Elementary Statically Indeterminate Structures (FE Civil)

NCEES FE Civil spec: Structural Engineering, E. Elementary statically indeterminate structures. FE Reference Handbook: Civil Engineering chapter, Structural Analysis section; Mechanics of Materials beam tables.

What the FE Civil Spec Covers

Structural Engineering topic E asks for elementary statically indeterminate structures. In practice that means beams with one or two redundants: propped cantilevers, fixed-fixed beams, two-span continuous beams, and simple frames where you can use superposition, a tabulated solution, or a short moment distribution. You will not be asked to run a full stiffness matrix by hand.

Key Concepts

Degree of indeterminacy is the number of unknown reactions (or member forces) beyond what the equilibrium equations can solve. A propped cantilever has 4 reactions and 3 equations, so it is indeterminate to the first degree.
Compatibility supplies the missing equations. Remove a redundant, compute the displacement at that point, then apply the redundant force so the displacement matches the real support (usually zero, or a known settlement).
Superposition works because the structure is linear elastic: deflections from separate loads simply add.
Stiffness and carryover: for a prismatic member with the far end fixed, the rotational stiffness is $4 E I / L$ and half of the applied end moment carries over to the far end. With the far end pinned, stiffness drops to $3 E I / L$ and nothing carries over.

Core Formulas

Compatibility at a removed support B:

Δ_{B, loads} - Δ_{B, redundant} = Δ_{B, actual} (usually 0)

Common results for a uniform load w on span L (constant EI):

Beam	Reactions	Key moments
Propped cantilever (fixed A, roller B)	$R_{A} = 5 w L /8, R_{B} = 3 w L /8$	$M_{A} = w L^{2} /8, M_{+, m a x} = 9 w L^{2} /128$
Fixed-fixed	$w L /2$ each end	$M_{end} = w L^{2} /12, M_{mid} = w L^{2} /24$
Two equal spans, continuous	$3 w L /8, 10 w L /8, 3 w L /8$	$M_{interior support} = w L^{2} /8$

Fixed-end moments for a midspan point load P are $P L /8$ at each end. Distribution factor at a joint:

D F_{i} = \frac{k _{i}}{\sum k}, k = \frac{4 E I}{L} (far end fixed), k = \frac{3 E I}{L} (far end pinned)

Where It Is in the FE Reference Handbook

Search the Civil Engineering chapter, Structural Analysis section, for beam stiffness and moment carryover and member fixed-end moments. The cantilever and simple beam deflection formulas you need for superposition are in the Mechanics of Materials chapter. Our moment distribution guide walks through the stiffness and carryover pages.

Worked Example: Propped Cantilever by Superposition

A beam is fixed at A and supported on a roller at B. Span L = 6 m, uniform load w = 10 kN/m, EI constant. Find the reactions and the fixed-end moment.

Step 1: Remove the roller and find the tip deflection

Δ_{B, w} = \frac{w L ^{4}}{8 E I} = \frac{10 ( 6 ) ^{4}}{8 E I} = \frac{1620}{E I}

Step 2: Deflection caused by the redundant R_B

Δ_{B, R} = \frac{R _{B} L ^{3}}{3 E I} = \frac{R _{B} ( 6 ) ^{3}}{3 E I} = \frac{72 R _{B}}{E I}

Step 3: Compatibility (roller does not move)

\frac{72 R _{B}}{E I} = \frac{1620}{E I} \Rightarrow R_{B} = 22.5 kN

This matches $3 w L /8 = 3 (10) (6) /8 = 22.5$ kN.

Step 4: Equilibrium for the fixed end

R_{A} = w L - R_{B} = 60 - 22.5 = 37.5 kN

M_{A} = R_{B} L - \frac{w L ^{2}}{2} = 22.5 (6) - \frac{10 ( 6 ) ^{2}}{2} = 135 - 180 = - 45 kN \cdot m

The fixed-end moment is 45 kN·m, hogging (top fiber in tension), which equals $w L^{2} /8$ . The largest positive moment occurs where shear is zero, 3L/8 = 2.25 m from B, and equals $9 w L^{2} /128 = 25.3$ kN·m. A simply supported span would have had 45 kN·m at midspan, so the fixity moved the peak moment to the support.

Two Quick Checks You Should Be Able to Do in a Minute

Two-span continuous beam

Two equal 8 m spans carry w = 15 kN/m over the full length. From the table, the interior reaction is

R_{B} = \frac{10 w L}{8} = 1.25 (15) (8) = 150 kN

and the hogging moment over the interior support is $w L^{2} /8 = 15 (64) /8 = 120$ kN·m. The end reactions are each $3 w L /8 = 45$ kN, and the three reactions add to the total load of 240 kN, a useful sanity check.

Distribution factors at a joint

A joint connects member 1 (EI, L = 4 m) and member 2 (EI, L = 6 m), both with fixed far ends. The stiffnesses are $4 E I /4 = 1.0 E I$ and $4 E I /6 = 0.667 E I$ , so

D F_{1} = \frac{1.0}{1.667} = 0.60, D F_{2} = \frac{0.667}{1.667} = 0.40

The shorter, stiffer member attracts more of any unbalanced moment. Distribution factors at a joint always add to 1.0.

Common Mistakes

Counting reactions wrong. A fixed support gives 3 unknowns in a plane, a pin 2 and a roller 1.
Mixing sign conventions between the load case and the redundant case. Draw both deflected shapes first.
Using 4EI/L for a member whose far end is pinned. It should be 3EI/L, with no carryover.
Forgetting that the maximum positive moment in a propped cantilever is not at midspan.
Treating a support settlement as zero displacement. If the support settles, set the compatibility equation equal to the settlement.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A horizontal prismatic beam is fixed at A and supported by a vertical roller at B, 4.0 m from A. A uniform load of 6.0 kN/m acts downward over the whole span. EI is constant and both supports stay at their original elevation. Using linear-elastic small-deflection theory, find the vertical reaction at B. See the worked solution or practice the Structural Analysis Methods deck
A propped beam is analyzed by removing its prop. Under the actual loads the released beam deflects 12 mm downward at the prop location. In the released beam, each 1 kN upward force at that location causes 3 mm upward displacement there. The actual prop settles 3 mm downward and stays in contact with the beam. The beam is linear elastic. Find the upward prop reaction R. See the worked solution or practice the Structural Analysis Methods deck
A continuous beam of two equal 6 m spans carries a uniformly distributed load of 12 kN/m over its full length. The reaction at the interior support is most nearly: See the worked solution or practice the Structural Basics deck
At joint J, two prismatic members have fixed far ends and the joint cannot translate. Member J1 has EI = 2EI₀ and length 4 m; member J2 has EI = EI₀ and length 6 m. The unbalanced moment at J is +24 kN·m (clockwise positive). In moment distribution, find the balancing moment assigned to member J1 at J. See the worked solution or practice the Structural Analysis Methods deck
A continuous prismatic beam spans A-B-C with AB = BC = 4 m and constant EI. A is a pin; B and C are rotationally free vertical supports, with C anchored so it can resist uplift. No support settles. Only span AB carries a downward uniform load of 6 kN/m; BC is unloaded. Using sagging positive, find the bending moment at B. See the worked solution or practice the Structural Analysis Methods deck
A planar propped cantilever is fixed at one end and supported by a vertical roller at the other. What is its degree of static indeterminacy? See the worked solution or practice the Structural Basics deck

Related Calculators and Examples

Frequently Asked Questions

Do I need moment distribution on the FE Civil exam?

Only at an elementary level. Expect one or two questions on stiffness, distribution factors or carryover, or a quick propped cantilever or two-span beam where a tabulated result or superposition is faster than a full moment distribution.

What is the fastest way to solve a propped cantilever on the FE?

Remove the prop, find the cantilever deflection at that point from a beam table, then find the prop force that pushes the tip back to zero deflection. For a uniform load this gives a prop reaction of 3wL/8, which is worth memorizing.

Why does an indeterminate beam have smaller moments than a simple beam?

The extra supports or fixed ends add stiffness and share the load. A fixed-fixed beam under uniform load has a midspan moment of wL squared over 24 and end moments of wL squared over 12, both smaller than the wL squared over 8 of a simply supported span.

Back to all Structural Engineering topics