Point-load internal forces

Expose and verify normal force, shear force, and bending moment using independent isolated-body calculations.

Interactive engineering simulation

Point-Load Diagram Explorer

Move a concentrated load and observe a vertical shear jump and continuous moment.

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)

and

dM/dx=V(x)

Load w(x)0.00 kN/m

Shear V(x)-8.00 kN

Moment M(x)44.00 kN·m

Beam length10 m

6–18

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Load magnitude20 kN

2–60

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Load position4.00 m

0.50–9.50

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Synchronized cursor x4.5 m

0.0–10.0

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Cursor shear

-8.00 kN

Cursor moment

44.00 kN·m

Left reaction

12.00 kN

Right reaction

8.00 kN

Maximum moment

48.00 kN·m at 4.00 m

Zero-shear locations

None

Piecewise shear equation

V(x)=12.000⟨x-0.00⟩⁰ + 8.000⟨x-10.00⟩⁰ - 20.000⟨x-4.00⟩⁰

Piecewise moment equation

M(x)=12.000⟨x-0.00⟩¹ + 8.000⟨x-10.00⟩¹ - 20.000⟨x-4.00⟩¹

\frac{dV}{dx}=-w(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.

Movable section cut Axial-force diagram Determinate-frame diagrams Gerber beam

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