MECHANICS OF MATERIALS · CHAPTER 12 — DEFLECTION OF BEAMS AND SHAFTS
Mechanics of Materials · Chapter 12

Deflection of Beams and Shafts

The elastic curve, four methods for slopes and deflections, and the solution of statically indeterminate beams.

Learning objectives

  • Relate internal bending moment, curvature, slope, and transverse displacement using a declared sign convention.
  • Select integration, discontinuity functions, moment-area, or superposition for a linear-elastic beam response.
  • Apply boundary and compatibility conditions to calculate slope, displacement, or a redundant reaction.
  • Account for changing flexural rigidity using a piecewise \(M/EI\) relation.
  • Audit a deflection prediction against units, support conditions, small-slope applicability, and serviceability intent.

12.1The Elastic Curve

The Elastic Curve

The deflected shape of a beam's centroidal axis. Deflections must often be limited — for stability, to protect brittle finishes (plaster, concrete), and most importantly because slopes and displacements supply the extra equations needed for statically indeterminate beams. Sketching rules: supports that resist a force restrict displacement; supports that resist a moment (fixed ends) restrict slope and displacement.

Reading the Curve from the Moment Diagram

By the beam sign convention, positive moment bends the beam concave up; negative moment concave down; a point of zero moment is an inflection point where the curvature reverses. The maximum deflection occurs either where the slope is zero or at an unsupported end — drawing the moment diagram first makes the elastic curve easy to sketch.

Moment–Curvature Relationship
\[ \frac{1}{\rho} = \frac{M}{EI} \]
  • \(\rho\) = radius of curvature at the point; \(EI\) = the flexural rigidity (always positive), the beam's resistance to bending deformation.

Physical meaning

Curvature is the rate at which a beam tangent turns with distance. A larger \(EI\) makes a given moment produce less turning; with \(v\) positive upward and sagging moment positive, this page uses \(EI v''=M\).

12.2Slope and Displacement by Integration

The exact curvature \(1/\rho = v''/[1 + (v')^2]^{3/2}\) makes the governing equation nonlinear, but design codes keep beam slopes small, so \((dv/dx)^2 \ll 1\) and the curvature reduces to \(v''\). With constant EI, three equivalent forms follow:

The Elastic Curve Differential Equations (Constant EI)
\[ EI\,\frac{d^2v}{dx^2} = M(x) \qquad EI\,\frac{d^3v}{dx^3} = V(x) \qquad EI\,\frac{d^4v}{dx^4} = w(x) \]
  • Starting from \(M(x)\) needs only two integrations (two constants); starting from \(w(x)\) needs four.
  • Boundary conditions: pin or roller → \(v = 0\); fixed end → \(v = 0\) and \(\theta = dv/dx = 0\).
  • Continuity conditions: where separate moment functions meet, slope and deflection from adjacent regions must match: \(\theta_1(a) = \theta_2(a)\), \(v_1(a) = v_2(a)\).

Problem-solving strategy

  1. Draw the free-body diagram and decide whether the target is a curve, slope, displacement, or compatibility condition.
  2. Declare \(x\), \(v\), and moment signs; obtain \(M(x)\) before choosing integration, moment-area, discontinuity functions, or superposition.
  3. List support and interface conditions before integrating.
  4. Use \(M/[E(x)I(x)]\) piecewise where stiffness changes.
  5. Audit endpoints, signs, units, small rotations, and the applicable serviceability limit.
EXAMPLE 1Wall-Bracket Tip Response by Integration
A horizontal steel wall bracket is a \(1.20\ \mathrm{m}\) cantilever carrying a \(2.50\ \mathrm{kN}\) downward tip load. Its constant \(EI=3.00\times10^6\ \mathrm{N\,m^2}\). Determine its elastic curve, free-end slope, and free-end displacement.

Recognition strategy

A single-load cantilever has a simple moment function and two known wall conditions, making direct elastic-curve integration the most transparent method.

Given information

  • \(L=1.20\ \mathrm{m}\), \(P=2500\ \mathrm{N}\), and constant \(EI\).
  • Set \(x=0\) at the wall; upward \(v\) and sagging \(M\) are positive.

Model and interpretation

The prismatic bracket is linear elastic with small rotation. Its moment is \(M(x)=-P(L-x)\), and the wall imposes \(v(0)=v'(0)=0\).

Step-by-step solution
1. Formulate. \[EI v''=-P(L-x).\]
2. Integrate. \[EI v'=-PLx+\frac{Px^2}{2}+C_1,\qquad EI v=-\frac{PLx^2}{2}+\frac{Px^3}{6}+C_1x+C_2.\]
3. Enforce the wall conditions. \(C_1=C_2=0\), hence \[v(x)=-\frac{Px^2(3L-x)}{6EI}.\]
4. Evaluate the end. \[\theta_L=-\frac{PL^2}{2EI}=-6.00\times10^{-4}\ \mathrm{rad},\qquad v_L=-\frac{PL^3}{3EI}=-4.80\times10^{-4}\ \mathrm{m}.\]

Verification

Units: \(PL^3/EI\) is \(\mathrm{m}\), and \(PL^2/EI\) is dimensionless radians.

Signs and direction: Negative values mean downward displacement and clockwise rotation under the declared convention.

Plausibility: \(0.480\ \mathrm{mm}\ll1.20\ \mathrm{m}\) and \(\lvert\theta_L\rvert\ll1\), consistent with small-slope theory.

Final result: \(\boxed{v_L=-0.480\ \mathrm{mm},\ \theta_L=-6.00\times10^{-4}\ \mathrm{rad}}\).

Engineering interpretation

The ideal beam prediction is modest, but wall connection flexibility and local deformation can add movement; they matter if the bracket locates a precision component.

12.3Discontinuity Functions

One Expression for the Whole Beam

With several loads, region-by-region integration multiplies the constants and continuity conditions. Discontinuity functions write the load or moment as a single expression valid over the whole beam, so only the boundary conditions remain. Macaulay functions handle distributed loads:

\[ \langle x - a\rangle^n = \begin{cases} 0 & x < a \\ (x - a)^n & x \ge a \end{cases} \quad (n \ge 0), \qquad \int \langle x - a\rangle^n dx = \frac{\langle x - a\rangle^{n+1}}{n+1} + C \]

Singularity functions (\(n = -1, -2\)) represent a concentrated force \(P\langle x - a\rangle^{-1}\) and couple \(M_0\langle x - a\rangle^{-2}\); they integrate by simply raising the exponent. In a moment expression, a force P applied at \(a\) contributes \(P\langle x - a\rangle^1\) and a couple contributes \(M_0\langle x - a\rangle^0\).

12.4Slope and Displacement by the Moment-Area Method

A semigraphical technique — ideal when the M/EI diagram is built of simple shapes. It uses the same assumptions as integration (small slopes, bending only).

Theorem 1 · Moment-Area (Slope)
\[ \theta_{B/A} = \int_A^B \frac{M}{EI}\,dx \]

The angle between the tangents at any two points of the elastic curve equals the area under the M/EI diagram between them — counterclockwise from tangent A to tangent B if the area is positive.

Theorem 2 · Moment-Area (Deviation)
\[ t_{A/B} = \int_A^B \frac{M}{EI}\,\bar{x}_A\,dx \]

The vertical deviation of point A from the tangent drawn at B equals the moment of the M/EI area between A and B, taken about A (\(\bar{x}_A\) = distance from A to the area's centroid). Note \(t_{A/B} \ne t_{B/A}\) in general. Deviations are measured perpendicular to the original axis; positive areas place A above the tangent.

12.5Method of Superposition

Adding Tabulated Deflections

The governing equation \(EI\,v'''' = w(x)\) is linear and assumes small deformations — the two requirements of superposition — so the deflection from several loads is the algebraic sum of the deflections from each load acting alone. Engineering handbooks tabulate slopes and deflections for standard cases; combining table entries solves most practical problems in a few lines. Frequently used entries:

\[ \text{Cantilever, end load: } v = \frac{PL^3}{3EI} \qquad \text{Simply supported, center load: } v = \frac{PL^3}{48EI} \qquad \text{Simply supported, uniform load: } v = \frac{5wL^4}{384EI} \]
EXAMPLE 2Platform Beam by Superposition
A simply supported \(4.00\ \mathrm{m}\) platform beam with \(EI=5000\ \mathrm{kN\,m^2}\) supports a \(3.00\ \mathrm{kN/m}\) uniform load and a \(10.0\ \mathrm{kN}\) midspan machine load. Find the midspan displacement.

Recognition strategy

Both standard loads act on the same linear, constant-stiffness support case and the response is wanted at their shared symmetry point, so tabulated midspan responses can be added.

Given information

  • \(L=4.00\ \mathrm{m}\), \(w=3.00\ \mathrm{kN/m}\), \(P=10.0\ \mathrm{kN}\), \(EI=5000\ \mathrm{kN\,m^2}\).
  • Downward displacement is reported negative.

Model and interpretation

Assume linear elasticity, small displacement, and no change in contact load as the beam moves. These assumptions make superposition valid.

Step-by-step solution
1. Select component cases. Use \(v_w=-5wL^4/(384EI)\) and \(v_P=-PL^3/(48EI)\).
2. Calculate the uniform-load part. \[v_w=-\frac{5(3.00)(4.00)^4}{384(5000)}=-0.00200\ \mathrm{m}.\]
3. Calculate the point-load part. \[v_P=-\frac{(10.0)(4.00)^3}{48(5000)}=-0.00267\ \mathrm{m}.\]
4. Superpose signed responses. \[v_C=v_w+v_P=-0.00467\ \mathrm{m}.\]

Verification

Units: Each numerator is \(\mathrm{kN\,m^3}\), so division by \(\mathrm{kN\,m^2}\) gives meters.

Signs and direction: Both loads bend the beam downward, so two negative contributions must add.

Plausibility: \(4.67\ \mathrm{mm}/4000\ \mathrm{mm}=1/857\), a small displacement relative to span.

Final result: \(\boxed{v_C=-4.67\ \mathrm{mm}\ \text{(downward)}}\).

Engineering interpretation

The point load contributes more than the distributed load. A serviceability decision must compare the calculated total with the project limit and include dynamic machine effects if they are relevant.

12.6Statically Indeterminate Beams and Shafts

Redundants and Degree of Indeterminacy

A beam is statically indeterminate when its unknown reactions outnumber the equilibrium equations. The extra reactions — those not needed for stable equilibrium — are redundants, and their count is the degree of indeterminacy: a propped cantilever (4 reactions, 3 equations) is indeterminate to the first degree; a two-span continuous beam to the second. Any reaction whose removal leaves the beam stable may be chosen as the redundant. The solution must satisfy equilibrium and compatibility simultaneously — by integration (12.7), moment-area (12.8), or superposition (12.9).

12.7Statically Indeterminate — Method of Integration

Extra Unknowns, Extra Conditions

Write \(M(x)\) including the unknown redundant reaction, integrate \(EI\,v'' = M\) twice, and note that the indeterminate beam has more boundary conditions than integration constants — exactly enough surplus to solve for the redundants too. A propped cantilever, for example, supplies three conditions (\(v = 0\) at both supports, \(\theta = 0\) at the wall) against two constants: the third condition determines the redundant.

12.8Statically Indeterminate — Moment-Area Method

Geometric Compatibility from M/EI Diagrams

Draw the M/EI diagram in terms of the unknown redundant — most conveniently as separate diagrams for each load and for the redundant (the superposition of diagrams keeps the shapes simple). Then enforce the support geometry through the theorems: e.g. for a propped cantilever, the deviation of the roller from the wall tangent must vanish, \(t_{B/A} = 0\), which yields the redundant directly.

12.9Statically Indeterminate — Method of Superposition

The Force Method for Beams

Remove the redundant to create the statically determinate, stable primary beam; let the applied loads deflect it by \(v_B\) at the redundant's location; then apply the redundant alone, deflecting the point back by \(v'_B\). Compatibility at the support requires the net displacement to vanish — an equation solved directly for the redundant (hence "force method"), with the remaining reactions following from equilibrium.

EXAMPLE 3Conveyor Shaft Serviceability Decision
A bearing-supported conveyor shaft is modeled as a simply supported \(1.50\ \mathrm{m}\) steel beam with a \(1.80\ \mathrm{kN}\) pulley load at midspan. Given \(E=200\ \mathrm{GPa}\), \(I=8.00\times10^{-6}\ \mathrm{m^4}\), and an alignment limit of \(1.00\ \mathrm{mm}\), decide whether bending deflection alone satisfies the limit.

Recognition strategy

The requested output is a serviceability decision at midspan for a standard simply supported center load. The tabulated elastic response is appropriate only after stating shaft, bearing, and load idealizations.

Given information

  • \(L=1.50\ \mathrm{m}\), \(P=1800\ \mathrm{N}\), \(E=200\times10^9\ \mathrm{Pa}\), \(I=8.00\times10^{-6}\ \mathrm{m^4}\).
  • Model bearings as simple supports and take downward deflection as negative.

Model and interpretation

The shaft is straight, prismatic, linear elastic, and lightly deflected. Bearing compliance, pulley overhang, torque-induced twist, and rotating-load dynamics are excluded from this bending-only screen.

Step-by-step solution
1. Establish rigidity. \[EI=(200\times10^9)(8.00\times10^{-6})=1.60\times10^6\ \mathrm{N\,m^2}.\]
2. Choose the response. For a centered load, \[v_C=-\frac{PL^3}{48EI}.\]
3. Compute deflection. \[v_C=-\frac{(1800)(1.50)^3}{48(1.60\times10^6)}=-7.91\times10^{-5}\ \mathrm{m}.\]
4. Compare the criterion. \(\lvert v_C\rvert=0.0791\ \mathrm{mm}<1.00\ \mathrm{mm}\), leaving \(0.921\ \mathrm{mm}\) under the stated static bending limit.

Verification

Units: \(PL^3/EI\) reduces to meters; the comparison is made in millimeters on both sides.

Signs and direction: The negative result is downward, consistent with the pulley load; the limit uses magnitude.

Plausibility: \(0.0791\ \mathrm{mm}/1500\ \mathrm{mm}\approx5.3\times10^{-5}\), so small-slope bending theory is self-consistent.

Final result: \(\boxed{v_C=-0.0791\ \mathrm{mm};\ \text{the bending-only screen passes}}\).

Engineering interpretation

The static beam model clears the alignment limit, not a complete shaft design. A professional release still checks bearing stiffness, actual load positions, fatigue, critical speed, torsional twist, tolerances, and the governing equipment specification.

Chapter recap

  • Use moment curvature to connect loading to shape, and impose geometric conditions at every support.
  • Select integration for custom functions, moment-area for simple \(M/EI\) geometry, and superposition only for linear compatible cases.
  • Before moving on: sketch the curve, state signs, apply boundary and compatibility conditions, treat variable \(EI\) explicitly, and compare a checked result with serviceability intent.
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