Question

CM 2315 Introduction to Mechanics for Construction NAME: DAVID HowpeDID: 1001902305 Instructions o WRITE YOUR NAME IN UPPERCASE LETTERS ON ALL PAGES (1) Follow the instructor to solve this problem step by step. fi Submit your work to the instructor at the end of class. (iv) Use a scientific calculator for calculations/computation. PRACTICE PROBLEM 5 [100 points] Draw the Free Body diagram, the sear diagram and the moment diagram for the uniformly loaded beam in Figure 1.1. w-6 kips/ft 18 ft Figure 1.1 STEP 1 Draw Free Body Diagram (30 points) Figure 1.2 Free Body diagram

Answer 1

6 (ft) 18 (ft) - 1 sec. (X1) → 2 sec (X2) Oy O -48 192 Mx -108

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:    HB = 0
ΣMA = 0:   The sum of the moments about the roller support at the point A: 
 - q1*24*(-6 + 24/2) + RB*18 = 0
ΣMB = 0:   The sum of the moments about the pin support at the point B:
 q1*24*(24 - 24/2) - RA*18 = 0
2. Calculate reaction of pin support at the point B:
RB = ( q1*24*(-6 + 24/2)) / 18 = ( 6*24*(-6 + 24/2)) / 18 = 48.00 (kipf)
3. Calculate reaction of roller support at the point A:
RA = ( q1*24*(24 - 24/2)) / 18 = ( 6*24*(24 - 24/2)) / 18 = 96.00 (kipf)
4. Solve this system of equations:
HB = 0 (kipf)
5. The sum of the forces about the Oy axis is zero:
ΣFy = 0:    - q1*24 + RA + RB =  - 6*24 + 96.00*1 + 48.00*1 = 0

Draw diagrams for the beam

Consider first span of the beam 0 ≤ x₁ < 6

Determine the equations for the shear force (Q):
Q(x1) =  - q1*(x1 - 0)
The values of Q at the edges of the span:
Q1(0) =  - 6*(0 - 0) = 0 (kipf)
Q1(6) =  - 6*(6 - 0) = -36 (kipf)
Determine the equations for the bending moment (M):
M(x1) =  - q1*(x1)2/2
The values of M at the edges of the span:
M1(0) =  - 6*(0 - 0)2/2 = 0 (kipf*ft)
M1(6) =  - 6*(6 - 0)2/2 = -108 (kipf*ft)

Consider second span of the beam 6 ≤ x₂ < 24

Determine the equations for the shear force (Q):
Q(x2) =  - q1*(x2 - 0) + RA
The values of Q at the edges of the span:
Q2(6) =  - 6*(6 - 0) + 96 = 60 (kipf)
Q2(24) =  - 6*(24 - 0) + 96 = -48 (kipf)

The value of Q on this span that crosses the horizontal axis. Intersection point:
x = 10
Determine the equations for the bending moment (M):
M(x2) =  - q1*(x2)2/2 + RA*(x2 - 6)
The values of M at the edges of the span:
M2(6) =  - 6*(6 - 0)2/2 + 96*(6 - 6) = -108 (kipf*ft)
M2(24) =  - 6*(24 - 0)2/2 + 96*(24 - 6) = 0 (kipf*ft)

Local extremum at the point x = 10:
M2(16) =  - 6*(16 - 0)2/2 + 96*(16 - 6) = 192 (kipf*ft)