Answer to Question #230782 in Civil and Environmental Engineering for Vasu

Question

Draw SFD and BMD for a simply supported beam AB of span 9 m. The loading data is
as follows
Full Span UDL = 25 kN/m
Point Load 2m from Support A = 15 kN.
Point Load 6m from Support A = 80 kN.

Answer 1

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 + P1*cos(30) = 0
ΣMA = 0:   The sum of the moments about a point A is zero: 
 - q1*4*(4/2) - P1*sin(30)*8 + RB*12 + M1 = 0
ΣMB = 0:   The sum of the moments about a point B is zero: 
 - RA*12 + q1*4*(12 - 4/2) + P1*sin(30)*4 + M1 = 0
2. Solve this system of equations: 
HA =  - P1*cos(30) =  - 20*0.8660 = -17.32 (kN)
Calculate reaction of roller support  about point B:
RB = ( q1*4*(4/2) + P1*sin(30)*8 - M1) / 12 = ( 6*4*(4/2) + 20*sin(30)*8 - 42) / 12 = 7.17 (kN)
Calculate reaction of pin support  about point A:
RA = ( q1*4*(12 - 4/2) + P1*sin(30)*4 + M1) / 12 = ( 6*4*(12 - 4/2) + 20*sin(30)*4 + 42) / 12 = 26.83 (kN)
3. The sum of the forces is zero: ΣFy = 0:    RA - q1*4 - P1*sin(30) + RB =  26.83 - 6*4 - 20*sin(30) + 7.17 = 0
Determine the equations for the axial force (N):
N(x1) = HA
N1(0) = 17.32 = 17.32 (kN)
N1(4) = 17.32 = 17.32 (kN)
Determine the equations for the shear force (Q):
Q(x1) =  + RA - q1*(x1 - 0)
Q1(0) =  + 26.83 - 6*(0 - 0) = 26.83 (kN)
Q1(4) =  + 26.83 - 6*(4 - 0) = 2.83 (kN)
Determine the equations for the bending moment (M):
M(x1) =  + RA*(x1) - q1*(x1)2/2
M1(0) =  + 26.83*(0) - 6*(0 - 0)2/2 = 0 (kN*m)
M1(4) =  + 26.83*(4) - 6*(4 - 0)2/2 = 59.33 (kN*m)
 Determine the equations for the axial force (N):
N(x2) = HA
N2(4) = 17.32 = 17.32 (kN)
N2(8) = 17.32 = 17.32 (kN)
Determine the equations for the shear force (Q):
Q(x2) =  + RA - q1*(4 - 0)
Q2(4) =  + 26.83 - 6*(4 - 0) = 2.83 (kN)
Q2(8) =  + 26.83 - 6*(4 - 0) = 2.83 (kN)
Determine the equations for the bending moment (M):
M(x2) =  + RA*(x2) - q1*(4 - 0)*[(x2 - 4) + (4 - 0)/2]
M2(4) =  + 26.83*(4) - 6*4*(0 + 2) = 59.33 (kN*m)
M2(8) =  + 26.83*(8) - 6*4*(4 + 2) = 70.67 (kN*m)