Question

Problem 3 (20 pts) Draw the shear and moment diagrams for the beam, and determine the shear and moment in the beam as functions of x for 0cxe4 ft, 4 ftcx<10 ft and 10 ft <x< 14 ft 250 lb 250 lb 150 lb/ft -4ft

Answer 1

Calculate the reactions at the supports of a beam

?Fx = 0:    HB = 0
?MA = 0:   The sum of the moments about a point A is zero: 
 P1*4 - q1*6*(6/2) + RB*6 - P2*10 = 0
?MB = 0:   The sum of the moments about a point B is zero: 
 P1*10 - RA*6 + q1*6*(6 - 6/2) - P2*4 = 0
2. Solve this system of equations: 
HB = 0 (lb)
Calculate reaction of pin support  about point B:
RB = ( - P1*4 + q1*6*(6/2) + P2*10) / 6 = ( - 250*4 + 150*6*(6/2) + 250*10) / 6 = 700.00 (lb)
Calculate reaction of roller support  about point A:
RA = ( P1*10 + q1*6*(6 - 6/2) - P2*4) / 6 = ( 250*10 + 150*6*(6 - 6/2) - 250*4) / 6 = 700.00 (lb)
3. The sum of the forces is zero: ?Fy = 0:    - P1 + RA - q1*6 + RB - P2 =  - 250 + 700.00 - 150*6 + 700.00 - 250 = 0

First span of the beam: 0 < x1 < 4

Determine the equations for the shear force (Q):
Q(x1) =  - P1
Q1(0) =  - 250 = -250 (lb)
Q1(4) =  - 250 = -250 (lb)
Determine the equations for the bending moment (M):
M(x1) =  - P1*(x1)
M1(0) =  - 250*(0) = 0 (lb-ft)
M1(4) =  - 250*(4) = -1000 (lb-ft)

Second span of the beam: 4 < x2 < 10

Determine the equations for the shear force (Q):
Q(x2) =  - P1 + RA - q1*(x2 - 4)
Q2(4) =  - 250 + 700 - 150*(4 - 4) = 450 (lb)
Q2(10) =  - 250 + 700 - 150*(10 - 4) = -450 (lb)

The value of Q on this span that crosses the horizontal axis. Intersection point:
x = 3
Determine the equations for the bending moment (M):
M(x2) =  - P1*(x2) + RA*(x2 - 4) - q1*(x2 - 4)2/2
M2(4) =  - 250*(4) + 700*(4 - 4) - 150*(4 - 4)2/2 = -1000 (lb-ft)
M2(10) =  - 250*(10) + 700*(10 - 4) - 150*(10 - 4)2/2 = -1000 (lb-ft)

Local extremum at the point x = 3:
M2(7) =  - 250*(7) + 700*(7 - 4) - 150*(7 - 4)2/2 = -325 (lb-ft)

Third span of the beam: 10 < x3 < 14

Determine the equations for the shear force (Q):
Q(x3) =  - P1 + RA - q1*(10 - 4) + RB
Q3(10) =  - 250 + 700 - 150*(10 - 4) + 700 = 250 (lb)
Q3(14) =  - 250 + 700 - 150*(10 - 4) + 700 = 250 (lb)
Determine the equations for the bending moment (M):
M(x3) =  - P1*(x3) + RA*(x3 - 4) - q1*(10 - 4)*[(x3 - 10) + (10 - 4)/2] + RB*(x3 - 10)
M3(10) =  - 250*(10) + 700*(10 - 4) - 150*6*(0 + 3) + 700*(10 - 10) = -1000 (lb-ft)
M3(14) =  - 250*(14) + 700*(14 - 4) - 150*6*(4 + 3) + 700*(14 - 10) = 0 (lb-ft)