Chapter 6 - Column.ppt

BNJ 20903 Lecture #07 By, Dr. Noraini Marsi University Tun Hussein Onn Malaysia (UTHM)

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Failure Theories

Strain Energy

Stress and Strain

Solid Mechanics

Shear Force and Bending Moment

Torsion

Thin cylinder

Column Stress and strain transformation

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



Long slender members subjected to an axial compressive force are called columns. The lateral deflection that occurs is called buckling.

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

The maximum axial load that a column can support when it is on the verge of buckling is called the CRITICAL LOAD, Pcr.



Any additional load will cause the column to buckle and therefore deflect laterally

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

When the bar are in the vertical position, the spring having the stiffness k, is unstretched.



FBD, the bar are displaced by pin at A which produce force F=k∆.

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

The spring will produce the force,



Restoring spring force become:



Applied load P will develops two horizontal components,



Since q is small, and tanq=q

F  k

 2

 q L

k qL F 2 P  P tan q x

2Px  2Pq

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•If the restoring force is greater than disturbing force

kq L  2 Pq 2

P  kL

Stable

equilibrium The force developed by the spring would be adequate to restore the bars back to their vertical position.

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Unstable

P  kL

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equilibrium If the load P is applied and a slight displacement occurs at A, the mechanism will tend to move out of equilibrium and not be restored to its original positions. Neutral

Pcr  kL

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equilibrium Any slight disturbance given to the mechanism will not cause it to move further out of equilibrium, nor will it be restored to its original position. Instead, the bars will remain in the deflected position.

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1.Column With Pin Support 





When the critical load Pcr is reached, the column is on the verge of becoming unstable, so that a small lateral force F will cause the column to remain in the deflected position when F is removed.

In order to determine the critical load and buckled shape of the column, the followed equation is used:Deflection y and internal moment M are in positive direction

d2y EI 2  M dx 8



•

With M=-Py Equation 3 is homogeneous, second order, linear differential equation

d2y EI 2  Py _________(1) dx d2y  P     y  0 _______(2) 2 dx  EI  or d2y 2   y  0 _________(3) 2 dx with

2 

P EI

P or      EI 

1 2

Using methods of differential equation or by direct substitution General solution is:

y  A cos (x )  B sin (x ) ____(4) 9



At x=0, y=0, then from equation (4), A=0 and equation (4) become as:y=B sin (x)_____(5)



At x=L, y=0, then from equation (5):0= B sin (L)



 

If B=0, It means no deflection occur in the column. Therefore, B0, but:sin (L)=0 (L)= , 2, 3, 4, …….n, if , n0 The smallest value of P is obtained when n=1, (L)=  or 1 2



P   .L    EI  so critical load for this column is:

2 EI Pcr  2 __________(6) L 10





A column will buckle about the principal axis of the cross section having the least moment of inertia (the weakest axis). As in picture, the column will buckle at the a-a axis not the b-b axis.

Pcr  

 

 2 EI L2

Pcr=critical or max axial load on the column just before it begins to buckle. This loads must not cause the stress in the column to exceed the proportional limit. E=modulus of elasticity material. I=least moment of inertia for the column crosssectional area.

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5.Column With Various Types of Supports Based on all Euler’s Formula for various types of supports, the formula can be written as:

 2 EI  2 EI Pcr   2 (KL) (L e ) 2

With K = constant depends on the end support types = 1, 2, 0.5, and 0.7

EI = column minimum stiffness (kNm2) L = Column actual length (m) Le = effective length P = Pcr=Buckling @ Critical Load (kN@MN)

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Effective Length :Le=KL

 2 EA  Pcr  2 KL r  2 EI  (KL) 2





 2 EI Pcr  L2



 EI (L e ) 2 2

2E  cr  2 KL r 2 E  2 L  e   r   

Pcr 

 2 E  A 2 KL r

 

 

 2 EI  2 L  1 2 EI  L2

Pcr   2 E  A Pcr  2 KL r

 

Pcr 



 Pcr 

 2 EI

0.5L 

2

 2 EI 2

0.25L 4 2 EI  L2

 2 E  A 2 KL r

 

 2 EI

2 L 2

 2 EI

4 L2 0.25 2 EI  L2

Pcr  

 2 EI

0.7 L 2

 2 EI

0.49 L2 2 2 EI  L2

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Column Buckling Stress

Buckling Stress   cr 

Buckling Load Cross  sec tional Area

Pcr A

 2 EI  1   cr   2  Le  A  I  Ar 2 where A  cross  sec tional area , r  smallest radius of gyration  2 E (Ar 2 )  1  2E  cr    2 2 Le  A   Le   r    Le is known as the slendernes s ratio r

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SLENDERNESS RATIO,L/r





Curve hyperbolic valid for critical stress below yield point. Eg:  Y  steel  250MPa

 cr   Y 



The smallest acceptance slenderness ratio for steel.(L/r=89) If (L/r>89) euler’s formula can be used however if (L/r<89) euler formula not valid.

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Critical Buckling Load : I

 d 

4

 d4

and K  1 for roller sup ported ends column.    42 64 Applying Euler ' s formula,

 d 4    21010  2 64   EI  Pcr   2 ( KL) 1(300)2 3

2

 d 4    21010   64  4(10) 3  2 (1)300 d  7.71mm  8mm 2

3

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Pcr 4(10) 3  cr    79.6MPa  Y  250MPa  2 A 8 4 Therefore , Euler ' s formula is valid

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The Secant Formula 







The Euler formula was derived with the assumptions that the load P is always applied through the centroid of the columns’s cross-sectional area and that the column is perfectly straight. This is quite unrealistic since manufactured columns are never perfectly straight.

In reality, columns never suddenly buckle; instead they begin to bend although ever so slightly , immediately upon application of the load. Therefore, load P will be applied to the column at a short eccentric distance e from the centroid of the cross section.

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Internal moment in the column:

M  Pe  y 

Differential equation for the deflection curve:

d2y EI 2  M dx

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When x =L/2, v = vmax MaximumDeflection :

P P y  C1sin x  C 2 cos xe EI EI Boundary Condition when x  0, y  0 C 2  e. when x  L, y  0 C1 





e 1  cos P EI L

y max  e sec P L   1 EI 2     if e  0, y max  0



sin P EI L

since :





1  cos P EI L  2 sin 2 and sin P EI L  2 sin C1  e tan





P EI L



P EI L

P EI L



2

2

cos

 P EI L

2



2 hence, the deflection curve written as; y  e tan P EI L sin P EI x  cos P EI x  1 2

 

 





 

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M  Pe  ymax  L  M  Pe sec P  EI 2  Max stress in the column is compressive

 max 

P Mc P Pec  P L   ;  max   sec  EI A I A I 2  

I  Ar 2  max 

P  ec  L P  1  sec    2 EA A r  2r 

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i. ii.

The W250x18 structural A-36 steel column is used to support a load of 4 kN. If the column is fixed at the base and free at the top, determine: The deflection at the top of the column due to the loading. The maximum stress in the column. E=210GPa, σy=250MPa.

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Solution (ii): sec tion

properties

A  2280mm2

 max

for W 250 18;

I x  22.5(106 )mm4

P  ec KL  1  2 sec ( A r 2r

rx  99.3mm d  251mm

P  ) EA 

P 4 103   1.75MPa A 2280  251  200  ec 2    2.5455  r2 99.32  KL P (2.0)(5000)  4(103 )   0.1455  3 2r EA 2(99.3)  210(10 )( 2280)   max  1.751  2.5455 sec(0.1455)

 max  6.252MPa Y

OK . 38



The W360 X 39 structural A-36 steel member is used as a column that is assumed to be fixed at its top and its bottom. If the 15 kN load is applied at an eccentric distance of 250 mm. Determine the MAXIMUM STRESS in the column. E = 210 GPa, σy = 250 MPa.

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Section

properties

A  4960mm 2 Yielding  max

about

for

W360  39 : 

d  363mm

I x  102(10 6 )mm 4

rx  143mm

x  x axis :

P  ec  KL P    1  2 sec  A  r 2 r EA  

K  0.5

 363  250   3 P 15(10 ) ec  2   2.2189   3.02MPa ; 2  A 4960 r (143) 2 KL P 0.5(5)(1000) 15(10 3 )   0.03317 3 2r EA 2(143) 210(10 )( 4960)

 max  3.021  2.2189 sec (0.03317)  max  9.72MPa  Y  250MPa

OK. 40

Exercise 2b 

Solve the problem if the column is fixed at its top and pinned at its bottom.

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Section

properties

A  4960mm Yielding  max 

2

about

for

W360  39 : 

d  363mm

I x  102(10 6 )mm 4

rx  143mm

x  x axis :

P  ec  KL P   1  sec  A  r 2 2 r EA  

K  0.7

 363  250  P 15(10 ) ec 2     3.02MPa ; 2   2.2189 A 4960 r (143) 2 3

KL P 0.7(5)(1000) 15(10 3 )   0.04644 2r EA 2(143) 210(10 3 )( 4960)

 max  3.021  2.2189 sec (0.04644)  max  9.73MPa  Y  250MPa

OK

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

A W360 x 45 structural A-36 steel column is pin connected at its ends and has a length L = 5 m. Determine the maximum eccentric load P that can be applied so the column does not buckle or yield. Compare this value with an axial critical load P applied through the centroid of the column. E = 210 GPa, σy = 250 MPa.

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Section

properties

A  5710mm

W360  45 : 

for

rx  146mm

2

I y  8.16(10 6 )mm 4

d  352mm

For a column pinned at both ends , K  1 : (KL) y  (KL) x  1(5)(1000)  5000mm Buckling

about

Critical

y  y axis : Applying

Euler

2

3

6

Pcr 68544   12.0MPa   y  250MPa A 5710 Yielding about x  x axis : Applying the

 cr 

 max 

 max  max

Formula ,

 EI  (210)(10 )(8.16)(10 )   68544 N  68.54kN 2 (KL) y (5000) 2 stress : Euler ' s formula is only valid if  cr   Y

P'  P  Pcr 

2

OK. sec ant

formula ,

P  ec  KL P   1  sec  A  r 2 2 r EA  

   352  150      68544 68544 2   5000      1 sec 3  5710  146 2 2 ( 146 ) ( 210 )( 10 )( 5710 )        26.9MPa   Y  250MPa OK

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

Solve the problem if the column is fixed connected at its ends.

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  

  

Mukasurat 556 Mechanics of material – johnson 9.51 9.52 9.53 9.54

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