solution - shape factor.doc

8/9/2019 Solution - Shape factor.doc

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NTNU

Norwegian University of Science and Technology.Faculty of Marine TechnologyDepartment of Marine Structures.

Problem One

a) The stress distribution over the cross-section is sketched in Figure 1. Since this is

an unsymmetrical cross-section, it is important to note that the position of the

neutral axis in a fully plastic state ill not be the same as at first yield state. The

neutral axis ill move toards the side of smaller stress value.

."

Top fbre yield Bottom fbre yield Elasto-plastic Fully Platic

σ < σ Y σ Y σ Y σ Y

−σ Y −σ Y −σ Y −σ Y

Figure 1. Stress #istributions.

" plastic hinge can be introduced hen a fully plastic state has been reached. This

is simply because at that point the section can not carry any more moment.

b) $ order to obtain the elastic and plastic section modulus, e need to calculate the

neutral axis for the elastic and fully plastic state. %ith reference to Figure &, the

neutral axis at the elastic state can be calculated as follos.

( ) ( ) ( )( )

ye = × × + × × + × ×× + × + ×

= ≅&'( &' 1& &'( &' 1'' 1)( 1) &*&

&'( &' &'( &' 1)( 1) 11++ 11+. mm

xercise

Solution

SIN 1048 Buckling and Collapse of StructuresShape factor and plastic analysis of beam #ate/anuary &((+ Signature /"m #istributed #ate #ue #ate

1



0n the other hand, the neutral axis at fully plastic state, is calculated such that the

area beteen the to sides of the cross-section compression and tension sides) are

e2ual. Therefore,

( )( ) ( )1)( 1) &'( &' &' &' &' &'( &'

** + + **

× + − − × = − × + ×

⇒ = ≅

y y

y

p p

p . mm

&'(

&'

&' &'(

1)

1)(

e p

.", elastic state

.", plastic state

Dimensions in mm

Figure 2. The !eutral "xis at lastic and 3lastic States.

The elastic section modulus is given by,

W I

y

nn= +)

here I nn is the moment of inertia about the neutral axis, and y is the longest distance

perpendicular to the neutral axis hich in this case is obviously to be toards the

upper flange. 4y using Figure &, e obtain,

I

W y

nn

e

= × × + × ×

+ × × + × ×

+ × × + × ×

= ×

= ×

−

= ×

1

1& &'( &' &'( &' 1(1

1

1& &' &'( &' &'( +1

1

1& 1)( 1) 1)( 1) 156

15* 1(

15* 1(

&7( 6 ' 1(

+ & + &

+ &

7 '

7

5 +

.

..

mm

mm

4y definition, the plastic section modulus is given by,

Z M p

Y

=

σ ')

here σ Y is the yield stress and M p is the plastic moment. From Figure +, e can rite

( ) M P y y p = + 8 5)

here,

&



( ) ( )

( )

( ) ( )

( )

( )

y

y

P AY Y Y

=

× × + × ×

× + ×

= ≅

=

× × + × ×

× + ×

=

= = × + × + × =

&' 17* 6+ 5 1)( 1) 165

&' 17* 1)( 1) 1+( ' 1+(

&' 5+ &) 5 &'( &' )5

&' 5+ &'( &' 57 (

& & 1)( 1) &' &'( &'( &' *('(

..

8.

.

mm

mm

σ σ

σ

Therefore,

( ) ( ) Z

M P y y p

Y Y

Y

Y

= =

+

=

+

= ×

σ σ

σ

σ

8.

*('( 1+( 571+ & 1(

5 +mm

&'(

&'

&' &'(

1)

1)(

77

Dimensions in mm

P

P

σ Y

−σ Y

'

Figure . 3lastic Section 9odulus.

c) The shape factor is given by,

α = =

Z

W 1 '.

This is a higher value as compared to those of typical $-profiles given in the

S:reide;s book, i.e 1.1 ∼ 1.2). $t can be inferred that the unsymmetrical $-profiles

have higher shape factors compared to symmetric ones. <onsidering the given

profile in this exercise, the factor ill decrease ith increasing si=e of the top

flange until symmetry has been reached.

d) The shape factor of a cross-section represents the reserve strength of that section

after first yield.

+



Problem Two

a) There are three possible collapse mechanisms for the given structure as shon in

Figure '. The associated collapse loads can be calculated from the principle of

virtual ork, 2uation ').

δ δ W W W W e i e i= ⇒ = )

P 0.5P

M p 0.5M p

& L>3 L>3 L>2 L>2

& L

Loaded Structure

Mechanism I

Mechanism II

Mechanism II I

w1

3θ

4θ

2θ

θ

3θ

2θ w

w

w2

Figure 4. 3ossible <ollapse 9echanism.

i) 9echanism $

( ) ( ) Pw M M w L

P M

L

p p

p

= + =

⇒ =

+ ( 5 &+

&

)

θ θ θ .

ii) 9echanism $$

( )(5 (5 (5 &&

)

. . . Pw M M w

L

P M

L

p p

p

= + =

⇒ =

θ θ θ

iii) 9echanism $$$

'



( ) ( ) Pw Pw M M w

L

w

L

P M

L

p p

p

1 &

1 &(5 + (5 '

+

&

)(

+ = + = =

⇒ =

. .θ θ θ

b) From the above results, it is natural to select the true collapse load from

mechanisms $ and $$ as,

P M

Lc

p=

)

#raing the moment diagram ill prove that this is the true collapse load if the

moment values do not exceed the plastic moment capacity on any span. That is,

M M x L

M M L x L

p

p

≤ ≤ <

≤ ≤ ≤

for

for

(

( 5 &.

<onsider mechanism $ and assume that the moment diagram has the shape shon

in Figure 5. Then,

R M

L

M

L

p p

1& +

+

&= =

( ) ( )

R M M

L

M xM

L

M

L x

M

L

p p p p p p

&

( 5

+

( 5

&

11

& &=

+

+

+

= +

. .

The moment e2uilibrium of the left span about point ") yields,

( 5 +&

(+ +

. M M

L

L R L R

M

L p

p p

− ⋅ + ⋅ = ⇒ =

and the moment e2uilibrium of span 4< about point 4) yields,

xM R L

x p = ⋅ ⇒ =+

&( 5.

Static e2uilibrium is satisfied by R R R P c1 & + 15+ + = .

5



p

(.5 M p

x M p

R+ R& R1

6M p> L 3M p> L

" 4

<

Figure !. 9oment #iagram.

solution - shape factor.doc

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