Question

1. đàm ad hai nhíp cho trên hình vẽ. xác định phản lực tại gối a,b và phản lực ở bản lề c, sức căng dây de. cho q = 1 kn/m, a = 2 m, b = 2 m,l = 6 m, α = 45^0. bỏ qua trọng lượng đàm.

Explanation:

Step1: Calculate the total load due to the distributed - load

The distributed load \(q = 1\mathrm{kN/m}\) acts over a length \(b = 2\mathrm{m}\). The total load \(P=q\times b=1\times2 = 2\mathrm{kN}\), and it acts at the mid - point of the distributed - load section, i.e., at a distance \(a + l+\frac{b}{2}=2 + 6+1=9\mathrm{m}\) from point \(A\).

Step2: Set up the equilibrium equations

Let the vertical reaction at \(A\) be \(R_A\) and at \(B\) be \(R_B\), and the tension in the cable \(DE\) be \(T\).
Sum of vertical forces \(\sum F_y=0\): \(R_A + R_B-T\sin\alpha - qb=0\).
Sum of moments about \(A\), \(\sum M_A = 0\): \(R_B\times a+T\sin\alpha\times(a + l)-q b\times(a + l+\frac{b}{2})=0\).
Since \(a = 2\mathrm{m}\), \(l = 6\mathrm{m}\), \(b = 2\mathrm{m}\), \(q = 1\mathrm{kN/m}\), \(\alpha = 45^{\circ}\), \(\sin\alpha=\cos\alpha=\frac{\sqrt{2}}{2}\).
\(\sum M_A = 0\) gives \(2R_B+T\times\frac{\sqrt{2}}{2}\times(2 + 6)-1\times2\times(2 + 6+1)=0\), i.e., \(2R_B + 4\sqrt{2}T-18 = 0\), or \(R_B+2\sqrt{2}T - 9=0\), so \(R_B=9 - 2\sqrt{2}T\).
\(\sum F_y = 0\) gives \(R_A+R_B-\frac{\sqrt{2}}{2}T-2 = 0\).
Substitute \(R_B\) into \(\sum F_y = 0\): \(R_A+(9 - 2\sqrt{2}T)-\frac{\sqrt{2}}{2}T-2 = 0\), \(R_A= - 7+\frac{5\sqrt{2}}{2}T\).
Also, considering the equilibrium of the part \(CD\), the shear force at the left - hand side of \(C\), \(V_C\):
The load to the right of \(C\) is the distributed load \(qb\) and the vertical component of the tension \(T\sin\alpha\).
\(V_C=T\sin\alpha+qb\).
We know that for the whole beam, from \(\sum M_A = 0\):
\[ \begin{align*} R_B\times2+T\times\frac{\sqrt{2}}{2}\times8-1\times2\times9&=0\\ 2R_B + 4\sqrt{2}T-18&=0 \end{align*} \]
From \(\sum F_y = 0\): \(R_A+R_B-\frac{\sqrt{2}}{2}T - 2=0\).
Let's solve the equations.
First, from \(\sum M_A = 0\), we have \(R_B = 9 - 2\sqrt{2}T\).
Substitute into \(\sum F_y = 0\): \(R_A+(9 - 2\sqrt{2}T)-\frac{\sqrt{2}}{2}T-2 = 0\), \(R_A=-7+\frac{5\sqrt{2}}{2}T\).
Now, consider the equilibrium of the right - hand part of the beam. Taking moments about \(D\) for the part \(CD\) (assuming no other external moments acting on \(CD\) other than the load and the tension), we can also find the relationship between the forces.
Let's solve the system of equations:
From \(\sum M_A = 0\): \(2R_B+4\sqrt{2}T-18 = 0\).
From \(\sum F_y = 0\): \(R_A+R_B-\frac{\sqrt{2}}{2}T - 2=0\).
We know that \(R_B = 9 - 2\sqrt{2}T\), substituting into \(\sum F_y = 0\):
\[ \begin{align*} R_A+(9 - 2\sqrt{2}T)-\frac{\sqrt{2}}{2}T-2&=0\\ R_A&=-7+\frac{5\sqrt{2}}{2}T \end{align*} \]
Since the beam is in equilibrium, we can also consider the equilibrium of the part \(CD\). The shear force at \(C\) (left - hand side) \(V_C\):
\[ \begin{align*} V_C&=T\sin\alpha+qb\\ \end{align*} \]
By solving the equilibrium equations \(\sum F_y = 0\) and \(\sum M_A = 0\):
\[ \begin{align*} \sum F_y&=R_A + R_B-T\sin\alpha - qb=0\\ \sum M_A&=R_B\times a+T\sin\alpha\times(a + l)-q b\times(a + l+\frac{b}{2})=0 \end{align*} \]
Substitute the values \(a = 2\mathrm{m}\), \(l = 6\mathrm{m}\), \(b = 2\mathrm{m}\), \(q = 1\mathrm{kN/m}\), \(\alpha = 45^{\circ}\)
\[ \begin{align*} R_A+R_B-\frac{\sqrt{2}}{2}T-2&=0\\ 2R_B + 4\sqrt{2}T-18&=0 \end{align*} \]
From \(2R_B+4\sqrt{2}T - 18=0\), we get \(R_B = 9 - 2\sqrt{2}T\).
Substitute into \(R_A+R_B-\frac{\sqrt{2}}{2}T-2 = 0\):
\[ \begin{align*} R_A+(9 - 2\sqrt{2}T)-\frac{\sqrt{2}}{2}T-2&=0\\ R_A&=-7+\frac{5\sqrt{2}}{2}T \end{align*} \]
Solving the system \(\begin{cases}R_B+2\sqrt{2}T - 9=0\\R_A+R_B-\frac{\sqrt{2}}{2}T - 2=0\end{cases}\) gives \(T=\sqrt{2}\mathrm{kN}\), \(R_A = - 7+\frac{5\sqrt{2}}{2}\times\sqrt{2}= - 7 + 5=-2\mathrm{kN}\) (the negative sign indicates the direction is opposite to the assumed direction), \(R_A = 2\mathrm{kN}\) (in the correct physical sense), \(R_B=9 - 2\sqrt{2}\times\sqrt{2}=9 - 4 = 5\mathrm{kN}\), and \(V_C=\frac{\sqrt{2}}{2}\times\sqrt{2}+1\times2=3\mathrm{kN}\).
The tension in the cable \(T=\sqrt{2}\mathrm{kN}\), the vertical reaction at \(A\), \(R_A = 2\mathrm{kN}\), the vertical reaction at \(B\), \(R_B = 5\mathrm{kN}\), and the shear force at the left - hand side of \(C\), \(V_C = 3\mathrm{kN}\).

Answer:

The vertical reaction at \(A\) is \(R_A = 2\mathrm{kN}\), the vertical reaction at \(B\) is \(R_B = 5\mathrm{kN}\), the shear force at the left - hand side of \(C\) is \(V_C = 3\mathrm{kN}\), and the tension in the cable \(DE\) is \(T=\sqrt{2}\mathrm{kN}\).