Continuous Beam

CalcpadCE worksheets in this section deal with the continuous beam — a single member spanning over three or more supports, where each interior support adds one degree of static indeterminacy.

The continuous-beam worksheet takes a vector of span lengths \(l\), a uniform load \(q\), and rectangular-section material and geometry properties. It builds the primary system by removing the interior supports, treats the redundant reactions as unknowns \(X_i\), and assembles the flexibility matrix \(\delta\) and load vector \(\Delta_F\) from bending and shear contributions through Area integrals along the full beam. Solving \(\delta\,X = -\Delta_F\) with clsolve gives the support reactions, and the bending moment, shear and deflection diagrams along the entire span follow by superposition of the primary-system response and the unit-load fields.

Span lengths, section dimensions and material constants are exposed at the top of the sheet, so different geometries and load levels can be checked without rewriting any of the analysis code.

Continuous Beam

Force-method analysis of a multi-span continuous beam under a uniform load \(q\), building the flexibility matrix from bending and shear integrals and solving for the redundant reactions, internal forces and deflections.

Code:

'<h4>Input data</h4>
'Span lengths -'l = [4; 7; 3; 5]*m
'Number of spans -'n = len(l)
#hide
x = vector(n)', '$Repeat{x.k = sum(first(l; k)) @ k = 1 : n}
#show
'Coordinates of supports -'x
'Total beam length -'L = x.n
'Load -'q = 10kN/m
'<h5>Material</h5>
'Elastic modulus -'E = 30GPa
'Poisson′s ratio - 'ν = 0.2
'Shear modulus -'G = E/(2*(1 + ν))
'<h5>Cross section</h5>
'Rectangular with dimensions:'b = 250mm', 'h = 500mm
'Area -'A = b*h
'Moment of inertia -'I = b*h^3/12
'Shear area -'A_Q = 5/6*b*h
'<h4>Solution</h4>
'The solution will be performed by the force method with a primary system - simply supported beam with internal supports removed and replaced by unknown forces <em>X</em><sub>i</sub>
'Bending moments
'- in section <em>a</em> due to unit force at distance <em>x</em> from the beginning of the beam:
M_1,max(x) = (x/L - 1)*x
M_1,a(a; x) = M_1,max(x)*switch(a < x; a/x; a ≡ x; 1; (L - a)/(L - x))
'- in section <em>a</em>, due to unit force <em>X</em><sub>i</sub>:
M_1(a; i) = M_1,a(a; x.i)
#hide
PlotHeight = 100', 'PlotWidth = 600
#show
$Plot{-M_1(x; 1) & -M_1(x; 2) & -M_1(x; 3) @ x = 0m : L}
'- due to external loads in primary system:
M_F(x) = q*x/2*(L - x)
$Plot{-M_F(x) @ x = 0m : L}
'Shear forces
'- in section <em>a</em> due to unit force at distance <em>x</em> from the beginning of the beam:
V_1,a(a; x) = if(a < x; x/L - 1; x/L)
'- in section <em>a</em>, due to unit force <em>X</em><sub>i</sub>:
V_1(a; i) = V_1,a(a; x.i)
$Plot{V_1(x; 1) & V_1(x; 2) & V_1(x; 3) @ x = 0m : L}
'- due to external loads in primary system:
V_F(x) = q*(L/2 - x)
$Plot{V_F(x) @ x = 0m : L}
'Number of unknowns by force method -'n_1 = n - 1
'Flexibility coefficients
δ(i; j) = $Area{M_1(x; i)*M_1(x; j)/(E*I) @ x = 0m : L} + $Area{V_1(x; i)*V_1(x; j)/(G*A_Q) @ x = 0m : L}|m/kN
Δ_F(i) = $Area{M_1(x; i)*M_F(x)/(E*I) @ x = 0m : L} + $Area{V_1(x; i)*V_F(x)/(G*A_Q) @ x = 0m : L}
δ = symmetric(n_1)','Δ_F = vector(n_1)
$Repeat{$Repeat{δ.(i; j) = δ(i; j) @ i = 1 : n_1} @ j = 1 : n_1}
$Repeat{Δ_F.i = Δ_F(i) @ i = 1 : n_1}
δ','Δ_F
'Calculation of the unknown forces <em>X</em><sub>i</sub>
X = -clsolve(δ; Δ_F)
'<h4>Results</h4>
'Bending moment diagram -'M(x) = M_F(x) + $Sum{M_1(x; i)*X.i @ i = 1 : n_1}
$Plot{-M(x) @ x = 0m : L}
#hide
M_min = vector(n_1)', 'M_max = vector(n)
$Repeat{M_min.i = M(x.i) @ i = 1 : n_1}
$Repeat{M_max.i = $Sup{M(x) @ x = x.i : x.i - l.i} @ i = 1 : n}
#show
M_max
M_min
'Shear force diagram -'V(x) = V_F(x) + $Sum{V_1(x; i)*X.i @ i = 1 : n_1}
$Plot{V(x) @ x = 0m : L}
#hide
V_min = vector(n)', 'V_max = vector(n)
$Repeat{V_min.i = $Inf{V(x) @ x = x.i : x.i - l.i} @ i = 1 : n}
$Repeat{V_max.i = $Sup{V(x) @ x = x.i : x.i - l.i} @ i = 1 : n}
#show
V_max
V_min
'Deflections
'- in section <em>a</em>, due to unit force <em>X</em><sub>i</sub>:
d_1(a; i) = $Area{M_1,a(x; a)*M_1(x; i)/(E*I) @ x = 0m : L} + $Area{V_1,a(x; a)*V_1(x; i)/(G*A_Q) @ x = 0m : L}|m/kN
'- due to external loads in primary system:
d_F(a) = $Area{M_1,a(x; a)*M_F(x)/(E*I) @ x = 0m : L} + $Area{V_1,a(x; a)*V_F(x)/(G*A_Q) @ x = 0m : L}
d(a) = d_F(a) + $Sum{d_1(a; i)*X.i @ i = 1 : n_1}|mm
$Plot{d(x) @ x = 0m : L}
'Maximum deflection -'d_max = $inf{d(x) @ x = 0m : L}|mm
'At a distance from the origin -'x_inf

Rendered Output:

Input data

Span lengths - ⃗l = [4; 7; 3; 5] · m = [4 m 7 m 3 m 5 m]

Number of spans - n = len ( ⃗l )  = 4

Coordinates of supports - ⃗x = [4 m 11 m 14 m 19 m]

Total beam length - L = ⃗x₄ = 19 m

Load - q = 10 kNm

Material

Elastic modulus - E = 30 GPa

Poisson′s ratio - ν = 0.2

Shear modulus - G = E2 ·  ( 1 + ν )  = 30 GPa2 ·  ( 1 + 0.2 )  = 12.5 GPa

Cross section

Rectangular with dimensions: b = 250 mm , h = 500 mm

Area - A = b · h = 250 mm · 500 mm = 125000 mm²

Moment of inertia - I = b · h³12 = 250 mm ·  ( 500 mm ) ³12 = 2604166667 mm⁴

Shear area - A_Q = 56 · b · h = 56 · 250 mm · 500 mm = 104167 mm²

Solution

The solution will be performed by the force method with a primary system - simply supported beam with internal supports removed and replaced by unknown forces X_i

Bending moments

- in section a due to unit force at distance x from the beginning of the beam:

M_1,max ( x )  = (xL − 1) · x

M_1,a ( a; x )  = M_1,max  ( x )  · {if a < x: ax
else if a ≡ x: 1
else: L − aL − x

- in section a, due to unit force X_i:

M₁ ( a; i )  = M_1,a  ( a; ⃗x_i ) 

- due to external loads in primary system:

M_F ( x )  = q · x2 ·  ( L − x ) 

Shear forces

- in section a due to unit force at distance x from the beginning of the beam:

V_1,a ( a; x )  = {if a < x: xL − 1
else: xL

- in section a, due to unit force X_i:

V₁ ( a; i )  = V_1,a  ( a; ⃗x_i ) 

- due to external loads in primary system:

V_F ( x )  = q · (L2 − x)

Number of unknowns by force method - n₁ = n − 1 = 4 − 1 = 3

Flexibility coefficients

δ ( i; j )  =   L∫0 m M₁  ( x; i )  · M₁  ( x; j ) E · I dx +   L∫0 m V₁  ( x; i )  · V₁  ( x; j ) G · A_Q dx

Δ_F ( i )  =   L∫0 m M₁  ( x; i )  · M_F  ( x ) E · I dx +   L∫0 m V₁  ( x; i )  · V_F  ( x ) G · A_Q dx

δ = symmetric ( n₁ )  = symmetric ( 3 )  = 000 000 000 , ⃗Δ_F = vector ( n₁ )  = vector ( 3 )  = [0 0 0]

$Repeat{$Repeat{δ_i,j = δ  ( i; j )  for i = 1...n₁} for j = 1...n₁} = 0.0011 m ∕ kN

$Repeat{⃗Δ_F.i = Δ_F  ( i )  for i = 1...n₁} = -0.161 m

δ = 0.000811 m ∕ kN0.00101 m ∕ kN0.000719 m ∕ kN 0.00101 m ∕ kN0.00174 m ∕ kN0.00133 m ∕ kN 0.000719 m ∕ kN0.00133 m ∕ kN0.0011 m ∕ kN , ⃗Δ_F = [-0.135 m -0.211 m -0.161 m]

Calculation of the unknown forces X_i

⃗X = -clsolve ( δ; ⃗Δ_F )  = [64.86 kN 53.48 kN 39.33 kN]

Results

Bending moment diagram - M ( x )  = M_F  ( x )  + n₁∑i= 1M₁  ( ⃗x; i )  · X_i

⃗M_max = [5.97 kNm 27.74 kNm -12.23 kNm 22.9 kNm]

⃗M_min = [-36.29 kNm -30.79 kNm -17.99 kNm]

Shear force diagram - V ( x )  = V_F  ( x )  + n₁∑i= 1V₁  ( ⃗x; i )  · ⃗X_i

⃗V_max = [10.93 kN 35.79 kN 19.26 kN 28.6 kN]

⃗V_min = [-29.07 kN -34.21 kN -10.74 kN -21.4 kN]

Deflections

- in section a, due to unit force X_i:

d₁ ( a; i )  =   L∫0 m M_1,a  ( x; a )  · M₁  ( x; i ) E · I dx +   L∫0 m V_1,a  ( x; a )  · V₁  ( x; i ) G · A_Q dx

- due to external loads in primary system:

d_F ( a )  =   L∫0 m M_1,a  ( x; a )  · M_F  ( x ) E · I dx +   L∫0 m V_1,a  ( x; a )  · V_F  ( x ) G · A_Q dx

d ( a )  = d_F  ( a )  + n₁∑i= 1d₁  ( a; i )  · ⃗X_i

Maximum deflection - d_max = $inf{d  ( x ) ; x ∈ [0 m; L]} = -1.42 mm

At a distance from the origin - x_inf = 7.56 m

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