What are the differential relationships between load, shear, and moment?

Three differential relationships govern all beam analysis and are the foundation for drawing shear force and bending moment diagrams: (1) dV/dx = −w(x) — the rate of change of shear force equals the negative of the distributed load intensity at that point. Where w = 0 (no load), shear is constant. Where w is constant (UDL), shear varies linearly. Where w varies linearly (triangular load), shear varies parabolically. (2) dM/dx = V(x) — the rate of change of bending moment equals the shear force at that point. Where V = 0, moment is at a local maximum or minimum. Where V is positive, moment increases. Where V is negative, moment decreases. (3) d²M/dx² = −w(x) — the curvature of the moment diagram equals the negative of the load intensity. These relationships enable the 'area method' for constructing diagrams: the change in shear between two points equals the negative area under the load curve between those points (ΔV = −∫w dx). The change in moment between two points equals the area under the shear curve (ΔM = ∫V dx). For a simply supported beam with UDL w: V(x) = wL/2 − wx (linear, starting positive, crossing zero at midspan, ending negative). The area under the V diagram from left support to midspan = (wL/2 × L/2)/2 = wL²/8, which equals the midspan moment. The slope of the moment diagram at any point equals the shear there — the moment diagram's maximum occurs where shear crosses zero (midspan). Using these relationships, you can construct the full SFD and BMD from the support reactions and load diagram without integrating the moment equation.

What are the SFD and BMD for a simply supported beam with a point load?

For a simply supported beam of span L with a concentrated point load P at distance a from the left support (b = L − a from the right): Reactions: RA = P×b/L, RB = P×a/L. SFD: from x = 0 to x = a, V(x) = +RA (constant, positive). At x = a, there's a step change downward of magnitude P — V drops from +RA to +RA − P = +P×b/L − P = −P×a/L. From x = a to x = L, V(x) = −RB (constant, negative). The magnitude of the step at the load equals P. Maximum shear = max(RA, RB). BMD: from x = 0 to x = a, M(x) = RA × x (linear, increasing from zero). At x = a, M = RA × a = Pab/L (maximum moment, directly under the load). From x = a to x = L, M(x) = RA × x − P × (x − a) (linear, decreasing to zero at x = L). For the common case where P is at midspan (a = b = L/2): RA = RB = P/2. V(x) = +P/2 (left half), −P/2 (right half). Mmax = P×L/4 at midspan. The moment diagram is a triangle peaking at PL/4. Example: P = 10 kips, L = 20 ft, a = 6 ft (b = 14 ft). RA = 10×14/20 = 7 kips, RB = 10×6/20 = 3 kips. V_left = +7 (0 to 6 ft), V_right = −3 (6 to 20 ft). Mmax at x=6 ft = 7 × 6 = 42 kip-ft. The step in the shear diagram is 10 kips, equal to the applied load. The slope of the moment diagram changes from +7 (positive slope) in the left segment to −3 (negative slope) in the right segment — the kink in the moment diagram at x = 6 ft reflects the concentrated load.

How do SFD and BMD change for a cantilever beam?

Cantilever beams are fixed at one end and free at the other — the fixed support provides both a vertical reaction and a moment reaction to maintain equilibrium. For a cantilever of span L with UDL w: the fixed support is conventionally placed at x = 0 (wall) and the free end at x = L. Reaction at fixed support: V_A = wL (upward), M_A = wL²/2 (counterclockwise, hogging at the support). SFD: V(x) = wL − wx, starting at +wL at the fixed end and decreasing linearly to 0 at the free end (no shear at a free end — no force is applied there). BMD: M(x) = wL × x − w×x²/2 − wL²/2 (or more directly, M(x) = −w×(L−x)²/2). Moment is maximum (negative/hogging) at the fixed support: M_A = −wL²/2, and decreases parabolically to zero at the free end (no moment at a free end — the beam can't develop moment where it's unsupported). For a cantilever with a point load P at the free tip: V(x) = +P (constant along the full length — shear is constant because no distributed load acts between the load and support). M(x) = −P × (L − x), varying linearly from M_A = −P×L at the fixed end to zero at the free end. The negative moment means hogging — tension on top, compression on bottom, opposite to a simply supported beam under gravity. Key differences from simply supported beams: (a) the maximum moment is at the support (not midspan), (b) shear is maximum at the support, (c) the free end has zero moment AND zero shear (unless a concentrated load is applied there). For a cantilever with both UDL and a tip load: superpose the diagrams — V_fixed = wL + P, M_fixed = wL²/2 + PL. The shape of the shear diagram is linear (UDL contribution) with an offset (point load contribution).

How are SFD and BMD constructed for an overhanging beam?

An overhanging beam extends beyond one or both supports, creating a mix of positive (sagging) and negative (hogging) moment regions. The overhanging portion behaves as a cantilever, with negative moment at the support and zero at the free end, while the portion between supports is similar to a simply supported beam but with the overhang's moment at the support imposing a negative starting point. Example: beam with supports at x = 0 (left) and x = L1 (right), with an overhang of length L2 beyond the right support, UDL w over the full length. Reactions: RA = w×(L1²/2 − L2²/2)/L1, RB = w×(L1+L2)²/(2×L1). (These come from moment equilibrium — the overhang creates uplift at the left support if L2 is long enough). SFD: starts at +RA at the left support, decreases linearly at rate w across the entire beam. V crosses zero somewhere between supports (the point of maximum positive moment). At the right support (x = L1), V = RA − w×L1 = −w×L2 (the overhang contributes shear). From x = L1 to x = L1+L2 (overhang), V continues decreasing to −w×(L1+L2) at the free end, where V must be zero — actually no: shear at the free end of an overhang under UDL is −w×(L1+L2), which is NOT zero — this is correct because the accumulated load on the overhang is w×L2, and that enters the right support as upward shear +RB. BMD: starts at 0 at left support, increases parabolically to a positive maximum where V = 0 between supports, then decreases to a negative value at the right support due to the overhang moment: M_right = −w×L2²/2. From x = L1 to L1+L2: the moment rises parabolically from −wL2²/2 to 0 at the free end. The negative moment region over the support requires top reinforcement in concrete or compression flange bracing in steel for the hogging zone.

What is the principle of superposition in SFD and BMD analysis?

The principle of superposition states that for a linear elastic structure, the total response (shear, moment, deflection) to multiple loads equals the sum of the responses to each load applied individually. This works because the governing differential equations (dV/dx = −w, dM/dx = V) are linear — doubling the load doubles the shear and moment everywhere. For constructing SFD and BMD: decompose a complex loading into simple standard cases, draw the diagram for each, and add them algebraically. Example: a simply supported beam with UDL w over the full span L PLUS a point load P at midspan. Case 1 (UDL only): V_UDL(x) = wL/2 − wx, M_UDL(x) = (wLx − wx²)/2, Mmax_UDL = wL²/8 at midspan. Case 2 (point load only): V_PL(x) = +P/2 (x L/2), M_PL max = PL/4 at midspan. Total: V(x) = V_UDL + V_PL, M(x) = M_UDL + M_PL. At midspan: Mmax_total = wL²/8 + PL/4. The total shear at the left support: V_max = wL/2 + P/2. The moment at the point load is found by evaluating M_UDL at x = L/2 (= wL²/8) plus M_PL at that point (= PL/4). This principle applies to any combination of point loads, distributed loads, and moments. Limitations: superposition is valid only for linear-elastic behavior (small deflections, no yielding, no contact nonlinearities). For elastic analysis of steel beams under service loads, superposition is always valid. For ultimate strength design, the load combinations in ASCE 7 already perform superposition at the load level — the combined effect of factored loads is computed by analyzing each load case separately and combining results using the load combination coefficients.

SFD & BMD Calculator — Shear Force & Bending Moment Diagrams

Shear Force Diagrams (SFD) and Bending Moment Diagrams (BMD) are the fundamental tools for visualizing internal forces in beams. Every structural engineer must be able to draw and interpret these diagrams.

This page covers the theory, formulas, sign conventions, and worked examples for SFD and BMD of common beam configurations.

What Are SFD and BMD?

Shear Force (V)

The shear force at any cross-section is the algebraic sum of all transverse forces acting on either side of that section. It represents the internal force that prevents one part of the beam from sliding vertically relative to the other.

Bending Moment (M)

The bending moment at any cross-section is the algebraic sum of the moments of all forces acting on either side of that section about the centroid of the cross-section. It represents the internal force that causes the beam to bend.

Key Relationships

These differential relationships govern all beam analysis:

dV/dx = -w(x) — the slope of the shear diagram equals the negative of the load intensity
dM/dx = V(x) — the slope of the moment diagram equals the shear force
d²M/dx² = -w(x) — the curvature of the moment diagram equals the negative of the load

Sign Convention

Shear force:

Positive (+) when the left side of the beam tends to move upward relative to the right side
Negative (-) when the left side tends to move downward

Bending moment:

Positive (+) causes sagging (concave upward, tension on bottom)
Negative (-) causes hogging (concave downward, tension on top)

SFD and BMD for Common Cases

Case 1: Simply Supported Beam — Point Load at Midspan

Setup: Beam of span L, point load P at midspan.

Reactions: RA = RB = P/2

SFD:

From x = 0 to x = L/2: V(x) = +P/2 (constant, positive)
From x = L/2 to x = L: V(x) = -P/2 (constant, negative)
At x = L/2: step change of magnitude P

BMD:

From x = 0 to x = L/2: M(x) = (P/2) × x (linear, increasing)
From x = L/2 to x = L: M(x) = (P/2) × x - P × (x - L/2) (linear, decreasing)
Mmax = PL/4 at x = L/2 (midspan)
M = 0 at both supports

Case 2: Simply Supported Beam — UDL

Setup: Beam of span L, uniformly distributed load w (force/length).

Reactions: RA = RB = wL/2

SFD:

V(x) = wL/2 - wx (linear, from +wL/2 to -wL/2)
V = 0 at x = L/2 (midspan)

BMD:

M(x) = (wL/2)x - wx²/2 (parabolic)
Mmax = wL²/8 at x = L/2

Case 3: Cantilever — Point Load at Free End

Setup: Cantilever of length L, fixed at left end, point load P at right (free) end.

Reactions at fixed end:

RA = P (upward)
MA = PL (counter-clockwise, hogging)

SFD:

V(x) = -P (constant throughout, negative)

BMD:

M(x) = -P × (L - x) (linear, from -PL at fixed end to 0 at free end)
Mmax = -PL at x = 0 (fixed support)

Case 4: Cantilever — UDL

Setup: Cantilever of length L, UDL w over full length.

Reactions at fixed end:

RA = wL (upward)
MA = wL²/2 (hogging)

SFD:

V(x) = -w(L - x) (linear, from -wL at fixed end to 0 at free end)

BMD:

M(x) = -w(L-x)²/2 (parabolic)
Mmax = -wL²/2 at x = 0 (fixed support)

Case 5: Fixed-Fixed Beam — Point Load at Midspan

Reactions: RA = RB = P/2

Fixed-end moments: MA = MB = PL/8 (hogging)

SFD:

Same as simply supported: V = +P/2 (left half), V = -P/2 (right half)

BMD:

At supports: M = -PL/8
At midspan: M = +PL/8
Net Mmax = PL/8 (half the simply supported moment)

Reading SFD and BMD — Quick Reference

Feature in Loading	Effect on SFD	Effect on BMD
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

Load Case	Max Shear	Max Moment	Moment Location
Point P at midspan	P/2	PL/4	Midspan
Point P at distance a from left	Pa/L (at A)	Pab/L	At load point
UDL w	wL/2	wL²/8	Midspan
Triangular 0 to w₀	w₀L/4	w₀L²/12√3	At x = L/√3
Two equal P at 1/3 pts	P	PL/3	Between loads

Cantilever Beam

Load Case	Max Shear	Max Moment	Location
Point P at free end	P	PL	Fixed end
UDL w	wL	wL²/2	Fixed end
Triangular 0 to w₀	w₀L/2	w₀L²/6	Fixed end

Fixed-Fixed Beam

Load Case	Support Moment	Midspan Moment	Max Shear
Point P at midspan	PL/8	PL/8	P/2
UDL w	wL²/12	wL²/24	wL/2

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

Calculate reactions using equilibrium (ΣFy = 0, ΣM = 0)
Divide the beam into segments between load points and supports
Cut a section at a general point x within each segment
Draw a free body diagram of either the left or right portion
Apply equilibrium to find V(x) and M(x) as functions of x
Plot the functions to create the SFD and BMD
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.

Beam Calculator — Interactive SFD, BMD, and deflection
Beam Capacity Calculator — Flexure, shear, LTB checks
Beam Deflection Calculator — 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.

SFD & BMD Calculator — Shear Force & Bending Moment Diagrams

What Are SFD and BMD?

Shear Force (V)

Bending Moment (M)

Key Relationships

Sign Convention

SFD and BMD for Common Cases

Case 1: Simply Supported Beam — Point Load at Midspan

Case 2: Simply Supported Beam — UDL

Case 3: Cantilever — Point Load at Free End

Case 4: Cantilever — UDL

Case 5: Fixed-Fixed Beam — Point Load at Midspan

Reading SFD and BMD — Quick Reference

SFD and BMD Formulas — Summary Table

Simply Supported Beam

Cantilever Beam

Fixed-Fixed Beam

Worked Example: Simply Supported Beam with Two Point Loads

Problem

Step 1: Reactions

Step 2: Shear Force Diagram

Step 3: Bending Moment Diagram

Step 4: Bending Stress Check

Step 5: Deflection

Drawing SFD and BMD — Step by Step Method

Frequently Asked Questions

Related Pages

Disclaimer