Simply Supported Beams (Shear & Moment Diagrams)

Simply supported beams (also know as pinned-pinned or pinned-roller) are the most common beams for both school and on the Professional Engineers exam. It is the starting point and the bread and butter of structural analysis. These beams have no leftover moment at the supports (a pinned connection cannot transfer moment). Therefore, as you will see below, the resulting Moment diagrams will always go to zero at both ends.

The Breakdown
  1. Pinned Roller Shear Moment Diagrams
    1. Defining the Variables
    2. Uniformly Distributed Load
    3. Triangular Load (apex at the support)
    4. Triangular Load (apex at the center)
    5. Partial Uniformly Distributed Load (wild)
    6. Partial Uniformly Distributed Load (at the end)
    7. Concentrated Load (at the center)
    8. Concentrated Load (wild)
    9. Two equal concentrated Loads (third points)
    10. Two equal concentrated Loads (wild)
  2. Fixed Pinned Shear Moment Diagrams
  3. Fixed Fixed Shear Moment Diagrams
  4. Fixed Free Shear Moment Diagrams

Defining the Variables

The variables for the Shear and Moment are defined below. We have also provided common units they will be given in. where:

R = Reactions (lbs, kips, kg)
Vx = Shear value at a distance 'x' along the beam (lbs, kips, kg)
Mx = Moment value at a distance 'x' along the beam (lb-ft, kip-ft, kip-in, kg-m)
Δx = Deflection value at a distance 'x' along the beam (in, ft, m)
Vmax = Maximum Shear Value
Mmax = Maximum Moment Value
Δmax = Maximum Deflection Value
P = The force of the concentrated load (kips, lbs, kg)
W = The total load acting on the beam (kips, lbs, kg)
w = The unit load acting on the beam (lbs/ft, kg/m)
l = the length of the beam (ft, m)
x = a distance along the beam from the designated end (ft, m)
E = the modulus of elasticity of the beam (ksi)
I = the Moment of Inertia of the beam (in4)

Note: Check your units! You don't want to use ft for length and then ksi for your modulus of elasticity (your answer will be off).

Uniformly Distributed Load

Shear Moment Diagram
Fig. 1: Simply Supported Beam with a Distributed Load
R = V = \frac{wl}{2}
V_x = w\left(\frac{l}{2}-x\right)
M_{max} = \frac{wl^2}{8}
M_x = \frac{wx}{2}\left(l-x\right)
{Δ_{max}} (@ center) = {{5wl^4}\over{384EI}}
{Δ_x} = {{wx}\over{24EI}}{\left(l^3-2lx^2+x^3\right)}

Triangular Load (apex at the support)

Shear Moment Diagram
Fig. 2: Simply Supported Beam with a Triangular Load (apex @ support)
R_1 = V_1 = \frac{W}{3}
R_2 = V_2 = V_{max} = \frac{2W}{3}
V_x = \frac{W}{3}-\frac{Wx^2}{l^2}
M_{max} (@ x =0.557*l) = \frac{2Wl}{9\sqrt{3}} = 0.128Wl
M_x = \frac{Wx}{3l^2}\left(l^2-x^2\right)
{Δ_{max}} (@ x=0.519*l) = 0.0130\frac{Wl^3}{EI}
{Δ_x} = \frac{Wx}{180EIl^2}\left(3x^4-10l^2x^2+7l^4\right)

Triangular Load (apex at the center)

Shear Moment Diagram
Fig. 3: Simply Supported Beam with a Triangular Load (apex @ center)
R = V = \frac{W}{2}
V_x (When x < l/2) = \frac{W}{2l^2}\left(l^2-4x^2\right)
M_{max} (@ center) = \frac{Wl}{6}
M_x = Wx \left(\frac{1}{2}-\frac{2x^2}{3l^2}\right)
{Δ_{max}} (@ center) = \frac{Wl^3}{60EI}
{Δ_x} (When x < l/2) = \frac{Wx}{480EIl^2} \left(5l^2-4x^2\right)^2

Partial Uniformly Distributed Load (wild)

Shear Moment Diagram
Fig. 4: Simply Supported Beam with a Partial Uniform Load (Wild)
R_1 = V_1 (max when a < c) = \frac{wb}{2l}(2c+b)
R_2 = V_2 (max when a > c) = \frac{wb}{2l}(2a+b)
V_x (when (a+b) > x >a) = R_1 - w(x-a)
M_{max} (at x = a + \frac{R_1}{w} ) = R_1\left(a+\frac{R_1}{2w}\right)
M_x (when x < a) = R_1x
M_x (when (a+b) > x > a) = R_1x - \frac{w}{2}(x-a)^2
M_x (when x > (a+b)) = R_2(l - x)

Partial Uniformly Distributed Load (at the end)

Shear Moment Diagram
Fig. 5: Simply Supported Beam with a Partial Distributed Load (at the end)
R_1 = V_1 = V_{max} = \frac{wa}{2l}(2l-a)
R_2 = V_2 = \frac{wa^2}{2l}
V_x (When x < a) = R_1 - wx
V_x (When x > a) = V_2
M_{max} (at x = \frac{R_1}{w} ) = \frac{R_1^2}{2w}
M_x (when x < a) = R_1x - \frac{wx^2}{2}
M_x (when x a a) = R_2(l-x)
Δ_{x} (when x<a) = \frac{wx}{24EIl}(a^2(2l-a)^2-2ax^2(2l-a)+lx^3)
Δ_{x} (when x><a) = \frac{wa^2(l-x)}{24EIl}(4xl-2x^2-a^2)

Concentrated Load (at the center)

Shear Moment Diagram
Fig. 6: Simply Supported Beam with a Concentrated Load @ Center
R = V = \frac{P}{2}
V_x = V
M_{max} = \frac{PL}{4}
M_x (when x < *l) = \frac{Px}{2}
Δ_{max} (@ pt of load) = {{Pl^3}\over{48EI}}
Δ_x (when x < *l) = {{Px^3}\over{48EI}}(3l^2-4x^2)

Concentrated Load (wild)

Shear Moment Diagram
Fig. 7: Simply Supported Beam with a Wild Concentrated Load
R_1 = V_1 = (V_{max} \hbox{when a < b)} = \frac{Pb}{l}
R_2 = V_2 = (V_{max} \hbox{when a > b)} = \frac{Pa}{l}
M_{max} (at point of load) = \frac{Pab}{l}
M_x (when x < a) = \frac{Pbx}{l}
Δ_{max} ( x = \sqrt{\frac{a(a+2b)}{3}} , when a > b) = \frac{Pab(a+2b)\sqrt{3a(a+2b)}}{27EIl}
Δ_a (at point of load) = \frac{Pa^2b^2}{3EIl}
Δ_x (when x < a) = \frac{Pbx}{6EIl}(l^2-b^2-x^2)

Two equal concentrated Loads (third points)

Shear Moment Diagram
Fig. 8: Simply Supported Beam with two Concentrated Loads @ Third Points
R = V = P
Mmax = Pa
M_x (when x < a) = Px
Δ_{max} (@ center) = \frac{Pa}{24EI}(3l^2 - 4a^2)
Δ_{max} (when a = l/3) = \frac{Pl^3}{28EI}
Δ_x (when x < a) = \frac{Px}{6EI}(3la-3a^2-x^2)
Δ_x (when (l-a) > x > a) = \frac{Pa}{6EI}(3lx-3x^2-a^2)

Two equal concentrated Loads (wild)

Shear Moment Diagram
Fig. 9: Simply Supported Beam with two wild Concentrated Loads
R1 = V1 (= Vmax when a < b) = \frac{P}{l}(l-a+b)
R2 = V2 (= Vmax when a > b) = \frac{P}{l}(l-b+a)
Vx (when a <x < (l-b)) = \frac{P}{l}(b-a)
M1 (= Mmax when a > b) = R_1a
M2 (= Mmax when a < b) = R_2b
Mx (when x < a) = R_1x
Mx (when a < x < (l-b)) = R_1x - P(x-a)

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