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Influence Lines for Beams and Trusses

Structural Analysis · FE Reference Handbook section

Structural Analysis
0 formulas
10 exam-style examples
~45 min
All Structural Analysis lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • An influence line shows the variation of an effect (reaction, shear and moment in beams, bar force in a truss) caused by moving
  • a unit load across the structure. An influence line is used to determine the position of a moveable set of loads that causes the

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Truss member force by the method of joints

A symmetric pin-jointed truss of span 12 m and height 4 m carries 60 kN at the apex. Find the force in each of the two inclined top members.

Given

  • Span=12m,height=4mSpan = 12 m, height = 4 m
  • Apexload=60kNApex load = 60 kN
  • Symmetric geometry

Find

Force in the inclined members

Start with the thinking

  • Symmetry means each inclined member carries half the vertical load.
  • Resolve at the apex joint using the member slope.
60 kNF cos θF cos θApex joint

Figure 1 — schematic for Truss member force by the method of joints

Step-by-step solution

  1. Member length

    L=(62+42)=7.21mL = \sqrt(6^{2} + 4^{2}) = 7.21 m
  2. Vertical component per member

    Fv=60/2=30.0kNFv = 60/2 = 30.0 kN
  3. Slope ratio

    L/height=7.21/4=1.803L/height = 7.21/4 = 1.803
  4. Member force

    F=Fv(L/height)=30.0(1.803)F = Fv(L/height) = 30.0(1.803)
  5. Result

    F=54.1kNcompressionineachinclinedmemberF = 54.1 kN compression in each inclined member
Answer:

54.1 kN compression

Why the other options are there

  • 30.0 kN (vertical component reported as member force)
  • 45.0 kN (horizontal projection used)

Reference: FE Reference Handbook — Statics — Trusses

Example 2
Influence line ordinate for a reaction

For a 20 m simple span, what is the influence-line ordinate for R_A when a unit load sits 6 m from B, and what is R_A for a 150 kN axle there?

Given

  • L=20mL = 20 m
  • Load position: 6 m from B (14 m from A)

Find

Ordinate and R_A

Start with the thinking

  • The influence line for R_A is a straight line from 1.0 at A to 0 at B.
  • Ordinate equals the distance from B divided by the span.

Step-by-step solution

  1. Ordinate

    η=(L−x)/LwithxmeasuredfromA\eta = (L - x)/L with x measured from A
  2. Substitute

    η=(20−14)/20=0.300\eta = (20 - 14)/20 = 0.300
  3. Reaction

    RA=Pη=150(0.300)R_A = P\eta = 150(0.300)
  4. Result

    RA=45.0kNR_A = 45.0 kN
  5. Check — R_B = 150 − 45.0 = 105 kN, consistent with the load sitting near B ✓

Answer:
η=0.300,RA=45.0kN\eta = 0.300, R_A = 45.0 kN

Why the other options are there

  • η = 0.700 (measured from the wrong end)
  • R_A = 75 kN (symmetric assumption)

Reference: FE Reference Handbook — Structural Analysis — Influence lines

Example 3
Reaction from influence-line ordinate — solve for reaction — Influence Lines for Beams and Trusses

A structural analysis problem uses Reaction from influence-line ordinate. Given moving load (P) = 16.0000 kip; influence ordinate (eta) = 0.4400, determine the reaction (R) in kip.

Given

  • movingload(P)=16.0000kipmoving load (P) = 16.0000 kip
  • influenceordinate(eta)=0.4400influence ordinate (eta) = 0.4400

Find

reaction (R), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Reaction from influence-line ordinate.
  • Everything except R is given, so isolate R 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.
  • Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=P×ηR = P \times \eta
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3

    Listthegivens:movingload(P)=16.0000kip,influenceordinate(eta)=0.4400List the givens: moving load (P) = 16.0000 kip, influence ordinate (eta) = 0.4400
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=7.0400 kipR = 7.0400\ \text{kip}
  6. Step 6 — Check: returning R = 7.0400 kip to

    R=P×ηR = P \times \eta

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

Answer:
R=7.0400 kipR = 7.0400\ \text{kip}

Why the other options are there

  • 14.0800 — kept a factor of two that cancels in the correct rearrangement.
  • 3.5200 — dropped that same factor in the other direction.
  • 7.7440 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 4
Reaction from influence-line ordinate — solve for moving load — Influence Lines for Beams and Trusses (2)

A structural analysis problem uses Reaction from influence-line ordinate. Given influence ordinate (eta) = 0.5100; reaction (R) = 37.8100 kip, determine the moving load (P) in kip.

Given

  • influenceordinate(eta)=0.5100influence ordinate (eta) = 0.5100
  • reaction(R)=37.8100kipreaction (R) = 37.8100 kip

Find

moving load (P), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Reaction from influence-line ordinate.
  • 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.
  • Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=P×ηR = P \times \eta
  2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

  3. Step 3

    Listthegivens:influenceordinate(eta)=0.5100,reaction(R)=37.8100kipList the givens: influence ordinate (eta) = 0.5100, reaction (R) = 37.8100 kip
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    P=74.1373 kipP = 74.1373\ \text{kip}
  6. Step 6 — Check: returning P = 74.1373 kip to

    R=P×ηR = P \times \eta

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

Answer:
P=74.1373 kipP = 74.1373\ \text{kip}

Why the other options are there

  • 148.3 — kept a factor of two that cancels in the correct rearrangement.
  • 37.0686 — dropped that same factor in the other direction.
  • 81.5510 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 5
Reaction from influence-line ordinate — solve for influence ordinate — Influence Lines for Beams and Trusses (3)

A structural analysis problem uses Reaction from influence-line ordinate. Given moving load (P) = 47.0000 kip; reaction (R) = 52.1600 kip, determine the influence ordinate (eta).

Given

  • movingload(P)=47.0000kipmoving load (P) = 47.0000 kip
  • reaction(R)=52.1600kipreaction (R) = 52.1600 kip

Find

influence ordinate (eta)

Start with the thinking

  • The governing relation printed in this handbook section is Reaction from influence-line ordinate.
  • Everything except eta is given, so isolate eta 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.
  • Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=P×ηR = P \times \eta
  2. Step 2 — Rearrange the relation so that eta stands alone on the left-hand side.

  3. Step 3

    Listthegivens:movingload(P)=47.0000kip,reaction(R)=52.1600kipList the givens: moving load (P) = 47.0000 kip, reaction (R) = 52.1600 kip
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    η=1.1098\eta = 1.1098
  6. Step 6 — Check: returning eta = 1.1098 to

    R=P×ηR = P \times \eta

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

Answer:
η=1.1098\eta = 1.1098

Why the other options are there

  • 2.2196 — kept a factor of two that cancels in the correct rearrangement.
  • 0.5549 — dropped that same factor in the other direction.
  • 1.2208 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 6
Reaction from influence-line ordinate — solve for reaction (case 2) — Influence Lines for Beams and Trusses (4)

A structural analysis problem uses Reaction from influence-line ordinate. Given moving load (P) = 60.0000 kip; influence ordinate (eta) = 0.3200, determine the reaction (R) in kip.

Given

  • movingload(P)=60.0000kipmoving load (P) = 60.0000 kip
  • influenceordinate(eta)=0.3200influence ordinate (eta) = 0.3200

Find

reaction (R), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Reaction from influence-line ordinate.
  • Everything except R is given, so isolate R 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.
  • Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=P×ηR = P \times \eta
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3

    Listthegivens:movingload(P)=60.0000kip,influenceordinate(eta)=0.3200List the givens: moving load (P) = 60.0000 kip, influence ordinate (eta) = 0.3200
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=19.2000 kipR = 19.2000\ \text{kip}
  6. Step 6 — Check: returning R = 19.2000 kip to

    R=P×ηR = P \times \eta

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

Answer:
R=19.2000 kipR = 19.2000\ \text{kip}

Why the other options are there

  • 38.4000 — kept a factor of two that cancels in the correct rearrangement.
  • 9.6000 — dropped that same factor in the other direction.
  • 21.1200 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 7
Reaction from influence-line ordinate — solve for moving load (case 2) — Influence Lines for Beams and Trusses (5)

A structural analysis problem uses Reaction from influence-line ordinate. Given influence ordinate (eta) = 0.6400; reaction (R) = 46.3500 kip, determine the moving load (P) in kip.

Given

  • influenceordinate(eta)=0.6400influence ordinate (eta) = 0.6400
  • reaction(R)=46.3500kipreaction (R) = 46.3500 kip

Find

moving load (P), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Reaction from influence-line ordinate.
  • 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.
  • Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=P×ηR = P \times \eta
  2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

  3. Step 3

    Listthegivens:influenceordinate(eta)=0.6400,reaction(R)=46.3500kipList the givens: influence ordinate (eta) = 0.6400, reaction (R) = 46.3500 kip
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    P=72.4219 kipP = 72.4219\ \text{kip}
  6. Step 6 — Check: returning P = 72.4219 kip to

    R=P×ηR = P \times \eta

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

Answer:
P=72.4219 kipP = 72.4219\ \text{kip}

Why the other options are there

  • 144.8 — kept a factor of two that cancels in the correct rearrangement.
  • 36.2109 — dropped that same factor in the other direction.
  • 79.6641 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 8
Reaction from influence-line ordinate — solve for influence ordinate (case 2) — Influence Lines for Beams and Trusses (6)

A structural analysis problem uses Reaction from influence-line ordinate. Given moving load (P) = 28.0000 kip; reaction (R) = 20.3800 kip, determine the influence ordinate (eta).

Given

  • movingload(P)=28.0000kipmoving load (P) = 28.0000 kip
  • reaction(R)=20.3800kipreaction (R) = 20.3800 kip

Find

influence ordinate (eta)

Start with the thinking

  • The governing relation printed in this handbook section is Reaction from influence-line ordinate.
  • Everything except eta is given, so isolate eta 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.
  • Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=P×ηR = P \times \eta
  2. Step 2 — Rearrange the relation so that eta stands alone on the left-hand side.

  3. Step 3

    Listthegivens:movingload(P)=28.0000kip,reaction(R)=20.3800kipList the givens: moving load (P) = 28.0000 kip, reaction (R) = 20.3800 kip
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    η=0.7279\eta = 0.7279
  6. Step 6 — Check: returning eta = 0.7279 to

    R=P×ηR = P \times \eta

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

Answer:
η=0.7279\eta = 0.7279

Why the other options are there

  • 1.4557 — kept a factor of two that cancels in the correct rearrangement.
  • 0.3639 — dropped that same factor in the other direction.
  • 0.8006 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 9
Reaction from influence-line ordinate — solve for reaction (case 3) — Influence Lines for Beams and Trusses (7)

A structural analysis problem uses Reaction from influence-line ordinate. Given moving load (P) = 31.0000 kip; influence ordinate (eta) = 0.5800, determine the reaction (R) in kip.

Given

  • movingload(P)=31.0000kipmoving load (P) = 31.0000 kip
  • influenceordinate(eta)=0.5800influence ordinate (eta) = 0.5800

Find

reaction (R), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Reaction from influence-line ordinate.
  • Everything except R is given, so isolate R 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.
  • Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=P×ηR = P \times \eta
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3

    Listthegivens:movingload(P)=31.0000kip,influenceordinate(eta)=0.5800List the givens: moving load (P) = 31.0000 kip, influence ordinate (eta) = 0.5800
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=17.9800 kipR = 17.9800\ \text{kip}
  6. Step 6 — Check: returning R = 17.9800 kip to

    R=P×ηR = P \times \eta

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

Answer:
R=17.9800 kipR = 17.9800\ \text{kip}

Why the other options are there

  • 35.9600 — kept a factor of two that cancels in the correct rearrangement.
  • 8.9900 — dropped that same factor in the other direction.
  • 19.7780 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 10
Reaction from influence-line ordinate — solve for moving load (case 3) — Influence Lines for Beams and Trusses (8)

A structural analysis problem uses Reaction from influence-line ordinate. Given influence ordinate (eta) = 0.1000; reaction (R) = 4.0600 kip, determine the moving load (P) in kip.

Given

  • influenceordinate(eta)=0.1000influence ordinate (eta) = 0.1000
  • reaction(R)=4.0600kipreaction (R) = 4.0600 kip

Find

moving load (P), in kip

Start with the thinking

  • The governing relation printed in this handbook section is Reaction from influence-line ordinate.
  • 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.
  • Structural Analysis items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=P×ηR = P \times \eta
  2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

  3. Step 3

    Listthegivens:influenceordinate(eta)=0.1000,reaction(R)=4.0600kipList the givens: influence ordinate (eta) = 0.1000, reaction (R) = 4.0600 kip
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    P=40.6000 kipP = 40.6000\ \text{kip}
  6. Step 6 — Check: returning P = 40.6000 kip to

    R=P×ηR = P \times \eta

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

Answer:
P=40.6000 kipP = 40.6000\ \text{kip}

Why the other options are there

  • 81.2000 — kept a factor of two that cancels in the correct rearrangement.
  • 20.3000 — dropped that same factor in the other direction.
  • 44.6600 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

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