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I'm reading about static determinacy of structures here.

A quote:

Therefore, one extra equilibrium equation is added to the system due to the introduction of the hinge: either ∑MABB=0 or ∑MBCB=0. Only one of these equations counts because the two equations are not independent. They are not independent because they both mean the same thing, that the moment at the hinge is zero. If I know the moment on one side of the cut is zero, then I know automatically as well that the moment on the other side of the cut must be zero. So, for each internal hinge in a structure, there is a single equation of condition: ec=1.

So, the introduction of a hinge gives us an extra equation, $\sum M_B=0$. This is called an equation of condition. The text mentions that only a sum of moments on one side of the hinge counts as a new equation, as the moment sums on either sides mean the same thing.

Let's say we have a frame like this:

enter image description here

The frame has a hinge 1/3 of the way along the top beam. So if the reaction forces at the supports are $A_y$, $A_x$, and $B_y$ (roller support at B), we could write moment about the hinge in two ways:

$$A_y \frac{L}{3} - A_x L = 0$$ or

$$B_y \frac{2L}{3} = 0$$

Here I have neglected the loadings on the frame, but they would just be included as constants in the equations.

So what I don't get is why these equations cannot be used as two new equations. The text says they must be dependent. Why?

I'm not asking specifically about this example, but rather in general. If we derive an equation for the moment on both sides of the hinge, we do indeed calculate the same thing, but we have two equations for the same thing. So why don't they count as two equations for static determinacy?

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  • $\begingroup$ Because the hinge pin links them. $\endgroup$
    – Solar Mike
    Mar 25 at 9:33
  • $\begingroup$ the hingle doesn't pass moments through it, so you have moments on each side of it. $\endgroup$
    – Tiger Guy
    Mar 25 at 14:17
  • $\begingroup$ @Tiger Guy So why not two independent equations then? $\endgroup$
    – S. Rotos
    Mar 25 at 14:28
  • $\begingroup$ I don't know about these definitions of numbers of equations, my experience is these things just work. A hinge will pass forces but not moments. The moments on each side will be related by the forces that pass through the hinge. $\endgroup$
    – Tiger Guy
    Mar 25 at 15:09

2 Answers 2

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The sum of moments about any point is 0. Since 0=0 is always true, the equation for moments about different points turn out to be the same equation! This can be better understood from an example. enter image description here

$$\sum M_A=0 \Rightarrow (L_1F_1+L_2F_2 + ...) - aB = 0 \tag{1}$$ $$\sum F=0 \Rightarrow B=A-(F_1+F_2+...) \tag{2}$$ $$\sum M_B=0 \Rightarrow ((L_1+a)F_1 + (L_2+a)F_2 + ...) -aA = 0 \tag{3}$$ Rearranging (3) and substituting (2) in the last step gives the identical equation as (1). Hence, (1) and (3) are the same equation, and both equations cannot be used to solve for the reaction forces. $$(L_1F_1+L_2F_2+...) +a(F_1+F_2+...)-aA=0 \\ (L_1F_1+L_2F_2+...) +a(F_1+F_2+...-A)=0 \\ (L_1F_1+L_2F_2+...) -a(A-(F_1+F_2+...))=0 \\ (L_1F_1+L_2F_2+...) -aB =0 $$

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It is confusing through reading the text, but it should be clear through the sketch below.

enter image description here

The system on the left has 4 unknowns, however, the presence of the hinge enables us to write the necessary 4 equations to solve the system.

But, if separating the integral frame into two subframes (similar to repeating the $\sum Mc = 0$ from each side of the hinge), and solving each frame with the equilibrium equations, $\sum Fx = 0$, $\sum Fy = 0$, and $\sum Mc = 0$, the results will be erroneous because of the dependency of internal forces (in this case, $V_{C_L}, V_{C_R}$) on each side of the hinge, albeit the overall moment about "C" is zero.

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