For the rigid jointed frame subjected to temperature variation as shown in the figure. Determine the horizontal deflection at A.

Assume α = 12 x 10^-6 /c depth of all member as 200 mm. Neglect effect of axial force.

Subject: Structural Analysis II

Topic: Deflection caused by temperature changes

Difficulty:Medium

To find horizontal deflection at A, apply fictitious or Dummy force P= 1 KN horizontal towards left.

1. Axial force(N) and Bending Moment(M)
   Diagram:

Applying C.O.E:

$ \Sigma \ M _D = 0 \ (+ve) $
$ -1 \times 3 - M_D =0 $
$ M_D = 3\ KN.m $
$ \Sigma F_y = 0 ( \uparrow +ve)$
$ V_D = 0$
$\Sigma F_x = 0 (\to +ve)$
$-1 +H_D =0$
$H_D=1 \ KN $
Consider port 'A B' :
$BM_B = 1 \times 1 = 1 \ KN$

Consider port BC:

Consider port CD: $BM_C = 1\ KN.m$
$BM_D = -3 KN.m $

2. Deflection at A: (Considering effect of axial forces)

(a) Member AB :$ (10^\circ C \to 30 \ ^\circ C)$
$ \therefore T_C =10 \ + \ \frac{(30-10)}{2} = 20 ^\circ C$
$\bigtriangleup t = 30 ^\circ C - 10 ^\circ C = 20 ^\circ C$
$ \frac{\bigtriangleup t}{n} = \frac{20}{0.2} =100 $
$ \overline{\rm A} = \frac{1}{2} \times 1\times 1 =0.5 m ^2$
$ \overline{\rm N}= 0 $

(b) Member BC :$ (10^\circ C \to 40 \ ^\circ C)$
$ \therefore T_C =10 \ + \ \frac{(40-10)}{2} = 25 ^\circ C$
$\bigtriangleup t = 40 ^\circ C - 10 ^\circ C = 30 ^\circ C$
$ \frac{\bigtriangleup t}{n} = \frac{30}{0.2} =150 $
$ \overline{\rm A} = 1 \times 4 =4\ m ^2$
$ \overline{\rm N}= 1 $

(c) Member CD :$ (10^\circ C \to 30 \ ^\circ C)$
$ \therefore T_C =10 \ + \ \frac{(30-10)}{2} = 20 ^\circ C$
$\bigtriangleup t = 30 ^\circ C - 10 ^\circ C = 20 ^\circ C$
$ \frac{\bigtriangleup t}{n} = \frac{20}{0.2} =100 $
$ \overline{\rm A} = \frac{-1}{2} \times 3\times 3 + \ \frac{1}{2} \times 1\times 1 =- 4 \ m ^2$
$ \overline{\rm N}= 0 $