buckling of strut report

8/18/2019 buckling of strut report

1/19

1.0 OBJECTIVE

i. To examine how shear force varies with an increasing point load

ii. To examine how shear force varies at the cut position of the beam for various

loading condition

iii. To establish the relationship between strut length (Columns length) and

buckling load.

iv. To study the effect of end condition on the buckling load

v. To inelastic buckling of a strut.

2.0 LEARNING OUTCOMES

i. Able to apply the engineering knowledge in practical application.

ii. To enhance technical competency in structure engineering through laboratory

application.

iii. Communicate effectively in group.

iv. To identify problem, solving finding out appropriate solution through

laboratory application.


2/19

3.0 INTRODUCTION

n this section we will determine the critical buckling load for a column

that is pin supported as shown in !ig. "a. The column to be considered is an ideal

column, meaning one that is perfectly straight before loading, is made of

homogeneous material, and upon which the load is applied through the centroid of

the cross section. t is further assumed that the material behaves in a linear#elastic

manner and that the column buckles or bends in a single plane.

n reality, the conditions of column straightness and load application are

never accomplished$ however, the analysis to be performed on an %ideal column&

is similar to that used to analy'e initially crooked columns or those having an

eccentric load application. These more realistic cases will be discussed later inthis chapter.

ince an ideal column is straight, theoretically the axial load could be

increased until failure occurs by either fracture or yielding of the material.

*owever, when the critical load er is reached, the column is on the verge of

becoming unstable, so that a small lateral force !, !ig. "b, will cause the column

to remain in the deflected position when ! is removed, !ig. "c. Any slight

reduction in the axial load from . will allow the column to straighten out, and

any slight increase in , beyond er, will cause further increases in lateral

deflection.

Figure 1

+


3/19

hether or not a column will remain stable or become unstable when

sub-ected to an axial load will depend on its ability to restore itself, which is based

on its resistance to bending. *ence, in order to determine the critical load and the

buckled shape of the column, we will apply d/v0dx/ 1 2(3), which relates the

internal moment in the column to its deflected shape, i.e.,

d/v0dx/ 1 2 (4)

5ecall that this e6uation assumes that the slope of the elastic curve is

small7 and that deflections occur only by bending. hen the column is in its

deflected position, !ig. +, the internal bending moment can be determined by

using the method of sections. The free#body diagram of a segment in the deflected

position is shown in !ig. +. *ere both the deflection v and the internal moment 2are shown in the positive direction according to the sign convention used to

establish 6. (4) umming moments, the internal moment is 2 1 #v8. Thus 6.

(4) becomes

d/v0dx/ 1 #v,

d/v0dx/ 9 (0)v 1 4 (")

This is a homogeneous, second#order, linear differential e6uation with

constant coefficients. t cans he shown by using the methods of differential

e6uations, or by direct substitution into 6. ("), that the general solution is

: 1 c"sin(;(0)x) 9 c+ cos (;(0)x ) (+)

The two constants of integration are determined from the boundary

conditions at the ends of the column. ince v 1 4 at x 1 4, then C+ 1 4. And since

v 4 at x 1


4/19

This e6uation is satisfied if C" 1 4$ however, then v 1 4, which is a trivial

solution that re6uires the column to always remain straight, even though the load

causes the column to become unstable. The other possibility is for

in((;0)

FIGURE 2

?r 1 n/>/0/0


5/19

This load is sometimes referred to as the uler load, after the wiss

mathematician x0


6/19

plates, etc., are better than sections that are solid and rectangular.

FIGURE 3 FIGURE 4

t is also important to reali'e that a column will buckle about the principal

axis of the cross section having the least moment of inertia (the weakest axis). !or

example, a column having a rectangular cross section, like a meter stick, as shown

in !ig. "=DB, will buckle about the aDa axis, not the b D b axis. As a result,

engineers usually try to achieve a balance, keeping the moments of inertia the

same in all directions. Eeometrically, then, circular tubes would make excellent

columns. Also, s6uare tubes or those shapes having y F x are often selected for

columns.

G


7/19

4.0 THEOR

." To predict the buckling load uler buckling formula is used. The critical

value in uler !ormula is the slenderness ratio, which is the ratio of the

length of the strut to its radius of gyration (


8/19

Curvature Quick derivation for curvature (1/R)

L


9/19

Meam e6uation

2 0 1 N 0 y 1 0 5

hen x 1 4 y 1 4 and 4 1 A cosO4 9 M.4 1 A therefore A 1 4

hen x 1 b , y 1 4 and so M sin Ob 1 4.

M cannot be 4 because there would be no deflection and no buckling which is contrary to

experience.

*ence sin Ob 1 4. therefore Ob 1 4, >, +>, = > etc

(0) b + 1 > +, .> +, P.> + etc


10/19

therefore 1 > + . 0 b+ , > + . 0 (b 0 +) +, > + . 0 (b 0 =) +,

As the moment of inertia 1 A.k + and the end force 1 N A. The formula can berewritten

' ( ) A ( * 2 E A + 2 , -2

T$ere%re

) ( * 2 E , /- , +2


11/19

.0 AARATUS

!ig "I 2achine ?f Muckling ?f truts

!ig =I truts !ig I Top Q Mottom Chuck

Rei5g

6e7er

5e!

8%57r%!

%i57 !%

guge


12/19

9.0 ROCEDURE

G." !it the bottom chuck to the machine and remove the top chuck (to give

two pinned ends). elect the shortest strut, number ", and measured the

cross section using the vernier provided and calculated the second moment

of area, , for the strut. /-3,12

G.+ Ad-ust the position of the sliding crosshead to accept the strut using the

thumbnut to lock off the slider. nsure that there is the maximum amount

of travel available on the handwheel threat to compress the strut. !inallytighten the locking screw.


13/19

G.= Carefully back#off the handwheel so that the strut is resting in the notch

but not transmitting any load. 5e'ero the forcemeter using the front panel

control.

G.". Carefully start to load the strut. f the strut begin to buckle to the left,%flick& the strut to the right and vice versa (this reduces any error

associated with the straightness of strut). Turn the handwheel until there is

no further increase in load (the load may peak and then drop as it settles in

the notches).

G.". 5ecord the final load in Table ". 5epeat with strut numbers +,=, and

ad-usting the crosshead as re6uired to fit the strut.

ANEL

CONTROL


14/19

:.0 RESULT AND CALCULATION

:.1 RESULT /ART 1

i55e;i55e<

S7ru7

Nu6-er

Le5g7$

/6

Bu8+!i5g L% /N

/E=>eri6e57

Bu8+!i5g L% /N

/T$e%r&

1,L2

/6;2

1 4.=+ # P.44 LL.L P.BB

2 4.=B # GL.44 GL."+ B.=4

3 4.+ # 4.44 ".=G .GB

4 4.B # .44 ".4" .= 4.+ # =+.44 ==." =.B4

T-!e 1

i55e;Fi=e<

S7ru7

Nu6-er

Le5g7$

/66

Bu8+!i5g L% /N

/E=>eri6e57

Bu8+!i5g L% /N

/T$e%r&

1,L2

/6;2

1 4.=4 # "LG.44 +4".++ "".""

2 4.= # "++.44 "B.L L."=

3 4.4 # L".44 ""=."P G.+

4 4. # G4.44 LP.= .P

4.4 # .44 B+. .44

T-!e 2

Fi=e;Fi=e<

S7ru7

Nu6-er

Le5g7$

/66

Bu8+!i5g L% /N

/E=>eri6e57

Bu8+!i5g L% /N

/T$e%r&

1,L2

/6;2

1 4.+L # 4B.44 G+.44 "+.BG2 4.== # =4.44 ==+.GP P."L

3 4.=L # +=+.44 +4.4 G.P=

4 4.= # "L=.44 "P.P ."

4.L # "=B.44 "B.+ .=

T-!e 3


15/19

:.2 CALCULATION

2oment of Area, 1 bd=

"+

1 4.4+ x (4.44+)/

"+

1 1.33=10;1164

uler Muckling /

/ x (GPx"4P ) x (".==x"4#"" )

4.=+/

1 ?? N

IN < IN CONDITION

Gi@e5 E ( 9 GN6;2 ( 69x10 N6;2

B ( 286 ( 0.026

D ( 0.286 ( 0.0026

L ( 0.326

<e6u


16/19


"+

1 4.4+ x (4.44+)/

"+

1 1.33=10;1164

uler Muckling /

/ x (GPx"4P ) x (".==x"4#"" ) 4.=4/

1 201.2: N

IN < FI CONDITION

Gi@e5 E ( 9 GN6;2 ( 69x10 N6;2

B ( 286 ( 0.026

D ( 0.286 ( 0.0026

L ( 0.306

<

<e6u


17/19


"+

1 4.4+ x (4.44+) /

"+

1 1.33=10;1164

uler Muckling /

/ x (GPx"4P ) x (".==x"4#"" ) 4.+=/

1 492 N

FI < FI CONDITION

Gi@e5 E ( 9 GN6;2

( 69x10

N6;2

B ( 286 ( 0.026

D ( 0.286 ( 0.0026

L ( 0.2?6

<e6

u

<


18/19

:.3 RESULT /ART 11

?.0 DISCUSSION

5eferring to the results from the calculation, we can conclude that, the

different between the theoretical and experimental results are very big for all of

the xperiment + and =. Thus, the percentage (R) of the difference between the

theoretical and experimental results are extremely big and high. !or experiment +

the ratio percentage is +B.R and for experiment = is "".R. !rom the experiment

done, we can notice that, the span with longer length will give us the bigger value

of deflection when the load is place at the mid span for both theoretical and

experimental results. hile for the span with shorter length, the deflection is

slightly small compare to the longer span.

!rom the experiment that we have done, we can conclude the buckling

load (J) are linear to values of 0


19/19

10.0 CONCLUSION

!rom this experiment, our group managed to determine the relationship

between span and deflection. n determining the deflections of the beams under

load, elastic theory is used. !rom the experiment and the results we get from this

experiment, we notice that, the span with longer length will give us the bigger

value of deflection when the load is place at the mid span for both theoretical and

experimental results. hile for the span with shorter length, the deflection is

slightly smaller compare to the longer span though the load used is same with the

longer one. Though the different between the theoretical and experimental results

are very big, but the deflection in the span also increase when the load is increase.

Thus, the conclusion is the buckling load (J) are influence to values of 0

Date post:	06-Jul-2018
Category:	Documents
Upload:	esyad-e-chad
View:	218 times
Download:	0 times

buckling of strut report

Documents