Engineering & Technology · Statics, Stress & Deformation
Bending, shear and beam behaviour
A simply supported beam has upward reactions at its ends and a downward central load. The whole beam balances, but a piece cut from its left half would not balance by those external forces alone. The material across the cut must transmit a force and a turning effect.
Use Forces, Energy & Simple Systems and Material Structure & Properties first. Resolve horizontal and vertical components and use simple ratios and powers. The chapter defines each new mechanics symbol before applying it; 3–4–5 triangles avoid a trigonometry prerequisite in the worked truss. All dimensions, loads, material parameters, limits and records are constructed teaching inputs, not equipment specifications, permissible loads or testing instructions.
Back to the chapterAfter reading, you can
Find internal shear and bending moment from a cut, and connect moment to one simple beam stress distribution.
Internal resultants hold each segment in equilibrium
At a section, replace the removed material by the combined actions it exerted: an axial force, transverse shear force V and bending moment M in a planar beam model. These are resultants of distributed stresses over the section. V is measured in N; M in N·m. They are not interchangeable stress quantities.
Define signs with a picture or words. Here positive V acts downward on the cut face of the left segment, while positive M acts counterclockwise there. Positive M represents sagging bending for this beam. Another book can use different signs consistently. The same physical loading must give matching forces once those conventions are translated.
Position matters even for a constant section
With a 100 N central load on a 2 m span, the reactions are 50 N each. For a left cut at x between 0 and 1 m, force balance gives V=50 N and moment balance gives M=50x N·m. To the right of the load, V=−50 N and M=50(2−x)N·m. The figure keeps these two units on separate plots.
The point load produces a 100 N jump in the shear diagram. Moment stays continuous here, reaching 50 N·m at the centre and zero at both simple supports. The straight connecting segment drawn at the shear jump indicates a discontinuity, not a range of forces spread along the beam. Distributed loading would produce different diagram shapes.
A bending stress law needs material and shape assumptions
For a homogeneous slender beam in small linear elastic bending, the normal stress magnitude at distance y from the neutral axis is |σ|=|My|/I. The neutral axis has zero bending normal strain in this model. For the rectangle used below, it passes through the centroid. Under sagging, the upper region is in compression and the lower region in tension.
I is the second moment of area about that bending axis, not the mass moment of inertia. A rectangle of width b and vertical height h has I=bh³/12. Its mm⁴ units matter. This bending law does not by itself check transverse shear stress, local bearing, large deformation, buckling or cracks. The structural question can require several checks at different positions.
Use the largest moment for a constructed rectangle
At the centre, M=50 N·m=50000 N·mm. Take b=20 mm and h=40 mm, so I=20×40³/12≈106667 mm⁴. The outer surface is y=20 mm from the neutral axis, giving |σ|max=50000×20/106667≈9.38 MPa. The sign differs above and below the axis.
For a supplied E=100000 N/mm², the same ideal beam’s midpoint deflection is PL³/(48EI). Using L=2000 mm gives 100×2000³/(48×100000×106667)≈1.56 mm. Stress and deflection are separate predictions. No allowable stress or permissible deflection has been supplied, so neither numerical result constitutes a pass decision.
Try the reasoning
Why make a cut after solving whole-beam equilibrium?
Show the answer and reasoning
The cut reveals the internal actions needed to balance a segment. Whole-body reactions alone do not describe their variation along the beam.
Does zero bending moment at a support imply every local stress is zero?
Show the answer and reasoning
No. Support contact and transverse shear can still produce stress not described by the simple bending expression.
Why does turning the rectangular section change its response?
Show the answer and reasoning
It changes area distribution about the bending axis and therefore I, even when total area remains unchanged.
What to carry forward
Use equilibrium to obtain internal actions, then apply a stress or deformation model with matching assumptions.
References
Teaching resources and research supporting this lesson’s explanations and examples.
Massachusetts Institute of Technology3 sources