A bar of varying section fixed at both ends is axially loaded as shown in fig. by the forces. Find the reaction at A and D and also the elongation in the bar AB and CD respectively.

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teamques10 ★ 68k

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For equilibrium,

$\sum F_x=0$

$R_A+R_D=100+200=300$ (1)

For rigid support/fixed end at both sides, then total elongation is zero.

$\delta l= (\delta l)_{AB}+(\delta l)_{BC}+(\delta l)_{CD}=0$

$\frac{R_A*1000}{2000*200*10^3}-\frac{100-R_A*1500}{3000*200*10^3}-\frac{R_D*2000}{4000*200*10^3}=0$

(+ve for tensile force, -ve for compressive forces)

$5R_A-5(100-R_A)-5R_D=0$

$5R_A-500-5R_A-5R_D=0$

$10R_A-5R_D=500$ (2)

On solving, (1) and (2), we get

$R_A=133.33KN$ $R_D=166.67KN$

$F_{BC}=33.33KN (T)$

$F_{BC}=100-R_A=100-133.33$

$F_{BC}=-33.33 KN$ (compressive)

$F_{BC}=33.33 KN$ (T)

$F_{CD}=R_D=166.67KN (C)$

Now, $(\delta l)_{AB}=\frac{133.33*1000}{2000*200*10^3}=0.333mm$

$(\delta l)_{CD}=\frac{166.67*2000}{4000*200*10^3}=0.417mm$

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