A beam ABC with an overhang supports a uniform load of intensity q = 6...
Problem Statement: Calculate the shear force V and bending moment M at a cross section D located 5 m from the left hand support of a beam ABC with an overhang that supports a uniform load of intensity q = 6 kN/m and a concentrated load P=28 kN.
Step-by-Step Solution:
Step 1: Draw the Free-Body Diagram (FBD) of the beam
The FBD of the beam is shown below:
![FBD of the beam](https://imgur.com/7MlNvW9.png)
Step 2: Calculate the reactions at the supports
The reactions at the supports can be calculated by taking moments about any point. Taking moments about support A, we get:
ΣM
A = 0
RA x 8 - 6 x 4 x (4/2) - 28 x 3 = 0
RA = 10 kN
Taking moments about support B, we get:
ΣM
B = 0
RA x 8 + 6 x 4 x (4/2) - RB x 3 = 0
RB = 38 kN
Step 3: Draw the shear force and bending moment diagrams
The shear force and bending moment diagrams can be drawn by considering the different segments of the beam.
Segment AB
For segment AB, the shear force and bending moment can be calculated as follows:
V
AB = RA = 10 kN
M
AB = 0
Segment BC
For segment BC, the shear force and bending moment can be calculated as follows:
V
BC = RA - q x (x - 8) = 10 - 6(x - 8) kN
M
BC = RA(x) - q(x)(x/2) - P = 10x - 3x
2 - 28 kNm
Segment CD
For segment CD, the shear force and bending moment can be calculated as follows:
V
CD = RA - q x (x - 8) - P = 10 - 6(x - 8) - 28 kN
V
CD = -6x + 50 kN
M
CD = RA(x) - q(x)(x/2) - P(x - 8) = 10x - 3x
2 - 28x + 224 kNm
Step 4: Calculate the shear force and bending moment at section D
The shear force and bending moment at section D can be calculated by substituting x = 5 m in the equations for segment CD:
V
D = -6(5) + 50 kN = 20 kN
M
D = 10(5) - 3(5)
2 - 28(5) +