L)=14.0000ftspan (L) = 14.0000 ft

Find

far reaction (Rb), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Simply supported beam reaction.
  • Everything except Rb is given, so isolate Rb 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:

    Rb=Pa/LR_b = P a / L
  2. Step 2 — Rearrange the relation so that Rb stands alone on the left-hand side.

  3. Step 3 — List the givens: point load (P) = 35.5000 kip, load location (a) = 17.5000 ft, span (L) = 14.0000 ft.

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

  5. Step 5 — Evaluate:

    Rb=44.3750 kipRb = 44.3750\ \text{kip}
  6. Step 6 — Check: returning Rb = 44.3750 kip to

    Rb=Pa/LR_b = P a / L

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

Answer:
Rb=44.3750 kipRb = 44.3750\ \text{kip}

Why the other options are there

  • 88.7500 — kept a factor of two that cancels in the correct rearrangement.
  • 22.1875 — dropped that same factor in the other direction.
  • 48.8125 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 3
Simply supported beam reaction — solve for point load — Equilibrium Requirements (2)

A statics problem uses Simply supported beam reaction. Given load location (a) = 4.0000 ft; span (L) = 16.0000 ft; far reaction (Rb) = 49.7400 kip, determine the point load (P) in kip.

Given

  • loadlocation(a)=4.0000ftload location (a) = 4.0000 ft
  • span(L)=16.0000ftspan (L) = 16.0000 ft
  • farreaction(Rb)=49.7400kipfar reaction (Rb) = 49.7400 kip

Find

point load (P), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Simply supported beam reaction.
  • Everything except P is given, so isolate P 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:

    Rb=Pa/LR_b = P a / L
  2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

  3. Step 3

    Listthegivens:loadlocation(a)=4.0000ft,span(L)=16.0000ft,farreaction(Rb)=49.7400kipList the givens: load location (a) = 4.0000 ft, span (L) = 16.0000 ft, far reaction (Rb) = 49.7400 kip
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    P=199.0 kipP = 199.0\ \text{kip}
  6. Step 6 — Check: returning P = 199.0 kip to

    Rb=Pa/LR_b = P a / L

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

Answer:
P=199.0 kipP = 199.0\ \text{kip}

Why the other options are there

  • 397.9 — kept a factor of two that cancels in the correct rearrangement.
  • 99.4800 — dropped that same factor in the other direction.
  • 218.9 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 4
Simply supported beam reaction — solve for load location — Equilibrium Requirements (3)

A statics problem uses Simply supported beam reaction. Given point load (P) = 44.5000 kip; span (L) = 34.0000 ft; far reaction (Rb) = 47.9400 kip, determine the load location (a) in ft.

Given

  • pointload(P)=44.5000kippoint load (P) = 44.5000 kip
  • span(L)=34.0000ftspan (L) = 34.0000 ft
  • farreaction(Rb)=47.9400kipfar reaction (Rb) = 47.9400 kip

Find

load location (a), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Simply supported beam reaction.
  • Everything except a is given, so isolate a 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:

    Rb=Pa/LR_b = P a / L
  2. Step 2 — Rearrange the relation so that a stands alone on the left-hand side.

  3. Step 3

    Listthegivens:pointload(P)=44.5000kip,span(L)=34.0000ft,farreaction(Rb)=47.9400kipList the givens: point load (P) = 44.5000 kip, span (L) = 34.0000 ft, far reaction (Rb) = 47.9400 kip
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    a=36.6283 fta = 36.6283\ \text{ft}
  6. Step 6 — Check: returning a = 36.6283 ft to

    Rb=Pa/LR_b = P a / L

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

Answer:
a=36.6283 fta = 36.6283\ \text{ft}

Why the other options are there

  • 73.2566 — kept a factor of two that cancels in the correct rearrangement.
  • 18.3142 — dropped that same factor in the other direction.
  • 40.2911 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 5
Simply supported beam reaction — solve for span — Equilibrium Requirements (4)

A statics problem uses Simply supported beam reaction. Given point load (P) = 17.0000 kip; load location (a) = 14.0000 ft; far reaction (Rb) = 48.6700 kip, determine the span (L) in ft.

Given

  • pointload(P)=17.0000kippoint load (P) = 17.0000 kip
  • loadlocation(a)=14.0000ftload location (a) = 14.0000 ft
  • farreaction(Rb)=48.6700kipfar reaction (Rb) = 48.6700 kip

Find

span (L), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Simply supported beam reaction.
  • 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.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Rb=Pa/LR_b = P a / L
  2. Step 2 — Rearrange the relation so that L stands alone on the left-hand side.

  3. Step 3 — List the givens: point load (P) = 17.0000 kip, load location (a) = 14.0000 ft, far reaction (Rb) = 48.6700 kip.

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

  5. Step 5 — Evaluate:

    L=4.8901 ftL = 4.8901\ \text{ft}
  6. Step 6 — Check: returning L = 4.8901 ft to

    Rb=Pa/LR_b = P a / L

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

Answer:
L=4.8901 ftL = 4.8901\ \text{ft}

Why the other options are there

  • 9.7802 — kept a factor of two that cancels in the correct rearrangement.
  • 2.4450 — dropped that same factor in the other direction.
  • 5.3791 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 6
Simply supported beam reaction — solve for far reaction (case 2) — Equilibrium Requirements (5)

A statics problem uses Simply supported beam reaction. Given point load (P) = 28.0000 kip; load location (a) = 4.0000 ft; span (L) = 22.0000 ft, determine the far reaction (Rb) in kip.

Given

  • pointload(P)=28.0000kippoint load (P) = 28.0000 kip
  • loadlocation(a)=4.0000ftload location (a) = 4.0000 ft
  • span(L)=22.0000ftspan (L) = 22.0000 ft

Find

far reaction (Rb), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Simply supported beam reaction.
  • Everything except Rb is given, so isolate Rb 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:

    Rb=Pa/LR_b = P a / L
  2. Step 2 — Rearrange the relation so that Rb stands alone on the left-hand side.

  3. Step 3 — List the givens: point load (P) = 28.0000 kip, load location (a) = 4.0000 ft, span (L) = 22.0000 ft.

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

  5. Step 5 — Evaluate:

    Rb=5.0909 kipRb = 5.0909\ \text{kip}
  6. Step 6 — Check: returning Rb = 5.0909 kip to

    Rb=Pa/LR_b = P a / L

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

Answer:
Rb=5.0909 kipRb = 5.0909\ \text{kip}

Why the other options are there

  • 10.1818 — kept a factor of two that cancels in the correct rearrangement.
  • 2.5455 — dropped that same factor in the other direction.
  • 5.6000 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 7
Simply supported beam reaction — solve for point load (case 2) — Equilibrium Requirements (6)

A statics problem uses Simply supported beam reaction. Given load location (a) = 5.5000 ft; span (L) = 28.0000 ft; far reaction (Rb) = 9.8000 kip, determine the point load (P) in kip.

Given

  • loadlocation(a)=5.5000ftload location (a) = 5.5000 ft
  • span(L)=28.0000ftspan (L) = 28.0000 ft
  • farreaction(Rb)=9.8000kipfar reaction (Rb) = 9.8000 kip

Find

point load (P), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Simply supported beam reaction.
  • Everything except P is given, so isolate P 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:

    Rb=Pa/LR_b = P a / L
  2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

  3. Step 3

    Listthegivens:loadlocation(a)=5.5000ft,span(L)=28.0000ft,farreaction(Rb)=9.8000kipList the givens: load location (a) = 5.5000 ft, span (L) = 28.0000 ft, far reaction (Rb) = 9.8000 kip
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    P=49.8909 kipP = 49.8909\ \text{kip}
  6. Step 6 — Check: returning P = 49.8909 kip to

    Rb=Pa/LR_b = P a / L

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

Answer:
P=49.8909 kipP = 49.8909\ \text{kip}

Why the other options are there

  • 99.7818 — kept a factor of two that cancels in the correct rearrangement.
  • 24.9455 — dropped that same factor in the other direction.
  • 54.8800 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 8
Simply supported beam reaction — solve for load location (case 2) — Equilibrium Requirements (7)

A statics problem uses Simply supported beam reaction. Given point load (P) = 27.5000 kip; span (L) = 33.0000 ft; far reaction (Rb) = 31.6700 kip, determine the load location (a) in ft.

Given

  • pointload(P)=27.5000kippoint load (P) = 27.5000 kip
  • span(L)=33.0000ftspan (L) = 33.0000 ft
  • farreaction(Rb)=31.6700kipfar reaction (Rb) = 31.6700 kip

Find

load location (a), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Simply supported beam reaction.
  • Everything except a is given, so isolate a 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:

    Rb=Pa/LR_b = P a / L
  2. Step 2 — Rearrange the relation so that a stands alone on the left-hand side.

  3. Step 3

    Listthegivens:pointload(P)=27.5000kip,span(L)=33.0000ft,farreaction(Rb)=31.6700kipList the givens: point load (P) = 27.5000 kip, span (L) = 33.0000 ft, far reaction (Rb) = 31.6700 kip
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    a=38.0040 fta = 38.0040\ \text{ft}
  6. Step 6 — Check: returning a = 38.0040 ft to

    Rb=Pa/LR_b = P a / L

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

Answer:
a=38.0040 fta = 38.0040\ \text{ft}

Why the other options are there

  • 76.0080 — kept a factor of two that cancels in the correct rearrangement.
  • 19.0020 — dropped that same factor in the other direction.
  • 41.8044 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 9
Simply supported beam reaction — solve for span (case 2) — Equilibrium Requirements (8)

A statics problem uses Simply supported beam reaction. Given point load (P) = 8.0000 kip; load location (a) = 7.5000 ft; far reaction (Rb) = 31.5200 kip, determine the span (L) in ft.

Given

  • pointload(P)=8.0000kippoint load (P) = 8.0000 kip
  • loadlocation(a)=7.5000ftload location (a) = 7.5000 ft
  • farreaction(Rb)=31.5200kipfar reaction (Rb) = 31.5200 kip

Find

span (L), in ft

Start with the thinking

  • The governing relation printed in this handbook section is Simply supported beam reaction.
  • 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.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Rb=Pa/LR_b = P a / L
  2. Step 2 — Rearrange the relation so that L stands alone on the left-hand side.

  3. Step 3 — List the givens: point load (P) = 8.0000 kip, load location (a) = 7.5000 ft, far reaction (Rb) = 31.5200 kip.

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

  5. Step 5 — Evaluate:

    L=1.9036 ftL = 1.9036\ \text{ft}
  6. Step 6 — Check: returning L = 1.9036 ft to

    Rb=Pa/LR_b = P a / L

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

Answer:
L=1.9036 ftL = 1.9036\ \text{ft}

Why the other options are there

  • 3.8071 — kept a factor of two that cancels in the correct rearrangement.
  • 0.9518 — dropped that same factor in the other direction.
  • 2.0939 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 10
Simply supported beam reaction — solve for far reaction (case 3) — Equilibrium Requirements (9)

A statics problem uses Simply supported beam reaction. Given point load (P) = 9.0000 kip; load location (a) = 3.0000 ft; span (L) = 24.0000 ft, determine the far reaction (Rb) in kip.

Given

  • pointload(P)=9.0000kippoint load (P) = 9.0000 kip
  • loadlocation(a)=3.0000ftload location (a) = 3.0000 ft
  • span(L)=24.0000ftspan (L) = 24.0000 ft

Find

far reaction (Rb), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Simply supported beam reaction.
  • Everything except Rb is given, so isolate Rb 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:

    Rb=Pa/LR_b = P a / L
  2. Step 2 — Rearrange the relation so that Rb stands alone on the left-hand side.

  3. Step 3 — List the givens: point load (P) = 9.0000 kip, load location (a) = 3.0000 ft, span (L) = 24.0000 ft.

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

  5. Step 5 — Evaluate:

    Rb=1.1250 kipRb = 1.1250\ \text{kip}
  6. Step 6 — Check: returning Rb = 1.1250 kip to

    Rb=Pa/LR_b = P a / L

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

Answer:
Rb=1.1250 kipRb = 1.1250\ \text{kip}

Why the other options are there

  • 2.2500 — kept a factor of two that cancels in the correct rearrangement.
  • 0.5625 — dropped that same factor in the other direction.
  • 1.2375 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

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