Fixed-Pinned Beams (Shear & Moment Diagrams)

Fixed-Pinned beams are common around the edges of a building. One side will retain no moment, and the other will be able to carry a moment force. Since a fixed connection is stronger than a pinned connection a majority of the force will attempt to travel in the direction of the fixed connection (this connection is stiffer) as is evident by the shear diagrams.

The Breakdown
  1. Pinned Roller Shear Moment Diagrams
  2. Fixed Pinned Shear Moment Diagrams
    1. Defining the Variables
    2. Uniformly Distributed Load
    3. Concentrated Load (at the center)
    4. Concentrated Load (wild)
  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: Fixed-Pinned Beam with a Distributed Load
R1 = V1 = \frac{3wl}{8}
R2 = V2 = Vmax = \frac{5wl}{8}
Vx = R_1 - wx
Mmax = \frac{wl^2}{8}
M1 (at x = 0.375*l) = \frac{9}{128}wl^2
Mx = R_1x - \frac{wx^2}{2}
Δmax (@ x = 0.4215*l) = {{wl^4}\over{185EI}}
Δx = {{wx}\over{48EI}}(l^3-3lx^2+2x^3)

Concentrated Load (at the center)

Shear Moment Diagram
Fig. 2: Fixed-Pinned Beam with a Concentrated Load @ Center
R1 = V1 = \frac{5P}{16}
R2 = V2 = Vmax = \frac{11P}{16}
Mmax (@ fixed end) = \frac{3Pl}{16}
M1 (@ pt of load) = \frac{5Pl}{32}
Mx (when x < 0.5*l) = \frac{5Px}{16}
Mx (when x > 0.5*l) = P\left(\frac{l}{2} - \frac{11x}{16}\right)
Δmax (at x = 0.4472*l) = \frac{Pl^3}{48EI\sqrt{5}}
Δx (at pt. of load) = \frac{7Pl^3}{768EI}
Δx (when x < 0.5*l)) = \frac{Px}{96EI}(3l^2-5x^2)
Δx (when x > 0.5*l) = \frac{P}{96EI}(x-l)^2(11x-2l)

Concentrated Load (wild)

Shear Moment Diagram
Fig. 3: Fixed-Pinned Beam with a Wild Concentrated Load
R1 = V1 = \frac{Pb^2}{2l^3}(a+2l)
R2 = V2 = \frac{Pa}{2l^3}(3l^2-a^2)
M1 (@ pt. of load) = R_1a
M2 (@ fixed end) = \frac{Pab}{2l^2}(a+l)
Mx (when x < a) =R_1x
Mx (when x > a) = R_1x - P(x-a)
Δmax (when a < 0.414*l @ x = l\frac{l^2 + a^2}{3l^2-a^2} ) = \frac{Pa}{3EI} \frac{(l^2-a^2)^2}{(3l^2-a^2)^2}
Δmax (when a > 0.414*l @ x = l\sqrt{\frac{a}{2l+a}} ) = \frac{Pab^2}{6EI} \sqrt{\frac{a}{2l+a}}
Δa (@ pt. of load) = \frac{Pa^2b^3}{12EIl^3}(3l+a)
Δx (when x < a) = \frac{Pb^2x}{12EIl^3}(3al^2-2lx^2-ax^2)
Δx (when x > a) = \frac{Pa}{12EIl^3}(l-x)^2(3l^2x-a^2x-2a^2l)

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