Mechanics of Materials · Chapter 6

Bending

Shear and moment diagrams, the flexure formula, and bending of unsymmetric, composite, and curved members.

Chapter Objectives

Construct shear and moment diagrams by the method of sections and by the graphical (slope/area) method.
Derive and apply the flexure formula σ = Mc/I for straight, symmetric, linear elastic beams.
Handle moments applied about arbitrary axes (unsymmetric bending) using principal axes.
Analyze composite, reinforced concrete, and curved beams; apply stress concentration factors; describe inelastic bending.

6.1Shear and Moment Diagrams

Beams

Slender members that carry loads applied perpendicular to their longitudinal axis — generally long, straight bars of constant cross section. Classified by support: simply supported (pin + roller), cantilevered (fixed + free), overhanging (ends extending past the supports). Floors, bridge decks, aircraft wings, axles, crane booms — even bones — act as beams.

Shear and Moment Diagrams

Loads make the internal shear \(V\) and bending moment \(M\) vary along the beam. Expressing \(V(x)\) and \(M(x)\) and plotting them gives the shear and moment diagrams, from which the maximum values — needed for design — are read directly. Because concentrated forces and couples create discontinuities, separate coordinates must be used for each region between load discontinuities.

Beam Sign Convention

By common engineering practice, positive directions are: distributed load acting upward; internal shear causing a clockwise rotation of the segment it acts on; internal moment causing compression in the top fibers, bending the segment so it "holds water." Opposite senses are negative.

Procedure for Analysis — V and M Functions

1. Support reactions. Find all reactions and resolve forces perpendicular to the beam's axis.

2. Section. For each region between discontinuities, section the beam at position \(x\), draw the free-body diagram of one segment with \(V\) and \(M\) shown in their positive sense.

3. Equilibrium. Sum forces for \(V(x)\); sum moments at the section for \(M(x)\). Plot the functions.

6.2Graphical Method for Constructing Shear and Moment Diagrams

Writing \(V(x)\) and \(M(x)\) for many-load beams is tedious. Equilibrium of a differential beam segment (away from concentrated loads) yields two differential relations that let the diagrams be sketched directly:

Slope Relations

\[ \frac{dV}{dx} = w(x) \qquad\qquad \frac{dM}{dx} = V \]

The slope of the shear diagram at each point equals the distributed-load intensity there.
The slope of the moment diagram at each point equals the shear there — so M is maximum or minimum where V = 0.

Area Relations

\[ \Delta V = \int w\,dx \qquad\qquad \Delta M = \int V\,dx \]

Change in shear between two points = area under the loading curve between them.
Change in moment between two points = area under the shear diagram between them.
Not applicable across points of concentrated force or couple.

Jumps at Concentrated Loads

\[ \Delta V = F \qquad\qquad \Delta M = M_0 \]

An upward force F makes the shear diagram jump up by F (downward → jump down).
A clockwise couple M₀ makes the moment diagram jump up by M₀ (counterclockwise → jump down).

6.3Bending Deformation of a Straight Member

Neutral Surface and Neutral Axis

A bending moment stretches the material on one side of a beam and compresses it on the other; between them lies the neutral surface, where longitudinal fibers do not change length. Grid lines on a bent bar show that horizontal lines curve while vertical lines remain straight but rotate. The intersection of the neutral surface with a cross section is the neutral axis (the z axis).

Three Kinematic Assumptions

1. The longitudinal axis (in the neutral surface) does not change length — it merely curves.

2. Cross sections remain plane and perpendicular to the longitudinal axis during deformation.

3. Lateral (Poisson) strains are neglected — the cross section keeps its shape.

The discussion assumes a prismatic beam, homogeneous material, a cross section symmetric about an axis, and moment applied perpendicular to that axis of symmetry.

Normal Strain Distribution in Bending

\[ \epsilon = -\frac{y}{\rho} \qquad\qquad \epsilon = -\left(\frac{y}{c}\right)\epsilon_{max} \]

\(y\) = distance above the neutral axis; \(\rho\) (rho) = radius of curvature of the longitudinal axis
The strain varies linearly with depth: zero at the neutral axis, extreme at the outermost fiber \(y = \pm c\). For positive M, fibers above the neutral axis shorten (−), fibers below elongate (+). This result is geometric — independent of material.

6.4The Flexure Formula

For linear elastic material, Hooke's law turns the linear strain variation into a linear stress variation: \(\sigma = -(y/c)\,\sigma_{max}\). Two equilibrium conditions then fix everything. The resultant force of the stress distribution must vanish, which requires \(\int_A y\,dA = 0\) — so the neutral axis passes through the centroid of the cross section. And the stress distribution's moment must equal M, which yields:

The Flexure Formula

\[ \sigma_{max} = \frac{M\,c}{I} \qquad\qquad \sigma = -\frac{M\,y}{I} \]

\(\sigma_{max}\) = maximum normal stress, at the point farthest from the neutral axis (\(y = c\))
\(M\) = internal moment about the neutral axis; \(I\) = moment of inertia of the cross section about the neutral axis
Signs: for positive M, positive y gives compression (−σ); negative y gives tension (+σ).

Bending stress distribution σ = −My/I over the depth of a beam: zero at the neutral axis, extreme values at the outer fibers.

FIG. 6-1 (original diagram)

6.5Unsymmetric Bending

Moment About a Principal Axis

The flexure formula actually holds for a cross section of any shape, provided the moment acts about a principal axis of inertia. Equilibrium imposes three conditions on the stress distribution — zero axial force, zero moment about y, and moment M about z — and the second is satisfied only if \(\int_A yz\,dA = 0\): the product of inertia must vanish, which defines the principal axes. For sections with an axis of symmetry, the principal axes are simply the symmetry axis and its perpendicular.

Moment Applied at an Angle — Superposition of Components

\[ \sigma = -\frac{M_z\,y}{I_z} + \frac{M_y\,z}{I_y} \qquad\text{with}\quad M_z = M\cos\theta,\;\; M_y = M\sin\theta \]

Resolve M into components along the principal axes, apply the flexure formula to each, and superpose. Tension positive.
\(y, z\) = point coordinates from the centroid; \(I_z, I_y\) = principal (maximum, minimum) moments of inertia.
Maximum tensile and compressive stresses occur at opposite corners of the section.

Orientation of the Neutral Axis

\[ \tan\alpha = \frac{I_z}{I_y}\,\tan\theta \]

Setting σ = 0 gives the neutral-axis line. Since \(I_z \ne I_y\) in general, \(\alpha \ne \theta\): the neutral axis is not perpendicular to the plane of loading unless the moment acts about a principal axis.

6.6Composite Beams

The Transformed-Section Method

Beams built from two or more materials (steel-reinforced wood, bimetallic strips) do not satisfy the single-material assumption of the flexure formula. The fix: transform the section into an equivalent one made of a single material. The transformation factor

\[ n = \frac{E_1}{E_2} \]

scales the width of material 1 so it behaves like material 2 (stiffer material → wider equivalent section; less stiff → narrower). The strain distribution remains linear through the depth; the flexure formula is applied to the transformed section about its centroid; and the stress computed in any transformed region must be multiplied back by n to recover the actual stress in the original material.

6.7Reinforced Concrete Beams

Concrete + Steel

Concrete is very weak in tension (and its tensile cracking is sudden), so concrete beams are reinforced with steel bars placed in the tension zone, as far from the neutral axis as cover allows. Design assumes the concrete below the neutral axis carries no stress — it is assumed cracked — so the section that "works" is the concrete compression block above the neutral axis plus the transformed steel area \(nA_{st}\) (with \(n = E_{st}/E_{conc}\)) below it.

The neutral-axis depth \(h'\) is found by requiring the first moment of the compression concrete about the neutral axis to balance that of the transformed steel — a quadratic in \(h'\) — after which the flexure formula on the transformed section gives the concrete compressive stress and (after scaling by n) the steel tensile stress.

6.8Curved Beams

When Curvature Matters

For sharply curved members — crane hooks, chain links, rings — the flexure formula no longer applies: fibers near the inner radius are much shorter than those near the outer radius, so equal rotations produce larger strains on the inside. Cross sections still remain plane, but the strain (and stress) varies hyperbolically with radius, and the neutral axis no longer passes through the centroid — it shifts toward the center of curvature. If the radius of curvature exceeds about five times the member's depth, the flexure formula is accurate to within ~7 % and is normally used instead.

Location of the Neutral Axis (Curved Beam)

\[ R = \frac{A}{\displaystyle\int_A \frac{dA}{r}} \]

\(R\) = radius of the neutral axis, measured from the center of curvature; \(A\) = cross-sectional area
\(r\) = radius to the area element dA. The integral is tabulated for common shapes (e.g. rectangle of width b between radii \(r_1, r_2\): \(\int dA/r = b\ln(r_2/r_1)\)).

The Curved-Beam (Winkler) Formula

\[ \sigma = \frac{M\,(R - r)}{A\,r\,(\bar{r} - R)} \]

\(M\) = internal moment about the neutral axis (positive if it tends to increase the radius of curvature, i.e. straighten the member)
\(\bar{r}\) = radius to the centroid; \(\bar{r} - R\) = distance between centroid and neutral axis (small — compute precisely)
Radial stresses also exist but are generally negligible for solid sections.

6.9Stress Concentrations

Bending Stress Concentrations

At sudden changes of cross section — shoulder fillets, notches, and holes in flexural members — the flexure formula underestimates the local peak stress. As for axial load and torsion, the remedy is a stress concentration factor K from experimental charts (functions of geometry ratios such as w/h and r/h). K matters for brittle materials and fatigue loading; ductile materials under static moment redistribute the peak by local yielding.

Maximum Bending Stress at a Discontinuity

\[ \sigma_{max} = K\,\frac{M\,c}{I} \]

\(Mc/I\) is evaluated at the smaller cross section at the discontinuity.

6.10Inelastic Bending

Plastic Moment and Shape Factor

If the moment exceeds the elastic capacity, the outer fibers yield first (at the yield moment \(M_Y = \sigma_Y I/c\)) and yielding then spreads inward at constant stress \(\sigma_Y\) (elastic perfectly plastic model). Two conditions always hold: strain still varies linearly with depth, and the stress distribution must produce zero axial force — which for a fully yielded section places the neutral axis so it divides the area into two equal halves — and a moment equal to M.

When the entire section has yielded, the beam carries its plastic moment. For a rectangle b × h:

\[ M_p = \sigma_Y\,\frac{bh^2}{4} = \frac{3}{2}\,M_Y \]

The ratio \(k = M_p/M_Y\) is the shape factor (1.5 for a rectangle, ≈1.1–1.2 for wide-flange shapes) — a measure of the reserve capacity beyond first yield. As with torsion, releasing a moment that caused plastic strain leaves residual stresses, found by superposing the elastic unloading distribution (−M applied elastically) on the plastic stress distribution; a permanent curvature also remains.

Problem-solving strategy

Draw the whole-member and cut free bodies before choosing a stress relation.
Declare axes and signs, then determine the internal resultant at the critical location.
Select a relation whose geometry and material assumptions match the member.
Check units, directions, strength, instability, and serviceability limits.

Physical meaning

The internal resultants describe how neighboring material resists the applied loading; stress is their local intensity, not the applied force.

FOUNDATIONMoment and stress in a shelf beam

A simply supported 3.0 m shelf beam carries a 4.0 kN point load at midspan. Its rectangular section is 60 mm wide by 120 mm deep. Determine the peak bending stress.

Recognition strategy

A transverse load first requires reactions and an internal moment diagram; then the straight-beam flexure formula converts the largest moment to normal stress.

Given information

The load is centered, \(L=3.0,\text{m}\), \(P=4.0,\text{kN}\), \(b=0.060,\text{m}\), and \(h=0.120,\text{m}\).

Model and interpretation

Use upward reactions positive and sagging moment positive. Assume a prismatic homogeneous beam, small curvature, plane sections, and elastic behavior.

Step-by-step solution

1. Symmetry and vertical force equilibrium give reactions of \(2.0,\text{kN}\) at each support.

2. A section just left of midspan gives \(M_{\max}=2.0(1.5)=3.0\,\text{kN}\!\cdot\!\text{m}\).

3. The centroidal section property is \(I=bh^3/12=8.64\times10^{-6},\text{m}^4\), with \(c=h/2=0.060,\text{m}\).

4. Using \(\sigma=-My/I\), the extreme-fiber magnitude is \(3000(0.060)/(8.64\times10^{-6})=20.8\,\text{MPa}\).

Verification

Units: \(\text{N}\!\cdot\!\text{m}^2/\text{m}^4\) reduces to pascals, confirming the stress calculation.

Signs and direction: Positive sagging places the top fiber in compression and the bottom fiber in tension under the stated convention.

Plausibility: Doubling depth would reduce this stress by four because section modulus for a rectangle scales with \(h^2\).

Final result: \(\boxed{|\sigma_{max}|=20.8,\text{MPa}}\)

Engineering interpretation

The result is normal stress, not shear force. Check both tension and compression allowables and deflection; bearing under the load and any notches can control a real shelf connection.

INTERMEDIATEMaximum moment under a partial load

A 5.0 m simply supported beam carries a uniform downward load of 3.0 kN/m over the left 3.0 m only. Determine the location and value of the maximum sagging moment.

Recognition strategy

A partial distributed load creates separate regions. The maximum moment occurs where the shear function changes sign, not automatically at midspan.

Given information

The resultant load is \(9.0,\text{kN}\) at \(1.5,\text{m}\) from the left support; no other loads act.

Model and interpretation

Take upward forces and sagging moments positive. Use a left-segment free body and represent the distributed load only where the cut lies within its footprint.

Step-by-step solution

1. Moment equilibrium about the left support gives \(R_B(5)-9(1.5)=0\), so \(R_B=2.70,\text{kN}\) and \(R_A=6.30,\text{kN}\).

2. For \(0\le x\le3\), the shear is \(V(x)=6.30-3.00x\,\text{kN}\), obtained from the cut free body.

3. Set \(V=0\) to locate the extremum: \(x=2.10,\text{m}\), which lies within the loaded interval and is admissible.

4. Integrating or taking moments gives \(M=6.30x-1.50x^2\); at \(2.10,\text{m}\), \(M_{max}=6.62,\text{kN}\!\cdot\!\text{m}\).

Verification

Units: Integrating \(\text{kN/m}\) gives kN for shear and integrating kN gives \(\text{kN}\!\cdot\!\text{m}\) for moment.

Signs and direction: Shear changes from positive to negative at \(2.10,\text{m}\), confirming a positive maximum moment rather than a minimum.

Plausibility: The moment returns to zero at the supports and the extremum lies closer to the more heavily loaded left side.

Final result: \(\boxed{M_{max}=6.62,\text{kN}\!\cdot\!\text{m}\text{ at }x=2.10,\text{m}}\)

Engineering interpretation

Moment-diagram shape is a diagnostic: load discontinuities change shear slope, whereas point loads jump shear. This result should feed flexural-stress and deflection checks at the identified critical section.

PROFESSIONALFatigue-sensitive bracket arm

A rectangular steel bracket arm is 25 mm wide and 80 mm deep at a shoulder. The internal sagging moment there is 1.6 kN m, and a chart-based bending concentration factor is \(K_t=1.6\). Determine the elastic peak stress estimate.

Recognition strategy

A shoulder is a local geometry discontinuity. Calculate nominal flexural stress on the reduced section, then apply the supplied elastic concentration factor.

Given information

Use \(M=1.6,\text{kN}\!\cdot\!\text{m}\), \(b=25,\text{mm}\), \(h=80,\text{mm}\), and the stated \(K_t\).

Model and interpretation

The local cut supplies the internal moment; use centroidal section properties and assume elastic behavior. This estimate is not a fatigue-life calculation.

Step-by-step solution

1. For the reduced rectangle, \(I=bh^3/12=25(80)^3/12=1.067\times10^6\,\text{mm}^4\).

2. The extreme distance is \(c=40,\text{mm}\), and convert the moment to \(1.6\times10^6,\text{N}\!\cdot\!\text{mm}\).

3. The nominal extreme stress is \(Mc/I=1.6\times10^6(40)/(1.067\times10^6)=60.0\,\text{MPa}\).

4. Applying the supplied local factor gives \(K_t\sigma_{\mathrm{nom}}=1.6(60.0)=96.0\ \text{MPa}\).

Verification

Units: Using \(\text{N}\!\cdot\!\text{mm}\), mm, and \(\text{mm}^4\) yields \(\text{N/mm}^2\), numerically MPa.

Signs and direction: The peak retains the tension or compression sense of the nominal extreme fiber; the factor magnifies magnitude only.

Plausibility: The local value is exactly 1.6 times nominal, and it must exceed the smooth-section result for a positive concentration factor.

Final result: \(\boxed{|\sigma_{peak}|=96.0,\text{MPa}}\)

Engineering interpretation

For fluctuating service, \(K_t\) alone is not a certified fatigue assessment: surface finish, notch sensitivity, mean stress, weld quality, load spectrum, and code-approved factors require review.

Chapter recap

Use an equilibrium free body to obtain the internal shear and moment before calculating stress.
State coordinates, signs, section properties, material assumptions, and the governing compatibility relation.
Before moving on: select the applicable model, calculate with consistent units, and audit strength and serviceability.