BEng Tech/NZDE Strength of Materials: Buckling of Beams Lab Report

This lab report investigates the buckling behavior of beams with varying slenderness ratios. The experiment involved compressing beams of different lengths using a materials testing machine and recording the critical failure load. The report includes detailed theoretical calculations using both Euler's and Rankine-Gordon's formulas to predict the critical failure load and stress. These theoretical values are then compared with the experimental results. The report also presents the experimental setup, data analysis, and MATLAB code used for generating graphs of force vs. slenderness ratio for each beam. The analysis includes calculating the slenderness ratio, critical failure load, and critical stress for each beam, comparing the experimental results with the theoretical predictions, and calculating the percentage error. The report concludes with a discussion of the results, highlighting the agreement and discrepancies between theory and experiment.

Student
Instructor
Analogue and digital electronics
Date

Slenderness ratio
S . R= le
k
Where le is the effective length , k is the radius of gyration
Radius of gyration is determined by
k = √ I
A
Where A is thecross sectional area of thebeam ,∧I is the second moment of inertia of the beam
Second moment of inertia of the beam is calculated using the expression below.
I = 1
12 a b3
Critical failure load using Euler formula is given by
Pcrit = π2 EA
( le /k )2
And critical stress using Euler is given by
Pcrit
A = π 2 E
( le/k ) 2 =σcrit
Critical failure load using Rankine-Gordon formula is given by
Pcrit = ( A σ y )
[ 1+a ( le
k ) 2
]
And critical stress using Rankine-Gordon formula is given by
Pcrit
A = ( σ y )
[1+a ( le
k )2
]=σcrit
Where σ y is yield stress of thematerial and constant (a) is found by the expression below.
a= σ y
π2 E
THEORETICAL CALCULATION OF THE PARAMETER.
1. Beam 2 (Compression)
Effective length le=15.15 mm

Radius of gyration k =2.39 mm
Slenderness ratio S . R= le
k =6.34
Critical failure load using Euler formula is given by
Pcrit = π2 EA
( le /k )2 = (2.11 ×109 Pa ) ( 48.93 ×1 0−6 m2 ) π 2
(6.34 )2
Pcrit =25,350 Pa
And critical stress using Euler is given by
Pcrit
A = π 2 E
( le/k ) 2 =σcrit
σ crit= ( 2.11 ×109 Pa ) π2
( 6.34 ) 2
σ crit=0.518 Gpa
Critical failure load using Rankine-Gordon formula is given by
Pcrit = ( A σ y )
[ 1+a ( le
k ) 2
]
Where a= σ y
π2 E = 42.7 ×106 Pa
π2 ( 2.11 ×1 09 Pa )
a=0.00205
Pcrit = ( ( 42.7 ×106 Pa ) ( 48.93 ×1 0−6 m2 ) )
[ 1+0.00205 ( 6.34 ) 2 ]
Pcrit =1930 Pa
And critical stress using Rankine-Gordon formula is given by
Pcrit
A = ( σ y )
[1+a ( le
❑ )2
]=σcrit
σ crit=¿ ( 42.7 × 106 )
[ 1+0.00205 ( 6.34 )2 ]
σ crit=¿ 0.03945Gpa

2. Beam 3 (Compression)
Effective length le=33.85 mm
Radius of gyration k =2.39 mm
Slenderness ratio S . R= le
k =14.16
Critical failure load using Euler formula is given by
Pcrit = π2 EA
( le /k )2 = (2.11 ×109 Pa ) ( 48.93 ×1 0−6 m2 ) π 2
(14.16 )2
Pcrit =5,082 Pa
And critical stress using Euler is given by
Pcrit
A = π 2 E
( le/k ) 2 =σcrit
σ crit= ( 2.11 ×109 Pa ) π2
( 14.16 )2
σ crit=0.10386 Gpa
Critical failure load using Rankine-Gordon formula is given by
Pcrit = ( A σ y )
[ 1+a ( le
k ) 2
]
Where a= σ y
π2 E = 42.7 ×106 Pa
π2 ( 2.11 ×1 09 Pa )
a=0.00205
Pcrit = ( ( 42.7 ×106 Pa ) ( 48.93 ×1 0−6 m2 ) )
[ 1+ 0.00205 ( 14.16 )
2 ]
Pcrit =1481 Pa
And critical stress using Rankine-Gordon formula is given by
Pcrit
A = ( σ y )
[1+a ( le
k )2
]=σcrit

σ crit=¿ ( 42.7 ×1 06 )
[ 1+0.00205 (14.16 )2 ]
σ crit=¿ 0.03026Gpa
3. Beam 4 (Compression)
Effective length le=60.1 mm
Radius of gyration k =2.39 mm
Slenderness ratio S . R= le
k =25.15
Critical failure load using Euler formula is given by
Pcrit = π2 EA
( le /k )2 = (2.11 ×109 Pa ) ( 48.93 ×1 0−6 m2 ) π2
(25.15 )2
Pcrit =1611 Pa
And critical stress using Euler is given by
Pcrit
A = π 2 E
( le/k ) 2 =σcrit
σ crit= ( 2.11 ×109 Pa ) π2
( 25.15 )2
σ crit=0.0329 Gpa
Critical failure load using Rankine-Gordon formula is given by
Pcrit = ( A σ y )
[ 1+a ( le
k ) 2
]
Where a= σ y
π2 E = 42.7 ×106 Pa
π2 ( 2.11 ×1 09 Pa )
a=0.00205
Pcrit = ( ( 42.7 ×106 Pa ) ( 48.93 ×1 0−6 m2 ) )
[ 1+ 0.00205 ( 25.15 )
2 ]

Pcrit =910 Pa
And critical stress using Rankine-Gordon formula is given by
Pcrit
A = ( σ y )
[1+a ( le
k )2
]=σcrit
σ crit=¿ ( 42.7 ×1 06 )
[ 1+ 0.00205 ( 25.15 ) 2 ]
σ crit=¿ 0.0186Gpa
EXPERIMENTAL RESULTS for Pcritical
Beam 2
Pcrit = F
A =2933 N
Beam 3(compression )
Pcrit = F
A =2853 N
Beam 3(Tensile )
Pcrit = F
A =1912 N
Beam 4
Pcrit = F
A =1342 N
Length(L) Area I k Le S.R Theoretical
Pcr
Experiment
a
Pcr
%error
30.3 48.93 279.5 2.39 15.15 6.34 2535 2933 16%
67.7 48.93 279.5 2.39 33.85 14.16 5082 2853 43%
67.7 48.93 279.5 2.39 33.85 14.16 5082 1912 62%
120.2 48.93 279.5 0.52 60.1 25.15 1611 1342 17%

GRAPHS
Beam 2 compression
Beam 3 Compression
Beam 4 compression

MATLAB CODE.
clear all
clc
E=210000000;
k1=2.39;
k2=2.39;
k3=2.39;
k4=0.52;
A=48.93;
%PInt=is Critical force, E is Modulus of Elaciticity of ABS,A is the cross
%sectional area, l is the effective lenght, k is radius of gyration
%and
%P=(A*pi^2*E)/(l/k1)^2;
% where (l/k) is the slenderness ration
%beam 2
%----------------------------------------------------------------------
%-------BEAM 2-------------------------------------------
x=[0.00289 0.00299 0.0031 0.00316 0.00326 0.00336 0.00344 0.00353 0.00363
0.00372 0.00384 0.00395 0.00407 0.00419 0.0043 0.00442 0.00455 0.00469
0.0048 0.0048]
%x=pi^2*E*[(2.39/2.89)^2 (2.39/2.99)^2 (2.39/3.1)^2 (2.39/3.16)^2
(2.39/3.26)^2 (2.39/3.36)^2 (2.39/3.44)^2 (2.39/3.53)^2 (2.39/3.63)^2
(2.39/3.72)^2 (2.39/3.84)^2 (2.39/3.95)^2 (2.39/4.07)^2 (2.39/4.19)^2
(2.39/4.3)^2 (2.39/4.42)^2 (2.39/4.55)^2 (2.39/4.69)^2 (2.39/4.8)^2
(2.39/4.8)^2]
%P=pi^2*E
y=[2348 2484 2629 2703 2806 2896 2933 2808 2144 1758 1665 1584 1480 1448
1402 1231 1190 1186 1175 1108]
figure,plot(x,y)
xlabel('le/k')
ylabel('Force N')
title('BEAM 2 COMPRESSION')
%----------------------------------------------------------
%-------------------BEAM 3--------------------------------
%Beam 3

x1=[0.01352 0.01367 0.01378 0.0139 0.01403 0.01419 0.01433 0.01446 0.01453
0.01463 0.01477 0.01491 0.01506 0.01517 0.0153 0.01545 0.01562 0.01579
0.01595 0.0161 0.01625 0.01632 0.01646 0.01656 0.01659 0.01663]
y1=[1465 1594 1694 1795 1908 2039 2179 2296 2359 2429 2543 2665 2759 2813
2853 2722 1766 1469 1323 1226 1160 1092 1066 869 778 732]
figure,plot(x1,y1)
title('BEAM 3 COMPRESSION')
xlabel('le/k')
ylabel('Force N')
%-------------------------------------------------------------
%-----------------BEAM 4----------------------------------
%----------------------------------------------------
%Beam 4
x2=[0.00292 0.00296 0.003 0.00303 0.00306 0.00313 0.0032 0.00326 0.00331
0.00336 0.00341 0.00344 0.00346 0.00348 0.00349 0.00351 0.00353 0.00355
0.00357 0.00358 0.0036]
y2=[1232 1250 1270 1290 1294 1320 1339 1342 1339 1332 1321 1306 1290 1278
1267 1258 1250 1242 1236 1229 1221]
figure,plot(x2,y2)
title('BEAM 4 COMPRESSION')
xlabel('le/k')
ylabel('Force N')