deflection in truss due to temperature &

fabricationEngr ,,,Muzammar cheema

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 :

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:

Problem Diagram :

KN4 KN4

KN6

HB CD

E

F

G

Am2 m2 m2m2

m3

m4

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

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:

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

∑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

∑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

∑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

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:

∆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

∆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 :