Deflection in truss due to temperature
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Transcript of Deflection in truss due to temperature
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deflection in truss due to temperature &
fabricationEngr ,,,Muzammar cheema
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In this cases truss member may change their length due to temperature. If we find the displacement of a truss joint due to temperature we use Eq.
∆ =∑ n∂∆TL • ∆ = External joint displacement due to
temperature.• ∆T = Change in temperature.• L = Length of member.• ∂ = coefficient of thermal expansion of member.• n = Internal virtual normal force in a truss caused
by external virtual unit load.
Temperature Errors :
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Error in fabricating the length of the member of a truss may occur. Also in same cases truss member must be made slightly longer or shorter in order to give the truss a camber. If a truss member is shorter or longer than intended, the displacement of a truss joint from its expected position can be determined from direct application of Eq written as:
∆ = ∑n∆L ∆ = External joint displacement caused by the fabrication error. n = internal virtual normal force in a truss member caused by
external virtual unit load. ∆L = Difference in length of the member from its intended size as
caused by a fabrication error.
Fabrication Errors:
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Problem Diagram :
KN4 KN4
KN6
HB CD
E
F
G
Am2 m2 m2m2
m3
m4
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Find vertical displacement of joint H of the given truss that shown in previous slide due to radiant heating from the member HC & GF is subjected to an increase in temperature of 150 F at HC & 130 F at GF the ∂ =
Also find vertical displacement of joint H due to fabrication error of shortening member AG 0.07mm and elongation member DE 0.0046mm.
Problem statement:
106.0 8x
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m2
KN4
HB CD
E
F
G
A m2 m2 m2m2
m3
m4
FAY KN1 FDY
Replace all external load and apply unit load at “h” in vertical direction:
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Find reaction due to virtual vertical unit load at joint “H” ∑Fx = 0FDX = 0∑FY = 0Fay + Fdy – 1 = 0 ∑Ma = 0(1X4) – Fdy (8) = 04 – 8FDy = 0Fdy = Fdy = 0.5KNPut in eq (a)0.5 – Fdy = 0 Fdy = 0.5KN
a
84
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∑Fy = 00.5 + FAGSin θ = 00.5 + FAGSin 56.30 = 00.5 + FAG 0.83 = 0FAG = -0.5/0.83FAG = 0.60 KN – c∑Fx = 0FAGcos θ + FAB = 0 tanθ = 3/2-0.60cos56.30 + FAB = 0 θ = 1.5 -0.33 + FAB = 0 θ = 56.30 FAB = 0.33KN - T
Joint A
FAB
FAy
FAG
A
tan 1
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∑Fy = 0FBG = 0∑Fx = 0 -FAB + FHB = 0 -0.33 + FHB = 0FHB = 0.33KN – T
Joint G :∑Fy = 0FGB + FGAsin56.30 + FGFsin63.43 = 0 0 + 0.60(0.83) + FGF(0.89) = 00.49 + FGF(0.89) = 0FGF = - 0.49/0.89FGF = 0.55KN – C
Joint B :FBHFAB
FBG
FGB
FGF
FGA
B
G
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∑Fy = 0-1 + FHF = 0FHF = 1KN – T∑Fx = 0-0.33 + FHC = 0FHC = 0.33KN – T
The truss is samitrecul SO,FDC = FABFDC = 0.33KN – TFDE = FAGFDE = 0.60KN – TFEF = FGFFEF = 0.55KN – C FCE = FBGFCE = 0KN
Joint H:
FHC
FHF
KN1
FHB
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AB,BH,HC, & CD = 2mBG,CE = 3mHE = 7m AG,DE =3.60m by using Pythagoras
theorem GF,EF = 4.47m by using Pythagoras
theorem
Find length of member:
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∆t = ∑n∂∆TL∆t = ∑{(0.33)( )(150)(2)} + {(-0.55)
( ) (130)(4.47)} ∆t = 0.000000594 – 0.00019∆t = 0.00018m∆t = 1.89mm
Put the value in equation for find displacement due to temperature :
106.0 8x
106.0 6x
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∆f = ∑n∆L∆f = ∑{(-0.60)(-0.00007)} + {(-0.60)(0.0000046)}∆f = 0.000042 – 0.00000276∆f = 0.000039m∆f = 0.039mmTOTAL DISPLACEMENT ∆ = ∆t + ∆f ∆ = 1.89 + 0.039 ∆ = 1.92mm
Put the value in equation for find displacement due to fabrication :