Structural Analysis

Syllabus

Statically determinate and indeterminate structures by the force and displacement methods; energy methods; influence lines; matrix methods (stiffness and flexibility); analysis of trusses, arches, beams and frames.

Key concepts and formulas

Static indeterminacy.

• Plane truss: D_s = m + r − 2j (m members, r reactions, j joints).
• Plane rigid frame: D_s = 3m + r − 3j, reduced by the releases at internal hinges.
• D_s = 0 means determinate (if stable); D_s > 0 means indeterminate; D_s < 0 means a mechanism.

Kinematic indeterminacy (degrees of freedom): for a plane rigid frame, D_k = 3j − r (members treated as axially rigid reduce this further).

Energy theorems.

• Strain energy in bending: U = ∫ M²/(2EI) dx; in axial loading: U = P²L/(2AE).
• Castigliano's theorem: the deflection under a load P is ∂U/∂P.
• Maxwell's reciprocal theorem: the deflection at A due to a unit load at B equals the deflection at B due to a unit load at A.
• Unit load method: δ = Σ (P·p·L)/(AE) for trusses, where p is the member force due to a unit load at the point and in the direction of the required deflection.

Slope-deflection and moment distribution.

• Fixed-end moments: central point load P gives PL/8 at each end; UDL w gives wL²/12.
• Member stiffness 4EI/L for a far end fixed, 3EI/L for a far end pinned. The carry-over factor is 1/2 to a fixed far end and 0 to a pinned one.
• Distribution factor at a joint = member stiffness / sum of stiffnesses at the joint.

Influence lines. An influence line shows how a response (reaction, shear or moment at a section) varies as a unit load moves across the structure. For a simply supported beam of span L, the influence line for the moment at a section at distance a from the left support is a triangle with peak ab/L at the section (b = L − a). By the Müller-Breslau principle, the influence line has the shape of the deflected structure when the corresponding restraint is released and given a unit displacement.

Three-hinged arch. It is determinate. For a parabolic three-hinged arch carrying a UDL over the whole span, the bending moment is zero everywhere. The horizontal thrust is H = wL²/(8h) for rise h.

Worked example

A propped cantilever (fixed at A, roller at B) of span L carries a UDL w. Find the prop reaction by compatibility (the deflection at the prop must be zero).

The deflection at B due to the UDL alone (as a cantilever) is wL⁴/(8EI) downwards; the prop reaction R gives RL³/(3EI) upwards. Setting the net deflection to zero gives R = 3wL/8, and the fixed-end moment is wL²/8.

L, w, EI = 5.0, 12.0, 1.0
R = 3 * w * L / 8
print(f"prop reaction R = 3wL/8 = {R:.3f}")

# Check compatibility: the UDL deflects the free end down by wL^4/(8EI), the prop pushes it up by RL^3/(3EI).
down, up = w * L ** 4 / (8 * EI), R * L ** 3 / (3 * EI)
print(f"deflection from the UDL = {down:.3f}, from the prop = {up:.3f}, net = {down - up:.3f}")
print(f"fixed-end moment = {w * L ** 2 / 2 - R * L:.3f} = wL^2/8 = {w * L ** 2 / 8:.3f}")

# Static indeterminacy of a plane truss: m + r - 2j
for m, r, j in [(9, 3, 6), (10, 3, 6), (11, 4, 7)]:
    print(f"truss m={m}, r={r}, j={j}: Ds = {m + r - 2 * j}")

prop reaction R = 3wL/8 = 22.500
deflection from the UDL = 937.500, from the prop = 937.500, net = 0.000
fixed-end moment = 37.500 = wL^2/8 = 37.500
truss m=9, r=3, j=6: Ds = 0
truss m=10, r=3, j=6: Ds = 1
truss m=11, r=4, j=7: Ds = 1

Practice questions