Cantilevers¶
CalcpadCE worksheets in this section deal with cantilevers — beams fixed at one end and free at the other, where the support has to carry the full bending moment and shear from every load applied along the span. They evaluate the bending moment and shear directly from the load parameters, covering the loadings encountered most often in practice.
The simplest sheets isolate a single action at the tip — a point force or a point moment — and recover the elementary \(M = F \, l\) and \(M = M_0\) results. The distributed-load variants cover a uniform load over the full length, a linearly varying load defined by intensities \(q_1\) and \(q_2\) at the ends, and a partial uniform load acting only on a stretch near the tip.
Beam length, load intensities and a probe coordinate \(x_1\) are exposed as inputs, so you can read \(M(x_1)\) and \(V(x_1)\) at any cross-section without touching the formulas.
Concentrated Force¶
Concentrated force \(F\) applied at the free end, giving the linear bending diagram \(M(x) = -F (l - x)\) and constant shear \(V = F\).
'<div style="max-width:180mm;">
'<img style="width:165pt;" alt="cantilever-beam-concentrated-force.png" class="side" src="../../Images/mechanics/beams/cantilever-beam-concentrated-force.png">
'<p><b>Input data</b></p>
'Beam Length -'l = ?'m
'Load -'F = ?'kN
#post
'<p><b>Internal forces</b></p>
'Bending -'M = F*l'kN·m
'Shear -'V = F'kN
'<p><b>Diagrams</b></p>
#hide
PlotWidth = 600
PlotHeight = 150
#show
'Calculate internal forces at'x_1 = ?'m
#pre
#post
'Bending
M(x) = -F*(l - x)
$Plot{-M(x) @ x = 0 : l}
M(x_1)'kN·m
'Shear
V(x) = F
$Plot{V(x) @ x = 0 : l}
V(x_1)'kN
#show
'</div> 5 10 1
Input data
Beam Length - l = 5 m
Load - F = 10 kN
Internal forces
Bending - M = F · l = 10 · 5 = 50 kN·m
Shear - V = F = 10 kN
Diagrams
Calculate internal forces at x1 = 1 m
Bending
M ( x ) = -F · ( l − x )
M ( x1 ) = M ( 1 ) = -40 kN·m
Shear
V ( x ) = F
V ( x1 ) = V ( 1 ) = 10 kN
Concentrated Moment¶
Concentrated moment \(M_0\) applied at the free end, producing a constant bending moment along the beam and zero shear.
'<div style="max-width:180mm;">
'<img style="width:165pt;" alt="cantilever-beam-concentrated-moment.png" class="side" src="../../Images/mechanics/beams/cantilever-beam-concentrated-moment.png">
'<p><b>Input data</b></p>
'Beam Length -'l = ?'m
'Load -'M = ?'kN·m
#post
'<p><b>Internal forces</b></p>
'Bending -'M = M'kN·m
'Shear -'V = 0'kN
'<p><b>Diagrams</b></p>
#hide
PlotWidth = 600
PlotHeight = 150
#show
'Calculate internal forces at'x_1 = ?'m
#pre
#post
'Bending
M(x) = -M
$Plot{-M(x) @ x = 0 : l}
M(x_1)'kN·m
'Shear
V(x) = 0
$Plot{V(x) @ x = 0 : l}
V(x_1)'kN
#show
'</div> 5 10 1
Input data
Beam Length - l = 5 m
Load - M = 10 kN·m
Internal forces
Bending - M = M = 10 kN·m
Shear - V = 0 kN
Diagrams
Calculate internal forces at x1 = 1 m
Bending
M ( x ) = -M
M ( x1 ) = M ( 1 ) = -10 kN·m
Shear
V ( x ) = 0
V ( x1 ) = V ( 1 ) = 0 kN
Linearly Distributed Load¶
Distributed load varying linearly between intensities \(q_1\) at the support and \(q_2\) at the tip, giving a cubic bending diagram and quadratic shear.
'<div style="max-width:180mm;">
'<img style="width:165pt;" alt="cantilever-beam-distributed-load-linear.png" class="side" src="../../Images/mechanics/beams/cantilever-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
'<p><b>Internal forces</b></p>
'Bending -'M = l^2*(q_1 + 2*q_2)/6'kN·m
'Shear -'V = (q_1 + q_2)*l/2'kN
'<p><b>Diagrams</b></p>
#hide
PlotWidth = 600
PlotHeight = 150
#show
'Calculate internal forces at'x_1 = ?'m
#pre
#post
q(x) = q_1 + (q_2 - q_1)*x/l
'Bending
M(x) = -M + V*x - (2*q_1 + q(x))*x^2/6
$Plot{-M(x) @ x = 0 : l}
M(x_1)'kN·m
'Shear
V(x) = V - (q_1 + q(x))*x/2
$Plot{V(x) @ x = 0 : l}
V(x_1)'kN
#show
'</div> 5 10 5 1
Input data
Beam Length - l = 5 m
Load - q1 = 10 kN/m, q2 = 5 kN/m
Internal forces
Bending - M = l2 · ( q1 + 2 · q2 ) 6 = 52 · ( 10 + 2 · 5 ) 6 = 83.33 kN·m
Shear - V = ( q1 + q2 ) · l2 = ( 10 + 5 ) · 52 = 37.5 kN
Diagrams
Calculate internal forces at x1 = 1 m
q ( x ) = q1 + ( q2 − q1 ) · xl
Bending
M ( x ) = -M + V · x − ( 2 · q1 + q ( x ) ) · x26
M ( x1 ) = M ( 1 ) = -50.67 kN·m
Shear
V ( x ) = V − ( q1 + q ( x ) ) · x2
V ( x1 ) = V ( 1 ) = 28 kN
Partially Distributed Load¶
Uniform load \(q\) acting on a stretch of length \(b\) adjacent to the free end, with the unloaded portion \(a = l - b\) kept clear.
'<div style="max-width:180mm;">
'<img style="width:165pt;" alt="cantilever-beam-distributed-load-partial.png" class="side" src="../../Images/mechanics/beams/cantilever-beam-distributed-load-partial.png">
'<p><b>Input data</b></p>
'Beam Length -'l = ?'m
'Load -'q = ?'kN/m
'Distances -'b = ?'m,'a = l - b'm
#post
'<p><b>Internal forces</b></p>
'Bending -'M = q*b*(a + b/2)'kN·m
'Shear -'V = q*b'kN
'<p><b>Diagrams</b></p>
#hide
PlotWidth = 600
PlotHeight = 150
#show
'Calculate internal forces at'x_1 = ?'m
#pre
#post
'Bending
M(x) = -M + V*x - q*(x - a)^2/2*(x > a)
$Plot{-M(x) @ x = 0 : l}
M(x_1)'kN·m
'Shear
V(x) = V - q*(x - a)*(x > a)
$Plot{V(x) @ x = 0 : l}
V(x_1)'kN
#show
'</div> 5 5 3 1
Input data
Beam Length - l = 5 m
Load - q = 5 kN/m
Distances - b = 3 m, a = l − b = 5 − 3 = 2 m
Internal forces
Bending - M = q · b · (a + b2) = 5 · 3 · (2 + 32) = 52.5 kN·m
Shear - V = q · b = 5 · 3 = 15 kN
Diagrams
Calculate internal forces at x1 = 1 m
Bending
M ( x ) = -M + V · x − q · ( x − a ) 22 · ( x > a )
M ( x1 ) = M ( 1 ) = -37.5 kN·m
Shear
V ( x ) = V − q · ( x − a ) · ( x > a )
V ( x1 ) = V ( 1 ) = 15 kN
Uniformly Distributed Load¶
Full-span uniform load \(q\) giving the textbook support actions \(M = q l^2 / 2\) and \(V = q l\).
'<img style="width:165pt;" alt="cantilever-beam-distributed-load-uniform.png" class="side" src="../../Images/mechanics/beams/cantilever-beam-distributed-load-uniform.png">
'Beam Length - 'l = ? 'm
'Load- 'q = ? 'kN/m
'<p><b>Internal forces</b></p>
'Bending - 'M = q*l^2/2'kN·m
'Shear - 'V = q*l'kN
5 5
Beam Length - l = 5 m
Load- q = 5 kN/m
Internal forces
Bending - M = q · l22 = 5 · 522 = 62.5 kN·m
Shear - V = q · l = 5 · 5 = 25 kN
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