Shear Force and Bending Moment Diagrams — Formulas & Examples

Shear force diagrams (SFD) and bending moment diagrams (BMD) are fundamental tools in structural analysis. This reference covers the key formulas and diagram shapes for common beam loading conditions.

Basic Sign Convention

Quantity Positive Negative
Shear force Upward on left face Downward on left face
Bending moment Sagging (concave up) Hogging (concave down)
Load direction Upward Downward

Key Relationships

The fundamental differential equations governing beams:

dV/dx = -w(x)          (shear rate = distributed load intensity)
dM/dx = V(x)           (moment rate = shear)
d²M/dx² = -w(x)        (curvature proportional to load)

Consequence: Under UDL, SFD is linear and BMD is parabolic. Under point loads only, SFD is step-wise and BMD is piecewise linear.

Simply Supported Beam — Point Load at Midspan

     P
     ↓
|------•------|
A              B

RA = RB = P/2

SFD:  +P/2 -------- -P/2   (step at load point)
BMD:  0 /-------\ 0        (peak = PL/4 at midspan)

Maximum moment: M_max = PL/4 (at load)

Simply Supported Beam — UDL

w (kN/m)
↓↓↓↓↓↓↓↓↓
|---------|
A         B

RA = RB = wL/2

SFD:  +wL/2 \   / -wL/2   (linear, zero at midspan)
BMD:  0 (----) 0           (parabola, peak = wL²/8 at midspan)

Maximum moment: M_max = wL²/8 (at midspan) Zero shear (location of max moment): x = L/2

Cantilever — Point Load at Free End

          P
          ↓
[========•
Fixed     Free

Reaction at fixed end: R = P (upward), M_fix = PL (hogging)

SFD:  P --------- P         (constant)
BMD:  0 (slope up) PL       (linear, max at fixed end)

Maximum moment: M_max = PL (at fixed support, hogging)

Cantilever — UDL

w (kN/m)
↓↓↓↓↓↓↓↓↓
[==========
Fixed      Free

Reaction: R = wL, M_fix = wL²/2

SFD:  wL --(linear)--> 0    (varies from wL at wall to 0 at tip)
BMD:  wL²/2 --(parabola)--> 0

Maximum moment: M_max = wL²/2 (at fixed support, hogging)

Propped Cantilever — UDL

w
↓↓↓↓↓↓↓↓↓
[==========]
Fixed      Pin

RA (pin) = 3wL/8
RB (fixed) = 5wL/8
M_B (fixed) = wL²/8 (hogging)

Point of zero shear: x = 3L/8 from pin
Max +ve moment: M = 9wL²/128 at x = 3L/8 from pin

Fixed-Fixed Beam — UDL

w
↓↓↓↓↓↓↓↓↓
[===========]
Fixed       Fixed

End moments: M_end = wL²/12 (hogging)
Midspan moment: M_mid = wL²/24 (sagging)
Reactions: R = wL/2

SFD: Linear, ±wL/2 at ends, 0 at center
BMD: wL²/12 (hogging) → wL²/24 (sagging) → wL²/12 (hogging)

Point of Contraflexure

The point where moment changes sign (M = 0) is the point of contraflexure. It marks where the beam transitions from hogging to sagging (or vice versa).

For fixed-fixed beam with UDL:

Maximum Moment Summary Table

Beam Type Loading M_max Location
Simply supported Point load P at midspan PL/4 Midspan
Simply supported UDL w wL²/8 Midspan
Cantilever Point load P at tip PL Fixed end
Cantilever UDL w wL²/2 Fixed end
Fixed-fixed Point load P at midspan PL/8 Midspan & ends
Fixed-fixed UDL w wL²/12 at ends, wL²/24 at midspan Ends
Propped cantilever UDL w wL²/8 at fixed end Fixed end

Fixed-end conditions significantly reduce midspan moments but introduce hogging moments at supports that must be designed for.

Moment Diagrams and Reinforcement

For reinforced concrete:

The moment diagram is the reinforcement demand diagram — more moment = more steel needed.

Numerical Example

Problem: Simply supported beam, L=6m, w=15kN/m + P=30kN at 2m from left support.

Reactions:

Shear at key points:

Zero shear: at x = 2 + 5/15 = 2.33m from left (max moment location)

Max moment: M = 65×2.33 - 15×2.33²/2 - 30×0.33 = 151.45 - 40.83 - 9.90 = 100.8 kN·m

→ Calculate beam reactions, SFD and BMD online →

Frequently Asked Questions

Where is the maximum bending moment in a simply supported beam with UDL? The maximum moment occurs at midspan (x = L/2) and equals wL²/8. This is where shear force equals zero — the zero-shear point always marks the maximum moment location. For UDL loading, the shear diagram is linear with +wL/2 at the left support decreasing to −wL/2 at the right support, crossing zero at exactly midspan.

How do I find the location of maximum moment for an off-center point load? Maximum moment occurs at the point load location. For a load P at distance a from the left support, the moment there equals R_A × a = (Pb/L) × a = Pab/L. The shear is constant between supports (with a step at the load), so the moment diagram is piecewise linear with a peak directly under the load.

What is a point of contraflexure and where does it occur? A point of contraflexure is where the bending moment equals zero and changes sign (from sagging to hogging or vice versa). For a fixed-fixed beam with UDL, contraflexure points occur at x = L/4 from each support — within these inner two points the beam sags, outside them it hogs. These points are structurally significant because a pinned connection at the contraflexure point introduces no additional moment.

Why does fixing a beam's ends reduce the midspan moment? End fixity introduces negative (hogging) moments at the supports that partially cancel the positive (sagging) moment at midspan. For a fixed-fixed beam with UDL, end moments = wL²/12 hogging, reducing midspan moment to wL²/24 — one-third of the simply supported wL²/8. However, the support connections must be designed for the full fixed-end moment; if they yield, the structure redistributes toward simply supported behavior.

What is the difference between sagging and hogging moments in reinforced concrete? Sagging (positive) moments put the bottom fiber in tension — reinforcement must be placed at the bottom of the beam. Hogging (negative) moments put the top fiber in tension — reinforcement must be placed at the top. Continuous beams have both: positive midspan moments requiring bottom steel and negative support moments requiring top steel. The moment diagram is the demand map for reinforcement placement.

BMD shapes for all common cases

The shape of the bending moment diagram is directly related to the loading type and support conditions. Understanding these patterns allows engineers to quickly sketch and verify moment diagrams.

Loading Case SFD Shape BMD Shape Key Feature
Simply supported, center point load Step (two zones) Triangle, peak at center Peak at load point, zero at ends
Simply supported, off-center point load Step (two zones) Triangle, peak at load Peak shifts toward closer support
Simply supported, UDL Linear ramp Parabola, peak at center Maximum where shear crosses zero
Simply supported, two symmetric PL Step (3 zones) Trapezoid between loads Flat top between equal point loads
Cantilever, tip point load Constant Linear (max at wall) Linear decrease from wall to tip
Cantilever, UDL Linear ramp Parabola (max at wall) Moment = wL²/2 at wall
Fixed-fixed, UDL Linear Parabola with end hogging Contraflexure at L/4 from each end
Fixed-fixed, center PL Step Triangle with end moments PL/8 at ends, PL/8 at center
Propped cantilever, UDL Linear Parabola, one end fixed M = wL²/8 at fixed end
Continuous, 2 equal spans, UDL Linear each span Sag + hog over support Support moment = wL²/8
Overhanging beam, UDL on span Linear Parabola + cantilever Zero moment at overhang tip

Sign convention clarification

Consistent sign convention is essential for correctly drawing and interpreting SFD and BMD:

Shear force convention:

Bending moment convention:

Load convention:

Practical tip: When drawing BMDs by hand, always start from the left end and work rightward. Calculate reactions first, then determine shear at key sections, then integrate (or use equilibrium) to find moments. The zero-shear point is always the location of maximum (or minimum) moment.

Moment diagram construction rules

Follow these rules to construct accurate moment diagrams:

  1. At simple supports (pins and rollers): The moment is zero unless an external moment is applied at the support.

  2. At fixed supports: The moment is non-zero and equals the fixed-end moment (hogging).

  3. At free ends (cantilever tips): The moment is zero unless an external moment is applied.

  4. Between point loads (no distributed load): The BMD is a straight line. The slope of the BMD equals the shear (dM/dx = V).

  5. Under uniform distributed load: The BMD is a parabola. The curvature is proportional to the load intensity (d²M/dx² = -w).

  6. At point load locations: The BMD has a change in slope (kink). The magnitude of the slope change equals the point load divided by the beam depth... actually, the slope change equals the magnitude of the change in shear at that point.

  7. At external moment locations: The BMD has a step change (jump) equal to the applied moment.

  8. Maximum moment location: Occurs where the shear force is zero (dM/dx = V = 0). For UDL on a simply supported beam, this is always at midspan.

Superposition examples

Superposition allows complex loading cases to be decomposed into simpler cases that are solved individually and then combined:

Example 1: Simply supported beam with UDL + point load Beam: L = 20 ft, w = 2 klf, P = 15 kips at midspan.

Case A (UDL alone): M_max = wL²/8 = 2 × 20²/8 = 100 kip-ft at midspan. Case B (PL alone): M_max = PL/4 = 15 × 20/4 = 75 kip-ft at midspan.

Combined: M_max = 100 + 75 = 175 kip-ft at midspan.

Reactions: R_A = wL/2 + P/2 = 20 + 7.5 = 27.5 kips each.

Example 2: Simply supported beam with two unequal point loads Beam: L = 24 ft, P1 = 10 kips at 6 ft from left, P2 = 20 kips at 16 ft from left.

By equilibrium: R_A × 24 - 10 × 6 - 20 × 16 = 0 R_A = (60 + 320)/24 = 15.83 kips R_B = 10 + 20 - 15.83 = 14.17 kips

Moment at P1: M1 = 15.83 × 6 = 95.0 kip-ft Moment at P2: M2 = 14.17 × 8 = 113.4 kip-ft (governs)

Example 3: Continuous beam using superposition For a two-span continuous beam with UDL on both spans, the support moment can be found by superposition of the fixed-end moments: M_support = wL²/8 (for equal spans with uniform load, this equals the negative moment at the middle support)

Relationship between load, shear, and moment diagrams

The three diagrams (load, shear, moment) are mathematically linked through differential calculus. Understanding this relationship allows verification and quick sketching:

Load diagram:       w(x)
                        |
                        | integrate
                        v
Shear diagram:      V(x) = -integral of w(x) dx + C1
                        |
                        | integrate
                        v
Moment diagram:     M(x) = integral of V(x) dx + C2

Key relationships:

Verification technique: After constructing the SFD and BMD, check that the change in moment from A to B equals the area under the SFD between A and B. If these do not match, there is an error in the diagram. This provides a powerful check on hand calculations.

Worked example: SFD and BMD for simply supported beam with UDL

Given: Simply supported steel beam (W16x36), span L = 24 ft. Dead load = 0.50 klf, live load = 0.80 klf. Factored load wu = 1.2(0.50) + 1.6(0.80) = 1.88 klf.

Step 1 — Reactions: R_A = R_B = wu × L / 2 = 1.88 × 24 / 2 = 22.56 kips.

Step 2 — Shear force diagram: At x = 0: V = +22.56 kips At x = L/2: V = 22.56 - 1.88 × 12 = 22.56 - 22.56 = 0 kips At x = L: V = 22.56 - 1.88 × 24 = 22.56 - 45.12 = -22.56 kips

The SFD is a straight line from +22.56 at the left support to -22.56 at the right support, passing through zero at midspan.

Step 3 — Bending moment diagram: M(x) = R_A × x - wu × x² / 2 At x = 0: M = 0 At x = 12 ft: M = 22.56 × 12 - 1.88 × 144 / 2 = 270.7 - 135.4 = 135.3 kip-ft At x = 24 ft: M = 22.56 × 24 - 1.88 × 576 / 2 = 541.4 - 541.4 = 0 kip-ft

The BMD is a parabola with M_max = 135.3 kip-ft at midspan.

Step 4 — Verification: Area under SFD from x=0 to x=12: (22.56 + 0)/2 × 12 = 135.4 kip-ft (matches M_max). Area under SFD from x=12 to x=24: (0 + (-22.56))/2 × 12 = -135.4 kip-ft (M returns to zero at right support).

Worked example: overhanging beam

Given: Beam with a pin support at A (x=0), roller support at B (x=20 ft), and a 6 ft overhang to the right of B. UDL wu = 1.5 klf on the entire 26 ft length.

Step 1 — Reactions: Sum moments about A: R_B × 20 - 1.5 × 26 × 13 = 0 R_B = 1.5 × 338 / 20 = 25.35 kips R_A = 1.5 × 26 - 25.35 = 13.65 kips

Step 2 — Shear at key points: x = 0: V = +13.65 kips x = 20 ft: V_left = 13.65 - 1.5 × 20 = -16.35 kips x = 20 ft: V_right = -16.35 + 25.35 = +9.0 kips (jump at support B) x = 26 ft: V = 9.0 - 1.5 × 6 = 0 kips (free end)

Step 3 — Zero-shear locations: Between A and B: V = 0 at x = 13.65/1.5 = 9.1 ft from A. Beyond B: V = 0 at x = 26 ft (free end).

Step 4 — Moment at key points: x = 0: M = 0 (pin support) x = 9.1 ft (zero shear): M = 13.65 × 9.1 - 1.5 × 9.1²/2 = 124.2 - 62.1 = 62.1 kip-ft (positive peak) x = 20 ft (support B): M = 13.65 × 20 - 1.5 × 20²/2 = 273.0 - 300.0 = -27.0 kip-ft (hogging) x = 26 ft (free end): M = 0

The BMD has a positive (sagging) peak of 62.1 kip-ft between the supports and a negative (hogging) moment of -27.0 kip-ft at support B. The overhang creates the negative support moment that reduces the positive midspan moment compared to a simply supported beam.

Run This Calculation

→ Beam Calculator — compute reactions, shear force diagrams, and bending moment diagrams for any loading pattern.

→ Beam Deflection Calculator — calculate deflections for simply supported, cantilever, and fixed beam configurations.

→ Continuous Beam Calculator — reactions, moments, and deflections for multi-span beams with fixed or pinned ends.

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Beam Design Methods

Lateral-Torsional Buckling

For beams that are not adequately braced against lateral movement and twist, the nominal moment capacity is governed by lateral-torsional buckling (LTB). The resistance depends on the unbraced length (Lb) relative to limit states:

Shear Design

Web shear strength depends on the panel aspect ratio and stiffener configuration. For unstiffened webs, the nominal shear capacity is:

Compact sections with low web slenderness (h/tw) can develop full shear yielding. Slender webs may require transverse stiffeners to develop adequate shear capacity.

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Frequently Asked Questions

What is the recommended design procedure for this structural element?

The standard design procedure follows: (1) establish design criteria including applicable code, material grade, and loading; (2) determine loads and applicable load combinations; (3) analyze the structure for internal forces; (4) check member strength for all applicable limit states; (5) verify serviceability requirements; and (6) detail connections. Computer analysis is recommended for complex structures, but hand calculations should be used for verification of critical elements.

How do different design codes compare for this calculation?

AISC 360 (US), EN 1993 (Eurocode), AS 4100 (Australia), and CSA S16 (Canada) follow similar limit states design philosophy but differ in specific resistance factors, slenderness limits, and partial safety factors. Generally, EN 1993 uses partial factors on both load and resistance sides (γM0 = 1.0, γM1 = 1.0, γM2 = 1.25), while AISC 360 uses a single resistance factor (φ). Engineers should verify which code is adopted in their jurisdiction.

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