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Shear Force &
Bending Moment
Diagrams

SFD · BMD · Structural Analysis
CE 2103 Mechanics of Solids Interactive
Saotul Umam · RUET · Roll 2300152
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What Are SFD & BMD?

Shear Force (V)

The algebraic sum of all transverse forces on either side of a cross-section. Represents the tendency of beam parts to slide relative to each other.

V(x) = ΣF_y (left of x)

Bending Moment (M)

The algebraic sum of moments of all forces about the cross-section. Represents the beam's tendency to bend under applied loads.

M(x) = ΣM (left of x)

SFD

A graph plotting V(x) along beam length. Shows where maximum shear occurs and how it varies across the span.

Shape: steps at point loads, linear under UDL

BMD

A graph plotting M(x) along beam length. Zero at free ends, maximum where shear = 0.

Shape: linear under point load, parabolic under UDL
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Sign Convention

Positive Shear (+V)

Left: upward force. Right: downward force. Causes clockwise rotation.

Negative Shear (−V)

Left: downward force. Right: upward force. Counter-clockwise rotation.

Positive Moment (+M)

sagging

Causes sagging — concave up. Bottom fibers in tension.

Negative Moment (−M)

hogging

Causes hogging — concave down. Top fibers in tension.

dV/dx = −w(x)  |  dM/dx = V(x)  |  d²M/dx² = −w(x)

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Equilibrium & Differential Relations

01
ΣFy = 0 → Find Reactions

Sum of vertical forces = 0. For SS beam with load P at distance a (span L):
R_A = P·b/L   R_B = P·a/L   where b = L−a

02
ΣMA = 0 → Verify Reactions

Take moments about A: R_B × L = P × a  ⟹  R_B = Pa/L. Independent check on reactions.

03
Cut-Section Method

Imaginary cut at x. Apply ΣFy = 0 and ΣM = 0 to the free body — this directly gives V(x) and M(x).

Differential Relations — How They Arise

Element dx: dV = −w·dx → dV/dx = −w
Moment eq: dM = V·dx → dM/dx = V
Combined: d²M/dx² = −w
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Support Types

Pin / Hinge

Resists H & V forces. Free to rotate.

R_x, R_y (2 reactions)

Roller

Resists V force only. Free to move H & rotate.

R_y only (1 reaction)

Fixed End

Resists all forces & moments. No movement.

R_x, R_y, M (3 reactions)

Simply Supported: Pin + Roller (determinate)  |  Cantilever: Fixed only  |  Propped: Fixed + Roller (indeterminate)

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Load Types

P

Point Load

Concentrated force. Sudden jump in SFD.

SFD: step change BMD: linear kink
w kN/m

UDL (Uniform)

Constant distributed load per unit length.

SFD: linear (slope=w) BMD: parabolic (2nd)
w₀

UVL (Triangular)

Varies linearly. Hydrostatic loading.

SFD: parabolic (2nd) BMD: cubic (3rd)
M₀

Point Moment

Concentrated couple. Jump in BMD.

SFD: no change BMD: step jump

Partial UDL

UDL over a portion only. Piecewise analysis.

SFD: piecewise linear BMD: piecewise parabolic
P₁P₂

Multiple Loads

Superposition applies. Cumulative SFD/BMD.

SFD: multiple steps BMD: piecewise linear
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SFD/BMD Interactive Simulator

Add loads → Calculate → Hover to inspect
Quick Presets
Beam Length
6 m
Support Type
Load Type
Position (m)
3.0 m
Magnitude (kN)
10 kN
Applied Loads
Load added!
Free Body Diagram
Shear Force Diagram (SFD)
Bending Moment Diagram (BMD)
Reactions
R_A—
R_B—
M_A—
Key Values
V_max—
V_min—
M_max—
M_min—
Zero V@—
Hover Inspector
Move cursor over SFD/BMD
x = —
V = —
M = —
Equations
Add loads and calculate.
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Worked: Simply Supported Beam

Problem Setup

SS beam, L = 6 m. Point load P = 12 kN at 2 m from A. UDL w = 3 kN/m over full span.

Step 1 — Reactions

ΣM_A = 0: R_B × 6 = 12 × 2 + 3 × 6 × 3

R_B = (24 + 54)/6 = 13 kN
R_A = (12 + 18) − 13 = 17 kN

Step 2 — SFD (0 ≤ x ≤ 2)

V(x) = R_A − w·x = 17 − 3x

V(0) = +17 kN, V(2⁻) = +11 kN

Step 3 — SFD (2 ≤ x ≤ 6)

V(x) = 17 − 12 − 3x = 5 − 3x

V(2⁺) = −1 kN, V(6) = −13 kN

Step 4 — M_max (Zero Shear)

5 − 3x = 0 → x = 3.67 m from A (1.67 m past load)

M_max ≈ 26.1 kN·m at x = 3.67 m
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Worked: Cantilever Beam

Problem Setup

Cantilever L = 4 m, fixed at A. Point load P = 8 kN at free end B. UDL w = 2 kN/m full length.

Step 1 — Reactions at A

ΣFy = 0: R_A = P + wL = 8 + 8 = 16 kN

M_A = P·L + w·L²/2 = 32 + 16 = 48 kN·m

Step 2 — SFD (from free end B)

V(x) = P + w·x = 8 + 2x

V(0) = 8 kN, V(4) = 16 kN (linear)

Step 3 — BMD (from free end B)

M(x) = −P·x − w·x²/2 (hogging throughout)

M(0) = 0, M(4) = −48 kN·m (parabolic)
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Key Rules & Summary

SFD Construction

  • Start from left end with zero or reaction value
  • At each point load P: jump by ±P
  • Under UDL (w): slope = −w (linear)
  • Under UVL: parabolic slope
  • Must close to zero at far end
  • dV/dx = −w → integrate w to get V

BMD Construction

  • Zero at free ends and simple supports
  • Point load → linear BMD (kink)
  • UDL → parabolic (2nd degree)
  • UVL → cubic (3rd degree)
  • M is max where V = 0
  • Point moment M₀: BMD jumps by M₀

Shape Relationships

  • Load order 0 → SFD order 1 → BMD order 2
  • +shear = BMD rising; −shear = BMD falling
  • Zero shear = BMD extremum
  • ∫V dx = M (area under SFD)
  • Sagging (+M): concave up, bottom in tension
  • Hogging (−M): concave down, top in tension

Common Mistakes

  • Forgetting equilibrium check (ΣFy ≠ 0)
  • Wrong sign at fixed-end moment
  • Not splitting at every load/support position
  • Missing UDL over full zone
  • Wrong side of cut to sum forces on
  • V = 0 ≠ M = 0

Key Identity: dM/dx = V  |  dV/dx = −w  |  Area under SFD = Change in BMD

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