MOD005790 Research Project: Beam Analysis and MATLAB Solution

This document presents a solution to a project brief requiring MATLAB code for beam analysis. The solution includes the calculation of reaction forces, shear force diagrams, bending moment diagrams, and the determination of maximum shear force and bending moment magnitudes along with their locations. The document provides the MATLAB code used for the analysis and presents the resulting diagrams and calculations. This project showcases the application of computational methods in solving structural engineering problems. Desklib offers a wide array of similar solved assignments and past papers for students seeking assistance with their studies.

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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