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Figures (12)

F. 1
F. 1

Deflection of a bi-metal strip while uniformly heated.

F. 2
F. 2

Deflection of a simply supported bi-metal strip.

F. 3
F. 3

Distribution of normal stresses on any cross section of bi-metal strip while uniformly heated.

F. 4
F. 4

Shearing stresses acting near the ends of bi-metal strip.

F. 5
F. 5

Equivalent section of homogeneous strip.

F. 6
F. 6

Bi-metal strip with clamped ends.

F. 7
F. 7

Stress distribution in bi-metal strip with clamped ends.

F. 8
F. 8

Bending of bi-metal strip with one end clamped and another simply supported.

F. 9
F. 9

Bending of curved bi-metal strip.

F. 10
F. 10

Bi-metal curved strip in a rigid frame.

F. 11
F. 11

Deflection curve of bi-metal strip.

F. 12
F. 12

Graphical determination of the temperature of operation of a bi-metal strip thermostat.

Tables (1)

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Table A

Equations (103)

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P 1 = P 2 = P
P h 2 = M 1 + M 2
M 1 = E 1 I 1 ρ M 2 = E 2 I 2 ρ
P h 2 = E 1 I 1 + E 2 I 2 ρ
a 1 ( t t 0 ) + P 1 E 1 a 1 + a 1 2 ρ = a 2 ( t t 0 ) P 2 E 2 a 2 a 2 2 ρ or
h 2 ρ + 2 ( E 1 I 1 + E 2 I 2 ) h ρ ( 1 E 1 a 1 + 1 E 2 a 2 ) = ( α 2 α 1 ) ( t t 0 )
1 ρ = ( α 2 α 1 ) ( t t 0 ) h 2 + 2 ( E 1 I 1 + E 2 I 2 ) h ( 1 E 1 a 1 + 1 E 2 a 2 )
a 1 a 2 = m , E 1 E 2 = n ,
I 1 = a 1 3 12 , I 2 = a 2 3 12
1 ρ = 6 ( α 2 α 1 ) ( t t 0 ) ( 1 + m ) 2 h ( 3 ( 1 + m ) 2 + ( 1 + m n ) ( m 2 + 1 m n ) )
a 1 = a 2 , m = 1 1 ρ = 24 ( α 2 α 1 ) ( t t 0 ) h ( 14 + n + 1 / n )
1 ρ = 3 2 ( α 2 α 1 ) ( t t 0 ) h
1 ρ = 48 33 ( α 2 α 1 ) ( t t 0 ) h
2 ρ δ = l 2 4 ,
δ = l 2 8 ρ
p max = P a + a 1 E 1 2 ρ
p max = 1 ρ ( 2 h a 1 ( E 1 I 1 + E 2 I 2 ) + a 1 E 1 2 )
E 1 = E 2 = E a 1 = a 2 = 1 2 h p max = E ρ h 4 ( 1 3 + 1 ) , or using ( b ) , p max = E 2 ( a 2 a 1 ) ( t t 0 )
a 2 a 1 = 4 × 10 6 t t 0 = 200 ° C
p max = 10800 lbs. / sq. in .
E 1 E 2 = n , a 1 a 2 = m c = ( a 1 + a 2 ) 2 2 + ( 1 1 n ) a 1 2 2 a 1 + a 2 n
c = h ( 1 + 1 / 4 ( n 1 ) ) n + 1
I = ( a 1 + a 2 ) 3 3 n + ( 1 1 / n ) a 1 3 3 ( a 1 + a 2 n ) c 2
I = h 3 ( 7 24 n + 1 24 ( 3 n + 1 ) 2 32 ( 1 + 1 / n ) )
I = h 3 ( 1 3 1 4 ) = h 3 12 = .0833 h 3
I = h 3 ( .2896 .2128 ) = .0768 h 3
I = .0573 h 3
E I = h 3 E 1 12 × 1 2 ( 1 + E 2 E 1 ) = h 3 E 1 12 × 1 + n 2 n
E I = E 1 h 3 1 × 1.85 2 = .0772 E 1 h 3
E I = h 3 E 1 12 × 3 4 = .0625 E 1 h 3
M 0 E I = 1 / ρ = 3 2 ( α 2 α 1 ) ( t t 0 ) h
E = E 1 + E 2 2 I = h 3 12
p 6 M 0 h 2 = 6 h 2 × 3 2 E I ( α 2 α 1 ) ( t t 0 ) h = 3 4 E ( α 2 α 1 ) ( t t 0 )
l 2 2 ρ = R l 3 3 E I
R = 3 2 E I l × 1 / ρ = 9 4 E I h l ( α 2 α 1 ) ( t t 0 )
M max = R l = 9 4 E I h ( α 2 α 1 ) ( t t 0 )
p max = 6 M max h 2 = 9 8 E ( α 2 α 1 ) ( t t 0 )
1 ρ 0 + 1 ρ = 3 2 ( α 2 α 1 ) ( t t 0 ) h
δ = l 2 8 ρ = l 2 8 ( 3 2 ( α 2 α 1 ) ( t t 0 ) h 1 ρ 0 ) = 3 16 l 2 h ( α 2 α 1 ) ( t t 0 ) δ 0
λ = 1 2 0 l ( d y d x ) 2 d x
y = δ 0 sin π x l
λ = δ 0 2 π 2 4 l
δ 0 2 π 2 4 l δ 1 2 π 2 4 l
P = ( δ 0 2 π 2 4 l δ 1 2 π 2 4 l ) E A l = E A π 2 4 l 2 δ 0 2 ( 1 δ 1 2 δ 0 2 ) or P l 2 E I π 2 = 3 δ 0 2 h 2 ( 1 δ 1 2 δ 0 2 )
δ s i n π x l
y + δ sin π x l
E I d 2 d x 2 ( y + δ sin π x l ) = P y
d 2 y d x 2 + P E I y = δ π 2 l 2 sin π x l
P E I = k 2
y = A sin k x + B cos k x + δ k 2 l 2 π 2 1 sin π x l
δ 1 = δ k 2 l 2 π 2 1
P l 2 E I π 2 = δ δ 1 + 1
P l 2 E I π 2 = ( 3 16 l 2 h δ 0 ( α 2 α 1 ) ( t t 0 ) 1 ) δ 0 δ 1 + 1
i = 3 16 l 2 h δ 0 ( α 2 α 1 ) ( t t 0 ) 1
P l 2 E I π 2 = y δ 0 δ 1 = x 3 δ 0 2 h 2 = a 3 16 l 2 h δ 0 ( α 2 α 1 ) ( t t 0 ) 1 = i
y = a ( 1 1 / x 2 )
y = i x + 1
d y d x = 2 a x 3
x 0 = δ 0 δ 1 = 3 a a 1
i = 2 a ( 1 / 3 1 / 3 a ) 3 / 2
t t 0 = 1 + 6 δ 0 2 h 2 ( 1 3 1 9 h 2 δ 0 2 ) 3 / 2 3 16 l 2 h δ 0 ( α 2 α 1 )
δ 2 = δ 1 k l 2 π 2
k l 2 π 2 P l 2 E I π 2 = 1 δ δ 2
P l 2 E I π 2 = 1 ( 3 16 l 2 h δ 0 ( α 2 α 1 ) ( t t 0 ) 1 ) δ 0 δ 2
P l 2 E I π 2 = 3 δ 0 2 h 2 ( 1 δ 2 2 δ 0 2 )
i = ( 3 16 l 2 h δ 0 ( α 2 α 1 ) ( t t 0 ) 1 )
δ 1 = .408 δ 0
δ 1 + δ 2 = 1.22 δ 0
t 2 t 1 = t t 0
t 1 = t 2 ( t t 0 )
3 16 l 2 h δ 0 ( α 2 α 1 ) ( t 2 t 0 ) = 2
( t 2 t 0 ) = 2 1 + i ( t t 0 )
l 1 t 0 = 1 i 1 + i ( t t 0 )
δ 0 2 π 2 4 l δ 1 2 π 2 4 l Δ
P = ( δ 0 2 π 2 4 l δ 1 2 π 2 4 l Δ ) E A l
Δ = P l E A 1
P = k E A π 2 4 l 2 δ 0 2 ( 1 δ 1 2 δ 0 2 ) where , k = A 1 A + A 1
P l 2 E I π 2 = k 3 δ 0 2 h 2 ( 1 δ 1 2 δ 0 2 )
k 3 δ 0 2 h 2 instead of 3 δ 0 2 h 2
k 3 δ 0 2 h 2 = 2
3 16 l 2 h δ 0 ( α 2 α 1 ) ( t t 0 ) = 1.272
t t 0 = 338 ° C
t 1 t 0 = 1 i 1 + i ( t t 0 ) 1 .272 1 + .272 × 338 = 193 ° C .
t 1 t 0 t t 0 = 1 i 1 + i
y = k a ( 1 1 x 2 )
y = i x + 1
d y d x = 2 k a x 3
k a ( 1 1 x 2 ) = 2 k a x 2 + 1
x 0 = δ 0 δ 1 = 3 k a k a 1
i = 2 k a ( 1 3 1 / 3 k a ) 3 / 2
t t 0 = 1 + 2 k a ( 1 3 1 / 3 k a ) 3 / 2 3 16 l 2 h δ 0 ( α 1 α 1 )
1 i 1 + i = .9 or i = .053
k a = k 3 δ 0 2 h 2 = 1.29
l 2 h δ 0 = 1 + i 3 16 ( t t 0 ) ( α 2 α 1 ) = 1.053 3 16 × 300 × 4 × 10 [ illegible ] = 4 × 1.053 × 10 4 9
l / h = 200 , we have , δ 0 = 9 / 1.053 h = 8.55 h
δ 0 δ 1 = 3.65
y = i x = 1 y = k a ( 1 1 / x 2 )
δ 0 δ 2 = 1.70
δ 1 + δ 2 = δ 0 ( 1 3.65 + 1 1.70 ) = .86 δ 0
3 δ 0 2 h 2 = 1.5
δ 0 h = .707
δ 1 δ 0 = .333
δ 2 δ 0 = .667