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Relaţiile pentru calculul ariilor suprafeţelor cecompun suprafeţele secţiunilor transversale ale

profilelor din oţel laminat la cald1. Profil L cu aripi egale (STAS 424 - 91)

= ⋅ (2 ⋅ − )= ⋅ (4 − )4= ⋅ (4 − )4

Aria suprafeţei secţiunii transversale a profilului

St = S1 + S2 - 2S3

r1

r1

g

ga-

ga

r

a

S3S2

S1

S3

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2. Profil L cu aripi inegale (STAS 425 - 80)

= ⋅ ( + − )= ⋅ (4 − )4= ⋅ (4 − )4

Aria suprafeţei secţiunii transversale a profilului

St = S1 + S2 - 2S3

r1

g

r1r

g

b

a

S3S2

S1

S3

a-g

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3. Profil I (STAS 565 - 86)

ht

b/4

db

r

R

Panta 14%

R

t

((b-

2d)/4

)*tg

()

t+((

b-2d

)/4)*

tg(

)

b/4

r

/2-

/2-

S1

S4 S3

S2

b

h

(b/4

)*tg

t-(b/

4)*t

g

(b-d)/2

d/2

db/2

(b-2d)/4

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Panta 14% = = arctan(0,14) = 7°58’10,597”= ⋅= ⋅ tan − 2 ⋅4 − − 2 ⋅4= [4 ⋅ − ⋅ tan( )] ⋅ ( − )8= ⋅ tan − 2 ⋅4 − − 2 ⋅4

Aria suprafeţei secţiunii transversale a profilului

St = S1 + 4(S2 + S3 - S4)

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4. Profil U (STAS 564 - 86)

db

t

h

b/2

r

R

Panta 8%

b/2

b-d

b

r

/2-

R

/2-

((b-

2d)/2

)*tg

()

t+((

b-2d

)/2)*

tg(

)

t

h

S1 S2 S3 S4

d

(b/2

)*tg

(b-2d)/2

t-(b/

2)*t

g

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Panta 8% = = arctan(0,08) = 4°34’23,117”= ⋅= ⋅ tan − 2 ⋅4 − − 2 ⋅4= [2 ⋅ − ⋅ tan( )] ⋅ ( − )2= ⋅ tan − 2 ⋅4 − − 2 ⋅4

Aria suprafeţei secţiunii transversale a profilului

St = S1 + 2(S2 + S3 - S4)

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4. Profil T (STAS 566 - 86)

a/4a

t

g

Panta 2%

Pan

ta 2

%

r1 r

r2

h

a/4aB1

B2t r

g

c1

c2

c3

c4

b2b2

h/2

h

b1

r

r1 r1

r2 r2

S3 S4 S1 S4 S3

S5

S6

S2

S5

S6

Panta 2%

Pan

ta 2

%

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Panta 2% = = arctan(0,02) = 1°8’44,746”

= −2= 2 + ⋅ tan( )4 −= ⋅ cos( ) − ⋅ sin( )cos ( ) − sin ( )= ⋅ cos( ) − ⋅ sin( )cos ( ) − sin ( )= + 2 ⋅ ⋅ sin( )= − ⋅ tan( )= 2 + ⋅ cos( )= ⋅ sin( ) + − ⋅ tan( )4= − ⋅ tan( )4= ⋅ cos( )

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= ⋅= ( + ) ⋅2= ( + ) ⋅2= ⋅ tan − 4 ⋅4 − − 4 ⋅4= ⋅ tan − 2 ⋅4 − − 2 ⋅4= ⋅ tan − 2 ⋅4 − − 2 ⋅4

Aria suprafeţei secţiunii transversale a profilului

St = S1 + S2 + 2[S3 + S4 -(S4 + S6)]