----------------- | ---------------------- | ---------------------------------- | | Point load | Step change (jump) | Kink (change of slope) | | UDL | Linear slope | Parabolic curve | | No load (between loads) | Constant (horizontal) | Linear (straight line) | | Concentrated moment | No change | Step change (jump) | | Zero shear force | — | Local maximum or minimum of moment | | Reaction force | Step change at support | Slope change at support |

SFD and BMD Formulas — Summary Table

Simply Supported Beam

Cantilever Beam

Fixed-Fixed Beam

Worked Example: Simply Supported Beam with Two Point Loads

Problem

A simply supported beam spans 24 ft (L = 288 in). Two point loads of P = 15 kips each are applied at the third points (8 ft from each support). Draw the SFD and BMD. E = 29,000 ksi, Ix = 484 in⁴ (W14x48).

Step 1: Reactions

By symmetry: RA = RB = 2 × 15 / 2 = 15 kips

Step 2: Shear Force Diagram

• From x = 0 to x = 8 ft: V(x) = +15 kips (constant)
• At x = 8 ft: step down by 15 kips → V = 0
• From x = 8 ft to x = 16 ft: V(x) = 0 (no shear between loads)
• At x = 16 ft: step down by 15 kips → V = -15 kips
• From x = 16 ft to x = 24 ft: V(x) = -15 kips (constant)

Step 3: Bending Moment Diagram

• From x = 0 to x = 8 ft: M(x) = 15x (linear, increasing)
  • At x = 8 ft: M = 15 × 8 = 120 kip-ft
• From x = 8 ft to x = 16 ft: M(x) = 15x - 15(x - 8) = 120 kip-ft (constant!)
• From x = 16 ft to x = 24 ft: M(x) = 15x - 15(x-8) - 15(x-16) = 15(24-x) (linear, decreasing)
  • At x = 24 ft: M = 0

Mmax = 120 kip-ft (constant between the two loads)

This is a characteristic SFD/BMD pattern: two equal loads at third points produce a constant (uniform) moment region between the loads. This is useful for plastic design and testing.

Step 4: Bending Stress Check

For W14x48 (Sx = 70.2 in³):

fb = Mmax / Sx = 120 × 12 / 70.2 = 20.5 ksi

Allowable (ASD): Fb = 0.66 × Fy = 0.66 × 50 = 33 ksi

20.5 < 33 ksi ✓

Step 5: Deflection

Using superposition for two point loads at third points:

Δmax ≈ 23PL³ / (648EI) (at midspan)

Δ = 23 × 15 × 288³ / (648 × 29000 × 484) = 23 × 15 × 23,887,872 / 9,109,632,000

= 8,251,316,160 / 9,109,632,000 = 0.906 in

L/360 = 288/360 = 0.80 in

0.906 > 0.80 in — does NOT pass L/360. Need a larger beam. A W14x53 (Ix = 577) or W16x50 (Ix = 660) would work.

Drawing SFD and BMD — Step by Step Method

1. Calculate reactions using equilibrium (ΣFy = 0, ΣM = 0)
2. Divide the beam into segments between load points and supports
3. Cut a section at a general point x within each segment
4. Draw a free body diagram of either the left or right portion
5. Apply equilibrium to find V(x) and M(x) as functions of x
6. Plot the functions to create the SFD and BMD
7. Check: SFD should return to zero at the free end; BMD should equal zero at simple supports

Frequently Asked Questions

What is the difference between SFD and BMD? SFD shows the internal shear force V(x) along the beam. BMD shows the internal bending moment M(x). They are related: dM/dx = V, meaning the slope of the BMD equals the shear force.

Where is the bending moment maximum? At the point where the shear force passes through zero (V = 0). For symmetric loading on simply supported beams, this is at midspan.

Why does the moment diagram jump at a concentrated moment? An external concentrated moment creates a discontinuity in the internal bending moment at that point. The shear force is not affected (only forces change shear, moments change moment).

What is the area method for drawing SFD and BMD? The change in shear between two points equals the negative area under the loading diagram: ΔV = -∫w dx. The change in moment equals the area under the shear diagram: ΔM = ∫V dx. This allows quick sketching without writing equations.

Can I use these diagrams for any material? Yes. SFD and BMD depend only on the loads, supports, and geometry — not the material. The same beam with the same loads has the same SFD and BMD whether it is steel, concrete, wood, or aluminum.

What causes a parabolic BMD? A uniformly distributed load (UDL) creates a parabolic bending moment diagram. Point loads create linear (triangular) moment diagrams.

Code References

• AISC 360-22 Chapter C -- Structural analysis requirements including second-order effects and notional loads
• AISC 360-22 Section C2.1 -- Direct analysis method requirements for member internal forces with P-Delta effects
• ASCE 7-22 Section 12.8 -- Equivalent lateral force procedure establishing required analysis methods for seismic design
• EN 1993-1-1 Section 5.2 -- Structural analysis for steel structures including global P-Delta and member P-delta effects

Try it now: Check your SFD and BMD with our free Beam Calculator calculator →

Related Pages

• Beam Calculator — Interactive SFD, BMD, and deflection
• Beam Capacity Calculator — Flexure, shear, LTB checks
• Steel Beam Deflection Check — L/360, L/240 deflection checks
• Beam Formulas — Complete beam formula reference
• Cantilever Beam — Cantilever formulas and examples
• Continuous Beam Tool — Multi-span analysis

Disclaimer

This is a calculation tool, not a substitute for professional engineering certification. All results must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) before use in construction, fabrication, or permit documents. The user is responsible for the accuracy of all inputs and the verification of all outputs.