Advanced Civil Engineering Handbook › Chapter 3

Chapter 3 of 14 · 43 formulas · 81 one-liners

Structural Analysis — formulas & one-liners

Indeterminacy, trusses, arches and cables, deflection methods, slope-deflection and moment distribution, influence lines, plastic analysis and matrix basics for AE/JE exams.

3.1 Static & Kinematic Indeterminacy

Static indeterminacy of a rigid-jointed plane frame

D_{s} = 3 m + r - 3 j - r_{r}

m = members; r = external reaction components; j = joints (incl. supports); r_r = releases

An internal hinge joining m' members gives r_r = m' - 1

Static indeterminacy of a plane truss

D_{s} = m + r - 2 j

m = members; r = reaction components; j = joints

Kinematic indeterminacy of a plane frame

D_{k} = 3 j - r

j = joints (incl. supports); r = restrained displacement components

Axial deformation included; if members are taken axially rigid, subtract the number of such members; add extra rotations at internal hinges

Kinematic indeterminacy of a plane truss

D_{k} = 2 j - r

j = joints; r = restrained displacement components

Static indeterminacy is D_s = m + r - 3j for a space truss and D_s = 6m + r - 6j - r_r for a rigid-jointed space frame.
If D_s = 0 a stable structure is statically determinate; D_s > 0 means indeterminate; D_s < 0 means a mechanism (unstable).
A joint has 2 degrees of freedom in a plane truss, 3 in a plane frame, 3 in a space truss and 6 in a space frame.
A perfect (just-rigid) plane truss satisfies m = 2j - 3; m < 2j - 3 is deficient and m > 2j - 3 is redundant.
A fixed-fixed beam has D_s = 3 (2 if only vertical loads are considered) and D_k = 0.
A propped cantilever has D_s = 1; a continuous beam of n spans on simple supports has D_s = n - 1 under vertical loads.
A three-hinged arch is statically determinate, a two-hinged arch has D_s = 1 and a fixed (hingeless) arch has D_s = 3.
A single closed ring or closed rectangular frame is internally indeterminate to the third degree (D_s = 3 per closed loop).
A single-bay portal frame with fixed bases has D_s = 3; with both bases hinged it has D_s = 1.
A cantilever has D_k = 3 at its free end (2 if axial deformation is neglected); a simply supported beam has D_k = 3 (2 if axial deformation is neglected).
The force (flexibility) method solves for D_s unknown redundants, while the displacement (stiffness) method solves for D_k unknown joint displacements.

3.2 Trusses, Arches & Cables

Profile of a symmetric parabolic arch

y = \frac{4 h x ( L - x )}{L ^{2}}

h = central rise; L = span; x = distance from left support; y = height of arch axis

Horizontal thrust: parabolic arch under UDL over the whole span

H = \frac{w L ^{2}}{8 h}

w = load per unit horizontal length; L = span; h = rise

Same for three-hinged and two-hinged parabolic arches; BM is zero everywhere (pure compression)

Horizontal thrust: three-hinged arch with point load W at the crown

H = \frac{W L}{4 h}

W = crown load; L = span; h = rise

Horizontal thrust: two-hinged parabolic arch with point load W at the crown

H = \frac{25 W L}{128 h}

W = crown load; L = span; h = rise

Assumes I = I₀ sec θ

Rise of crown of a three-hinged arch due to temperature rise

δ h = \frac{( L ^{2} + 4 h ^{2} ) α T}{4 h}

α = coefficient of thermal expansion; T = temperature rise; L = span; h = rise

Horizontal tension in a cable under UDL (supports level)

H = \frac{w L ^{2}}{8 d}

w = load per unit horizontal length; L = span; d = central dip (sag)

Maximum tension in a cable under UDL (supports level)

T_{ma x} = H^{2} + (\frac{w L}{2})^{2}

H = horizontal tension; wL/2 = vertical reaction at each support

Occurs at the supports; minimum tension = H at the lowest point

Length of a parabolic cable (supports level)

S \approx L + \frac{8 d ^{2}}{3 L}

L = span; d = central dip

Deflection of a truss joint by the unit load method

Δ = \sum \frac{P k L}{A E}

P = member force due to actual loads; k = member force due to unit load at the joint; L, A = member length and area

At an unloaded joint with only two non-collinear members, both members are zero-force members.
At an unloaded joint with three members of which two are collinear, the third member carries zero force.
In the method of joints not more than two unknown member forces should act at the joint being solved; in the method of sections not more than three members should be cut.
Under gravity loading a Pratt truss has its diagonals in tension and verticals in compression, while a Howe truss has diagonals in compression and verticals in tension.
The method of tension coefficients is used for the analysis of space trusses.
The Williot-Mohr diagram gives the displacements of truss joints graphically.
Eddy's theorem: the bending moment at any section of an arch equals H times the vertical intercept between the linear arch (thrust line) and the arch axis.
A three-hinged arch is determinate, so temperature change, support settlement or lack of fit produce no stresses in it.
A two-hinged parabolic arch carrying UDL on half the span has horizontal thrust H = wL²/(16h), half of the full-span value.
A two-hinged semicircular arch with a point load W at the crown has horizontal thrust H = W/π.
A two-hinged semicircular arch of radius R carrying UDL w per unit horizontal length over the whole span has H = 4wR/(3π).
A temperature rise T in a two-hinged parabolic arch (I = I₀ sec θ) induces a horizontal thrust H = 15EI₀αT/(8h²).
A cable carrying a load uniformly distributed along the horizontal span hangs as a parabola; under its own weight along its length it hangs as a catenary.
The horizontal component of cable tension is constant along the cable and is inversely proportional to the central dip.
With supports at different levels, the maximum cable tension occurs at the higher support.
A suspension bridge with a three-hinged stiffening girder is statically determinate; with a two-hinged stiffening girder it is indeterminate to the first degree.

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3.3 Deflection: Moment-Area, Conjugate Beam & Energy Methods

Moment-area theorem I (change of slope)

θ_{B} - θ_{A} = \int_{A}^{B} \frac{M}{E I} d x

M/EI = curvature; θ = slope

Equals the area of the M/EI diagram between A and B

Moment-area theorem II (tangential deviation)

t_{B / A} = \frac{A _{A B} x ˉ _{B}}{E I}

A_AB = area of BMD between A and B; x̄_B = distance of its centroid from B

Deviation of B from the tangent drawn at A

Unit load (virtual work) method for beams and frames

δ = \int \frac{M m}{E I} d x

M = BM due to actual loads; m = BM due to a unit load at the point and in the direction of δ

Castigliano's second theorem

δ_{i} = \frac{\partial U}{\partial P _{i}}

U = total strain energy; P_i = load at point i; δ_i = deflection at i along P_i

Cantilever with point load W at the free end

δ = \frac{W L ^{3}}{3 E I}, θ = \frac{W L ^{2}}{2 E I}

δ, θ = deflection and slope at the free end

Cantilever with UDL w over the full length

δ = \frac{w L ^{4}}{8 E I}, θ = \frac{w L ^{3}}{6 E I}

δ, θ = deflection and slope at the free end

Simply supported beam with central point load W

δ = \frac{W L ^{3}}{48 E I}, θ = \frac{W L ^{2}}{16 E I}

δ at midspan; θ at the supports

Simply supported beam with UDL w over the span

δ = \frac{5 w L ^{4}}{384 E I}, θ = \frac{w L ^{3}}{24 E I}

δ at midspan; θ at the supports

In the conjugate beam method the real beam's M/EI diagram is applied as load; shear in the conjugate beam gives slope and its bending moment gives deflection of the real beam.
In the conjugate beam a fixed end becomes a free end and a free end becomes a fixed end.
Simple end supports remain simple supports in the conjugate beam; an internal hinge becomes an internal support and an internal support becomes an internal hinge.
Maxwell's reciprocal theorem: the deflection at i due to a unit load at j equals the deflection at j due to a unit load at i (δ_ij = δ_ji).
Betti's law: the work done by one load system through the displacements caused by a second system equals the work done by the second system through the displacements caused by the first.
A cantilever with a moment M at the free end has end deflection ML²/(2EI) and end slope ML/EI.
A fixed-fixed beam with a central point load deflects WL³/(192EI) at midspan, one-fourth of the simply supported value.
A fixed-fixed beam with UDL deflects wL⁴/(384EI) at midspan, one-fifth of the simply supported value.
In a simply supported beam with one point load anywhere on the span, the maximum deflection always lies within 0.0774L of midspan.
The principle of superposition does not apply to strain energy, because strain energy is a quadratic function of load.

3.4 Slope-Deflection, Moment Distribution & Three-Moment Equation

Slope-deflection equation

M_{A B} = M_{F A B} + \frac{2 E I}{L} (2 θ_{A} + θ_{B} - \frac{3Δ}{L})

M_FAB = fixed-end moment; θ_A, θ_B = end rotations; Δ = relative transverse displacement of ends

Clockwise moments and rotations positive

Modified slope-deflection equation (far end B hinged)

M_{A B} = (M_{F A B} - \frac{M _{F B A}}{2}) + \frac{3 E I}{L} (θ_{A} - \frac{Δ}{L})

M_FAB, M_FBA = fixed-end moments with sign; θ_A = rotation at A; Δ = relative displacement

Rotational (absolute) stiffness of a member

k = \frac{4 E I}{L} (far end fixed), k = \frac{3 E I}{L} (far end hinged)

k = moment to rotate the near end by unit angle; EI = flexural rigidity; L = length

Distribution factor of a member at a joint

D F = \frac{k}{\sum k}

k = stiffness of the member; Σk = sum of stiffnesses of all members meeting at the joint

Fixed-end moment due to relative settlement Δ of the ends

M = \frac{6 E I Δ}{L ^{2}} (both ends fixed), M = \frac{3 E I Δ}{L ^{2}} (one end hinged)

Δ = relative settlement; EI = flexural rigidity; L = span

Clapeyron's three-moment equation (EI constant, supports at the same level)

M_{A} L_{1} + 2 M_{B} (L_{1} + L_{2}) + M_{C} L_{2} = \frac{6 a _{1} x ˉ _{1}}{L _{1}} + \frac{6 a _{2} x ˉ _{2}}{L _{2}}

M_A, M_B, M_C = support (hogging) moments; a₁, a₂ = areas of free BMDs of spans L₁, L₂; x̄_1 from A, x̄_2 from C

For a span with UDL w, 6ax̄/L = wL³/4

The carry-over factor of a prismatic member is 1/2 when the far end is fixed and 0 when the far end is hinged.
The sum of distribution factors at a joint is 1; the DF at a fixed support is 0 and at a free hinged end is 1.
For a member bent in symmetric double curvature about a line of symmetry the effective stiffness is 2EI/L; for antisymmetric deformation it is 6EI/L.
The lateral (translational) stiffness of a member is 12EI/L³ when both ends are fixed and 3EI/L³ when one end is hinged.
The slope-deflection method was introduced by G.A. Maney (1915) and the moment distribution method by Hardy Cross (1930).
In Kani's method the rotation factor of a member at a joint is -1/2 times its distribution factor.
Sway occurs in a portal frame when the loading, geometry, member stiffnesses or support conditions are unsymmetrical; an extra shear (sway) equation is then needed.
A two-span continuous beam of equal spans L with UDL w has a hogging moment of wL²/8 at the central support and a central reaction of 5wL/4.

3.5 Fixed-End Moments & Standard Beam Results

FEM: UDL over the whole span

M_{F} = \frac{w L ^{2}}{12}

w = load per unit length; L = span

Hogging at both ends; midspan sagging moment = wL²/24

FEM: central point load

M_{F} = \frac{W L}{8}

W = point load; L = span

Midspan sagging moment is also WL/8

FEM: eccentric point load

M_{A} = \frac{W a b ^{2}}{L ^{2}}, M_{B} = \frac{W a ^{2} b}{L ^{2}}

a = distance of load from A; b = distance from B; L = a + b

FEM: triangular load, zero at A rising to w at B

M_{A} = \frac{w L ^{2}}{30}, M_{B} = \frac{w L ^{2}}{20}

w = peak intensity at B; L = span

Larger moment at the heavier end

A couple M applied at the midspan of a fixed beam produces fixed-end moments of M/4 at each end, both in the same rotational sense.
A fixed beam with UDL has points of contraflexure at 0.211L from each support.
A fixed beam with a central point load has points of contraflexure at L/4 from each support.
A propped cantilever with UDL has a prop reaction of 3wL/8 and a fixed-end moment of wL²/8.
In a propped cantilever with UDL the maximum sagging moment is 9wL²/128, at 3L/8 from the prop, and the point of contraflexure is at L/4 from the fixed end.
A propped cantilever with a central point load W has a prop reaction of 5W/16 and a fixed-end moment of 3WL/16.

3.6 Influence Lines & Rolling Loads

ILD ordinate for BM at section C of a simply supported beam (under C)

y_{ma x} = \frac{a b}{L}

a, b = distances of C from left and right supports; L = span

ILD is a triangle with apex at C

ILD ordinates for shear at section C of a simply supported beam

- \frac{a}{L} (just left of C), + \frac{b}{L} (just right of C)

a, b = distances of C from left and right supports; L = span

Value of a function from its influence line

S = \sum W_{i} y_{i} + w \times A_{I L D}

W_i = point loads; y_i = ILD ordinates under them; w = UDL intensity; A_ILD = area of ILD under the UDL

Absolute maximum BM due to a UDL of length a shorter than span L

M_{ma x} = \frac{w a ( 2 L - a )}{8}

w = intensity; a = loaded length; L = span

Occurs at midspan with the load centred on the span

Müller-Breslau principle: the influence line of a reaction or internal force is the deflected shape obtained by removing that restraint and applying a unit displacement in its direction.
Influence lines for statically determinate structures are made of straight lines, while those for indeterminate structures are curved.
The ILD for bending moment at the fixed end of a cantilever is a triangle with ordinate equal to the span at the free end.
For a UDL longer than the span, the maximum BM at any section occurs when the whole span is loaded; it equals w·a·b/2.
For a UDL shorter than the span, the BM at a section is maximum when the section divides the load in the same ratio as it divides the span.
For a train of point loads, the BM at a section is maximum when the average load per unit length on the left of the section equals that on the right.
The absolute maximum BM under a train of loads occurs under a load placed so that midspan bisects the distance between that load and the resultant of all loads on the span.
The absolute maximum BM due to a single rolling point load W on a simply supported beam is WL/4, at midspan.
For a UDL longer than the span, the maximum positive shear at a section occurs when the load covers the portion from the section to the right support.

3.7 Plastic Analysis

Plastic moment of resistance

M_{p} = f_{y} Z_{p}

f_y = yield stress; Z_p = plastic section modulus

Shape factor

S = \frac{M _{p}}{M _{y}} = \frac{Z _{p}}{Z _{e}}

M_y = yield moment; Z_e = elastic section modulus

Plastic section modulus

Z_{p} = \frac{A}{2} (\overset{y}{ˉ}_{1} + \overset{y}{ˉ}_{2})

A = total area; ȳ_1, ȳ_2 = centroids of the areas above and below the equal-area axis, measured from it

Collapse load: propped cantilever with UDL

w_{c} = \frac{11.66 M _{p}}{L ^{2}}

M_p = plastic moment; L = span

Sagging hinge at 0.414L from the prop

Collapse load: fixed beam with eccentric point load

W_{c} = \frac{2 M _{p} L}{a b}

a, b = distances of the load from the supports; L = span

Shape factor is 1.5 for a rectangle, 16/(3π) ≈ 1.70 for a solid circle, 2.0 for a diamond (square with a vertical diagonal) and 2.34 for a triangle.
Shape factor of a thin-walled circular tube is 4/π ≈ 1.27, and of a rolled I-section about its major axis about 1.12 to 1.15.
For a rectangular section Z_p = bd²/4 while Z_e = bd²/6.
The plastic neutral axis divides the section into two equal areas, while the elastic neutral axis passes through the centroid.
A plastic hinge is a fully yielded zone that rotates freely at a constant moment M_p.
Collapse loads of simply supported beams: 4M_p/L for a central point load and 8M_p/L² for UDL.
Collapse loads of fixed beams: 8M_p/L for a central point load and 16M_p/L² for UDL.
A propped cantilever with a central point load collapses at W_c = 6M_p/L; a cantilever with an end load collapses at M_p/L.
A fixed-base portal frame of height h with uniform M_p fails in the sway mechanism at a lateral load of 4M_p/h.
Complete collapse of a structure with static indeterminacy D_s needs D_s + 1 plastic hinges; partial collapse can occur with fewer.
Static (lower bound) theorem: a load for which a safe, statically admissible moment distribution exists is less than or equal to the true collapse load.
Kinematic (upper bound) theorem: the load computed from any assumed mechanism is greater than or equal to the true collapse load.
Uniqueness theorem: a load that is both statically admissible and corresponds to a mechanism is the true collapse load.
The number of independent mechanisms equals N - D_s, where N is the number of possible plastic hinge locations; the basic types are beam, sway, gable and joint mechanisms.
In a simply supported rectangular beam at collapse, the yielded (plastic) zone extends over L/3 for a central point load and L/√3 for a UDL.

3.8 Matrix Methods: Basics

Relation between stiffness and flexibility matrices

[K] = [F]^{- 1}

K = stiffness matrix; F = flexibility matrix (same coordinates)

Stiffness matrix of an axial (truss) element in local axes

[k] = \frac{A E}{L} [1 - 1 - 1 1]

A = area; E = modulus of elasticity; L = length

Rotational stiffness matrix of a beam element (end rotations only)

[k] = \frac{2 E I}{L} [2112]

EI = flexural rigidity; L = length

Translations of the ends restrained

The flexibility (force) method takes redundant forces as unknowns; the stiffness (displacement) method takes joint displacements as unknowns.
The order of the flexibility matrix equals the degree of static indeterminacy, and the order of the stiffness matrix equals the degree of kinematic indeterminacy.
A stiffness coefficient k_ij is the force at coordinate i due to a unit displacement at coordinate j with all other displacements held at zero.
Stiffness and flexibility matrices are symmetric (from Maxwell's reciprocal theorem) and their diagonal elements are always positive.
The stiffness matrix of an unsupported structure is singular (determinant zero) until boundary conditions are applied.
A plane truss element has a 4 × 4 global stiffness matrix, a plane frame element 6 × 6 and a space frame element 12 × 12.

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