Beam Analysis

Assumptions and Conventions

Every beam carries internal forces that tell engineers where it will bend or break. Your job is to find them.

What you’re solving for:

Shear Force (V) The internal vertical force at a cross-section — one part of the beam trying to slide past the other. Bending Moment (M) The internal twisting effect that bends the beam. Where M is largest, the beam is most likely to fail.

Sign convention (use this everywhere):

Forces
↑ Upward = positive (+)
↓ Downward = negative (−)

Moments
↺ Counter-clockwise = positive (+)
↻ Clockwise = negative (−)

Assumptions we make:

The beam is straight and horizontal. Deflections are small. The material is uniform throughout. We ignore stretching along its length. These simplifications keep the math manageable and work for 90% of real designs.

Phase 1: Observe

Before any math, identify what you’re looking at. Find every support and every load on the beam.

Vocabulary for this phase:

Pin Support Resists horizontal and vertical movement. Allows rotation. Provides 2 reactions (H_A, V_A). Roller Support Resists vertical movement only. Can slide sideways. Provides 1 reaction (V_B). Fixed Support Resists everything: horizontal, vertical, and rotation. Provides 3 reactions (H_A, V_A, M_A). Point Load A single concentrated force at one location. Measured in Newtons (N). Distributed Load (UDL) A load spread over a length. Measured in N/m. Total force = intensity × length.

What to do: Look at the beam diagram. Count each type of support and each type of load. Write them down before moving on.

Phase 2: Determinacy

Can you solve this beam with equilibrium alone? Count reactions and check.

Vocabulary for this phase:

Reaction A force or moment provided by a support to keep the beam from moving. Equilibrium All forces and moments balance to zero — the beam stays still. Statically Determinate All reactions can be found using just the 3 equilibrium equations.

The rule:

R = 3 → solvable

Count unknowns per support:
Pin → 2 unknowns (H + V) | Roller → 1 unknown (V) | Fixed → 3 unknowns (H + V + M)

Why 3? In 2-D we have exactly 3 equilibrium equations: ∑F_x = 0, ∑F_y = 0, ∑M = 0. If R = 3, those three equations give us three unknowns — one unique answer. If R > 3, we need extra methods (the beam is indeterminate).

What to do: Add up the unknowns from all supports. If the total is 3, proceed to Phase 3. If not, the beam requires advanced methods.

Phase 3: Reactions

Use the three equilibrium equations to solve for every support reaction.

Vocabulary for this phase:

Moment Arm The perpendicular distance from a force to a pivot point. Moment = Force × distance. Resultant Force A single equivalent force replacing a distributed load: w × length, acting at the center of the loaded region.

The three equations:

∑F_x = 0 — horizontal forces cancel
∑F_y = 0 — vertical forces cancel (up = down)
∑M_point = 0 — moments about a point cancel

Strategy — start with moments:

Pick a pivot point where the most unknowns act — those forces have zero moment arm and drop out. For simply supported beams, pivot at the pin to isolate V_B. For cantilevers, pivot at the wall to isolate M_A.

Simply Supported recipe:
1. ∑M_A = 0 → solve V_B 2. ∑F_y = 0 → solve V_A 3. ∑F_x = 0 → H_A = 0

Cantilever recipe:
1. ∑F_y = 0 → V_A = total load 2. ∑M_A = 0 → solve M_A 3. ∑F_x = 0 → H_A = 0

Overhanging recipe:
1. ∑M_A = 0 → solve V_B (at the roller, not at the end) 2. ∑F_y = 0 → solve V_A (can be negative!) 3. ∑F_x = 0 → H_A = 0

Distributed load shortcut: Replace the UDL with a single force = w × length, acting at the midpoint of the loaded region (not the center of the beam).

Common mistakes: Wrong sign on a moment arm (CW vs CCW). Forgetting to replace a distributed load with its resultant. Placing the resultant at the beam center instead of the load center. Not checking that reactions add up to the total load.

Phase 4: Shear Diagram

Build V(x) by walking left to right across the beam. At every key position, figure out what the shear is.

Vocabulary for this phase:

V(x) Shear force at position x along the beam. V⁻ / V⁺ Shear just before vs. just after a point event (load or reaction). A point load causes a jump between these. Segment A stretch of beam between two key positions (supports, loads, load boundaries). Within a segment, the loading pattern is consistent.

How shear behaves:

Start: V(0) = left reaction (upward = positive).

Between key positions:
• No load in the segment → V stays constant (flat line)
• UDL of w N/m in the segment → V changes linearly (sloped line, slope = w)
At a key position:
• Downward point load P → V jumps down by P
• Upward reaction R → V jumps up by R

Common mistakes: Starting V at zero instead of the left reaction. Forgetting the jump at a point load. Drawing a slope where there’s no distributed load (V should be flat). Confusing V⁻ (before the jump) with V⁺ (after the jump).

Phase 5: Moment Diagram

Build M(x) using the area under the shear diagram. Each moment value builds on the last.

Vocabulary for this phase:

Sagging Beam bends concave-up (smile shape). Positive moment. Tension on the bottom. Hogging Beam bends concave-down (frown shape). Negative moment. Tension on the top. ΔM Change in moment between two positions = area under the shear diagram between those positions.

The core rule:

M(next) = M(previous) + area under V between the two positions

How to find the area:

• Constant V (flat shear) → rectangle: area = V × Δx → linear moment
• Linear V (sloped shear under UDL) → trapezoid: area = (V_start + V_end) / 2 × Δx → parabolic moment

Key facts:

• M = 0 at pin/roller ends and at free ends.
• M is maximum at fixed walls (cantilevers) and where V = 0.
• Moment is continuous — no sudden jumps at point loads (unlike shear).
• Maximum moment occurs where shear crosses zero.

Common mistakes: Drawing straight lines under a UDL (moment is parabolic there). Expecting M = 0 at a fixed support (it’s maximum there). Using V⁺ (after a jump) instead of V⁻ (before) when calculating area. Moment not returning to zero at the far support.

Phase 6: Check Your Work

Run these three checks every time. If any one fails, go back and find the error.

☐ Do reactions add up?
V_A + V_B should equal the total downward load. ☐ Does shear return to zero?
After adding the last support reaction, V must be 0. ☐ Does moment return to zero?
M must be 0 at pin/roller ends and at free ends of cantilevers.

If your answer is wrong, check whether you…
• Mixed up the sign on a force or moment (upward = +, downward = −)
• Forgot to include a load or reaction somewhere
• Used V⁺ instead of V⁻ when calculating ΔM
• Drew a straight moment line where it should be curved (under a UDL)

Worked Examples

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

1. Reactions

∑M_A = 0 → R_B × 6 = 10 × 2 → R_B = 3.33 kN ↑
∑F_y = 0 → R_A + 3.33 = 10 → R_A = 6.67 kN ↑

2. Shear (left → right)

x = 0: V = +6.67 kN
0–2 m: constant +6.67 kN
At 2 m: drops by 10 → −3.33 kN
2–6 m: constant −3.33 kN

3. Moment

x = 0: M = 0
x = 2: M = 0 + 6.67 × 2 = +13.33 kN·m
x = 6: M = 13.33 + (−3.33 × 4) = 0 ✓

4. Checks

✓ 6.67 + 3.33 = 10 kN | ✓ V → 0 | ✓ M starts/ends at 0 | ✓ M_max at x = 2

R_AX

∑F_x = 0 — No horizontal loads.
0 + R_AX = 0
R_AX = 0 N

R_BY

Method: ∑M_A = 0 to eliminate R_AX and R_AY.
R_BY × 6 m (CCW)
Load moments: 10 kN × 2 m = 20 kN·m (CW) → Total CW = 20 kN·m
−20 + R_BY × 6 = 0 → R_BY = 20 / 6
R_BY = 3.33 kN

R_AY

Method: ∑F_y = 0
Total downward = 10 kN | R_BY = 3.33 kN ↑ | R_AY = ? ↑
−10 + 3.33 + R_AY = 0 → R_AY = 10 − 3.33
R_AY = 6.67 kN

V(0)

Cut just right of x = 0. Only force to the left: R_AY ↑
V(0) = +R_AY = +6.67 kN

V(2⁻) then V(2⁺)

V(0) = +6.67 kN | Δx = 2 m | No loads → V constant
V(2⁻) = +6.67 kN
Point load P₁ = −10 kN acts here.
V(2⁺) = 6.67 + (−10) = −3.33 kN

V(6⁻) then V(6⁺)

V(2) = −3.33 kN | Δx = 4 m | No loads → V constant
V(6⁻) = −3.33 kN
Roller R_BY = +3.33 kN brings shear to zero.
V(6⁺) = −3.33 + 3.33 = 0 kN ✓

M(0)

Pin support → no moment resistance.
M(0) = 0 kN·m

M(2)

M(0) = 0 | V_start = +6.67 | Δx = 2.00 m
M(2) = M(0) + V_start × Δx = 0 + 6.67 × 2 = +13.33 kN·m

M(6)

M(2) = 13.33 | V_start = −3.33 | Δx = 4.00 m
M(6) = 13.33 + (−3.33) × 4 = 13.33 − 13.33 = 0 kN·m ✓

✓ Reactions = total load | ✓ V → 0 | ✓ M starts/ends at 0 | ✓ M_max = 13.33 at x = 2 (V changes sign)

Cantilever beam, L = 8 m, fixed wall at A, 1200 N downward point load at the free end (x = 8 m).

1. Reactions

∑F_x = 0 → H_A = 0
∑F_y = 0 → V_A = 1200 → V_A = 1200 N ↑
∑M_A = 0 → M_A + 1200 × 8 = 0 → M_A = −9600 N·m

2. Shear

x = 0: V = +1200 N (wall reaction)
0–8 m: constant +1200 N (no loads between)
x = 8: V = 0 N (free end)

3. Moment

x = 0: M = −9600 N·m (wall reaction moment)
x = 8: M = −9600 + 1200 × 8 = 0 ✓

4. Checks

✓ V_A = 1200 N = total load | ✓ M = 0 at free end | ✓ M_max = 9600 at wall

Simply-supported beam, L = 8 m, UDL of 50 N/m downward from x = 2 to x = 6 m. Total distributed force = 50 × 4 = 200 N at center x = 4 m.

1. Reactions

∑M_A = 0 → V_B × 8 = 200 × 4 → V_B = 100 N ↑
∑F_y = 0 → V_A = 200 − 100 → V_A = 100 N ↑

2. Shear

x = 0: V = +100 N | x = 2: V = +100 N (constant, no load)
x = 4: V = 100 + (−50)(2) = 0 N (linear under UDL — V crosses zero!)
x = 6: V = 0 + (−50)(2) = −100 N | x = 8: V = −100 N → 0 ✓

3. Moment

x = 0: M = 0 | x = 2: M = 0 + 100 × 2 = 200
x = 4: M = 200 + (100+0)/2 × 2 = 300 N·m (trapezoid → parabolic curve)
x = 6: M = 300 + (0+(−100))/2 × 2 = 200 | x = 8: M = 200 + (−100)(2) = 0 ✓

4. Checks

✓ 100 + 100 = 200 N | ✓ V → 0 | ✓ M starts/ends at 0 | ✓ M_max = 300 at x = 4 (V = 0)

R_AX

∑F_x = 0 — No horizontal loads.
R_AX = 0 N

R_BY

Method: ∑M_A = 0 to eliminate R_AX and R_AY.
R_BY × 8 m (CCW)
Distributed load: 50 N/m × 4 m = 200 N (resultant at center: (2+6)/2 = 4 m from A)
Load moment: 200 N × 4 m = 800 N·m (CW) → Total CW = 800 N·m
−800 + R_BY × 8 = 0 → R_BY = 800 / 8
R_BY = 100 N

R_AY

Method: ∑F_y = 0
Total downward = 200 N | R_BY = 100 N ↑ | R_AY = ? ↑
−200 + 100 + R_AY = 0 → R_AY = 200 − 100
R_AY = 100 N

V(0)

Cut just right of x = 0. Only force to the left: R_AY ↑
V(0) = +R_AY = +100 N

V(2)

V(0) = +100 N | Δx = 2 m | No loads → V constant
V(2) = V(0) = +100 N

V(4) — inside UDL region

V(2) = +100 N | Δx = 2 m
UDL w₁ = −50 N/m acts over this segment.
V(4) = V(2) + w₁ × Δx = 100 + (−50) × 2
V(4) = 100 − 100 = 0 N ← shear crosses zero → moment is at an extreme here

V(6) — end of UDL region

V(4) = 0 N | Δx = 2 m
UDL w₁ = −50 N/m continues.
V(6) = 0 + (−50) × 2 = −100 N

V(8⁻) then V(8⁺)

V(6) = −100 N | Δx = 2 m | No loads → V constant
V(8⁻) = −100 N
Roller R_BY = +100 N brings shear to zero.
V(8⁺) = −100 + 100 = 0 N ✓

M(0)

Pin support → no moment resistance.
M(0) = 0 N·m

M(2)

M(0) = 0 | V_start = +100 | Δx = 2.00 m | Constant V → rectangle
M(2) = 0 + 100 × 2 = 200 N·m

M(4) — under UDL (trapezoidal area)

M(2) = 200 | V_start = +100 | V_end = 0 | Δx = 2.00 m
ΔM = (V_start + V_end) / 2 × Δx = (100 + 0) / 2 × 2 = 50 × 2 = 100
M(4) = 200 + 100 = 300 N·m ← maximum moment (V = 0 here)

M(6) — under UDL (trapezoidal area)

M(4) = 300 | V_start = 0 | V_end = −100 | Δx = 2.00 m
ΔM = (0 + (−100)) / 2 × 2 = −50 × 2 = −100
M(6) = 300 + (−100) = 200 N·m

M(8)

M(6) = 200 | V_start = −100 | Δx = 2.00 m | Constant V → rectangle
M(8) = 200 + (−100) × 2 = 200 − 200 = 0 N·m ✓

✓ Reactions = total load | ✓ V → 0 | ✓ M starts/ends at 0 | ✓ M_max = 300 N·m at x = 4 (V = 0)
Key pattern: linear shear under UDL → trapezoidal area → parabolic moment curve

✓ Analysis Complete!

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