48610 Chap.12 Bending Stress in Beams
Bending Stress in Beams
A beam does not fail because of the load; it fails because of the internal bending moment the load produces, and that varies along the length. The Week 9 bending exercises supply a shear force and bending moment diagram alongside each beam for exactly that reason: the stress formula needs one number, the bending moment at the section being checked, and the diagram is where it comes from.
Three facts make those diagrams readable.
The shear steps at every point load by the size of that load; the bending moment is the running area under the shear diagram, so it peaks where the shear crosses zero; and between point loads with no distributed load the shear is constant and the moment is a straight line.
When a beam bends, fibres on one face shorten and those on the other lengthen, and between them lies the neutral axis, which for a uniform elastic material passes through the centroid of the section.
Stress varies linearly with distance from that axis, from zero at the axis to a maximum at the extreme fibre, which is why material near the axis carries almost nothing and why tube, channel and I sections exist.
For a symmetric section there is one maximum stress; for a T, an L or an unequal channel the two extreme fibres are at different distances, so both stresses must be computed and the larger compared with the material limit. Once a section has a section modulus, the maximum stress for any moment is a single division.
What this chapter covers
- 01
Why the internal bending moment rather than the load decides failure
- 02
Shear force stepping at each point load
- 03
The bending moment as the area under the shear diagram
- 04
The moment peaking where the shear crosses zero
- 05
Two quick checks on a supplied diagram
- 06
The neutral axis passing through the centroid
- 07
Stress proportional to distance from the neutral axis
- 08
Section modulus as the figure of merit for a cross section
- 09
Consistent units: newton millimetres, millimetres, and megapascals
- 10
Unsymmetric sections and the governing fibre
- 11
Tension against compression in brittle and ductile materials
Is the T section strong enough as a cantilever?
- +1Find the largest bending moment. For a cantilever with one end load it occurs at the built in end and equals load times length, that is 250 times 0.500, which is 125 newton metres, or 1.25 times ten to the fifth newton millimetres in consistent units.
- +1Identify the two extreme fibre distances. The centroid is 50.57 mm above the base, so the bottom fibre is 50.57 mm away and the top fibre is 72.0 minus 50.57, that is 21.43 mm away.
- +1Apply the formula at each fibre. The bottom gives 1.25 times ten to the fifth times 50.57 divided by 7.607 times ten to the fifth, which is 8.31 MPa; the top gives the same calculation with 21.43, which is 3.52 MPa.
- +1Say which is which. The load bends the cantilever downward, so the top surface stretches and the bottom is squeezed: 3.52 MPa tension at the top of the flange and 8.31 MPa compression at the bottom of the web. The governing magnitude is the bottom one, furthest from the neutral axis.
- +1Compare with the material and state the section modulus for reuse. Against roughly 310 MPa the peak of 8.31 MPa leaves a factor of about 37, so bending strength is not the limiting consideration here. The section modulus for the bottom fibre is 7.607 times ten to the fifth divided by 50.57, that is 15 043 cubic millimetres.
Key terms
- Bending moment
- The internal moment at a section of a beam, obtained from the bending moment diagram. It is the single quantity from the loading that the stress calculation needs.
- Shear force diagram
- A plot of the internal shear along a beam, stepping by each point load. The bending moment is its running area, so the moment peaks where the shear crosses zero.
- Neutral axis
- The layer within a bending beam that neither shortens nor lengthens. For a uniform elastic material it passes through the centroid of the cross section.
- Extreme fibre
- The material furthest from the neutral axis on one face of a section, where the bending stress reaches its maximum on that face.
- Section modulus
- The second moment of area divided by the distance to the extreme fibre. It converts a bending moment directly into a maximum stress and is the figure of merit for comparing sections.
- Megapascal
- A stress of one newton per square millimetre. Working consistently in newtons and millimetres produces stresses in this unit without any further conversion.
- Hogging
- Bending in which the upper fibres of a member are in tension, as happens at the built in end of a loaded cantilever, as opposed to sagging where the lower fibres stretch.
Bending Stress in Beams FAQ
How do I find where the bending moment is largest?
From the shear force diagram. The moment is the running area under the shear, so it rises while the shear is positive and falls while it is negative, which means it peaks where the shear crosses zero. On a simply supported beam with one point load that crossing is at the load.
Two further checks are worth making on any diagram: the shear must start at one reaction and close at the other, and the moment must be zero at a simple support with nothing beyond it.
Which fibre do I check on an unsymmetric section?
Both, then compare. A T or an L has its centroid off mid depth, so the two extreme fibres are at different distances and carry different stresses in the ratio of those distances. For a ductile material such as steel or aluminium the larger magnitude usually governs.
For a brittle material such as cast iron, which is far weaker in tension than in compression, the governing fibre may be the one with the smaller stress if that is the tensile side, so say which fibre is in tension as well as giving the number.
Why is section modulus more useful than the second moment?
Because it answers the question directly. Once Z is known the maximum stress is just the moment divided by Z, so comparing two candidate sections is one division each rather than a full calculation. It also makes the depth effect explicit: the second moment grows with depth cubed while the distance to the extreme fibre grows linearly, so Z grows roughly with depth squared.
That is the arithmetic behind every deep, thin webbed beam, and behind the fact that a flat bar turned on edge is far stiffer than the same bar lying flat.
What assumptions does the bending formula rely on?
Three, and the project report asks you to state the simplifications you made. The material is assumed linearly elastic, so stress is proportional to strain. Plane sections are assumed to stay plane as the beam bends, which is what makes the stress distribution a straight line rather than a curve.
And the beam is assumed to bend about a principal axis of the section, which for the sections in this subject means an axis of symmetry or the centroidal axis of a symmetric profile. All three hold comfortably for the members in a small prototype.
Assessment move
Practise the chain rather than the formula. Take a loading, find the reactions, sketch the shear and moment diagrams, compute the centroid and second moment of the section, then the stress at both extreme fibres, and finish by saying which fibre is in tension and how the number compares with the material.
Doing the whole chain three or four times is worth more than doing twenty stress substitutions, because in the quiz and in the report the work is in connecting the steps. Keep the section modulus of any section you compute, so that a later change of load costs one division.
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