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Shear-force and bending-moment diagrams2D

Shear-force diagram

Build discontinuity-aware load, shear, and moment diagrams with synchronized analytical relationships.

Open the complete lesson
Interactive engineering simulation

Shear-Force Diagram Builder

Predict shear jumps and distributed-load slopes before revealing the diagram.

Tension-positive N · sagging-positive M
Sign convention: positive tension. On the left cut face, positive shear acts downward and positive moment counterclockwise; arrows reverse on the right face. Downward distributed load is positive, so dV/dx=w(x)dV/dx=-w(x) and dM/dx=V(x)dM/dx=V(x).
20.0 kN
Shear V(x)-4.35 kN
Left reaction: jump from 0.00 to 15.65 kNPoint load: jump from 15.65 to -4.35 kNRight reaction: jump from -20.55 to 0.00 kN15.6-20.6x=4.50 m
Beam length
10 m
m
618

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Load magnitude
20 kN
kN
260

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Load position
4.00 m
m
0.509.50

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Synchronized cursor x
4.5 m
m
0.010.0

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Cursor shear
-4.35 kN
Cursor moment
60.40 kN·m
Left reaction
15.65 kN
Right reaction
20.56 kN
Maximum moment
62.58 kN·m at 4.00 m
Zero-shear locations
None
Piecewise shear equation
V(x)=15.645⟨x-0.00⟩⁰ + 20.555⟨x-10.00⟩⁰ - 20.000⟨x-4.00⟩⁰ - 3.600⟨x-5.50⟩¹ + 3.600⟨x-10.00⟩¹
Piecewise moment equation
M(x)=15.645⟨x-0.00⟩¹ + 20.555⟨x-10.00⟩¹ - 20.000⟨x-4.00⟩¹ - 1.800⟨x-5.50⟩² + 1.800⟨x-10.00⟩²
dVdx=w(x)\frac{dV}{dx}=-w(x)
dMdx=V(x)\frac{dM}{dx}=V(x)
Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.