Question

4. Use the Beam Loading Diagrams. Draw the Shear Diagram and Bending Moment Diagram. Indicate the factored Reactions, the maximum Mf (kNm) and the maximum Vf (kN) wf-15kN/m 10m Rf= Rf- /2 Vf 13 Shear Diagram (kN) Mf 13 Bending Moment Diagram (kNm)

Answer 1

Calculation of the reactions at the supports of a beam

1. 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.
ΣFx = 0:    HA = 0
ΣMA = 0:   The sum of the moments about the pin support at the point A: 
 - (U1right *10/2) * (10 - (1/3)*10) + RB*10 = 0
ΣMB = 0:   The sum of the moments about the roller support at the point B:
 - RA*10 + (U1right *10/2) * (10 - 10 + (1/3)*10) = 0
2. Calculate reaction of roller support at the point B:
RB = ( (U1right *10/2) * (10 - (1/3)*10)) / 10 = ( 75 * 6.67) / 10 = 50.00 (kN)
3. Calculate reaction of pin support at the point A:
RA = ( (U1right *10/2) * (10 - 10 + (1/3)*10)) / 10 = ( 75 * 3.33) / 10 = 25.00 (kN)
4. Solve this system of equations:
HA = 0 (kN)
5. The sum of the forces about the Oy axis is zero:
ΣFy = 0:    RA - (U1right *10)/2 + RB =  25.00*1 - (15*10)/2 + 50.00*1 = 0

Draw diagrams for the beam

Consider first span of the beam 0 ≤ x₁ < 10

Determine the equations for the axial force (N):
N(x1) = HA
The values of N at the edges of the span:
N1(0) = 0 = 0 (kN)
N1(10) = 0 = 0 (kN)
Determine the equations for the shear force (Q):
Q(x1) =  + RA - (U1right *(x - 0)/10*(x - 0))/2
The values of Q at the edges of the span:
Q1(0) =  + 25.00 - (15*(0 - 0)/10*(0 - 0))/2 = 25 (kN)
Q1(10) =  + 25.00 - (15*(10 - 0)/10*(10 - 0))/2 = -50 (kN)

The value of Q on this span that crosses the horizontal axis. Intersection point:
x = 5.77
Determine the equations for the bending moment (M):
M(x1) =  + RA*(x1) + (U1right *(x - 0)/10*(x - 0))/2*(x - 0)*(1/3)
The values of M at the edges of the span:
M1(0) =  + 25.00*(0) - (15*(0 - 0)/10*(0 - 0))/2*(0 - 0)*(1/3) = 0 (kN*m)
M1(10) =  + 25.00*(10) + (15*(10 - 0)/10*(10 - 0))/2*(10 - 0)*(1/3) = 0 (kN*m)

Local extremum at the point x = 5.77:
M1(5.77) =  + 25.00*(5.77) - (15*(5.7732558139535 - 0)/10*(5.77 - 0))/2*(5.77 - 0)*(1/3) = 96.23 (kN*m)

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