Simply Supported Beams¶

CalcpadCE worksheets in this section collect the standard load cases for the simply supported beam — a span resting on a pin at one end and a roller at the other. The system is statically determinate, so the reactions and internal forces follow from equilibrium alone and depend only on the geometry and the applied load.

A single point force and a point moment at distance \(a\) from the left support cover the basic concentrated cases, while a series of equal concentrated forces at constant spacing \(a = l / (n + 1)\) models a row of identical loads such as wheel groups or hangers. For distributed actions, the uniform load reproduces the classroom \(M_{max} = q l^2 / 8\), \(V_{max} = q l / 2\) result, the linearly varying load covers the trapezoidal and triangular shapes via end intensities \(q_1\) and \(q_2\), and the partial uniform load acts on a stretch of length \(b\) between offsets \(a\) and \(c = l - a - b\).

Span, load magnitudes, distances and a probe coordinate \(x_1\) are exposed as inputs, so \(M(x_1)\) and \(V(x_1)\) can be picked off the bending and shear diagrams at any cross-section.

Uniformly Distributed Load¶

Full-span uniform load \(q\), recovering the textbook results \(M_{max} = q l^2 / 8\) and \(V_{max} = q l / 2\).

Code:

'<div style = "max-width:180mm;">
'<img style="width:190pt;" alt = simply-supported-beam-distributed-load-uniform.png" class="side" src = "../../Images/mechanics/beams/simply-supported-beam-distributed-load-uniform.png">
'<p><b>Input data</b></p>
'Beam length -'l = ?'m
'Load -'q = ?'kN/m
#post
'<p><b>Internal forces</b></p>
'Bending -'M_max = q*l^2/8'kN·m
'Shear -'V_max = q*l/2'kN
'<p><b>Diagrams</b></p>
#hide
PlotWidth = 600
PlotHeight = 150
#show
'Calculate internal forces at'x_1 = ?'m
#pre

#post
'Bending
M(x) = q*x*(l - x)/2
$Plot{-M(x) @ x = 0 : l}
M(x_1)'kN·m
'Shear
V(x) = q*(l/2 - x)
$Plot{V(x) @ x = 0 : l}
V(x_1)'kN
#show
'</div>5    10  1

simply-supported-beam-distributed-load-uniform.png" — Rendered Output:

Linearly Distributed Load¶

Trapezoidal load defined by end intensities \(q_1\) and \(q_2\), locating the in-span peak from the zero of \(V(x)\) and falling back to the uniform and triangular special cases.

Code:

'<div style = "max-width:180mm;">
'<img style="width:190pt;" alt = "simply-supported-beam-distributed-load-linear.png" class="side" src = "../../Images/mechanics/beams/simply-supported-beam-distributed-load-linear.png">
'<p><b>Input data</b></p>
'Beam length -'l = ?'m
'Load -'q_1 = ?'kN/m,'q_2 = ?'kN/m
#post
Δq = q_2 - q_1'kN/m
'<p><b>Internal forces</b></p>
#if '<!--'q_1 ≡ q_2'-->
    'Bending -'M_max = q_1*l^2/8
#else if '<!--'q_1 ≡ 0'-->
    'Distance -'x_max = l/sqr(3)'m
    'Bending
    M_max = Δq*l^2/(9*sqr(3))'kN·m
#else
    k = Δq/q_1
    'Distance
    x_max = l/k*(sqr(1 + k + k^2/3) - 1)'m
    'Bending
    M_max = q_1*x_max*(l - x_max)/2 + Δq*x_max*l*(1 - (x_max/l)^2)/6'kN·m
#end if
'Shear
V_A = l*(q_1/2 + Δq/6)'kN
V_B = l*(q_1/2 + Δq/3)'kN
'<p><b>Diagrams</b></p>
#hide
PlotWidth = 600
PlotHeight = 150
#show
'Calculate internal forces at'x_1 = ?'m
#pre



#post
'Bending
M(x) = q_1*x*(l - x)/2 + Δq*x*l*(1 - (x/l)^2)/6
$Plot{-M(x) @ x = 0 : l}
M(x_1)'kN·m
'Shear
V(x) = q_1*(l/2 - x) + Δq*(l/6 - x^2/(2*l))
$Plot{V(x) @ x = 0 : l}
V(x_1)'kN
#show
'</div>5    5   15  1

Partially Distributed Load¶

Uniform load \(q\) acting on a stretch of length \(b\) between offsets \(a\) and \(c = l - a - b\), with a guard against an invalid load length when \(a + b > l\).

Code:

'<div style = "max-width:180mm;">
'<img style="width:190pt;" alt = "simply-supported-beam-distributed-load-partial.png" class="side" src = "../../Images/mechanics/beams/simply-supported-beam-distributed-load-partial.png">
'<p><b>Input data</b></p>
'Beam length -'l = ?'m
'Load -'q = ?'kN/m
'Distances -'a = ?'m,'b = ?'m
#post
c = l - (a + b)'m
#if '<!--'c < 0'-->
    '<p class="err">Invalid load length</p>
#else
    '<p><b>Internal forces</b></p>
    'Distance -'x_max = a + b*(b/2 + c)/l'm
    'Bending
    M_max = q*(x_max^2 - a^2)/2'kN·m
    'Shear
    V_A = q*b*(b/2 + c)/l'kN
    V_B = -q*b*(b/2 + a)/l'kN
    '<p><b>Diagrams</b></p>
    #hide
    PlotWidth = 600
    PlotHeight = 150
    #show
    'Calculate internal forces at'x_1 = ?'m
    #pre

    #post
    'Bending
    M(x) = V_A*x - q*(x - a)^2/2*(x > a) + q*(x - a - b)^2/2*(x > a + b)
    $Plot{-M(x) @ x = 0 : l}
    M(x_1)'kN·m
    'Shear
    V(x) = V_A - q*(x - a)*(x > a) + q*(x - a - b)*(x > a + b)
    $Plot{V(x) @ x = 0 : l}
    V(x_1)'kN
#end if
#show
'</div>5    10  1   2   1.5

Concentrated Force¶

Single point force \(F\) at distance \(a\) from the left support, with reactions \(V_A = F (l - a) / l\), \(V_B = -F a / l\) and the bending peak \(M_{max} = F (l - a) a / l\) at the load.

Code:

'<div style = "max-width:180mm;">
'<img style="width:190pt;" alt = "simply-supported-beam-concentrated-force.png" class="side" src = "../../Images/mechanics/beams/simply-supported-beam-concentrated-force.png">
'<p><b>Input data</b></p>
'Beam length -'l = ?'m
'Load -'F = ?'kN
'Distance -'a = ?'m
#post
'<p><b>Internal forces</b></p>
'Bending
M_max = F*(l - a)*a/l'kN·m
'Shear
V_A = F*(l - a)/l'kN
V_B = -F*a/l'kN
'<p><b>Diagrams</b></p>
#hide
PlotWidth = 600
PlotHeight = 150
#show
'Calculate internal forces at'x_1 = ?'m
#pre

#post
'Bending
M(x) = V_A*x - F*(x - a)*(x > a)
$Plot{-M(x) @ x = 0 : l}
M(x_1)'kN·m
'Shear
V(x) = V_A - F*(x > a)
$Plot{V(x) @ x = 0 : l}
V(x_1)'kN
#show
'</div>5    10  2   1

Concentrated Forces¶

Series of \(n\) equal forces \(F\) spaced at \(a = l / (n + 1)\), with \(M_{max}\) taken at mid-span and switching between the odd-\(n\) and even-\(n\) closed forms.

Code:

'<div style = "max-width:180mm;">
'<img style="width:190pt;" alt = "simply-supported-beam-forces.png" class="side" src = "../../Images/mechanics/beams/simply-supported-beam-forces.png">
'<p><b>Input data</b></p>
'Beam length -'l = ?'m
'Load -'F = ?'kN
'Number of forces -'n = ?
#post
'Number of spacings -'n_1 = n + 1
'<p><b>Internal forces</b></p>
#if '<!--'n%%2 ≡ 0'-->
    'Bending -'M_max = (n_1^2 - 1)*F*l/(8*n_1)'kN·m
#else
    'Bending -'M_max = n_1*F*l/8'kN·m
#end if
'Shear -'V_max = n*F/2'kN
'<p><b>Diagrams</b></p>
#hide
PlotWidth = 600
PlotHeight = 150
#show
'Calculate internal forces at'x_1 = ?'m
#pre

#post
a = l/n_1'm
n(x) = max(ceiling(x/a) - 1; 0)
x_F(x) = n(x)*(x  - (n(x) + 1)*a/2)
'Bending
M(x) = V_max*x - F*x_F(x)
$Plot{-M(x) @ x = 0 : l}
M(x_1)'kN·m
'Shear
V(x) = V_max - F*n(x)
$Plot{V(x) @ x = 0 : l}
V(x_1)'kN
#show
'</div>5    10  2   1

Concentrated Moment¶

Concentrated moment \(M\) applied at distance \(a\), giving a constant shear \(V = M / l\) and a bending diagram with a step of \(M\) at the load section.

Code:

'<div style = "max-width:180mm;">
'<img style="width:190pt;" alt = "simply-supported-beam-concentrated-moment.png" class="side" src = "../../Images/mechanics/beams/simply-supported-beam-concentrated-moment.png">
'<p><b>Input data</b></p>
'Beam length -'l = ?'m
'Load -'M = ?'kN·m
'Distance -'a = ?'m
#post
'<p><b>Internal forces</b></p>
'Bending
M_a = M/l*a'kN·m
M_b = M/l*(l - a)'kN·m
'Shear
V = M/l'kN
'<p><b>Diagrams</b></p>
#hide
PlotWidth = 600
PlotHeight = 150
#show
'Calculate internal forces at'x_1 = ?'m
#pre

#post
'Bending
M(x) = -V*x + M*(x > a)
$Plot{-M(x) @ x = 0 : l}
M(x_1)'kN·m
'Shear
V(x) = V
$Plot{V(x) @ x = 0 : l}
V(x_1)'kN
#show
'</div>5    10  2   1

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