Why does Q jump at the point load?

Across an **infinitesimal** segment carrying a **concentrated downward** force, equilibrium of the left segment requires the internal shear to **drop by P** so vertical forces still balance.

Why is M a parabola under UDL?

With **w** constant, **Q** decreases **linearly** along the span. Since **dM/dx = Q** in this sign convention, integrating gives **M** as a **quadratic** (parabola).

Does changing axial N affect Q or M here?

**No.** For a **straight horizontal** beam with **vertical** loads only, flexural resultants decouple from a **uniform axial** term in this linear model. The axial slider is for **conceptual** practice stacking effects.

Why might my textbook use opposite signs for M or V?

**Sign conventions differ** by country and author. Here **sagging M** is **positive** and **Q** is chosen to match that derivative relation. Always state your convention when comparing numbers.

Beam Q, M & N Diagrams

Internal force diagrams (shear Q, bending moment M, axial force N) are the standard way to read how a straight beam carries loads to supports. This page uses a simply supported beam (pin at A, roller at B) of fixed span L = 6 m with a downward point load at a user-chosen position and a uniformly distributed load (UDL) over the full span. Reactions follow from static equilibrium; Q(x) is built from vertical forces to the left of a cut (with a step at the point load); M(x) integrates the shear pattern—parabolic under UDL, kinked under a point force—with sagging moment taken positive. N is modeled as a uniform axial term you set separately: vertical loads alone do not induce axial force in a horizontal beam, but the flat N diagram illustrates superposition when axial effects from columns or temperature are added in a larger model. The sketch is schematic (no deflection shape, shear deformation, or beam width).

Who it's for: Introductory statics and strength-of-materials students learning to sketch SFD/BMD and to separate flexural and axial resultants.

Key terms

Shear force diagram
Bending moment diagram
Axial force
Simply supported beam
Point load
Uniformly distributed load (UDL)
Static equilibrium
Sagging moment sign

How it works

For a straight beam, internal shear Q (here V), bending moment M, and axial force N describe section resultants. This page uses a simply supported span with a point load at a chosen position and a uniform distributed load over the whole span. Q jumps down at the point force; under UDL it slopes linearly. M is parabolic under UDL and piecewise between kinks; we take sagging M > 0. N is independent of vertical loads in this model and appears as a flat line for a uniform axial term (useful when superposing with a truss or column action).

Key equations

ΣF_y = 0, ΣM_A = 0 → R_A, R_B

Q(x) from vertical forces to the left of the cut

dM/dx = Q (with sign convention used here)

Frequently asked questions

Why does Q jump at the point load?: Across an infinitesimal segment carrying a concentrated downward force, equilibrium of the left segment requires the internal shear to drop by P so vertical forces still balance.
Why is M a parabola under UDL?: With w constant, Q decreases linearly along the span. Since dM/dx = Q in this sign convention, integrating gives M as a quadratic (parabola).
Does changing axial N affect Q or M here?: No. For a straight horizontal beam with vertical loads only, flexural resultants decouple from a uniform axial term in this linear model. The axial slider is for conceptual practice stacking effects.
Why might my textbook use opposite signs for M or V?: Sign conventions differ by country and author. Here sagging M is positive and Q is chosen to match that derivative relation. Always state your convention when comparing numbers.

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