situation 1 - for the loading shown, knowing that beams ac and bd have the same flexural rigidity, find the…

situation 1 - for the loading shown, knowing that beams ac and bd have the same flexural rigidity, find the following:\n1. the vertical reaction at support b (in kn). answer: \n2. the moment reaction at fixed support a (in kn - m). answer: \n3. the vertical reaction at point c (in kn). answer:

situation 1 - for the loading shown, knowing that beams ac and bd have the same flexural rigidity, find the following:\n1. the vertical reaction at support b (in kn). answer: \n2. the moment reaction at fixed support a (in kn - m). answer: \n3. the vertical reaction at point c (in kn). answer:

Answer

Explanation:

Step1: Analyze beam AC

Beam AC is a cantilever - beam with a uniformly distributed load (w = 9\ kN/m) over a length (L_{AC}=0.6\ m). The total load on beam AC is (W=w\times L_{AC}=9\times0.6 = 5.4\ kN), and it acts at the mid - point of the uniformly distributed load, i.e., at (x = 0.3\ m) from A.

Step2: Take moments about point A for beam AC

The moment due to the uniformly distributed load about A is (M_{A - AC}=W\times0.3=5.4\times0.3 = 1.62\ kN - m) (clock - wise).

Step3: Consider the interaction at C

Let the vertical reaction at C be (V_C) and at B be (V_B). Since the beams AC and BD have the same flexural rigidity and considering the equilibrium of beam BD, we know that the vertical reaction at C due to beam BD is related to the reaction at B. For beam BD, taking moments about B, if we assume the vertical reaction at C is (V_C) and the length (L_{BC}=0.5\ m) and (L_{CD}=0.5\ m). For the overall equilibrium of the structure, consider the vertical forces. Let's first consider beam AC. The vertical reaction at A, (V_A) must balance the load on AC and the reaction from C. Taking moments about A for the whole structure: Let (V_B) be the vertical reaction at B. The moment due to the load on AC and the reactions at B and C. The load on AC creates a clock - wise moment. The reaction (V_B) creates a counter - clockwise moment about A with a lever arm of (1\ m) (distance from A to B), and (V_C) creates a counter - clockwise moment about A with a lever arm of (0.6\ m).

  1. For the vertical reaction at support B:
    • Consider the equilibrium of vertical forces and moments. The total load on AC is (F = 9\times0.6=5.4\ kN).
    • Taking moments about A for the whole structure: (\sum M_A = 0). Let (V_B) be the vertical reaction at B and (V_C) be the vertical reaction at C.
    • The load on AC creates a clock - wise moment (M_{load}=9\times0.6\times0.3). The reaction (V_B) creates a counter - clockwise moment (M_{V_B}=V_B\times(0.6 + 0.5+0.5)) and (V_C) creates a counter - clockwise moment (M_{V_C}=V_C\times0.6).
    • Also, from the equilibrium of beam BD, (V_C = V_B) (due to symmetry and equal flexural rigidity).
    • (\sum M_A=9\times0.6\times0.3-V_B\times1.6 - V_B\times0.6=0).
    • (1.62-V_B\times(1.6 + 0.6)=0).
    • (1.62 - 2.2V_B=0).
    • (V_B=\frac{1.62}{2.2}\approx0.736\ kN).
  2. For the moment reaction at fixed support A:
    • The moment reaction at A is due to the load on AC and the reactions at B and C.
    • The load on AC creates a clock - wise moment (M_{load}=9\times0.6\times0.3 = 1.62\ kN - m). The reactions at B and C create counter - clockwise moments. Since (V_B = V_C\approx0.736\ kN), the total counter - clockwise moment due to B and C about A is (V_B\times(0.6 + 0.5+0.5)+V_C\times0.6).
    • (M_A=9\times0.6\times0.3-0.736\times(1.6 + 0.6))
    • (M_A = 1.62-0.736\times2.2)
    • (M_A=1.62 - 1.62=0\ kN - m) (due to the balance of moments).
  3. For the vertical reaction at point C:
    • Since (V_C = V_B) (from the equilibrium of beam BD and equal flexural rigidity of AC and BD), (V_C\approx0.736\ kN).

Answer:

  1. (0.736)
  2. (0)
  3. (0.736)