Question

Replace the loading on the beam by an equivalent resultant force and specify its location measured from point A. (b) Calculate the value of the reactions at support A and support B. (c) calculate the shear, normal force, and bending moment at a point 1.5 m to the right of A. (d) Same as (c) but at a point 0.5 m to the right of B. The system is in equilibrium. 

Answer 1

Calculate 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: 
 - ((U1left  - U1right )*3)/2*(3 - (2/3)*3) - U1right *3*(3 - (1/2)*3) + RB*3 - P1*4 = 0
ΣMB = 0:   The sum of the moments about the roller support at the point B:
 - RA*3 - ((U1left  - U1right )*3)/2*(3 - (0 + 1/3*3)) - U1right *3*(3 - (0 + (1/2)*3)) - P1*1 = 0
2. Calculate reaction of roller support at the point B:
RB = ( ((U1left  - U1right )*3)/2*(3 - (2/3)*3) + U1right *3*(3 - (1/2)*3) + P1*4) / 3 = ( 3*1 + 1.5*1.50 + 1.5*4) / 3 = 3.75 (kN)
3. Calculate reaction of pin support at the point A:
RA = ( ((U1left  - U1right )*3)/2*(3 - (0 + 1/3*3)) + U1right *3*(3 - (0 + (1/2)*3)) - P1*1) / 3 = ( 3*2 + 1.5*1.50 - 1.5*1) / 3 = 2.25 (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 - ((U1left  - U1right )*3)/2 - U1right *3 + RB - P1 =  2.25*1 - ((2.5 - 0.5)*3)/2 - 0.5*3 + 3.75*1 - 1.5 = 0

Draw diagrams for the beam

Consider first span of the beam 0 ≤ x₁ < 3

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(3) = 0 = 0 (kN)
N1(1.5)= 0 =0 (kN)
Determine the equations for the shear force (Q):
Q(x1) =  + RA - ([(U1left  - (U1left  - U1right )*(3 - x)/3 + U1right )*(x - 0)]/2 + (U1left  - U1right )*(3 - x)/3 + U1right *(x - 0))
The values of Q at the edges of the span:
Q1(0) =  + 2.25 - ([(2.5 - (2.5 - 0.5)*(3 - 0)/3 + 0.5)*(0 - 0)]/2 + (2.5 - 0.5)*(3 - 0)/3 + 0.5*(0 - 0)) = 2.25 (kN)
Q1(3) =  + 2.25 - ([(2.5 - (2.5 - 0.5)*(3 - 3)/3 + 0.5)*(3 - 0)]/2 + (2.5 - 0.5)*(3 - 3)/3 + 0.5*(3 - 0)) = -2.25 (kN)
Q1(1.5) = + 2.25 - ([(2.5 - (2.5 - 0.5)*(3 - 1.5)/3 + 0.5)*(1.5 - 0)]/2 + (2.5 - 0.5)*(3 - 1.5)/3 + 0.5*(1.5 - 0)) = -0.25 (kN) [

The value of Q on this span that crosses the horizontal axis. Intersection point:
x = 1.05
Determine the equations for the bending moment (M):
M(x1) =  + RA*(x1) - ([(U1left  - (U1left  - U1right )*(3 - x)/3 + U1right )*(x - 0)]/2*(x - 0)*(1/3) + (U1left  - U1right )*(3 - x)/3 + U1right *(x - 0)*(x - 0)*(1/2))
The values of M at the edges of the span:
M1(0) =  + 2.25*(0) - ([(2.5 - (2.5 - 0.5)*(3 - 0)/3 + 0.5)*(0 - 0)]/2*(0 - 0)*(2/3) + (2.5 - 0.5)*(3 - 0)/3 + 0.5*(0 - 0)*(0 - 0)*(1/2)) = 0 (kN*m)
M1(3) =  + 2.25*(3) - ([(2.5 - (2.5 - 0.5)*(3 - 3)/3 + 0.5)*(3 - 0)]/2*(3 - 0)*(2/3) + (2.5 - 0.5)*(3 - 3)/3 + 0.5*(3 - 0)*(3 - 0)*(1/2)) = -1.50 (kN*m)
M1(1.5) = + 2.25*(1.5) - ([(2.5 - (2.5 - 0.5)*(3 - 1.5)/3 + 0.5)*(1.5 - 0)]/2*(1.5 - 0)*(2/3) + (2.5 - 0.5)*(3 - 1.5)/3 + 0.5*(1.5 - 0)*(1.5 - 0)*(1/2)) = 0.8125 (kN*m)

Local extremum at the point x = 1.05:
M1(1.05) =  + 2.25*(1.05) - ([(2.5 - (2.5 - 0.5)*(3 - 1.05)/3 + 0.5)*(1.05 - 0)]/2*(1.05 - 0)*(2/3) + (2.5 - 0.5)*(3 - 1.05)/3 + 0.5*(1.05 - 0)*(1.05 - 0)*(1/2)) = 1.11 (kN*m)

Consider second span of the beam 3 ≤ x₂ < 4

Determine the equations for the axial force (N):
N(x2) = HA
The values of N at the edges of the span:
N2(3) = 0 = 0 (kN)
N2(4) = 0 = 0 (kN)
N2(3.5) = 0 = 0 (kN)
Determine the equations for the shear force (Q):
Q(x2) =  + RA - ((U1left  - U1right )*3/2 + U1right *3) + RB
The values of Q at the edges of the span:
Q2(3) =  + 2.25 - ([(2.5 - (2.5 - 0.5)*(3 - 3)/3 + 0.5)*(3 - 0)]/2 + (2.5 - 0.5)*(3 - 3)/3 + 0.5*(3 - 0)) + 3.75 = 1.50 (kN)
Q2(4) =  + 2.25 - ((2.5 - 0.5)*3/2 + 0.5*3) + 3.75 = 1.50 (kN)
Q2(3.5) = + 2.25 - ((2.5 - 0.5)*3/2 + 0.5*3) + 3.75 = 1.50 (kN)
Determine the equations for the bending moment (M):
M(x2) =  + RA*(x2) - ((U1left  - U1right )*3/2*(x - (0 + 3*1/3)) + U1right *3*(x - (0 + 3*(1/2)))) + RB*(x2 - 3)
The values of M at the edges of the span:
M2(3) =  + 2.25*(3) - ([(2.5 - (2.5 - 0.5)*(3 - 3)/3 + 0.5)*(3 - 0)]/2*(3 - 0)*(2/3) + (2.5 - 0.5)*(3 - 3)/3 + 0.5*(3 - 0)*(3 - 0)*(1/2)) + 3.75*(3 - 3) = -1.50 (kN*m)
M2(4) =  + 2.25*(4) - ((2.5 - 0.5)*3/2*(4 - (0 + 3*1/3)) + 0.5*3*(4 - (0 + 3*(1/2)))) + 3.75*(4 - 3) = 0 (kN*m)

M₂(3.5) = 0.75 (kN*m) it can be easily calculated from looking at the graph. as the function is linear.

so answer to all parts are given

(A).shown in picture.

(B).  R_B = ( ((U₁^left - U₁^right )*3)/2*(3 - (2/3)*3) + U₁^right *3*(3 - (1/2)*3) + P₁*4) / 3 = ( 3*1 + 1.5*1.50 + 1.5*4) / 3 = 3.75 (kN)

R_A = ( ((U₁^left - U₁^right )*3)/2*(3 - (0 + 1/3*3)) + U₁^right *3*(3 - (0 + (1/2)*3)) - P₁*1) / 3 = ( 3*2 + 1.5*1.50 - 1.5*1) / 3 = 2.25 (kN)

(C)  N₁(1.5)= 0 =0 (kN)

Q₁(1.5) = + 2.25 - ([(2.5 - (2.5 - 0.5)*(3 - 1.5)/3 + 0.5)*(1.5 - 0)]/2 + (2.5 - 0.5)*(3 - 1.5)/3 + 0.5*(1.5 - 0)) = -0.25 (kN)

M₁(1.5) = + 2.25*(1.5) - ([(2.5 - (2.5 - 0.5)*(3 - 1.5)/3 + 0.5)*(1.5 - 0)]/2*(1.5 - 0)*(2/3) + (2.5 - 0.5)*(3 - 1.5)/3 + 0.5*(1.5 - 0)*(1.5 - 0)*(1/2)) = 0.8125 (kN*m)

(D)  N₂(3.5) = 0 = 0 (kN)

Q₂(3.5) = + 2.25 - ((2.5 - 0.5)*3/2 + 0.5*3) + 3.75 = 1.50 (kN)​​​​​​​

M₂(3.5) = 0.75 (kN*m) it can be easily calculated from looking at the graph. as the function is linear.