Beam Dimensions Under Deflection And Material Limits

Why Beam Dimensions Must Be Chosen Together

A rectangular beam can use little material yet still bend too much under its load. Increasing its depth is an efficient way to reduce bending deflection, but depth may be limited by the available space. Increasing width also adds stiffness, though it uses material differently. A useful design therefore balances material use against the mechanical requirements and the permitted dimensions.

We will analyze a uniform beam with a rectangular cross-section. The span, load, and material are specified; the design variables are the cross-section’s width and depth. The goal is to minimize material while meeting a deflection limit and, when required, an allowable bending-stress limit.

Model assumptions: The beam is simply supported at its ends and carries a point load at its midpoint. It is made of a homogeneous, linearly elastic material, and its deflection is small enough for Euler–Bernoulli beam theory. The depth is the vertical dimension, so bending occurs about the cross-section’s horizontal centroidal axis.

Deflection and Material Use

Let \(b\) be the beam’s width, \(h\) its depth, \(L\) its span, \(P\) the applied midpoint load, and \(E\) the material’s Young’s modulus. For a rectangular cross-section, the second moment of area about the horizontal centroidal axis is \(I=bh^3/12\). Under the stated loading and support conditions, the maximum deflection occurs at midspan and is \(PL^3/(48EI)\).

\delta_{\max,\mathrm{actual}} = \frac{PL^3}{48EI} = \frac{PL^3}{4Ebh^3}.

Suppose the permitted deflection is \(\delta_{\mathrm{lim}}>0\). The requirement is that the actual maximum deflection not exceed this limit. Rearranging the formula gives a lower bound on the product \(bh^3\).

\frac{PL^3}{4Ebh^3}\leq \delta_{\mathrm{lim}} \quad\Longleftrightarrow\quad bh^3\geq K, \qquad K=\frac{PL^3}{4E\delta_{\mathrm{lim}}}.

Here \(K\) has units of length to the fourth power, matching \(bh^3\). If the beam has uniform material and constant span, its volume is \(Lbh\). Minimizing the material volume is therefore equivalent to minimizing the cross-sectional area \(bh\).

Definition: In this design problem, the material-use objective is the cross-sectional area \(A=bh\). For a fixed span \(L\) and uniform material, the beam volume is \(V=Lbh\), so both objectives have the same minimizing dimensions.

Adding an Allowable-Stress Limit

A stiffness requirement limits deflection; it does not by itself ensure that the beam’s bending stress is acceptable. For the same simply supported beam with a midpoint load, the maximum bending moment is \(M_{\max}=PL/4\). The rectangular section modulus is \(S=bh^2/6\), and the maximum bending stress is \(M_{\max}/S\). Thus an allowable stress \(\sigma_{\mathrm{lim}}>0\) requires:

\frac{3PL}{2bh^2}\leq \sigma_{\mathrm{lim}} \quad\Longleftrightarrow\quad bh^2\geq J, \qquad J=\frac{3PL}{2\sigma_{\mathrm{lim}}}.

The units of \(J\) are length cubed, matching \(bh^2\). The stress constraint is optional when the design specification does not impose a strength limit. When both limits apply, the width must satisfy both inequalities.

How to Find the Least-Material Design

Let the permitted dimensions be \(b_{\min}\leq b\leq b_{\max}\) and \(h_{\min}\leq h\leq h_{\max}\), where all bounds are positive. At any fixed depth \(h\), the smallest width that meets the lower bounds from deflection and stress is the largest of the required widths. If no stress limit applies, omit the stress term.

b_{\mathrm{req}}(h) = \max\left( b_{\min}, \frac{K}{h^3}, \frac{J}{h^2} \right), \qquad A(h)=h\,b_{\mathrm{req}}(h).

A depth is feasible only if \(b_{\mathrm{req}}(h)\leq b_{\max}\) and \(h\) lies within its permitted range. These conditions give the feasible depth interval:

h_{\mathrm{low}} = \max\left( h_{\min}, \left(\frac{K}{b_{\max}}\right)^{1/3}, \left(\frac{J}{b_{\max}}\right)^{1/2} \right), \qquad h_{\mathrm{high}}=h_{\max}.

Omit the term involving \(J\) if there is no stress limit. If \(h_{\mathrm{low}}>h_{\mathrm{high}}\), the constraints cannot be met by any permitted dimensions. Otherwise, the optimum can be found by identifying where the width bound \(b_{\min}\) takes over from the mechanical requirements.

Theorem: When the feasible depth interval is nonempty, define \(h_0=\max\big((K/b_{\min})^{1/3},(J/b_{\min})^{1/2}\big)\), omitting the stress term when it does not apply. The minimizing depth is \(h_*=\min(h_{\mathrm{high}},\max(h_{\mathrm{low}},h_0))\), and the minimizing width is \(b_*=b_{\mathrm{req}}(h_*)\).

Why the Optimization Rule Works

Multiplying the required width by the depth expresses the area as the largest of one increasing term and the mechanical terms that decrease with depth:

A(h) = \max\left( b_{\min}h,\, \frac{K}{h^2},\, \frac{J}{h} \right).

The term \(b_{\min}h\) increases with \(h\). The deflection term \(K/h^2\) and the stress term \(J/h\) decrease with \(h\). The depth \(h_0\) is the point by which both decreasing terms have fallen to at most \(b_{\min}h\): these comparisons are equivalent to \(h\geq (K/b_{\min})^{1/3}\) and \(h\geq (J/b_{\min})^{1/2}\). Hence, below \(h_0\), at least one mechanical term controls the area, and the area decreases as depth increases. Above \(h_0\), the width minimum controls and the area increases with depth. The least area is therefore at \(h_0\), unless a feasible interval endpoint is closer on the relevant side.

Find the mechanical requirements.
Compute \(K\) from the deflection limit and, if applicable, \(J\) from the stress limit.

Check feasibility.
Calculate \(h_{\mathrm{low}}\); if it exceeds \(h_{\max}\), no permitted depth can satisfy the requirements.

Locate the best depth.
Clamp \(h_0\) to the feasible interval to obtain \(h_*\).

Choose the least permitted width.
Set \(b_*=b_{\mathrm{req}}(h_*)\), then verify the constraints and units.

Worked Examples

Worked Example: Minimum Area Under a Deflection Limit

A beam has span \(L=2\,\mathrm{m}\), midpoint load \(P=1200\,\mathrm{N}\), and modulus \(E=200\,\mathrm{GPa}=200\times10^9\,\mathrm{N/m^2}\). Its permitted maximum deflection is \(3\times10^{-5}\,\mathrm{m}\). The bounds are \(0.05\leq b\leq0.12\,\mathrm{m}\) and \(0.15\leq h\leq0.30\,\mathrm{m}\). There is no separate stress limit.

First calculate the deflection constant:

K = \frac{(1200\,\mathrm{N})(2\,\mathrm{m})^3} {4(200\times10^9\,\mathrm{N/m^2})(3\times10^{-5}\,\mathrm{m})} = \frac{9600}{24\times10^6}\,\mathrm{m^4} = 0.0004\,\mathrm{m^4}.

The depth must be at least \(0.15\,\mathrm{m}\) and at least \((K/b_{\max})^{1/3}=(0.0004/0.12)^{1/3}\,\mathrm{m}\), approximately \(0.1494\,\mathrm{m}\). Thus the feasible depth interval is \([0.15,0.30]\,\mathrm{m}\). The crossover depth is \(h_0=(K/b_{\min})^{1/3}=(0.0004/0.05)^{1/3}=0.20\,\mathrm{m}\), which lies in that interval.

At \(h_*=0.20\,\mathrm{m}\), the required width is \(b_*=\max(0.05,0.0004/0.20^3)\,\mathrm{m}=0.05\,\mathrm{m}\). The area is \(A_*=0.05(0.20)=0.010\,\mathrm{m^2}\), and the volume is \(LA_*=0.020\,\mathrm{m^3}\).

Check the deflection directly: \(bh^3=(0.05)(0.20)^3=0.0004\,\mathrm{m^4}\), so \(\delta=9600/(4(200\times10^9)(0.0004))\,\mathrm{m}=3\times10^{-5}\,\mathrm{m}\), exactly the limit. Both dimensions meet their bounds.

Worked Example: Deflection and Stress Limits Meet at the Optimum

Keep the beam, deflection limit, and dimension bounds from the previous example. Now also impose an allowable bending stress of \(\sigma_{\mathrm{lim}}=1.8\,\mathrm{MPa}=1.8\times10^6\,\mathrm{N/m^2}\). The stress constant is:

J = \frac{3(1200\,\mathrm{N})(2\,\mathrm{m})} {2(1.8\times10^6\,\mathrm{N/m^2})} = 0.002\,\mathrm{m^3}.

The deflection crossover is \((K/b_{\min})^{1/3}=0.20\,\mathrm{m}\). The stress crossover is \((J/b_{\min})^{1/2}=(0.002/0.05)^{1/2}=0.20\,\mathrm{m}\). Therefore \(h_0=0.20\,\mathrm{m}\), and it is feasible. At that depth, \(K/h^3=0.05\,\mathrm{m}\) and \(J/h^2=0.05\,\mathrm{m}\), so \(b_*=0.05\,\mathrm{m}\).

The deflection check is the same as in the first example. For stress, \(bh^2=(0.05)(0.20)^2=0.002\,\mathrm{m^3}\), giving \(\sigma=7200/(2(0.002))\,\mathrm{N/m^2}=1.8\times10^6\,\mathrm{N/m^2}=1.8\,\mathrm{MPa}\). Both mechanical limits are met at equality, and the dimensions satisfy their bounds.

Worked Example: An Upper Depth Bound Forces the Optimum

Use the same deflection requirement and width bounds as in the first example, but restrict the depth to \(0.15\leq h\leq0.16\,\mathrm{m}\). The crossover depth remains \(0.20\,\mathrm{m}\), above the permitted interval. Since area decreases with depth below the crossover, the best feasible choice is the largest permitted depth, \(h_*=0.16\,\mathrm{m}\).

The minimum required width is \(b_*=K/h_*^3=0.0004/(0.16)^3=0.0004/0.004096=0.09765625\,\mathrm{m}\). This lies between \(0.05\,\mathrm{m}\) and \(0.12\,\mathrm{m}\). The area is \(0.16(0.09765625)=0.015625\,\mathrm{m^2}\).

Check the stiffness requirement: \(b_*h_*^3=(0.09765625)(0.004096)=0.0004\,\mathrm{m^4}=K\). Substitution into the beam formula therefore gives \(\delta=3\times10^{-5}\,\mathrm{m}\), the specified limit. The width and depth are both within bounds, so this is a feasible design and the rule selects the upper depth bound correctly.

Common Pitfall: Dropping the Factor in the Second Moment

The rectangular section has \(I=bh^3/12\), not \(I=bh^3\). If the factor \(1/12\) is omitted, the resulting deflection constraint is too weak by a factor of 12. For the first example, that error would use \(K=0.0004/12=0.0000333\ldots\,\mathrm{m^4}\). At the smallest dimensions \(b=0.05\,\mathrm{m}\), \(h=0.15\,\mathrm{m}\), it would falsely conclude that \(bh^3=0.00016875\,\mathrm{m^4}\) exceeds the required amount.

With the correct formula, however, \(bh^3=0.00016875\,\mathrm{m^4}<0.0004\,\mathrm{m^4}\). The actual deflection is \(3\times10^{-5}(0.0004/0.00016875)\,\mathrm{m}\), approximately \(7.11\times10^{-5}\,\mathrm{m}\), which exceeds the \(3\times10^{-5}\,\mathrm{m}\) limit. Check the section formula before optimizing: a small modeling error changes the feasible designs and can lead to an unsafe choice.

Check Your Understanding

Use the beam model and optimization rule to answer these questions.

Why does minimizing \(bh\) also minimize material volume when the span and material are fixed?
Which product of width and depth is constrained by a deflection limit, and which by an allowable-stress limit?
What does it mean physically if \(h_{\mathrm{low}}>h_{\max}\)?
How does the optimal depth change when the crossover depth lies above the permitted depth interval?
Why is it incorrect to use \(I=bh^3\) for a rectangular cross-section?