Effect of Compressive Axial Load on Deflection of the Beam Lab Report

Aim 

To determine the effect of compressive axial load due to transverse load on deflection of the beam 

Objection

1. Experimentally determine the deflection in beam with vertical end
2. Experimentally determine the deflection in beam with oblique end
3. Theoretically determine the deflection in beam with vertical end
4. Theoretically determine the deflection in beam with oblique end

Introduction

Beam
A structural member who is designed to resist the bending stresses produce in it when an external load is applied on it.

Simply supported beam
Types of beam which has support available at its both ends is called simply support beam

Overhanging beam
Types of beam which has support available at its both ends but has some of its area extended over one of its ends is called overhanging beam

Cantilever beam
Type of beam which has support available only at one of its ends and other ends is free is called cantilever beam

I beam
Type of beam which has cross section shape like English alpha bet I is called I beam

Procedure

Vertical end condition
1. first step in this experiment is to fix the vertical end of the apparatus with the beam
2. After that check the apparatus for any damages or errors present in it. if there is any of them make sure they are removed before experiment starts
3. Third step is attached hanger to the beam at its central point. This is important to attach the hanger in the middle as the apparatus has deflection measuring arrangement only at that point
4. After that add weight of 5 N and check the reading of the beam deflection
5. Repeat the above step for different weight by adding 10 N each time and check the deflection reading for each of them
6. When loaded up to 100 N, start removing weight and check deflection for that point too.

Oblique end condition
1. first step in this experiment is to fix the oblique end of the apparatus with opposite end of the beam
2. After that check the apparatus for any damages or errors produce due to the vertical end condition test. if there is any of them make sure they are removed before oblique end condition experiment starts
3. Third step is attached hanger to the beam at its central point. This is important to attach the hanger in the same spot as done for the vertical end condition test in order to ckect the true effect of oblique end condition
4. After that add weight of 5 N and check the reading of the beam deflection
5. Repeat the above step for different weight by adding 10 N each time and check the deflection reading for each of them.
6. When loaded up to 100 N, start removing weight and check deflection for that point too.

Readings

Force on beam	Loaded deflection mm oblique	Unloaded deflection  mm oblique	Loaded deflection mm vertical	Unloaded deflection  mm vertical	Average deflection oblique	Average deflection mm vertical
5	1	1.5	0	1	1.25	0.5
15	4	3.5	3	4	3.75	3.5
25	6	5.5	5	5	5.75	5
35	8	7.5	7	7	7.75	7
45	11	10	10	10	10.5	10
55	13	12.5	12	12	12.75	12
65	16	15	14	15	15.5	14.5
75	18	17	17	17	17.5	17
85	21	20.5	20	20	20.75	20
95	23.5	23.5	23	23	23.5	23
105	27	26	26	26	26.5	26

 Calculation

Support Reactions

sum of forces in x direction is equa to zero

A-F+B=0

A=F-B

sum of moment is equal to zero

A*0.8=F*0.4

B=(15*0.4)/0.8=7.5 N

A=F-B

A=15-7.5=7.5 N

Shear force

Shear force Before B

Shear force  = A=7.5 N

Shear force After b

SShear force  = A– F=7.5-15= -7.5 N

Bending moment

At A

Bending Moment = A*L =7.5*0 = 0 Nm

At B

Bending Moment = F*L =1 5*0.4 =6 Nm

At C

Bending Moment = B*L =75*0 = 0 Nm

Beam Deflection

moment of inertia=1/12 a*b^3

moment of inertia=1/12*50*〖125〗^3

moment of inertia=8*〖10〗^(-9)  m^4

modulus of elasticity=-(Force*〖Length〗^3)/(48*deflection*moment of inertia)

modulus of elasticity=-(15*〖0.8〗^3)/(48*0.003*8*〖10〗^(-9) )=6666666667 Pa=6.67 GPa

The actual value of modulus of elasticity for aluminium is 68 GPa which is far too much as compared to the experimental value of modulus of elasticity of aluminium which is only 6.67 GPa. The difference between values of modulus of elasticity is may be due to faults in apparatus or due to the errors made by operator during experimentation and observation.

Compressive force 

Compressive load when load is 5 N

C S =(Moment*b/2)/(moment of inertia)

C S =(6*0.0125/2)/(8*〖10〗^(-9) )=4687500 Pa

C S=Force/Area

4687500=Force/(5*125)

F=2929.68N

Moment equation

M=nornmal load bending+axial compressive load bending

Moment=F*L/2+Fc*deflection

M/EI=(〖 d〗^2 y)/(dx^2 )  

EI (〖 d〗^2 y)/(dx^2 )  =F*L+Fc*deflection

Integrating

dy/dx  =1/EI(F*L^2/2+Fc*deflection*L+C_1

Integrating

y =1/EI(F*L^3/6+Fc*deflection*L^2/2+C_1*L+C_2

Put x=0 y= 0

C_2=0

Put x = 0.4 y = 0

0=15*〖0.4〗^3/6+〖0.4*C〗_1

C1 =0.4

Deflection at 0.4 meter due to load of 15 N

70*〖10〗^9*8*〖10〗^(-9)*y =15*〖0.4〗^3/6+0.14*0.4+2929*y 〖0.4〗^2/2

560y-234y =0.216

y =0.216/326

y=6.63*〖10〗^(-4)  m

y =0.66 mm 

Discussion

The graph of deflection made against load is shown below and according to that graph the deflection of oblique end condition is much more than that of vertical end condition. According to graph the initial loading which is from 0 to 15 N does not show much difference in deflection but as the graph moves from 15 to 100 N load the deflection of oblique end condition increase much more rapidly than that of the vertical end condition. So from this it can be concluded that the axial compressive load due to oblique end condition has considerable effect on deflection of beam.