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: Understand the setup. Bar BCD is fixed at B and bar FG is fixed at G. Both bars have the same axial rigidity \(EA = 20000\) kN. Segment BC = 5 m, segment CD = 5 m, segment FG = 5 m, and there is a small \(2\) mm gap between the free end D of the first bar and the free end F of the second bar. Load \(P = 20\) kN pushes point C towards D. We first need to check whether this load is enough to close the 2 mm gap.
Step 2: Check if the gap closes. Imagine bar FG is not there yet (D still free). Since D is a free end, the segment CD carries no force at all (nothing sits beyond it to pull on it), so the entire load P is carried only by segment BC, in tension. \[ N_{BC} = P = 20 \text{ kN} \] Elongation of BC, and hence the displacement of D (since CD does not stretch further, it just rides along with C): \[ \delta_D = \frac{N_{BC} L_{BC}}{EA} = \frac{20 \times 5}{20000} = 0.005 \text{ m} = 5 \text{ mm} \] Since \(5\) mm is more than the \(2\) mm gap, D actually travels through the gap and presses against F. So the gap DOES close, and the problem becomes statically indeterminate: both walls B and G now share the load.
Step 3: Set up the force method (remove support at G). Release the support at G, so that end G is temporarily free. In this released structure, under P alone, D moves \(5\) mm as found above. It uses up the \(2\) mm gap, and since the released bar FG is now completely unsupported (a free rigid body with no stiffness of its own until G is restrained), the remaining \(5 - 2 = 3\) mm just carries straight through as a rigid shift of the whole FG bar. So the displacement of G in the released structure due to P alone is: \[ \delta_P = 5 - 2 = 3 \text{ mm} = 0.003 \text{ m} \]
Step 4: Find the flexibility at G. Once contact is made, a unit force applied at free end G (with B fixed) travels in series through FG, then through the closed contact, then through CD and BC, all the way to B. All three 5 m segments then carry the same force, so this is exactly like one straight bar of length \(5+5+5 = 15\) m under an end load: \[ f_{GG} = \frac{L_{total}}{EA} = \frac{15}{20000} = 0.00075 \text{ m/kN} = 0.75 \text{ mm/kN} \]
Step 5: Apply compatibility at G. In the real structure G cannot move at all (it is a fixed wall), so the net displacement at G, from P and from the actual reaction \(R_G\) together, must equal zero: \[ \delta_P - f_{GG} R_G = 0 \] \[ 3 = 0.75 \, R_G \implies R_G = 4 \text{ kN} \]
Step 6: Use overall equilibrium to get R_B. Both wall reactions resist the same applied load P (the load pushes D into F, and both B and G hold the assembly back against that push), so: \[ R_B + R_G = P = 20 \text{ kN} \] \[ R_B = 20 - 4 = 16 \text{ kN} \]
Final Answer: The horizontal reaction at B has a magnitude of 16 kN. \[ \boxed{R_B = 16 \text{ kN}} \]
