Material Strength2

    

(    ) 

∑F

i

= 0 ⇒ ∑ σ i dA = 0 → ∫ σdA = 0 A

 ' (    )* + .             !"# $%  . , ,-.  ,12 y + ,-. /0  σ M = − ∫ (σdA) y

. 3  $% ,-. ""0  4# /0  5  6 (  0  6 %7"8

:        ! 3      9!  : 3 /1; < = + /1; < 9- % (  % 1> = %6 . @ (0 3 3 " @' 0 6  9 !(   6 ( >14 : .";  (

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A    

ε u = − y tan θ ⇒ δ u = − y ???

e ε = Lim u = ∆x → 0 ∆x

∆x − y tan θ θ y = − y Lim = − y Lim P = − Lim ∆x P ∆x → 0 ∆x → 0 ∆x ∆x → 0 ∆x

ε=

−y P

: # $ %&' ( )' ' * +, ε=

σ E

⇒

−y σ −E = ⇒σ =( )y P E P E

∫ σdA = 0 ⇒ ∫ (− P ) ydA = 0 : 0 B  % A

−

A

E ydA = 0 → ∫ ydA = 0 ' 2!# C8 D + E"F  B  % P ∫A

y=

∑Ay ∑A

i i i

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.";  ( 9 #   -  C :0%

G    

M = ∫(

M =

E 2 y dA P ∫A

M =−

σ =−

−E 2 1 ) y dA = − ∫ Ey 2 dA P P

σ y

Mc I

I

⇒

⇒

M =

EI P

σ =−

⇒

R {


‫ا‬

=

1 M = P EI

My I

. 0 #    6 ( < .";  (   %.0*: C

) . H , I 1#:0  /"( 

σ max = −

M max =

MC I

wl 2 1 × 52 = = 3.125 t.m = 3.125 × 105 8 8

kg.cm

. ,  ' (  .";  ( ( 9 K2  %)

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M    

"; (

∫ ydA = 0 A

σ max =

I=

bh3 30(50) 2 = = 3.125 × 105 12 12

C=

h 50 = = 25 2 2

3.125 × 105 × 25 ⇒ σ max = 25 3.125 × 105

cm4

→ σ max = ±25

kg cm 2

  σ a = 1400 M = σ a S xx ⇒

kg cm 2

kg cm 2

< 4 /"( % 

wl 2 εσ S 8 × 1400 × 81.9 = σ a S x ⇒ w = a2 x = 8 l 900 2

w = 5.73

σ a = 1400

kg cm kg cm 2

= .573 , w =1

ton m ton m

(N

, lmax = ?

1

1

wl 2  8σ .S  2  8 × 1400 × 81.9  2 M = σ a .S x ⇒ = σ a .S x → l =  a x  =   ⇒ l = 303 cm 8 10  w   

σa =

M M wl 2 10(400) 2 →S = = = ⇒ S x = 142.86cm3 Sx σ a 8σ a 8 × 1400 INP

160

117

INP

180

161

(

cm3 cm3

(@

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O    

wl 10 × 400 V 2 % /"( τ = ⇒ τ max = 2 = ⇒ AV h.tv 14 × 0.5

: ) . 3(  I (L ) 3 6 %.0* . , σ a = 50

kg cm 2

0 < 4 /"( % .  ) .  H 

σa = − l=

Mc Wl 2c =− I eI

8 × 3.125 × 105 × 50 10 × 25

⇒

⇒

l2 = −

8Iσ a wc

lmax = 707 cm

:) . σ =?

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P    

M max =

."; ( ∫ ydA = 0

→y=

wl 2 200 × 2 2 = = 100 kg.m 8 8

∑Ay ∑A i

i

i

A

I = ∫ y 2 dA = ∑ I oi + ∑ Ai d ?? =

=

30 × 5 × 2.5 × +3 × (40 − 5) × 22.5 = 10.73 cm 30 × 5 + 3 × 35

30 × 53 3 × 353 2 + 30 × 5(10.73 − 2.5) + + 3 × 35( 22.5 − 90.73) 2 12 12

= 35737 cm 4

σ =

− MC 100 × 100 × 10.73 =− ⇒ ‫ 
ري‬ I 35737

σ =−

100 × 100 × (−40 − 10.73) ⇒ ‫آ‬ 35737

σ = −30.02

σ = 14.19

kg cm 2

kg cm 2

:) .  A = 18.2cm 2    I 4 3 S=  I xx = 573 cm → S xx = 81.9 cm C    4 3  I yy = 25.2 cm → S yy = 107 cm

σ max = ? (Q8

 5+ 0   

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R    

σ =

M wl 2 M × (400) 2 = = ⇒ σ = 2442 S 8S xx 8 × 81.9

τ max = 250.6

kg cm 2

kg cm 2

: / % 0-. @ %0V) U S8% ‘  0 ‘  91>  S3T < #   6   ' ! %  -1 . , W ( @13 #:4( S3T +  6 8X %3T  -3  6 < !#   ' !  % %  -2 . #+T  ,  %Y 1F+

[0 %6 < 9 * ' % /"( \ % % %7"8  ' :16  +% 3 %Y ,(  % ZG .  "0  %-F 1F 6  3 F   P  σ p = − A     σ ′ = ± MY N  I

σt = −

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P MY ± A I

]    

σ

A

= −

2000 18.2

1 × 10 5 × −

573

14 2

= −1331

Kg cm

4

,

σB =

− 2000 18.2

10 5 × +

14 2

573

= +1111

kg cm

: #:1( H, I  318T+   ,12 %6 /"( P=10 ton < :Q8 H   2( \ B  < !# P=Pu  I < (^ H,I 'K 90 )B %

_( P=1.2Pu < (N

ES =2.1×106 Eall=7×105

σall=800 σs =2400

kg cm2 kg cm2 kg cm2 kg cm2

δ = ? + σ all = 500  "0 $% % ( P = Fs + Fall Fl F l ∆S = ∆Pl → S = all → Es AS Es AS

 FS + Fal = 10 ton  ⇒   F = 1.44 F all  S

(Q8

Fall = 4.1ton   F = 5.9 ton  S

(^ }?

  2( [ "#[8T Fall = σ a Al = 800 × 25 = 2000 = 20ton

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`    

  2(   Fu = σ s . AS = 2400 × 12 = 28.8 ton ≤ 1.44 × 20 = 25.8 s

."   2(  a8 b# +%6 Pu = 20 + 28.8 = 48.8ton

(N P = 1.2 × 48.8 = 58.56ton

(, 58.5b>48.8 ;%I)., ,# S3  9! %

_( F = 500 × 25 = 12.5ton Fas = 28.8 > 1.44 × 12.5 = 18ton

( Pu = 12.5 + 18 = 30.5ton P2 = 1.2 Pu = 1.2 × 30.5 = 36.6ton Fs 2 = 36.6 − 12.5 = 24.1ton FL 24100 × 200 δ = = ⇒ δ = 0.19cm AE 2.1 × 106 × 12

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($
    ! " #)
 
      

c    

12 3-

42 ' 5 6 7 

N ;    ,8 *  -  F(  91>  0  +% 3  [X( /1;<  ; ,8 *  C :0%  0 p +% 3  , ) = :0% < N ; +% 3 .91>  ., :0% < : 1; %7"8 @+d= "0 %Y e%> M=P. e σ =

P M .y ± A I

:  , %% e%>  3 %6  /"( σ =

P P.e y + A I

.  9 * /"( ' % % + 2 < σ =

P P.e y + =0 A I

⇒

y=

−I Ae

, % 'T  /"( 0 , ; 8 =

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Z  %#<  0 +% 3 b# P 0 = 0 6  ' 3 8 = #  " ,d> .";   # /6 0 e ,#:0 % < N+%; % .  % Z     .";     .   b#:3 C :0%  .";  \  # /#: e% +  6; + C :0% <    6 < !# +  P :0% < N ; 0 +% 3 %Y 3 0 =  . ,% 6; f   6 )* ' :16  /1; 33 % e%>  6 )* 1;  6%7"8  6 /# 13 ez , ey   P %Y 3 f DKg %  . f #  .  "6; ,P.ey P.ez %% h (%(  >  z , y :  , %% z, y f DKg  e%>  <  3  "#T % 1> /"( σ =

P P.ez P.e y + + A Iy Iz

:) . C 88 %6 , @ 9 !( B 3    88 + ' iF  ABC ^  .  d = 27.3 cm F ; % + I = 8820 cm 4

Kg8 + K , A = 103.9 cm 2 

H = 1.8,  L = 2.4m  P = 1350kg $%   ^    + 0  6 /"( %.0*

.  "0  5

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A    

:9*

M B = 1620kgm

' ( AB = H 2 + L2 = 1.8 2 + 2.4 2 = 3m cos α =

H 1.8 = = 0.6 AB 3

, sin α = 0.8

P 2

A < X  1;%7"8 M = R A sin αx = ( )(0.8) x = 0.4 Px = 540 x .   B 3 1;%7"8%.0* ⇒ M B = 540(3) = 1620kgm

 , %%+ , Y AB)B%%  +% 3 ⇒ P = R A cos α =    B   ^     + 0  S"( %.0* σ =

P My P cos α M B y + =− + A I 2A I

   B  3   (    /"( %.0* (σ c )max

(σ c )max

=−

P cos α 2

P cos α =− − 2A

(1350)(0.6) (162000)(27.3) − = 254.6 (2)(103.4) (2)(8820)

− P cos α , %% B  3 K(  (  0 /"( %.0* (σ t ) = + 2A

d MB( ) 2 I

kg cm 2

d MB( ) 2 = 246.8 I

kg cm 2

:) .

/"( %.0* . , @ QD3 mm   9! %     b#  C .  "0 ^ 2* p   %Y ,(   @ bI0      + 0 @

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G    

%% y  < z    1> /"(  % /1; ,( y  )*  # :  , σ=

P σ= + a (a )   2

P Mz + A I

pa z 2p z 4 = 2 (1 + 12 ) a 3 1 a (a )( 3 ) a 12

a    0 /"( %.0* 4

:  , %% +    z = (σ t ) max = σ ( z = a4 ) =

8P a2

a 4

:  , %% +    z = −      /"( %.0* a 4p (σ c ) max = σ ( z = − ) = − 2 4 a

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M    

:) . : < "( ->  # f Dg  "0  5  %#< 9! I  K26 I Z = 27346 cm 4 , I y = 1544 cm 4

e1 = −

e2 = −

A = 94.84 cm 4

Iz 27346 = = 13.96cn Ay (94.84)(20.65) Iy Az

=

1544 = 1.81cm (94.84)(9)

2e2 6 7 81 } .    3.62 + 27.92    <8 b# : K26 #% "

2e

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O    

:) . @ K;  :#%0 ; S73 % @  e%>  0 # " C8 D <  * # "0  + # %  q   k ;.    γ =1.8 t/m3

# " jDg '<+. ,

, 1 .5t/m2 %% #  5  'T %.0* + "0 %

_( ;  # l (  0 H  "0 ^ 2* #  5   1 " + 1#:0     6 /"(. F  L=1m

) # < K12

'(   ,12 # mmm "0 [ ,% %a3   @%B @% ( b# fD .   % / 3 K(  SK3  0 i; '<+ %Y ,(  @%B % ( # e%>    @% k ;  + . , 6;  /"( #%K  #  5

σ =

P M ± A S

Q1 + Q2 %% @ F ,12 '<+ : P

Q1 = 12 h(b − a ) Lγ = 12 (4.5)(2 − 0.8)(1)(1.8) = 4.86 T = c1 Q2 = haLγ = (4.5)(0.8)(1)(1.8) = 6.48 T = c2 P = Q1 + Q2 = 4.86 + 6.48 = 11.34 T

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P    

0 , Z  )* k ;  "#T%+ Q2 + Q1  6 '<+ %7"8 l14  %% M 1; %7"8 ."0  -> #  5 e%>  C :0%< M = Q1 (a +

b−a b b a q h h − ) − Q2 ( − ) + max ( ) = (4.86)(0.2) − (6.48)(0.6) + 16 (1.5)(4.5) 2 = 2.15 T .M 3 2 2 2 2 3 −P M Lb 2 #    S = + A = bl = 2m 2 'I σ = ± A 6 6

σ =−

11.34 2.15 ± = −5.67 ± 3.21 t / m 2 2 0.62

.   B 3  1 "    /"(+ A 3  1#:0    /"( 1#:0    /"( :

σ max = −5.67 − 3.21 = −8.88

1 "    /"( : σ min = −5.67 + 3.21 = −2.46

t m2

t m2

: ) .  6 /"( #<(

=( ,2. , @ @ ' 3 ..   "K  b# <%K b# f -   .   @ @ ' 3 <%(  ^T C 0 K+ B-A     K jDg '<+. # 0 $% :4 % ( b# fD n 0  % 1> ,SF   % (  "S5 . , h=! %K % ( 3  0 AO

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R    

H = V = 12 γh × h × 1  1  H = V = ( )(2.7)(10) (2.7)(1) = 36.4 KN  2    1  W = P =  (1.8)(3)(1) 25 = 67.5 kn  2  

M 0 = H (0.9) − w(0.3) = 36.4)(0.9) − (67.5)(0.3) = 12.5 KN .m

σ =

− w MC 67.5 12.5 × .9 + =− + = −37.5 + 23.2 = −14.3 KN / m 2‫ 
ري‬ 1.82 A I 1.8 × 1 12

σA = −

w MC − = −37.5 − 23.2 = −60.7 KN / m2‫ 
ري‬ A I

:) .  0 M   ,;"!# @%K2   b# 91( % 100∗150 mm I % ( b# ^ 2* 3 6 o+   1; /"( %.0* .  % @ K  %KG 3 6  ( 3 . (  "0 %a3 p% % ( '<+ <) . "0

=(  .";  ( 9  ' 16 %+  "0

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]    

M =

WL 4 × 3 = = 1.5 KN .m 8 8

IZ =

1 (100)(150)3 = 2.81 × 107 12

σ A = +3.47 + 3 = 6.47

cos 30 } 3 M Z = M cosα = 1.5( ) = 1.3 KN .m 2

1 (150)(100)3 = 1.25 × 107 mm 4 12 M z y M Y Z 1.3 × 106 σ =± ± = IZ Iy

Iy =

N mm 2

‫آ‬

σ B = +3.47 − 3 = 0.47

N mm 2

‫آ‬

σ c = −3.47 − 3 = −6.47

N mm 2

‫ 
ري‬

σ ِ = −3.47 + 3 = −0.47

N mm 2

‫آ‬

:) . .  "0 -  C,B,A bI0 ,# S3  q:F   %YX 1#:0   6 /"( ( % -  3   #  S"( [ 1()

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`    

(32 7 3-  6 )32 7  - 809

:F ( ""0 %Y .. q:F < AB C +  #  0 # 6 /"(

=( , ^ () . .3 1   ) =(  bI0

F1 = 3 cos α

F2 = 2 cos α F3 = 2 sin α

F4 = 1sin α

→+

∑ Fx(Fn) = 0 N = F1 cos α − F2 sin α − F3 cos α + F4 sin α ⇒ N = 1.29 N

∑ Fs = 0

S = F1 sin α + F2 cos α − F3 sin α − F4 cos α ⇒ S = 2.12 N

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Ac    

σ X = 1.29 N

∑F X′=0

τ x = 2.12 N

N = F1 cos θ − F2 sin θ − F3 cos θ − F4 sin θ = 0

σ x′ dA = σ x cos 2 θ dA + τxy dA cos θ sin θ + τxy dA sin θ cos θ + σ y dA sin 2 θ

 "!  p*  dA %B < σ x′ = σ x

(1 + cos 2θ ) (1 − cos 2θ ) + τxy sin 2θ +σ y 2 2

%7# r   b# < "( 9#-( σ x′ =

σ x +σ y

τ x′y′ = − σ y′ =

2

+

σ x −σ y 2

σx +σ y 2

−

σ x −σ y 2

cos 2θ + τ xy sin 2θ

sin 2θ + τ xy cos 2θ

σ x −σ y 2

cos 2θ − τ xy sin 2θ

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(1) (A) (G)

A    

.#T  , (G) r +  6  % θ = 90 + θ , θ ο  F () r ."0  W ,SF o , 9- 3 ' 16  σ x′ , σ y′ ,τx ′y ′ σ x′ + σ y′ = σ x + σ y = σ 1 + σ 2 = , Y

(1 + 3) ⇒

(:') ( min, Max )  -

.  1#:0  σ x′ \ θ  I q<  σ x′ =

σ x +σ y 2

+

σ x −σ y 2

cos 2θ + τ xy sin 2θ

 6 /"(  6 % % + 2  9 * 4 K3 + #% 7 sK θ  ,-23 ()r < %  ."#T  ,  ( min, Max )  σ x −σ y dσ x′ =− 2 sin θ + 2τ xy cos 2θ = 0 dθ 2

2θ1′

τ xy  tan 2θ = (σ x − σ y )  2

2θ1′′ = 180 + 2θ1′

θ ′ = θ  ⇒ .3 pdK; 90 0 %7#!#  1 " + 1#:0  /"(   θ ′′ = 90 + θ ′ 1  1

.(31> 6% S"( f ) ."#   6 /"(  1 " + 1#:0   S"(

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AA    

:' 3-

(PRINCIPAP) σ x′ = σ 1

Max

σ y′ = σ 2

Min

???? A =

(σ x − σ y ) 2 2

+ τ 2 xy

(σ x − σ y ) 2

cos 2θ = (

σ x −σ y 2

sin 2θ =

(II) ) 2 + τ xy 2

τ xy σ x −σ y   

2

2

(I)

 + τx 2 y 

.# (3) + (1)  II + I o+ 7# F  (σ )

max x′ min

= σ1




σ2 =

σx +σ y 2

σ x −σ y ±  2 

2

  + τ xy 2 

. "0  @ K  #<    K3 #+< % 

 3- ;2'0< τ x′y′ = −

σ x −σ y 2

sin 2θ + τ xy cos 2θ

(2)

. 6% % + 2  sK  + #% 7 sK \ θ  ,-23 (2)  < % 

dτ x′y′ dθ

=0⇒

(σ x − σ y ) tan 2θ = −

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2

τ xy

=−

σ x −σ y 2τ xy

AG    

 F tu + 2   S"( f   9* #+%.0* % /"( f   #+< .  

WD %.0* %  S"( τ

max min

σ x−σ = ±  2 

2

y

  + τ xy 2 

(3)

:  3%% ""0  91> %.0* %  6 /"( f  +  0 1   6 /"( σ′ =

σ x +σ y 2

!"# %7  6; @%16   /"( b#  16 % /"( %.0*  #% " σ x +σ y = 0

\  6 % (G)   σ y , σ x 14 2 43

‫ا‬

% 

‫
ي‬

: @ 73T  % + 2 τ xy \ 6 % (3)      6 /"( %   τ max =

σ1 − σ 2 2

.  % % /"( % % 'T  (σ 2 , σ 1 )   S"( 0    0=−

0=−

σ x −σ y 2

σ x −σ y 2

sin 2θ + τ xy cos 2θ

tan 2θ + τ xy → tan 2θ =

τ xy → σ x −σ y 2

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AM    

., % 'T %  S"( 0  

., K% % 60KN 0 +% 3 %Y ,( \850mm2  C     () . . "0  5     ,-23 30 0 r#+<    +   % + 1> /"((Q8 ., I % /"( 1#:0  (^

θ = 120 = 90 + 30 ⇒ 2θ = 240

(Q8 (9*  'T # @ '+ F 30 0   \ f DKg @ 7K  ,-23 "#7%  ., θ = 30 0 σx =

70.6 }

σ x′ =

}0

σ x +σ y 2

P 60 × 10 3 = = 70.6 A 850mm 2

70.6 }

+

N ← ( /"() # /"( 6x ,SF mm 2

}0

σ x −σ y 2

cos 2θ + τ xy sin 2{θ = 17.65 1> { 240ο

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0

AO    

70.6 }

τ x′y′ = −

σ x −σ y

τ max

τ max =

}0

2

}0

sin 2{θ + τ xy cos 2{θ = 30.6 N 240ο

σ x −σ y =  2 

70.6 = 33.3 2

240ο

mm 2

(%)

2

  + τ xy  {0

N

(^

mm 2

.#+T ,     6 /"( + f SF () .

σx =σy = 0

.# % /"( o+ 3% %% σ y ‫ و‬σ x 4"# 

}0

σ 1 ,σ 2 =

tan 2θ =

σ x +σ y 2

τ xy }

}0

0

}0

σ x +σ y ±  2 

= ∞ = tan 90 0

σ x −σ y 2

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σ 1 = τ xy 2    + τ xy 2 ⇒ σ 1 , σ 2 = ±τ xy →   σ = −τ xy  2

2θ ′ = 90 → θ ′ = 45 0 1  1   2θ ″ = 270 0 → θ ″ = 135 0 1 {  1 180+ 90

AP    

σ x′ =

σ x +σ y 2

+

σ x −σ y 2

cos 2θ + τ xy sin 2{θ

2× 45ο

σ x′= τ xy sin 90 = +τ xy

.#+T , θ = 60 0  /"( () .

σ x = −30

σ x′ =

σx +σ y 2

+

σ x −σ y 2

cos 2θ + τ xy sin 2θ

σ y = +20

τ xy = −10

σ x′ =

− 30 + 20 − 30 − 20 N + cos120 − 10 sin 120 = −1.16 2 2 mm

θ = 60

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AR    

 - =8'

R = τ max

σ x −σ y = OM = ON ′ ⇒ R =  2 

2

  + τ xy 2 

σ 1 = σ ave + R → ‫
آ‬ σ 2 = σ ave − R →  σ x ≠ 0   τ max K+ σ x′ = σ ave =

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σx +σ y 2

A]    

- 0>  ?<

σ 1 > σ 2 > σ 3 :  %

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A`    

9!   @ @ ' 3 "( ,8 * % () . #< 2   @%# (Q8  "0  5    6 /"( (^ #+T ,  v%   /"( + %.0* % /"( (N   /"( 0   @# 9!  F( 

6x K  0 , ,1 F+ % (,-.), 0     6 %> pd; C q:F  '+  9# 1(  ,1  F+ ' 16 % %YX % /"(  %#<+    , p%B  + ( σ x ,−τ xy ) f DKg  x 4 K3  (,-.) ,>  %  % 

σ y′ = −10

σ x = 50 X

Y

− τ xy = −40

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τ xy′ = 40

Gc    

σ ave =

σ x +σ y 2

σ x −σ y R =  2  R = 50 N

=

50 − 10 N = 20 2 mm 2

2

  50 − (−10) 2  + τ xy 2 =  2  

  + 40 2 ⇒ 

mm 2

R = 30 2 + 40 2 ⇒ R = 50

σ 1 = σ max = σ ave + R = 20 + 50 = 70

N mm 2

σ 2 = σ min = σ ave − R = 20 − 50 = −30

N mm 2

40 → θ = 26.6 0 30 2θ = 53.13 tan 2θ =

τ max = R = 50

N 1#:0  % /"( mm 2

σ ′ = σ ave = 20

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N   /"( mm 2

G    

, ^ %#< 9!  @ @ ' 3  /"( ,8 * % () .   6 /"( +  f  (Q8  q:F % %YX /"(  6 8X (^ pd;  \@ @ C q:F '+ < ,  F 300 1 ,>   6 %> .#T 

σ ave =

σx +σ y

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2

=

100 + 60 = 80 N mm 2 2

GA    

CF = OF − OC = 100 − 80 = 20

N mm 2

R = (CF ) 2 + ( FX ) 2 = (20) 2 + (48) 2 = 52

tan 2θ =

− 48 = 2.4 → 2θ 1 = −67.4 0 20

N mm 2

, θ1 = −33.7 0

σ max = OA = OC + CA = 80 + 52 = 132

N mm 2

σ min = OB = OC − BC = 80 − 52 = 28

N mm 2

σ x′ = ok = OC − KC = 80 − (52) cos 52.6 = 48.4

N mm 2

σ y′ = ol = OC + CL = 80 + (52) cos 52.6 = 111.6

N mm 2

φ = 180 − 60 − 67.4 = 52.6 0 τ x′y′ = k x′ = 52 sin 52.6 0 = 41.3

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N mm 2

GG    

. "0   + /"( ,8 * %   @%# () . . "0

=(    θ = 20 0 % 0 %D"> lde +   τ θ′ , σ θ′ ,τ θ , σ θ  6 /"(

σ x = −840 σ y = 280 θ = 20 0 → 2θ = 40 0 σ ave =

σx +σ y 2

=

280 − 840 kg = −280 2 2 cm

R = 840 − 280 = 560

kg cm 2

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GM    

τ θ′ = R sin 40 = 560 × sin 40 = 354.46

kg cm 2

σ ave + R = σ max σθ =

σx +σ y

+

2

σ x −σ y 2

cos 2θ + τ xy sin 2θ

σ ave − R = σ min σθ =

τθ = −

kg − 840 + 280 − 840 − 280 cos 40 = −708.98 2 + 2 2 cm

σ x −σ y 2

sin 2θ + τ xy cos 2θ =

σ θ′ = σ θ ′′ =

σ x +σ y 2

−

σ x −σ y 2

− 840 − 280 sin 40 = +359.96 2

cos 2θ + τ xy sin 2θ

− 840 + 280 − 840 − 280 − cos 40 = 148.98 2 2

:[%( ' # 5 . "0 -   @%# =  ,8 *   6 /"( () . (σ 3 = 0) 6  % % + 2 %3T   K3 σ 3 K+

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GO    

σ ave =

50 + 30 = 40 2

X =

R = (10 2 ) + 20 2 = 22.36 ≅ 22..4

N mm 2

Y=

50 − 20

30 20

σ 1 = σ ave + R = 40 + 22.4 = 62.4 σ 2 = σ ave − R = 40 − 22.4 = 17.6

 =8' @7  < 9!  F(  ( θ a")   6  + 6 /"( ( Q8 .  @%0 '+ 45 0 @<3  0 %D"> +   6 /"( (^ 1#:0  %  S"( (N . 6 ' 3 @%0  5 '+ % ">  6 9! +  w# K3

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GP    

σ x = 1120 σ y 420 τ xy = 280

2τ xy

Q8 )

2 × 280 tan 2θ = = = 0.8 σ x − σ y 1120 − 420

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2θ = 38.65    0 0 θ 1 = 19 .32 = 19 .20′   θ 2 = 90 + 19.32 = 109 0.20′

GR    

‫اﻝ‬

2  σ +σ y σ x −σ y  σ , σ = x  + τ xy 2 ±   1 2 2 2     1218.21 kg 2 ← σ1  1120 + 420   1120 − 420  cm 2 2 σ σ = ± + = , 280     1 2 2 2   321.79 kg 2 ← σ 2  cm  

max σ x min

σy

1120 + 420 1120 − 420 kg  + cos90 { + 280 sin 90 = 1050 ‫) ب‬σ θ = 2 2 cm 2 2θ    }90 τ = σ x − σ y sin 2θ − τ cos2θ = 350 kg xy  θ 2 cm 2   ‫ب‬  ′ 1120 + 420 1120 − 420 cos 270 + 280 sin 270 = 440 +  σ θ = 2 2   θ = 90 + 45 = 135 ⇒    1120 − 420  τ θ′ = sin 270 − 280 cos 270 = −350 2      2θ = 270

2θ = 38.56 0 = 38 0.33′  2 × 280 ‫ ج‬tan 2θ = = = 0.8 →  σ x − σ y 1120 − 420 2θ = π + 38.56 = 128.65 = 123 0.40 → θ = 64 0.20′  2 2τ xy

# + σθ =

τθ =

σ x +σ y

+

2

σ x −σ y

τ max =

2

2

cos 2θ + τ xy sin 2θ = 770 kg

sin 2θ − τ xy cos 2θ = 448 kg

σ1 − σ 2 2

σ x −σ y

=

cm 2

cm 2

1218.21 − 321.79 = 448.21 kg 2 = 44.8 N cm mm 3 2

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G]    

2θ = −51.34 = π − (51.34) = 128.65 → θ = 64.32 0 1120 − 420 − 1  tan 2θ = − = → 2 × 280 0.8  ο 2θ = 128.65 + 180 → θ = 154.32

%  .%  % h  S6  !(  x  210 c f%* F  :   b#() . ^ 2* pq 9#     % + 1>  6 /"(  # /#: 930 c  f%* F . "0

 'T > 4( h#%e +

α = 11.7 × 10 −6 c −1

  : (%* v 2-3 h#%e . "0 $% E = 2.1 × 10 6 kg

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cm 2

G`    

   K2 p%B + 'I 8+ %0   5 )B  #< δ = lα∆T @<3   
PL AEδ ⇒P= AE L

θ = 30 0 ⇒ 2θ = 60 0

σX

P = = A

AEδ

L = Eδ = E ( Lα∆T ) = Eα∆T A L L

or ε = δ l ⇒ σ = E. ε{ = Eα∆T ε =α∆T

εX = 0 =

or

σX E

+ α∆Τ ⇒ σ x = − Eα∆Τ

σ x = Eα∆T = −(2.1 × 10 6 )(11.7 × 10 −6 )(93 0 − 210 ) = −1769 kg }0

1> /"( σ x′ = σ x′ =

σ x +σ y 2

cm 2

= 176.9 N

mm 2

}0

+

σ x −σ y 2

cos2{θ + τ xy sin 2θ 60ο

− 1769 1769 + cos 60 = −1326.75 kg 2 = −132.67 N cm mm 2 2 2

"( τ x′y′ =

}0 −σ x −σ y

2

}0 + 1769 sin 2{θ + τ xy cos 2θ = sin 60 0 = 766 kg 2 = +76.6 N cm mm 2 2 60

i θ = 30 + 90 9- y2   θ = 30 4"#

: ) .

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Mc    

    6 /"( ,SF +  .   3 b#  + + 6 /"( ,SF+   "0

=( : 3

dA %% AB C ,* 2.3 F+ % /"(AB   C +

:#T  ,  ( /"() σ 1 (  2#3  s K OAB%D"> ) =( 8 =

(

)

F1 dA = 2(2 p)dA cos 60 0 + 2 3 p dA cos 30 0 ⇒ σ 1 = 5 p1

dA %mmm% :mmm 3CB+ OACmmm ,* 2mmm Emmm5 ldmmme  + 2mmmK OABzmmm. 'mmmI .3 F+ % /"( OH.  

  K  ) =( 8 =

σ2

 3 3 1  + 3 pdA  = 0 ⇒ σ 2 = P dA − 2 PdA  2 2  2 

9!m %D"> +    6 /"(+ "     +   6   +  #% " ., @ {g %#<

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M    

+  m% /"( .  @ 0 15t +% 3    @%#  C     () . . "0 ^ 2*    % ."0 <+ 4( 600 kg

τ=

VQ = It

cm 2

< # -3 } < b# | 6

R2 R ×4 2 3 = 600 → V × 4 = 600 4 R 3πR 2 π × 2R 4

V ×π

%m 50KN m  m +% 3 ,( 25mm e+ %       b# () . 300 m#+<  m +%m 3 %Y o;  ,-23 0   +   % + 1>  S"(. . "0  5 6  9 !( σ = −20 Mpa τ = −94.6

. % 9! s  C /"(  %D">  "0

=( %6 @%# < @ K  (Q8   6  + S"( ., @%0 '+ 450 @<3  0   +  S"((^ .1#:0  %  6 /"( (~

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MA    

σ θ = 770 % 0  % σ θ  + c @%# :0% (9* σ θ = 770

%   C A l = + C :0%  @%# E .   A

3 9 τ θ = −230

C A = 448kg / cm 2

σ 2 = 770 − 448 = 322, σ 1 = 770 + 448 = 1218   S"( %6  @%# € % ZQ8

   190,200   + x   #+< #% " 2θ = 38 0 ,40′ #+< + θ = 19 0 ,20′ + 90 0 = 109 0 20′

Dm3  m %6 +  , @%0 '+ 450 @<3  0   +  6 /"( (^ .  {g 2θ = 90 0.  D m3 f DmKg < .m  m 90 ο − 38 0 40′ = 510 20′ %% σ θ  + CD o;  #+< :m 3 D ′ m3 m6 /"m(.m"#T Em , τ θ = 350 kg

cm 2

.(N 9! "3 ) .   τ θ = 350

, σ θ = 1050 kg

kg cm 2

cm 2

, σ θ ′ = 440

 S"( kg cm 2

9! "3 ) @

=( E ′, E v 3   B%  + 1#:0  %  6 /"( (~ .(

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MG    

:    A

> 4( " h  + "0 +% 5 k6 '3  < % ( @  0 "K26 W  K+ %#< o+ :K!3 .  bI0  ; % (

-6' – C- 7 -6' – 2 B'7 > 4( "" 9! % _( .";  ( ( % ( 3 QKg  6 (   :y

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MM    

/1;  % ( < = b# 9! % _(

y

∆u = − y∆θ  K% %a3   p%B  y ,-. ,SF %  x

 8+ )B @"6 ' 3 ∆s € )B #% "  13 3%0 3 | 6 ,( ab C D′C ′ + A′B′  +  F  6 (

∆u

lim ∆s ∆S → 0

= − y lim ∆S → 0

du =ε ds

∆θ ∆s

#

du dθ = −y ds ds

 % .";< y    0   % ( <  ( ; /3%0 ' 16  ρ∆θ  %B< ∆s =  ∆θ = ∆ s  p

du ds = dθ − y ds σ

ds

}ρ }E ∆θ dθ 1 −ε = = =k= lim ds p y ∆S → 0 ∆s

(q "3 l = E!>) ( 5 0):K my I

}

M

1

*ρ

=−

σ Ey

M

→*

1

ρ

=−

M → EI

d 2v M =− 2 EI dx

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M

1

*ρ

=

d 2V dx 2

MO    

*σ = −

εy ρ

<+ @mT ,  {8 ; /1; <  3  6 9! %

_( < @ K  o : o+ :K!3  m6 9!m %m

_( '%m0 %a"% ., @#% %a"% S3T  i% <  3  6 9! %

_( #  S3T  0  F+ 6 # ".K  8+ , /g ,# e %.0  i%<  3 .,% %a3   %  6 9! %

_(

 DE '   A  /2<  'F8 >

1

ρ

=

d 2v dx 2    dv  1 +   12dx3     

 2    

3

= 2

v ′′

[1 + (v′) ]

3 2 2

⇒

1

ρ

≈

d 2v * dx 2

:q "3 .  ,-.  6y ,1  'T %=( 0 , ,-. K+ "" b#

. $% ,-.  p%B  V %  ) . %  $> m # v ,d>  % % 

d 2v M =− 2 dx EI

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MP    

d 2v M =− 2 dx EI

→

M = − EI

d 2v = − EI dx 2

% V = − EI

} v′′

‫" !م‬

dm = −V dx

d 3v = − EIv′′ dx 3

(@%K2  )+% 3 q = − EI

d 4v = − EIv ′′′′ dx 4

 B8' % @  !( Z v ( 0) = 0   x = 0  θ = dv = 0  dx

‫" م‬ #‫ﺵ‬

(!Kn Z D)@  @  !(ZA @  !( 9 x = 0 ‫" م‬  → V (0) = 0 # h  x = 0 → M = 0 
:  K  #<  6 h  % ( > 4( "" 0 (

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MR    

tan θ = V ′

k=

1

ρ

=

θ = Arc tan v ′

,

dθ d ( Arc tan v ′) dx = ds dx ds ds  dv  = 1 + ( ) 2  dx  dx 

ds 2 = dx 2 + dv 2 1

2

[

= 1 + (V ′) 2

d v ′′ ( Arc tan v ′) = dx 1 + (v ′) 2 (1), (2) ⇒ K =

1

ρ

=

dθ = ds

]

1

2

( 2)

v ′′

[1 + (v′) ]

3 2 2

(3)

, q = 0   :0%1K  % 

% @ K  %‚< f   > 4( U 6%‚( 9‚23%‚ f  = 4"‚  ) * :‚6‚

v = > 4( "" ' ƒ %

_(

dv = v′ = > 4( "" h  dx ‫" !م‬ } d 2v M = EI 2 = EI v′′ = /1; dx

θ=

d 3v = EIv′′ = i% dx 3 d 4v q = EI 4 = EIv′′′′ =   dx

V = EI

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M]    

) .   + EI , Y 1; Kg + L  @%B % ( b#  4( "" r

=( ,2

v(0) = 0  <% o#%  θ (0) = 0 

1; %7"8 EI h  EI EI

M ( L) = + M 1  ← x = L  <% v%  V ( L) = 0 

d 2v = M = M1 dx 2

dv = M 1 x + c3 → EIv′(0) = C3 = 0 dx

dv = M 1 x → EIv = 1 M 1 x 2 + c4 + c3 x 2 dx EIV (0) = C 4 = 0 ⇒ v = M 1 x

2

2 EI

 ,1  > 4( "" ' ! %

_( 06  ' 3 @T ,  4 K3 ,-. ,d> .    ' # +

M1L + 2  4( h  .6  „ X=L  v #%K : EI

. 

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M`    

) .  m"# 13  91* + "0  ; %3T   ,3+  +  %K 12 ( T 91* % x 1= /"m( m  50cm( mT m; m; %+ % 91* '%0 ;  +20mm ( T ! ( E st = 2.1 × 10 5 N

mm 2

  13 -  #T  F ( T  0 1#:0 

3 #%(    y = 10mm

σ=

− yE P

> 4( "" $% 0 6  ' 3 9-  (1) 8 =  (3) 8 = 2#  :K!3 .   *+  2#   ( V ′ )2 '13 %3 p%  ) = ,g(

: ) . E I %m ( 1; "" ., $+%  5 p%B  ,;"!# @%K2   ,( @  % ( b# % ( > 4( ""

=( , ^ ., , Y 'T 3 6 )B [ 1(  (.  

ϖ =0 q=0

f 'T   -3 :0%1K   % ( b# % )

[+ F 9 23%# 8 = < @ K  (Q8 [ SI F 9 23%# 8 = < @ K  :^

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Oc    

RA + BB = −ω0 L ↵− ∑ M

A

⇒ −ω0 L L + RB( L) = 0 2

=O

EI

ω οLx ω 0 x 2 d 2v = M = − 2 2 dx 2

EI

dv ω 0 Lx 2 ω 0 x 3 = − + C3 dx 4 6

EIV =

( A)

ω 0 Lx 3 12

−

ω0 x 4 24

RB = wο

L 2

}0 + C3 x + C 4

 6  %  C3  A   x =0 } v(0) # EIv(0) = 0 ⇒ C4 = 0

EIv( L) = 0 ⇒ C 3 = −

ω 0 L3 24

+ ω0 x 4 + ω0 L4 − ω0 3 + C3 x = +c L⇒ v = ( Lx − 2 L3x + x 4 ) ( B ) 3 24 24 24 EI at x = L / 2 ⇒ vmax = Copyright by:www.Afshinsalari.com

5w0 L4 384 EI

O    

 SI F 9‚23%‚ 8 = < @ K  9* ^) EL

d 4v d 3v = q = −ω 0 → EL 3 = −ω 0 x + C1 dx 4 dx EL

d 2v ω0 x 2 + c x + c = − 1 2 2 dx 2

, EIv′′(0) = 0, M (0) = 0 , or , or ELv′′( L) = 0 = M ( L) = 0

<% i+ 

=−

ω 0 L2

2

+ C1 L ⇒ C1 =

ω0L

⇒ C2 = 0

2

 B    # ' % 6  „ x = L 2  ' ! %

_( %.0* :#T  ,  ' !%

_( %.0*    Q8 ,8 * s  y2 8 -3 v max =

5ω 0 L4

384 EI

DE ' G > 

:) . . "0  5  %#< … % b# + % %2!# % (  S"0+

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OA    

@%K2   q : EI (1) V = EI

d ′′v = −ω 0 dx ′′

d 3v = −ω0 x + C1 dx 3

(2) ."; %7"8 M = EI (3) h  θ : EI

d 2v ω x2 + C x + C =− 0 1 2 2 2 dx

dv ω x 3 + C1 x 3 + C x + C =− 0 2 3 2 2 dx

(4)[% %

_( v : EIv = − ω0 x 24 + C1 x 6 + C2 x 2 + C3 x + C4 ⇓ 4

3

2

v = 0 ⇒ C 4 = 0  x = 0 dv θ = dx = 0 ⇒ C3 = 0

4  ω 0 L4 L3 L2 + C1 + C2 =0 v = 0 ⇒ − 24 6 2  X = L  2 ω 0 L2 M = 0 − ⇒ 2 + C1 L + C 2 = 0 

)S4 A 8 = A : C1 =

5WL 8

+ C2 =

− WL2 8

V = dv ω = v′ = θ = 0 dx EI

ω0  x 4

5 Lx 3 L2 x  + − −  [% %

_( 8 = EI  24 8 × 6 16 

 x 3 5 Lx 2 L2  −  − + 16 8  6

 x2  x1 = α 5 Lx L2  v′ = 0 → x + −  = x = β 16 8   6  2

" %7# + ,-. !# ^FA , )- 9  ,-. ^F

 x = o ⇒

M ( 0) = M A = −  

i% V = + wo − x + V ( L) = Rb =

L2ω0 8

5ω 0 L 5L  → V (0) = − Ra =  8  8

3ω0 L 8

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OG    

. "0  5  q ,;"!# @%K2   %Y ,( @  % ( b# > 4( "" ., % 3 6 o+   4# "" h  M =

qLx qx 2 − 2 2

†I @  !( < x    1; %7"8

d 2v − M qLx qx 2 dv qlx 2 qx 3 ′ ′ = ⇒ EI V = − ⇒ EI = − ± + C1 EI 2 2 dx 4 6 dx 2 V′ = 0

(x = L 2 ‫ ⇒ ) در‬V ′(12 ) = 0 ⇒ C ( A)

1

. 6 %A 8 =   C1  EIv ′ = .#%  )%7 K3 A< EIV =

=−

( A)

qL3 24

− qLx 2 qx 3 qL3 ± + 4 6 24

− qLx 3 qx 3 qL3 x ± + + c2 12 24 24

V = (0) = 0 ⇒ C 2 = 0 ⇒ % ( > 4( "" 8 = V =

v =0

x =0 

qx (− L3 ± X 3 µ 2LX 2 ) ( B) 24 EI

m B m8 =  x = 1 2 ' %m m+ m -  % ( 3 6 o+  ( δ ) 1#:0  ' ! %

_( #T  , δ = Vmax

EI

5qL4 = 384 EI

d 3v = qx + C1 dx 3

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d 4v EI = 4 = q %7# #%B dx EIv′′ =

qx 2 + C1 x + C {2 2 0

OM    

"   % % (  SK3 +  1;  6%7"8 + 6 ' !%

_( EIv′ =

qx 3 x2 + c1 + c2 x + c3 6 2

⇒ v ( 0) = v ( L ) = 0 v (0) = v(l ) = 0

EIv = ⇒

c1 =

qx 4 x3 x2 + c1 + c2 + c3 x + c4 { 24 6 2 0

− ql 2

EIv = −

,

c2 = 0

,

c3 =

ql 3 24

c4 = 0

qx 4 ql 3 x qlx 3 + + +0 24 24 12

:) . @%K2   ,( 0  @%B % ( 1#:0  h  + 1#:0  ' ! %

_( + > 4( "" .#+T  ,    ,;"!#

 − q( L − x) 2 M = 2  V = q ( L − x)  6 7v′′8 2 d 2v EI ( 2 ) = M = −q ( L − x) 2 dx

}v′ 3 qL3 dv EI ( ) = M = −q ( L − x) + c1 , % %% h  @  !(  V ′(0) = 0 ⇒ C1 = 6 dx 6

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OO    

qx (3L2 − 3Lx + x 2 ) → 6 EI qx 2 v= (6 L2 − 4 Lx + x 2 ) + C2 24 EI

(A) v′ = ( B)

EIv =

q (l − x) 4 + C1 x + C2 24

&‫'
دﻝ‬  → v = v(0) = 0 @  !( 9 * v: ⇒ c 2 = 0  

qx 2 (6l 2 − 4l x + x 2 ) 24 EI

X=L "#:7# F  0    % ( 
_( #T  , (B)

δ = v( L) =

qL4 8 EI

( A)

θ = v′( L) =

qL3 6 EI

(. 9   'F8 > @7 EI

d 4v d 3v = q → EI = qx + C1 dx 4 dx 3

(k )

:%#< f  * v%    % x =L  % +% 3 'I d 3v = 0 ⇒ C1 = − qL →  6    "3    +  6  % K 8 = dx 3

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OP    

() . "0  91(  q0 %.0* f  ..   0   @  % ( ' !%

_( "" 8 =  "0

=( ) = , #+% 3 ..   "#T% L

qoL

B   F 2 2 B @  !( <

∑M RA =

q0 L 6

B

= 0 → R A ( L) =

+

⇒ ∑ Fy ↑ = 0

RB = q0 L

EI =

d 2v =M dx 2

M ( x) =

R A + RB =

2

− q0 L

6

= q0 L

q0 L q L ⇒ RB = 0 2 3

3

q0 L q L q 1 x x x − ( x)( )(q 0 )( ) = 0 x − 0 x 3 2 2 L 3 6 6L

EI

q d 2 v q0 L = x = 0 x3 2 6 6L dx

EI

dv q 0 L 2 q 0 4 = x − x + C1 dx 12 24 L

EIv =

qo L L ( ) 3 2

q 0 l 3 q0 5 x − x + c1 x + c2 36 120l

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OR    

q0 L q x = →y= 0 y x L − qοL qx x x x+ ο × × +M =0 6 L 2 3

* o#% v(0) = 0

(1),

v( L) = 0

( 2)

(1) ⇒ C2 = 0 q0 L4 q0 L5 − + C1 L = 0 36 120 L q0 x (−3 x 4 + 10 L2 x 2 + 7 L4 ) v= 360 EIL

( 2) ⇒

: C1 =

7 q0 L3 360

[ SI -(% 9 23%# 8 = < @ K  [+ #%B

EI

q x q d 4v =q= 0 = 0 x 4 l l dx

EI

d 3v q0l 2 = x + C1 2l dx 3

EI

d 2 v q0 3 = x + C1 X + C 2 dx 2 6l

EI

dv q0 4 1 = x + C1 x 2 + C2 x + C3 2 dx 24l

EIV =

()

}0 }0 q0 5 q0 l x + C1 x 3 + 1 C 2 x 2 + C 3 x + C 4 2 120l 6

* o#% v(0) = 0 v′′(0) = 0

3  →

C4 = 0

1  → C2 = 0

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(A) (G)

O]    

q0 L4 q0 L4 7 − + C3 L = 0 → C3 = q0 L3 120 6 × 6 360 2 q0 L qL 1 v′′( L) = 0  → + C1L → C1 = 0 6 6 3 v ( L) = 0  →

(G)   %#  "#:7# F ⇒ V =

q0 x (3 x 4 − 10 L2 x 2 + 7 L4 ) 360 EIL

:) .

=(    M ο :0%1K %7"8 b# %Y ,( 0   @  % ( ' ! %

_( "" 8 = . "0

←+

∑M A = 0

→ RB ( L) + M ο = 0 − Mο L M → RA + RB = 0 → RA = ο L M ο = − RB L

+

∑ Fy ↑ = 0

→ RB =

0≤ x≤a ⇒ M = RA x =

a≤x≤L

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Mο x L

Mο  L x M = M( x −1)  L

0≤ x ≤a 0≤ x ≤ L

O`    

⇒ M = RA x − M ο =

Mο x x − M ο = M ο ( − 1) L L

  13  v2 + v1 h (%(  CB + AC  SK12   ' ! %

_( "" 8 =   EI     EI 

M d 2v =M = οx 2 L dx

0≤ x≤a

d 2v x = M = M ο ( − 1) 2 L dx

a≤x≤L

 }v1′  dv M 0≤ x≤a  EI 1 = ο x 2 + C1 2L  dx   v2′  }  EI dv 2 = M ο ( 1 x 2 − L ) + C a≤x≤L x 2  dx L 2

M0 3   EIV1 = L x + C1 x + C 3    M 1 1  EIV2 = 0 ( x 3 − Lx 2 ) + C 2 x + C 4 L 6 2 

0≤ x≤a

a≤x≤L

:* o#% 1 → C3 = 0 v1 (0) = 0 

v2 (l ) = 0

( 2)

v1 (a ) = v2 (a )

(3)

M M ′ v1 (a ) = v2 (a ) → (4) ⇒ 0 a 2 + C1 = 0 ( 1 a 2 − La ) + C2 2 2L L C1 = C 2 + M 0 a

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(5)

Pc    

( 2) ⇒

M 0 −1 3 1 3 ( L + L ) + C 2 L + C 4 = 0 ⇒ C 4 = −C 2 L + 1 M 0 L2 2 6 3 L (3) ⇒

( 6)

M0 3 M a + C1a = 0 ( 1 a 3 − 1 La 2 ) + C 2 a + C 4 6 2 6L L

C1 a = − 1 M 0 a 2 + C 2 a + C 4 2

(7 )

  4 K3 7 ( 5 o+ 9* < C1 = −

M0 (3a 2 + 2 L2 − 6 La ) 6L

C2 = −

M0 (3a 2 + 2 L2 ) 6L

C4 =

M 0a2 2

v1 =

M0x (6aL − 3a 2 − 2 L2 − x 2 ) 6 LEI

v2 =

M 0 x  1 2 1 3 3a 2 + 2 L2 a2 L  x+  Lx − x −  LEI  2 6 6 2 

0≤x≤a

a≤x≤L



() 9! % Q8 g C 3 ' ! %

_( 0 ,8 *  + % ( 3 6 o+ , M0 %7"8 0 K+ .   ( a = 1 2 )   (A)9! , 6g3 ' ! %

_( c 3

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P    

:) .    P   ,( i
_( "" 8 = . "0 -  : 3  
_( .#+T ,

}v′′ d 2v EL 2 = M = − p ( L − x) dx }v′ dv θ : EI = − P ( Lx − 1 2 x 2 ) + C1 dx }0 }0 EIV = − P ( Lx 2 − 1 x 3 ) + C1 x + C 2 2 3

* o#% V ( 0) = 0 ⇒ V ′(0) = 0 V =

δ = V ( L) =

⇒

C2 = 0 C1 = 0

Px 2 ( x − 3L ) 6 EI

pL2 − pL3 ( L − 3L ) = 6 EI 3EI PL2

θ = V ( L) = − p( L( L) − 1 2 L2 ) = − PL2 + P 2 L2 = − 2 EI

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PA    

:) . +  % 9#  !n @  !( A 3  0  ABC% ( AB ,12 ' ! %

_( "" % ( ' ! %

_( 9 23%# 8 = < @ K   -  ..    ,( BC ,12  ., I i+ 40  AB 3 6  ' ! %

_( 1#:0  . "0

=(

∑ MA = 0 ⇒ LR ∑F

y

B

=

qL L qL ( L + ) : RB = 4 3 3

= 0 ⇒ R A cos 60 0 − RB +

EIv′′ = − M = − RA cos 600 x = −

qL qL = 0 : RA = 4 6

qL x 12

0≤ x≤L

qL 2 x + C1 24 qL 3 EIv = − x + C1 x + C2 72 EIv′ = −

v(0) = 0 (1)

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v( L) = 0

( 2)

(1) ⇒ C 2 = 0

PG    

(2) ⇒ − EIv =

qL4 + C1 L = 0 72

: C1 =

qLx 2 (L − x2 ) 72

qL3 72

(3)

. 6  % % + 2 v sK 1#:0  ' ! %

_( 9

=( % V′ =

ql ( L2 − 3x 2 ) = 0 72 EI

:x =

L

3

.#T  ,  1#:0  ' ! %

_( (3) 8 = x = vmax =

L

3

"#:7# F 

qL4 108 3EI

.   [ SI -(% 9 23%# 8 = < @ K  y2 # 9* %7# #%B EI

d 4v = −q = 0 dx 4

d 3v EI 3 = C1 dx

EIv′ = 1 C1 x 2 + C2 x + C3 2

EIv′′ = C1 x + C2 EIv = 1 C1 x 3 + 1 C2 x 2 + C3 x + C4 → C4 = 0 6 2

:* o#% C2 =0

v′′(0) = v (l ) = 0

} v′′ (0) = 0

qL  qL  L  EIv′′( L) = − M B = −   = C1 L ⇒ C1 = 12  4  3 

' !%

_( "" 8 = 43T <+ % )%7K3 , Y  SI W * v%  SI < @ K  #T  ,  (3) 8 =

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PM    

HF  7 % ' '6  7' #' Pcr =

π 2 EI L2

=

π 2 EAr 2 L2

#:7# F )%  %3T   %K10 [0 %6  "0  ^ 2*  + %6 I y + I x I -  % :I σ cr   @  3 3% /"( \  3%    v% /"(   "0  σ cr =

Pcr π 2 EAr 2 = A AL2

r=

σ cr =

I ;%I l = A

π 2E  L   r

2

"# 'K %n h#%e 

%n h#%e 9  1>   /"( 13 E = 2 × 10 5

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N mm 2

L ,-23 r

PO    

+% 42' 5 

:0%  P 9#-( %

σ max =

P Mc + A I

M max = PVmax + Pe = P(Vmax + e)

σ max =

P  (Vmax + e)c  1+  A  r2 

Vmax = e(sec

  P L Vmax = e sec( ) − 1 EI 2  

P=

Pcr FS

π 2

P − 1) Pcr

FS =

Pcr P

 P  ec P L  ) σ max = 1 + 2 sec( A r EI 2      π P  P  ec σ 1 + 2 sec   max =  A r 2 Pcr   P 1 = Pcr F .S

:) . @ 9 !( 3 @ 9DK %7#!#  # 6    0 9!I  + < 3K ( %K# ) I x = 863.28 cm 4 + A = 22.94 cm 2 B  < b# %6  C . ,

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PP    

.    ( %K =e ) I y = 71.09 cm 4 , I 'K 3%      36ocm 'T )B + @  'K  S6  !( E = 2.1 × 10 6

Kg cm 2

I x = I x′ = 2 I X = 2(863.28)1726.56 cm 4

[

] [

]

I Y ′ = 2 I y + Ad 2 = 2 71.09 + 22.94(5 2 ) = 1289.18 cm 4 I min = 1289.18

Le = L = 360 cm

@  @  !( % + p cr =

p all =

π 2 EI min Le

2

=

(3.14) 2 × 2.1 × 10 6 × 1289.18 = 206171 Kg (360) 2

Pcr = 82.4 2.5

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TON

PR    

 S6  !( + , $+% 2.5m )+ 50 × 100  C  I 'K b# : ) . . , @ $% @  'T p%B + .   σ y = 44 × 10 6

N N , E = 13 × 10 9 2 %  2 m m

'K %n h#%e Z 'K 3%   ZA '  F /"(  3% /"( ,-23 ZG . # 13 91( 2.5 ' " 1B h#%e  3(  'K 0 < 4   %.0*ZM

1 2 hb 12

r=

I A

r=

1 hb 2 b2 b b 50 12 = = = = mm hb 12 12 2 3 2 3

I Y = I min =

%n h#%e =

2) Pcr =

π 2 EA  Lc     r 

2

=

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LC 2.5 × 10 3 = = 173.2 50 r 2 3

π 2 (13 × 10 9 )(0.1 × 0.05)m 2 (173.2) 2

= 2138.50 N

P]    

Pcr 21385  6 N 2 σ cr = A = (0.1 × 0.05) = 4.277 × 10 m  3) σ 4.277 × 10 6  cr = = 0.0972 44 × 10 6  σ y

4) Pall = Pmax =

σ cr = 9.7% σ y

Pcr 21385 = = 8.56 kN FS 2.5

/3 10 6  $%  %  'I % 13 % /3 10  ,+ | 6 /0 ,( ‡> . %  % ;  F  ' 0 %Y    K

N N %   %K A 'K b#:) . \ E = 0.125 × 10 5 2 mm mm 2

/"() σ all = 12

:'K  = K , 6=2.5 + (< 4 ( 3  0 100   (Q8 . "0 91(  ( 3  0 200   (^ P = 100

F .S =

Pcr =

L = 2m

KN

π 2 EI L2

π 2 EI L2

→ Pcr = 2.5(100) = 250 KN

→I =

Pcr L2 (250 × 10 3 )(2 × 10 3 ) 2 = = 8106 × 10 3 mm 4 π 2E π 2 (0.125 × 10 5 )

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P`    

I=

1 4 a 12

→

1 4 a = 8106. × 10 3 → a = 99.3mm ≅ 100mm 12

 P 100 × 10 3 N N σ = = = 10 < 12  2 A 100 × 100 mm mm 2  'K   /"( )%K"0   3 σ = P = 12 = 100 × 10 → a = 91.3m  A A = a2

1 &()

. "0  ^ gK3  %KI0a : K!3 ‰ŠŠ*‰ŠŠ=  ., )- 9  ‰ŠŠ*‰ŠŠ  8 , < 4 /"( < %K!I0 @T , /"( 'I ^) P = 200 KN

Pcr = 2.5(200) = 500 KN

I=

Pcr L2 (500 × 10 3 )(2 × 10 3 ) = = 16210 × 10 3 mm 4 2 2 5 π E π (0.125 × 10 )

I=

1 4 1 a ⇒ a 4 = 76210 × 10 3 → a = 118.1mm 12 12

, < 4 /"( < %K : /"(  'I σ=

A=

P 200 × 10 = = 14 .34 A = a (118 .1) 2 3

P

σ all

=

N mm 2

200 × 10 3 = 16667 mm 2 12

  13 )- 9  @T ,m @<3 € % #   'K m Cmmmm + .   13

=( 'T   ,+ 

a 2 = 16667 ⇒ a = 129 .1 mm

.    h " 130*130mm  b#

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Rc    

@ K;  \ @ @ f Dg  B „  3 b# < %K 2.45 ) AB 'K : ) . . , %  B%   /"(+ < 4    \ A ' " 1B h#%e %8+  < @ K  ( Q8 .  "0 ^ 2* W 'K % %K   ` ,#:0% < N+%;  @T , Q8 ,12  0 < 4    "0 $% (^  @  8( %.0*   /"( + 'K    ' ! %

_( \ f #  % + 'K .  "0  5  'K

A = 2284mm 2

I = 3.33 × 10 6 mm 4

r = 38.18mm

c = 50mm


π 2 EI

σ all =

^)

Le

2

=

π 2 (2.1 × 10 5 )(3.33 × 10 6 ) (4900) 2

Pall 143.75 × 10 3 = = 62.44 A 2284



π



2

:0% < N ;   Vmax = e sec(

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= 287456 N = 287.5 KN

N mm 2

  π P   ) − 1 = 19 sec × 0.707  − 1 Pcr    2 

R    

Vmax = 19(2.252 − 1) = 23.79

σm =

π P  ec 1 + 2 sec( A r 2

σ m = 155.64

π  P  143.75 × 10 3  19 × 50 ) = sec( ) 1 + 2 Pcr  2284 2 2   (38.18)

N mm 2

:) . o#% $%  .   !( ,0%* 9  % n  S6  !( % D SK3 +  !#    \H %0 6;  4# /3 10    ∆t f%* F  e I > 4(  K  )T @# .  "0 $% α    (%* v 2-3 h#%e ∆L = Lα∆Τ =

FL AE

: F = AEα∆Τ

( F = Pcr ) .   Pcr 3%   %% F "0  3 10   K+ Pcr =

π 2 EI L2

= F = AEα∆Τ ⇒ ∆Τ =

π 2I AαL2

:) . +% !"# $%  , @ 9 !( ' 2!# E"F +   b#    + < ABC  5 %;  P   # %K  '(  θ #+< I ,( . % f 'T  ‡> /3 10 %Y  5%* Kg# (0 < θ <

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π 2

) H %0 + 5%; %

RA    

. ""0 3 10 ' :16 BC,AB  + 0  6; 1#:0  K+ P   FAB = (Pcr ) AB =

π 2 EI = P cos θ ( L cos β ) 2

FCB = ( Pcr )CB =

π 2 EI = P sin θ ( L sin β ) 2

:  4 K3  + %B '%0  2( < tan θ = cot 2 β

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⇒ θ = Arc tan(cot 2 β )

RG    

:) .  S6  !(  SK3 +  2.5 × 5 cm  = + 9 K2 %     'K b# %8+ )% 'T % 0 'K L )B #%( @ (0 . ,      b# ,( + @  H , I    W  3%   % .  "0 $% E = 2.1 × 10 6

Kg cm 2

σ PL = 2100

Kg    h "( * cm 2

H  6; I 'T  3%   /"(   120cm )B  'K %  – ^ σ Pl  6  % σ cr  F 

'K  3% /"( σ cr =

π 2 EI L2 = = Aσ Pe

Pcr π 2 EI = A AL2

1 × 5 × 2.5 3 ) 12 = 5140.42 (2.5 × 5)(2100)

π 2 (2.1 × 10 6 )(

L = 71.71 cm

  @ K %8+ )% < L = 120cm > 71.70 'I

^)

σ cr

6 4I‫اق‬-‫ﺡ‬ 7 48 1 π 2 (2.1 × 10 6 )( × 5 × 2.5) Pcr π 2 EI kg N 12 = 750 = 75 = = = 2 2 2 A mm AL 2.5 × 5 × 120 cm

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RM    

:) .  "0 ^ 2*   BC ‡> '%0 3 10 z>  0  P 3%   9!  F(  ( 9D B,C)

→+

∑M

A

⇒

Pcr = FBC

PL − FBC cos 45 L = 0 2

⇒

FBC =

Pcr 2

=

→ FBC =

π 2 EI ( 2 L) 2

P 2 : Pcr =

π 2 EI 2 L2

:) . I p   3%  . % f 'T  ‡> /3 10 %Y  @<  Kg# +% !"# $%  .  $% β=30. #+< .   

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RO    

. ""0 3 10 6  ‡>  %6 0 ,g# 6; +% /3 10 %Y  K+ : @< 

FAD = FCD : # ' ( ,-23 P = FBD + 2 FCD cos β (1) ( Pcr ) AD = ( Pcr ) CD =

1) ⇒ Pcr =

π 2 EI 2

L

π 2 EI Le = 0.7 L

} L ( e )2 cos β

+2

π 2 EI L 2 ( ) cos β

=

20.19 EI π 2 EI ( Pcr ) BD = L 2 L2 ( ) cos β

cos β = 36.1

EI L2

:) . P=4050kg :0% < N ;   ,( 1.80m )B + 5cm × 5cm  =+ %   3K $%  .    2.5m %% + %  6% < !# K  e :0% < N+%;   @% .  #

 'T     /"( %.0*   @   S6  !(  SK3 +  'K !"# .    E = 2.1 × 10 6

Kg    > 4( h#%e cm 2

 mmmm P   +% 3 %Y mm3 E  "0 mmm $%mmm 6; fm X mmmmm )mmmmm* 'mmmmmK /1; e = EO = 2.5cm ,%

   x  )* ABC z. + K %% + x  )* %  + K

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RP    

3

AC (OB) I x = 2( I ABC ) x = 2 =2 12

5 2

)3

12

= 52.08 cm 4

P M max P Pe KL + = + sec A S A S 2

σ max =

S=

5 2(

Ix OB

=

52.08 = 14.73 cm 3 5 2

K=

σ max =

P = EI x

4050 6.04 = 6 (2.1 × 10 )(52..08) 1000

4050 4050 × 2.5 6.04 × 180 Kg + sec = 162 + 805.13 = 967.13 5×5 14.73 2000 cm 2

:) . EI 'mmmmmmK + % ( ."; , - H "0  3 10 CD 'K P   I ,( 9  9! . , %% + %6 )B +    'K % + F   +% 3 K mmmm "0  mmmmm

=mmmm(  CD 3% %%  'T Emmmmmmmmm .  6 mmm %mmm CD 'mmmK

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RR    

C 3  6 ' ! %

_(  <  v%  "0 %a3 p% 'K  9! %

_( < %  .   K3 %#< fD

δ =δ′     

δ ′, δ

δ =δ′⇒

 5 PL4  δ = 384 EI  

,

δ′=

FL3 48 EI

5 PL4 FL3 5 PL = ⇒F= 384 EI 48 EI 8

. #T , Pcr "=# P   3%   6 % + 2 ( Pcr ) cd   F +% 3 %  ) * 5 PL π 2 EI = 8 (0.7 ???)2

:

Pcr = 32.3

EI L3

:) .   . , @ 9 !( I 9* 3  + K + L ) ' 2!#  ‡> < W 9!  5 %; .  "0  5   'T  ‡> < !# /3 10 %Y  5 %; Kg# +% z>  0  Pcr 3%

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R]    

P P  1 → F1 cos 30 = : F1 = ∑ F y = 0  2 3    P P 1 ∑ Fx = 0  → F2 cos 60 = : F1 =  2 3 3

P  2 → F3 = F1 = ‫ 
ري‬ ∑ Fy = 0  3    P 2 ∑ Fx = 0  → F4 = ( F1 + F3 ) cos 60 =  3

.   

P 3

‫ 
ري‬

‫آ‬

‫آ‬

S3T 16   +% 3 + "     ,( ,CD.BC,AB 6‡>

+% K+ 5%; + "   3%  6‡> S3T  %6 , %% : 3 ‡>  # )B 'I . % F1 = F4 =

Pcr 3

=

π 2 EI L2

π 2 EI

3%   %% S3T ; +% 3 0 :# 

L2 :

Pcr =

3π 2 EI L2

:) . z>  0

3%  "   I 9* 3  + K  716 9#V 9!  5%;  ‡> . "0  5   'T  ‡> < b# /3 10 HHHHH 5%; Kg# +%

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R`    

1 ∑ Fy = 0  → F2 cos 45 = P : F2 = 2 P    1 → F1 = F2 cos 45 = P ∑ Fx = 0  2 ∑ Fx = 0  → F3 cos 45 = P :    2 → F4 cos 45 = P ∑ Fy = 0 

‫ 
ري‬

‫آ‬

F3 = 2 P

‫ 
ري‬

‫آ‬

S3T 16   +% 3+ "     ,(

AE,BE,CE,DE  6mmm‡> #% "

; K+ 5%;. ""0  3 10 6  %6 , ' 2!# : 3 S3T )B 'I."   2 P .%

π 2 EI (

F 2 = F3 = 2 Pcr =

L 2

)

π 2 EI (

L 2

3%  %% S3T ; +% 3 0 :#% 

2

)2

: Pcr =

2π 2 EI L2

:) . b# %Y ,( +  % BE,AD,AC  6 'K + % 9! s  W '< - 2F +" fD 6 'K  .    0.1W + 2 (   # 8:8< <  3 )  +% 3 : @<  (   3(  I W '<+ %.0* .    @ @ ' 3  =  8 ; ( H :#%3 +% /3 10 @#5 %Y  .    2.1 × 10 6

kg > 4( h#%e    6 'K E"F cm 2

H"K26 # 5 S3K #T ,+  

=+ /3 10 

= %a3<

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]c    

      o S3T % +   "   D SK3 +  6 'K 'I + →

∑M + →

∑F

x

A

= 0 ⇒ FAC cos 60 − 0.1w − 4 FBE = 0 → FBE = 0.495 N

= 0 ⇒ FAC cos 60 − 0.1w = 0 → FAC = 0.2 w

+

↑ ∑ Fy = 0 ⇒ FAC cos 30 + FAD + FBE − w = 0

;

FAD = 0.33 w

 , !1 /3 10 FBE>FAD +    ' 2!# 'K  %6  0 #  F(   "0  ^ 2*  'K + # < b# %6 /3 10   #% " . 6 „ AC # BE  S3K . ,% 6; f ( Q =e  ) Z  )* 6 'K # /3 10 \ IZ =

12(8) 3 − 9(5) 3 = 418.25 cm 4 12

( Pcr ) BE =

π 2 EI Z L2

=

42.81 t > 0.495 w

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π 2 (2.1 × 10 6 )(418.25) (450) 2 ⇒ w = 86.81 t

= 42.81 t

]    

( Pcr ) AC =

π 2 EI Z L 2 ( ) cos 30

=

32.11 t > 0.2 w

π 2 (2.1 × 10 6 )(418.25) (450

⇒

)2 cos 30

= 32.11t

w =< 160.55 t

%K10 86.48ton < #  w :#%3 +% /3 10 @#5 %Y   : @<  !"# % #% " . %0 6; 3 10 BE 'K % 86.48t %% Œ3 "I  

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