Bending Stress Calculator
Free Bending stress Calculator for structural engineering projects. Enter dimensions to get material lists and cost estimates.
Bending Stress Calculator
Calculator
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Formula: sigma = M * c / I = M / S
Worked example โ Bending stress = 6.25 ksi (6,250 psi)
Formula
sigma = M * c / I = M / S
Bending stress sigma equals the bending moment M multiplied by the distance c from the neutral axis to the extreme fiber, divided by the moment of inertia I. Equivalently, sigma equals M divided by the section modulus S, where S = I/c. The moment must be in consistent units (kip-inches when I is in in^4) to get stress in ksi.
Worked Examples
Example 1: Rectangular Timber Beam
Problem:Find the bending stress in an 8x12 inch timber beam carrying a moment of 100 kip-ft.
Solution:I = 8 * 12ยณ / 12 = 1,152 inโด c = 12 / 2 = 6 in S = 1,152 / 6 = 192 inยณ sigma = (100 * 12) / 192 = 6.25 ksi = 6,250 psi
Result:Bending stress = 6.25 ksi (6,250 psi)
Example 2: Steel Beam Check
Problem:A steel beam with S = 50 inยณ carries a moment of 80 kip-ft. Check against Fb = 24 ksi.
Solution:sigma = M / S = (80 * 12) / 50 = 19.2 ksi Utilization = 19.2 / 24 = 80% 19.2 < 24, so the beam is adequate.
Result:Bending stress = 19.2 ksi, 80% utilized
Frequently Asked Questions
What is bending stress and how is it calculated?
Bending stress is the normal stress induced at a point in a beam due to an applied bending moment. It is calculated using the flexure formula: sigma = M * c / I, where M is the bending moment, c is the perpendicular distance from the neutral axis to the point of interest (usually the extreme fiber), and I is the moment of inertia about the neutral axis. The stress varies linearly from zero at the neutral axis to a maximum at the outermost fibers. The ratio I/c is called the section modulus S, so sigma = M/S is an equivalent form.
What is the section modulus and why is it useful?
The section modulus S equals I divided by c, where I is the moment of inertia and c is the distance to the extreme fiber. It combines the geometric properties of the cross section into a single value that directly relates moment to stress through sigma = M/S. A larger section modulus means lower bending stress for the same moment. Engineers often select beams by comparing the required section modulus M/Fb to available steel shapes listed in the AISC manual.
What is an allowable bending stress for steel and wood?
For structural steel with Fy = 50 ksi, AISC allows a bending stress of 0.66 * Fy = 33 ksi for compact sections in ASD, or uses a resistance factor of 0.90 in LRFD giving phi-Mn = 0.90 * Fy * Z. For dimensional lumber, the National Design Specification (NDS) provides reference bending values Fb ranging from about 750 to 2,400 psi depending on the species and grade, which are then adjusted by multiple factors for size, duration, moisture, and other conditions.
What happens if bending stress exceeds the yield strength?
If bending stress reaches the yield strength, the outermost fibers begin to yield plastically while the interior fibers remain elastic. As the moment increases further, yielding progresses inward toward the neutral axis, forming a plastic hinge. The fully plastic moment Mp equals Fy times the plastic section modulus Z. For steel beams, the shape factor Z/S typically ranges from 1.1 to 1.5. Beyond Mp, the beam cannot carry additional moment and undergoes large rotations, potentially leading to collapse if not properly braced.
How does the cross-section shape affect bending stress?
The cross-section shape determines the moment of inertia I and the extreme fiber distance c, which together control the bending stress. I-beams and wide-flange shapes are efficient because they concentrate material at the flanges far from the neutral axis, maximizing I while minimizing weight. A rectangular section has I = bh-cubed/12 and S = bh-squared/6. Doubling the depth quadruples I and doubles S, making deeper beams much more efficient in bending.
Where is the neutral axis and how do I locate it on an unsymmetrical section?
In elastic bending the neutral axis passes through the centroid of the cross-sectional area, and it is the line where bending stress is zero. For a symmetric shape such as a rectangle or a wide-flange it sits at mid-depth, so c is half the depth. For a tee, channel or built-up section, divide the shape into rectangles, multiply each area by the distance from its own centroid to a common reference edge, sum those products and divide by the total area. The two distances to the extreme fibers then differ, and the larger one governs.
What is the difference between the elastic section modulus S and the plastic section modulus Z?
S applies while the whole section is elastic and stress varies linearly to a peak at the extreme fiber, giving first yield at a moment of Fy times S. Z applies once the entire cross-section has yielded, with a uniform tensile block and a uniform compressive block, and it gives the plastic moment Fy times Z. The ratio Z over S is the shape factor: about 1.5 for a rectangle and roughly 1.1 to 1.2 for a typical wide-flange. Allowable stress checks use S; strength design of compact steel shapes uses Z.
How do bending and shear stress vary along a beam?
They peak in different places, so both must be checked. On a simply supported beam under a uniform load, shear is largest at the supports and zero at midspan, while the bending moment is zero at the supports and largest at midspan. Bending stress is greatest at the extreme fibers where shear stress is zero; transverse shear stress follows VQ over It and is greatest at the neutral axis where bending stress is zero. Short, deep beams and beams with heavy loads near supports are the ones where shear governs.
What is lateral-torsional buckling and why can it govern before the flexure formula does?
The compression flange of a beam behaves like a slender column, and if it is not braced sideways it can displace laterally and twist the section while the tension flange stays put. The beam then fails at a moment well below the value that would produce the allowable bending stress. Capacity depends on the unbraced length between points of lateral restraint, so a floor beam with a diaphragm fastened to its top flange is fully braced, while a bare beam spanning open air is not. Narrow, deep sections are the most vulnerable.
How do I find the bending moment M to enter into the flexure formula?
M comes from the load and support arrangement, not from the section. A simply supported beam with a uniform load w over span L has a maximum moment of w times L squared divided by 8 at midspan. The same beam with a single central point load P has P times L divided by 4. A cantilever of length L carries P times L at the fixed end under a tip point load, or w times L squared divided by 2 under a uniform load. Continuous beams need an analysis, and their support moments often exceed the span moments.
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Reviewed for accuracy by Abdullah, Technical Content Specialist ยท Editorial policy
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