MATLAB Code for Beam - Bending Moment and Shear Force Diagram

This document provides a MATLAB code for generating the Bending Moment and Shear Force Diagram for a beam. It also includes the calculation of the maximum shear force and bending moment magnitude, along with their respective locations on the beam.

Element 011 Project Brief
Service Team
Research & Computational
Methods
Project Brief
MOD005790

Element 011 Project Brief
Description
You need to write a MATLAB code for the beam presented in Figure 1. Table 1 outlines details of the
beam. Your code must generate the following FOUROUTPUTS:
1. Bending Moment Diagram
2. Shear Force Diagram
3. Maximum Shear Force magnitude and and its location along the beam
4. Maximum Bending Momenmt magnitude and its locationalong the beam
Figure 1
Table 1
P1 (kN) P2 (kN) Span
L(m)
α
Sum of your student number
digits
0.4 * (Sum of your
student number digits) 6 30o
Tips
 You need to follow all the instructions discussed in the class to write, improve and run your code.
 Presentation of the results is very important.

Element 010 Project Brief SEM 1 – 2018/19
Solution:
P1 (kN) P2 (kN) Span
L(m)
α
35 14 6 30o
Here,
P1 is 35 KN
P2 is 14 KN
L = 6 m
Now,
Resolving all the inclined loads:
Load P2 is having:
Horizontal component:P2h: cos 30 ×14=12.124 KN
Vertical Component P2v: sin 30 ( 14 ) =7 KN
Now,
we need to find the reaction forces: Ra and Rb
Take moment about A
Rb ( 6 )= ( P 1 ) 2+P 2 v ( 4 )
Rb= 35 ( 2 ) +(7 (4))
6 =16.333 KN
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A B

Element 010 Project Brief SEM 1 – 2018/19
Now,
Ra+Rb = Total Vetrical loadings
Ra = 42-16.333 = 25.667 KN
Ans:
Ra: 25.667 KN
Rb: 16.333 KN
Shear ForceDiagram:
Shear force can be calculated as:
S.F at A = +Ra = +25.667 KN
S.F at P1 = SF at A−35=25.667−35=−9.333 KN
S.F at P2: S . F at P 1−7 = −9.333−7=−16.333 KN
S.F at B : −16.333 KN
Fig1. Shows Shear Force Diagram
Magnitute of max shear force is 25.667 KN and it acts at the point A to P1. (Johnson, R.P., 2018.)
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Element 010 Project Brief SEM 1 – 2018/19
Bending Moment:
BM at A is MA = 0
BM at P1 is Mp1 = Ra x 2 = 51.334 KN.m
BM at P2 is Mp2 = Ra x 4 −P 1(2) = 25.667 ( 4 )− ( 35 ( 2 ) ) =32.668 KN . m
BM at B is MB = 0
Fig2. Shows Bending Moment diagram
Maximum Bending Moment Magnitude is 51.334 KN.m and it acts at the point P1. (Thomson,
W., 2018)
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Element 010 Project Brief SEM 1 – 2018/19
MATLAB Code:
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Element 010 Project Brief SEM 1 – 2018/19
References:
Johnson, R.P., (2018). Composite Structures of Steel and Concrete: beams, slabs, columns and
frames for buildings. John Wiley & Sons.
Thomson, W., (2018). Theory of vibration with applications. CrC Press.
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