Mechanics of Materials · Chapter 11

Design of Beams and Shafts

Selecting beam sections for bending and shear, shaping fully stressed beams, and sizing shafts under combined bending and torsion.

Chapter Objectives

Distinguish strength, stiffness, stability, and connection requirements in a member selection.
Construct internal-resultant diagrams and select a beam from required section modulus.
Check shear, local buckling, bearing, and serviceability after a bending selection.
Relate a varying bending-moment diagram to a practical fully stressed beam profile.
Size a circular shaft from combined bending and torque while recognizing fatigue-sensitive details.

11.1Basis for Beam Design

Design for Strength

Strength design starts with equilibrium, not a catalog shape. A free-body diagram and cuts along the member give the internal shear \(V\), bending moment \(M\), axial force \(N\), and torque \(T\). The usual flexure and shear formulas idealize a straight, prismatic, homogeneous member that remains linear elastic and has small deformation. They predict nominal stresses away from load introduction, holes, weld toes, and abrupt section changes. A candidate also needs adequate lateral restraint, bearing detail, connection capacity, and the deflection limits developed in Chapter 12.

Stress Trajectories

At outer fibers, bending produces its largest normal stress; near the neutral axis, transverse shear is often largest. Principal-stress directions trace paths through a loaded member. In brittle materials, tensile cracking and reinforcement layout are strongly related to those paths, but trajectories do not replace a cracking, anchorage, or code check. Local compression beneath a concentrated load is a bearing problem, not the same as the average flexural stress.

Physical meaning

The section modulus \(S=I/c\) measures how effectively area is placed away from the neutral axis. Increasing depth is usually more efficient than adding equal area near that axis; however, a deeper web can introduce local buckling or fabrication constraints.

Problem-solving strategy

Draw the complete loading and support free-body diagram, then locate peaks in \(V\), \(M\), and \(T\).
Choose the governing limit state: bending, web shear, bearing, deflection, buckling, or shaft fatigue.
Use consistent force-length units to obtain required section properties before selecting dimensions.
Recalculate nominal stresses in the selected section and inspect the critical detail rather than only the span.
Record restraints, load combinations, safety factors, and code checks that the classroom model omits.

11.2Prismatic Beam Design

For many slender beams, bending governs first, but that is a result to verify rather than an assumption. At each candidate section, take \(M_{\max}\) from the moment diagram and compare the extreme-fiber nominal stress with the allowable or factored resistance basis being used.

Required Section Modulus

\[ S_{\mathrm{req}} = \frac{M_{\max}}{\sigma_{\mathrm{allow}}} \]

\(M_{\max}\) is an internal moment, while \(\sigma_{\mathrm{allow}}\) is a material/design limit; use one consistent design format.
For a simple shape, solve \(S=I/c\). For a standard shape, choose a tabulated \(S\) at least as large as required, then check the actual material, orientation, self-weight, and unbraced length.

Shear Check

\[ \tau_{allow} \ge \frac{V\,Q}{I\,t} \qquad\text{(wide-flange shortcut: } \tau_{avg} \approx \frac{V_{max}}{A_{web}}\text{)} \]

The shortcut is an average web stress, not the peak distribution. It is useful only where the web carries most shear and should be supplemented by web yielding, buckling, and load-introduction checks.

Wood, Steel, and Built-Up Sections

Actual, not nominal, timber dimensions govern section properties. Rolled steel properties are tabulated, but the designation alone does not establish bending, shear, lateral-torsional buckling, or connection resistance. In built-up members, moving material into flanges raises \(I\), while welds or fasteners must transfer the shear flow \(q=VQ/I\).

Procedure for Analysis — Beam Design

First establish reactions and internal-resultant envelopes; then use bending to establish a preliminary section and shear to test its web or full section. Finally inspect local bearing, lateral restraint, deflection, durability, and the connection load path. A stress calculation does not certify a member.

Physical meaning

Shear transfers load between neighboring beam slices, whereas moment creates the tensile and compressive couple across a section. A member can pass one check and fail the other, particularly near supports or concentrated loads.

11.3Fully Stressed Beams

Nonprismatic Beams

A prismatic beam is fully stressed in bending only where its moment is critical. Varying the area can make the nominal outer-fiber stress approach the target along the span, saving mass in cast members, haunches, or girders with cover plates. The simple flexure model is most credible for gradual tapers; abrupt changes add stress concentration, fabrication detail, and local stability concerns.

Shape of a Fully Stressed Beam

\[ S(x) = \frac{M(x)}{\sigma_{allow}} \]

Express \(M\) by position, then translate the resulting \(S(x)\) into a manufacturable geometry with minimum thickness and transition radii.
Shear, bearing, buckling, vibration, and connection requirements can impose a minimum section where bending would predict very little material.

11.4Shaft Design

Combined Bending and Torsion

Resolve gear or pulley forces in two perpendicular planes, draw a moment diagram in each, and combine them as \(M=\sqrt{M_y^2+M_z^2}\) for a circular shaft. A torque diagram gives the internal \(T\). At the surface, bending supplies \(\sigma=Mc/I\) and torsion supplies \(\tau=Tc/J\). Rotating shafts commonly see fully reversed bending and steady or fluctuating torque, so a static yield check is not a fatigue assessment; shoulders, keyways, and surface finish matter.

Shaft Radius — Maximum Shear Stress Theory (Solid Circular Shaft)

\[ \tau_{allow} = \sqrt{\left(\frac{Mc}{2I}\right)^2 + \left(\frac{Tc}{J}\right)^2} \qquad\Rightarrow\qquad c = \left(\frac{2}{\pi\,\tau_{allow}}\sqrt{M^2 + T^2}\right)^{1/3} \]

This is a static Tresca screening expression for a solid circular shaft, using \(J=2I=\pi c^4/2\).
Choose a failure theory appropriate to the material and loading history; fatigue design needs concentration, notch-sensitivity, and mean-stress information.

EXAMPLE 1Rectangular Beam Preliminary Size

A simply supported timber beam has a maximum internal bending moment of \(8.0\ \text{kN}\!\cdot\!\text{m}\). Its actual depth must be twice its width, and allowable bending stress is \(8.0\ \text{MPa}\). Find the minimum actual width from bending alone.

Recognition strategy

The requested geometry follows from a known internal moment and an allowable normal stress, so section modulus is the direct preliminary design tool.

Given information

\(M_{\max}=8.0\ \text{kN}\!\cdot\!\text{m}\), \(h=2b\), and \(\sigma_{\mathrm{allow}}=8.0\ \text{MPa}\).

Model and interpretation

Use a prismatic rectangular, linear-elastic beam with bending about its strong centroidal axis. Positive sagging moment puts the bottom fiber in tension; only stress magnitude is needed for sizing.

Step-by-step solution

1. Obtain the required section modulus. The nominal flexure criterion gives \(S_{\mathrm{req}}=M_{\max}/\sigma_{\mathrm{allow}}=8.0\times10^3/(8.0\times10^6)=1.00\times10^{-3}\ \text{m}^3\).

2. Express the rectangle property. With \(h=2b\), \(S=bh^2/6=b(2b)^2/6=(2/3)b^3\), which places depth efficiently away from the neutral axis.

3. Solve the bending condition. Setting \((2/3)b^3=1.00\times10^{-3}\ \text{m}^3\) gives \(b^3=1.50\times10^{-3}\ \text{m}^3\) and \(b=0.1145\ \text{m}\).

4. State a practical preliminary size. The corresponding depth is \(h=0.229\ \text{m}\); selection must round up to available actual timber dimensions before separate shear and deflection checks.

Verification

Units: Moment divided by stress has units \(\text{N}\!\cdot\!\text{m}/(\text{N}/\text{m}^2)=\text{m}^3\), matching \(S\); the cube root therefore returns a length.

Signs and direction: The sagging-moment sign changes which face is tensile, not the required magnitude; timber grade checks must still use the relevant tension and compression limits.

Plausibility: Substitution gives \(S=(2/3)(0.1145)^3=1.00\times10^{-3}\ \text{m}^3\), so the nominal stress returns to \(8.0\ \text{MPa}\).

Final result: \( \boxed{b_{\min}=115\ \text{mm},\quad h_{\min}=229\ \text{mm}} \)

Engineering interpretation

This is a bending-only lower bound. Timber shear parallel to grain, bearing at supports, moisture effects, lateral restraint, and the serviceability deflection limit can require a larger member.

EXAMPLE 2Gradually Tapered Cantilever

A cantilever of length \(2.4\ \text{m}\) carries \(6.0\ \text{kN}\) downward at its free end. A steel plate beam has constant width \(b=80\ \text{mm}\), allowable bending stress \(120\ \text{MPa}\), and a gradual rectangular taper. Determine its bending-based depth at the fixed end and at \(x=1.2\ \text{m}\) measured from the free end.

Recognition strategy

A varying moment and prescribed width signal a fully stressed profile: find \(M(x)\), set its section modulus equal to \(M(x)/\sigma_{\mathrm{allow}}\), then solve for depth.

Given information

\(P=6.0\ \text{kN}\), \(L=2.4\ \text{m}\), \(b=0.080\ \text{m}\), and \(\sigma_{\mathrm{allow}}=120\ \text{MPa}\).

Model and interpretation

Cut the cantilever a distance \(x\) from the free end. The internal moment magnitude is \(M(x)=Px\); a gradual taper permits use of nominal flexure, while the constant shear remains \(P\).

Step-by-step solution

1. Form the required modulus function. The free-body diagram of the outboard segment gives \(M(x)=6000x\ \text{N}\!\cdot\!\text{m}\), so \(S(x)=6000x/(120\times10^6)\).

2. Link modulus to depth. A rectangle has \(S=bh^2/6\), hence \(h(x)=\sqrt{6M(x)/(b\sigma_{\mathrm{allow}})}\), showing that depth varies with \(\sqrt{x}\).

3. Evaluate at the fixed end. At \(x=2.4\ \text{m}\), \(h=\sqrt{6(6000)(2.4)/(0.080)(120\times10^6)}=0.0949\ \text{m}\).

4. Evaluate the interior station. At \(x=1.2\ \text{m}\), the moment is half as large and \(h=\sqrt{6(6000)(1.2)/(0.080)(120\times10^6)}=0.0671\ \text{m}\).

Verification

Units: In \(6M/(b\sigma)\), the units are \(\text{N}\!\cdot\!\text{m}/[\text{m}(\text{N}/\text{m}^2)]=\text{m}^2\), so the square root correctly gives depth.

Signs and direction: The cantilever moment is hogging under the stated load; its sign identifies the tensile face, while the symmetric rectangular section uses its magnitude for depth.

Plausibility: Halving \(x\) should reduce depth by \(1/\sqrt{2}\); \(67.1/94.9=0.707\), confirming the expected fully stressed scaling.

Final result: \( \boxed{h(2.4\ \text{m})=94.9\ \text{mm},\quad h(1.2\ \text{m})=67.1\ \text{mm}} \)

Engineering interpretation

The mathematical profile trends to zero at the free end although shear remains \(6.0\ \text{kN}\). Fabrication, minimum plate thickness, connection geometry, and shear/buckling checks establish the real end depth.

EXAMPLE 3Static Shaft Screening Diameter

At a shoulder just inboard of a pulley, a solid ductile shaft has resultant bending moment \(M=1.80\ \text{kN}\!\cdot\!\text{m}\) and torque \(T=1.20\ \text{kN}\!\cdot\!\text{m}\). Using a static Tresca allowable shear stress of \(55\ \text{MPa}\), determine the minimum smooth-shaft diameter before keyway and fatigue factors are applied.

Recognition strategy

The surface sees simultaneous bending normal stress and torsional shear. For a solid circular ductile shaft under the stated static criterion, the combined Tresca diameter expression is appropriate.

Given information

\(M=1800\ \text{N}\!\cdot\!\text{m}\), \(T=1200\ \text{N}\!\cdot\!\text{m}\), and \(\tau_{\mathrm{allow}}=55\times10^6\ \text{Pa}\).

Model and interpretation

Use a smooth solid circular section at the critical shoulder. The resultant bending moment comes from perpendicular-plane diagrams; this preliminary model excludes stress concentration and cyclic loading.

Step-by-step solution

1. Combine the loading measures. The Tresca relation uses \(R=\sqrt{M^2+T^2}=\sqrt{1800^2+1200^2}=2163\ \text{N}\!\cdot\!\text{m}\).

2. Write the radius criterion. For a solid circle, \(c=[2R/(\pi\tau_{\mathrm{allow}})]^{1/3}\), obtained from the surface bending and torsion stresses with \(J=2I\).

3. Calculate the minimum radius. Substitution gives \(c=[2(2163)/(\pi(55\times10^6))]^{1/3}=0.02924\ \text{m}\).

4. Convert to diameter and select upward. The smooth theoretical diameter is \(d=2c=58.5\ \text{mm}\); a final nominal size requires the later detail and fatigue review.

Verification

Units: The quantity \(R/\tau\) has \((\text{N}\!\cdot\!\text{m})/(\text{N}/\text{m}^2)=\text{m}^3\), so its cube root is a physically valid radius.

Signs and direction: Reversing torque or bending changes stress-element directions but not this static magnitude; the actual rotating stress history is essential for the subsequent fatigue calculation.

Plausibility: Torsion alone would require a smaller diameter than combined loading, and the cubic relation means modest diameter increases substantially reduce nominal stress.

Final result: \( \boxed{d_{\min}=58.5\ \text{mm}\ \text{for the smooth static model}} \)

Engineering interpretation

Do not release this diameter for a keyed rotating shaft. A shoulder fillet, keyway concentration, fatigue strength, reliability, bearings, critical speed, and applicable design standard can control the final diameter.

Chapter recap

Use section cuts to turn applied loading into internal \(M\), \(V\), and \(T\) before choosing a section.
Select by section modulus, then check shear and the local, stability, connection, and serviceability limit states.
Before moving on, verify that you can derive a tapered profile and identify why a rotating shaft needs fatigue review.