For the rigid jointed frame subjected to temperature variations as shown in the figure. Determine total deflection at 'C'. Assume $ \ \alpha = 12 \times 10 ^{-6} $ and depth of all members 400m. Neglect the effect of axial force.

To find horizontal deflection at D, apply dummy force P=1 KN. [Horizontal towards right]. $ (i) \ Axial \ force \ (\overline{\rm N}) \ and \ bending \ moment \ (\overline{\rm M}) \\ (image 7) \\ Applying \ C.O.E:\\ \Sigma M_A=0 \\ x ^2 - M_A = 0\\ M_A = 2 KN.m\\ \Sigma F_Y = 0 \ (\uparrow +ve)\\ \Sigma F_X = 0 \ (\rightarrow +ve)\\ -H_A + 1 =0\\ H_A = 1 KN \\ Consider port AB: \\ (image 8 )\\ B.M_A = -2KN.m\\ B.M_B= 1 \times 5 - 2 \ = 3 \ KN.m \\ Consider port BC:\\ \\ ( image 9)\\ Consider port CD:\\ \\image 10\\ $ (ii) \ Deflection \ of \ D \ [Considering \ effect \ of \ axial \ force] \ $ (a) Member AB: $ t_c = 20 + ( \frac{30-20}{2} ) = 25 ^\circ C\ \triangle t = 30-20 = 10 ^\circ C\ \frac {\triangle t }{n} = \frac {10}{0.4} = 25\ A= - \frac{1}{2} \times 2 \times 2 + \frac{1}{2} \times 3 \times 3 = 2.5 m^2\ \overline{\rm N} = 0\ (b) Member \ BC :\ t_c = 15 + ( \frac{25-15}{2} ) = 20 ^\circ C\ \triangle t = 25-15 = 10 ^\circ C\ \frac {\triangle t}{n} = \frac {10}{0.4} = 25\ \overline{\rm N} = 1 KN\ A = 3 \times 4 = 12 m^ 2\ (c) Member \ CD :\ t_c = 15 + ( \frac{30-15}{2} ) = 22.5 ^\circ C\ \triangle t = 30-15 = 15 ^\circ C\ \frac {triangle t}{n} = \frac {15}{0.4} = 37.5\ \overline{\rm N} = 0 KN\ A = \frac{1}{2} \times 3 \times 3 = 4.5 m^ 2\ $ Using Mohr's Equation, $ \ \Sigma \overline{\rm P_j}.\delta _j = \alpha_t \ \Sigma[\frac{\triangle t}{n} .\overline{\rm A} +t_c \overline{\rm N_L} ]\ 1 \delta_D = 12 \times 10^{-6} [(25 \times 2.5) + 0 + ( 25 \times 12 + 20 \times 4 \times 1 ) + (37.5 \times 4.5)] \ \delta_D = 12 \times 10 ^{-6} [62.5 +380 +168.75]\ \delta_D = 12 \times 10 ^{-6} [611.25]\ \delta_D = 7.335 \times 10 ^{-3}\ \delta_D = 7.335 mm \rightarrow $ \\ $ (iii) \ Deflection \ at \ D \ [Neglecting \ effect \ of \ axial \ force]\ \delta_D = \alpha_t . \Sigma [ \frac{\triangle t}{h} . A ]\ \delta_D = 12 \times 10 ^{-6} [62.5 +380 +168.75]\ \delta_D = 12 \times 10 ^ {-6} [531.25]\ \delta_D = 6.375 \times 10 ^{-3} mm\ \delta_D = 6.375m(\rightarrow) $ Answer