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
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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