Linearly elastic, homogeneous, uniform bars BCD and FG shown in the figure have fixed supports at B and G, respectively. For both the bars, axial rigidity is \(20000\) kN. A gap of \(2\) mm exists between D and F prior to application of any load (i.e. \(P = 0\)). Small deformation and infinitesimal strain assumptions are valid for the given bars.
(Figure not to scale)
The magnitude of the horizontal reaction (in kN) at B after application of the axial force P of \(20\) kN at C is (rounded off to the nearest integer).
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First check whether the load alone (ignoring the gap) would push D more than 2 mm - if it does, the gap closes and both B and G share the load, making it an indeterminate problem.
Step 1: Model each segment as a spring and write node equations. Since all three segments (BC, CD, FG) have the same length (5 m) and the same $EA = 20000$ kN, each behaves as a spring of stiffness $k = EA/L = 20000/5 = 4000$ kN/m. Let $u_C$, $u_D$, $u_F$ be the rightward displacements of C, D, F (with $u_B = u_G = 0$ since both walls are fixed).
Step 2: Confirm contact happens. Before contact, D is a free end, so spring CD carries zero force, meaning $u_D = u_C$. Node C equilibrium (spring BC pulling C back, spring CD carrying zero net pull, load P pushing forward) gives $k\,u_C = P$, so $u_C = 20/4000 = 0.005$ m $= 5$ mm, and $u_D = 5$ mm too. Since this exceeds the 2 mm gap, contact is confirmed, exactly as expected physically.
Step 3: Write the four equations for the contact state. Once D presses on F, let $N_c$ be the compressive contact force between them. The equilibrium equations are: Node C: $-k u_C + k(u_D - u_C) + P = 0$ Node D: $k(u_C - u_D) = N_c$ Node F: $N_c = k\,u_F$ Contact condition (gap fully closed): $u_D - u_F = 0.002$ m
Step 4: Solve the system. From the node F equation, $u_F = N_c/k$. Using the contact condition, $u_D = N_c/k + 0.002$. Substituting into the node D equation and then into the node C equation and simplifying gives: \[ N_c = \frac{P - 0.002k}{3} = \frac{20 - 0.002(4000)}{3} = \frac{20 - 8}{3} = 4 \text{ kN} \] Then $u_F = 4/4000 = 0.001$ m, $u_D = 0.001 + 0.002 = 0.003$ m, and from the node C equation, $u_C = 0.004$ m.
Step 5: Recover the reactions. The contact force $N_c$ passes straight through the FG bar to the wall at G, so $R_G = N_c = 4$ kN. The force in segment BC equals $k\,u_C$ (the stretch in BC times its stiffness), which is exactly the reaction needed to hold that stretched spring at the wall: \[ R_B = k\,u_C = 4000 \times 0.004 = 16 \text{ kN} \] This matches the equilibrium check $R_B + R_G = 16 + 4 = 20$ kN $= P$. \[ \boxed{R_B = 16 \text{ kN}} \]
