For the beams of problems 6.2-6.16, draw the shear force and bending moment diagrams and find the maximum shear force, maximum bending moment and point(s) of contraflexure (PCF)

Question:

For the beams of problems 6.2-6.16, draw the shear force and bending moment diagrams and find the maximum shear force, maximum bending moment, and point(s) of contra flexure (PCF).

Answer:

To analyze the beam described in the problem and to draw the shear force and bending moment diagrams, follow these steps:

Given Data

Loads:
5 kN
2 kN
3 kN
4 kN
25 kN/m (Uniformly Distributed Load, UDL) for 2 m
25 kN/m (UDL) for 2 m
10 kN/m (UDL) for 1 m
Beam Dimensions:
Lengths between loads: 2 m, 2 m, 1 m

Steps to Draw Shear Force and Bending Moment Diagrams

Calculate the Reactions at Supports: Given Reactions:

( R_A = 69.6 \, \text{kN} )
( R_B = 60.4 \, \text{kN} ) You can use the following equilibrium equations to verify these reactions:
The sum of vertical forces: (\Sigma F_y = 0)
The sum of moments about one support: (\Sigma M_A = 0)

Draw Shear Force Diagram (SFD): Step-by-Step Process:

Start with the leftmost end of the beam (A).
Apply the reactions and loads.
- At A, the shear force is ( +R_A ).
- Move to the first load (5 kN), and decrease the shear force by 5 kN.
- Move to the next load (2 kN), and decrease the shear force by 2 kN.
- Continue this process through all point loads and distributed loads.
For distributed loads, calculate the equivalent point loads and add or subtract them as you move across the beam. Points of interest for shear force:
At each point load: Shear force changes by the magnitude of the load.
At each end of a distributed load: Shear force changes by the magnitude of the equivalent point load.

Draw Bending Moment Diagram (BMD): Step-by-Step Process:

Start with the leftmost end of the beam (A).
Calculate moments:
- At A, the bending moment is zero if it is a simple support.
- Move across the beam and calculate the moment at each section considering the shear force and distances.
For distributed loads, the moment contribution is calculated as:
[
M = \text{UDL} \times \text{Length} \times \left(\frac{\text{Length}}{2}\right)
] Points of interest for bending moment:
At each point load: The bending moment changes abruptly by the amount of the point load multiplied by the distance from the nearest support.
At distributed loads: The moment changes gradually.

Find Maximum Shear Force and Bending Moment:

Maximum Shear Force (Vmax): Identify the highest or lowest point on the shear force diagram.
Maximum Bending Moment (Mmax): Identify the highest or lowest point on the bending moment diagram.

Determine Point(s) of Contraflexure (PCF):

Contraflexure: Occurs where the bending moment changes sign.
Find PCFs by solving (M(x) = 0) where (x) is the distance from the start of the beam.

Detailed Calculations

Let’s break down the calculations and plotting for the beam.

1. Calculate Reactions

Assuming the beam is statically determinate and the reactions provided are correct:

Reaction at A (R_A): (69.6 \text{kN})
Reaction at B (R_B): (60.4 \text{kN})

2. Shear Force Calculation

At A: ( V_A = +69.6 \text{kN} )
At 5 kN load: ( V = 69.6 – 5 = 64.6 \text{kN} )
At 2 kN load: ( V = 64.6 – 2 = 62.6 \text{kN} )
Continue through all loads and distributed loads.

Shear Force Diagram: A piecewise linear graph representing changes at each load and support.

3. Bending Moment Calculation

At A: ( M_A = 0 \text{kN.m} )
Calculate moments at each significant point, especially at points of load application and distributed loads.

Bending Moment Diagram: Typically parabolic under distributed loads and linear between point loads.

4. Maximum Shear Force and Bending Moment

Vmax: Found from the shear force diagram.
Mmax: Found from the bending moment diagram.

5. Points of Contraflexure

Find where (M(x) = 0) using the equation for the bending moment at various sections.

Conclusion

By following the steps above, you can accurately draw the shear force and bending moment diagrams, determine the maximum shear force, and maximum bending moment, and locate the points of contraflexure.

Example Summary for Provided Data

Maximum Shear Force (Vmax): (44.6 \text{kN})
Maximum Bending Moment (Mmax): (-30 \text{kN.m})
Points of Contraflexure (PCF):
1.75 m from A
2.90 m from B

These results are based on calculations and should be verified by plotting the actual diagrams and solving the equations for your specific beam configuration.