1. **Problem Statement:** Determine the reactions at supports A and B for an A-36 steel rod of diameter 50 mm, initially at temperature $T_1=80^\circ C$, cooled to $T_2=20^\circ C$, with an axial force $P=200000$ N applied at the midpoint. 2. **Given Data:** - Diameter $d=50$ mm = 0.05 m - Length of each segment $L=0.5$ m - Initial temperature $T_1=80^\circ C$ - Final temperature $T_2=20^\circ C$ - Axial force $P=200000$ N - Modulus of elasticity for A-36 steel $E=200\times10^9$ Pa (typical value) - Coefficient of thermal expansion for steel $\alpha=12\times10^{-6} / ^\circ C$ 3. **Formulas and Concepts:** - Thermal strain due to temperature change: $\varepsilon_{thermal} = \alpha \Delta T = \alpha (T_2 - T_1)$ - Mechanical strain due to axial force: $\varepsilon_{mechanical} = \frac{\sigma}{E} = \frac{F}{AE}$ - Total strain in each segment is sum of mechanical and thermal strains. - Since the rod is fixed at both ends and force $P$ is applied at midpoint, reactions at A and B are equal in magnitude but opposite in direction to maintain equilibrium. 4. **Calculate cross-sectional area:** $$ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.05}{2}\right)^2 = \pi (0.025)^2 = 0.0019635 \text{ m}^2 $$ 5. **Calculate temperature change:** $$ \Delta T = T_2 - T_1 = 20 - 80 = -60 ^\circ C $$ 6. **Calculate thermal strain:** $$ \varepsilon_{thermal} = \alpha \Delta T = 12 \times 10^{-6} \times (-60) = -0.00072 $$ 7. **Set up equilibrium and compatibility:** - Let $R_A$ and $R_B$ be reactions at A and B. - The axial force $P$ acts at midpoint, splitting the rod into two segments. - Each segment experiences force $R_A$ or $R_B$ plus thermal contraction. - Since the rod is fixed, total elongation is zero. 8. **Calculate mechanical strain in each segment:** - For left segment: force is $R_A - P$ (since $P$ acts to the right) - For right segment: force is $R_B - P$ (since $P$ acts to the left) - But by symmetry and equilibrium: $R_A = R_B = R$ 9. **Write strain compatibility:** $$ \varepsilon_{total} = \varepsilon_{mechanical} + \varepsilon_{thermal} = 0 $$ 10. **Mechanical strain in each segment:** $$ \varepsilon_{mechanical} = \frac{R}{AE} $$ 11. **Total strain zero implies:** $$ \frac{R}{AE} + \varepsilon_{thermal} = 0 \Rightarrow R = -AE \varepsilon_{thermal} $$ 12. **Calculate reaction force:** $$ R = -0.0019635 \times 200 \times 10^9 \times (-0.00072) = 282,888 \text{ N} $$ 13. **Adjust for applied force $P$ at midpoint:** - The applied force $P=200,000$ N acts at midpoint, so reactions must balance both thermal contraction and applied force. - Reaction at A: $R_A = R + \frac{P}{2} = 282,888 + 100,000 = 382,888$ N - Reaction at B: $R_B = R + \frac{P}{2} = 382,888$ N 14. **Final answer:** $$ R_A = R_B = 382,888 \text{ N} $$ Thus, the reactions at supports A and B are approximately 383 kN each, directed to resist the applied force and thermal contraction.