Beam Analysis

Key Vocabulary

Beam Horizontal member resisting vertical loads and moments. Support Connection point: pin, roller, or fixed. Reaction Force/moment from support countering loads. Shear Force (V) Internal force perpendicular to beam axis. Bending Moment (M) Internal twisting effect causing bending. Point Load Concentrated force at a point. Distributed Load Uniform or varying load over length. Equilibrium Sum of forces/moments = 0.

Why We Make These Simplifications

Real beams are complicated, but for most school problems and many real designs, these assumptions work great and keep the math doable.

The beam is straight and horizontal

We only apply straight-down (vertical) loads — no sideways or angled forces.

Deflections are small

The beam bends just a little bit, so our equations stay accurate and simple.

Material is consistent

The beam is made of the same material all the way through (no sudden changes in strength).

No stretching or squeezing along its length

We ignore forces that would compress or stretch the beam lengthwise (axial forces).

We neglect shear deformation

We assume the beam bends without the cross-section warping much — this is the basic beam theory you’ll use most often.

Statically determinate

The supports give us exactly enough unknowns to solve using just three equilibrium equations (∑Fₓ=0, ∑Fᵧ=0, ∑M=0).

Real-world note: Bridges, floor beams, and diving boards are designed using these same ideas. When a beam gets very deep or short, engineers use more advanced models — but 90% of the time, these simplifications are plenty!

Identifying Components

Before analyzing a beam, correctly identify its basic parts:

Pin Support Triangle with hatching. Resists horizontal (R_AX) and vertical (R_AY) forces — 2 reactions. Roller Support Circle on hatched surface. Resists vertical force only (R_BY) — 1 reaction. Fixed Support Wall-like hatch. Resists horizontal (R_AX), vertical (R_AY), and moment (M_A) — 3 reactions. Point Load Arrow at a specific position. Distributed Load Uniform load over a segment.

Tip: Examine the FBD to count supports and loads.

Static Determinacy

A plane beam is statically determinate (solvable using only equilibrium) when the number of unknown reactions equals the number of available equations.

R = 3

Rnumber of reaction components (forces + moments)

R = 3 → determinate (solvable)
R < 3 → unstable (mechanism)
R > 3 → indeterminate (needs additional methods)

Why 3? We have exactly 3 equilibrium equations for a 2D beam: ∑F_x=0, ∑F_y=0, ∑M=0.

Unknowns per support type:
Pin/Hinge → 2 unknowns (H + V)
Roller → 1 unknown (V)
Fixed → 3 unknowns (H + V + M)

Note about pins: You can take moments about a pin (the two forces there drop out of the equation). But the pin itself provides no moment resistance — that's why M = 0 at pins in the moment diagram.

Propped cantilever: A fixed end (3 unknowns) + roller (1 more) → R=4 total. That extra support makes it stiffer but harder to solve with just equilibrium.

Sign Conventions

Signs help us know the direction of forces and moments.
Positive (+) and negative (-) depend on the direction.

Forces:
+x = → (to the right)
-x = ← (to the left)
+y = ↑ (upward)
-y = ↓ (downward)

Moments (rotation):
+ = ↺ (counterclockwise)
- = ↻ (clockwise)

Finding Reactions: Moment Equilibrium

The fastest way to eliminate unknowns is to sum moments about a point where unknowns act — those unknowns drop out of the equation.

∑M_point = 0

How to choose where to take moments:

Pick a point where the most unknowns act. Forces passing through that point produce zero moment and vanish from the equation.
For a simply supported beam, take moments about one support to immediately find the reaction at the other support.
For a cantilever or propped cantilever, taking moments about the fixed end eliminates H_A and V_A (but M_A stays in the equation).

Moment of a force = Force × perpendicular distance

A point load P at distance d from the moment point: M = P × d.
A distributed load w over length ℓ: replace it with a single resultant force (w×ℓ) acting at the centroid (midpoint for uniform loads). Then take that resultant’s moment.
Sign: forces that would rotate the beam counterclockwise about the point are positive (+); clockwise are negative (−).

Example — Simply Supported (Problem 1)
L = 10 m, P = 800 N downward at x = 4 m. Pin at A (x = 0), roller at B (x = 10).

Take moments about A (eliminates V_A and H_A):
∑M_A = 0 → V_B·10 − 800·4 = 0 → V_B = 320 N

Example — Cantilever (Problem 2)
L = 8 m, P = 1200 N downward at x = 8 m. Fixed at A (x = 0).

Take moments about A:
∑M_A = 0 → M_A + V_A·0 − 1200·8 = 0 → M_A = 9600 N·m
(This is the reaction moment; the internal moment at the wall is −9600 N·m.)

Distributed load tip: A uniform load of 50 N/m over 4 m → resultant = 200 N acting at the midpoint of the loaded region. Use this single force in your moment equation.

Finding Reactions: Force Equilibrium

Every beam must be in static equilibrium — all forces and moments balance to zero. This section walks through the complete process of using equilibrium to solve for unknown reactions.

The Big Idea:

If a beam is sitting still (not accelerating or spinning), then every force pushing it one way must be balanced by forces pushing it the other way, and every torque trying to spin it must be balanced by opposing torques. This gives us three equations we can write down and solve.

The three equilibrium equations (2-D):

∑F_x = 0 — all horizontal forces cancel (left = right)
∑F_y = 0 — all vertical forces cancel (up = down)
∑M_{any point} = 0 — all moments about any chosen point cancel (CW = CCW)

Step 0 — Draw a Free-Body Diagram (FBD):

The FBD is the foundation of every statics problem. Remove the supports and replace them with unknown reaction forces:

Isolate the beam — mentally “cut” it free from its supports:
Pin / hinge → H_A (horizontal) + V_A (vertical) — 2 unknowns
Roller → V_B (perpendicular to rolling surface) — 1 unknown
Fixed wall → H_A + V_A + M_A (moment) — 3 unknowns
Show all applied loads — point loads as single arrows; distributed loads as a line of arrows with a connecting bar. Label every magnitude and position.
Assume positive directions — draw reactions in their assumed positive direction (↑ for V, → for H, ↺ for M). If the algebra returns a negative value, the actual direction is simply opposite to what you assumed.

Step 1 — Choose a moment point wisely:

The key strategy is to eliminate unknowns from the moment equation by summing moments about a point where unknown forces act (their moment arms become zero).

For simply supported beams: take ∑M_A = 0 about the pin. This eliminates both H_A (zero arm) and V_A (zero arm), leaving only V_B to solve for.
For cantilevers: take ∑M_A = 0 about the fixed end. This eliminates H_A and V_A, leaving M_A as the only unknown.

Why moments first? Because a cleverly chosen pivot eliminates the most unknowns in a single equation. Force sums (∑F_x, ∑F_y) keep all unknowns — you can only use them easily after you’ve already found some reactions.

Step 1b — Handle distributed loads in moment equations:

You cannot take the moment of a distributed load directly. Instead, replace it with a single resultant force:

Resultant = w × ℓ (intensity × loaded length)
Acts at the centroid of the loaded region (midpoint for uniform loads)

Example: 50 N/m over 4 m → resultant = 200 N, acting at the midpoint of the loaded span.

Step 2 — Solve force equilibrium:

∑F_y = 0: set all upward forces equal to all downward forces. With one reaction already known from Step 1, this directly gives the remaining vertical reaction.
∑F_x = 0: with vertical-only loading (which is every problem in this tool), H_A = 0 immediately.

Step 3 — Verify your answers:

Take moments about a different point (e.g., ∑M_B). Plug in all your solved reactions. If the sum equals zero, your work is correct.
Sanity check: the support closer to the load should carry a larger share of the total force.
Check: all upward reactions should sum to the total downward load.

Simply Supported — step-by-step recipe:

∑M_A = 0 → solve for V_B (all loads × their distances from A = V_B × L)
∑F_y = 0 → V_A = total downward load − V_B
∑F_x = 0 → H_A = 0 (no horizontal loads)
Verify: ∑M_B = 0?

Cantilever — step-by-step recipe:

∑F_y = 0 → V_A = total downward load
∑M_A = 0 → M_A(reaction) = sum of (each load × its distance from A)
Internal moment convention: M_A is negative (hogging) when loads pull the beam downward away from the wall
∑F_x = 0 → H_A = 0

Overhanging Beam (R = 3, determinate):

Pin at A (left end) → R_AX, R_AY. Roller at B (interior) → R_BY. Free end extends beyond the roller.
∑M_A = 0: Note that R_BY acts at the roller position (not at the beam end).
∑F_y = 0: R_AY can be negative (downward!) if the overhang load is large.
At the free end: V = 0 and M = 0 — use these as verification checks.

Worked Example — Problem 1 (SS, P = 800 N at 4 m, L = 10 m)

FBD unknowns: V_A, H_A at pin; V_B at roller.

1. ∑M_A = 0: V_B·10 − 800·4 = 0 → V_B = 320 N
2. ∑F_y = 0: V_A + 320 − 800 = 0 → V_A = 480 N
3. ∑F_x = 0: H_A = 0
4. Verify ∑M_B: 480·10 − 800·6 = 4800 − 4800 = 0 ✓

Sanity check: Load is at 4 m (closer to A), so V_A (480 N) > V_B (320 N). ✓

Worked Example — Problem 3 (SS, w = 50 N/m from 2–6 m, L = 8 m)

FBD unknowns: V_A, H_A at pin; V_B at roller.
Resultant: 50 × 4 = 200 N at centroid x = (2+6)/2 = 4 m.

1. ∑M_A = 0: V_B·8 − 200·4 = 0 → V_B = 100 N
2. ∑F_y = 0: V_A + 100 − 200 = 0 → V_A = 100 N
3. ∑F_x = 0: H_A = 0
4. Verify: Load is symmetric about midpoint → V_A = V_B. ✓

Worked Example — Problem 4 (Cantilever, w = 60 N/m over 0–4 m, L = 6 m)

FBD unknowns: V_A, H_A, M_A at fixed wall.
Resultant of distributed load: 60 × 4 = 240 N at centroid x = 2 m.

1. ∑F_y = 0: V_A − 240 = 0 → V_A = 240 N
2. ∑M_A = 0: M_A(rxn) − 240·2 = 0 → M_A(rxn) = 480 N·m (CCW)
Internal moment at wall: M_A = −480 N·m (hogging)
3. ∑F_x = 0: H_A = 0

Worked Example — Problem 5 (SS, P = 400 N at 4 m + w = 20 N/m from 2–8 m, L = 10 m)

FBD unknowns: V_A, H_A at pin; V_B at roller.
Resultant of dist load: 20 × 6 = 120 N at centroid x = (2+8)/2 = 5 m.
Total load: 400 + 120 = 520 N.

1. ∑M_A = 0: V_B·10 − 400·4 − 120·5 = 0
V_B·10 = 1600 + 600 = 2200 → V_B = 220 N
2. ∑F_y = 0: V_A + 220 − 520 = 0 → V_A = 300 N
3. ∑F_x = 0: H_A = 0
4. Verify ∑M_B: 300·10 − 400·6 − 120·5 = 3000 − 2400 − 600 = 0 ✓

Worked Example — Problem 9 (Overhanging, P₁ = 300 N at 3 m, P₂ = 150 N at 8 m, L = 9 m, roller at 6 m)

FBD unknowns: R_AX, R_AY at pin (A, x = 0); R_BY at roller (B, x = 6 m). (R = 3)

1. ∑M_A = 0: R_BY·6 − 300·3 − 150·8 = 0
R_BY·6 = 900 + 1200 = 2100 → R_BY = 350 N
2. ∑F_y = 0: R_AY + 350 − 300 − 150 = 0 → R_AY = 100 N
3. ∑F_x = 0: R_AX = 0
4. Verify ∑M_B: 100·6 − 300·3 + 150·2 = 600 − 900 + 300 = 0 ✓

Note: The overhanging load (P₂ at x = 8) creates a moment about A that is larger per newton than the interior load, because its moment arm (8 m) exceeds the roller’s moment arm (6 m).

Distributed load shortcut: Replace with a single resultant force = w × length, acting at the centroid (midpoint for uniform loads).
Example: 100 N/m over 8 m → resultant = 800 N at x = 4 m (center of loaded span).

Common mistakes:

Forgetting to include all reactions in the force sum
Wrong sign on a downward load (should be negative in ∑F_y)
Not replacing a distributed load with its resultant for moment calculations
Placing the resultant at the wrong centroid — it goes at the center of the loaded span, not the center of the beam
Confusing H_A with V_A
Using internal moment sign when the equilibrium equation uses the reaction moment
Forgetting the verification step — always check with ∑M about a different point

Dividing the Beam into Segments

Split the beam wherever the loading changes so each segment has a simple expression for V and M.

Mark every support, point load, start/end of a distributed load, and applied moment.
Create segments between consecutive markers (e.g. 0–2 m, 2–6 m).
Within each segment, V changes only due to distributed loads.

Pitfall: Missing the end of a distributed load as a segment boundary, or treating a point load as affecting only one side.

Shear Force Diagram Rules

The Shear Force Diagram shows how internal shear changes along the beam, constructed left to right.

Positive Shear (V): Left part pushed upward relative to the right part.

Key behavior: Reactions act upward, so they contribute positive shear. Downward loads cause negative changes.

Start at left end: V = left reaction (upward reaction → positive jump in V).
Between supports with only point loads → V is constant (horizontal line) within each segment.
At a downward point load P → V drops abruptly by exactly P. This is a sudden, vertical jump down on the diagram.
At an upward reaction R → V jumps abruptly upward by exactly R.
Downward Uniformly Distributed Load of intensity w → straight line sloping downward, slope = −w.
At a free (unsupported) end: V = 0 (no support means no shear).
At a fixed support: V equals the total of all loads the support must resist.

Common mistakes: Forgetting the sudden drop at a point load; wrong slope direction under Uniformly Distributed Load; starting V at zero instead of the left reaction; confusing constant shear (point loads only) with linearly changing shear (distributed loads).

Bending Moment Diagram Rules

The Bending Moment Diagram shows how internal bending varies along the beam.

Positive Moment (M): Sagging appears as a smile shape with tension on bottom and compression on top. Downward loads always produce positive bending moment.

M = 0 at pinned and roller supports (simply-supported ends). Not at fixed supports.
At a free (unsupported) end: M = 0 (nothing to create a moment).
At a fixed support: M is at its maximum value; the fixed end carries the accumulated moment from all loads.
No jumps in moment: Unlike shear, the moment diagram changes continuously with no sudden jumps at point loads. Moment changes linearly between key points (for point-load-only segments).
Slope of Bending Moment Diagram at any point = value of shear V at that point.
Under a Uniformly Distributed Load → Bending Moment Diagram is a parabolic curve (not a straight line!).
ΔM between two points = area under the Shear Force Diagram between those points.
Max moment occurs where V = 0 or changes sign.

Common mistakes: Drawing Bending Moment Diagram as straight lines under Uniformly Distributed Load (must be curved); expecting M = 0 at a fixed support (it is maximum there, not zero); expecting jumps in moment at point loads (moment is always continuous).

Common Pitfalls by Phase

FBD (Free Body Diagram):

Forgetting to show a support reaction.
Drawing a load in the wrong direction.
Missing an applied couple/moment.

Calculating Reactions:

Wrong sign on a moment arm (CW vs CCW).
Choosing a moment point that still leaves two unknowns.
Arithmetic errors when solving simultaneous equations.

Shear Force Diagram:

Forgetting the sudden drop at a point load (shear changes abruptly at point loads).
Wrong slope under Uniformly Distributed Load (should be negative for downward load).
Starting V at zero instead of the left reaction (upward reactions create positive shear).
Expecting shear to change between supports when only point loads are present (it stays constant in those segments).

Bending Moment Diagram:

Drawing Bending Moment Diagram as straight lines under Uniformly Distributed Load (must be parabolic).
Expecting jumps in moment at point loads (moment is always continuous with no jumps).
Expecting M = 0 at a fixed support (M is maximum there; M = 0 is at pin/roller/free ends).
Bending Moment Diagram not returning to zero at pin/roller supports for simply-supported beams.

Check Your Work

Use these quick checks to catch errors before you finish:

Simply-supported beams: Bending Moment Diagram must start and end at zero (M = 0 at pin and roller supports).
Cantilevers: M = 0 at the free end; max |M| at the fixed support. V = 0 at the free end.
Moment diagram is always continuous with no jumps, even at point loads. Changes are smooth and linear (for point loads) or curved (for distributed loads).
Area under Shear Force Diagram between two points must equal the change in Bending Moment Diagram (ΔM).
Maximum moment occurs where shear = 0 or changes sign.
Every point load causes an exact, abrupt jump in shear with no rounding errors.
Shear is constant between point loads when no distributed load is present.
All signs must match the sign conventions.

Final sanity check: Sum of upward reactions = sum of all downward loads. For fixed supports, the reaction shear and moment are the maximum values that carry all applied loads.

Full Worked Example

Simply-supported beam, L = 6 m, 10 kN downward point load at 2 m from A.

1. Reactions

ΣM_A = 0 → R_B × 6 = 10 × 2 → R_B = 3.33 kN ↑
ΣF_y = 0 → R_A + 3.33 = 10 → R_A = 6.67 kN ↑

2. Segments

Segment 1: 0–2 m | Segment 2: 2–6 m

3. Shear Force Diagram (left → right)

At x = 0: V jumps up to +6.67 kN (upward reaction = positive shear)
0–2 m: constant +6.67 kN (no loads in this segment)
At 2 m: drops abruptly by 10 kN → −3.33 kN
2–6 m: constant −3.33 kN (no loads in this segment)

4. Bending Moment Diagram (left → right)

At x = 0: M = 0 (pin support → M starts at zero)
At x = 2: M = 0 + (6.67 × 2) = +13.33 kN·m (linear increase, no jump)
At x = 6: M = 13.33 + (−3.33 × 4) = 0 ✓ (roller support → M ends at zero)
Note: Moment changes continuously with no jumps even at the point load. Downward load produces positive moment.

5. Checks

✓ Bending Moment Diagram starts and ends at zero (pin and roller supports)
✓ Shear is constant in each segment (point loads only, no distributed load)
✓ Moment changes continuously withg no jumps at the point load
✓ Area under Shear Force Diagram = change in Bending Moment Diagram (6.67 × 2 = 13.33; −3.33 × 4 = −13.33)
✓ Max moment at x = 2 m where V changes sign
✓ Downward load produces positive moment (sagging)

✓ Analysis Complete!

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