Analysis of Statically Determinate Beams, Columns, Trusses and Frames

NCEES FE Civil spec: Structural Engineering, A. Analysis of statically determinant beams, columns, trusses, and frames. FE Reference Handbook: Statics chapter; Mechanics of Materials chapter (Beams); Civil Engineering chapter, Structural Analysis section.

What the FE Civil Spec Covers

Structural Engineering section A asks you to analyze structures that can be solved with equilibrium alone: simply supported, cantilever and overhanging beams, axially loaded columns, plane trusses, and simple frames. Typical questions ask for a support reaction, the shear or moment at a point, the maximum moment, or the force in one truss member. These are fast points if your free body diagrams are clean.

Key Concepts

Free body diagram first. Replace every support with its reactions: a roller gives one force, a pin gives two, a fixed support gives two forces and a moment.
Distributed loads act through their resultant for reaction calculations: a uniform load w over length a is a force wa at the middle of a.
Shear and moment relationships. The slope of the shear diagram equals the load intensity, and the slope of the moment diagram equals the shear. Maximum moment occurs where shear passes through zero.
Trusses are pin-connected two-force members loaded only at joints, so every member is in pure tension or compression.
Frames have at least one multi-force member. Take apart the frame at its pins and solve each piece, using equal and opposite pin forces.

Core Formulas

Equilibrium in the plane:

\sum F_{x} = 0, \sum F_{y} = 0, \sum M = 0

Load, shear and moment relationships (x measured left to right, w positive downward):

\frac{d V}{d x} = - w (x), \frac{d M}{d x} = V (x)

So the change in moment between two points equals the area under the shear diagram:

M_{2} - M_{1} = \int_{x_{1}}^{x_{2}} V d x

Common maximum moments worth memorizing (simply supported span L):

M_{ma x} = \frac{w L ^{2}}{8} (uniform load), M_{ma x} = \frac{P L}{4} (midspan point load), M_{ma x} = \frac{P ab}{L} (point load at a)

Where It Is in the FE Reference Handbook

Equilibrium and truss methods are in the Statics chapter. Shear and moment sign conventions and the beam formulas are in the Mechanics of Materials chapter under Beams. The Civil Engineering chapter, Structural Analysis section, adds influence lines and moving load sets. Search "shearing force and bending moment" to find the sign convention quickly.

Columns and Frames in Section A

A determinate column question usually reduces to axial load: add up the tributary loads from each floor above, or find the column force from the beam reactions it supports. For a simple frame such as an L-shaped bent fixed at its base, write equilibrium for the whole frame to get the base reactions, then cut through the member at the point of interest. The internal forces at a cut in a frame member are an axial force, a shear and a moment, just like a beam, so the same sign discipline applies.

Worked Example: Overhanging Beam

Beam ABC has a pin at A (x = 0) and a roller at B (x = 8 m). It overhangs to a free end C at x = 10 m. A uniform load of 10 kN/m acts on span AB, and a 20 kN point load acts downward at C. Find the reactions, the maximum positive moment, and the moment at B.

Step 1: Reaction at B (moments about A)

\sum M_{A} = 0 : 8 R_{B} - (10) (8) (4) - (20) (10) = 0

R_{B} = \frac{320 + 200}{8} = 65 kN

Step 2: Reaction at A (vertical forces)

R_{A} = (10) (8) + 20 - 65 = 35 kN

Step 3: Location of zero shear

Shear starts at +35 kN at A and drops 10 kN per meter. It reaches zero at x = 35/10 = 3.5 m. Just left of B the shear is 35 minus 80 = minus 45 kN; the 65 kN reaction lifts it to +20 kN, which stays constant to C.

Step 4: Maximum positive moment

M (3.5) = 35 (3.5) - \frac{10 ( 3.5 ) ^{2}}{2} = 122.5 - 61.25 = 61.25 kN \cdotp m

Step 5: Moment at B

Cut at B and look right (only the 20 kN load on the 2 m overhang):

M_{B} = - (20) (2) = - 40 kN \cdotp m

Check from the left: 35(8) minus 10(8)(4) = 280 minus 320 = minus 40 kN·m. Both sides agree, so the answers are R_A = 35 kN, R_B = 65 kN, maximum positive moment 61.25 kN·m at 3.5 m, and maximum negative moment 40 kN·m at B.

Common Mistakes

Placing the resultant of a distributed load at the wrong point, or forgetting the overhang load in the moment sum.
Reporting the moment at midspan as the maximum when the loading is not symmetric. Find where V = 0.
Missing the negative moment over an interior support on overhanging beams.
Mixing sign conventions between the shear diagram and the moment diagram.
In truss problems, assuming a member is in compression without checking the sign of the answer.

Next, see how these structures deflect in deflection of beams, trusses and frames, or check whether a structure can be solved by statics at all in determinacy and stability.

Practice Problems

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

A simply supported beam spans 8 m from pin A to roller B. Two downward point loads act at 2 m and 6 m from A: 12 kN and 20 kN, respectively. Neglect beam self-weight. What is the vertical reaction at B? See the worked solution or practice the Beam Analysis deck
A simply supported beam spans 10 m. A 40 kN point load acts 4 m from the left support A. What is the maximum bending moment? See the worked solution or practice the Beam Analysis deck
A simply supported beam of span 10 m carries a single 50 kN point load 3 m from the left support. The maximum bending moment is most nearly: See the worked solution or practice the Structural Basics deck
A planar L-shaped frame has a 4 m vertical column fixed at its base and a 6 m horizontal beam rigidly connected at the top. The beam carries a uniformly distributed load of 5 kN/m over its full length. The moment reaction at the fixed base is most nearly: See the worked solution or practice the Structural Basics deck
A simply supported Pratt truss spans 30 m in five 6 m panels with a height of 4 m. Vertical loads of 40 kN act at the two top-chord joints located 6 m from each support. Using the method of sections, the force in the bottom-chord panel at midspan is most nearly: See the worked solution or practice the Trusses deck

Related Calculators and Examples

Frequently Asked Questions

What does statically determinate mean on the FE Civil exam?

A structure is statically determinate when all reactions and internal forces can be found from the equilibrium equations alone. In 2D you have three equations (sum of horizontal forces, sum of vertical forces, sum of moments), plus one extra equation for each internal hinge.

Where is the maximum bending moment in a beam?

The maximum moment occurs where the shear diagram crosses zero, or at a point load or support where the shear changes sign. On an overhanging beam, check both the largest positive moment in the span and the negative moment over the support.

Should I use the method of joints or the method of sections for a truss?

Use the method of sections when the question asks for one or two members in the middle of a truss: cut through at most three unknown members and take moments about the point where two of them intersect. Use joints when the member is near a support or a simple joint.

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