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Shear-force and bending-moment diagrams2D

Synchronized load-shear-moment cursor

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

Open the complete lesson
Interactive engineering simulation

Load–Shear–Moment Synchronized Cursor

Use one cursor across w(x), V(x), and M(x) and verify the differential relationships.

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 kN18.0 kN·m
Load w(x)0.00 kN/m
7.60.0x=4.50 m
Shear V(x)-0.34 kN
Left reaction: jump from 0.00 to 19.66 kNPoint load: jump from 19.66 to -0.34 kNRight reaction: jump from -30.04 to 0.00 kN19.7-30.0x=4.50 m
Moment M(x)54.46 kN·m
Applied couple: jump from 39.25 to 57.25 kN·m57.30.0x=4.50 m
Beam length
10 m
m
618

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

Load magnitude
20 kN
kN
260

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

Load position
4.00 m
m
0.509.50

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

Clockwise applied couple
18 kN·m
kN·m
-4040

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

Synchronized cursor x
4.5 m
m
0.010.0

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

Cursor shear
-0.34 kN
Cursor moment
54.46 kN·m
Left reaction
19.66 kN
Right reaction
30.04 kN
Maximum moment
57.15 kN·m at 7.21 m
Zero-shear locations
None
Piecewise shear equation
V(x)=19.658⟨x-0.00⟩⁰ + 30.042⟨x-10.00⟩⁰ - 20.000⟨x-2.80⟩⁰ - 3.200⟨x-4.50⟩¹ - 0.400⟨x-4.50⟩² + 7.600⟨x-10.00⟩¹ + 0.400⟨x-10.00⟩²
Piecewise moment equation
M(x)=19.658⟨x-0.00⟩¹ + 30.042⟨x-10.00⟩¹ - 20.000⟨x-2.80⟩¹ - 1.600⟨x-4.50⟩² - 0.133⟨x-4.50⟩³ + 3.800⟨x-10.00⟩² + 0.133⟨x-10.00⟩³ + 18.000⟨x-7.20⟩⁰
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.