Deflection Calculation of Beam

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Introduction

Beam is the basic support to any structure which includes building, bridges and vehicles like cars, boats and airplanes. Beam provides strength to structure by taking the entire load applied on structure. Load applied can be either due to the weight of the structure or the working load applied during operation/working of vehicle [1]. There are several types of beam available based on the types of the supports available to beam and shape of the beam. Based to type of support there are three types available which include fixed ends, pined ends and roller ends. Based on the types of shape there are three basic types which include square beam, rectangular beam and I beam [2].
Beam in the simplest form are called the simply supported beam which have two support present at both ends on beam. When load is applied on beam, it produces moment in mean which try to rotate beam around its supports[3]. Due to the reaction of beam shear force is produce in beam which is perpendicular to length of the beam. Efficiency of the beam depends on its ability to resist the bending moment and shear forces produce in it during working [3].
Performance of the beam depends on the three basic things, one its material, second its shape and third its length between end to end [2]. Material of the beam provides the modulus of elasticity, a material property which decides how much a beam will bend when a certain load is applied on it. Shape of the beam provides the moment of inertia of beam which shows the beam ability to resist change in its state/ shape of rest. Length of the beam will define the moment produce by the load applied. Longer the beam higher will be the moment produce and higher the chances of beam failure under small load [4].

Aims

Aim of this work is to analyse the working of a simply supported beam

Objective

To achieve the aim of this lab work, the following objectives of should be achieve in mention sequence

1.      To understand the basic working of  the simply supported beam
2.      To experimentally find the deflection of beam with one point load and different cross section
3.      To experimentally find the deflection of beam with two point loads and different cross section
4.      To calculate the deflection of beam with one point load and different cross section
5.      To calculate the deflection of beam with two point loads and different cross section
6.      Compare the both results and analyse the effect of different loads and cross sections


Experimental Procedure

In order to find the deflection of beam through experiment, the following procedure can be adopted
1.      The first step in this experiment is to setup the apparatus which includes, installing the beam, attaching the weight hanger and adjusting the deflection measuring instrument
2.      After that add weight on the weight hanger to apply external load on beam
3.      Note the deflection of beam using deflection measuring instrument. Take deflection reading at different point. For this experiment note deflection reading at every 100 mm distance starting from left end.
4.      Repeat the above procedure for different loads and different beams

 Theoretical work

1.      Find the reaction of supports of beam for a given load applied during the experiment
2.      Establish the moment equation of simply supported beam for each loading case
3.      Drive the equation of slope and deflection using Macaulay’s theory
4.      Find the deflection using the equation of deflection at the point where experimental deflection was noted.
5.      Compare both results by plotting a comparative graph.
6.      Discuss the results

Experimental Results

Shape and Size
Steel 1*3.2 mm thick *19mm wide (ref A-3/4'*1/8')
steel -745 mm long 1*6.4 mm thick * 13 mm wide (rif l-1/2'* 1/4')
Beam 
beam 1
beam 2
beam 2
beam 2
Material
steel
steel
steel
steel
position of left support
500mm
500mm
500mm
500mm
position of right support
1200mm
1200mm
1200mm
1200mm
load 1 (N)
10 N
10N
10N
10N
position of load 1
850mm
650mm
50mm
650mm
load 2
X
10N
X
10N
position of load 2
X
950mm
X
950mm

           
Beam 1
Beam 2
Beam 1
Beam 2
Distance along length
Deflection in mm
Distance along length
Deflection in mm
Distance along length
Deflection in mm
Distance along length
Deflection in mm
x
Y
X
Y
x
Y
x
Y
-
-
500
0
-
-
-
-
500
0
600
-6.98
500
0
500
0
600
-4.92
700
-12.38
600
-1.25
600
-1.09
700
-8.76
800
-15.27
700
-2.35
700
-2.01
800
-11.18
850
-15.74
800
-3.07
800
-2.53
900
-11.42
900
-15.5
900
-3.08
900
-2.54
1000
-9.33
1000
-12.67
1000
-2.43
1000
-2.13
1100
-5.28
1100
-7.15
1100
-1.35
1100
-1.14
1200
0
1200
0
1200
0
1200
0


Theoretical Calculation

Beam 1

As per specification provided for experiment, beam one is made of mild steel and has the length of about 700 mm between fixed ends. Cross section of beam is 3.2*19 mm as thickness*width. Moment of inertia of this rectangular cross section beam can be calculated as [4]

I=1/12 width*thickness^3

I=1/12*19*〖3.2〗^3

I=0.0834*19*32.768

I=51.92 mm^4

Young modulus of mild steel is 21*〖10〗^10  N/m^2 which is equal to21*〖10〗^4  N/mm^2. 

So 
EI=21*〖10〗^4*51.92=10.90*〖10〗^6  Nmm^2

Load of about 10 N is applied on beam one at the distance of 350 mm from its left ends. Due to this the reaction forces of beam can be calculated as follow

Sum of all forces in y axis direction is zero.

Ra+Rb=10 N

Ra=10-Rb

Moment at point A is consider zero. 

Rb*distance from A=Froce*distance from A

Rb*700=10*350

Rb=(10*350)/700

Rb=5 N

Put this value in above equation of forces

Ra=10-5=5 N


Equation of moment for beam one is mention below
EI*  (d^2 y)/(dx^2 )=Ra*x-F(x-a) 

EI*  (d^2 y)/(dx^2 )=5*x-10(x-a) 

Integrating above equation

EI*  dy/dx=〖5x〗^2/2-(10(x-a)^2)/2+C1

Integrating once again

EI*y=(5x^3)/6-(10(x-a)^3)/6+C1*x+C2
Put x = 0 and y = 0

EI*0=(5*0^3)/6-(10(0-350)^3)/6+C1*0+C2

Ignoring any bracket containing negative value 

C2=0

Put x = 700 and y = 0
EI*0=(5*〖700〗^3)/6-(10(700-350)^3)/6+C1*700+0

C1*700= -  (5*〖700〗^3)/6+(10(350)^3)/6

C1*700= -385833333+71458333.3

C1*700= -314375000

C1=-449107.14

So equation of deflection became
EI*y=(5x^3)/6-(10(x-350)^3)/6-449107.14*x

y=((5x^3)/6-(10(x-350)^3)/6-449107.14*x)/EI

Put x = 0 to find reaction at 500 mm
y=((5〖*0〗^3)/6-(10(0-350)^3)/6-449107.14*0)/EI
Ignoring negative terms 

y=0

Put x = 100 to find reaction at 600 mm

y=((5〖*100〗^3)/6-(10(100-350)^3)/6-449107.14*100)/EI


y=(833333.33-0-44910714.29)/EI

y=-4.04 mm

Put x = 200 to find reaction at 700 mm

y=((5〖*200〗^3)/6-(10(200-350)^3)/6-449107.14*200)/EI


y=(6666666.67-0-89821428)/EI

y=-7.62 mm

Put x = 300 to find reaction at 800 mm

y=((5〖*300〗^3)/6-(10(300-350)^3)/6-449107.14*300)/EI


y=(22500000-0-134732142)/EI

y=-12.36 mm

For a simply supported beam with one point load present at the half length of beam, the deflection equation is valid only up to the centre point. After that same equation can be used from other end to check deflection of beam.
 
Beam 2

As per specification provided for experiment, beam two is made of mild steel and has the length of about 700 mm between fixed ends. Cross section of beam is 3.2*19 mm as thickness*width. Moment of inertia of this rectangular cross section beam can be calculated as

I=1/12 width*thickness^3

I=1/12*19*〖3.2〗^3

I=0.0834*19*32.768

I=51.92 mm^4

Young modulus of mild steel is 21*〖10〗^10  N/m^2 which is equal to21*〖10〗^4  N/mm^2. 

So 
EI=21*〖10〗^4*51.92=10.90*〖10〗^6  Nmm^2

Two loads of 10 N are applied on beam, one at the distance of 150 mm and second at 450 mm from its left ends. Due to this the reaction forces of beam can be calculated as follow

Sum of all forces in y axis direction is zero.

Ra+Rb=10+10 N

Ra=20-Rb

Moment at point A is consider zero. 

Rb*distance from A=Froce*distance from A

Rb*700=10*150+10*450

Rb=(10*150+10*450)/700

Rb=8.571 N

Put this value in above equation of forces

Ra=20-8.571=11.42 N


Equation of moment for beam one is mention below
EI*  (d^2 y)/(dx^2 )=Ra*x-F1(x-a)-F2(x-b)

EI*  (d^2 y)/(dx^2 )=11.42*x-10(x-150)-10(x-450) 

Integrating above equation

EI*  dy/dx=〖11.42x〗^2/2-(10(x-150)^2)/2-(10(x-450)^2)/2+C1

Integrating once again

EI*y=(11.42x^3)/6-(10(x-150)^3)/6-(10(x-450)^3)/6+C1*x+C2

Put x = 0 and y = 0

EI*0=(11.42*0^3)/6-(10(0-150)^3)/6-(10(0-450)^3)/6+C1*0+C2

Ignoring any bracket containing negative value 

C2=0

Put x = 700 and y = 0
EI*0=(11.42*〖700〗^3)/6-(10(700-150)^3)/6-(10(700-450)^3)/6+C1*700+0

C1*700= -(11.42*〖700〗^3)/6+(10(550)^3)/6+(10(250)^3)/6

C1*700= -652843333.3+277291666.7+26041666.67

C1*700= -349510000

C1=-499300

So equation of deflection became
EI*y=(11.42x^3)/6-(10(x-150)^3)/6-(10(x-350)^3)/6-499300*x

y=((11.42x^3)/6-(10(x-150)^3)/6-(10(x-350)^3)/6-499300*x)/EI

Put x = 0 to find reaction at 500 mm
y=((11.42〖*0〗^3)/6-(10(0-150)^3)/6-(10(0-350)^3)/6-499300*0)/EI


Ignoring negative terms 

y=0


Put x = 100 to find reaction at 600 mm
y=((11.42〖*100〗^3)/6-(10(100-150)^3)/6-(10(10-350)^3)/6-499300*100)/EI


Ignoring negative terms 

y=(1903333.3-0-0-49930000)/10900000


y=-4.40 mm

Put x = 200 to find reaction at 700 mm

y=((11.42〖*200〗^3)/6-(10(200-150)^3)/6-(10(20-350)^3)/6-499300*200)/EI


Ignoring negative terms 

y=(15226666.67-208333.33-0-99860000)/10900000

y=184701664.7/10900000

y=-7.78 mm

Put x = 300 to find reaction at 800 mm

y=((11.42〖*300〗^3)/6-(10(300-150)^3)/6-(10(300-350)^3)/6-499300*300)/EI


Ignoring negative terms 

y=(51390000-5625000-0-149790000)/10900000


y=-9.54 mm

Put x = 350 to find reaction at 850 mm

y=((11.42〖*350〗^3)/6-(10(350-150)^3)/6-(10(350-350)^3)/6-499300*350)/EI


Ignoring negative terms 

y=(81605416.67-13333333.33-0-174755000)/10900000

y=-184701664.7/10900000

y=-9.76 mm


Put x = 400 to find reaction at 900 mm

y=((11.42〖*400〗^3)/6-(10(400-150)^3)/6-(10(400-350)^3)/6-499300*400)/EI


Ignoring negative terms 

y=(121813333-26041666.7-208333.33-199720000)/10900000

y=-104156667/10900000

y=-9.55 mm
Put x = 500 to find reaction at 1000 mm

y=((11.42〖*500〗^3)/6-(10(500-150)^3)/6-(10(500-350)^3)/6-499300*500)/EI


Ignoring negative terms 

y=(237916667-71458333.3-5625000-249650000)/10900000

y=-88816666/10900000

y=-8.14 mm


Put x = 600 to find reaction at 11000 mm

y=((11.42〖*600〗^3)/6-(10(600-150)^3)/6-(10(600-350)^3)/6-499300*600)/EI


Ignoring negative terms 

y=(411120000-151875000-26041666.7-299580000)/10900000

y=-66376667/10900000

y=-6.089 mm
 
Beam 3

As per specification provided for experiment, beam three is made of mild steel and has the length of about 700 mm between fixed ends. Cross section of beam is 6.4 * 13 mm as thickness*width. Moment of inertia of this rectangular cross section beam can be calculated as

I=1/12 width*thickness^3

I=1/12*13*〖6.4〗^3

I=283.98 mm^4

Young modulus of mild steel is 21*〖10〗^10  N/m^2 which is equal to21*〖10〗^4  N/mm^2. 

So 
EI=21*〖10〗^4*283.98=59.63*〖10〗^6  Nmm^2

Load of about 20 N is applied on beam one at the distance of 350 mm from its left ends. Due to this the reaction forces of beam can be calculated as follow

Sum of all forces in y axis direction is zero.

Ra+Rb=20 N

Ra=20-Rb

Moment at point A is consider zero. 

Rb*distance from A=Froce*distance from A

Rb*700=20*350

Rb=(20*350)/700

Rb=10 N

Put this value in above equation of forces

Ra=20-10=10 N


Equation of moment for beam one is mention below
EI*  (d^2 y)/(dx^2 )=Ra*x-F(x-a) 

EI*  (d^2 y)/(dx^2 )=10*x-20(x-a) 

Integrating above equation

EI*  dy/dx=〖10x〗^2/2-(20(x-a)^2)/2+C1

Integrating once again

EI*y=(10x^3)/6-(20(x-a)^3)/6+C1*x+C2
Put x = 0 and y = 0

EI*0=(10*0^3)/6-(20(0-350)^3)/6+C1*0+C2

Ignoring any bracket containing negative value 

C2=0

Put x = 700 and y = 0
EI*0=(10*〖700〗^3)/6-(20(700-350)^3)/6+C1*700+0

C1*700= -  (10*〖700〗^3)/6+(20(350)^3)/6

C1*700= -571666666.7+142916666.7

C1*700= -428750000

C1=-612500

So equation of deflection became
EI*y=(5x^3)/6-(10(x-350)^3)/6-612500*x

y=((5x^3)/6-(10(x-350)^3)/6-612500*x)/EI

Put x = 0 to find reaction at 500 mm
y=((5〖*0〗^3)/6-(10(0-350)^3)/6-612500*0)/EI
Ignoring negative terms 

y=0

Put x = 100 to find reaction at 600 mm

y=((10〖*100〗^3)/6-(20(100-350)^3)/6-612500*100)/EI


y=(1666666.67-0-61250000)/EI

y=-59583333.6/59630000

y=-0.99 mm

Put x = 200 to find reaction at 700 mm

y=((10〖*200〗^3)/6-(20(200-350)^3)/6-612500*200)/EI


y=(13333333.33-0-122500000)/EI

y=-1.83 mm

Put x = 300 to find reaction at 800 mm

y=((10〖*300〗^3)/6-(20(300-350)^3)/6-612500*300)/EI


y=(45000000-183750000)/EI

y=-2.32 mm

For a simply supported beam with one point load present at the half length of beam, the deflection equation is valid only up to the centre point. After that same equation can be used from other end to check deflection of beam.
 
Beam 4

As per specification provided for experiment, beam two is made of mild steel and has the length of about 700 mm between fixed ends. Cross section of beam is 3.2*19 mm as thickness*width. Moment of inertia of this rectangular cross section beam can be calculated as

I=1/12 width*thickness^3

I=1/12*13*〖6.4〗^3

I=283.98 mm^4

Young modulus of mild steel is 21*〖10〗^10  N/m^2 which is equal to21*〖10〗^4  N/mm^2. 

So 
EI=21*〖10〗^4*283.98=59.63*〖10〗^6  Nmm^2

Two loads of 10 N are applied on beam, one at the distance of 150 mm and second at 450 mm from its left ends. Due to this the reaction forces of beam can be calculated as follow

Sum of all forces in y axis direction is zero.

Ra+Rb=10+10 N

Ra=20-Rb

Moment at point A is consider zero. 

Rb*distance from A=Froce*distance from A

Rb*700=10*150+10*450

Rb=(10*150+10*450)/700

Rb=8.571 N

Put this value in above equation of forces

Ra=20-8.571=11.42 N


Equation of moment for beam one is mention below
EI*  (d^2 y)/(dx^2 )=Ra*x-F1(x-a)-F2(x-b)

EI*  (d^2 y)/(dx^2 )=11.42*x-10(x-150)-10(x-450) 

Integrating above equation

EI*  dy/dx=〖11.42x〗^2/2-(10(x-150)^2)/2-(10(x-450)^2)/2+C1

Integrating once again

EI*y=(11.42x^3)/6-(10(x-150)^3)/6-(10(x-450)^3)/6+C1*x+C2

Put x = 0 and y = 0

EI*0=(11.42*0^3)/6-(10(0-150)^3)/6-(10(0-450)^3)/6+C1*0+C2

Ignoring any bracket containing negative value 

C2=0

Put x = 700 and y = 0
EI*0=(11.42*〖700〗^3)/6-(10(700-150)^3)/6-(10(700-450)^3)/6+C1*700+0

C1*700= -(11.42*〖700〗^3)/6+(10(550)^3)/6+(10(250)^3)/6

C1*700= -652843333.3+277291666.7+26041666.67

C1*700= -349510000

C1=-499300

So equation of deflection became
EI*y=(11.42x^3)/6-(10(x-150)^3)/6-(10(x-350)^3)/6-499300*x

y=((11.42x^3)/6-(10(x-150)^3)/6-(10(x-350)^3)/6-499300*x)/EI

Put x = 0 to find reaction at 500 mm
y=((11.42〖*0〗^3)/6-(10(0-150)^3)/6-(10(0-350)^3)/6-499300*0)/EI


Ignoring negative terms 

y=0


Put x = 100 to find reaction at 600 mm
y=((11.42〖*100〗^3)/6-(10(100-150)^3)/6-(10(10-350)^3)/6-499300*100)/EI


Ignoring negative terms 

y=(1903333.3-0-0-49930000)/(59630000)


y=-0.805 mm

Put x = 200 to find reaction at 700 mm

y=((11.42〖*200〗^3)/6-(10(200-150)^3)/6-(10(20-350)^3)/6-499300*200)/EI


Ignoring negative terms 

y=(15226666.67-208333.33-0-99860000)/59630000

y=84841666.66/59630000

y=-1.42 mm

Put x = 300 to find reaction at 800 mm

y=((11.42〖*300〗^3)/6-(10(300-150)^3)/6-(10(300-350)^3)/6-499300*300)/EI


Ignoring negative terms 

y=(51390000-5625000-0-149790000)/59630000


y=-1.57 mm


Put x = 400 to find reaction at 900 mm

y=((11.42〖*400〗^3)/6-(10(400-150)^3)/6-(10(400-350)^3)/6-499300*400)/EI


Ignoring negative terms 

y=(121813333-26041666.7-208333.33-199720000)/59630000

y=-104156667/59630000

y=-1.74 mm
Put x = 500 to find reaction at 1000 mm

y=((11.42〖*500〗^3)/6-(10(500-150)^3)/6-(10(500-350)^3)/6-499300*500)/EI


Ignoring negative terms 

y=(237916667-71458333.3-5625000-249650000)/59630000

y=-88816666/59630000

y=-1.48 mm


Put x = 600 to find reaction at 11000 mm

y=((11.42〖*600〗^3)/6-(10(600-150)^3)/6-(10(600-350)^3)/6-499300*600)/EI


Ignoring negative terms 

y=(411120000-151875000-26041666.7-299580000)/59630000

y=-66376667/59630000

y=-1.11mm
 

Discussion

In order to understand the working of simply supported beam two beams made of steel having different cross section areas were used in this experiment. For two beams a set of experiment was conducted using single point load in first case and two point loads in second case. Deflection of beam was noted for each case at different length of 100 mm apart, starting from the left end on beam. Theoretical calculations were made based on the Macaulay’s theory to find deflection of beam at required length. Comparison of the experimental and theoretical calculation for each case is shown in graph section. According to graphs it can be concluded that in almost every case the value of experimental deflection much more than that of the theoretical deflection. Difference in deflection can be due to lower value of modulus of elasticity of material as compared to book, manufacturing defect in beam used for experiment, error introduced during experiment due to lack of experience of worker or due to the use of faulty deflection measuring instruments. Comparing the different case show that increase in number of load from single point load to two point loads on same beam, increase the deflection in beam as seen in beam one and two case. Comparing beam two and beam four result show that increase the moment of inertia of beam by using beam with bigger cross section area decrease the deflection of beam. Comparing results of beam three and beam four shows that decrease in load magnitude decrease the deflection in beam. 

Conclusion
A series of experiment was conducted on simply supported beam in order to understand its working. From the experimental and theoretical result obtained it can concluded that increase in number of load increase the deflection of beam, increase in magnitude of load increase the deflection of beam, increase in moment of inertia of beam decrease the deflection of beam and Experimental deflection in beam is always greater than the theoretical deflection on beam,

Talha Atta is an automotive expert and one of the finest technical content writers from Pakistan. Talha owns WheelWisely.com and is Chief Editor at AutoGlobes.com. He has covered extremely granular content about structures, manufacturing process, automotive, materials and renewable energy. Connect with him on Upwork or check out more of his work on LinkedIn.

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