Question

Draw the shear and moment diagrams for the beam. 

Answer 1

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 = 0
ΣFy = 0:    RA - (U1right *4)/2 - (U2left *4)/2 = 0;
ΣMA = 0:    MA - (U1right *4/2) * (4 - (1/3)*4) - (U2left *4/2) * (8 - (2/3)*4) = 0;
3. Solve this system of equations: 
HA = 0 (kN)
 RA =  (U1right *4)/2 + (U2left *4)/2 =  (80*4)/2 + (80*4)/2 = 320.00 (kN)
 MA =  (U1right *4/2) * (4 - (1/3)*4) + (U2left *4/2) * (8 - (2/3)*4) =  = 1280.00 (kN*m)
4. Verification equation of equilibrium about the free end of the beam: 
 - 8*RA + MA + (U1right *4/2) * (8 - 4 + (1/3)*4) + (U2left *4/2) * (8 - 8 + (2/3)*4) =  - 8*320.00 + 1280.00 + (U1right *4/2) * (8 - 4 + (1/3)*4) + (U2left *4/2) * (8 - 8 + (2/3)*4) = 0

Draw diagrams for the beam

Consider first span of the beam 0 ≤ x₁ < 4

Determine the equations for the shear force (Q):
Q(x1) =  + RA - (U1right *(x - 0)/4*(x - 0))/2
The values of Q at the edges of the span:
Q1(0) =  + 320 - (80*(0 - 0)/4*(0 - 0))/2 = 320 (kN)
Q1(4) =  + 320 - (80*(4 - 0)/4*(4 - 0))/2 = 160 (kN)
Determine the equations for the bending moment (M):
M(x1) =  + RA*(x1) - MA + (U1right *(x - 0)/4*(x - 0))/2*(x - 0)*(1/3)
The values of M at the edges of the span:
M1(0) =  + 320*(0) - 1280.00 - (80*(0 - 0)/4*(0 - 0))/2*(0 - 0)*(1/3) = -1280 (kN*m)
M1(4) =  + 320*(4) - 1280.00 + (80*(4 - 0)/4*(4 - 0))/2*(4 - 0)*(1/3) = -213.33 (kN*m)

Consider second span of the beam 4 ≤ x₂ < 8

Determine the equations for the shear force (Q):
Q(x2) =  + RA - (U1right *4)/2 - ([(U2left  - U2left *(8 - x)/4)*(x - 4)]/2 + U2left *(8 - x)/4*(x - 4))
The values of Q at the edges of the span:
Q2(4) =  + 320 - (80*(4 - 0)/4*(4 - 0))/2 - ([(80 - 80*(8 - 4)/4)*(4 - 4)]/2 + 80*(8 - 4)/4*(4 - 4)) = 160 (kN)
Q2(8) =  + 320 - (80*4)/2 - ([(80 - 80*(8 - 8)/4)*(8 - 4)]/2 + 80*(8 - 8)/4*(8 - 4)) = 0 (kN)
Determine the equations for the bending moment (M):
M(x2) =  + RA*(x2) - MA + (U1right *4)/2*(x - 4 + (1/3)*4) + ([(U2left  - U2left *(8 - x)/4)*(x - 4)]/2*(x - 4)*(2/3) + U2left *(8 - x)/4*(x - 4)*(x - 4)*(1/2))
The values of M at the edges of the span:
M2(4) =  + 320*(4) - 1280.00 + (80*(4 - 0)/4*(4 - 0))/2*(4 - 0)*(1/3) - ([(80 - 80*(8 - 4)/4)*(4 - 4)]/2*(4 - 4)*(2/3) + 80*(8 - 4)/4*(4 - 4)*(4 - 4)*(1/2)) = -213.33 (kN*m)
M2(8) =  + 320*(8) - 1280.00 + (80*4)/2*(8 - 4 + (1/3)*4) + ([(80 - 80*(8 - 8)/4)*(8 - 4)]/2*(8 - 4)*(2/3) + 80*(8 - 8)/4*(8 - 4)*(8 - 4)*(1/2)) = 0 (kN*m)