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Internal forces in structural members2D

Determinate-frame diagrams

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

Open the complete lesson

Interactive engineering simulation

Determinate Frame Internal-Force Diagrams

Display N, V, and M for both members of a rigid determinate frame.

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)

.

Beam BC normal NkN

Beam BC shear VkN

Beam BC moment MkN·m

Column AB normal NkN

Column AB shear VkN

Column AB moment MkN·m

Aₓ

-8.00 kN

Aᵧ

8.00 kN

Cᵧ

12.00 kN

Force residual

0.00e+0

Moment residual

0.00e+0

Frame span10 m

m

6–18

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

Column height5.0 m

m

2.0–10.0

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

Vertical beam load20 kN

kN

2–60

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

Horizontal joint load8 kN

kN

0–30

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

Vertical-load position4.00 m

m

0.50–9.50

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

\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 Point-load internal forces Gerber beam