1
$\begingroup$

What are the difference between these two ? Here's my notes , The author just stated the formula without explaining much . Can someone explain further what 's their difference ? gg

$\endgroup$
  • 2
    $\begingroup$ Frankly, who cares? As the notes say, you solve all of them the same way. Whoever produced that book can't even write grammatical English sentences, so don't expect to learn much from it! $\endgroup$ – alephzero Jan 25 '17 at 5:38
  • $\begingroup$ @alephzero can you help to explain what is external intermediate and internal indeterminate ? $\endgroup$ – kelvinmacks Jan 25 '17 at 5:52
  • $\begingroup$ If you consider ground as another member of the structure then there is no difference between the two. $\endgroup$ – cent Jan 25 '17 at 19:54
1
$\begingroup$

For starters, @alephzero's comment is entirely true: there is, in real practice, no difference between them. Also, you seem to be using a garbage textbook. Do yourself a favor and throw it away. You can probably learn more from videos on Youtube.

Now, to actually answer your question, here's the basic difference between them: using only the global equilibrium equations, in externally indeterminate structures, you can't even solve for the support reactions, while in internally indeterminate structures you can find the reactions but not the internal forces in the members.

Externally indeterminate structures

These structures have too many supports. By this, I mean that you cannot calculate the reactions for such structures with just the global equilibrium equations. Take, for instance, the following continuous beam (with nodes $A$, $B$ and $C$):

enter image description here

Let's try to calculate the reactions:

$$\begin{align} \sum F_x &= 0 \therefore A_x = 0 \\ \sum F_y &= A_y + B_y + C_y - 1\cdot(5+8) = 0 \\ \sum M_A &= 5B_y + (5+8)C_y - 1\cdot(5+8)\cdot\left(\frac{5+8}{2}\right) = 0 \end{align}$$

So, we know that $A_x$ (the horizontal reaction from the first support) is equal to zero. But we still have three unknowns ($A_y$, $B_y$, and $C_y$) and only two equations left. We're stuck.

To solve this, we need to create one more equation: the fact that the beam's rotation on the left side of $B$ (the middle support) is equal to the rotation on the right side of $B$. This is called a compatibility equation, which is the sort of equation necessary to solve indeterminate structures.

Internally indeterminate structures

These structures have just enough supports, but the structure itself is too complex. You can therefore calculate the support reactions with the global equilibrium equations, but not the members' internal forces. Take, for instance, the following truss (with supports $A$ and $B$):

enter image description here

We can easily find the reactions:

$$\begin{align} \sum F_x &= 0 \therefore A_x = 0 \\ \sum M_A &= 5B_y - 1\cdot3 = 0 \therefore B_y = \frac{3}{5} \\ \sum F_y &= A_y + B_y - 1 = 0 \therefore A_y = \frac{2}{5} \\ \end{align}$$

Now, let's try to find the axial forces in each bar. Let's start with node $A$ (the bottom left node with the support), where $N_1$, $N_2$, and $N_3$ are the axial forces in the vertical, diagonal and horizontal bars, respectively.

$$\begin{align} \sum F_{x,A} &= N_2\cos45 + N_3 = 0 \\ \sum F_{y,A} &= N_1 + N_2\sin45 = 0 \end{align}$$

Three unknowns, but only two equations. Can't be done. And you can do this for all the nodes, but you'll still end up with more unknowns than equations.

So you once again need to create additional compatibility equations, such as the fact that the deflection of all the bars at a given node must be equal (for instance, the vertical, diagonal and horizontal bars at $A$ must all end up with a deflection of zero).

$\endgroup$
  • $\begingroup$ So , the main point of externally indeterminate condition is the structure has too many external support , while for the internally indeterminate , the point is the structure have just enough supports, but the structure itself is complex??? $\endgroup$ – kelvinmacks Jan 25 '17 at 15:43
  • $\begingroup$ @kelvinmacks: You basically just copy-pasted what I wrote, so yes. $\endgroup$ – Wasabi Jan 25 '17 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.