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Beam stresses and failure criteria

The reactions and diagrams tell you which internal forces act at each section; this topic answers the remaining question: can the material take it? To do that, you go from the forces ( $N$ , $V$ , $M$ ) to the stresses at each point of the section, combine them into an equivalent stress with a failure criterion and compare it with the strength of the material. That is what the Analysis tab (safety factor) and the Design tab (required size) of the beam simulator do.

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Stresses in the section

The bending moment stretches the fibers on one side of the neutral axis and shortens those on the other, and the strain grows linearly with the distance $y$ from the neutral axis. For an elastic material the normal stress is linear too; if there is also an axial force $N$ , it adds uniformly:

σ (y) = \frac{N}{A} - \frac{M y}{I} σ_{m \overset{a}{ˊ} x} = \frac{∣ M ∣ c}{I} = \frac{∣ M ∣}{S}

$y$ is measured upward from the neutral axis, which passes through the centroid; $c$ is the distance to the farthest fiber and $I$ the second moment of area about the neutral axis. With the simulator's convention ( $M$ positive puts the bottom fiber in tension), the minus sign gives compression on top and tension at the bottom when $M > 0$ .

The shear force produces shear stresses that are not uniform: they are zero at the extreme fibers (a free surface carries no shear) and largest at the neutral axis. They come from the shear formula,

τ = \frac{V Q}{I t}

where $Q$ is the first moment, about the neutral axis, of the area above (or below) the point and $t$ the width of the section at that point.

Rectangular section with

M > 0

: the normal stress is linear and the shear stress parabolic.

Section	$I$	$S = I / c$	$τ_{max}$ (at the neutral axis)
Rectangular $b \times h$	$\frac{b h ^{3}}{12}$	$\frac{b h ^{2}}{6}$	$\frac{3 V}{2 A}$
Round, diameter $d$	$\frac{π d ^{4}}{64}$	$\frac{π d ^{3}}{32}$	$\frac{4 V}{3 A}$
I-section	from the section tables	from the tables	$\approx \frac{V}{h _{w} t _{w}}$ (the web carries almost all the shear)

The section modulus

S

sums up in one number how well a shape resists bending: that is why I-sections put the material far from the neutral axis. For the same area, a deeper section has a much larger

S

(it grows with

h^{2}

Plane stress and principal stresses

At each point of the beam a normal stress $σ$ (along the axis) and a shear stress $τ$ act together. It is a plane stress state with $σ_{y} = 0$ , and rotating the element you find two directions without shear, with the principal stresses:

σ_{1, 2} = \frac{σ}{2} \pm (\frac{σ}{2})^{2} + τ^{2} τ_{m \overset{a}{ˊ} x} = (\frac{σ}{2})^{2} + τ^{2}

Since $σ_{y} = 0$ , in a beam always $σ_{1} \geq 0 \geq σ_{2}$ : there is one principal tension and one principal compression. Mohr's circle (the Mohr's circle tool) draws exactly this calculation.

You do not need to check every point. At the critical section two are enough: the extreme fiber (A), with the largest $σ$ and no shear, and the neutral axis (B), with the largest $τ$ and only the $σ = N / A$ of the axial force. In I-sections it is worth also looking at the flange–web junction, where $σ$ and $τ$ are both large.

Stress elements at the two points that are checked.

Criteria for ductile materials

A ductile material (steel, aluminum) fails when it yields: it deforms permanently. The tensile test gives the yield strength $S_{y}$ with a single stress; for a combined state you need a criterion that says which combination of $σ$ and $τ$ is equivalent to that test:

von Mises (distortion energy): $σ_{v m} = σ_{1}^{2} - σ_{1} σ_{2} + σ_{2}^{2} = σ^{2} + 3 τ^{2}$ . It is the one that best matches test data.
Tresca (maximum shear): $σ_{T} = σ_{1} - σ_{2} = σ^{2} + 4 τ^{2}$ . It always gives an equal or larger stress, so it is more conservative.

FS = \frac{S _{y}}{σ _{eq}}

Both agree when there is only $σ$ (the extreme fiber) and differ most in pure shear: von Mises predicts yielding at $τ = S_{y} / 3 \approx 0.577 S_{y}$ and Tresca at $τ = 0.5 S_{y}$ , 15 % less. In the figure, the von Mises ellipse encloses the Tresca hexagon and touches it at its six vertices.

Criteria for brittle materials

A brittle material (cast iron, ceramics, concrete) does not yield: it fractures. It is also much stronger in compression ( $S_{u c}$ ) than in tension ( $S_{u t}$ ), so a single number is not enough. With the principal stresses ordered, $σ_{A} \geq σ_{B}$ (in a beam, $σ_{A} \geq 0 \geq σ_{B}$ ), the usual criteria are:

Criterion	Safety factor $n$	Comment
Maximum normal stress	$n = min (\frac{S _{u t}}{σ _{A}}, \frac{S _{u c}}{∣ σ _{B} ∣})$	the simplest; it can be unsafe with tension and compression together
Coulomb–Mohr	$\frac{1}{n} = \frac{σ _{A}}{S _{u t}} - \frac{σ _{B}}{S _{u c}}$	the most conservative
Modified Mohr	if $∣ σ_{B} ∣ \leq σ_{A}$ : $n = \frac{S _{u t}}{σ _{A}}$ ; otherwise $\frac{1}{n} = \frac{( S _{u c} - S _{u t} ) σ _{A}}{S _{u c} S _{u t}} - \frac{σ _{B}}{S _{u c}}$	the best fit to cast-iron test data

Safe regions of each criterion in the plane of the principal stresses. On the right, the fourth quadrant, where every point of a beam lies.

Practical consequence: in a beam of brittle material the tension fiber almost always governs. That is why cast-iron beams often have unsymmetric (T) sections with the flange on the tension side: that fiber ends up closer to the neutral axis and its stress drops.

Safety factor and sizing the section

Checking a beam means computing the safety factor of a given section. Designing it is the reverse: choosing the smallest size for a required safety factor. With the allowable stress $σ_{allow} = S_{y} / FS$ and a section where bending governs,

S \geq \frac{∣ M ∣ _{m \overset{a}{ˊ} x}}{σ _{perm}} ⟹ d = 3 \frac{32 ∣ M ∣ _{m \overset{a}{ˊ} x} FS}{π S _{y}} (secci \overset{o}{ˊ} n circular)

The value of $FS$ is set by the design code or the problem statement; it reflects the uncertainty in loads, material and model.
The calculated size is rounded up to a practical size, and the actual safety factor is recomputed with it.
Then check what the formula does not look at: shear at the supports, deflection and, if there is compression, buckling.

The simulator's Design tab does all this for six shapes at once (round, tube, square, rectangular…) and compares their weight: for the same strength, hollow sections weigh much less.

Common mistakes

Using the second moment of area about the base instead of the neutral axis (the one through the centroid).
Combining the maximum $σ$ with the maximum $τ$ at the same point: in a rectangular section they do not occur together.
Using von Mises or Tresca with a brittle material: they would compare with a «yield strength» that does not exist and do not tell tension from compression.
Mixing units: with $M$ in kN·m, $c$ in mm and $I$ in mm⁴, convert $M$ to N·mm ( $\times 1 0^{6}$ ) to get MPa.
Rounding the design size down, or not recomputing the safety factor with the final size.

Worked examples

Example 1Steel cantilever: von Mises and Tresca

A 1 m A36 steel cantilever (

S_{y} = 250

MPa), with a rectangular section 100 mm wide and 200 mm deep, carries 40 kN at its free end. Find the stresses at the extreme fiber and at the neutral axis of the critical section, the von Mises and Tresca stresses, and the safety factor.

Solution

Critical section: the fixed end, where $∣ M ∣ = 40 \cdot 1 = 40$ kN·m and $V = 40$ kN.
Properties: $I = \frac{100 \cdot 20 0 ^{3}}{12} = 66.67 \times 1 0^{6}$ mm⁴, $c = 100$ mm, $A = 20000$ mm².
Point A (extreme fiber): $σ = \frac{40 \times 1 0 ^{6} \cdot 100}{66.67 \times 1 0 ^{6}} = 60$ MPa (tension on top, compression at the bottom) and $τ = 0$ . So $σ_{v m} = σ_{T} = 60$ MPa.
Point B (neutral axis): $σ = 0$ and $τ = \frac{3 V}{2 A} = \frac{3 \cdot 40 000}{2 \cdot 20 000} = 3$ MPa, so $σ_{v m} = 3 \cdot 3 = 5.196$ MPa and $σ_{T} = 2 \cdot 3 = 6$ MPa.
Point A governs: $FS = \frac{250}{60} = 4.17$ with both criteria. In a long solid beam shear hardly matters.

σ_{max} = 60

MPa,

τ_{max} = 3

MPa;

σ_{v m} = σ_{T} = 60

MPa at A and

FS = 4.17

Solve it in Simuveliq

In Engineering → Beams, length 1 m; in Supports keep a single Fixed support at 0.
In Section and material: material A36 steel, section type Solid rectangle and standard shape 100 × 200 mm; failure criterion von Mises (ductile).
In Loads, a 40 kN + Point load downward (↓ −y) at $x = 1$ .
Read the Max normal stress, von Mises · Tresca and Safety factor cards. In the procedure, the «Failure criteria: von Mises and Tresca» step has the table with points A and B (the simulator takes A at the bottom fiber, in compression: for a ductile material it makes no difference).

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Example 2The same cantilever in cast iron

Repeat the previous example with ASTM 30 gray cast iron (

S_{u t} = 214

MPa,

S_{u c} = 752

MPa, the simulator's values). Use the three criteria for brittle materials.

Solution

The stresses do not depend on the material: $\pm 60$ MPa at the extreme fibers and $τ = 3$ MPa at the neutral axis.
Top fiber (tension): $σ_{A} = 60$ , $σ_{B} = 0$ . All three criteria give $n = \frac{214}{60} = 3.57$ .
Bottom fiber (compression): $σ_{A} = 0$ , $σ_{B} = - 60$ . All three give $n = \frac{752}{60} = 12.5$ : the iron takes 3.5 times more in compression.
Neutral axis: $σ_{A, B} = \pm 3$ MPa. Maximum normal stress and modified Mohr ( $∣ σ_{B} ∣ \leq σ_{A}$ ): $n = 214/3 = 71.3$ ; Coulomb–Mohr: $\frac{1}{n} = \frac{3}{214} + \frac{3}{752}$ , $n = 55.5$ .

The tension fiber governs:

n = 3.57

with any of the three criteria.

Solve it in Simuveliq

With the model of example 1, change the material to Gray cast iron ASTM 30 (brittle) and the failure criterion to Modified Mohr (brittle).
The Equivalent stress and Safety factor cards give 60 MPa and 3.57.
The «Failure criterion for a brittle material» step of the procedure shows the table with the three criteria at points A and B; try changing the criterion and you will see the result does not change, because there is no compression at A.

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Example 3Diameter of a round bar

A solid round A36 steel bar, simply supported over a 3 m span, carries 10 kN/m along its whole length. Find the smallest diameter for a safety factor of 2 with von Mises.

Solution

Maximum moment, at midspan: $M = \frac{w L ^{2}}{8} = \frac{10 \cdot 3 ^{2}}{8} = 11.25$ kN·m. There $V = 0$ , so there is only bending.
Allowable stress: $σ_{allow} = \frac{250}{2} = 125$ MPa, so $S \geq \frac{11.25 \times 1 0 ^{6}}{125} = 90000$ mm³.
$\frac{π d ^{3}}{32} = 90000 \Rightarrow d = 97.14$ mm. Practical size: 98 mm.
With 98 mm: $σ = \frac{32 \cdot 11.25 \times 1 0 ^{6}}{π \cdot 9 8 ^{3}} = 121.8$ MPa and $FS = \frac{250}{121.8} = 2.05$ .
Shear at the supports: $V = 15$ kN and $τ_{max} = \frac{4 V}{3 A} = \frac{4 \cdot 15 000}{3 \cdot 7543} = 2.65$ MPa, far below; and there $M = 0$ .

d_{req} = 97.14

mm; with the practical size of 98 mm,

FS = 2.05

Solve it in Simuveliq

Length 3 m, Pin at 0 and Roller at 3; a uniform + Distributed load from 0 to 3 with 10 kN/m; material A36 steel.
Open the Design tab: section shape Solid circle, Minimum FS 2 and criterion von Mises (ductile).
Read Required size (97.14 mm), Use (practical size) (98 mm) and Achieved FS (2.05). «How it was calculated» repeats the formula above, and the table compares the other shapes: the rectangular tube is the lightest.

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Beam stresses and failure criteria

Stresses in the section

Plane stress and principal stresses

Criteria for ductile materials

Criteria for brittle materials

Safety factor and sizing the section

Common mistakes

Worked examples

Example 1Steel cantilever: von Mises and Tresca

Solution

Solve it in Simuveliq

Example 2The same cantilever in cast iron

Solution

Solve it in Simuveliq

Example 3Diameter of a round bar

Solution

Solve it in Simuveliq

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