Loading study handbook…

Deflection

5% of exam

Beam and frame deflections by double integration, moment-area, conjugate-beam, and virtual-work methods; standard deflection formulas for common loadings; and serviceability deflection limits.

4 concepts

F. Deflection

Deflection by Virtual Work (Unit Load)

The unit-load (virtual-work) method for the deflection of any truss joint ($\Sigma nNL/AE$) or beam/frame point ($\int mM/EI\,dx$), and how to choose the virtual system.

Almost every deflection problem on the PE — a truss chord that sags too far, a frame joint that drifts, the tip of a bracket — yields to one idea: put a dummy unit load where you want the answer, and let the work it does equal the internal work of the real deformations. Virtual work is the most general deflection tool you carry into the exam: it handles trusses, beams, frames, mixed axial-and-bending members, even temperature and fabrication-misfit effects, all with the same bookkeeping. The NCEES PE Civil Reference Handbook gives it directly in §4.1.4 (Truss Deflection by Unit Load Method) and §4.1.5 (Frame Deflection by Unit Load Method), and this is where the most reliable partial credit on the deflection area lives — set the table up right and the answer falls out.

The principle: a unit load that does the work

Apply a single virtual force of magnitude one (a unit load) at the point and in the direction of the deflection you want. The external virtual work it does,

1\cdot\Delta

, must equal the internal virtual work stored as the virtual internal forces ride through the real deformations. For an axially loaded member that real deformation is

NL/AE

; for a flexural member it is the bending strain over the length. Because the external load is exactly one, the work integral becomes the deflection itself — that is the whole trick, and it is why the method is sometimes called the unit-load method.

1\cdot\Delta = \sum n\,\delta_{\text{real}} \quad\Longrightarrow\quad \Delta = \sum \int (\text{virtual force})\,(\text{real deformation})

Truss joints: the $\Sigma nNL/AE$ table

For a pin-jointed truss every member carries only axial force, so the real elongation of member

i

is

N_iL_i/A_iE_i

and the virtual member force is

n_i

from the unit load. The joint displacement is the sum over all members. Tension is positive in both the real force

N

and the virtual force

n

; when both are tension (or both compression) the product

nN

is positive and that member adds to the deflection. The exam-day discipline is to build a column table — member,

n

,

N

,

L

,

A

, then the product

nNL/AE

— and sum the last column.

\Delta = \sum_{i=1}^{m} \frac{n_i\,N_i\,L_i}{A_i E_i}

Beams and frames: the $\int mM/EI\,dx$ integral

For members in bending, replace the axial term with the flexural one:

m(x)

is the bending-moment diagram from the unit load and

M(x)

is the moment diagram from the real loads, integrated over each member and divided by

EI

. A unit point force gives a deflection; a unit couple (unit moment) gives a rotation. If either the real loads or the unit load produce no moment in a member, that member drops out of the sum — a free tip of a frame or an unloaded leg contributes nothing.

\Delta = \sum_i \int_0^{L_i} \frac{m_i(x)\,M_i(x)}{E I_i}\,dx

Choosing the virtual system

The virtual structure must be the same geometry and supports as the real one, but it carries only the dummy unit load — nothing else. Want a vertical deflection? Apply a unit vertical force there. Want a horizontal drift? Apply a unit horizontal force. Want a rotation or relative rotation? Apply a unit couple. For a truss, you solve the virtual structure once (by joint or section) for the

n_i

, solve the real structure once for the

N_i

, then multiply. Picking the unit load that matches the desired displacement is the single most important setup decision; get the direction wrong and a positive answer that should be a sag comes out backwards.

Temperature and misfit by the same machinery

The unit-load method shines because the real deformation need not come from load at all. For a temperature change, the real member elongation is

\alpha\,\Delta T\,L

; for a fabrication error it is simply the misfit

\Delta L

. You still apply the same unit load to get the

n_i

, then multiply each

n_i

by its real

\Delta L_i

and sum. This is why a truss with no external load can still deflect, and why the handbook lists load, temperature, and misfit together under one equation.

\Delta = \sum n_i\,(\Delta L)_i,\qquad (\Delta L)_i = \frac{N_iL_i}{A_iE_i}\ \text{or}\ \alpha_i(\Delta T)_iL_i\ \text{or}\ \text{misfit}

Exam strategy

Decide first what you want — vertical deflection, horizontal drift, or a rotation — and apply the matching unit load (force for a translation, couple for a rotation) in that exact direction. For a truss, build the member table and keep tension positive in both

n

and

N

; a negative sum means the joint moves opposite to your unit load. For beams and frames, sketch both

m(x)

and

M(x)

and integrate piecewise — or use a diagram-multiplication shortcut when both are simple shapes. Watch units religiously: with

A

in

\text{in}^2

,

E

in

\text{ksi}

, and

L

in

\text{in}

, the product

nNL/AE

comes out in inches. The most common loss of points is a units slip, so fix your units before you sum the column.

Key equations

External virtual work = internal virtual work

1\cdot\Delta = \sum n\,\delta_{\text{real}}

The defining statement: a unit load does external work $1\cdot\Delta$ equal to the internal work of virtual forces through real deformations.

Truss joint deflection

\Delta = \sum \dfrac{n\,N\,L}{A\,E}

$n$ = member force from unit load, $N$

Beam/frame deflection

\Delta = \sum \int_0^{L} \dfrac{m(x)\,M(x)}{E I}\,dx

Real member elongation (load)

(\Delta L) = \dfrac{N\,L}{A\,E}

Axial deformation of a truss member under the real load; the real-deformation term inside the truss sum.

Real member elongation (temperature)

(\Delta L) = \alpha\,\Delta T\,L

$\alpha$ = coefficient of thermal expansion, $\Delta T$ = temperature change. Substitutes for $NL/AE$

Rotation by a unit couple

\theta = \sum \int_0^{L} \dfrac{m_\theta(x)\,M(x)}{E I}\,dx

Combined axial + flexural work

\Delta = \sum \dfrac{n\,N\,L}{A\,E} + \sum \int_0^{L}\dfrac{m\,M}{E I}\,dx

Diagram multiplication (Vereshchagin)

\int_0^L \dfrac{m\,M}{EI}\,dx = \dfrac{A_M\,\bar{m}}{EI}

Worked examples

Vertical deflection of a truss joint

Problem. A steel truss ( $E = 29{,}000\,\text{ksi}$ ) has all members with area $A = 4\,\text{in}^2$ . To find the vertical deflection at a lower-chord joint, a unit vertical load is applied there. The four loaded members carry real forces and virtual forces as tabulated; a fifth member directly under the unit load carries no real force. Members 1 and 2: $n = 0.667$ , $N = 20\,\text{kip}$ , $L = 180\,\text{in}$ each. Members 3 and 4: $n = -0.833$ , $N = -25\,\text{kip}$ , $L = 150\,\text{in}$ each. Member 5: $n = 1.0$ , $N = 0$ , $L = 240\,\text{in}$ . Find the vertical deflection.

Solution. Compute

AE = (4)(29{,}000) = 116{,}000\,\text{kip}

for every member. Form

nNL

for each member and sum: - Members 1, 2:

(0.667)(20)(180) = 2401\,\text{kip}\cdot\text{in}

\Delta = \frac{\sum nNL}{AE} = \frac{11{,}050}{116{,}000} = 0.0953\ \text{in}

Midspan deflection of a simple beam by virtual work

Problem. A simply supported beam spans $L = 20\,\text{ft}$ and carries a uniform load $w = 2\,\text{kip/ft}$ . The section has $I = 300\,\text{in}^4$

Tip deflection of a cantilever frame leg

Problem. A cantilever member of length $L = 10\,\text{ft}$ carries a vertical point load $P = 5\,\text{kip}$ at its free tip. With $I = 200\,\text{in}^4$

Common pitfalls

•Applying the unit load in the wrong direction or at the wrong point: the method returns the displacement only where, and along which line, the unit load acts. Want a rotation? Use a unit couple, not a unit force.
•Sign confusion in the truss sum. Keep tension positive in BOTH $n$ and $N$ ; two compression members still give a positive $nN$ product. A negative total simply means the joint moves opposite to the assumed unit-load direction.
•Mixing units in $nNL/AE$ . With $A$ in $\text{in}^2$ , $E$ in ksi, and $L$ in inches the answer is in inches; if $L$ is in feet you are off by a factor of 12. Convert $L$ before summing.
•Forgetting in-versus-ft in the beam term: $M = wL^2/8$ and $5wL^4/384EI$ need $L$ in inches and $w$
•Dropping a member that has zero real force but is still part of the structure — that is fine, it contributes nothing — but never drop a member that carries force in either the real or the virtual system.
•Citing a code edition or a steel-design table for what is purely an analysis method. Virtual work is a Reference Handbook §4.1.4/§4.1.5 method; the deflection itself needs no AISC/ACI strength check.
•Using $S$ (elastic section modulus) where $I$ (moment of inertia) belongs. Deflection scales with $I$ , not $S$ ; $S$ is for stress, not deformation.

References

NCEES PE Civil Reference Handbook — §4.1.4 Truss Deflection by Unit Load Method
NCEES PE Civil Reference Handbook — §4.1.5 Frame Deflection by Unit Load Method
Hibbeler, Structural Analysis — Virtual-work / unit-load derivations and worked truss and frame examples

Standard Deflection Formulas & Serviceability Limits

The AISC Table 3-23 closed-form beam deflections (5wL⁴/384EI and friends), superposition of load cases, and the L/240–L/360 serviceability limits from IBC Table 1604.3.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

Moment-Area & Conjugate-Beam Methods

The two moment-area theorems (area and first moment of the M/EI diagram) and the conjugate-beam analogy that turns slope and deflection into shear and moment of a fictitious beam.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

Deflection by Double Integration

Integrating $EI\,y'' = M(x)$ twice and applying boundary conditions to recover the elastic curve, slope, and deflection of a beam.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

= member force from real loads (tension +),

L

= length,

A

= area,

E

= modulus. Result is the displacement at and along the unit load.

$m(x)$ = moment from a unit load (force for deflection, couple for rotation); $M(x)$ = moment from real loads; integrate over each member.

to get thermal deflection by the same unit-load sum.

Apply a unit moment at the point of desired rotation; $m_\theta(x)$ is the resulting virtual-moment diagram.

Frames with significant axial force keep both terms; usually the bending term dominates and the axial term is dropped.

When $m(x)$ is linear, the integral equals the area $A_M$ of the real $M$ diagram times the ordinate $\bar{m}$ of the virtual diagram at the centroid of $A_M$ , over $EI$ .

each

\to 4802

. - Members 3, 4:

(-0.833)(-25)(150) = 3124\,\text{kip}\cdot\text{in}

each

\to 6248

(both forces compression, product positive). - Member 5:

(1.0)(0)(240) = 0

. Sum

\sum nNL = 4802 + 6248 + 0 = 11{,}050\,\text{kip}\cdot\text{in}

(with

L

in in).

\Delta = \dfrac{\sum nNL}{AE} = \dfrac{11{,}050}{116{,}000} = 0.0953\,\text{in}

downward. Sanity check: units are

(\text{kip})(\text{in})/(\text{kip}) = \text{in}

, and a ~0.1 in sag on an 80-kip-class truss is reasonable.

\Delta \approx 0.0953\,\text{in}\ \downarrow

.

and

E = 29{,}000\,\text{ksi}

. Using a unit load at midspan, find the midspan deflection by virtual work, and confirm it matches the standard closed form.

Solution. Real moment:

M(x) = \frac{wL}{2}x - \frac{w}{2}x^2

. Virtual moment from a unit midspan load:

m(x) = \frac{x}{2}

for

0 \le x \le L/2

(symmetric, so integrate half and double). Carrying out

\Delta = \frac{2}{EI}\int_0^{L/2} m\,M\,dx

gives the well-known result

\Delta = \dfrac{5wL^4}{384EI}

. Evaluate in consistent units — convert

w = 2\,\text{kip/ft} = 0.1667\,\text{kip/in}

and

L = 240\,\text{in}

:

\Delta = \frac{5(0.1667)(240)^4}{384(29{,}000)(300)} = 0.828\,\text{in}.

Sanity check:

L/360 = 240/360 = 0.667\,\text{in}

, so this beam exceeds a typical live-load limit — plausible for a deliberately flexible section.

\Delta = 0.828\,\text{in}\ \downarrow

.

\Delta = \frac{5wL^4}{384EI} = \frac{5(0.1667)(240)^4}{384(29{,}000)(300)} = 0.828\ \text{in}

and

E = 29{,}000\,\text{ksi}

, use a unit tip load to find the vertical tip deflection.

Solution. Measure

x

from the free tip. Real moment

M(x) = -Px

and virtual moment from a unit tip load

m(x) = -x

, so

mM = Px^2

.

\Delta = \dfrac{1}{EI}\int_0^L Px^2\,dx = \dfrac{PL^3}{3EI}

. With

L = 120\,\text{in}

:

\Delta = \frac{(5)(120)^3}{3(29{,}000)(200)} = \frac{8.64\times10^6}{1.74\times10^7} = 0.497\,\text{in}.

Sanity check: matches the AISC Table 3-23 cantilever formula

PL^3/3EI

exactly, and units

(\text{kip})(\text{in}^3)/(\text{ksi}\cdot\text{in}^4) = \text{in}

.

\Delta = 0.497\,\text{in}\ \downarrow

.

\Delta = \frac{PL^3}{3EI} = \frac{5(120)^3}{3(29{,}000)(200)} = 0.497\ \text{in}

in kip/in to pair with

E

in ksi and

I

in

\text{in}^4

.