Length of each member of truss is 2 m. Members AB and CD are subjected to a temperature rise of 20°C and coefficient of thermal expansion is $12 \times (10^{- 6}) / ° C$ . Then vertical deflection at joint E will be (in mm)

0.42

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0.55

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0.25

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Solution

The correct option is C 0.55
Apply a unit load at joint E

$δ_{E} = Σ K (L α Δ T) . . . (1)$

$A t j o i n t A; Σ F_{y} = 0$

$\Rightarrow 0.5 - K_{A B} sin 60^{\circ} = 0$

$∴ K_{A B} = \frac{1}{\sqrt{3}} (c o m p r e s s i v e)$

$A l s o; k_{A B} = K_{C D} (D u e t o s y m m e t r y)$

From (1)

$∴ δ_{E} = (\frac{1}{\sqrt{3}} \times 2 \times 12 \times 10^{- 6} \times 20) \times 2 m = 0.55 m m$

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Length of each member of truss is 2 m. Members AB and CD are subjected to a temperature rise of 20°C and coefficient of thermal expansion is 12×(10−6)/°C. Then vertical deflection at joint E will be (in mm)

The correct option is C 0.55Apply a unit load at joint E δE=ΣK(LαΔT)...(1) At joint A; ΣFy=0 ⇒0.5−KABsin60∘=0 ∴KAB=1√3(compressive) Also; kAB=KCD(Duetosymmetry) From (1) ∴δE=(1√3×2×12×10−6×20)×2m=0.55mm

Length of each member of truss is 2 m. Members AB and CD are subjected to a temperature rise of 20°C and coefficient of thermal expansion is $12 \times (10^{- 6}) / ° C$ . Then vertical deflection at joint E will be (in mm)