## 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

### 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).

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)

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)

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)

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)

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

### Concentrated Load (at the center)

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 < ½*l) = \frac{Px}{2} Δ_{max} (@ pt of load) = {{Pl^3}\over{48EI}} Δ_x (when x < ½*l) = {{Px^3}\over{48EI}}(3l^2-4x^2)

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)

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)

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 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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