﻿ Determining the Shear Force and Bending Moment Equations of Cantilever Beam

## Example 3. Cantilever beam calculation carrying a uniformly distributed load and a concentrated load #### Calculate the reactions at the supports of a beam

`A beam is in equilibrium when it is stationary relative to an inertial reference frame. The following conditions are satisfied when a beam, acted upon by a system of forces and moments, is in equilibrium:1. The fixed support is located at point A (on the left). A fixed support will resist translation in all directions and rotation (moment) - HA, RA, MA.2. The sum of the forces and moment about any point is zero: ΣFx = 0, ΣFy = 0, ΣMA = 0.ΣFx = 0:    HA = 0ΣFy = 0:    RA - q1*1.8 - P1 = 0;ΣMA = 0:    MA - q1*1.8*(1.8/2) + M1 - 3*P1 = 0;3. Solve this system of equations: HA = 0 (kN) RA =  q1*1.8 + P1 =  2*1.8 + 7 = 10.60 (kN) MA =  q1*1.8*(1.8/2) - M1 + 3*P1 =  2*1.8*(1.8/2) - 19 + 3*7 = 5.24 (kN*m)4. Verification equation of equilibrium about the point B (on the right): - 3*RA + MA + q1*1.8*(1.2+1.8/2) + M1 + 0*P1 =  - 3*10.60 + 5.24 + 2*1.8*(1.2+1.8/2) + 19.00 + 0*7 = 0`

#### Draw diagrams for the beam

First span of the beam: 0 ≤ x1 < 1.8
`Determine the equations for the shear force (Q):Q(x1) =  + RA - q1*(x1 - 0)Q1(0) =  + 10.60 - 2*(0 - 0) = 10.60 (kN)Determine the equations for the bending moment (M):M(x1) =  + RA*(x1) - MA - q1*(x1)2/2M1(0) =  + 10.60*(0) - 5.24 - 2*(0 - 0)2/2 = -5.24 (kN*m)`
Second span of the beam: 1.8 ≤ x2 < 2.4
`Determine the equations for the shear force (Q):Q(x2) =  + RA - q1*(1.8 - 0)Q2(1.80) =  + 10.60 - 2*(1.8 - 0) = 7 (kN)Q2(2.40) =  + 10.60 - 2*(1.8 - 0) = 7 (kN)Determine the equations for the bending moment (M):M(x2) =  + RA*(x2) - MA - q1*(1.8 - 0)*[(x2 - 1.80) + (1.80 - 0)/2]M2(1.80) =  + 10.60*(1.80) - 5.24 - 2*1.8*(0 + 0.90) = 10.60 (kN*m)M2(2.40) =  + 10.60*(2.40) - 5.24 - 2*1.8*(0.60 + 0.90) = 14.80 (kN*m)`
Third span of the beam: 2.4 ≤ x3 < 3
`Determine the equations for the shear force (Q):Q(x3) =  + RA - q1*(1.8 - 0)Q3(2.40) =  + 10.60 - 2*(1.8 - 0) = 7 (kN)Q3(3) =  + 10.60 - 2*(1.8 - 0) = 7 (kN)Determine the equations for the bending moment (M):M(x3) =  + RA*(x3) - MA - q1*(1.8 - 0)*[(x3 - 1.80) + (1.80 - 0)/2] - M1M3(2.40) =  + 10.60*(2.40) - 5.24 - 2*1.8*(0.60 + 0.90) - 19 = -4.20 (kN*m)M3(3) =  + 10.60*(3) - 5.24 - 2*1.8*(1.20 + 0.90) - 19 = 0 (kN*m)`

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